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Chapter 3 Mendel’s Laws And Their Deviation Summary

WBBSE Class 10 Mendel’s Laws Overview

• Mendel performed experiments on pea plants to study inheritance patterns. He chose seven characters of pea plants like stem height, seed colour and shape, pod colour and shape, flower colour and position.
• Each character has a dominant and a recessive phenotype. He performed a monohybrid cross with a single character and a dihybrid cross with two characters.
• Mendel derived his first law i.e. the law of segregation from the monohybrid cross. This law states that characters never get mixed but rather get segregated in the second generation.

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• Mendel’s second law i.e. the law of independent assortment came from the dihybrid cross. This law states that characters not only segregate but rather are expressed in all possible combinations.
• Deviation from Mendel’s principle is exhibited by certain organisms in terms of certain characters. The inheritance of flower colour of Mirabilis jalapa (4 o’clock plant) is one such example where the F₁ hybrid organism shows an intermediate phenotype of two parents is and none of the parental phenotypes is completely dominant over the other.

Chapter 3 Mendel’s Laws And Their Deviation Long Answer Type Questions

Question 1. Explain the two conclusions of Mendel’s monohybrid cross. Rough coat (RR) is dominant and smooth coat (rr) is a recessive character for guinea pigs. If two heterozygous rough-coated guinea pigs are crossed, what will be the types of offspring in F₁ generation?

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Answer:

Two conclusions of Mendel’s monohybrid cross Analysing the results of the monohybrid cross of garden pea plants, Mendel reached two conclusions. These are popularly known as the law of dominance and the law of segregation.

1. Law of dominance:

In a cross between two parents that are pure for contrasting traits, only one of the two expresses itself in the first filial generation. This is controlled by a dominant factor that helps to express the dominant trait by suppressing the recessive one. Here, the expressed one is the dominant trait and the suppressed one is the recessive trait.

2. Law of segregation:

This law states that during the formation of gametes, two alleles controlling each character, move apart due to the separation of the homologous chromosomes during meiosis. Thus, each gamete receives only one allele of each character on a random basis.

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Monohybrid cross between heterozygous rough-coated guinea pigs

Let us mark the rough coat (dominant) as ‘RR’ and the smooth coat (recessive) as ‘rr

Therefore, the genotype of a heterozygous rough-coated guinea pig should be denoted as ‘Rr. The cross between two heterozygous rough-coated Rr guinea pigs is schematically represented here.

From this cross, we get two different phenotypes, 75% or 3/4th part of the progeny are rough-coated and the remaining 25% or 1/4th part are smooth-coated. Therefore, the phenotypic ratio is- Rough coat: Smooth coat = 3:1.

The genotypic ratio of this cross is- Pure rough coat: Hybrid rough coat: Pure smooth coat = 1:2:1.

Checkerboard of F₁ generation

Question 2. In a cross between two pea plants bearing, pure violet [dominant] and pure white [recessive] flowers, what will be the ratio of pure violet and pure white-flowered plants in the F₂ generation?
Answer:

The ratio of pure violet and pure white-flowered plants in F₂ generation

Let us represent the dominant pure violet flower character as (W) and the recessive pure white flower character as (w).

In a cross between the pure violet (W) and pure white (w) flowered pea plants, all Fj progeny grow violet flowers. This is due to the dominant nature of the violet flower colour. If self¬pollination is allowed in the F₁ plants, two different phenotypes will appear in the F₂ generation.

Among the F₂ progeny, 3/4th parts of plants grow violet flowers and the remaining 1/4th part of plants bear white flowers. Of these 3/4th parts of plants beaming violet flowers, 1/4th part of plants grow pure violet and 2/4th parts of plants grow hybrid violet flowers.

Therefore, among the F₂ progeny, 1/4th part of plants (25%) grow pure violet, 2/4th parts of plants (50%) grow hybrid violet and the remaining 1/4th part of plants (25%) grow pure white flowers. The cross is schematically represented here with a checkerboard.

Checker board of F₂ generation

The phenotypic and genotypic ratio of pea plant of F₂ generation

Therefore, the ratio of violet and white flowered pea plants (phenotypic ratio) is 3: 1. The ratio of pure violet, hybrid violet and pure white flowered pea plants of the F₂ generation (genotypic ratio) is 1: 2: 1. The ratio of pure violet and pure white flowered plants of the F₂ generation is 1: 1.

Mendel’s Law of Segregation Explained

Question 3. Explain with a checkerboard, how mendel reached the conclusion to his monohybrid cross experiment.
Answer :

The conclusion of Mendel’s monohybrid cross

Garden pea plants have seven different genetic characteristics, each having two distinct opposite traits.

For the monohybrid cross experiment, Mendel selected only one character, that is the height of the garden pea plant. The two opposite traits are denoted as “IT for pure tall and ‘tt’ for the pure dwarf. The steps of the experiment are mentioned here.

1. The first filial generation:

Mendel collected Pure tall plant Pure dwarf plant pollens from a flower of a pure tall pea plant and placed it on the stigma of some emasculated flowers (flowers, artificially. converted into unisexual female flowers by removing the anthers) of dwarf pea plants.

The progeny plants produced from this cross were the first filial or F₁ generation, all of which were tall in phenotype.

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2. The second filial generation:

Mendel then allowed self-pollination among the F1 plants and got the F₂ progeny with two different phenotypes. 75% of these were tall and 25% were dwarf. Hence the phenotypic ratio is Tall: Dwarf = 3:1., Among the F₂ plants, 25% were pure tall, 50% were hybrid tall and 25% were pure dwarf.

Therefore, the genotypic ratio of a monohybrid cross is Pure tall: Hybrid tall: Pure dwarf = 1:2:1.

The cross is schematically represented here with checkerboard.

Checker board of F₂ generation

The phenotypic and genotypic ratio of pea plant of F₂ generation

Conclusion:

The experimental result shows that during hybridisation the parental factors, which come together in F₁ generation, do not get mixed. During gamete formation of F₁ generation, the factors segregate from each other and each gamete carries one factor. Thus, by analysing the result of the monohybrid cross, Mendel framed the law of segregation.

Question 4. Explain the monohybrid cross of an animal and draw a conclusion from this experiment.
Answer :

The monohybrid cross of an animal :

Monohybrid cross of an animal and its conclusion Monohybrid cross can be performed on guinea pig. The black coat (BB) of a guinea pig is a dominant trait over the white coat (bb).

In a cross between pure black and white coated guinea pigs, all F₁ offsprings are black (Bb). These hybrid black F₁ guinea pigs give birth to F₂ offspring with different phenotypes, black and white. Out of this 75% is black and 25% is white. Therefore the phenotypic ratio is 3:1. Among F₂ offspring, three different genotypes are found.

These are pure black (BB), hybrid black (Bb) and pure white (bb) in the ratio of 1:2:1.

The monohybrid cross of a guinea pig is schematically represented here

Checker board of F₂ generation

The phenotypic and genotypic ratio of the guinea pig of F₂ generation

Conclusion:

The result of the monohybrid cross in between black and white guinea pigs maintains absolute similarity with the Mendelian monohybrid cross and supports the law of segregation.

Question 5. Describe. Mendel’s dihybrid cross experiment and explain the conclusion drawn from it.
Answer :

Dihybrid cross experiment:

For his dihybrid cross experiment, Mendel selected two different characters of garden pea plants, each having two distinct contrasting traits. The two characters are—

1. Colour of  seed or cotyledon and
2. The shape of seed. For colour of seed, the contrasting traits are yellow (YY) and green (yy), where yellow one is dominant over green. For the shape of the seed, the opposite traits are round (RR) and wrinkled (rr), where the round is dominant over wrinkled.

1. First filial generation:

Mendel crossed pure round and yellow seed-bearing pea plants (RRYY) with wrinkled and green seed-bearing pea plants (rryy). From this cross, he got the F₁ hybrids with yellow cotyledons and round seeds (YyRr).

2. Second filial generation:

Mendel allowed self-pollination among the F1 plants and F₂ offspring were produced. These were of four different phenotypes—round yellow, wrinkled yellow, round green, and wrinkled green in the ratio of 9:3:3:1. By analysing the genotypes of these F₂ plants, Mendel found nine different combinations in the ratio of 1:2:2:4; 1:2:1:2:1.

Mendel’s dihybrid cross experiment is schematically represented here.

Checkerboard of F₂ generation

Phenotype and genotype of F₂ generation of Mendelian dihybrid cross

Conclusion:

From the results of this experiment, Mendel came to the following conclusions.

1. The factors, responsible for any of the cotyledon colours and shapes of the seeds, do not get mixed in the F₁ progeny. All four factors were segregated during the gamete formation of the F₁ plants.
2. The gametes, which carry the colour and shape regulating factors of pea seeds in a segregated manner, unite independently in all possible combinations to form the F₂ progeny. Based on these conclusions, Mendel framed his second law of heredity or the law of independent assortment.

Question 6. Mention the seven pairs of contrasting characters of garden pea plants as selected by Mendel.
Answer :

Seven pairs of contrasting characters of garden pea plant:

For his experiment on heredity, Mendel selected seven pairs of opposite characters of garden pea plants, which are mentioned below in a table.

Question 7. Mention the causes of Mendel’s success in his experiments on heredity with garden pea plants. Which types of gametes unite to form a male baby?
Answer :

Causes of Mendel’s success in his experiments on heredity with garden pea plants

1. Mendel selected the pure breed of pea plants for a particular character after two years of continuous successive self-breeding.
2. The seven pairs of factors (genes), responsible for the selected contrasting traits, are present on seven separate sets of homologous chromosomes. This feature helps in the independent assortment of the factors.
3. Pea plants are easy to breed, the life cycle is short and therefore, Mendel could study several generations in a short time. Pea flowers are bisexual, and naturally self-pollinating, but Mendel successfully manipulated it for artificial cross¬pollination. Fie was also successful in preventing any undesired pollination.
4. Mendel, though selected seven traits of the pea plant, but worked on one or two at a time. This helped him in the collection and analysis of the data.
5. The characters, selected by Mendel, were sharply contrasting and showed complete dominance.
6. Mendel used many samples at a time for breeding to get a large number of progeny, which helped him for statistical analysis of the data and to get the correct ratio between two or more contrasting characters.

Gametes are required to produce a male baby

Union of an ovum with 22 autosomes and an X chromosome (22A+X) and a sperm with 2 autosomes and a Y chromosome (22A+Y) can produce a male baby with a genotype (44A+XY).

Question 8. What is a hybrid organism? What will be the phenotype of the F₁ offspring from a cross between a hybrid black and a pure white guinea pig? Explain your answer.
Answer :

Hybrid organism

The progeny, produced from a cross between parents bearing two contrasting traits with respect to a single or more character, is called a hybrid organism.

Checker board of F₁ generation

Hybridisation experiment on guinea pig

In a cross between hybrid black (Bb) and pure pHHH white (bb) guinea pigs, .two different b ‘ phenotypes will be seen among the F1 offsprings. 50% of the offspring will have “b ~ black coat and the remaining 50% will be white coated. Therefore, the phenotypic ratio will be Black: White = 1:1. The cross is schematically represented here.

Explanation:

The genotype of the hybrid black guinea pig is ‘Bb’. Therefore, this parent produces two different types of gametes, ‘B’ and ‘b! The other parent, a pure white guinea pig is homozygous recessive having a genotype of ‘bb! This parent produces only ‘b’ type of gametes. Union of ‘B’ and ‘b’ gametes produce hybrid black ‘Bb’ offspring.

On the other hand, the fertilisation between two ‘b’ gametes from two parents gives rise to pure white ‘bb’ offspring. The checker board of the F1 generation of this cross clearly shows that the proportion of hybrid black and pure white offspring is equal.

Mendel’s Law of Independent Assortment

Question 9. The black coat of the guinea pig is dominant over the white coat. If a cross is performed between two hybrid black guinea pigs, what will be the result in the F1 generation? Which types of gametes unite to form a female baby?
Answer :

Crossing of hybrid black guinea pig

Let the allele for the dominant black coat colour be ‘B’ and the recessive white coat colour be ‘b’. So, the genotype of a hybrid guinea pig can be denoted as ‘Bb’.

A hybrid black-coated guinea pig may produce two different types of gametes ‘B’ 1 and ‘b’. Therefore, the F₁ progeny produced from a cross between two hybrid black guinea pigs will have two different phenotypes. 75% of them will be black and the remaining 25% will be white.

Hence, the phenotypic ratio for Black :

white is 3:1. If the genotypes of F₁ progeny are analysed on a checkerboard, we will see three different types—25% of it is pure black, 50% is hybrid black and the remaining 25% is pure white. Therefore, the genotypic ratio for pure black: hybrid black: pure white is 1:2:1.
The cross and the checkerboard is shown here.

checker board of F₁, generation

Gametes are required to produce a female baby

A Union of an ovum with 22 autosomes and an X chromosome (22A+X) and a sperm with 22 autosomes and an X chromosome (22A+X) can produce a female baby with genotype 44A+XX.

Question 10. Describe a dihybrid cross-experiment on an animal (guinea pig) and analyse the result.
Answer :

Dihybrid cross-experiment on animal

In the dihybrid cross experiment guinea pig, the coat colour and fur character are considered. The two opposite traits for coat colour are black and white. The opposite traits for fur characters are rough and smooth. The black coat (BB) and rough fur (RR) are dominant over the white coat (bb) and smooth fur (rr) respectively.

1. First filial generation:

First filial generation:

A cross is done between a black and rough-furred generation male and a white and smooth-furred female guinea pigs to get the F₁ offspring. Eventually, all these hybrid offsprings are with black coats and rough fur (BbRr).

2. Second filial generation:

The F1 offspring are allowed to breed among themselves. They produce four different types of phenotypes. 9 animals of which are black-rough furred, 3 animals are black-smooth furred, 3 animals are white-rough furred and remaining 1 animal is white-smooth furred.

Therefore, the phenotypic ratio of this dihybrid cross is 9:3:3:1.

The F₁ hybrids produce four different types of gametes— BR, Br, bR and br. These four different varieties of male and female gametes unite with each other in all possible combinations to produce F₂ generation with nine different types of genotypes in the ratio of 1:2:2:4:1:2:1:2:1.

The checkerboard for the above-mentioned cross is represented here.

Checker board of F₂ generation

Phenotype and genotype of F₂ generation of Mendelian dihybrid cross

Conclusion:

The result obtained from this experiment on the dihybrid cross of guinea pigs is the same as that of the Mendelian dihybrid cross of garden pea plants. This experimental result confirms that the law of independent assortment is also applicable to the heredity of the animals like guinea pigs.

Examples of Mendelian Traits

Question 11. Why did Mendel select garden pea plants for his experiments on heredity? Or, Mention the main causes of the selection of pea plants sativum) by Mendel for his experiments on heredity.
Answer :

Causes of Selection of pea plants by Mendel

1. Pea flower is naturally self-pollinating. The stamens and pistils always remain closed inside the petals, therefore, cross-pollination by insects and air does not occur. However, these flowers can be manipulated for artificial cross-pollination for experimental purposes.
2. The bisexual pea flowers can be emasculated and transformed into female flowers by simply cutting the stamens and removing the anthers.
3. Pea plants have distinct . contrasting phenotypic features and the characters remain unchanged for generations.
4. The life cycle of pea plants is short, hence, a number of generations can be studied within a few years. The dormancy of seed is brief, therefore, a new generation can grow quickly.
5. The hybrids of pea plants are fertile and these are suitable for both selling and crossing to produce the next generations.
6. Each pea plant produces several seeds, therefore, analysis of the result can be done more accurately.

Question 12. Describe the experimental works of Mendel on garden pea plants.
Answer :

Mendel’s work on garden pea plant

1. Mendel first visually selected seven distinct contrasting characters of the pea plant.
2. To become sure of the homozygosity of the parental generation, he allowed the selfing of selected character-bearing plants for several successive generations.
3. Before crossing, Mendel gently opened the petal cover to expose the stamens and cut the anthers before they mature. By this process, he emasculated the bisexual flowers to prevent undesired self-pollination.
4. For crossing, Mendel collected pollens from a selected flower from a parental plant with a fine brush and placed it delicately on the stigma of the emasculated flowers of the other parental plant, bearing the opposite trait. He performed reciprocal crossing between two parental pea plants.
5. After crossing, he covered the cross-pollinated flower with fine muslin cloth to prevent any undesired pollination.
6. He collected seeds from the resultant fruits of the cross and sowed them to get the generation.
7. Mendel meticulously studied the concerned characters of each generation and statistically analysed the result to reach a conclusion.

Through this process, Mendel performed separate experiments on different sets of contrasting traits before framing his famous laws of heredity.

Question 13. What is incomplete dominance? Explain with a suitable example.
Answer :

Incomplete dominance

In a cross between two pure contrasting trait-bearing parents, sometimes none of the traits dominates completely over the other so the phenotypic expressions of both the parents’ characters are subdued partially in the F₁ generation. This event is treated as incomplete dominance.

Explanation of incomplete dominance

Among plants, incomplete dominance is clearly noticed in four o’clock (Mirabilis jalapa) plants. In this plant, two contrasting flower colours are red and white. Let us mark the pure red flower as ‘RR’ and pure white as ‘. When these two contrasting flower-bearing plants are crossed, the F₁ generation grows neither red nor white flowers, instead, they grow pink flowers.

Here, phenotypic expressions of flower colour of both parents are subdued to some extent. When these F₁ plants are self-pollinated, three different phenotypes are produced. These are- 25% red, 50% pink and 25% white. The genotypes of red, pink and white flower-bearing plants are ‘RR, ‘Rr’ and ‘rr. Here, both phenotypic and genotypic ratios are identical, i.e., 1: 2: 1, which is a deviation from the Mendelian monohybrid cross.

Checker board of F₂ generation

The phenotypic and genotypic ratio of four o’clock flower of F₂ generation

Question 14. Distinguish between—monohybrid & dihybrid cross, homozygous and heterozygous state.
Answer :

Differences between monohybrid and dihybrid cross

Differences between homozygous and heterozygous state

Question 15. State the differences between pure and hybrid organisms. Mention differences between complete and incomplete dominance.
Answer :

Differences between pure and hybrid organisms

Differences between complete and incomplete dominance

Deviations from Mendel’s Laws

Question 16. Colour of seed and shape of seed of a pea plant—taking these two characters Mendel performed a dihybrid cross. Write the genotypes of pea plants having yellow and round seeds produced in the F₂ generation of this experiment. State the law of Independent Assortment of Mendel.
Answer :

The genotypes of yellow and round seeds of F₂ generation are—

1. YYRR
2. YYRr
3. YyRR
4. YyRr.

Law of independent assortment

Mendel derived this law from the results of his dihybrid cross-experiment. It states that—each pair of alleles segregates independently of other pair of alleles during gamete formation.

In other words, the allele a gamete receives for one gene does not influence the allele received for another gene.

Question 17. State the opposite traits for each of the characters of flowers of the pea plant chosen by Mendel for his experiment. “For developing the scientific ideas on heredity the experiments performed by Mendel on pea plants are epochmaking.”—Mention three reasons behind his success in performing these experiments.
Answer :

Opposite traits for the characters of flowers of pea plant chosen by Mendel

Three reasons behind success of Mendel’s experiment

1. Mendel concentrated only on one or two characters at a time and kept accurate records. This prevented confusion and puzzle.
2. The pea plant is an annual and self-pollinated plant. So plants are homozygous. Therefore the pure line of pea plants were easily available.
3. The genes controlling characters chosen by Mendel are located on different chromosomes. This prevents the linkage and gave accurate results.

Question 18. With the help of a checker board show the types of offspring that might be produced in a cross between a hybrid black guinea pig and a pure white guinea pig. State the law of segregation as proposed by Mendel.
Answer :

Monohybrid cross in Guineapig

The black and white guinea pig will be produced by the cross in 1:1 ratio.

Checkerboard

Phenotypic ratio – Black: White – 1: 1
Genotypic ratio – Bb: bb – 1: 1

Law of segregation

The law of segregation states that the two alleles for a heritable character segregate (separate from each other) during gamete formation and end up in different gametes.

Question 19. Tabulate three pairs of dominant-recessive traits of pea plant as selected by Mendel. State the first law of Mendel as deduced from the experiment of a monohybrid cross.
Answer :

Three pairs of dominant-recessive traits of pea plant as selected by Mendel

Question 20. Show the result of hybridisation till F₂ generation between a pure black (BB) coarse-haired (RR) guinea pig and a pure white (bb) smooth-haired (rr) guinea pig by a checkerboard. State the conclusion one can reach from this hybridisation experiment.
Answer :

Dihybrid cross in Guineapig

F₁………… Hybrid black and coarse-haired guinea pig.

Checkerboard of F₂ generation

Phenotype and genotype of F₂ generation

In a cross between pure black and coarse-haired guinea pig and pure white and smooth-haired guinea pig all the F₁ individuals will be black and coarse-haired guinea pig because these are dominant traits.

Each of the F₁ individuals can produce four types of gametes and their cross produce four types of individuals in a ratio of 9:3:3:1.

Conclusion

The conclusion that can be reached from this hybridisation experiment i.e. —when two organisms with two or more characters with their opposite phenotypes are crossed then the characters not only get separated but remain expressed in all possible combinations independent to each other.

Non-Mendelian Inheritance Patterns

Chapter 3 Mendel’s Laws And Their Deviation Short Answer Type Questions

Question 1. Mention the Mendelian laws of heredity.
Answer :

Mendel proposed two laws of heredity, they are-

1. Law of Segregation and
2. Law of independent assortment.

Question 2. How did Mendel explain the occurrence of all tall pea plants in the F₁, generation of his monohybrid cross experiment?
Answer :

In the monohybrid cross between pure tall and pure dwarf pea plants, all progeny of the F₁ generation came out as tall.

From this observation, Mendel concluded that the tall character is a dominant one, which was expressed by the suppression of the recessive dwarf character. From this finding, Mendel framed a law of heredity, known as the law of dominance.

Question 3. Which type of offspring will come out of a Which type of monohybrid cross between a pure black and pure white guinea pig? Why does it happen?
Answer :

In a cross between a pure black and a pure white guinea pig, all F₁ offspring will be hybrid black.

In guinea pigs, the black coat colour is dominant over the white coat colour. Therefore, the gene for black coat colour will express itself by suppressing the gene for white coat colour.

Question 4. Which law of heredity did Mendel frame from the results of his famous monohybrid cross experiment? State the law.
Answer :

Law of heredity from monohybrid cross:

From the results of the monohybrid cross experiment on garden pea plants, Mendel framed the law of segregation.

Law of segregation:

During hybridisation, the factors of each character are transmitted to the F₁ generation from their parents but do not get mixed. During gamete formation, these factors segregate from each other, so that each gamete receives only one factor for each character.

Question 5. What is a test cross?
Answer :

Test cross:

To determine the genotype of an individual exhibiting the dominant phenotype of a trait, it is crossed with an individual that is homozygous recessive for that trait. This cross is known as the test cross.

Example- To test the genotype of the tall F, pea plants in Mendel’s monohybrid cross, these can be crossed with homozygous recessive dwarf (tt) pea plants.

Question 6. What is a back cross?
Answer :

Back cross:

A cross between any of the parents or any individual with a genotype, identical to any of the parents, and a member of the F₁, generation or any individual with a similar genotype to that of the F1 progeny is known as the back cross.

Example- A cross between “TT” and “Tt’ and ‘tt’ and ‘T’ are examples of backcrosses with respect to Mendel’s monohybrid cross.

Question 7. Explain: ‘Every test cross is a back cross’.
Answer :

‘Every test cross is a back cross’:

Across between any of the parents ‘TT’ or ‘tt’ and a member of the F, generation “Tt’ is a back cross. In the Mendelian monohybrid cross, one of the parents is homozygous recessive (tt).
When this recessive homozygous (tt) parent is used for back cross with a F₁ progeny (Ti), it becomes a test cross.

Therefore, every test cross may be treated as a back cross

Question 8. What is incomplete dominance?
Answer :

Incomplete dominance:

Incomplete dominance is a genetic event where one of the alleles does not dominate completely over the other, but both the alleles express themselves partially when both are present together in the hybrid.

Question 9. What is a checkerboard? Why is it known as the Punnett square?
Answer :

Checkerboard:

 

Checker board:

The tabular representation, of a genetic cross show- ing all genotypes of offspring, produced from every possible combination of gametes, is known as checkerboard.

Punnett Square:

Geneticist Reginald C Punnett first represented a cross in tabular form. Therefore, checker board is also known as the Punnett Square, after his name.

Punnett Squares for Monohybrid Crosses

Question 10. Give an example of incomplete dominance in a plant and an animal.
Answer :

Example of incomplete dominance in a plant:

From a cross between red and white flowered four o’clock (Mirabilis jalapa) plants, we get pink flowers in F, generation.

Example of incomplete dominance in an animal:

If the pure black Andalusian hen is crossed with a pure white cock, the F, chicks will be blue in colour.

Question 11. Mention the contrasting traits of any two characters of the garden pea plant.
Answer :

Two characteristics of garden pea plants are height and flower colour. Opposite traits of height are tall and dwarf. Opposite traits of flower colour are violet and white.

Question 12. What will be the phenotypes of the two pea plants, when their genotypes are ‘TT’ and tt? Mention the chromosomal distribution of normal human males and females.
Answer :

Phenotypes of ‘TT’ and ‘tt’:

The phenotype of “TT” will be tall and ‘tt’ will be dwarf.

Chromosomal distribution of humans:

Chromo- somal distribution of normal human males is 44A+XY and that of females is 44A+XX.

Question 13. The genotype of a sweet pea plant bearing round and yellow seeds is ‘RrYy’. Which types of gametes may be produced from this plant?
Answer :

From a pea plant with the genotype ‘RrYy, four different types of gametes may be produced. These are ‘RY ‘Ry, ‘ry’ and ‘ry.

Question 14. In a dihybrid cross experiment, the black and rough-coated male is crossed with a white and smooth-coated female guinea pig. What will be the phenotypes of the F₂ progeny?
Answer :

In a cross between a black and rough-coated male with a white and smooth-coated female guinea pig, the F, progeny will be of four types.

These are-

1. Black and rough-coated,
2. Black and smooth coated,
3. White and rough coated and
4. White and smooth coated.

Question 15. What is a genome?
Answer :

Genome:

All the genes present in the haploid set of chromosomes of an individual organism are collectively known as the genome. The genome is carried through the chromosomes of a gamete from an individual parent to its offspring.

Question 16. Mention some human characters and their respective dominant and recessive traits.
Answer :

Question 17. How did Mendel explain the combinations of different dominant and recessive factors in F₂ generation of a dihybrid cross?
Answer :

In a dihybrid cross, all different types of gametes, carrying various combinations of dominant and recessive factors, unite randomly in all possible combinations to create the F₂, offspring.

During this process, no factor affects the expression of any other. Mendel explained this random distribution of factors as an independent assortment.

Question 18. What are the limitations of law of segregation?
Answer :

The limitations of law of segregation:

1. When one character is controlled by more than one gene then in that case, segregation may take place in a different ratio other than monohybrid cross.
2. In case of incomplete dominance segregation takes place but not in a ratio of monohybrid cross.

Question 19. What is checker board?
Answer :

Checker board:

Checker board is a method of expressing different combinations of genotypes through a tabular representation of the genetic cross. It was discovered by Reginald Punnet.

Question 20. Dwarf plants are always pure’-Justify it.
Answer :

Dwarf plants are always pure

Dwarf phenotype is a recessive phenotype. In the presence of a dominant allele, the recessive one is not expressed. Therefore recessive character is always expressed in homozygous. So dwarf plants are always homozygous or pure.

Question 21. State with an example how dominant trait is expressed in the experiment of hybridisation.
Answer :

The expression of the dominant trait can be explained with the example of hybridisation between pea plants bearing white and purple-coloured flowers.

In the case of peas, the purple flower colouration is controlled by a dominant gene (designated here as ‘P’), while the white colouration is controlled by a recessive gene (designated here as ‘p’).

When these (PP and pp) were bred to create the first generation, the offsprings were pea plants bearing purple flowers, with a genotype Pp. So, although both alleles were passed down, the recessive white colour alleles were masked by the dominant purple colour alleles

Question 22. What would be the ratio of phenotype and genotype of F2 generation in a monohybrid experiment in case of incomplete dominance?
Answer :

F₂ generation

Genotypic ratio
– RR: RW : WW
– 1: 2: 1

Question 23. What are all probable gametes to be produced from the hybrid plant YyRr produced at F₁ generation in a dihybrid cross experiment?
Answer :

The probable gametes to be produced from the hybrid plant YyRr at F₁, generation in a dihybrid cross experiment are four types which are YR, Yr, YR, yr.

Question 24. To establish the law of segregation in case of pea plants, you are asked to select characters. Write two such characters having opposite traits.
Answer :

Question 25. Different genotypes produce the same phenotype-Justify the statement in the form of a table by taking an example from the result of the dihybrid cross of pea plant.
Answer :

Dominant phenotypes are expressed in both homozygous and heterozygous forms. So a dominant phenotype may have two genotypes one is homozygous and one is heterozygous yellow seeded plants have a single phenotype with yellow seed colour but its genotype is either YY or Yy.

Question 26. State the limitations of the Law of segregation.
Answer :

The limitations of the Law of segregation are-

1. When a character is controlled by more than one pair of alleles, then though segregation occurs, it may not follow the Law of Segregation.
2. In the case of incomplete dominance, though segregation occurs, the phenotypic ratio deviates from that of Mendel’s monohybrid cross.

Question 27. State the limitations of the Law of independent assortment.
Answer :

The limitations of the Law of independent assortment are-

1. Often the genes on the same. chromosomes have a tendency of being transmitted together. This is called linkage. In this case, Law of independent assortment is not followed.
2. Sometimes, during gametogenesis, homologous chromosomes may pass to the same gamete. This is called non-disjunction. This phenomenon can not be explained by the Law of independent assortment.

Question 28. Differentiate between monohybrid and dihybrid cross.
Answer :

Difference between monohybrid and dihybrid cross:

Chapter 3 Mendel’s Laws And Their Deviation Multiple Choice Questions B Answers [Mcq]

Question 1. The laws of heredity were framed by

1. Hugo de Vries
2. Darwin
3. Lamarck
4. Mendel

Answer: 4. Mendel

Question 2. Mendelian concept deviates in

1. Law of segregation
2. Law of dominance
3. Law of incomplete dominance
4. Law of independent assortment

Answer: 3. Law of incomplete dominance

Question 3. A genetic cross between hybrid offspring and one of the homozygous parental types is known as

1. Back cross
2. Test cross
3.  Reciprocal cross
4.  Final cross

Answer: 1. Back cross

Question 4.In a cross between red (WW) and white (ww) flower-bearing evening primrose plants, all plants of F₁ generation bear pink (Ww) flowers. It is an example of

1. Hybridisation
2. Dominance
3. Incomplete dominance
4. Mutation

Answer: 3. Incomplete dominance

Question 5. A cross is performed between white (recessive) and violet (dominant) flower-bearing garden pea plants. What would be the percentage of white flower-bearing plants F2 generation?

1. 100%
2. 25%
3. 50%
4. 75%

Answer: 2. 25%

Question 6. What would be the genotypic ratio of the offspring from a test cross of an organism having an AaBb genotype?

1. 1:1:1:1
2. 1:2:1
3. 3:1
4. 9:3:3:1

Answer: 1. 1:1:1:1

Question 7. The findings of Mendel on heredity and the laws of inheritance are collectively known as

1. Mendelism
2. Monohybrid cross
3. Genetics
4. Dihybrid cross

Answer: 1. Mendelism

Question 8. The genotype of a pure dwarf pea plant is

1. TT
2. Tt
3. tt
4. Ttt

Answer: 3. tt

Question 9. How many pairs of contrasting traits of plants did Mendel select for carrying out his experiments on heredity?

1. One pair
2. Two pairs
3. Seven pairs
4. Nine pairs

Answer: 3. Seven pairs

Question 10. As per Mendel’s experiment on monohybrid cross, a cross is performed between a heterozygous tall (Tt) and a homozygous dwarf (tt) garden pea plant. What would be the percentage of tall pea plants in the first filial (F₁,) generation?

1. 25%
2. 50%
3. 75%
4. 100%

Answer: 4. 100%

Question 11. The phenotypic ratio in F₂ generation of Mendel’s monohybrid cross is

1. 1:2:1
2. 3:1
3. 9:3:3:1
4. 1:1

Answer: 3. 3:1

Question 12. The phenotypic ratio in F₂ generation of Mendel’s dihybrid cross is—

1. 1:2:1
2.  3:1
3. 9:3:3:1
4. 1:1

Answer: 3. 9:3:3:1

Question 13. What would be the percentage of dwarf pea plants in F₁ generation resulting from a cross between two hybrid tall pea plants?

1. 25%
2. 50%
3. 75%
4. 100%

Answer: 1. 25%

Question 14. Dwarf pea plants are always—

1. Homozygous
2. Heterozygous
3. Hybrid tall
4. Hemizygous

Answer: 1. Homozygous

Question 15. Name the Mendelian law that is derived from the dihybrid cross.

1. Law of segregation
2. Law of incomplete dominance
3. Law of dominance
4. Law of independent assortment

Answer: 4. Law of independent assortment

Question 16. A cross is performed between two hybrid tall pea plants. What would be the percentage of tall and dwarf offspring respectively, in the F₁ generation of this cross?

1. 50% and 25%
2. 25% and 25%
3. 75% and 25%
4. 50% and 50%

Answer: 3. 75% and 25%

Question 17. The pea plants, produced from the seeds resulting from a cross between hybrid tali (Tt) and pure dwarf (tt) pea plants will be—

1. All tall
2. All dwarf
3. 50% tall and 50% dwarf
4. 75% tall and 25% dwarf

Answer: 3. 50% tall and 50% dwarf

Question 18. The numerical ratio of pure black and pure white offspring obtained in the F₂ generation of a monohybrid cross between pure black (BB) and pure white (bb) guinea pig, is—

1. 1:1
2. 1:2
3. 2:1
4. 3:1

Answer: 1. 1:1

Question 19. The numerical ratio of pure white and hybrid black offsprings obtained in the F₂ generation of a monohybrid cross between pure black (BB) and pure white (bb) guinea pig, is—

1. 1:1
2. 1:2
3. 3:1
4. 2:1

Answer: 2. 1:2

Question 20. The ratio of pure genotypes produced in F₂ generation of a monohybrid cross is—

1. 1:1
2. 1:2
3. 1:3
4. 3:1

Answer: 1. 1:1

Question 21. The percentage of hybrid tall offspring obtained from a cross between two hybrid tall pea plants is—

1. 25%
2. 50%
3. 75%
4. 100%

Answer: 2. 50%

Question 22. The ABO blood group in humans is an example of— a”

1. Dominance
2. Codominance
3. Polygene
4. Incomplete dominance

Answer: 2. Codominance

Question 23. Mendel’s Law of segregation is derived from—

1. Monohybrid cross
2. Dihybrid cross
3. Hybridisation
4. Variation

Answer: 1. Phenotypic

Question 24. The 1:2:1 ratio in the F₂ generation of Mendel’s monohybrid cross is—

1. Phenotypic
2. Genotypic
3. Homozygotic
4. Heterozygotic

Answer: 2. Genotypic

Question 25. Self-pollination is possible in garden pea flowers because it is a—

1. Male flower
2. Female flower
3. Unisexual flower
4. Bisexual flower

Answer: 4.  Bisexual flower

Question 26. Which of the following indicates the test cross of Mendel’s monohybrid experiment?

1. Tt x TT
2. Tt x Tt
3. Tt x tt
4. TT x TT

Answer: 3. Tt x tt

Question 27. In which of the following cases, both phenotypic and genotypic ratios are 1:2:1 in F₂ generation of a monohybrid cross?

1. Complete dominance
2. Incomplete dominance
3. Super dominance
4. Mendelian dominance

Answer: 2. Incomplete dominance

Question 28. The allelic state or the genetic constitution of an organism is known as—

1. Allelomorph
2. Gene
3. Genotype
4. Phenotype

Answer: 3. Genotype

Question 29. Which of the following is a dominant character?

1. Stem height-Dwarf
2. Shape of seed-Wrinkled
3. Cotyledon colour-Yellow
4. Flower colour-White

Answer: 3. Cotyledon colour-Yellow

Question 30. Which of the following is a recessive character?

1. Wrinkled seed
2. Purple flower
3. Yellow coloured seed
4. Axillary flower

Answer: 1. Wrinkled seed

Question 31. o’ is a symbol of—

1. Bisexual
2. Male
3. Female
4. None of the above

Answer: 2. Male

Question 32. Which of the following cross will produce a 1:1 ratio?

1. TTxtt
2. TTxTt
3.  tTtxTt
4. Ttxtt

Answer: 4. Ttxtt

Question 33. How many genotypes are possible of a tall pea plant?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 34. How many individuals out of 500 will be heterozygous in the cross Tt x Tt?

1. 100
2. 250
3. 350
4. 500

Answer: 2. 250

Question 35. How many types of genotypes are obtained in Mendel’s monohybrid cross in the F₂ generation?

1. 1
2. 2
3. 3
4. 4

Answer: 3. 3

Question 36. A test cross is done between—

1. F1x F1
2. F1 x dominant parent
3. F1 x recessive parent
4. None of the above

Answer: 3. F1 x recessive parent

Question 37. The phenotypic ratio of offspring obtained from a cross between a homozygous (bb) and heterozygous (Bb) individual—

1. 3:1
2. 2:1
3. 1:1
4. 1:2

Answer: 3.  1:1

Question 38. How many types of gametes will be produced from YyRr?

1. 1
2. 2
3. 3
4. 4

Answer: 4.  4

Question 39. How many types of gametes will be produced from the genotype YYRR?

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 40. How many types of genotypes are obtained in Mendel’s dihybrid cross F2 generation?

1.  4
2. 6
3. 8
4. 9

Answer: 4. 9

Question 41. Which of the following traits in the pea plant is recessive?

1. Wrinkled seed
2. Yellow coloured seed
3. Purple coloured flower
4. Axial flower

Answer: 1. Wrinkled seed

Question 42. How many types of gametes are formed from pea plants having genotype YyRr?

1. 1
2. 4
3. 2
4. 3

Answer: 2. 4

Question 43. Which of the following is the genotypic ratio generation of Mendel’s monohybrid cross?

1. 1:2:1
2. 3:1
3. 9:3:3:1
4. 2:1:2

Answer: 1. 1:2:1

Question 44. What would be the phenotypic ratio in the F2 generation of a monohybrid cross in case of incomplete dominance?

1. 3:1
2. 2:1:1
3. 9:3:3:1
4. 1:2:1

Answer: 4. 1:2:1

Question 45. Asses how many types of gametes are produced from the pea plant having the genotype RRYY.

1. One type
2. Four types
3. Two types
4. Three types

Answer: 1. One type

Question 46. Identify which of the following is a dominant trait—

1. Length of the stem—Dwarf
2. The shape of the seed—Wrinkled
3. Colour of the cotyledon—Yellow
4. Colour of the flower—White

Answer: 3. Colour of the cotyledon—Yellow

Question 47. Select which of the following two genotypes are responsible for the expression of the phenotype wrinkled yellow in pea plants—

1. RRYY and rryy
2. RRYy and RrYy
3. RRyy and Rryy
4. rrYY and rrYy

Answer: 4. rrYY and rrYy

Question 48. Identify the genotype of a guinea pig having black colour and rough hair—

1. BbRr, BBRr
2. BBrr, Bbrr
3. bbRR, bbRr
4. bbrr, bbRr

Answer: 1.  BbRr, BBRr

Question 49. Decide which of the following two were selected by Mendel as recessive traits—

1. Colour of flower—purple, the position of the flower—axial
2. Length of stem—dwarf, form of ripe seed—wrinkled
3. Form ol ripe seed—round, colour of seed—yellow
4. Position of flower—axial, length of stem—tall

Answer: 2. Length of stem—dwarf, form of ripe seed—wrinkled

 

Chapter 3 Mendel’s Laws And Their Deviation VSAQs

Question 1. Write down the scientific name of the garden pea plant.
Answer: Pisum sativum

Question 2. What is the phenotypic ratio in F₂ generation of a monohybrid cross?
Answer: The phenotypic ratio in F₂ generation of a monohybrid cross is, Dominant: Recessive=31.

Question 3. What is the genotypic ratio in F₂ generation of a monohybrid cross?
Answer: The genotypic ratio in F₂ generation of a monohybrid cross is—Pure dominant : Hybrid: Pure recessive = 1:2:1.

Question 4. Give an example of a dominant character of guinea pig.
Answer: Black coat colour is a dominant character of guinea pig.

Question 5. In a monohybrid cross, the phenotypic and genotypic ratio of F2 generation are 3:1 and 1:2:1 respectively. What does this observation indicate?
Answer: This observation indicates the complete dominance of the concerned character.

Question 6. What does ‘Tt’ indicate in Mendel’s experiment?
Answer: In Mendel’s experiment, ‘Tt’ indicates the genotype of a hybrid tall pea plant.

Question 7. Which type of pea plant is denoted by ‘TT’ in Mendel’s experiment?
Answer: In Mendel’s experiment, ‘TT’ indicates a pure tall pea plant.

Question 8. Which type of pea plant is denoted by ‘tt’ in Mendel’s experiment?
Answer: In Mendel’s experiment, ‘tt’ indicates a pure dwarf pea plant.

Question 9. Mention a character and its opposite traits found in guinea pig.
Answer: Coat colour is a character of guinea pig, which has two opposite traits—black and white.

Question 10. Which is the dominant colour of flowers of garden pea plants?
Answer: The dominant colour of flowers of garden pea plant is violet.

Question 11. Which ratio indicates the external features of the progeny of a hybrid cross?
Answer: Phenotypic ratio

Question 12. In which journal Mendel’s paper on the heredity of garden pea plants was published?
Answer: Mendel’s paper on the heredity of garden pea plants was published in the Journal of Natural History Society at Brunn.

Question 13. What does the ratio of 3:1 indicate in genetics?
Answer: The ratio of 3:1 indicates the phenotypic ratio of the F₂ progeny of the monohybrid cross.

Question 14. What does the ratio ail :2:1 indicate in genetics?
Answer: The ratio of 1:2:1 indicates the genotypic ratio of the F₂ progeny of monohybrid cross

Question 15. What does the ratio of ot9:3:3:1 indicate in genetics?
Answer: The ratio of 9:3:3:1 indicates the phenotypic ratio ot the F₂ progeny of dihybrid cross.

Question 16. Which Mendelian law is derived from the result ot a monohybrid cross experiment?
Answer: The law of segregation is derived from the result ol Mendel s monohybrid cross experiment.

Question 17. Which Mendelian law is derived from the result of a dihybrid cross experiment?
Answer: The law of independent assortment is derived from the result of Mendel’s dihybrid cross experiment.

Question 18. Who first designed the checker board?
Answer: Reginald C Punnett, a British geneticist, first designed the checker board.

Question 19. What does Mendelian factors mean?
Answer: Mendelian factors are the alleles of genes, which are responsible for expression of various hereditary characters

Question 20. What is mulatto?
Answer:

Mulatto

The medium complexioned offspring, bom to pure black and pure white parents, is called mulatto

Question 21. In case of guinea pig whether the phenotype of the two genotypes bbRR and bbRr is same?
Answer: Yes, the phenotype of the two genotypes bbRR and bbRr will be the same in case of guinea pig.

Question 22. Write the phenotypic ratio obtained in the F2 generation of Mendel’s dihybrid cross experiment.
Answer: 9:3:3:1

Question 23. Write the genotypic ratio obtained at 2nd filial generation in the monohybrid experiment done by Mendel.
Answer: 1:2:1

Question 24. Which law did Mendel conclude from his dihybrid cross experiment?
Answer: Law of independent assortment.

Impact of Environment on Genetic Expression

Chapter 3 Mendel’s Laws And Their Deviation Fill In The Blanks

Question 1. An individual which carries both dominant and recessive alleles is called _______ individual.
Answer: Hybrid

Question 2. Mendel’s first law of heredity is known as the law of ______
Answer : Segregation

Question 3. Mendel’s second law of heredity is known as the law of _______
Answer: Independent assortment

Question 4. In a monohybrid cross between red and white flower-bearing evening primrose plants, the phenotypic ratio of the F₂ progeny is ______
Answer: 1:2:1

Question 5. The parental generation of a monohybrid cross is marked as ______ generation.
Answer : P

Question 6. Mendel performed ______ pollination between the flowers of pure tall and pure dwarf pea plants.
Answer: Cross

Question 7. To get F₂ generation, Mendel allowed ______ pollination among F1 plants.
Answer: Self

Question 8. The yellow colour of cotyledons of pea plants is a ______character.
Answer: Dominant

Question 9. Hybridisation between two traits of a single character is known as ______ cross.
Answer: Monohybrid

Wbbse Class X Life Science

Question 10. Permanent and heritable change in the number and structure of gene or chromosome is known as _______
Answer: Mutation

Question 11. Mendel selected ______ number of characters of pea plant for his experiment.
Answer: Seven

Question 12. Different ______ of an individual may show the same phenotype.
Answer: Genotypes

Question 13. The white and smooth fur of a guinea pig is a __________ character.
Answer: Recessive

Question 14. The ratio of TT:Tt:tt in monohybrid cross is ______
Answer: 1:2:1

 

Chapter 3 Mendel’s Laws And Their Deviation State True Or False

Question 1. A temporary change in the number and structure of a gene or chromosome is known as mutation.
Answer: False

Question 2. Mendel selected seven pairs of opposite characters of the garden pea plant for his experimental observations.
Answer: True

Wbbse Class X Life Science

Question 3. The scientific name of evening primrose is Mirabilisjalapa.
Answer: False

Question 4. In F₂ generation of Mendel’s dihybrid cross, both the phenotypic and genotypic ratio is 9: 3:3:1.
Answer: False

Question 5. In Mendel’s monohybrid cross, all F₁ offsprings have pure dominant character.
Answer: False

Question 6. In genetics, cross denotes the union of a male and a female gamete.
Answer: True

Question 7. Round seed is a recessive character.
Answer: False

Question 8. Each character selected by Mendel was controlled by multiple alleles.
Answer: False

Question 9. In pea plant, both selfing and cross-pollination can be done.
Answer: True

Question 10. Mendel obtained 75% tall plant in F₁ generation of a monohybrid cross.
Answer: False

Wbbse Class 10 Life Science Solutions

Question 11. Back cross always means test cross.
Answer: False

Question 12. In the Evening primrose plant incomplete dominance is seen.
Answer: True

Common Misconceptions About Mendel’s Laws

Question 13. Recessive character is always expressed in homozygous conditions.
Answer: True

Wbbse Class X Life Science

Question 14. The dominant phenotype may be both homozygous or heterozygous.
Answer: True

Question 15. In his monohybrid cross experiment, Mendel obtained 75% pure tall pea plants in the first filial generation.
Answer: False

Question 16. Mendel used the term ‘gene’ while describing his experiments related with heredity.
Answer: False

 

Chapter 3 Mendel’s Laws And Their Deviation Match The Columns

Question 1.

Answer: 1. B, 2. C, 3. E, 4. F, 5. A, 6. G

Question 2.

Answer: 1. B, 2. F, 3. D, 4. E, 5. A, 6. G

 

Chapter 3 Mendel’s Laws And Their Deviation Find The Odd One Out

Question 1. Grefcn colour of the pod, White colour of the flower, Round shape of seed, Axial position of the flower
Answer: The white colour of the flower

Question 2. Crossing over, Meiosis, Genetic recombination, Mitosis
Answer: Mitosis

Question 3. Law of segregation, Law of independent assortment, Law of use and disuse, Law of dominance
Answer: Law of use and disuse

Wbbse Class X Life Science

Question 4. 9:3:3:1, independent assortment, 3:1 BbSs x BbSs
Answer: 3:1

Question 5. TT, Tt, tt, rr
Answer: rr

Wbbse Class 10 Life Science Solutions

Question 6. Round wrinkled, Yellow wrinkled, Green wrinkled, Tall
Answer: Tall

Question 7. Purple flower x white flower Yellow seed x green seed, Round seed x wrinkled seed, Yellow Round Seed x green wrinkled seed
Answer: Yellow Round seed x Green Wrinkled seed

Question 8. Purple flower, White flower, Green pea seed, Axial flower
Answer: White flower

 

Chapter 3 Mendel’s Laws And Their Deviation Fill In The Blanks By Looking At The First Pair

Question 1. Monohybrid cross: Law of segregation:: Dihybrid cross: ______
Answer: Law of independent assortment

Question 2. Purple flower: Dominant character:: White flower:______
Answer: Recessive character

Wbbse Class X Life Science

Question 3. Monohybrid phenotype ratio : 3:1 :: Dhybridphenotypic ratio :______
Answer: 9:3:3:1

Question 4. Mendelian inheritence: Pea plant :: Mendel’s deviation:______
Answer: Mirabilis Jalapa

Question 5. Mendel’s first law: Law of segregation:: Mendel’s second law: ______
Answer: Law of independent assortment

Question 6. Cross of P generation : Cross-pollination :: Cross of F₁ generation:______
Answer: Self-pollination

Question 7. Axial flower: Dominant Character:: Terminal flower: ______
Answer: Recessive character

 

Chapter 3 Mendel’s Laws And Their Deviation Among The Four Concepts Given, Three Of Them Belong To One. Find That

Question 1. Monohybrid cross, Dihybrid cross, Hybridisation technique, Crossbreeding
Answer: Hybridisation technique

Question 2. TT x tt, Monohybrid Cross, Tt x Tt, BB x bb
Answer: Monohybrid Cross

Wbbse Class 10 Life Science Solutions

Question 3. Green Round, F₂ phenotype, Yellow Wrinkled, Yellow Round
Answer: F₂ Phenotype

Question 4. Mirabilis jalapa, Pink flower, Incomplete dominance, 1:2:1
Answer: Incomplete dominance

Question 5. Purple flower, Tall plant, Yellow seed, Dominant Character
Answer: Dominant character

Chapter 3 Some Genetic Diseases Summary

WBBSE Class 10 Genetic Diseases Overview

• There are some genetic diseases like-Thalassemia, Haemophilia, and Colour blindness seen in the human population.
• Thalassemia is an autosomal recessive disease. It is mainly of two types-a and B thalassemia. On the basis of severity, it is of two types-Highly severe thalassemia major and relatively less severe thalassemia minor.
• In thalassemia, globin protein synthesis is disturbed resulting in Anaemia. Due to the high accumulation of iron, deformities in bone are generated.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

• Haemophilia is an X-linked disease in which due to the absence of blood clotting factor, blood clotting does not take place.
  Haemophilia on the basis of severity is of three types-

1. Light haemophilia [Blood clotting factor 5-50%]
2. Moderate haemophilia [Blood clotting factor 1-5%]
3. Severe haemophilia [Blood Clotting factor <1%]

• On the basis of the type of blood clotting factor that is absent, haemophilia is of 3 types- Haemophilia A, haemophilia B or Christmas disease and haemophilia C.
• Colour blindness is another X-linked recessive disease where the affected person can not identify blue, green or red colour.
• Cololur blindness is of three types-Protanopia [a person can not recognise red colour], Deuteranopia [a person can not recognise green colour], and Tritanopia [a person can not recognise Blue colour]. The inheritence pattern of haemophilia, thalassemia and colour blindness can be calculated by a genetic cross.
• Genetic counselling is important to prevent genetic disease. Through genetic counselling, the chance percentage of a disease can be calculated.

Chapter 3 Some Genetic Diseases  Long Answer Type Questions

Question 1. What is thalassemia? Briefly describe different types of thalassemia.

Answer:

Thalassemia:

The genetic blood disorder, in which any of the globin peptide chains of haemoglobin is not synthesised at all or produced in much less quantity than normal, is known as thalassemia.

In this disease, the quantity of haemoglobin is reduced, and RBC becomes smaller and abnormally shaped.

Wbbse Class X Life Science

Types of thalassemia:

Two types of thalassemia are seen in humans. These are-

1. α-thalassemia and
2. β-thalassemia.

1. α-thalassemia:

In this type of thalassemia, synthesis of the α-globin peptide chain of haemoglobin is either reduced or stopped. In the 16th homologous chromosome pair, there are two genes (HBA1 and HBA2) and their two sets of alleles to control a-globin peptide synthesis.

If a mutation occurs in one set of alleles, a-thalassemia. minor occurs. When both sets of alleles are mutated, the most severe, a-thalassemia major occurs.

2. β-thalassemia:

If the synthesis of the β-globin peptide chain of haemoglobin is reduced or stopped in the body, a person suffers from β-thalassemia. A pair of alleles of the HBB gene is located on the 11th homologous pair of chromosomes of a human cell responsible for β-globin peptide synthesis.

In case of mutation in one of the alleles, β-thalassemia minor occurs. If both the alleles undergo mutation, β-thalassemia major occurs. American physician Thomas Benton Cooley discovered β-thalassemia, hence this disease is also known as Cooley’s anaemia.

Common Genetic Disorders in Humans

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WBBSE Class 10 History Multiple Choice Questions

 

Question 2. Mention the symptoms of thalassemia.
Briefly describe the cause of thalassemia.

Answer:

Symptoms of thalassemia:

1. Severe anaemia due to reduced synthesis of haemoglobin.
2. Rapid breakdown of RBC leads to the release and accumulation of iron in different organs, resulting in their dysfunction.
3. Excessive proliferation of bone marrow leads to deformity of bones, especially of the face and skull.
4. Enlargement of the liver and spleen occurs. It leads to jaundice, stunted growth and a weakened immune system.

Cause of thalassemia:

Wbbse Class X Life Science

Thalassemia occurs due to mutation in the globin-synthesising genes. The haemoglobin molecule is composed of two a and β-globin peptide chains. The genes for synthesis of a and β-globin are located on the 16th and 11th autosomes of a human cell.

The mutant a and β-globin genes hinder the synthesis of respective globin chains, resulting into abnormal haemoglobin that leads to thalassemia.

Question 3. What is haemophilia? Briefly describe the different types of haemophilia.

Answer:

Haemophilia:

Haemophilia is a genetic disorder, caused by an X chromosome-linked mutant gene, in which bleeding time from any wound is so prolonged that a person suffers from a life-threatening level of blood loss.

Mutation in a single gene on X-chromosome causes this disease in males, while a mutation in two genes on both the X-chromosomes will cause haemophilia in females.

Types of haemophilia

In humans, two types of haemophilia are seen, these are-

1. Haemophilia A or classical haemophilia:

In case of deficiency of a blood coagulating factor, Factor VIII or anti-haemophilic factor (AHF) in plasma, this serious type of haemophilia occurs. In our country, 80% of haemophilic persons suffer from this particular disease.

2. Haemophilia B or Christmas disease:

Deficiency of a blood coagulating factor called factor IX or Christmas factor or plasma thromboplastin component (PTC) in plasma, causes this type of haemophilia. This is not as deadly as haemophilia A.

In India, 20% of haemophilic persons suffer from haemophilia B. This disease was first detected in a person named Stephen Christmas. Therefore, this is also known as Christmas disease.

Question 4. Mention the symptoms and causes of haemophilia.

Answer:

Symptoms of haemophilia:

Based on intensity, haemophilia may be classified into three different types-mild, moderate and severe haemophilia, which have different symptoms, as mentioned below.

1. Symptoms of mild haemophilia:

The symptoms of mild haemophilia remain suppressed for a long time. With ageing, the problem of blood clotting appears slowly. The problem of bleeding becomes prominent during surgical operations or major injury.

2. Symptoms of moderate haemophilia:

Wbbse Class X Life Science

Symptoms of moderate haemophilia appear since birth. The patients suffer from prolonged bleeding. Internal haemorrhage occurs in bone joints. Swelling of joints and joint stiffness are also seen.

3. Symptoms of severe haemophilia:

Patients suffering from severe haemophilia often bleed for no apparent reason, known as spontaneous bleeding. In the case of severe haemophilia, profuse internal haemorrhage occurs in bone joints resulting in to serious joint deformity (hemarthrosis).

Besides, haemorrhage also occurs from nasal mucosa, within the skull, which may result in to paralysis, even death of the patient.

Causes of haemophilia:

The causes of haemophilia A and haemophilia B are mentioned below.

1. Cause of Haemophilia A or classical haemophilia:

Deficiency of blood coagulating factor, called factor VIII or anti-haemophilic factor (AHF) in blood.

2. Cause of Haemophilia B or Christmas disease:

Deficiency of blood coagulating factor, called plasma thromboplastin component (PTC) or factor IX in blood.

Question 5. Mention the types and causes of colour blindness.

Answer:

Types of colour blindness:

Different types of colour blindness are seen in the human population.

1. Protanopia or red colour blindness:

In this disease, an individual cannot detect red colour. They see red as black or deep brown, orange-yellow-green as different shades of yellow and violet as blue.

2. Deuteranopia or green colour blindness:

In the case of deuteranopia, one cannot detect green colour. Deutaronopes cannot detect red, orange and yellow properly but their colour vision is better than protanopes.

3. Tritanopia or blue colour blindness:

In blue colour blindness, a person cannot detect blue colour.

Cause of colour blindness:

Colour detection in human eye depends upon the cone cells of the retina. There are three types of cone cells in human retina, which are sensitive to red, green and blue colours.

These cells contain separate photopsin pigments to sense the wavelengths of red, green and blue lights. Synthesis of red and green photopsin pigments are controlled by X-linked genes, whereas, the blue photopsin synthesis is controlled by an autosomal gene. In case of mutation in these genes, respective colour blindness occurs.

Mendelian Inheritance and Genetic Diseases

Question 6. What is meant by genetic counselling? Discuss the role of genetic counselling in combating thalassemia.

Answer:

Genetic counselling:

By analysing the history of genetic diseases of family members and performing some genetic tests, the probability of any genetic disorder being transmitted to the future generations can be predicted.

Based on these investigations, a pre-marriage suggestion is provided to a couple by the genetic counsellor so that they can have healthy offspring. This is known as genetic counselling.

Thalassemia and genetic counselling:

Thalassemia is an inherited genetic disease. Therefore, genetic counselling is needed to combat this disease. In this disease, haemoglobin is synthesised in insufficient amounts and thus oxygen transportation is hampered.

As a result, huge quantities of iron accumulate in the heart, liver and endocrine glands of the body. Therefore, genetic screening of haemoglobin synthesising genes must be done before marriage in families with histories of thalassemia.

Using this data, the probability of children having thalassemia can be reduced. With the help of genetic counselling, the probable percentages of offspring having thalassemia major or minor from thalassemia minor parents can be determined.

Thus genetic counselling plays very important role in combating thalassemia.

Wbbse Class X Life Science

Question 7. How is genetic counselling performed? Mention the importance of genetic counselling.

Answer:

Method of genetic counselling:

Generally, a genetic counsellor can offer genetic counselling. This counselling includes the following steps-

1. At first, the counsellor collects the history of genetic diseases of different family members and related bloodlines. The collected data is then analysed to detect the genealogical trend of any disease.
2. Next, the cells or blood samples of the concerned individual are collected and tested in laboratories to find any genetic abnormality.
3. If any person is found with a genetic abnormality, the same tests are performed on his or her partner.
4. After analysing the genetic status of both partners, the chance of inheritance of any disease and its expected intensity are estimated.
5. Now, the couple is explained about the genetic status of their expected baby, either before their marriage or before conceiving.
6. If a foetus is found positive to any serious genetic disease, the parents are advised for medical term of pregnancy.

 

 

Importance of genetic counselling:

Occurrence of thalassemia, haemophilia and some other genetic diseases may be determined by genetic counselling. Genetic counselling has been proven effective if-

1. Both or any of the partners carry thalassemia-causing genes.
2. The parents are aged, especially when the mother is in her forties.
3. Genealogical lines show any history of thalassemia or other genetic diseases.
4. Foetal death occurs or the baby is born with a genetic disorder.

Question 8. Differentiate between thalassemia and haemophilia.

Answer:

Differences between thalassemia and haemophilia:

Life Science Class 10 Wbbse

Causes of Genetic Disorders

Question 9. What symptoms are expressed in a child affected with thalassemia? What suggestions are given at the time of genetic counselling in such cases?

Answer:

Symptoms of thalassemia affected child:

1. Continuous fatigue or weakness,
2. Shortness of breath,
3. Yellowing of skin and eyes,
4. Slow growth and abdominal swelling,
5. Facial bone deformities and anaemia,
6. Dark colouration of urine

The suggestions are given at the time of genetic counselling

1. Not to conduct marriage between carrier male and female.
2. Not to conduct marriage between thalamic male and normal or carrier female or vice-versa.
3. Must rest for genetic diseases before marriage or child planning

Question 10. A colour-blind female married a normal male. Judge the probability of colour blindness among their children in the first filial generation. Show with the help of a cross how the first law of Mendel is deviated in the case of the Four o’clock plant in F₂ generation.

Answer:

Inheritance of colour blindness:

Life Science Class 10 Wbbse

 

 

Checker board of F₁ generation:

In the F, generation all the female born will be carrier and all the male born will be colour blind.

Deviation from Mendel’s first law:

 

Checker board of F₂ generation

Phenotypic ratio- Red: Pink: White = 1:2:1

Genotypic ratio-RR: RW: WW = 1:2:1
In the F₁, generation the red and white colour is not segregated in a 3:1 ratio rather the phenotypic ratio is 1:2: 1. The phenotypic and genotypic ratios both are the same i.e. 1: 2: 1: Thus in this case Mendel’s 1st law is clearly deviated.

Symptoms of Common Genetic Diseases

Chapter 3 Some Genetic Diseases Short Answer Type Questions

Question 1. Name two autosomal and two sex-linked hereditary diseases.
Answer:

Autosomal

1. Two autosomal hereditary diseases are- thalassemia and albinism.

Two sex-linked hereditary diseases

1. Two sex-linked hereditary diseases are- haemophilia and colour blindness.

Question 2. What is thalassemia?
Answer:

Thalassemia:

Thalassemia is an autosomal hereditary disease caused due to abnormal synthesis of globin peptide chains of haemoglobin. The formation of haemoglobin in a thalassemic patient is either stopped or reduced.

RBC of a thalassemic patient is smaller in size and of shorter life span (the normal life span of RBC is 120 days) and the patient suffers from severe anaemia.

Question 3. What is thalassemia major?
Answer:

Thalassemia major:

When both a and B-globin chains of haemoglobin are incompletely formed due to genetic mutation in both a. and ẞ-globin genes located on the 16th and 11th chromosomes respectively, the disease that occurs is known as thalassemia major.

Question 4. What is haemophilia?
Answer:

Haemophilia:

Haemophilia is a hereditary haemorrhagic disease, caused by an X-linked recessive gene. In this disease, blood lacks the capacity to coagulate and a patient may bleed to death even due to a small cut.

This disease occurs due to a deficiency of any one of the two blood-coagulating factors- factor VIII and factor IX.

Question 5. What is thalassemia minor?
Answer:

Thalassemia minor:

When any one of a or β-globin chains of haemoglobin is incompletely formed due to mutation i.e., there is one defective allele, the disease that occurs, is called thalassemia minor.

Question 6. What are the different types of haemophilia occurring in man?
Answer:

The different types of haemophilia occurring in man:

Generally, two types of haemophilia occur in men. These are haemophilia A and haemophilia B. Besides, there is another variety of haemophilia, seen among a few people. This is known as haemophilia C.

Genetic Counseling for Inherited Disorders

Question 7. What is the cause of haemophilia A?
Answer:

The cause of haemophilia A:

In the case of a specific recessive gene mutation in the X chromosome, a blood-coagulating factor, namely Factor VIII or anti-haemophilic factor (AHF) is not synthesised in the body. A deficiency of AHF prevents blood coagulation and causes haemophilia A.

Life Science Class 10 Wbbse

Question 8. Why is haemophilia more common among males?
Answer:

Haemophilia more common among males because

Haemophilia is an X-linked recessive genetic disorder. It is more common among males because they inherit only one X-chromosome. Since males carry one X-chromosome and one Y-chromosome, they develop symptoms of the disease if the single X-chromosome carries the recessive mutant allele.

Question 9. Why a carrier female of haemophilia A gene does not get affected by the disease?
Answer:

Somatic cells of a female carry two X chromosomes. When one of the two X chromosomes carries the recessive allele for haemophilia but the other carries the normal dominant allele, the female becomes a carrier.

This normal dominant allele suppresses the recessive haemophilia allele and prevents the female from being haemophilic. Therefore, a female, carrying a single recessive allele for haemophilia is not affected by haemophilia A.

Question 10. What is the cause of haemophilia B?
Answer:

The cause of haemophilia B:

In the case of a specific recessive gene mutation in the X chromosome, a blood-coagulating factor, namely factor IX or Christmas factor is not synthesised in the body. Absence or deficiency of this factor prevents blood coagulation and causes Haemophilia B.

Question 11. What is crisscross inheritance?
Answer:

Crisscross inheritance:

The transmission of a genetic character from father to daughter and from mother to son is known as crisscross inheritance. All X-linked characters show crisscross inheritance.

In this type of inheritance, the following inheritance pattern is seen-Father → Daughter → Grandson → Great granddaughter.

Question 12. Name few genetic disorders that can be prevented by undergoing genetic counselling.
Answer:

Genetic counselling is considered by couples before marriage to prevent the occurrence of any genetic disorder in their children. Examples of genetic diseases that can be prevented by undergoing genetic counselling are thalassemia, haemophilia, colour blindness, etc.

Life Science Class 10 Wbbse

Question 13. What is colour blindness?
Answer:

Colour blindness:

Colour blindness is a genetic disorder in which a person cannot detect red, green or blue colour. Red-green colour blindness occurs due to X-linked recessive mutation, but blue colour blindness is an autosomal recessive disorder.

Question 14. What is monochromacy?
Answer:

Monochromacy:

In the human retina, there are three types of colour-sensitive cone cells, which can detect red, Checkerboard green and blue colours separately. If 2 or 3 types of pigmented cone cells are absent the colour, and detecting ability is totally lost.

Such a person can view the world in black and white. This rare type of colour blindness is known as monochromacy.

Question 15. What is dichromacy?
Answer:

Dichromacy:

If out of the three types of cone cells, two remain active and one inactive, a person cannot detect that particular single colour. This type of colour blindness is called dichromacy.

The common red, green or blue colour blindness are an example of dichromacy.

X-Linked Genetic Disorders Explained

Question 16. Write down the symptoms of Thalassemia.
Answer:

Symptoms of Thalassemia

1. Tiredness and fatigue.
2. Malformation of bone, especially bone of the face.
3. Blackish urine.
4. The yellowish colouration of skin and retardation of growth.

Question 17. Which of the three organs of the human body are damaged due to thalassemia?
Answer:

Liver, spleen and heart of human is damaged in thalassemia due to high accumulation of iron.

Question 18. Why haemophilia is more expressed in males than females?
Answer:

Haemophilia is a sex-linked recessive disorder whose gene is present on X chromosome. Males have only one X chromosome so if it fears the allele for haemophilia then it is always expressed.

But females have two X chromosomes. So for be haemophilic, each of the two must have the allele for chromosomes haemophilia, otherwise the disease is not expressed in females, so haemophilia is more expressed is females than in males.

Wbbse Class 10 Life Science Solutions

Question 19. If the mother is a haemophilic carrier and the father is haemophilic then what will be the nature of spring? Show it with a cross.
Answer:

 

 

Checker board

 

The nature of the offspring: 50%-haemophilic, 25%-Carrier, 25%-Normal.

Question 20. What will be the genotype of the father of a haemophilic girl?
Answer:

As both the chromosomes of a haemophilic girl contain an allele for colour blindness. So the father must have the X chromosome with an allele for colour blindness. He also possesses a Y chromosome. So his genotype is XCY.

Autosomal Recessive vs Dominant Disorders

Question 21. If the mother is colour blind then the son is always colour blind-Justify.
Answer:

A colour-blind mother always has two X chromosomes, each with an allele for colour blindness, i.e. genotype is XCXC. A son gets his X chromosome from the mother and his Y chromosome from the father.

So he will get an Xc chromosome from his mother and his genotype will be XCY. Thus he will be colour-blind.

Question 22. What is genetic counselling?
Answer:

Genetic counselling:

Genetic counselling is a process through which by analyzing the history of genetic disease Xin a family and by a genetic test, a prediction is made about the chances of the offspring to be normal or with the disease.

Wbbse Class 10 Life Science Solutions

Question 23. What is the importance of genetic counselling?
Answer:

The importance of genetic counselling:

1. It prevents the birth of child with a genetic disease.
2. It can help in calculating the chances of genetic disease of a child.

Question 24. Show with the help of the cross, how colour blindness is inherited.
Answer:

Colour blindness is a genetic disorder, caused by an X chromosome-linked mutant gene. Mutation in a single gene causes this disease in males, while a mutation in two genes on both the X- chromosomes will cause colour blindness in females.

So, when a colour-blind woman is married to a normal man, their sons will be colour-blind while daughters will be carriers of colour blindness. The cross is schematically represented below.

 

 

Checker board

 

Question 25. A daughter is born to a woman carrier for the colour-blind disease who married a colour-blind man. What would be the probability of expression of colour blindness in that girl child? Analyse your answer.
Answer:

Wbbse Class 10 Life Science Solutions

 

 

Checker board

 

Colour blind girl =  ½ x 100 = 50%

There will be a 50% chance of the daughter being colour-blind. This is because the X chromosome that she will obtain from her father will definitely bear an allele for colour blindness.

However, the X chromosome that she will obtain from her mother may either bear the allele for colour blindness or that for normal vision. So, she has a 50% chance of being colour-blind.

Question 26. Write the names of two genetic diseases which are expressed in human populations.
Answer:

Two genetic diseases expressed in the human population are-

1. Thalassemia and
2. Haemophilia.

Question 27. Give your opinion about probable suggestions which can be given to a pair of contenders before marriage in order to prevent the spread of a genetic disease from the society already known to you.
Answer:

The pair of contenders should be suggested to test whether they are carriers or have any particular disease. If both are a carrier for an autosomal disease or one is a carrier and the other have the disease then the marriage should be cancelled.

If they both are normal or if one is normal and the other is a carrier then they can marry.

Wbbse Class 10 Life Science Solutions

Question 28. One day students read an article in the newspaper on thalassemia and were very scared to know the fate of a thalassemic patient. Write what kind of measure they can take to eradicate this disease from the population.
Answer:

The measure they can take to eradicate this disease from the population are-

1. Genetic counselling before marriage and not to conduct marriage between carrier male and female or affected and normal individual.
2. If genetic counselling is not done before marriage then it must be done before conceiving a child and not to conceive in sensitive cases.

 

Chapter 3 Some Genetic Diseases MCQs

Question 1. A sudden and permanent change in genes is called—

1. Adaptation
2. Inversion
3. Mutation
4. Recombination

Answer: 3. Mutation

Question 2. A symptom of thalassemia is—

1. Scurvy
2. Anaemia
3. Rickets
4. Malaria

Answer: 2. Anaemia

Question 3. Anti-haemophilic globulin factor is known as—

1. Factor VII
2. Factor VIII
3. Factor IX
4. Factor X

Answer: 2. Factor VIII

Question 4. Plasma thromboplastin antecedent is also known as—

1. Factor VII
2. Factor VIII
3. Factor XI
4. Factor X

Answer: 3. Factor XI

Question 5. Thalassemia minor is also known as—

1. Thalassemia trait
2. Thalassemia hydrops
3. Cooley’s anaemia
4. Thalassemia child

Answer: 1. Thalassemia trait

Wbbse Class 10 Life Science Solutions

Question 6. Which of the following is a wrong statement about thalassemia?

1. It is a sex chromosome-linked trait
2. It is a hereditary disease
3. In this disease, structural deformity of haemoglobin is seen
4. In this disease, iron is deposited in the blood

Answer: 1. It is a sex chromosome-linked trait

Question 7. The disease, where a patient faces the problem of haemorrhage is known as—

1. Leukaemia
2. Haemophilia
3. Thalassemia
4. Colour blindness

Answer: 2. Haemophilia

Question 8. Thalassemia major is also known as—

1. Cooley’s anaemia
2. Cooley’s syndrome
3. Down’s syndrome
4. Turner’s syndrome

Answer: 1. Cooley’s anaemia

Question 9. The human chromosome that carries the n-globin gene is—

1. 11th
2. 12th
3. 14th
4. 16th

Answer: 4. 16th

Question 10. The human chromosome that carries the y3-globin gene is—

1. 11th
2. 12th
3. 14th
4. 16th

Answer: 1. 11th

Question 11. The colours, which a colour-blind person is unable to detect, are—

1. Violet-yellow
2. Indigo-blue
3. Red-green
4. Orange-yellow

Answer: 3. Red-green

Question 12. An example of autosomal hereditary disease is—

1. Haemophilia
2. Colour blindness
3. Thalassemia
4. Scurvy

Answer: 3. Thalassemia

Wbbse Class 10 Life Science Solutions

Question 13. Which of the following is a sex-linked hereditary disease?

1. Anaemia
2. Scurvy
3. Thalassemia
4. Haemophilia

Answer: 4. Haemophilia

Question 14. An example of X-linked inherited disease is—

1. Colour blindness
2. Night blindness
3. Thalassemia
4. Malaria

Answer: 1. Colour blindness

Question 15. The type of colour blindness, in which a person cannot detect green colour, is—

1. Haemophilia
2. Thalassemia
3. Protanopia
4. Deuteranopia

Answer: 4. Deuteranopia

Question 16. Which of the following diseases occurs mostly among males?

1. Red-green colour blindness
2. Thalassemia
3. Cancer
4. Tuberculosis

Answer: 1. Red-green colour blindness

Question 17. The gene causing colour blindness is located on—

1. X chromosome
2. Y chromosome
3. Z chromosome
4. M chromosome

Answer: 1. X chromosome

Question l8. The other name of blood coagulating factor VIII is—

1. ABB
2. HBA
3. PTA
4. AHF

Answer: 4. AHF

Class 10 Life Science Wbbse

Question 19. The disease, caused due to abnormal structure of the globin peptide chain, is—

1. Haemophilia
2. Thalassemia
3. Malaria
4. Scurvy

Answer: 2. Thalassemia

Question 20. Christmas disease is—

1. Haemophilia A
2. Colour blindness
3. Royal haemophilia
4. Haemophilia B

Answer: 4. Haemophilia B

Question 21. The other name of blood coagulating factor IX is—

1. AHF
2. HBA
3. HBB
4. PTA

Answer: 4. PTA

Question 22. To whom should a person approach to get pre¬marriage suggestions to avert any genetic disease in his or her offspring?

1. Psychologist
2. Genetic counsellor
3. Physician
4. Surgeon

Answer: 2. Genetic counsellor

Question 23. Genetic counselling is advisable for—

1. Thalassemia
2. Malaria
3. Hepatitis
4. Diarrhoea

Answer: 1. Thalassemia

Class 10 Life Science Wbbse

Question 24. Deficiency of which of the following factors causes haemophilia?

1. Factor VIII
2. Factor IX
3. Factor III
4. Factors VIII And IX

Answer: 4. Factors VIII And IX

Question 25. The type of haemophilia carried by Queen Victoria was—

1. Haemophilia A
2. Haemophilia B
3. Haemophilia C
4. Haemophilia D

Answer: 1. Haemophilia A

Question 26. Criss-cross inheritance occurs in case of—

1. Colour blindness
2. Diabetes
3. Anaemia
4. Scurvy

Answer: 1. Colour blindness

Question 27. The main problem in haemophilia is—

1. Disability of colour detection
2. Internal haemorrhage
3. Internal blood coagulation
4. High fever and inflammation

Answer: 2. Internal haemorrhage

Question 28. The plasma protein, which is not synthesised in haemophilia, is—

1. Albumin
2. Globulin
3. Fibrin
4. Chitin

Answer: 3. Fibrin

Question 29. The disease in which the function of the liver and heart is hampered due to iron accumulation—

1. Haemophilia
2. Anaemia
3. Thalassemia
4. Blood Cancer

Answer: 3. Thalassemia

Question 30. Which of the following is not true for thalassemia?

1. It is a sex-linked disease
2. This is a genetic disease
3. In this disease, haemoglobin is not formed properly
4. Iron is accumulated in this disease

Answer: 1. It is a sex-linked disease

Question 31. The reason behind Cooley’s anaemia is

1. Mutation
2. Adaptation
3. HbS
4. HbF

Answer: 1. Mutation

Class 10 Life Science Wbbse

Question 32. Cooley’s anaemia is—

1. α thalassaemia major
2. α thalassemia minor
3. β thalassemia major
4. β thalassemia minor

Answer: 3. β thalassemia major

Question 33. If both the parents are carriers of thalassemia then the chance percentage of their child to be thalassemic is—

1. 100%
2. 75%
3. 50%
4. 25%

Answer: 4. 25%

Question 34. Which of the following is not controlled by the autosomal gene of humans?

1. Roller tongue
2. Haemophilia
3. Thalassemia
4. Attached ear hole

Answer: 2. Haemophilia

Question 35. The factor responsible for haemophilia A is—

1. VI
2. VII
3. VIII
4. IX

Answer: 3. VIII

Question 36. The haemophilia Known as classic haemophilia is—

1. Haemophilia A
2. Haemophilia C
3. Haemophilia B
4. None of the above

Answer: 1. Haemophilia A

Question 37. If a haemophilic male marries a homozygous normal female then what will be the chances of a haemophilic child?

1. 0%
2. 25%
3. 50%
4. 75%

Answer: 1. 0%

Question 38. If a carrier haemophilic female marries a normal male then what will be the chances of their of their child to be haemophilic?

1. 0%
2. 25%
3. 50%
4. 75%

Answer: 2. 25%

Question 39. Blue blindness is called—

1. Presbyopia
2. Tritanopia
3. DuteranopiaProtanopia

Answer: 2. Tritanopia

Class 10 Life Science Wbbse

Question 40. The genotype of a colour-blind male is—

1. X^C+Y
2. X^CX^C
3. X^CY
4. X^CY^C

Answer: 3. X^CY

Question 41. The phenotype of Xc+Xc will be—

1. Normal female
2. Normal male
3. Carrier female
4. Colour blind female

Answer: 3. Carrier female

Question 42. If a male is colour blind then the chance of his son to be colour blind by his gene—

1. 0%
2. 50%
3. 75%
4. 100%

Answer: 1. 0%

Question 43. The probability of haemophilic girl children born to haemophilia carrier mother and normal father is—

1. 75%
2. 50%
3. 100%
4. 0%

Answer: 4. 0%

Question 44. Assess from the following, the probable genotype of parents having haemophilic son and normal daughter-

1. H || h, h
2. H || H, H
3. H || H, h
4. H || h, H

Answer:  3. H || H, h

 

Chapter 3 Some Genetic Diseases Answer In A Single Word Or Sentence

Question 1. Who coined the term ‘thalassemia’ and when?
Answer: Whipple and Bradford coined the term ‘thalassemia, in the year 1932.

Question 2. What percentage of iron is present in haemoglobin?
Answer: Haemoglobin contains 0.34% iron.

Question 3. Name the group of diseases caused due to defective haemoglobin.
Answer: A group of diseases caused due to defective haemoglobin is called haemoglobinopathy.

Question 4. Which protein is produced in reduced quantity in a thalassemia patient?
Answer: Haemoglobin, the conjugated protein, is produced in reduced quantity in a thalassemia patient.

Question 5. Which metal accumulates in the body of a thalassemia patient due to the rapid destruction of red blood cells?
Answer: Huge quantity of iron accumulates in the body of a thalassemia patient due to the rapid destruction of red blood cells.

Class 10 Life Science Wbbse

Question 6. Which organs of the body are mostly affected due to accumulation of excessive iron?
Answer: The heart, liver, spleen and endocrine system are mostly affected due to the accumulation of excessive iron.

Question 7. How much oxygen is carried by normal haemoglobin?
Answer: 1 g of normal haemoglobin carries 1.34 ml. of oxygen.

Question 8. What is the cause of the malformation of α and β- chains in the haemoglobin of a thalassemia patient?
Answer: Due to mutation in α and β-globin genes, α and β-peptide chains in the haemoglobin of a thalassemia patient are malformed.

Question 9. What is α thalassemia?
Answer:

α thalassemia

The type of thalassemia, caused due to incomplete formation of α-chains of haemoglobin, is known as thalassemia.

Question 10. What is β thalassemia?
Answer:

β thalassemia

The type of thalassemia, caused due to incomplete formation of β-chains of haemoglobin, is known as β thalassemia.

Question 11. In which disease, bleeding does not stop due to non-coagulation of blood?
Answer: In haemophilia, bleeding does not stop due to the non-coagulation of blood.

Question 12. Name the disease caused due to deficiency of clotting factor VIII in the blood.
Answer: Haemophilia A is caused due to deficiency of clotting factor VIII in the blood.

Question 13. Name the disease caused due to deficiency of clotting factor IX in blood.
Answer: Haemophilia B is caused due to deficiency of clotting factor IX in the blood.

Question 14. What is the other name of the haemorrhagic disease?
Answer: The other name of haemorrhagic disease is haemophilia.

Question 15. Which type of disease is haemophilia B?
Answer: Haemophilia B is an X chromosome-linked recessive hereditary disease.

Question 16. Why does haemophilia occur?
Answer: Haemophilia occurs due to a recessive mutation in an X-linked gene.

Impact of Genetic Mutations on Health

Question 17. In which chromosome of the human cell does the haemophilia gene occur?
Answer: Haemophilia gene occurs in X chromosome of a human cell.

Question 18. What is the other name of haemophilia A?
Answer: The other name of haemophilia A is classical haemophilia.

Question 19. Name the disease-causing gene, which was first carried by Queen Victoria of Britain.
Answer: Queen Victoria of Britain was the first to carry the disease-causing gene of haemophilia B.

Question 20. Who are called protanopes?
Answer: People who are suffering from protanopia and hence, cannot detect red colour, are called protanopes.

Question 21. Who are called deuteranopes?
Answer: People who are suffering from deuteranopia and hence, cannot detect green colour, are called deuteranopes.

Question 22. Who are called tritanopes?
Answer: People who are suffering from tritanopia and hence, cannot detect blue colour, are called tritanopes.

Wbbse Class 10 Life Science

Question 23. Which type of male baby is expected from a colour blind mother and a normal father?
Answer: A colour-blind mother and a normal father will have a colour-blind son.

Question 24. Which type of thalassemia is predominant in India?
Answer: α thalassemia

Question 25. Which type of haemophilia is not a sex-linked disease?
Answer: Haemophilia C is caused by an autosomal mutation and hence, not a sex-linked disease.

Question 26. What is the reason behind the name ‘thalassemia’?
Answer: The disease thalassemia was first discovered in the people living near the mediterranean sea. In the Latin language sea is called ‘thalassa’ and so the disease is called thalassemia.

Question 27. Which chromosomes contain genes for the production of α and β chains of globin protein?
Answer: α chain-chromosome 11 and β chain- chromosome 16

Question 28. Give two examples of X-linked inheritence.
Answer: Haemophilia and colour blindness

Question 29. How much oxygen is carried by 1 gm of haemoglobin?
Answer: 1 gm haemoglobin carries 1.34 ml 0₂,

Question 30. Which disease is also known as bleeder’s disease?
Answer: Haemophilia

Question 31. What is cyanopia?
Answer:

Cyanopia

Sensitivity to blue light is called cyanopia.

Question 32. Which type of chromosome in humans carries the gene responsible for the disease thalassemia?
Answer: Autosome in human carries the gene responsible for the disease thalassemia.

Question 33. What is the cause of expression of haemophilia disease only at homozygous conditions?
Answer: Recessive characters are always expressed in homozygous conditions. Haemophilia is a recessive character. So it is expressed only at homozygous conditions.

 

Chapter 3 Some Genetic Diseases Fill In The Blanks

Question 1. The genetic disease, which was expressed for the first time in the British royal family, is ________.
Answer: Haemophilia

Question 2. Thalassemia is a ________ disease.
Answer: Genetic

Question 3. If the synthesis of ________ is hampered, a person develops severe anaemia.
Answer: Haemoglobin

Question 4. Regular blood transfusion increases ________ content in the body of a thalassemia patient.
Answer: Iron

Question 5. Haemophilia is a/an ________ chromosome-linked genetic disease.
Answer: X

Wbbse Class 10 Life Science

Question 6. Almost ________ % of haemophilia is caused by the deficiency of AHF.
Answer: 80

Question 7. Haemophilia B or ________ disease is caused by the deficiency of blood clotting factor IX.
Answer: Christmas

Question 8. The colour-detecting cells of the retina are known as ________ cells.
Answer: Cone

Question 9. The gene responsible for blue colour blindness is a/an ________ gene.
Answer: Autosomal

Question 10. Only ________ can be carriers in case of genetic diseases.
Answer: Females

Question 11. In ________ disease O2 transport capacity diseases.
Answer: Thalassemia

Question 12. ________ is an example of a sex-linked disease.
Answer: Haemophilia

Question 13. In ________ chromosome of humans ẞ globin gene is present.
Answer: 11th

Question 14. Cooley’s anaemia is ________.
Answer: ẞ Thalassemia major

Question 15. The cells of the eye associated with colour recognition is ________.
Answer: Cone cell

Question 16. All ________ will be colour-blind if the mother is colour-blind and the father is normal.
Answer: Sons

Question 17. A disease in the human population caused by a recessive gene located in the ‘X’ chromosome is ________.
Answer: Haemophilia

Question 18. ________ is a disease created by a sex-linked gene.
Answer: Haemophilia

 

Chapter 3 Some Genetic Diseases State True Or False

Question 1. Malaria, AIDS. hepatitis etc., are examples of genetic diseases.
Answer: False

Question 2. In thalassemia, bone marrow proliferates at an excessive rate.
Answer: True

Question 3. In thalassemia, the synthesis of globin protein is hampered.
Answer: True

Question 4. Haemophilia is an autosome-linked genetic disorder.
Answer: False

Question 5. Haemophilia A is caused by the restricted synthesis of factor VIII.
Answer: True

Wbbse Class 10 Life Science

Question 6. Haemophilia B is caused by the restricted synthesis of factor X.
Answer: False

Question 7. A protanope is unable to detect red colour.
Answer: True

Question 8. The gene responsible for blue colour blindness is located in the 8th chromosome.
Answer: False

Question 9. The pre-marriage suggestion to a couple for having a baby without any genetic disorder is known as genetic counselling.
Answer: True

Question 10. In an anaemic patient, oxygen transport is seriously hampered.
Answer: True

Question 11. Night blindness is an X chromosome-linked disease.
Answer: False

Question 12. The blood coagulation process is hindered in a haemophilic patient.
Answer: True

Question 13. Rod cells of the human retina help in colour detection.
Answer: False

Question 14. In the case of serious haemophilia, blood contains 1% more blood-coagulating factors than normal.
Answer: False

Question 15. 5-8% male of the total global population are colour-blind.
Answer: True

 

Chapter 3 Some Genetic Diseases Match The Columns

Question 1.

Answer: 1-E; 2-A; 3-B; 4-C; 5-D; 6-G

Question 2.

Answer: 1-E; 2-A; 3-B; 4-F; 5-C; 6-G

Question 3.

Answer: 1-B; 2-A; 3-D; 4-C; 5-F; 6-E

Question 4.

Answer: 1-F; 2-C; 3-D; 4-A; 5-B; 6-G

 

Chapter 3 Some Genetic Diseases Find The Odd One Out

Question 1. Widow’s peak, Rolling tongue, Haemophilia, Thalassemia
Answer: Haemophilia

Question 2. Protanopia, Deuteranopia, Hypermetropia, Tritanopia
Answer: Hypermetropia

Question 3. Hepatomegaly, Splenomegaly, Iron accumulation, Haemophilia
Answer: Haemophilia

Wbbse Class 10 Life Science

Question 4. Bleeding, Swelling of the joint, Royal disease, Protanopia
Answer: Protanopia

Question 5. Protanopia, Deuteranopia, Tritanopia, Haemophilia
Answer: Haemophilia

Question 6. Thalassemia, Albinism, Haemophilia, Colour blindness
Answer: Thalassemia

Question 7. Haemophilia, Colour blindness, Thalassemia, Myopia
Answer: Myopia

Question 8. Protanopia, Factor IX, Antihaemophilic factor, PTA
Answer: Protanopia

 

Chapter 3 Some Genetic Diseases Fill In The Blanks By Looking At The First Pair

Question 1. Haemoglobin deficiency: Thalassemia :: Blood coagulation disorder: _________
Answer: Haemophilia

Question 2. Protanopia: Red colour blindness Deuteranopia: _________
Answer: Green colour blindness

Question 3. Colour blindness: Sex-linked :: Rolling tongue: _________
Answer: Autosomal

Question 4. Continuous bleeding: Haemophilia:: Iron accumulation: _________
Answer: Thalassemia

Question 5. Splenomegaly Increase of spleen:: Hepatomegaly: _________
Answer: Increase of liver

Question 6. Thalassemia: Autosomal disease: Colour blindness: _________
Answer: Sex-linked disease

Question 7. Factor VIII: AHF:: Factor IX:_________
Answer: PTC

Question 8. Classic haemophilia Haemophilia A:: Christmas disease: _________
Answer: Hemophilia B

 

Chapter 3 Some Genetic Diseases Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Protanopia, Deuteranopia, Tritanopia, Colour blindness
Answer: Colour blindness

Question 2. Colour blindness, Genetic disorder, Thalassemia, Haemophilia
Answer: Genetic disorder

Question 3. Colour blindness, Haemophilia, Albinism, Sex-linked trait
Answer: Sex-linked trait

Wbbse Class 10 Life Science

Question 4. a globin, ẞ globin, Heme, Haemoglobin
Answer: Haemoglobin

Question 5. Thalassemia, Haemophilia, Genetic counselling, Colour blindness
Answer: Genetic counselling

Question 6. Protanopia, Deuteranopia, Tritanopia Colour blindness
Answer: Colour blindness

 

Chapter 3 Some Genetic Diseases Advanced Questions And Answers

Question 1. What are Mendelian and Non-Mendelian traits? Give example.
Answer:

Mendelian traits

The character which is controlled by one pair of alleles located on the same locus is called a Mendelian character. For example, the stem height of a pea plant.

Non-Mendelian traits

The character which is controlled by more than two alleles or more than two genes located in different loci are called Non-Mendelian traits.

Example-human height.

Question 2. What are multiple alleles?
Answer:

Multiple alleles

When there are more than two alleles of a gene then they are called multiple alleles. The gene responsible for human blood.group is a gene with three alleles-I^A, I^B and i.

Question 3. What is a polygenic character?
Answer:

Polygenic character

A character which is controlled by two or more than two genes is called a polygenic character. Example-Human body colour. It is a deviation from Mendelism.

Question 4. Write down some methods of treatment for thalassemia.
Answer:

Methods of treatment for thalassemia

1. Blood transfusion,
2. Iron chelation,
3. Bone marrow transplantation from a normal man.

Question 5. In which countries or regions of the world is thalassemia seen more?
Answer: Southeast Asia, India, Middle East and Africa.

Question 6. In which country or regions of the world is B thalassemia seen more?
Answer: Mediterranean region, Northern Africa, Western Asia.

Question 7. What is haemoglobin H disease?
Answer:

Haemoglobin H disease

Haemoglobin H disease is a form of alpha thalassemia in which reduced formation of a Globin chain takes place due to mutation in its gene. It leads to moderately severe anaemia.

Chapter 2 Continuity Of Life Reproduction Summary

WBBSE Class 10 Reproduction Overview

• Reproduction is an important event for the existence of living organisms. Reproduction is mainly of two types-asexual reproduction and sexual reproduction. Vegetative reproduction is a special type of asexual reproduction.
• The formation of new offspring from somatic cells without fertilization is called asexual reproduction and the formation of new offspring by fusion of gametes is called sexual reproduction.
• The formation of the new individual directly from the segment of parent body is called vegetative reproduction. Asexual reproduction is mediated by single-parent cells, whereas in sexual reproduction two parent cells need to participate.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

• In sexual reproduction, due to meiotic division, variation occurs in new offspring, so offspring with new characteristics will be produced.
• Asexual reproduction may be of various types like fission [Amoeba] budding [Hydra, Yeast], fragmentation [Spirogyra], regeneration [Planaria], and sporulation [Moss, fern]
• In vegetative reproduction, a part of the plant cell can produce new plantlets. It is of two types-natural and artificial. In the natural process, shoot, root, or leaf can give rise to new plantlets, whereas in the artificial process cutting, grafting or micropropagation can mediate the generation of new plantlets.
• The alternation of a haploid and a diploid generation in the life cycle of an organism is called the alternation of generations. It is most distinct in fern.

Chapter 2 Continuity Of Life Reproduction Long Answer Type Questions

Question 1. Mention the common features of reproduction. Mention the significance of reproduction. Or, Why reproduction is important for an organism?

Answer:

Common features of reproduction:

Different phases of reproduction involve mitosis, meiosis, or both the processes of cell division. From gametogenesis to the embryonic development of a progeny, cell division is required in every phase.

1. The molecular basis of reproduction is DNA replication.
2. An offspring is formed through the union of gametes, germination of spores, or division of vegetative cells.

Significance of reproduction:

The significance of reproduction is as follows.

1. Increasing population size:

By means of reproduction, new individuals are born. As a result, the number of individuals in a population increases and the race of a species is maintained.

2. Maintaining the flow of genes:

By reproduction, the genes of a species flow from one generation to the next maintaining the hereditary characteristics.

3. Maintaining ecological balance:

Natural or unnatural death of individuals reduces the population size of a species. By reproduction, this loss is replenished in nature. This process keeps the ratio of different species in an ecosystem constant and thus maintains the ecological balance.

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4. Variation:

By sexual reproduction, different genetic variations come into existence. These variations assist any species to adapt in its surroundings and finally help in the evolution of new species.

Question 2. Briefly describe the different types of reproduction.

Answer:

Types of reproduction:

Two types of reproduction are seen in the living world. These are-

1. Asexual reproduction and
2. Sexual reproduction.

1. Asexual reproduction:

The type of reproduction, in which the formation of gametes and their fertilization do not occur, instead, certain body cells or specialized cells, like spores, divide to form the progeny, is known as asexual reproduction.

In this process, only one parent is required. All parental characters are inherited by the offspring. All the offspring become, genotypically and phenotypically identical to their mother.

For example Bacteria, several protists like Amoeba, and some fungi like Mucor, Penicillium, etc. perform asexual reproduction.

2. Sexual reproduction:

The type of reproduction, that involves the union of two different types of gametes to produce a zygote and the development of a new progeny from it, is known as sexual reproduction. Most of plants and animals perform sexual reproduction.

This process requires a male and a female member of the same species or a single organism with two different types of gametangia to produce male and female gametes. In this process, the union of male and female gametes takes place.

As a result, the progeny gets genotypic and phenotypic characteristics different from its parents. Sexual reproduction essentially requires meiotic cell division for gamete formation and for keeping the chromosome number of a species constant.

Example: Sexual reproduction occurs in all higher animals and plants such as frogs, men, birds, angiosperms, etc.

Life Science Class 10 Wbbse

Question 3. Describe the five processes of asexual reproduction with one suitable example for each.

Answer:

Processes of asexual reproduction:

The following types of asexual reproduction can be seen in organisms-

1. Fission:

In most of the unicellular organisms two or more offspring are produced through mitosis or amitosis. This process is called fission. Some example are given below.

1. Amoeba:

Binary fission:

In favourable conditions, the nucleus of Amoeba undergoes amitotic division and gives of two nuclei. Its cytoplasm divides longitudinally creating a furrow and forms two offsprings.

Multiple fission:

In unfavorable conditions, the pseudopodia of Amoeba are destroyed and the body gets encapsulated by a cyst wall. Inside the cyst, numerous small spores are formed through multiple fission of nucleus and cytoplasm.

This type of multiple fission is called sporulation. When the condition becomes favorable, spores come out rupturing the cyst wall and giving rise to new Amoeba offspring.

2. Plasmodium:

In the case of Plasmodium, multiple fission takes place in two stages-Schizont and sporont. In the stomach of female Anopheles mosquito, numerous Plasmodium offsprings are born by the multiple fission called schizogony and sporogony is the schizont and sporont stages respectively.

The Plasmodium offsprings born through schizogony and sporogony are called merozoite and sporozoite respectively.

2. Budding:

Budding is a type of asexual reproduction in which a new organism develops from an outgrowth or bud of the parent body due to cell division at one particular site. Some examples of budding are given below.

Yeast:

Due to the unequal division of the yeast parent cell, a small outgrowth or bud is formed which contains the parent’s nucleus. Later on the bud separates from the parental body and develops into new yeast.

In special cases, the bud of yeast divides many times in the torula stage through the Tormulation process and forms mycelium.

Hydra:

In the case of Hydra, the bud is formed outside the parent body (exogenous bud). After the formation of the oral aperture, tentacles, etc. when the offspring matures, it gets detached from the parent body.

3. Fragmentation:

The type of asexual reproduction where the parent body is fragmented into two or more pieces and each fragment turns into new offspring, is called fragmentation.

Example:

The filamentous body of Spirogyra, an alga gets fragmented due to water currents or external damage. Each fragment undergoes mitotic cell division and gives rise to new offspring.

4. Regeneration:

The type of asexual reproduction where a mere body part of a parent organism creates a new offspring, is called regeneration. It is also called morphallaxis.

Life Science Class 10 Wbbse

Example:

If any body part of Planaria, a flatworm gets detached, it forms a new offspring. This process also can be seen in Hydra.

5. Sporulation:

Sporulation is a type of asexual reproduction in which unicellular spores are produced in moss, fern and fungi, they are dispersed and give birth to new offspring. Examples of sporulation are given below.

Fungi:

In fungi, spores with different structure and with or without motility can be seen. These spores germinate in favourable conditions and new fungi are born. Different types of spores are motile zoospores, non-motile aplanospores, thick-walled chlamydospores, oidia, and conidia formed from fragments of filaments, sporangiospores formed in sporangium.

Moss and Fern:

Spores are formed from the sporophytic tissue of the sporophytic plant body of moss. In the sporangium of sporophytic plant body of fern, spores with similar or different shapes are formed.

Types of Reproduction in Living Organisms

Question 4. Describe the mechanism of fission in Plasmodium with a diagram.

Answer:

Mechanism of fission in Plasmodium:

Plasmodium is a unicellular parasitic protist. It causes malaria in human beings. It goes through a number of multiple divisions to complete its life cycle. In human liver cells, this type of division is called schizogony.

The stage of Plasmodium undergoing schizogony is known as schizont. A schizont undergoes several successive karyokineses to produce many nuclei in it. Then each nucleus undergoes cytokinesis and liberates as a tiny daughter cell, called merozoites.

By this multiple fission, several hundreds of merozoites are formed, which again attack neighbouring liver cells and propagate in the same manner.

Inside the mosquito gut, the oocyst or sporont phase undergoes a similar type of multiple fission to produce sporozoites. This type of division is called sporogony.

Question 5. Briefly describe the asexual reproduction of Bryophyta, Pteridophyta, and fungi through spore formation.

Answer:

Bryophytes:

Bryophytes have two phases in their life cycle-the sporophytic (2n) and gametophytic (n) phases. The diploid sporophyte produces spores through meiosis in the specialized structure called sporangium.

All the spores are look alike i.e. homosporous. The sporangia are located in the capsule part of the sporophyte. The spores (n) are aplanospores, endogenous and when fall on a suitable substratum produce new gametophytes.

Pteridophyta:

In pteridophytes, spores (n) are formed by sporophytes (2n) in specialized sporangium. The spores may be homosporous or heterosporous. Homosporous spores are alike and produce a bisexual gametophyte.

Heterospores are of two types with microspores producing male gametophytes and megaspores producing female gametophyte. Spores are endogenous and haploid in nature as they are produced through meiosis from the spore mother cell.

Life Science Class 10 Wbbse

Fungi:

The spores in fungi may be endogenous in origin when produced in sporangia or may be exogenous origin like conidia produced by cutting the tip of fungi. In fungi, the endogenous spores are formed inside sporangia through mitosis.

The hyphal tip swells forming the sporangia. The nuclei undergo free nuclear division. Each nucleus with cytoplasm and membrane becomes a spore. The spores on suitable substratum germinate to produce new hypha.

Asexual Reproduction Methods

Question 6. What is meant by vegetative propagation? Briefly describe the processes of natural vegetative propagation. Part Question, ‘The adventitious leaf bud plays significant role in natural vegetative propagation of plant’-Evaluate the validity of the statement with a proper example.

Answer:

Vegetative propagation:

The asexual reproduction, in which any portion of the vegetative body of certain plants separates out from the mother’s body and finally grows into a new individual by mitotic cell division, is known as vegetative propagation.

Processes of natural vegetative propagation:

Vegetative propagation occurs by the growth of different parts of plants naturally, which are mentioned below.

1. By leaves:

Certain plants like Bryophyllum, Bigonia, etc. develop buds along the edges of their leaves. These are called leaf buds. These buds grow adventitious roots from their base.

When such a leaf detaches from the plant body and comes in contact with the soil, each of the buds grows into an individual daughter plant.

2. By roots:

Roots of sweet potato, pointed gourd, etc. grow adventitious buds. These are called root buds. When detached from the root, these buds grow into new daughter plants.

3. By stem:

Vegetative propagation by stem occurs by two different types of modified stems-

1. Underground modified stems:

The tuber of potato, bulb of onion, the rhizome of ginger, turmeric, etc. are underground stems, modified for storage of food. These organs have buds, which may grow into daughter plants.

2. Sub-aerial modified stems:

Nodes of sub-aerial modified stems of Mentha, Marsilea, Oxalis, Centella, water hyacinth, Chrysanthemum, etc. plants grow adventitious roots. When detached from the mother plant, the rooted branches grow as a daughter plant.

Question 7. Mention the advantages and disadvantages of vegetative propagation.

Answer:

Advantages of vegetative propagation:

1. Identical character-bearing plants can be produced by vegetative propagation.
2. By this process, several daughter plants may be produced from a single mother plant within a brief period of time.
3. Vegetative propagation can be practiced in any season throughout the year.
4. Desired characteristics of the mother plant remain intact in the daughter plants.

Disadvantages of vegetative propagation:

1. Daughter plants from vegetative propagation do not have any newer characters.
2. The daughter plants formed by this type of reproduction do not show any variation. Therefore, these plants cannot adapt themselves to the changed environment.
3. Vegetative propagation does not provide any chance for evolution of a new species. Hence, there is a chance of extinction.

Question 8. How is layering technique used for artificial vegetative propagation of certain plants? How is the “gootee” prepared for artificial vegetative propagation in certain plants?

Answer:

Use of layering technique for artificial vegetative propagation:

In the layering technique, a comparatively young and tender twig is bent, such that the nodal portion of it touches the soil. Now, the portion of the branch, touching the ground, is covered with a layer of moistened soil and then a brick or stone is placed on it.

The layered portion is watered regularly. After a few weeks, adventitious roots develop from that portion of the stem under the soil.

The branch is then chopped off from the mother plant and the twig with roots is planted elsewhere to grow further as a progeny. Layering is applicable to plants like lemon, jasmine, wood-apple etc.

Preparation of “Gootee” for artificial vegetative propagation:

In this artificial vegetative propagation technique, a strong and stiff twig of a shrub or tree is selected to prepare a Gootee. First, a few inches of the bark is peeled off with a knife.

Now this portion is covered with a muddy mixture of organic manure and soil. Then the soil layer is wrapped with moistened rag or coir. An additional layer of polythene sheet may be wrapped to prevent quick drying.

After a few weeks, adventitious roots grow there and appear on the surface of the goatee. Now, the twig is chopped off and used for plantation. Gootee is applicable in plants like mango, guava, litchi, etc.

Life Science Class 10 Wbbse

Question 9. Briefly describe the methods of cutting and grafting for artificial vegetative propagation.

Answer:

Methods of artificial vegetative propagation:

Artificial vegetative propagation of plants is done in different ways, which are mentioned below.

1. Cutting:

In this process, 20- 30 cm long twigs of certain plants are cut up from the mother plant. These are then planted in moistened soil.

Within a week or so, adventitious roots develop from the nodes and the twig grows as an individual plant. Cutting is successfully practiced upon the rose, china rose, drumstick plants, etc.

2. Grafting:

Grafting is the process of artificial vegetative propagation by which younger branches or buds of a superior variety of plants are affixed with an older and healthy stem of the same species to grow better varieties of plants in a short time.

It is of two types:

1. Scion grafting and
2. Bud grafting.

1. Scion grafting:

In scion grafting, stems of two different varieties of plants are attached together. Of these two, the stem of the older one is chopped off leaving a portion of it above the ground. This rooted portion is called the stock.

A younger stem branch with superior features is affixed with the stock, which is called the scion. The end of the scion is cut with a sharp knife to give it a shape of a wedge.

The top or side of the stock is also cut to allow the scion to fit there perfectly. This region is wrapped with a moist rag or coir. After some days, the scion joins to the stock and grows further.

Life Science Class 10 Wbbse

2. Bud grafting:

In this process, axillary bud from a superior variety of plants is sliced off. Now a T-shaped slit is made on the bark of a healthy quick-growing plant of the same species. The bud is then fitted in the slit and is tied with string.

Finally, the bud joins to the stem and grows as a branch with superior features. Grafting is applicable for plants like mango, guava, litchi, etc.

Question 10. Classify sexual reproduction based on the nature of gametes. Name the different types of asexual reproduction with example.

Answer:

Types of sexual reproduction:

Based on the nature of gametes, sexual reproduction can be classified into three types.

These are

1. Anisogamy: It occurs by the union of two morphologically non-identical gametes.
2. Isogamy: It occurs by the union of two morphologically and physiologically identical gametes.
3. Oogamy: This type of sexual reproduction occurs by the union of a smaller motile and a larger non-motile gamete.

Types of asexual reproduction:

In the living world, different types of asexual reproduction are seen.

These are

1. Fission (Amoeba, Plasmodium),
2. Budding (Yeast, Hydra),
3. Fragmentation (Spirogyra),
4. Sporulation (Moss, fern, fungi),
5. Regeneration (Planaria).

Question 11. Mention the advantages and disadvantages of sexual reproduction.

Answer:

Advantages of sexual reproduction:

1. By sexual reproduction, the genetic characters of the parents are inherited by the offspring and this event is carried forward in the following generations as well.
2. In sexual reproduction, the recombination of genes takes place, which results in to the formation of new genetic variations.
3. New genetic variations, which develop among the progeny, help them to adapt efficiently with the changing environment.
4. Due to better adaptability, the progeny of sexual reproduction come out as a winner in the struggle for existence. They get the favour of natural selection and finally take part in the evolution of new species.

Disadvantages of sexual reproduction:

1. Sexual reproduction is a slow process. Therefore, it takes a longer time to produce offspring.
2. The availability of two different types of gametes is essential, which may not always be available.
3. The success rate of fertilization depends upon a large number of male gametes, which may not always be available.

Question 12. Mention the advantages and disadvantages of asexual reproduction. 

Answer:

Advantages of asexual reproduction:

1. Asexual reproduction involves only one parental member, so it is an easier process.
2. By this process, numerous offspring are produced at a time.
3. This process needs very little energy.
4. This is a very simple process and takes a very short time to complete.
5. The parental features are retained within the progeny without any change.
6. All progeny grow in close vicinity to its mother. Therefore, the progeny adapt easily to the known environment.

Disadvantages of asexual reproduction:

1. Meiosis does not occur in asexual reproduction. Therefore, recombination of genes does not occur.
2. Variations do not occur in the offspring of asexual reproduction.
3. The chance of adaptability of these offspring is very low. Therefore, there is a possibility of their extinction.

Question 13. State the principle of micropropagation. Distinguish between asexual and sexual reproduction. 2+3 Or, Distinguish between asexual and sexual reproduction on the basis of the following features.

1. Number of parents,
2. Nature of offspring.

Answer:

Principle of micropropagation:

Certain plant cells have a unique ability to grow into a full plant body by means of cell division. This unique ability of plant cells is known as totipotency. Micropropagation is a tissue culture technique, which depends upon totipotency.

The word ‘micro’ means ‘very small. In micropropagation, very small portions are collected from vegetative parts of a plant and are cultured in a proper culture medium inside the laboratory to produce saplings.

Life Science Class 10 Wbbse

Differences between asexual and sexual reproduction

Question 14. Distinguish between vegetative propagation and asexual reproduction.

Answer:

Differences between vegetative propagation and asexual reproduction

Question 15. What is meant by the alternation of generations? Mention the importance of meiosis in the alternation of generations.

Answer:

Alternation of generations:

The cyclic alternation between haploid gametophytic and diploid sporophytic generations in the life cycle of any sexually reproducing organism is called the alternation of generation.

Example:

In plants like moss, fern, etc., and protists like Paramoecium, Monocystis, etc. distinct alternation of generation is seen.

Importance of mitosis in alternation of generations:

During the alternation of generation, the cyclic rotation continues between haploid (n) and diploid (2n) generations. Diploid spore mother cells divide by meiosis to give rise to haploid spores.

These spores germinate to grow as haploid (n) gametophytes. The gametophytes carry separate gametangia to produce haploid male and female gametes. These gametes fertilise to form diploid zygotes to initiate the sporophytic generation.

That is why meiosis is important for the alternation of generations.

Question 16. With the help of a schematic diagram represent the alternation of generations in fern. Similar Question, Represent the alternation of generation in fern with a suitable schematic diagram.
Or,
Depict the alternation of generations in a fern with the help of a flowchart.

Answer:

Alternation of generations in fern:

In the life cycle of ferns, two distinct phases, namely the sporophytic (2n) phase, and gametophytic (n) phase, are present. The sporophytic phase is the predominant one, which is perennial and conspicuous.

Within the sporangia, diploid spore mother cells undergo meiosis to produce numerous haploid spores. With the formation of spores, the haploid gametophytic generation sporophyte starts. These spores germinate to form gametophytes.

In ferns, the gametophytes are short-lived and inconspicuous. The gametophytes develop male and female gametangia, named as antheridia and archegonia respectively.

Within antheridia, haploid male gametes are produced and female gametes originate in archegonia. These gametes fertilize to form diploid zygospores to mark the onset of the next sporophytic generation.

Sexual Reproduction Explained

Question 17. Briefly describe the micropropagation technique. Mention its advantages. Or, How micropropagation is done?

Answer:

Technique of micropropagation:

The stepwise technique of micro- propagation are as follows-

1. At first, a disease-free plant is selected and a portion from the plant body is cut off with a sharp, well-sterilized instrument. This plant part is called an explant.
2. The explants is sterilized by treating it with 70% ethyl alcohol for 1 minute. and kept in 10% hypochlorite solution for 15 minutes for further sterilization.
3. The explant is then placed in a culture medium containing sucrose and a mixture of growth regulators like auxin and cytokinin.
4. The containers are placed on agitators in a completely sterile laboratory room in ambient temperature.
5. After a few days, a cluster of undifferentiated cells grow from the explant, these are called callus. Later, the cells of the callus divide and differentiate to form tiny somatic embryos or embryoids.
6. By controlled treatment of hormones, the embroids are grown as small plantlets. These are then transferred to sterilized soil to grow more. As the plantlets grow as saplings these are planted in natural soil for further growth.

Advantages of micropropagation:

1. By micropropagation, a large number of saplings can be formed in a small place.
2. The saplings, produced by this process are healthy and disease free.
3. By this process, desired varieties of plants can be produced at any season throughout the year.
4. The plants, which are unable to perform sexual reproduction, can be propagated by micropropagation.

Chapter 2 Continuity Of Life Reproduction Short Answer Type Questions

Question 1. Mention two characteristic features of sexual reproduction.

Answer:

Two characteristic features of sexual reproduction are as follows-

1. In sexual reproduction, two different gametes, i.e., male and female gametes, unite to form the offspring.
2. The progenies are born with variations, which in the long run, help in the formation of new and evolved species.

Question 2. How is a definitive nucleus formed during the sexual reproduction of angiospermic plants?

Answer:

Within the embryo sac of ovule of flowers in angiospermic plants, two haploids (n) polar nuclei unite together. As a result, a diploid (2n) definitive nucleus is formed.

Polar nucleus (n) + Polar nucleus (n) = Definitive nucleus (2n)

Question 3. What is syngamy?

Answer:

Syngamy:

Syngamy is a the complete and permanent union of two different gametes, i.e., male and female gametes, outside the gametangia. Syngamy is found in all sexually reproducing organisms.

Question 4. What is oogamy?

Answer:

Oogamy:

Oogamy is a type of sexual reproduction, where a small and motile male gamete fertilizes a large and non-motile female gamete. This is noticed among some algae like Oedogonium, Chara, etc., and in all higher plants and animals.

Question 5. What is isogamy?

Answer:

Isogamy:

Isogamy is a type of sexual reproduction where two morphologically and physiologically identical gametes unite to form a zygote. Isogamy occurs in certain lower organisms like Chlamydomonas, Monocystis, etc.

Question 6. What is anisogamy?

Answer:

Anisogamy:

Anisogamy is a type of sexual reproduction where two morphologically different gametes unite to form the zygote. This type of sexual reproduction is seen in certain species of Chlamydomonas.

Question 7. What is meant by asexual reproduction?

Answer:

Asexual reproduction:

The type of reproduction, in which the formation of male and female gametes and their fertilization do not occur. Instead, certain body cells or specialized cells like spores divide to form the progeny. This is known as asexual reproduction.

Example of Asexual reproduction:

Fungi, like yeast and certain lower animals perform asexual reproduction.

Question 8. What is meant by asexual reproduction by fission?

Answer:

Asexual reproduction by fission:

For propagation, most unicellular organisms undergo two or multiple number of cell divisions to form two or more offspring, either through mitosis or amitosis. This type of asexual reproduction is called fission.

Example of Asexual reproduction by fission:

Amoeba and Plasmodium perform asexual reproduction by binary and multiple fissions respectively.

Question 9. What is budding?

Answer:

Budding:

In certain lower organisms, a small portion of the vegetative body swells out as an outgrowth and finally, it gets detached from the mother body and grows as a new progeny. This type of asexual reproduction is called budding.

Example of Budding:

Budding is commonly seen in yeast and Hydra.

Importance of Reproduction for Species Continuity

Question 10. What are the different steps of budding in yeast?

Answer:

The different steps of budding in yeast:

Initiation of bud as outgrowth → Division of parent nucleus into two nuclei → Migration of one nucleus into bud→ Wall formation between parent cell and bud→ Detachment of bud and formation of new cell

Question 11. What is meant by sporulation?

Answer:

Sporulation:

In certain lower plants, numerous, typically thick-walled cells or spores are produced in sporangia. On attaining maturity, these spores are liberated from the mother plant and fall on the ground and finally germinate to form the new progeny.

This type of asexual reproduction is known as sporulation.

Example of Sporulation: Sporulation occurs in fungi, moss, fern, etc.

Question 12. What is multiple fission?

Answer:

Multiple fission:

Multiple fission is a type of asexual reproduction in which the parent nucleus undergoes free nuclear division forming a huge number of nuclei. Each nuclei absorb little ammount of cytoplasm and forms a membrane around it to form a new daughter cell.

Example of Multiple fission: Amoeba.

Question 13. What is meant by regeneration?

Answer:

Regeneration:

The process of asexual reproduction in which, a new progeny is formed from a fragment of the parent’s body, is known as regeneration.

Example of Regeneration: Planaria propagates by regeneration.

Question 14. What is meant by fragmentation?

Answer:

Fragmentation:

Filaments of certain lower plants often tear apart into fragments due to mechanical impact. By mitotic cell division, all these fragments may grow as individual progenies.

This type of vegetative propagation is known as fragmentation, for example-Spirogyra performs asexual reproduction by fragmentation.

Question 15. What is binary fission?

Answer:

Binary fission:

Binary fission is a type of asexual reproduction found in some prokaryotes and some unicellular eukaryotes in which the parent cell divides amitotically into two daughter cell.

Example of Binary fission: Amoeba.

Question 16. What are the different type of fission found in Amoeba?

Answer:

The different type of fission found in Amoeba:

Amoeba is an unicellular eukaryote which reproduces asexually by two methods binary fission and multiple fission.

Amoeba produces two daughter cells from parent cells at a time by binary fission and numerous daughter cells by free nuclear division through multiple fission.

Question 17. What are zoospores?

Answer:

Zoospores:

Certain asexual spores are capable of locomotion with the help of cilia or flagella. These spores are called zoospores.

Example of Zoospores:

Spores of Volvox, Chlamydomonas, etc. are zoospores.

Question 18. What are aplanespores?

Answer:

Aplanespores:

The non-motile asexual spores, which do not possess any locomotory organ, are known as aplanospores.

Example of Aplanespores: Spores of Mucor, Penicillum, etc. are aplanospores.

Question 19. What is meant by zygospore?

Answer:

Zygospore:

In certain lower organisms, two different gametes unite to form a diploid cell by conjugation, which develops a thick protective is known as a zygospore.

Example of Zygospore:

Spirogyra, cell wall to form a typical spore. This type of spore Oedogonium have zygospores.

Question 20. What is propagule?

Answer:

Propagule:

The plant part from which a new plant is formed during reproduction is called the propagule. Bud is a propagule of asexual reproduction and seed is a propagule of sexual reproduction.

Question 21. What are the processes of asexual reproduction in spirogyra and planaria?

Answer:

Spirogyra mediates their reproduction by fragmentation, whereas planaria make it by regeneration.

Question 22. What is gametophyte and sporophyte?

Answer:

Gametophyte and sporophyte:

The plant gamete which is produced from spore and produces gamete is called gametophyte. The plant which is produced from a zygote and produces a spore is called a sporophyte.

Question 23. What is meant by artificial vegetative propagation?

Answer:

Artificial vegetative propagation:

In the practice of agriculture and horticulture, vegetative propagation of different plants is done under human manipulation. This is known as artificial vegetative propagation.

Question 24. What is stolen?

Answer:

Stolen:

The internodes of certain herbs grow to form arches which remain above the ground. However, their nodes remain affixed with the soil by adventitious roots.

This type of sub-aerial modified stem is known as stolon, which is seen in plants like Mentha, strawberry, etc.

Question 25. Give examples of a few plants, which naturally perform vegetative propagation.

Answer:

Many plants naturally perform vegetative propagation. These include sweet potato, potato, pointed gourd, onion, Bryophyllum, etc.

Question 26. What is meant by stock and scion?

Answer:

Stock and scion:

In grafting, stems of two different varieties of plants are attached together. Of these two, the stem of the older plant is chopped off leaving a portion of it above the ground.

This rooted portion is called the stock and the younger stem branch with superior features which is affixed with the stock, is called the scion.

Question 27. Why grafting is not possible in paddy or maize plants.

Answer:

During grafting, stems of two different varieties of plant are attached together at the cambium region. So, for the grafting process, cambium plays a major role.

But paddy and maize, are monocot in nature and they do not contain cambium. So due to a lack of cambium, grafting is not possible in these plants.

Question 28. Which type of root is seen in sweet potato and what is its function?

Answer:

Fleshy tuberous adventitious roots are seen in sweet potatoes. The function of the roots of sweet potatoes is to store food. It also helps in vegetative propagation.

Question 29. Distinguish between cutting and grafting.

Answer:

Distinguishing features between cutting and grafting are-

Question 30. What is micropropagation?

Answer:

Micropropagation:

The process by which tissues, cells, protoplasts or organs are collected from plants and cultured in laboratories within proper nutritive medium to produce progeny plants, is known as micropropagation.

By means of micro- propagation large number of identical progeny plants, can be produced within a short period.

Question 31. What is totipotency?

Answer:

Totipotency:

The inherent ability of any cell to divide and develop into a full-grown plant under proper environmental conditions outside the living body is known as totipotency.

Due to this unique ability, small segments of plant tissue can be cultured within the laboratories to produce tiny saplings from it.

Question 32. Mention two distinctive features of micropropagation.

Answer:

Two distinctive features of micropropagation are as follows-

1. Micropropagation is a tissue culture technique by which several clones of plants can be produced vegetatively.
2. By this process, the number of desired varieties of plants can be increased within a short time.

Question 33. Mention the disadvantages of micropropagation.

Answer:

The disadvantages of micropropagation-

1. In micropropagation, no genetic variations are developed. Therefore, the plants produced in this process have lesser adaptability.
2. Due to a lack of variation, micropropagation reduces the chance of evolution.

Question 34. What is meant by tissue culture?

Answer:

Tissue culture:

The biotechnological process in which tissues, collected from different organisms are cultured in the laboratory within a proper nutritive culture medium, under a suitable sterile environment, is known as tissue culture.

This process can form sprouts within laboratories from small plant tissues.

Question 35. Which plant hormones are mixed with a special tissue culture medium and why?

Answer:

Auxin and cytokinin are the hormones which are mixed in a special tissue culture medium. These two hormones regulate the growth of the cultured tissues. Therefore, these two growth regulators are used in tissue culture medium.

Comparative Study of Asexual and Sexual Reproduction

Question 36. What are explants?

Answer:

Explants:

The small plant parts, tissues, or cells, which are collected from plant bodies for culturing in laboratories, are called explants. The meristematic tissue-rich portions of plants, like apical buds, root tips, and tubers are mostly collected as explants.

Question 37. What is a callus?

Answer:

Callus:

While growing in sterile culture media, mitosis starts in explants. After some time, a cluster of undifferentiated cells are produced from the cells of explants.

This cluster of cells is known as a callus. The cells of the callus are totipotent in nature, which may divide to form a full-grown plant.

Question 38. What is meant by caulogenesis?

Answer:

Caulogenesis:

The development of stem from a callus, within a tissue culture medium, is called caulogenesis. This is a tissue differentiation process, which is initiated by a higher concentration of auxin and cytokinin in the culture medium.

Question 39. What is meant by histogenesis?

Answer:

Histogenesis:

The development of a root from a callus, within a tissue culture medium, is called histogenesis. This is a tissue differentiation process, which is initiated by a lower concentration of auxin and cytokinin in the culture medium.

Question 40. What is meant by histogenesis?

Answer:

Histogenesis:

The development of xylem and phloem tissues by the differentiation of a callus within a tissue culture medium is called histogenesis. This is a tissue differentiation process, which is regulated by auxin and cytokinin in the culture medium.

Question 41. What is an alternation of generations?

Answer:

Alternation of generations:

The alternation of a haploid (n) and a diploid (2n) generation in the life cycle of an organism showing sexual reproduction is called the alternation of generations.

Example: Fern.

Question 42. Mention the importance of the alternation of generations.

Answer:

The importance of the alternation of generations:

The cyclic rotation of haploid and diploid generations helps in keeping the chromosome number constant. The alternation of generations also helps in maintaining the genetic identity of any organism for generations.

Chapter 2 Continuity Of Life Reproduction Very Short Answer Type Questions Multiple Choice Questions And Answers [Mcq]

Question 1. Which of the following organisms performs reproduction by fission?

1. Plasmodium
2. Hydra
3. Fern
4. Planaria

Answer: 1. Plasmodium

Question 2. Which of the following organisms performs reproduction by budding?

1. Amoeba
2. Plasmodium
3. Hydra
4. Planaria

Answer: 3. Hydra

Question 3. Amoeba performs—

1. Fission
2. Budding
3. Fragmentation
4. Sporulation

Answer: 1. Fission

Question 4. Yeasts reproduce by—

1. Fragmentation
2. Sporulation
3. Regeneration
4. Budding

Answer: 4. Budding

Question 5. Which of the following organisms does not perform reproduction by spore formation?

1. Moss
2. Planaria
3. Fern
4. Fungi

Answer: 2. Planaria

Question 6. Fragmentation is seen in—

1. Hydra
2. Yeast
3. Plasmodium
4. Spirogyra

Answer: 4. Spirogyra

Question 7. Which of the following organisms performs reproduction by regeneration?

1. Hydra
2. Spirogyra
3. Planaria
4. Amoeba

Answer: 3. Planaria

Question 8. Adventitious bud formation occurs in the leaves of which of the following plants?

1. Sweet potato
2. Potato
3. Water hyacinth
4. Bryophyllum

Answer: 4. Bryophyllum

Question 9. Which of the following organisms performs reproduction by sporulation?

1. Fungi
2. Planaria
3. Hydra
4. Amoeba

Answer: 1. Fungi

Question 10. An example of the juicy root is—

1. Water hyacinth
2. Sweet potato
3. Potato
4. Bryophyllum

Answer: 2. Sweet potato

Question 11. Which of the following organisms shows both asexual as well as sexual reproduction?

1. Hydra
2. Amoeba
3. Paramoecium
4. Horse

Answer: 3. Paramoecium

Question 12. Grafting is a technique of—

1. Sexual reproduction
2. Asexual reproduction
3. Vegetative propagation
4. Parthenogenesis

Answer: 3. Vegetative propagation

Question 13. Reproduction in plants by cutting method is a type of—

1. Artificial vegetative propagation
2. Asexual reproduction
3. Sexual reproduction
4. Natural vegetative propagation

Answer: 1. Artificial vegetative propagation

Question 14. The reproductive stage, produced by the conjugation of similar gametes, is called

1. Zygote
2. Zygospore
3. A zygospore
4. Oospore

Answer: 2. Zygospore

Question 15. Which of the following performs reproduction by conjugation?

1. Spirogyra
2. Mucor
3. Moss
4. Fern

Answer: 1. Spirogyra

Question 16. Fern performs reproduction by—

1. Budding
2. Fragmentation
3. Spore formation
4. Regeneration

Answer: 3. Spore formation

Question 17. Fertilized ovum is known as—

1. Spermatozoon
2. Zygote
3. Zygospore
4. Spore

Answer: 2. Zygote

Question 18. An unfertilized egg, taking part in parthenogenesis, is known as—

1. Zygote
2. Zygospore
3. A zygospore
4. Oospore

Answer: 3. Azygospore

Question 19. The process of ovum formation is called—

1. Spermatogenesis
2. Oogenesis
3. Parthenogenesis
4. Gametogenesis

Answer: 2. Oogenesis

Question 20. The process of sperm formation is known as—

1. Spermatogenesis
2. Oogenesis
3. Parthenogenesis
4. Gametogenesis

Answer: 1. Spermatogenesis

Question 21. The Union of two identical gametes is called—

1. Isogamy
2. Anisogamy
3. Oogamy
4. Syngamy

Answer: 1. Isogamy

Question 22. The process of reproduction, in which offspring are formed by the division of body cells without the union of gametes, is called—

1. Sexual reproduction
2. Vegetative propagation
3. Asexual reproduction
4. Parthenogenesis

Answer: 3. Asexual reproduction

Question 23. In which of the following organisms sexual dimorphism is seen?

1. Cockroach
2. Earthworm
3. Amoeba
4. Plasmodium

Answer: 1. Cockroach

Question 24. External fertilization takes place in which of the following animals?

1. Human
2. Fish
3. Deer
4. Horse

Answer: 2. Fish

Question 25. Which of the following is an example of a hermaphroditic animal?

1. Cockroach
2. Snake
3. Earthworm
4. Toad

Answer: 3. Earthworm

Question 26. The formation of offspring from unfertilized egg is known as—

1. Paedogenesis
2. Parthenogenesis
3. Parthenocarpy
4. Oogenesis

Answer: 2. Parthenogenesis

Question 27. The process of asexual reproduction in bacteria is known as—

1. Binary fission
2. Fragmentation
3. Multiple fission
4. Mitosis

Answer: 1. Binary fission

Question 28. Which of the following plants performs reproduction by means of root buds?

1. Rose
2. Dahlia
3. Pointed gourd
4. Bryophyllum

Answer: 3. Pointed gourd

Question 29. The animal in which unfertilized eggs develop into offspring is—

1. Toad
2. Honey bee
3. Pigeon
4. Fish

Answer: 2. Honey bee

Question 30. Spore formation, from the spore mother cells in ferns, occurs by—

1. Meiosis
2. Amitosis
3. Mitosis
4. Binary fission

Answer: 1. Meiosis

Question 31. Sexual phase of alternation of generation is called—

1. Gametophytic phase
2. Sporophytic phase
3. Gamete
4. Parthenogenesis

Answer: 1. Gametophytic phase

Question 32. The asexual phase of alternation of generation in a plant is known as—

1. Gametophytic phase
2. Sporophytic phase
3. Embryo
4. Oogenesis

Answer: 2. Sporophytic phase

Question 33. The haploid phase of the alternation of generation in fern is—

1. Gametophytic
2. Sporophytic
3. Parthenocarpic
4. Both 1 and 2

Answer: 1. Gametophytic

Question 34. The sporophytic phase of the alternation of generation in fern is—

1. Haploid (n)
2. Diploid (2n)
3. Triploid (3n)
4. Quadruploid (4n)

Answer: 2. Diploid (2n)

Question 35. Vo I vox reproduces by—

1. Sexual reproduction
2. Asexual reproduction
3. Vegetative propagation
4. Parthenogenesis

Answer: 2. Asexual reproduction

Question 36. The modified sub-aerial stem of water hyacinth, which helps it in vegetative propagation, is called—

1. Sucker
2. Offset
3. Runner
4. Stolon

Answer: 2. Offset

Question 37. The buds, present on the margins of leaves of Bryophyllum, are called—

1. Leaf buds
2. Epiphytic buds
3. Terminal buds
4. Lateral buds

Answer: 1. Leaf buds

Question 38. Artificial vegetative propagation involving stock and scion is known as—

1. Cutting
2. Gootee
3. Grafting
4. Layering

Answer: 3. Grafting

Question 39. Artificial vegetative propagation which is applied for china rose and marigold, is—

1. Cutting
2. Gootee
3. Grafting
4. Layering

Answer: 1. Cutting

Question 40. The plant that performs vegetative propagation by bulbil is—

1. Sweet potato
2. Oxalis
3. Air potato (Dioscorea)
4. Dahlia

Answer: 3. Air potato (Dioscorea)

Question 41. A hermaphrodite animal that can perform self-fertilization is—

1. Toad
2. Liver fluke
3. Ascaris
4. Salamander

Answer: 2. Liver fluke

Question 42. A plant hormone, provided externally in tissue culture medium is—

1. Thyroxine
2. Artificial auxin
3. Insulin
4. Glucagon

Answer: 2. Artificial auxin

Question 43. Method of cloning is an example of—

1. Micropropagation
2. Genetic engineering
3. Biotechnology
4. All of these

Answer: 4. All of these

Question 44. The artificial reproduction in which culture medium is required is called—

1. Cutting
2. Grafting
3. Micropropagation
4. All of these

Answer: 3. Micropropagation

Question 45. Other than Planaria, which of the following animals shows regeneration?

1. Spongilla
2. Amoeba
3. Pogonatum
4. Paramoecium

Answer: 1. Spongilla

Question 46. The event of growth of any organism within the uterus is known as—

1. Alternation of generation
2. Vivipaiy
3. External fertilisation
4. Internal fertilisation

Answer: 2. Vivipaiy

Question 47. The physiological process by which an organism produces its own replica is known as—

1. Respiration
2. Circulation
3. Excretion
4. Reproduction

Answer: 4. Reproduction

Question 48. In grafting, the cut branch of the superior plant is called a ‘scion’. The inferior rooted portion, on which it is fixed, is called—

1. Bark
2. Stem
3. Stock
4. Bulb

Answer: 3. Stock

Question 49. The cutting, by which sugarcane plants can be propagated, is collected from—

1. Root
2. Stem
3. Leaf
4. All of these

Answer: 2. Stem

Question 50. A complete and permanent union of two identical gametes is called—

1. Isogamy
2. Anisogamy
3. Syngamy
4. Heterogamy

Answer: 3. Syngamy

Question 51. union of two male and female gametes of unequal size and shape is called—

1. Isogamy
2. Anisogamy
3. Syngamy
4. Heterogamy

Answer: 3. Anisogamy

Question 52. The presence of an organ, which is essential to distinguish between male and female of any unisexual organism, is called—

1. Reproductive organ
2. Respiratory organ
3. Excretory organ
4. Sensory organ

Answer: 1. Reproductive organ

Question 53. The continuous alternation of haploid and diploid generations in the life cycle of certain organisms is called—

1. Alternation of generation
2. Reproduction
3. Conjugation
4. Transformation

Answer: 1. Alternation of generation

Question 54. The cell division, by which the zygotes of higher plants and animals develop into full-grown organisms, is called—

1. Meiosis
2. Mitosis
3. Both 1 and 2
4. Amitosis

Answer: 2. Mitosis

Question 55. The fertilization between male and female gametes of higher organisms is known as—

1. Isogamy
2. Anisogamy
3. Oogamy
4. Syngamy

Answer: 3. Oogamy

Question 56. The first object produced from explants during micropropagation is—

1. Callus
2. Stock
3. Scion
4. Spore

Answer: 1. Callus

Question 57. The unequal-sized spores produced during asexual reproduction are called—

1. Parthenocarpic
2. Anthers
3. Homospores
4. Heterospores

Answer: 4. Heterospores

Question 58. Spores which are produced within sporangium, are called—

1. Sporangiospores
2. Aplanospores
3. Zoospores
4. Zygospores

Answer: 1. Sporangiospores

Question 59. An animal, capable of reproducing by sexual, asexual, and vegetative means, is—

1. Spirogyra
2. Hydra
3. Spirogyra
4. Hydra

Answer: 3. Spirogyra

Question 60. An animal,which performs asexual, asexual, and vegetative means is

1. Spirogyra
2. Hydra
3. Spirogyra
4. Hydra

Answer: 4. Hydra

Question 61. An aquatic organism that performs internal fertilization is—

1. Mrigel fish
2. Shark
3. Rohu fish
4. Catla fish

Answer: 2. Shark

Question 62. An example of an ovoviviparous animal is—

1. Rohu fish
2. Prawn
3. Shark
4. Whale

Answer: 3. Shark

Question 63. A plant that shows viviparous germination

1. Garan
2. Papaya
3. Mango
4. Pumpkin

Answer: 1. Garan

Question 64. The formation of offspring from any cut-off organs of a plant is called—

1. Development
2. Regeneration
3. Transformation
4. Reconstruction

Answer: 2. Regeneration

Question 65. The chromosome number of sperm is—

1. Diploid
2. Triploid
3. Haploid
4. Tetraploid

Answer: 3. Haploid

Question 66. The chromosome number of a zygote is—

1. 2n
2. n
3. 3n
4. 4n

Answer: 1. 2n

Question 67. The chromosome number of the ovum is—

1. Diploid
2. Triploid
3. Pentaploid
4. Haploid

Answer: 4. Haploid

Question 68. The capability of any tissue cell to form an entire plant body is called—

1. Alternation of generation
2. Parthenogenesis
3. Totipotency
4. Transformation

Answer: 3. Totipotency

Question 69. Which of the following pair is correct?

1. Budding—Yeast
2. Fragmentation—Earthworm
3. Spore formation—Amoeba
4. Regeneration—Dryopoteris

Answer: 1. Budding—Yeast

Question 70. Which one of the following statements is correct regarding sexual reproduction?

1. Haploid gamete formation is essential in sexual reproduction
2. Sexual reproduction depends only on mitosis
3. In sexual reproduction, offspring could be produced from a single parental organism.
4. The offsprings produced in sexual reproduction are genetically identical with a parental organism.

Answer: 1. Haploid gamete formation is essential in sexual reproduction

Question 71. Select the correct pair and write it—

1. Multiple fission—Hydra
2. Fragmentation—Spirogyra
3. Regeneration—Fern
4. Budding—Planaria

Answer: 2. Fragmentation—Spirogyra

 

Chapter 2 Continuity Of Life Reproduction Answer In A Single Word Or Sentence

Question 1. What is the process of formation of similar or dissimilar character-bearing offspring from parent organism called?
Answer: Reproduction

Question 2. What is the main cause of reproduction?
Answer: To maintain the race of a species

Question 3. What is the unit of sexual reproduction?
Answer: Gamete

Question 4. In which organ the spores for asexual reproduction of a plant are produced?
Answer: Sporangium

Question 5. What is produced after the union of sperm and ovum, during sexual reproduction?
Answer: A zygote is produced.

Question 6. Name a dioecious or unisexual plant.
Answer: Papaya plant

Wbbse Class X Life Science

Question 7. Name a bisexual but monoecious plant.
Answer: Mango plant

Question 8. Name a bisexual flower.
Answer: China rose (Hibiscus sp.)

Question 9. Name a unisexual flower.
Answer: Flower of pumpkin is a unisexual flower.

Question 10. Which is the unit of asexual reproduction of a plant?
Answer: Spore is the unit of asexual reproduction of a plant.

Question 11. Which type of reproduction is performed by Amoeba?
Answer: Amoeba performs asexual reproduction.

Question 12. Name a plant, which shows alternation of generation in its life cycle.
Answer: Selaginella (a fern)

Question 13. Name an animal, which shows an alternation of generations in its life cycle.
Answer: Obelia (a cnidarian)

Common Reproductive Strategies in Animals

Question 14. Name an animal, which performs reproduction by conjugation.
Answer: Paramoecium (a protist)

Question 15. Name an animal, which can reproduce by parthenogenesis.
Answer: Honey bee

Question 16. Name an oviparous animal.
Answer: Bird

Question 17. Name a viviparous animal.
Answer: Human

Question 18. Name the event of the union of male and female gametes in sexual reproduction.
Answer: Fertilisation

Wbbse Class X Life Science

Question 19. Name two hormones used for micropropagation.
Answer: Auxin and cytokinin

Question 20. In which type of reproduction new genetic features may appear in the offspring?
Answer: In sexual reproduction, new genetic features may appear in the offspring.

Question 21. By which process does Spirogyra perform sexual reproduction?
Answer: Spirogyra performs sexual reproduction by conjugation.

Question 22. Which type of offspring are produced in asexual reproduction?
Answer: The offspring produced in asexual reproduction, become genetically identical to its parents, i.e., no genetic variation occurs.

Question 23. How does the body fragment of a Planaria grow into a fully formed one?
Answer: By regeneration.

Question 24. How do you denote the union of a male and a female gamete of identical shape and size?
Answer: Isogamy.

Question 25. What is heterogamy?
Answer: Heterogamy is the union of two gametes with non-identical sizes and shape.

Question 26. How do you refer to the union of two gametes, produced in the body of the same animal?
Answer: The union of two gametes, produced in the body of the same organism, is called self-fertilization.

Question 27. How do you refer to the union of two gametes, produced in two different animals?
Answer: The union of two gametes, produced in two different animals, is called cross-fertilization.

Question 28. Name the type of fertilization that occurs in an open environment, outside the animal body.
Answer: External fertilization

Wbbse Class X Life Science

Question 29. Which phase of life cycle starts in an organism with the formation of haploid spores?
Answer: Gametophytic generation starts with the formation of haploid spores, in the life cycle of an organism.

Question 30. Name an organism, which performs asexual reproduction by multiple fission.
Answer: Plasmodium

Question 31. Name the process of fertilization of male and female gametes in the sexual reproduction of higher organisms.
Answer: The process of fertilization of male and female gametes in the sexual reproduction of higher organisms is known as oogamy.

Question 32. Upon which type of cell division does sexual reproduction depend directly?
Answer: Meiotic cell division

Question 33. Which type of cell division occurs during the gametogenesis of sexually reproducing higher organisms?
Answer: In higher organisms, meiotic cell division occurs during gametogenesis.

Question 34. Which type of cell division occurs during the sporogenesis of ferns and mosses?
Answer: Meiotic cell division

Question 35. Which type of cell division occurs during spore formation in Mucor and Penicillium?
Answer: Mitosis

Question 36. Which type of reproduction helps in creating variations among offspring?
Answer: Sexual reproduction

Wbbse Class X Life Science

Question 37. Name one organism that can reproduce through a motile spore.
Answer: Chlamydomonas

Question 38. Name one organism that can reproduce through a nonmotile spore.
Answer: Mucor

Question 39. Name fungi that reproduce through conidia.
Answer: Penicillium

Question 40. Give one example of a fleshy not, helping in vegetative reproduction.
Answer: Sweet potato

Question 41. Which plant vegetatively reproduces through stolon?
Answer: Mentha

Question 42. Which plant reproduces through a leaf?
Answer: Bryophyllum

Question 43. Who discovered micropropagation?
Answer: Frederick Campion Steward

Wbbse Class 10 Life Science Solutions

Question 44. Micropropagation is based on which property of a cell?
Answer: Totipotency

Question 45. What is a plantlet?
Answer: The small plants produced through micropropagation are called plantlets.

Question 46. Name one organism that reproduces through Conjugation.
Answer: Paramoecium

Question 47. What is the torula stage?
Answer: The bud of yeast looks like the animal Torula hence this stage is called the torula stage.

Question 48. What is syngamy?
Answer: Syngamy is the fusion of two different gametes.

Question 49. What is rhizogenesis?
Answer: Rhizogenesis is the production of roots in callus culture.

Question 50. Name one animal showing, vegetative, asexual, and sexual reproduction.
Answer: Hydra sp.

 

Chapter 2 Continuity Of Life Reproduction Fill In The Blanks

Question 1. A progeny comes from _________ generation.
Answer: Parental

Question 2. The individual of a new generation that originates from the parental generation is known as _________.
Answer: Offspring

Question 3. The number of a population _________ by reproduction.
Answer: Increases

Question 4. In sexual reproduction, _________ originate in a population.
Answer: Variations

Question 5. In asexual reproduction, only _________ parent is required.
Answer: One

Question 6. The union of male and female gametes is known as _________.
Answer: Fertilisation

Question 7. Based on the nature of gametes, sexual reproduction can be classified into _________ types.
Answer: Three

Wbbse Class 10 Life Science Solutions

Question 8. The sexual reproduction, in which the sizes of male and female gametes are unequal, is called _________.
Answer: Anisogamy

Question 9. The sexual reproduction, in which the sizes of male and female gametes are identical, is called _________.
Answer: Isogamy

Question 10. The sexual reproduction, in which the male gametes are small and motile, whereas the female gamete is large and non-motile, is called _________.
Answer: Oogamy

Question 11. Daughter cells of Plasmodium, produced by schizogony, are called _________.
Answer: Merozoites

Question 12. Daughter cells of Plasmodium, produced by sporogony, are called _________.
Answer: Sporozoites

Question 13. Planaria can perform asexual reproduction by _________ of its body fragments.
Answer: Regeneration

Question 14. Numerous _________ are produced from the sporogenous tissue of moss.
Answer: Spores

Question 15. Bryophyllum performs vegetative propagation by the on _________ its leaf margins.
Answer: Buds

Question 16. _________ is a type of tissue culture technique.
Answer: Micropropagation

Question 17. Spore-producing plants are known as _________ plants.
Answer: Sporophytic

Question 18. The _________, used in the grafting process, carries the superior characteristics of a plant.
Answer: Scion

Question 19. The diploid cell produced by fertilisation is _________.
Answer: Zygote

Question 20. Spore is formed within _________.
Answer: Sporangium

 

Chapter 2 Continuity Of Life Reproduction State True Or False

Question 1. The production of new offspring by a parental generation is called reproduction.
Answer: True

Question 2. Like nutrition, respiration or excretion, reproduction is also an essential life process of the living organisms.
Answer: False

Question 3. The unit of sexual reproduction is the gamete.
Answer: True

Question 4. The unit of vegetative propagation is spore.
Answer: False

Question 5. Amoeba divides by mitosis.
Answer: False

Question 6. Plasmodium reproduces by fragmentation.
Answer: False

Question 7. Yeast reproduces by spore production.
Answer: False

Question 8. Micropropagation is none other than a technique of tissue culture.
Answer: True

Question 9. Potato reproduces by stem buds.
Answer: True

Wbbse Class 10 Life Science Solutions

Question 10. Bryophyllum reproduces by root buds.
Answer: False

Question 11. Water hyacinth performs vegetative propagation by stolon.
Answer: False

Question 12. Birds show distinct alternations of generation.
Answer: False

Question 13. Scion, used in grafting, bears superior character.
Answer: True

Question 14. Apomixis is a process by which certain organisms can produce their progeny, without- out fertilization.
Answer: True

Question 15. The haploid phase of the alternation of generation in an organism is none other than the sexual phase.
Answer: True

Question 16. Fertilization is the union of male and female gametes.
Answer: True

Question 17. Planaria performs reproduction by regeneration.
Answer: True

Question 18. Paramoecium reproduces by budding.
Answer: False

Question 19. Plasmodium propagates by multiple fission.
Answer: True

Question 20. The most common process of sexual reproduction of animals and plants is syngamy.
Answer: True

 

Chapter 2 Continuity Of Life Reproduction Match The Columns

Question 1.

Answer: 1. E, 2. D, 3. B, 4. A, 5. F, 6. G

Question 2.

Answer: 1. E, 2. C, 3. A, 4. B, 5. D, 6. G

Question 3.

Answer: 1. B, 2. A, 3. D, 4. C, 5. F, 6. E

Chapter 2 Continuity Of Life Reproduction Find The Odd One Out

Question 1. Cutting of rose plant, Grafting of lemon plant, Micropropagation of carrot plant, Formation of leaf bud of Bryophyllum plant.
Answer: Formation of leaf bud of Bryophyllum plant.

Question 2. Schizogony of Plasmodium, Zoospore formation of Phytophthora, Conidiospore, formation of Penicillium, Budding of yeast.
Answer: Budding of yeast

Question 3. Oidia, Conidia, Zoospore, Gamete.
Answer: Gamete

Question 4. Zoospore, Aplanospore, Gamete, Embryo.
Answer: Embryo

Wbbse Class 10 Life Science Solutions

Question 5. Sponge, Hydra, Yeast, Sporulation
Answer: Sporulation

Question 6. Regeneration, Fragmentation, Budding, Syngamy.
Answer: Syngamy

Chapter 2 Continuity Of Life Reproduction Fill In The Blanks By Looking At The First Pair

Question 1. Leaf bud: Bryophyllum:: Adventitious root bud: ________.
Answer: Sweet potato

Question 2. Spirogyra: Fragmentation:: Yeast: ________.
Answer: Buddings

Question 3. Regeneration: Asexual reproduction:: Micropropagation: ________.
Answer: Vegetative reproduction

Question 4. Bryopyllum: Leaf bud :: Water hyacinth: ________.
Answer: Offset

Question 5. Binary fission: Amoeba:: Regeneration: ________.
Answer: Planaria

Wbbse Class 10 Life Science Solutions

Question 6. Grafting: Mango:: ________: Chinarose.
Answer: Cutting

Question 7. Heterogamete: Anisogamy:: Isogamete:: ________.
Answer: Anisogamy

Question 8. Motile spore: Zoospore :: Non motile spore: ________.
Answer: Aplanopore

Chapter 2 Continuity Of Life Reproduction Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Adventitious root bud, Offset of water hyacinth, Natural vegetative propagation, Leaf bud of Bryophyllum.
Answer: Natural vegetative propagation

Question 2. Cutting of rose twigs, Grafting of lemon plant, Artificial vegetative propagation, Layering of Chrysanthemum branch.
Answer: Artificial vegetative propagation

Question 3. Asexual reproduction, Sporulation, Fragmentation, Budding.
Answer: Asexual reproduction

Question 4. Cutting, Grafting, Micropropagation, Artificial Asexual reproduction.
Answer: Artificial asexual reproduction

Question 5. Gametophyte, Sporopyte, Fertilization, Alternation of generation.
Answer: Alternation of generation

Question 6. Cutting, Grafting, Vegetative reproduction; Micropropagation.
Answer: Vegetative reproduction

Question 7. Spore, Conidia, Sporangiospore, Oidia
Answer: Spore

Question 8. Asexual propagation of animals, Schizogony of Plasmodium, Binary fission of Amoeba, Regeneration of Planaria.
Answer: Asexual propagation of animals

Chapter 4 Evidence For The Theory Of Evolution Summary

WBBSE Class 10 Evidence for Evolution Overview

• There are various compelling evidences of evolution and one of them is fossil. Fossils provide solid evidence for the fact that the ancestors of an organism were not the same as those found today. Fossils show the progression of the evolution of an organism.
• Highly detailed fossil records have been discovered that depict the evolution of modern horses. The actual sequence of species in the evolution of horses is-Eohippus (first ancestor)→ Mesohippus (intermediate horse) → Merychippus → Pliohippus → Equus (Modern horse).
• In the course of evolution, horses developed larger bodies, longer limbs, longer necks and longer and slimmer heads. Reduction in the number of toes and development of a third toe, reduction in the number of teeth and concentration on incisor and molar teeth are a few other features observed in the line of evolution of horses.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Life Science Class 10 Wbbse

• The organs with the same basic structural design and origin but with different functions are called homologous organs.
• Example: Flipper of a whale, the hand of a human, patagium of the bat. Homologous organs are the result of divergent evolution.
• Divergent evolution refers to the process by which interbreeding species diverge into two or more evolutionary groups mainly because of adaptation to different environmental conditions.
• The organs with different origins but the same functions are called analogous organs.
• Example: Wings of a bird and insect. Analogous organs develop because of convergent evolution. Convergent evolution is the process by which organisms, not closely related, independently evolve similar traits in order to adapt to similar environments.
• The organ having no apparent functions within an organism, but was active in the ancestors of that organism, is called a vestigial organ.
• Example: vermiform appendix and coccyx of human, wings of ostriches, pistinode of Cassia, etc.
• Structures of hearts of vertebrates show gradual complexity and thus point out towards evolution. Embryos of all the vertebrates shows pharyngeal gill cleft and myotome muscle indicating their interrelationship.

Chapter 4 Evidence For The Theory Of Evolution LAQs

Question 1. Explain the importance of fossils in finding the evolutionary lineage of horse.

Answer:

Importance of fossils in finding the evolutionary lineage of horse

Fossils play very important role in finding the evolutionary lineage of any group of organisms. In case of horse’s evolution also, fossils played a very important role.

Palaeontologists have discovered the fossils of all the ancestors of horse, which helped the scientists to draw a complete evolutionary line of the modern horse from its earliest ancestor.

Life Science Class 10 Wbbse

It is now known that the earliest ancestor of horse, Eohippus, emerged about 55 million years ago. It was adapted in forest environment with its shorter legs consisting multiple digits.

Gradually, grassland developed and horses were adapted in the new environment by gradually acquiring changes in their physical structure. The evolutionary lineage of horse is-

Eohippus → Mesohippus → Merychippus → Pliohippus → Equus

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By comparative analysis of fossil structures, scientists found the following features-

1. Only 11 inches tall Eohippus has evolved into 60 inches tall modern horse, Equus.
2. The hind legs have become more muscular to facilitate fast running.
3. The number of digits in both limbs has reduced gradually and hoofs have developed.
4. The length of the neck has increased.
5. Sizes of cranium and cerebrum have increased significantly.
6. The crowns of teeth have become higher to help in chewing of grass.

From this fossil study, the scientists have revealed almost all the adaptive features of the ancestral horses with the changing environment and presented the evolutionary lineage of modern horse from its ancestors.

Question 2. Briefly describe the evolution of horse.

Answer:

Evolution of horse:

The evolutionary lineage of horses has become clear today because palaeontologists have discovered fossils of all the ancestors of the modern horses.

The oldest ancestor of a horse was Eohippus, from which the modern horse, Equus, has come into being after crossing three other stages of evolution. Descriptions of the fossils of all ancestors of horses are mentioned below.

Homologous Chromosomes Definition

Fossil Evidence for Evolution

Question 3. How do homologous organs support organic evolution?

Answer:

Homologous organs in support of organic evolution:

Homologous organs support organic evolution to a great extent. This is explained as below.

1. Homologous organs of animals:

The basic skeletal structures of forelimbs of whale, bat, bird, cat. and man are same. All possess humerus, radius-ulna, carpals, metacarpals and phalanges.

However, these animals use their forelimbs in different ways to adapt to separate environments. Whale uses its forelimbs for swimming, but birds and bats use it for flying. Cat walks and captures prey with its forelimb and man uses it for grasping and eating.

Homologous

These evidences strongly indicate that all vertebrates have evolved from the same ancestor but separated from each other to adapt in separate environments.

2. Homologous organs in plants:

The thorn of the wood apple tree, stem tendril of the passion flower plant, phylloclade of cactus, etc. are modifications of the same organ but these perform different functions.

The above-mentioned examples prove that these organisms evolved from the same ancestor but with time, their organs have changed their shapes and functions to adapt in separate environments.

 

Question 4. How do analogous organs support organic evolution?

Answer:

Analogous organs in support of organic evolution:

Analogous organs support organic evolution. This is explained below.

1. Analogous organs in animals:

Wings of birds and wings of insects are used for flying. However, structurally these two are different. Wings of birds are modifications of the forelimbs but the wings of insects are lateral folds of the cuticle of the thorax.

The sting of a scorpion and the sting of a honeybee are functionally identical, however, the sting of a scorpion is the modification of a telson and the sting of a honeybee is a modified ovipositor.

2. Analogous organs in plants:

The leaf tendril of the pea plant and the stem tendril of the passion flower plant are functionally similar. However, the former one is the modification of terminal leaflets and the latter one is the modified branch of the stem.

Homologous

Analogous organs indicate the adaptation of distantly related organisms in identical environments. Therefore, analogous organs provide strong support in favour of organic evolution.

Question 5. Distinguish between homologous and analogous organs. Mention differences between amphibian and reptilian hearts.

Answer:

Distinguishing features between homologous organs and analogous organs

Life Science Class 10 Wbbse

Differences between amphibian and reptilian heart

Question 6. How do vestigial organs favour organic evolution?

Answer:

Vestigial organs as evidence for organic evolution:

1. In many herbivores, like rabbits, guinea pigs, etc. a big saccular caecum is present at the junction of the large and small intestine. It gives shelter to symbiotic bacteria to facilitate their cellulose digestion. However, human food contains less cellulose. Therefore, this organ remains as a functionless, tiny worm-like projection in a man called a vermiform appendix.
2. The tail of the monkey acts as a balancing organ in arboreal locomotion. In humans, tail is of no need, hence it is present as a tiny coccyx bone at the end of the spine.
3. The nictitating membrane is active in amphibians but non-functional in humans. Muscles of pinnae are useless in humans but functional in cattle. Transverse abdominal muscles are essential for cheetahs but of no use in human beings.
4. Wings of birds are essential organs but in flightless birds (ostrich, kiwi, emu, etc.), wings become vestigial.
5. Scale leaves of the rhizome of ginger and turmeric, and staminodes of Cassia are examples of vestigial organs in plants.

 

Homologous and Analogous Structures Explained

The presence of vestigial organs in certain organisms proves that in the course of evolution, those have become functionless and reduced but in ancestral forms, they were prominent and active. Therefore, vestigial organs can be treated as good evidence in favour of organic evolution.

Question 7. Explain the comparative anatomical evidence in support of organic evolution.
Or,
Explain the comparative anatomy of vertebrate hearts and forelimbs in support of organic evolution. Part question, How the comparative anatomy of hearts act as proof in favour of evolutionary theory?

Answer:

Comparative anatomical evidence in support of organic evolution:

A comparative study on related groups of organisms, clearly reveals the basic structural similarity among few of their organs. It also indicates the evolutionary trend of those organisms.

Here, comparative studies on some organs of different groups of vertebrates are mentioned below.

Homologous

1. Heart of vertebrates:

The basic structure of the hearts of fishes, amphibians, reptiles, birds and mammals is similar. All of them have an auricle as the blood-receiving chamber and a ventricle as the chamber that expels blood from the heart. However, in the course of evolution, the hearts of different vertebrates encountered some changes.

The heart of fish has one auricle and one ventricle and thus carries only deoxygenated blood along a single circuit. Amphibians have two auricles and one ventricle in their heart. Here, oxygenated and deoxygenated blood get mixed in the ventricle.

Apart from two auricles, the reptilian heart has a partially divided ventricle, in which partial mixing of oxygenated and deoxygenated blood occurs (exception-crocodile has a four-chambered, double-circuit heart).

Hearts of birds and mammals have two auricles and two ventricles. Therefore oxygenated and deoxygenated blood flow through separate circuits.

Conclusion:

The above-mentioned evidence proves that advanced and complex forms of animals have evolved from primitive and simple forms. The gradual development of the heart and circulatory system of different vertebrates point towards the evolutionary trend as mentioned below.

Fishes → Amphibians → Reptiles → Birds and mammals

Life Science Class 10 Wbbse

2. Forelimbs of different vertebrates:

The basic skeletal structures of the forelimbs of whales, bats, birds, horses and men are all alike. All possess humerus, radius-ulna, carpals, metacarpals and phalanges.

However, these animals use their forelimbs in different ways to adapt in different environments. As whale uses its forelimbs for swimming, birds and bats use them for flying. The horse runs with the help of their forelimbs and man use them for grasping.

Conclusion:

This evidence strongly indicates that all vertebrates have evolved from a single ancestor but diverged from each other to adapt in separate environments.

Question 8. Briefly describe the embryological evidence in favour of organic evolution.

Answer:

Embryological evidence in favour of organic evolution:

Comparative and chronological study of the embryos of fishes, amphibians, reptiles, birds and mammals reveals that all these groups of vertebrate organisms have many similarities in the shape and structure of the embryos for a period of time.

This observation directly indicates that all these different groups of organisms have evolved from a single ancestor.

Vestigial Organs and Evolution

The similarities are-

1. Embryos of all these forms have developed from a unicellular zygote.
2. The zygotes of these organisms have divided mitotically to form blastocysts.
3. In the next phase of development, these blastocysts transform into gastrula.
4. All these forms are triploblastic.

A comparative study of these vertebrates reveals two main structural similarities among them.

Homologous

1. Presence of pharyngeal gill clefts and internal gill pouches:

All these groups of animals have pharyngeal gill clefts at the lateral sides of their heads in their initial embryonic states. The endothelium of the pharynx infolds to form two laterally placed gill pouches in all these embryos.

The gill clefts and gill pouches jointly form the pharyngeal organ. In the course of maturity, fishes develop gills from this pharyngeal organ, but in other terrestrial organisms, the pouches either give rise to other structures or disappear.

In humans, the clefts develop into thyroid, parathyroid and thymus glands.

2. Tail-like structure and myotome muscles:

Embryos of all these classes have a tail-like structure and laterally placed myotome muscles. For example, a frog bears a tail in its larval stages but not in its adult stage.

Similarly, in human beings, the tail becomes vestigial in the form of a coccyx after embryonic development. Further, in addition to fishes, myotome muscles are also found in human beings, in a specific segment of the spinal cord where the spinal nerve innervates.

These above-mentioned similarities indicate towards same ancestral lineage of these vertebrate classes and also support the concept of organic evolution.

Chapter 4 Evidence For The Theory Of Evolution SAQs

Question 1. Furnish some evidence in favour of Darwinism.
Answer:

The pieces of evidence in favour of Darwinism is as follows-

1. Stick insects and leaf insects take the shape of dry branches and dry leaves respectively to camouflage in the plants to avoid predation. These typical shapes are naturally selected to help them survive on the earth.
2. The length of the suckers of nectar-eating insects is so adjusted that can reach the nectar glands of their favourite flowers. This typical length of the sucker is naturally selected to let the insects get their food easily.

Question 2. What are fossils?
Answer:

Fossils:

Fossils are the petrified remains, impressions traces of extinct primitive organisms, preserved via natural processes within the layers of ancient sedimentary rocks.

Example-Fossils of dinosaurs, Archaeopteryx, etc.

Question 3. What are living fossils?
Answer:

Living fossils:

Certain primitive plants and animals are still surviving in the recent world without any significant characteristic change throughout the course of evolution. These organisms are called living fossils.

Example-Limulus (animal), Ginkgo biloba (plant), etc.

Embryological Evidence for Evolution

Question 4. What are missing links?
Answer:

Missing links:

While drawing the evolutionary lineage of any group of organisms, it is often noticed that there are certain gaps in it due to the extinction of certain connecting species.

These extinct species are called missing links. Fossil of Archaeopteryx, for instance, is the missing link between reptiles and birds.

Question 5. Mention the importance of fossils as evidence of organic evolution.
Answer:

Fossils provide strong evidence in favour of organic evolution in the following ways-

1. Fossils provide the idea about the size, shape, food habit and approximate age of a fossilised organism.
2. By studying fossils, scientists can acquire knowledge about the nature of flora and fauna present in the environment of that time.
3. Fossils can trace the ancestral history of any modern species, thus depicting the evolutionary lineage.

Wbbse Class 10 Life Science Solutions

Question 6. Distinguish between fossil and living fossil
Answer:

The distinguishing features between fossil and living fossils are-

Question 7. What is palaeontology?
Answer:

Palaeontology:

Palaeontology is a branch of biology, in which fossils are studied to gain knowledge about the characteristic features of ancient animals. Palaeontology helps to draw the evolutionary trend of the related organisms.

Moreover, study of palaeontology helps the scientists to find out the lineage of emergence of a new species from it’s ancestors.

Question 8. Write in chronological sequence the names of four main fossil ancestors during the course of evolution of the horse.
Answer:

In chronological sequence, the four main fossil ancestors during the course of evolution of the horse are-Eohippus → Mesohippus → Merychippus → Pliohippus → Equus (modern horse)

Question 9. Write four important features modified in the evolution of horse.
Answer:

Four important features modified in the evolution of horse are-

1. Gradual increase in height,
2. Decrease in number of digits in both forelimbs and hindlimbs and formation of hoof,
3. Increase in the size of crown of molar and pre- molar teeth,
4. Increase in volume of cerebral hemisphere.

Question 10. Distinguish between Eohippus and Equus.
Answer:

The distinguishing features between Eohippus and Equus are-

Wbbse Class 10 Life Science Solutions

Molecular Evidence Supporting Evolution

Question 11. What are homologous organs?
Answer:

Homologous organs:

The organs of different organisms, having similar origin and fundamental structures but with differences in appearance and functions, are called homologous organs.

Example: Hands of humans and wings of birds.

Question 12. Which anatomical feature proves forelimbs of amphibians and wings of birds as homologous organs?
Answer:

Both the forelimbs of amphibians and wings of birds have same skeletal elements, i.e. humerus, radius-ulna, carpal, metacarpal and phalanges. This anatomical similarity proves that these two organs are homologous in nature.

Question 13. Mention two features of internal structures on the basis of which ‘flippers of whales and ‘wings of birds’ are considered homologous organ.
Answer:

Two features of internal structures on the basis of which ‘flippers of whale’ and ‘wings of birds’ are considered homologous organ are-

1. The basic skeletal structures of flipper of whales and wings of birds are same. Both possess humerus, radius- ulna, carpals, metacarpals and phalanges.
2. Well developed flexor and extensor muscles are present in flipper of whale and also in wings of birds.

Question 14. What is divergent evolution?
Answer:

Divergent evolution:

To adapt in dissimilar environments, certain organs, having same origin and structure, may take different shapes to perform separate functions in descendants coming from a common ancestral form. This type of adaptive evolution is called divergent evolution.

Example:

Homologous organs like forelimbs of horse, paddle of whale and hands of man are structurally similar but functionally different, which indicates divergent evolution.

Wbbse Class 10 Life Science Solutions

Question 15. What are analogous organs?
Answer:

Analogous organs:

The organs of different organisms having different origin and fundamental structures, but with somewhat similar appearances and identical functions are called analogous organs.

Example: Patagium of bats and wings of insects.

Question 16. Give examples of analogous organs in animals.
Answer:

Examples of analogous organs in animals

The wings of birds and insects are examples of analogous organs in animals. Both these animals use the wings for flight, but structurally these are completely different.

Question 17.

1. Structure and function
2. Indicating the nature of evolution

Based on the above two features, establish the concept of analogous organs with the help of proper example.
Answer:

Analogous organs are those organs which are different in structure and origin but perform similar functions.

Analogous organs indicate convergent evolution. For example, wings of bird, wings of insect and patagium of bat perform same function, i.e., they help the organisms to fly but they have different structures.

Wings of birds are modification of forelimbs, wings of insects are outgrowth of insects’ exoskeleton and patagium of bat is actually fold of skin between forelimbs and hind limbs.

All these structures point towards the fact that in same environment, different structures may evolve to perform same function. This further points toward convergent evolution.

Question 18. What is meant by convergent evolution?
Answer:

Convergent evolution:

Distantly related groups of animals or plants, inhabiting in identical surroundings for a long period, may develop analogous organs to adapt themselves in that particular environment.

This adaptive evolution is known as convergent evolution. Example-The terminal leaflets of pea plants and some branches of pumpkin plants are modified into tendrils.

Question 19. Distinguish between homologous and analogous organs.
Answer:

The distinguishing features between homologous and analogous organs are-

 

Question 20. What is parallel evolution?
Answer:

Parallel evolution:

Two or more descendants from the same ancestor, inhabiting in identical surroundings may develop similar traits to adapt themselves in that particular environment. This adaptive evolution is known as parallel evolution.

Example:

Deer and antelopes have same ancestors, they thrive in similar environments and both have split hoofs for fast running.

Wbbse Class 10 Life Science Solutions

Question 21. What are vestigial organs?
Answer:

Vestigial organs:

Vestigial organs are the organs which have become reduced and non-functional due to prolonged disuse. These organs were previously functional and well-built in ancestral forms.

The muscles of the pinnae, vermiform appendix, and nictitating membrane of the eye are some vestigial organs in humans.

Question 22. Write the names of one each vestigial organ present in the vertebral column and alimentary canal in the human body.
Answer:

In the human body, the vestigial organ present in the vertebral column is the coccyx and that present in the alimentary canal is the appendix.

Question 23. Distinguish between the amphibian heart and the mammalian heart.
Answer:

The distinguishing features between the amphibian heart and the mammalian heart are-

 

Question 24. Mention the biogenetic theory of organic evolution.
Answer:

The biogenetic theory of organic evolution:

Biogenetic theory of evolution or the theory of recapitulation, proposed by scientist Ernst Haeckel, states that ontogeny repeats phylogeny. It means the embryonic phases of a species resemble its evolutionary trend.

Question 25. Briefly mention the role of Ernst Haeckel in the field of comparative embryology.
Answer:

German scientist Ernst Haeckel explained the organic evolution under the light of comparative embryology and thus confirmed Darwinian concept on evolution.

He compared three embryonic stages of eight different vertebrates and reached to a conclusion that ontogeny repeats phylogeny. This means, the embryonic stages of animals resemble their evolutionary trend.

 

Natural Selection as Evidence of Evolution

Question 26. How do principles of Von Baer, on comparative embryology, support the organic evolution?
Answer:

To support the organic evolution in 1828, German scientist Von Baer had put forward certain principles on embryology. According to him-

1. General species characters appear first in its early embryonic stages.
2. With advance- ment of embryonic development, typical and specific characters of species appear in the embryos.

Question 27. What are pharyngeal gill clefts?
Answer:

Pharyngeal gill clefts:

In any of the stages of their life, chordates. possess several parallel clefts at two lateral sides of the pharynx. In aquatic forms, these clefts transform into gills.

In terrestrial forms, these ridges develop certain glands, which are known as pharyngeal gill clefts.

 

Chapter 4 Evidence For The Theory Of Evolution MCQs

Question 1. The stings of honeybees and wasps are—

1. Homologous organs
2. Analogous organs
3. Vestigial organs
4. Both 1 and 2

Answer: 2. Analogous organs

Question 2. The branch of science, dealing with the study of fossils, is called—

1. Palaeontology
2. Palaeogeography
3. Herpetology
4. Embryology

Answer: 1. Palaeontology

Question 3. Organs with similar structure and origin but different functions are called—

1. Homologous organs
2. Analogous organs
3. Vestigial organs
4. Connecting organs

Answer: 1. Homologous organs

Question 4. The feature of analogous organs is—

1. Different in origin and functions are also different
2. Different in origin but functions are the same
3. Indicates divergent evolution
4. Identical in origin and structure

Answer: 2. Different in origin but functions are the same

Question 5. A vestigial structure in the human body is—

1. Humerus
2. Caecum
3. Vermiform appendix
4. Trachea

Answer: 3. Vermiform appendix

Question 6. Which of the following is not vestigial in humans?

1. Tail bone
2. Appendix
3. Nail
4. Nictitating membrane

Answer: 3. Nail

Question 7. The connecting link between reptiles and mammals is—

1. Archaeopteryx
2. Platypus
3. Peripatus
4. Octopus

Answer: 2. Platypus

Question 8. The fossilised animal, Archaeopteryx, is treated as the missing link between which of the following two groups?

1. Reptiles and mammals
2. Birds arid mammals
3. Amphibians and mammals
4. Reptiles and birds

Answer: 4. Reptiles and birds

Question 9. The heart, carrying deoxygenated blood, is present in—

1. Toad
2. Fish
3. Snake
4. Crocodile

Answer: 2. Fish

Question 10. The tendril of pea and passion flower plants are—

1. Homologous organs
2. Vestigial organs
3. Analogous organs
4. Growth organs

Answer: 3. Analogous organs

Question 11. Horses evolved on earth approximately—

1. 40 million years ago
2. 65 million years ago
3. 100 million years ago
4. 55 million years ago

Answer: 4. 55 million years ago

Question 12. An example of a missing link is—

1. Equus
2. Archaeopteryx
3. Eohippus
4. Ostrich

Answer: 2. Archaeopteryx

Question 13. The modern horse evolved approximately—

1. 5 million years ago
2. 1 million years ago
3. 15 million years ago
4. 20 million years ago

Answer: 2. 1 million years ago

Question 14. Name an animal which is a living fossil.

1. Earthworm
2. Rohu fish
3. Limulus
4. Garden lizard

Answer: 3. Limulus

Question 15. The height of Eohippus was—

1. 11-12 inches
2. 51-52 inches
3. 71-72 inches
4. 81-82 inches

Answer: 1. 11-12 inches

Question 16. Name a plant which is a living fossil.

1. Cooksonia
2. Rhynia
3. Pinus
4. Ginkgo biloba

Answer: 4. Ginkgo biloba

Question 17. Dawn horse is more commonly called as

1. Eohippus
2. Merychippus
3. Mesohippus
4. Pliohippus

Answer: 1. Eohippus

Question 18. Which is known as a ‘ruminating horse’?

1. Eohippus
2. Pliohippus
3. Mesohippus
4. Merychippus

Answer: 4. Merychippus

Question 19. The intermediate horse is—

1. Pliohippus
2. Mesohippus
3. Merychippus
4. Equus

Answer: 2. Mesohippus

Question 20. The horse with a single digit is—

1. Mesohippus
2. Merychippus
3. Pliohippus
4. Equus

Answer: 4. Equus

Question 21. The oldest ancestor of the horse was seen in—

1. Eocene era
2. Miocene era
3. Pliocene era
4. Oligocene era

Answer: 1. Eocene era

Question 22. The 24 inches tall ancestor of the modern horse, which was seen in the Oligocene era, is—

1. Eohippus
2. Mesohippus
3. Pliohippus
4. Merychippus

Answer: 2. Mesohippus

Question 23. The horse of the Miocene era, having three digits in both the fore and hindlimbs, is—

1. Merychippus
2. Mesohippus
3. Eohippus
4. Pliohippus

Answer: 1. Merychippus

Question 24. The height of Merychippus was—

1. 100 cm
2. 60 cm
3. 28 cm
4. 160 cm

Answer: 1. 100 cm

Question 25. The reptilian heart has two auricles and a partly divided ventricle. An exception is seen in—

1. Snake
2. Garden lizard
3. Tortoise
4. Crocodile

Answer: 4. Crocodile

Question 26. The modern horse is called—

1. Eohippus
2. Mesohippus
3. Pliohippus
4. Equus

Answer: 4. Equus

Question 27. Who discovered the fossils showing the evolution of horse?

1. Darwin
2. Lamarck
3. Wales
4. Charles Marsh

Answer: 4. Charles Marsh

Question 28. Which of the following denotes a vestigial organ of plants?

1. Stamen
2. Leaf
3. Staminode
4. None of the above

Answer: 3. Staminode

Question 29. Which of the following has vestigial wings?

1. Pigeon
2. Hornbill
3. Vulture
4. Kiwi bird

Answer: 4. Kiwi bird

Question 30. “Ontogeny repeats phytogeny”—Who proposed this?

1. Mendal
2. Lamarck
3. Darwin
4. Haeckel

Answer: 4. Haeckel

Question 31. Number of chambers in the heart of reptiles—

1. Two
2. Three
3. Incompletely four
4. Four

Answer: 3. Incompletely four

Question 32. Which of the following sequence is right?

1. Pishes → Amphibians → Birds and mammals → Reptiles
2. Amphibians → Fishes → Reptiles → Birds and mammals
3. Birds and mammals → Fishes → Amphibians → Reptiles
4. Fishes → Amphibians → Reptiles → Birds and mammals

Answer: 4. Fishes → Amphibians → Reptiles → Birds and mammals

Question 33. Which of the following changes did not occur during the course of the evolution of the horse?

1. Lengthening of limbs
2. Increase in the length and thickness of all digits in limbs
3. Increase in the length and thickness of only the third digit in limbs
4. Increase in size of the whole body.

Answer: 2. Increase in the length and thickness of all digits in limbs

 

Chapter 4 Evidence For The Theory Of Evolution Answer In A Single Word Or Sentence

Question 1. What type of organs are the flippers of whales, the forelimbs of horses and the hands of man?
Answer: Homologous organs

Question 2. What type of organs are the wings of birds and wings of insects?
Answer: Analogous organs

Question 3. Which type of evolution is indicated by homologous organs?
Answer: Divergent evolution

Question 4. Which reptile has a four-chambered heart?
Answer: Crocodile

Question 5. Which group of animals has a venous heart?
Answer: Fish

Question 6. What is the scientific name of the modern horse?
Answer: Equus caballus

Question 7. Give an example of a vertebrate living fossil.
Answer: Tuatara (Sphenodon sp.)

Question 8. From the point of view of origin, which organ of the horse shows similarity with the wings of birds?
Answer: Forelimbs

Question 9. Certain organisms possess characteristics of two different groups of organisms. What are these organisms called?
Answer: Connecting link

Question 10. Certain organs have become non-functional in the course of evolution. What are these called?
Answer: Vestigial organs

Question 11. Which digit of a horse is modified into a hoof?
Answer: The third digit

Question 12. Which is the vestigial organ of an ostrich?
Answer: Wings

Question 13. Name an egg-laying mammal.
Answer: Duck-billed platypus

Question 14. Name the oldest ancestor in the evolutionary path of the horse.
Answer: Eohippus

Artificial Selection and Its Role in Evolution

Question 15. How many digits were there in the fore and the hindlimbs of an Eohippus?
Answer: The forelimbs and the hindlimbs of an Eohippus had four and three functional digits respectively.

Question 16. Which type of fossils are examined under the microscope?
Answer: Microfossils

Question 17. Name the connecting link between gymnosperms and angiosperms.
Answer: Gnetum

Question 18. On attaining maturity, what happens to the embryonic gill pouches and gill clefts in mammals?
Answer: On attaining maturity, the embryonic gill pouches and gill clefts in mammals transform into thyroid, parathyroid and thymus glands.

Question 19. Who proposed the concept, ‘ontogeny repeats phylogeny’?
Answer: Scientist Ernst Haeckel

Question 20. What is Archaeopteryx?
Answer: Archaeopteryx is the missing link between reptiles and birds.

Question 21. Which digit of the horse is transformed into a hoof?
Answer: The third digit

Question 22. Which is the vestigial organ of the kiwi?
Answer: Wings

Question 23. Which teeth of a human is vestigial?
Answer: The third molar teeth

Question 24. When was Merychippus originated?
Answer: Miocene

Question 25. How many chambers are there in a bird’s heart?
Answer: Four

Question 26. Write down the names of two vestigial organs of plants.
Answer: Staminode of Cassia and scale leaf present in the rhizome of ginger.

Question 27. What is the alternative name of biogenetic law?
Answer: Recapitulation theory

 

Chapter 4 Evidence For The Theory Of Evolution Fill In The Blanks

Question 1. The petrified remains or impression of any organism, kept for ages in the layers of sedimentary rocks are called ________.
Answer: Fossils

Question 2. Coccyx is a ________ organ of the human body.
Answer: Vestigial

Question 3. Among the ancestors of the horse, hoofs developed first in ________.
Answer: Merychippus

Question 4. The hoof of modern horse is the transformation of the digit number ________ of their ancestors.
Answer: Three

Question 5. Homologous organs evolve due to ________ evolution.
Answer: Divergent

Question 6. Analogous organs in different types of organisms, provide evidence of ________ evolution.
Answer: Convergent

Question 7. Phylloclade of cactus and tuber of potato are examples of two ________ organs.
Answer: Homologous

Question 8. The leaf tendril of pea plant and stem tendril of the passion plant are examples of ________ organs.
Answer: Analogous

Question 9. The appendix is a ________ organ of the human body.
Answer: Vestigial

Question 10. The heart of fish carries deoxygenated blood only, therefore, it is known as ________ heat.
Answer: Venous

Question 11. Hearts of birds and mammals have ________ chambers.
Answer: Four

Question 12. One living fossil animal is ________.
Answer: Limuliis

Question 13. The number of digits on the forelegs of Merychippus were ________.
Answer: 3

Question 14. The heart of fish is ________ chambered.
Answer: Two

Question 15. According to Haeckel, ontogeny repeats ________.
Answer: Phylogeny

Question 16. The name of the modern horse is ________.
Answer: Equus

Question 17. The earliest ancestor in the evolution of the horse is ________.
Answer: Eohippus

 

Chapter 4 Evidence For The Theory Of Evolution State True Or False

Question 1. Although a crocodile is a reptile, it has a four-chambered heart.
Answer: True

Question 2. The reptilian heart has a partly segmented ventricle.
Answer: True

Question 3. The sequence of vertebrate evolution can be represented as, fishes → amphibians → reptiles → birds and mammals.
Answer: True

Question 4. The wings of birds and insects are examples of analogous organs.
Answer: True

Question 5. The organs of different organisms, having same origin and fundamental structures but performing different functions, are called analogous organs.
Answer: False

Question 6. The hands of humans and the wings of birds are analogous organs.
Answer: False

Question 7. The Stem tendril of the passion flower plant and the thorn of the wood apple tree are examples of homologous organs.
Answer: True

Question 8. The branch of science that deals with the study of fossils to find out the line of evolution of any organism, is known as palaeontology.
Answer: True

Question 9. Staminodes of Cassia are its vestigial organs.
Answer: True

Question 10. Mesohippus is the earliest ancestor of the modern horse.
Answer: False

Question 11. Among the ancestors of horses, hoofs developed first in Merychippus.
Answer: True

Question 12. Pliohippus is called a ruminating horse.
Answer: False

Question 13. The pinnae are vestigial organs of the human body.
Answer: True

 

Chapter 4 Evidence For The Theory Of Evolution Match The Columns

Question 1.

Answer: 1. C, 2. E, 3. A, 4. D, 5. B, 6. F

Question 2.

Answer: 1. C, 2. F, 3. A, 4. E, 5. B, 6. D

Question 3.

Answer: 1. B, 2. A, 3. E, 4. C, 5. D, 6. G

 

Chapter 4 Evidence For The Theory Of Evolution Find The Odd One Out

Question 1. Forelegs of the horse, Wings of a pigeon, Wings of the bat, Wings of a butterfly
Answer: Wings of butterfly

Question 2. Eohippus, Merychippus, Mesohippus, Mammoth
Answer: Mammoth

Question 3. Eohippus, Dinosaur Archaeopteryx, Crocodile,
Answer: Crocodile

Question 4. Coccyx, Spleen, Appendix, Nictitating membrane
Answer: Spleen

Question 5. Wings of a bird, Forelegs of the horse, Hands of a human, Tail of a whale
Answer: Tail of a whale

Question 6. Adaptation in a particular environment, Analogous organ, Convergent evolution, Divergent evolution
Answer: Divergent evolution

 

Chapter 4 Evidence For The Theory Of Evolution Fill In The Blanks By Looking At The First Pair

Question 1. Amphibian heart: Mixed blood:: Reptilian heart: ___________
Answer: Partially mixed blood

Question 2. Vermiform appendix and nictitating membrane of man: Vestigial organ:: Wings of birds and insects: ___________
Answer: Analogous organs

Question 3. Biogenetic law: Ernst Haeckel:: Mutation theory: ___________
Answer: Hugo de Vries

Question 4. Eocene: Eohippus :: Oligocene: ___________
Answer: Mesohippus

Question 5. Homologous Organ: Divergent evolution Analogous organ: ___________
Answer: Convergent evolution

Question 6. Mammals Four chambered heart:: Amphibians: ___________
Answer: Three-chambered heart

Question 7. Functionally different: Homologous organ:: Difference in origin:___________
Answer: Analogous Organ

Question 8. Eohippus: Oldest ancestor of horse:: Equus: ___________
Answer: Modern horse

 

Chapter 4 Evidence For The Theory Of Evolution Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Flippers of whales and hands of humans, Structurally similar but functionally different organs, Homologous organs, Wings of birds and bats
Answer: Homologous organs

Question 2. Reduction in the number of toes, Increasing running ability, Lengthening of legs, Development of single hard hoof all
Answer: Increasing running ability

Question 3. Pliohippus, Fossils of horse, Merychippus, Mesohippus
Answer: Fossils of horse

Question 4. Staminode, Wings of kiwi, Pistilode, Vestigial organs
Answer: Vestigial organ

Question 5. Human heart, Heart of a bird, Double circuit heart, Heart of a frog
Answer: Double-circuit heart

Question 6. Concept of evolution, the Concept of ancient environment, the Importance of fossils, Concept of the origin of an organism
Answer: The importance of fossil

Chapter 4 Theories Of Organic Evolution Summary

Theory Of Organic Evolution

• There are various theories regarding evolution, of which Lamarckism and Darwinism are the most significant.
• In the year 1809, Lamarck published his book ‘Philosophie Zoologique’ where he proposed and explain his theories regarding the evolution of new species.
• The main features of the theory of Lamarck are-

1. Internal vital force
2. New needs culminate in the formation of the new organ
3. Law of use and disuse of an organ
4. Inheritance of acquired characters
5. Formation of new species

• German biologist August Weismann conducted an experiment where he removed the tails of mice repeatedly over 22 generations and reported that no mice were born in consequence without a tail or even with a shorter tail.
• This experiment opposed the theory of inheritance of acquired characters of Lamarck. Charles Robert Darwin sailed around the world, as a naturalist, aboard the HMS Beagle and this experience and studies that he made, helped him to propose the theory of evolution which was published in his book, ‘Origin of Species by Means of Natural Selection’.
• The salient features of the theory of Darwin are-

Life Science Class 10 Wbbse

1. Prodigality of production
2. Limited food and shelter on earth
3. Struggle for existence
4. Variations
5. Survival of the fittest
6. Natural selection
7. Origin of new species

• Few limitations of Darwin’s theory concern the origin of variation, the origin of new characters, the arrival of the fittest, etc.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Chapter 4 Theories Of Organic Evolution Long Answer Type Questions

Question 1. Briefly describe the Lamarckian concept of evolution. Part question, Describe two major tenets of Lamarck’s evolutionary theory.

Answer:

Lamarckian concept of evolution:

In 1809, Lamarck proposed the concept of organic evolution in his famous book, titled ‘Philosophie Zoologique. This concept is popularly known as Lamarckism.

Lamarckism is based on the following basic propositions.

Wbbse Class 10 Life Science Solutions

WBBSE Class 10 Theories of Organic Evolution Overview

1. Internal vital force:

An internal vital force or energy in all life forms helps in the development of the body and organs.

2. Development of new organs according to need:

With the change in environment, a need develops among the organisms for morphological and behavioural change. Newly developed organs help the organisms to adapt in the changing environment.

3. Law of use and disuse:

Any organ, which is used more, grows in size and strength, whereas, that which is used less, reduces in size, becomes weak and finally becomes extinct.

Example of use of organ:

Ancestors of giraffes had short necks. To collect the leaves of taller plants, they stretched their necks continuously. This conscious effort made their necks long and eventually, giraffes with longer necks evolved.

Example of disuse of organ:

The ancestors of snakes had limbs. While adapting to the fossorial habitat, their limbs were left unused. Finally, limbs. of snakes were lost during the course of evolution.

4. Inheritance of acquired characters:

Any change in the characteristics of an organism, which an organism achieves by conscious effort, is inherited by its offspring.

5. Formation of new species:

The characters acquired during the lifetime are passed onto the progeny and after a number of generations, new species is produced.

To get leaves from tall trees, ancestors of giraffes stretched their necks. By this process, one might have lengthened it a little, which was inherited directly by its offspring.

This process continued for several thousand generations and finally completely different-looking long-necked giraffes emerged as a new species from their ancestors.

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WBBSE Class 10 History Very Short Answer Questions	WBBSE Solutions for Class 10 Physical Science and Environment
WBBSE Class 10 History Multiple Choice Questions

 

Wbbse Class 10 Life Science Solutions

Question 2. What is meant by the ‘Inheritance of acquired characters’? Explain with an example. Mention the demerits in the Lamarckian concept of evolution.

Answer:

Inheritance of acquired characters:

According to Lamarck, any feature that organisms acquire during its conscious effort to adapt in to an environment is inherited by its offspring. This event is known as the ‘Inheritance of acquired characters.

Example:

The necks of ancient giraffes were short. With a continuous effort to reach the leaves of taller trees, they stretched to lengthen their necks. The little change achieved by this process in one generation was inherited by the next.

This process continued for several successive generations and gradually long-necked giraffes have arrived as a new species.

Demerits of Lamarckism:

1. The law of use and disuse is not universally accepted. In some cases, active organs may reduce in size in the course of evolution.
2. Only internal vital force cannot develop a new organ. An aquatic animal cannot become a terrestrial animal only by internal desire.
3. Continuous use of any organ never always makes an organ stronger. The heart, for instance, never becomes larger and stronger with time.
4. Inheritance of acquired characters is not always evident. No somatic change is inherited by the next generation, it may occur only in case of any change in the germ cells.

Darwin’s Theory of Natural Selection Explained

In this connection, Weismann’s experiment may be cited. He cut the tails of newborn rats for 22 successive generations; however, tail-less rats were never born.

Similarly, Drosophila flies were reared in complete darkness for 60 successive generations, but no blind fly was born. Based on this observations, Weismann strongly opposed the Lamarckian concept of inheritance of acquired characters.

Question 3. Explain the Darwinian concept on organic evolution. Part Questions, Discuss the following three events as mentioned in the evolutionary theory of Darwin:

1. The prodigality of production,
2. Origin of variation,
3. Natural selection

Answer:

Darwinian concept on organic evolution:

In 1859, Charles Darwin explained the modern concept of organic evolution in his famous book ‘On. the Origin of Species by Means of Natural Selection, which has become famous as Darwinism or as the theory of natural selection. The basic thoughts of Darwinism are mentioned below.

1. Prodigality of production:

Darwin opined that all organisms increase their population in a geometric rate. For example, a pond snail lays 6 million eggs in a year. A female salmon fish releases 30 million eggs in a reproductive season.

2. Limited food and shelter on earth:

The space on the earth and supply of food do not increase in proportion to the rate of increase in population. Therefore, a scarcity of food and shelter occurs in the living world.

3. Struggle for existence:

Continuous increase in population size and scarcity of food and shelter results in to conflicts among the organisms for survival. Darwin referred to it as the struggle for existence. This struggle is of three types-

1. Intraspecific struggle-The conflict among the members of a same species population for food, shelter and mating partners.
2. Inter-specific struggle-The- conflict among the members of different species for resources and shelter.
3. Environmental struggle- This struggle is for surviving different natural calamities, like drought, flood, earthquake, etc.

Wbbse Class 10 Life Science Solutions

Lamarckism vs Darwinism

4. Variation:

Darwin indicated that no two individuals of the same species are identical. There must be some differences between them, which he stated as variations.

5. Survival of the fittest:

Darwin mentioned two different types of variations-favourable and unfavourable. Favourable variations help an organism to adapt to changing environment but unfavourable variations cannot help in their adaptation.

The individuals with unfavourable variations face a defeat in the struggle for existence and become extinct in the course of time. Whereas, the organisms with favourable variations win that struggle and survive on Earth.

Scientist Herbert Spencer denoted this event as the ‘survival of the fittest.

6. Natural Selection:

According to Darwin, organisms with favourable variations are selected by nature for survival because they are fittest to face their surroundings. Darwin explained this event as ‘natural selection.

The naturally selected forms thrive successfully and propagate very fast to increase their population.

7. Origin of new species:

Accumulation of many favourable variations in any group of organisms for generations makes their descendants widely different from their ancestors. Finally, these completely changed descendants emerge as new species.

Question 4. What is natural selection? Mention the drawbacks of Darwinism. Part question, Explain the process of natural selection as proposed by Darwin with the help of a suitable example.

Answer:

Natural selection:

Natural selection means the selection of a suitable organism in an environment by nature itself. According to Darwin, organisms acquire either favourable or unfavourable variations during their struggle for existence.

Those with favourable variations are selected by nature. They can adapt perfectly with its environment and come out victorious in the struggle for life. They thrive successfully and propagate quickly.

On the other hand, the organisms, which have unfavourable variation, do not get the favour of natural selection and cannot adapt themselves in their environment. The population of these forms reduces gradually and finally, becomes extinct.

For example, in an ecosystem, some giraffes have long necks and others have short necks. If a situation arises, where there is a scarcity of low-lying shrubs, the giraffes with short necks would not get enough food. After a few generations, all the giraffes would have long necks.

 

Evidence Supporting Organic Evolution Theories

Drawbacks of Darwinism:

1. Darwin mentioned the role of variation in evolution but could not explain the sources of it.
2. Darwin emphasised on small variations but in reality, small variations play no role in the origin of species.
3. Darwin supported the concept of survival of the fittest, however, he could not explain the cause of the emergence of the fittest.
4. According to Darwin, excessive development of any organ facilitates evolution but it is seen that over-specialisation often pushes a species towards extinction.
5. Darwin did not differentiate between somatic and reproductive variations.
6. During natural calamities, even the fittest organisms also die, where natural selection does not operate.

Question 5. What is ‘survival of the fittest’? Explain how do certain organs become vestigial giving examples of few vestigial organs in animals and plants.

Answer:

Survival of the fittest:

Here, the word ‘fittest’ means, the most competent individual or group among a number of individuals or groups. On the other hand, ‘survival’ means coming out victorious in the struggle for existence.

During the struggle for existence in a hostile environment, certain organisms acquire some favourable variations, which help them to survive in the struggle. However, there are some other organisms, which have unfavourable variations, cannot overcome environmental barriers and gradually become extinct.

Therefore, the fittest organisms win the battle for survival and thrive successfully. Herbert Spencer denoted this event as ‘survival of the fittest’.

 

The emergence of vestigial organs:

Certain organs, which were once active and useful in ancestors, may become non-functional and weak in their descendants due to continuous disuse in the changed environment. Thus, certain active organs transform into vestigial and inactive organs.

1. Example of vestigial organs in animals:

Muscles of the pinna, the nictitating membrane of the eye, vermiform appendix are a few examples of vestigial organs in man. The pelvic girdle bones of a whale, and wings of ostrich are some vestigial organs of other animals.

2. Example of vestigial organs in plants:

Scale leaves of underground-modified stems, sterile gynoecia of coconut, staminodes of Cassia, and scale-like leaves of Cuscuta are some vestigial organs in plants.

Question 6. Explain the process of lengthening of giraffe’s neck under the light of Lamarckism and Darwinism.

Answer:

Explanation of lengthening of giraffe’s neck:

Lamarck and Darwin, the two pioneer figures of evolutionary science, explained the event of gradual lengthening of the giraffe’s neck in two different ways.

1. Lamarckian explanation:

Lamarck believed that the necks of the ancestors of giraffes were almost similar to that of recent horses. To reach the leaves of taller trees they used to stretch their necks continuously.

In this process, members of every generation lengthened their necks to a little extent. This changed feature was inherited by the next generation and this process continued for several successive generations.

Finally, after thousands of generations, the long-necked giraffes have emerged as a new species.

2. Darwinian explanation:

According to Darwin, in ancient days members of the giraffe population had various lengths of necks. Those with longer necks could reach the leaves of taller trees and hence got more food than the ones with shorter necks.

The long-necked giraffes became healthier and thrived successfully. On the other hand, the short-necked giraffes became weaker and reproductively incapable due to a shortage of adequate food.

Gradually, the number of these giraffes went down and at the end they became extinct. However, the long-necked giraffes came out victorious in the struggle for existence and survived on earth as a new species.

Types of Evolutionary Theories

Question 7. Distinguish between Lamarckism and Darwinism.

Answer:

Distinguishing features between Lamarckism and Darwinism

 

Chapter 4 Theories Of Organic Evolution Short Answer Type Questions

Question 1. What made Lamarck famous?

Answer:

Lamarck :

Lamarck is famous for his two theories on organic evolution.

These are-

1. Law of use and disuse and
2. Law of inheritance of acquired characters.

Question 2. What do you mean by the ‘law of use and disuse’? Or, Mention the ‘law of use and disuse’ in relation to organic evolution.

Answer:

‘Law of use and disuse’:

According to Lamarck, any organ that is used more, becomes stronger and well built, whereas Put emphasis on three different types of struggles for existence; intra- specific struggle, inter-specific struggle and struggle any unused organ becomes weak and reduces in the course of time. This theory is known as the ‘law of use and disuse.

For example, flightless birds have evolved from their flying ancestors. As the ancestor birds do not use wings, their wings become vestigial structures.

Role of Adaptation in Organic Evolution

Question 3. What is meant by ‘inheritance of acquired characters’?

Answer:

‘Inheritance of acquired characters’:

According to Lamarck, the change in characteristics, which an organism acquires in the course of its adaptation to a changing environment, is inherited by its offspring.

These characteristics are also carried to successive generations. This proposition is known as the ‘inheritance of acquired characters.

Question 4. What is meant by germplasm theory?

Answer:

Germplasm theory:

According to Weismann, the inheritable features are never carried by the somatic cells from – parents to offspring. Only the germ cells carry these characteristics from one generation to the next. This concept is known as the germplasm theory.

Question 5. What is Neo-Lamarckism?

Answer:

Neo-Lamarckism:

A number of modern scientists, viz., Waddington, Packard, Spencer and others modified the original theory of Lamarck under the light of the interaction between life and the environment. This modified form is known as Neo-Lamarckism.

Question 6. Why was Charles Darwin famous?

Answer:

Charles Darwin:

In 1859, Charles Darwin published his concept on organic evolution in his book, titled ‘On the Origin of Species by Means of Natural Selection. In his book, he explained the role of natural selection on the origin of new species from an older one.

This theory of natural selection made Darwin famous in the field of biology.

Question 7. What is meant by a struggle for existence?

Answer:

A struggle for existence:

To survive on the earth, every individual organism has to face competition for food, shelter and mating partners. In this context, it comes across either with the members of the same species or that of different species.

Apart from this, it has to fight against different natural adversities, like floods, droughts, volcanic eruptions, forest fires, etc. Charles Darwin described all these conflicts as the struggle for existence.

Question 8. How many types of the struggle for existence did Darwin suggest?

Answer:

Darwin suggested three different types of struggle for existence among the organisms on Earth.

These are-

1. Intra-specific struggle (competition among the members of the same species),
2. Inter-specific struggle (competition among the members of different species) and
3. Environmental struggle or the struggle against nature (struggle against different natural calamities).

Question 9. What is an intra-specific struggle?

Answer:

Intra-specific struggle:

Members of the same species often get involved in conflict with each other for food, shelter and mating partners. Darwin explained this type of conflict as an intra-specific struggle.

Example: Dogs often fight among themselves for food and mating partners.

Molecular Evidence for Evolutionary Theories

Question 10. What is an inter-specific struggle?

Answer:

Inter-specific struggle:

Members of different species often fight with each other for food and shelter. Darwin explained this type of conflict as an inter-specific struggle.

Example: Dogs fighting with cats for the same food, sparrows fighting with common myna for food and shelter.

Question 11. Tigers are extinct but there are many deer in the jungle. A few tigers are brought from another sanctuary and released in such a jungle. Imagine and write the names of different kind of struggles for existence those tigers have to be involved in.

Answer:

The tigers, brought from another sanctuary, have to face mainly two types of struggle for existence.

These are-

Inter-specific struggle:

They have to get involved in an inter-specific struggle with deer for food.

The struggle against nature:

Firstly, the tigers will face difficulty in adjusting with the new environment and secondly, as the number of deer is high in that jungle, the jungle will be devoid of bushes which generally provide hiding space to tigers during hunting sessions.

Question 12. A good number of Tilapia fishes are released in a pond having only different indigenous fish species grown naturally. Think and write which types of a struggle for existence Tilapia fishes have to face in order to survive.

Answer:

The Tilapia fishes have to face three types of struggle for existence in the pond where they are released.

These are

Intra-specific struggle:

It is the struggle among all the newly released tilapia fishes for food and habitat

Inter-specific struggle:

It is the struggle between Tilapia and other naturally grown indigenous fishes for food and habitat.

Struggle with the environment:

Life Science Class 10 Wbbse

The Tilapia fishes struggle with the condition of the pond where they are newly released. Water content, Temperature of water, amount of dissolved oxygen, poisonous substances present in that pond water, etc. are the obstacles that they struggle against, to overcome.

Question 13. What is the variation?

Answer:

Variation:

The distinctive feature that separates one individual of a species from the other members of the same species is called variation. The difference in the colour of the human eye is an example of variation.

Artificial Selection as Evidence of Evolution

Question 14. What are favourable and unfavourable variations?

Answer:

Favourable variations:

According to Darwin, the variations, which help an organism to stay alive in the struggle for existence, are known as favourable variations.

Unfavourable variations:

According to Darwin, certain variations do not help an organism to survive in its struggle for existence but rather push them towards extinction. These variations are called unfavourable variations.

Question 15. How did Darwin apply the evidence of comparative embryology in his theory of evolution?

Answer:

Charles Darwin applied the concept of Von Baer’s comparative embryology in his evolutionary thought. According to him, embryo-logical similarity indicates the same ancestral lineage.

He also opined that the special features, which develop in the embryonic state, help an animal to adapt in its environment.

Question 16. What is mutation? Who is the proponent of the mutation theory?

Answer:

1. Mutation:

Any sudden, stable and inheritable change in a chromosome or gene that may or may not have a direct impact on the phenotype of an individual, is called a mutation.

2. Proponent: Hugo de Vries is the proponent of the mutation theory.

Question 17. What is the synthetic theory of evolution?

Answer:

Synthetic theory of evolution:

Modern scientists have found certain drawbacks of the Darwinian theory of evolution. Without completely discarding it, they added several new concepts to the basic natural selection theory under the light of modern scientific facts.

The compilation of all these concepts has developed a new theory of evolution, popularly known as the synthetic theory of evolution.

Question 18. Mention the he four main foundations of the synthetic theory of evolution.

Answer:

The synthetic theory of evolution is built on four main foundations.

These are-

1. Gene mutation,
2. Chromosomal aberration,
3. Recombination of genes and
4. Natural selection.

Question 19. What is Neo-Darwinism?

Answer:

Neo-Darwinism:

Modern scientists, viz. Morgan, Haldane, Dobjhansky and others modernised the original Darwinian theory of evolution by incorporating the concepts of variation, isolation, mutation and genetics into it.

This renewed and updated Darwinian theory of evolution is known as Neo-Darwinism.

Life Science Class 10 Wbbse

Chapter 4 Theories Of Organic Evolution Very Short Answer Type Questions Multiple Choice Questions And Answers [Mcq]

Question 1. Which of the following can be explained by the theory of natural selection?

1. Over specialisation
2. Discrete variation
3. Survival of the fittest
4. Vestigial organ

Answer: 3. Survival of the fittest

Question 2. The author of the famous book, ‘Philosophic Zoologique’ was—

1. Darwin
2. Weismann
3. Lamarck
4. Hugo de Vries

Answer: 3. Lamarck

Question 3. The author of the famous book, ‘On the Origin of Species by Means of Natural Selection was—

1. Lamarck
2. Hugo de Vries
3. Darwin
4. Spencer

Answer: 3. Darwin

Question 4. Who first used the statement ‘struggle for existence’?

1. Darwin
2. Lamarck
3. Mendel
4. Weismann

Answer: 1. Darwin

Question 5. Who propounded the theory of natural selection?

1. Lamarck
2. Darwin
3. Mendel
4. Weismann

Answer: 2. Darwin

Question 6. Who proposed the concept that organisms are evolved from mutation? Or, Who proposed the theory of mutation?

1. Osborn
2. Cuvier
3. Weismann
4. Hugo de Vries

Answer: 4. Hugo de Vries

Question 7. Determine the term related to the theory of Lamarck.

1. Struggle for existence
2. Origin of variation
3. Inheritance of acquired characters
4. Natural selection

Answer: 3. Inheritance of acquired characters

Question 8. The Lamarckian concept of inheritance of acquired characters was opposed by—

1. Osborn
2. Cuvier
3. Darwin
4. Weismann

Answer: 4. Weismann

Question 9. Who proposed the germplasm concept?

1. Goldsmith
2. Spencer
3. Hugo de Vries
4. Weismann

Answer: 4. Weismann

Question 10. Which one of the following is the intraspecific struggle for the same food? .

1. The struggle between vulture and hyaena
2. The struggle between the eagle and the kite
3. The struggle among the members of rohu fish in the pond
4. Struggle between Egret and Kingfisher

Answer: 3. Struggle among the members of rohu fish in the pond

Life Science Class 10 Wbbse

Question 11. Identify which of the following indicates interspecific struggle—

1. A struggle among the members of guppy fish for feeding on mosquito larvae
2. The struggle between snakes and owl
3. Struggle within a group of deer for feeding on grass at the same location
4. The struggle between tigers in a jungle for preying on deer.

Answer: 2. Struggle between snakes and owl

Question 12. Parthenium is an exotic species in our country. Other indigenous species cannot survive in such places where in grows. This establishes one of the postulates of Darwin’s theory. Identify the postulate—

1. Intraspecific struggle
2. Interspecific struggle
3. Struggle with environment
4. Origin of new species

Answer: 2. Interspecific struggle

Chapter 4 Theories Of Organic Evolution Answer In A Single Word Or Sentence

Question 1. Who is known as the father of evolution?
Answer: Charles Robert Darwin

Question 2. Who proposed the mutation theory?
Answer: Hugo de Vries

Question 3. What is the title of the famous book, authored by Darwin, featuring organic evolution?
Answer: On the Origin of Species by Means of Natural Selection

Question 4. On which ship, did Darwin set off for his tour around the world?
Answer: Darwin set off for a five years long tour (1831-1836) around the world on the ship named HMS (Her Majesty’s Ship) Beagle.

Question 5. Name the island, where Darwin found different types of finches and various other special birds.
Answer: Galapagos island

Question 6. Who proposed the theory of use and disuse?
Answer: Lamarck

Question 7. Which theory did Weismann propose in the year 1892?
Answer: The germplasm theory of evolution

Question 8. What is the name of the struggle, faced by different species, in search of food and shelter?
Answer: Inter-specific struggle

Life Science Class 10 Wbbse

Question 9. According to Darwin, what was the nature of the growth rate of a natural population?
Answer: According to Darwin, any natural population grows at a geometric rate.

Question 10. Name a scientist who advocated Neo- Lamarckism.
Answer: Herbert Spencer

Question 11. What are the main criteria of survival of the fittest?
Answer: Suitable variation and natural section

Question 12. Write down one limitation of Darwinism.
Answer: Darwin can not explain the reason of the formation of variation.

Common Misconceptions About Evolutionary Theories

Chapter 4 Theories Of Organic Evolution Fill In The Blanks

Question 1. For food, shelter and mating partners, members of the same species involve in _______ struggle.
Answer: Intra-specific

Question 2. Members of different species involved in _______ struggle for food and shelter.
Answer: Inter-specific

Question 3. Darwin described _______ as the ‘game of nature’.
Answer: Mutation

Question 4. Darwin had put special emphasis on small and unstable _______in the origin of a new species.
Answer: Variation

Question 5. According to _______, any organism in its lifetime is inherited by its successive generations.
Answer: Lamarck

Question 6. _______ opined that, the use and disuse of any organ results in to the physical change of the organ.
Answer: Lamarck

Question 7. _______ was the first to use the term ‘evolution.
Answer: Spencer

Question 8. It is now proven that the inheritance of acquired characters occurs only by_______ cells.
Answer: Germ

Question 9. Darwin observed the structure of the beak of _______ bird.
Answer: Finch

Question 10. Darwin termed the difference between two organisms as_______.
Answer: Variation

Chapter 4 Theories Of Organic Evolution State True Or False

Question 1. Darwin authored the famous book, ‘Origin of Life’.
Answer: False

Question 2. Lamarck proposed the concept of survival of the fittest.
Answer: False

Question 3. According to Lamarck, natural selection is the main driving force for organic evolution.
Answer: False

Question 4. According to Lamarck, the neck of the giraffe is long due to the use of organs.
Answer: True

Life Science Class 10 Wbbse

Question 5. According to Darwin, organisms reproduce at a geometric ratio.
Answer: True

Question 6. The concept of survival of the fittest is proposed by Lamarck.
Answer: False

Question 7. Darwin gives importance to mutation in his theory.
Answer: False

Question 8. Weismann opposed the theory of Lamarck.
Answer: True

Question 9. The fundamental tenet of the evolutionary theory of Darwin is the process of natural selection.
Answer: True

Chapter 4 Theories Of Organic Evolution Match The Columns

Question 1.

Answer: 1. C, 2. F, 3. E, 4. B, 5. A, 6. D

Question 2.

Answer: 1. E, 2. D, 3. A, 4. F, 5. C, 6. B

Chapter 4 Theories Of Organic Evolution Find The Odd One Out

Question 1. Law of use and disuse, Law of independent assortment, Theory of natural selection, Law of inheritance of acquired characters
Answer: Law of independent assortment

Question 2. Weismann, Hugo de Vries, Goldsmith, Darwin
Answer: Darwin

Question 3. Herbert Spencer, Charles Darwin, Gregor Johan Mendel, Jean Baptist L’amarck
Answer: Gregor Johan Mendel

Life Science Class 10 Wbbse

Question 4. Lamarkism, Darwinism, Haldane-Oparin hypothesis, Mendelism
Answer: Mendelism

Question 5. Use and disuse of organs, Inheritance of acquired character, Formation of new organs within organisms out of a need
Answer: Natural Selection

Question 6. Natural selection, Struggle for existence, Variation, Use and disuse of an organ
Answer: Use and disuse of an organ

Chapter 4 Theories Of Organic Evolution Fill In The Blanks By Looking At The First Pair

Question 1. Inheritance of acquired characters: Theory of Lamarck:: Survival of the fittest:________
Answer: The theory of Darwin

Question 2. On the Origin of Species by Means of Natural Selection: Charles Darwin:: Philosophie Zoologique: ________
Answer: Jean Baptist Lamarck

Question 3. Lamarckism: Inheritance of acquired characters:: Darwinism: ________
Answer: Natural selection

Life Science Class 10 Wbbse

Question 4. Neo-Lamarckism: Herbert Spencer :: Neo- Darwinism:________
Answer: Hugo de Vries

Chapter 4 Theories Of Organic Evolution Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Fighting among members of the same species, Prey-predator interaction, Struggle for existence, Trying to survive against natural calamities
Answer: Struggle for existence

Question 2. Lamarckism, Use and disuse of organs, Inheritance of acquired characters, Formation of organs within organisms from the need
Answer: Lamarckism

Question 3. Drought, Heavy rain, Competition for food, Struggle for Existence
Answer: Struggle for existence

Question 4. HMS Beagle, Natural selection, Darwin, On the Origin of Species by Means of Natural selection
Answer: Darwin

Question 5. Darwinism, Lamarckism, Mutation theory, Evolution theory
Answer: Evolution theory

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Summary

WBBSE Class 10 Evolution Overview

• Evolution is the change in the heritable characteristics of the organisms over successive generations. Complex organisms are produced from simple organisms through evolution.
• Various concepts and theories are proposed regarding the origin of life, of which the theory of Oparin and Haldane is mostly accepted. Earth formed around 4.6 billion years ago.
• At that time, the temperature of earth was 5000-6000°C and the environment was reducing in nature. Gradually, organic compounds developed from inorganic compounds via a process called chemogeny.
• According to Oparin and Haldane, life originated in the ancient seas, that contained hot water and a large amount of organic monomers and polymers.
• Scientist Haldane named it ‘hot dilute soup! Oparin and Haldane propounded the idea of coacervate, which was described as a colloidal, membrane-bound, organic matter containing particles, formed within a hot dilute soup and acting as a precursor of life.
• Sidney Fox proposed the theory of microsphere and according to him, life originated from microsphere. The first cell or protocell originated about 3.7 billion years ago in the hot, ancient seas.
• In 1953, Miller and Urey conducted an experiment where, in the presence of water vapour, they combined methane, ammonia and molecular hydrogen at the ratio of 2:2:1 and pulsed that in a controlled environment with an electrical discharge.
• The experiment was a success as simple organic molecules, including amino acids, had formed in that set-up, thus supporting the hypotheses of Oparin and Haldane.
• The main events of the evolution of life are-Origin of earth-Origin of life-Origin of unicellular organisms-Origin of photosynthetic bacteria-Origin of multicellular organisms-Origin of aquatic vertebrates-Evolution of plants on land-Origin of four-legged terrestrial vertebrates.

Wbbse Class 10 Life Science Solutions

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Long Answer Type Questions

Question 1. Show the major evolutionary events with the help of a chart.
Or,
With the help of arrow signs, show the major evolutionary events as occurred gradually after the origin of life.

Answer:

The major evolutionary events:

Darwin’s Theory of Natural Selection Explained

Question 2. Describe the concept of chemosynthesis of life.

Answer:

Chemosynthesis of life:

Wbbse Class 10 Life Science Solutions

The chemosynthesis of life occurred in four phases. These are described below.

1. Phase 1-Origin of simple organic compounds:

In the primitive ocean, several inorganic substances reacted with each other to synthesize simple organic compounds, like acetylene, methane, ethylene, etc. In these reactions ultraviolet rays, lightning, extreme heat, etc.

supplied the requisite energy. Later, these organic substances reacted with water vapour to synthesize some hydrocarbon compounds, such as acetaldehyde.

2. Phase 2-Synthesis of complex organic compounds:

The interaction of these simple organic compounds later gave rise to complex organic matters, like carbohydrates (glucose, fructose, etc.), fatty acids and glycerol, amino acids (alanine glycine, etc) etc.

These complex compounds then reacted with each other to synthesize more complex and larger organic molecules, like simple proteins, nucleotides and lipids.

Finally, nucleic acids and complex proteins were produced in the hot water of the primitive ocean.

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3. Phase 3-Formation of coacervates:

In the course of time, the organic macromolecules interacted to produce numerous spherical, double membrane-bound, divisible, colloidal microparticles, which were denoted as coacervates by Oparin.

According to Oparin and Haldane, these microparticles were composed of several components like proteins, nucleic acids, carbohydrates, etc. which were important for the origin of life on Earth.

Sidney Fox opined that… several protein molecules united in the primitive ocean to produce tiny proteinaceous particles called microspheres. According to him, life originated on Earth from microspheres.

4. Origin of protocells:

Scientists assumed that the tiny coacervates engulfed nucleic acids and proteins from the environment, enlarged in size and finally took the shape of protocells. These protocells are thought to be the earliest forms of life on Earth.

RNA was the genetic material of these protocells. These were heterotrophic in nature. In the course of organic evolution, chemosynthetic and photosynthetic forms of life emerged on Earth.

With the emergence of photosynthetic organisms, free oxygen was liberated into the primitive atmosphere.

Wbbse Class 10 Life Science Solutions

Question 3. What is hot dilute soup? Give a brief description of the ‘biogeny’ of the ‘evolution of life’ on earth.

Answer:

Hot dilute soup:

Before the origin of life on earth, several simple organic compounds reacted with each other to synthesize complex organic matter, like amino acids, nucleotides, lipids, carbohydrates, etc.

All these compounds remained suspended in hot water of the primitive ocean. Scientist Haldane described this mixture as ‘hot dilute soup. Sometimes it is also called ‘prebiotic soup. Scientists assumed that life on earth originated in this ‘hot dilute soup’.

Biogeny or evolution of life on Earth:

About 3.7 billion years back, life originated on Earth in the form of protocells. The successive events during the origin of life on Earth are mentioned below.

1. Origin of protocell:

By the process of chemosynthesis, coacervates were formed in the primitive ocean. Later, nucleic acids and protein molecules fused with coacervates to form protocells.

2. Origin of autotrophs:

The primitive organisms, that evolved from the protocells were chemo-heterotrophs. Later came the chemoautotrophs and then emerged the photosynthetic prokaryotic life forms.

With the emergence of photosynthetic organisms, free oxygen was released into the primitive atmosphere. Thus, the character of the atmosphere changed from reducing to oxidising one.

3. Origin of eukaryotes:

About 1.5 billion years ago, unicellular, aerobic eukaryotes evolved on Earth. In the course of evolution, multicellular complex organisms came into being from those unicellular forms.

Types of Evolutionary Adaptations

Question 4. Explain the important events of organic evolution.

Answer:

Important events of organic evolution:

The whole event, starting from the origin of life on Earth to the ongoing organic evolution, it has always been a very complex process.

By assembling the concepts and theories of different scientists, we have developed some ideas on the origin of life on Earth and the trend of organic evolution, which are mentioned below.

1. Origin of Earth:

Earth is the third planet in the solar system. It has originated from the solar nebula about 4.5 to 5.0 billion years back.

2. Origin of Life:

About 3.5 to 3.7 billion years ago, life originated on Earth. Two scientists, Haldane and Oparin, have put forward a clear idea on the origin of life in the water of primitive oceans by successive reactions among several inorganic and organic chemicals.

3. Origin of unicellular organisms:

The first living forms on earth were the primitive unicellular forms, which were the primitive heterotrophic prokaryotes. From the prokaryotes, emerged the unicellular eukaryotes.

4. Origin of photosynthetic prokaryotes:

From the primitive heterotrophic prokaryotes, there evolved some photosynthetic prokaryotes. With the emergence of photosynthetic forms, the production of free oxygen started and the primitive reducing atmosphere transformed into an oxidising one.

5. Emergence of multicellular organisms:

In the course of evolution, multicellular organisms emerged on Earth. These were more complex and advanced than their ancestors.

6. Emergence of different invertebrates and fish-like vertebrates:

With gradual advancement, different invertebrates came into being. After the emergence of higher invertebrates, there came some fish-like vertebrates.

7. Emergence of terrestrial plants:

Primitive plants originated in water. In the course of evolution, terrestrial plants emerged. These newly arrived terrestrial plants got themselves adapted to this new environment slowly and thereafter several advanced terrestrial plants emerged on Earth.

8. Emergence of terrestrial animals:

While living in an aquatic environment, some fish-like forms have developed some organs suitable for terrestrial living. Gradually some of these forms, for instance; amphibians, rejected the aquatic life and started dwelling on the land.

Later, different advanced terrestrial forms, like mammals and birds, emerged on Earth.

Wbbse Class 10 Life Science Solutions

Question 5. Describe the brief outline of the chemical origin of life mentioning Miller and Urey’s experiment. Similar question, Briefly describes the experiment conducted by Miller and Urey on the origin of life.

Answer:

The experiment of Miller and Urey:

Based on the concept of chemosynthesis of life, proposed by Oparin and Haldane, Stanley Miller and Harold Urey performed an experiment in 1953 in the laboratory of Chicago University.

1. Procedure:

They prepared a highly heat-resistant instrument with two inter-connected chambers of unequal diameters. The larger chamber was fitted with two electrodes and was filled with a mixture of hydrogen, methane and ammonia gases.

The outlet pipe of the larger chamber was fitted with a cooling device and a tap which was used to tap down the contents from it. The smaller chamber was used as a boiler and a connecting pipe was fitted with it to supply vapour to the larger chamber.

Powerful electric sparks were generated in the larger chamber to mimic the lightning-like situation (that prevailed in the primitive atmosphere). The water of the smaller chamber was boiled by heating it externally, thus the newly formed water vapour is pushed into the larger chamber.

The machine was run continuously for one week and the product from the larger chamber was tapped down after condensation.

 

Life Science Class 10 Wbbse

2. Observation:

After analysing the product, scientists found a number of amino acids, viz. glycine, alanine, glutamic acid, aspartic acid, etc. in it.

3. Conclusion:

From the result, Miller and Urey concluded that in the primitive, free oxygen-lacking, reducing atmosphere, different inorganic matter reacted with each other to synthesize organic substances.

In this reaction, heat, lightning, ultraviolet ray and other cosmic rays provided the energy that was needed. The organic matter, thus synthesized, acted as the building blocks of cells and finally helped in the origin of life.

Significance of Miller and Urey’s experiment:

Miller and Urey’s experiment supports the abiogenetic theory of the origin of life. According to this theory, life was originated from abiotic or inorganic substances around 3.7 billion years ago.

Evidence of Evolution in Fossils

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Short Answer Type Questions

Question 1. Define life.

Answer:

Life:

Life is the condition that differentiates a living organism from a non-living object. Life shows the signs of growth, reproduction, metabolic activities, excitability, adaptability, ageing and senescence.

Life Science Class 10 Wbbse

Question 2. What do you mean by organic evolution? Or, What is meant by evolution?

Answer:

Organic evolution:

Organic evolution is the process of slow and steady changes in the characteristics of organisms. These changes result in to the origin of complex and higher descendent varieties or species from simpler and primitive ancestral types. Organic evolution is the cause of biodiversity on the earth.

Question 3. What do you mean by the cosmogenic origin of life?

Answer:

The cosmogenic origin of life:

Swiss author, Erik Von Daniken, in his number of books, claimed that life was introduced to Earth from other planets of other galaxies. This pseudo-scientific concept is known as the theory of the cosmogenic origin of life.

 

 

Question 4. What is meant by the cyanogen theory of the origin of life on Earth?

Answer:

The cyanogen theory of the origin of life on Earth:

According to German scientist Pflüger, when the earth was cooling down, carbon and nitrogen combined to form a complex compound, named cyanogen.

It is a protein-like compound, which in the course of time produced protoplasm. This concept is known as the cyanogen theory of the origin of life.

Life Science Class 10 Wbbse

Question 5. What is meant by the spontaneous creation of life or abiogenesis?

Answer:

The spontaneous creation of life or abiogenesis:

The Greek philosopher Aristotle and famous Belgian chemist, Van Helmont believed that life has originated on earth spontaneously from non-living matters.

This concept on the origin of life is known as the spontaneous creation of life or abiogenesis or biopoiesis.

Artificial Selection and Its Role in Evolution

Question 6. What is Chemogeny?

Answer:

Chemogeny:

Chemogeny is the origin of life, on earth through the ongoing synthesis of complex compounds by successive chemical reactions of different elements and simpler compounds.

Question 7. How did free oxygen evolve in Earth’s atmosphere?

Answer:

Free oxygen evolve in Earth’s atmosphere:

In primitive Earth, the pioneer life forms were chemoheterotrophs. In the course of evolution, these organisms were gradually transformed into photoautotrophs. These aerobic photosynthetic organisms released free oxygen in the primitive atmosphere by performing photosynthesis.

Question 8. What are coacervates?

Answer:

Coacervates:

According to Alexander Oparin, in the organic compound-rich hot water of the primitive ocean, many spherical colloidal droplets were formed.

These tiny droplets were composed of several amino acids, simple carbohydrates and lipid molecules. He denoted these colloidal droplets as coacervates.

 

Molecular Evidence for Evolution

Question 9. Write down the characteristics of coacervate.

Answer:

The characteristics of coacervate are

1. It is a large organic compound, composed of carbohydrates, amino acids, proteins, etc.
2. Coacervate was covered by a lipid membrane.
3. Coacervate was able to absorb nutrients from the surroundings.

Question 10. What are microspheres?

Answer:

Microspheres:

According to Sidney Fox, in the primitive ocean, many amino acid molecules united to form a double-layered lipid membrane-bound, tiny, divisible, spherical droplets.

These are called microspheres, which are assumed to be the precursor of primitive cells. Sidney Fox opined that life has originated from microspheres, not from coacervates.

Life Science Class 10 Wbbse

Question 11. Write down the characteristics of the microsphere.

Answer:

The characteristics of a microsphere are-

1. Microsphere originates from the polypeptide.
2. Microsphere is more durable than coacervate.
3. Microsphere is covered by a double membrane.
4. It is capable of bud formation and can use ATP as a source of energy.
5. It has self propagative properties.

Question 12. Distinguish between coacervate and microsphere.

Answer:

The distinguishing features between coacervate and microsphere are-

 

Question 13. What are protobionts?

Answer:

Protobionts:

The earliest form of life originated on earth in the primitive ocean in form of a double-layered lipid membrane-bound, organic molecule-containing structure. These primitive rudimentary cell-like structures are called protobionts.

Scientists believe that prokaryotic cells have emerged from these protobionts. Example-Microspheres.

Question 14. What is the RNA world hypothesis?

Answer:

RNA world hypothesis:

A line of thought states that the first genetic material present in the protocell was RNA. Later, in the course of evolution, DNA molecules were formed from RNA by biochemical transformation. This concept is known as the RNA world hypothesis.

Question 15. What are naked genes?

Answer:

Naked genes:

In the second phase of Chermogenesis, a number of free nucleotides joined together to form polynucleotide or nucleic acid chains. These are called naked genes. These naked genes are nothing but the RNA molecules.

Question 16. What are proteinoids?

Answer:

Proteinoids:

At high temperatures and in the presence of ultraviolet rays, many protein molecules denature and form amino acids. With a gradual decrease in temperature, these amino acid molecules polymerise to form tiny protein-like globules, called proteinoids.

Scientists believe that the proteinoids played some role in the origin of primordial cells.

Life Science Class 10 Wbbse

Question 17. What are the characteristics of protocells?

Answer:

The characteristics of protocells are-

1. Protocell contains nucleic acid.
2. Protocell is covered by a bilayered membrane.
3. Protocell has the ability to produce replicas by division.

Question 18. Write the name of the reactants used and one organic compound formed in the experiment of Miller and Urey in connection with the chemical origin of life.

Answer:

In the experiment of Miller and Urey in connection with the chemical origin of life, the reactants used were methane (CH₄), ammonia (NH₃), hydrogen (H₂) and water vapour (H₂O) experiment was amino acids, viz. glycine, alanine, and the organic compound formed in the glutamic acid, aspartic acid, etc.

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Very Short Answer Type Questions Multiple Choice Questions And Answers [Mcq]

Question 1. The theory of the origin of life was first proposed by —

1. Pasteur
2. Oparin
3. Darwin
4. Mendel

Answer: 2. Oparin

Question 2. Scientist Stanley Miller was able to synthesize simple amino acids from a mixture of—

1. H₂, O₂, N₂ and H₂O
2. H₂, NH₃, CH₄ and water vapour (H₂O)
3. NH₃, CH₄, HCN
4. N₂, NH₃, HCN and 0₂

Answer: 2. H₂, NH₃, CH₄ and water vapour (H₂O)

Question 3. Miller and Urey, in their experiment, were able to synthesize some preliminary constituents necessary for the creation of life. Identify the ones which were amino acids among them—

1. Lactic acid, Acetic acid
2. Urea, Adenine
3. Glycine, Alanine
4. Formic acid, Acetic acid

Answer: 3. Glycine, Alanine

Question 4. The most important organic matter for the origin of life is—

1. Nucleic acid
2. Nucleoside
3. Amino acid
4. Protein

Answer: 1. Nucleic acid

Question 5. The gas which was absent in the environment ^ during the origin of life is—

1. Hydrogen
2. Oxygen
3. Methane
4. Ammonia

Answer: 2. Oxygen

Question 6. Who was the first to disapprove the doctrine of spontaneous generation of life?

1. Louis Pasteur
2. Alexander Oparin
3. Aristotle
4. Miller

Answer: 1. Louis Pasteur

Question 7. The cause of biodiversity on earth is—

1. Metabolism
2. Locomotion
3. Evolution
4. Adaptation

Answer: 3. Evolution

Question 8. Where did life originate on Earth?

1. In Ocean
2. On land
3. In space
4. Under soil

Answer: 1. In  Ocean

Question 9. The cellular organisation appeared first in—

1. Nucleoproteins
2. Coacervates
3. Bionts
4. Protobionts

Answer: 2. Coacervates

Question 10. The chemosynthesis of life was clearly demonstrated by—

1. Lamarck and Darwin
2. Miller and Urey
3. Plato and Dante
4. Starling and Bayliss

Answer: 2. Miller and Urey

Question 11. The gradual development of a modern organisms from primitive ones is called—

1. Variation
2. Evolution
3. Adaptation
4. Behaviour

Answer: 2. Evolution

Question 12. The study of development of an organism in its life-span is called—

1. Organic evolution
2. Ontogeny
3. Mutation
4. Special creation

Answer: 2. Ontogeny

Question 13. Which of the following means—changes in the organisms for a prolonged time?

1. Adaptation
2. Accommodation
3. Evolution
4. Reproduction

Answer: 3. Evolution

Question 14. The term ‘evolution’ was coined by—

1. Lamarck
2. Darwin
3. Hugo de Vries
4. Herbert Spencer

Answer: 4. Herbert Spencer

Question 15. The unit of studying evolution is—

1. DNA
2. RNA
3. Protein
4. Species

Answer: 4. Species

Question 16. Life originated on Earth about

1. 3.7 billion years ago
2. 2.7 billion years ago
3. 1.1 billion years ago
4. 5.5 billion years ago

Answer: 1. 3.7 billion years ago

Question 17. According to the opinion of scientists, the prebiotic environment of the earth that led to the origin of life was like this—

1. Hot dilute soup is the seawater
2. Cold dilute soup in the seawater
3. Cold dilute soup in the river water
4. Hot dilute soup in the groundwater

Answer: 1. Hot dilute soup is the seawater

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Answer In A Single Word Or Sentence

Question 1. What is the approximate age of the earth?
Answer: About 4.6 billion years

Question 2. How many years ago did life originate on the earth?
Answer: About 3.7 billion years ago

Question 3. Which was the first nucleic acid on Earth?
Answer: RNA (Ribonucleic acid)

Question 4. In which form was carbon present in the primitive earth?
Answer: In the primitive earth, carbon was present in reduced form as hydrocarbons.

Question 5. Name a few hydrocarbons that were present in the primitive earth.
Answer: Methane, acetylene, ethylene, etc.

Question 6. Which concept states that life has originated from non-living substances?
Answer: The concept of abiogenesis

Evolutionary Mechanisms: Natural Selection vs Genetic Drift

Question 7. What will you call a nucleic acid-carrying coacervate?
Or,
What is formed when coacervate unites with nucleic acid?
Answer: Protocell

Question 8. Which end product did Miller and Urey get from their experiment? Similar question, Write down the names of two amino acids formed in the experiment of Miller and Urey.
Answer: Amino acids, viz. alanine, glycine, etc.

Question 9. Who is known as the father of evolution?
Answer: Charles Darwin

Question 10. Write two reasons of evolution.
Answer:

Two reasons of evolution

Mutation and adaptation

Question 11. What is biogeny?
Answer:

Biogeny

The chain of events, responsible for the origin of life and its evolution, is called biogeny.

Question 12. Who first introduced the concept of coacervate?
Answer: Alexander Oparin and JBS Haldane

Question 13. Who proposed the concept of a microsphere?
Answer: Sidney Fox

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Fill In The Blanks

Question 1. The process of emergence of complex organisms from simple ones by slow, continuous and irreversible changes is called ________ evolution.
Answer: Organic

Question 2. According to Greek philosopher ________, living forms have emerged on earth by the transformation of inanimate objects.
Answer: Aristotle

Question 3. The term ‘hot dilute soup’ was coined by ________.
Answer: Haldane

Question 4. The chemical basis of the origin of life is called ________.
Answer: Chemogeny

Question 5. The spherical aggregates of microdroplets of lipids that originated in the ‘hot dilute soup’ are known as ________.
Answer: Coacervates

Question 6. ________ was the aggregate of some large colloidal molecules during the early phase of the origin of life
Answer: Microsphere

Question 7. The first nucleic acid, produced in the primitive ocean was ________.
Answer: RNA

Question 8. Miller and ________ first proved the theory of chemogeny of life in a laboratory.
Answer: Urey

Question 9. The most acceptable theory on organic evolution was proposed by________.
Answer: Darwin

Question 10. The book titled ‘Origin of Life’ was authored by ________.
Answer: Oparin

Question 11. According to Sidney Fox, an organism originates from ________.
Answer: Microsphere

Question 12. The alternative name of hot dilute soup is ________.
Answer: Prebiotic

Question 13. The organic compound formed in the experiment of Miller and Urey was ________.
Answer: Amino acid

Question 14. On earth, life was originated in ________.
Answer: Primitive ocean

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life State True Or False

Question 1. Protocells originated by the association of nucleic acid with coacervates.
Answer: True

Question 2. The materials necessary for the origin of life was present on the landmass of the primitive earth.
Answer: False

Question 3. Miller and Urey experimentally proved the chemosynthetic origin of life on Earth.
Answer: True

Question 4. Darwin coined the term ‘evolution.
Answer: False

Stages of Human Evolution

Question 5. Miller proposed the theory of coacervate.
Answer: False

Question 6. The first formed life was photosynthetic.
Answer: False

Question 7. According to some scientists, the dust particles of the universe condensed to from earth.
Answer: True

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Match The Columns

Question 1.

Answer: 1. D, 2. F, 3. A, 4. E, 5. C, 6. B

Question 2.

Answer: 1. C, 2. D, 3. B, 4. A, 5. F, 6. G

Common Misconceptions About Evolution

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Find The Odd One Out

Question 1. Amino acid, Fatty acid, Nucleic acid, Glucose
Answer: Nucleic acid

Question 2. DNA, RNA, Protein, Fatty acid
Answer: Fatty acid

Question 3. Acetylene, Methane, Ammonia, Ethylene
Answer: Ammonia

Question 4. Alexander Oparin, JBS Haldane, Jean Baptist Lamarck, Sidney Fox
Answer: Jean Baptist Lamarck

Question 5. CH₄, NH₃, H₂, H₂S
Answer: H₂S

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Fill In The Blanks By Looking At The First Pair

Question 1. Coacervate : Alexander Oparin Microsphere:_________
Answer: Sidney Fox

Question 2. Oparin: Coacervate :: Haldane: _________
Answer: Hot dilute soup

Question 3. Protocell formation: Biogeny:: Formation of the compound from inorganic-organic compound: _________
Answer: Chemogeny

 

Chapter 4 Concept Of Evolution Major Evolutionary Events Origin Of Life Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Abiogenesis, Coacervate, Microsphere, Hot dilute soup
Answer: Abiogenesis

Question 2. Double membrane covering, Ability to use ATP, Microsphere, Protenoid derivative
Answer: Microsphere

Question 3. Carbohydrate formation, Fat formation, Formation of an organic compound, Amino acid formation
Answer: Formation of organic compound

Question 4. Stages of development of life, Origin of unicellular. organisms, Origin of- photosynthetic bacteria, Evolution of plants on land
Answer: Stages of development of life

Chapter 2 Growth And Development Summary

WBBSE Class 10 Growth and Development Overview

• Growth is an important and distinct characteristic of life. Growth is an anabolic process through which the size, volume and dry weight of organisms is increased. Development is the increasing complexity in organisms.
• There are three phases of growth, viz-

1. A phase of cell division,
2. A phase of cell enlargement and
3. A phase of cell differentiation.

• In the phase of cell division, the number of cells increases. The volume of protoplasm increases during the phase of cell enlargement, whereas, cells metamorphose into the tissue, organ through a phase of cell differentiation.
• The physical, mental and emotional changes those are evident with the course of time is called development in human. In human life, there are five phases of development, viz.-

1. Newborn,
2. Childhood,
3. Adolescence,
4. Matured phase and
5. Old age or late mature phase.

Wbbse Class 10 Life Science

Chapter 2 Growth And Development Long Answer Type Questions

Question 1. What is growth? Explain different types of growth in living organisms.

Answer:

Growth:

Growth is an anabolic process by which the size and dry weight of any living organism increases permanently and irreversibly, by means of cell division.

Read and Learn More WBBSE Solutions for Class 10 Life Science And Environment

Types of growth in living organisms:

In living organisms, three different types of growth can be noticed. These are-

1. Vegetative growth:

By this process, a unicellular zygote divides consecutively to give rise to a fully formed multicellular organism. For example, the events of germination of seeds, growth of saplings, and formation a full-grown plant are treated as vegetative growth.

2. Regenerative growth:

The damaged and injured parts of any living body are repaired or healed by this process of growth.

3. Reproductive growth:

This process of growth helps any organism to become reproductively capable. By this growth process, the reproductive organs of young organisms attain maturity and start producing gametes. In plants, this growth process is seen in flowers, fruits and seeds.

WBBSE Solutions for Class 10 History	WBBSE Solutions for Class 10 Geography and Environment
WBBSE Class 10 History Long Answer Questions	WBBSE Solutions for Class 10 Life Science And Environment
WBBSE Class 10 History Short Answer Questions	WBBSE Solutions for Class 10 Maths
WBBSE Class 10 History Very Short Answer Questions	WBBSE Solutions for Class 10 Physical Science and Environment
WBBSE Class 10 History Multiple Choice Questions

 

Question 2. Name three main phases of growth. Briefly describe those phases.

Answer:

Phases of growth:

The growth of multicellular organisms can be divided into three phases.

These are-

1. The phase of cell division,
2. The phase of cell enlargement and
3. The phase of cell differentiation.

Description of the phases of growth:

Wbbse Class 10 Life Science

The three different phases of growth are described below.

1. Phase of cell division:

In this phase, cells divide mitotically to increase in number in a living body. Cells of apical and lateral meristem divide to make a plant body grow in length and width.

In animals, a unicellular zygote divides repeatedly and undergoes through different stages, like morula, blastula, and gastrula of a multicellular embryo to give rise to a fully grown animal body.

2. Phase of cell enlargement:

In this phase, the newly divided cells absorb water and synthesise different cellular components to increase the protoplasmic volume. By this process, the cells become larger and the tissues formed by these cells grow in volume irreversibly.

In some tissues, secretory materials accumulate in the intercellular spaces to increase the volume of that region, as we can see in bones and cartilage.

 

3. Phase of cell differentiation:

In this phase, the fully grown daughter cells slowly modify and gain certain specialised features. These typical cells then form different tissues and tissue systems to perform specific functions.

Finally, these tissue systems form different organs and organ systems of a living body.

Wbbse Class 10 Life Science

Question 3. Briefly describe the five stages of human development.

Answer:

Phases of human development:

According to WHO human development includes five different stages which are described below.

1. Newborn or neonatal phase:

This phase of development lasts for the first one month after the birth of a baby. During this phase, the baby remains asleep most of the time and shows a very high rate of metabolism.

In this phase, a baby can detect light, the direction of sound, primary colours and above all, can recognise the mother’s face. Body weight may reduce during the first few days but soon it resumes. For newborn babies, laughing and crying are the main signs of communication.

2. Childhood:

This phase of life starts from the second month and continues up to ten years. During this period, physical, mental and emotional development occurs at a faster rate. Motor activities increase and limbs grow quicker than other parts of the body.

Power to memorise, the ability of reading, writing, drawing and problem-solving improve gradually. A child can express fear, anger and excitement but finer emotions are not expressed perfectly.

3. Adolescence:

This is the intermediate phase between childhood and adulthood. The span between the age of 10 to 19 years (according to WHO) is treated as adolescence. The age of adolescence is not fixed for all. It varies with gender, environment, nutrition, geographical location, etc.

During this phase, several hormones come into action to regulate the growth and functioning of sexual organs. An adolescent boy and girl go through significant changes in physical, mental, emotional and social aspects.

From this phase of life, sexual attraction towards the opposite gender emerges. Mature gamete formation starts during this phase. Concentration, thinking skills, analytical ability and intelligence grow very fast in adolescent individuals.

At this age, they dare to take risks. Therefore, adolescence is treated as the ‘period of turmoil.

4. Adulthood or matured stage:

The age between 19 to 60 years is denoted as adult age. This phase is characterised by maturity, self-confidence, practical, purposeful and self-directed attitude. In this period, physical growth reduces quickly and ultimately stops.

At this age, a sense of responsibility towards family and society grows significantly. An adult person gains experiences of life and mostly follows moral values.

Wbbse Class 10 Life Science

5. Old age:

The age beyond 60 years is treated as old age. In this age, cellular activities gradually slow down. Visual and auditory abilities get weakened. Old age is characterised by thinning and greying of hair.

Skin gets heavily wrinkled and muscles become weak. Bones gradually degenerate and joints lose their movability which are the consequences of different bone diseases like osteoporosis and osteoarthritis. People at old age lose memory and become mentally fatigued.

 

 

Question 4. Distinguish between the growth and development processes in human beings.

Answer:

Difference between human growth and development processes

Types of Growth in Living Organisms

Question 5. Mention the changes that occur in adolescent males and females.

Answer:

Changes in adolescent males and females:

Several changes occur in adolescent males and females. These are mentioned below.

In male:

1. The male reproductive organs become fully grown to produce gametes and secrete sex hormones.
2. Development of the endocrine glands present all over the body is completed.
3. Body weight and height increase at a faster rate.
4. Sweat and sebaceous glands get stimulated. Hypersecretion of these glands results in oily facial skin, thereby causing acne.
5. Beards grow on the face, hairs develop on the chest, armpit and pubic region.
6. Muscles and bones become stronger and the voice becomes husky.

 

 

In female:

1. Female reproductive organs become adolescence completely developed and start producing gametes and secreting sex hormones.
2. The beginning of the menstrual cycle is the indication of the onset of puberty.
3. Mammary glands grow in size during this time.
4. Body weight and height increase in females during adolescence.
5. Additional fat below the skin gives lustre to it.
6. Deposition of extra fat in hip and thigh region gives the body a feminine shape.
7. Hyperactivity of sweat and sebaceous glands makes the face oily, due to which acne grows.
8. Hairs grow on armpits and pubic regions.

 

Chapter 2 Growth And Development Short Answer Type Questions

Question 1. What is meant by development?

Answer:

Development:

Development is the collective changes, including growth and other transformations, which occur continuously since the formation of a zygote until the death of any organism. It is a quantitative change in a multicellular organism.

Question 2. ‘Development is impossible without growth’- Explain.

Answer:

‘Development is impossible without growth’:

Growth refers to an increase in the measurable, physical size of the whole or any of definite part of an organism. Whereas, development refers to the quantitative as well as the qualitative changes in the organism as a of whole.

Life of any multicellular organism starts with a unicellular zygote which increases in cell number by cell division and increases complexity at the tissue or organ level by cell differentiation.

The development also can be defined as the increase of complexity. impossible without growth. So we can easily say that ‘development is impossible without growth’.

Wbbse Class 10 Life Science Solutions

Question 3. Write the differences between plant growth and animal growth.

Answer:

The differences between plant growth and animal growth are-

Factors Affecting Growth and Development

Question 4. What is meant by human development?

Answer:

Human development:

Human development is the collective irreversible changes in a physical, mental and emotional state, that any human being encounters from its birth until death.

By this developmental process, a human baby grows into an adult and finally reaches the old age.

Question 5. What is meant by human growth?

Answer:

Human growth:

Human growth is the spontaneous and irreversible change in size, volume, height and weight of a human body with age. By this process, a newborn baby transforms into a fully grown human being.

Question 6. What is meant by positive growth?

Answer:

Positive growth:

When anabolic activities increase in comparison to catabolic activities, the dry weight of the body increases. This change is known as positive growth.

Example: Positive growth is noticed during a grand period of growth in all living beings.

Question 7. What is meant by negative growth?

Answer:

Negative growth:

When catabolic activities increase in comparison to anabolic activities, the dry weight of the body reduces. This change is known as negative growth.

Example: Negative growth is seen during the initiation of seed germination and during senescence.

Wbbse Class 10 Life Science Solutions

Wbbse Class 10 Life Science Solutions

Question 8. Mention different types of growth depending on nature.

Answer:

Depending on nature, growth is of three types.

These are-

1. Vegetative
2. Regenerative growth and
3. Reproductive growth.

Question 9. What is meant by the phase of cell division of ant by the phase of cell division of growth?

Answer:

The phase of cell division of ant by the phase of cell division of growth:

The phase of growth, in which an organism grows physically only by means of mitotic cell division, is known as the phase of cell division of growth.

Example: Growth of an embryo from the zygote, growth of shoot and root apices in plants, etc.

Stages of Plant Growth Explained

Question 10. What is meant by the phase of cell enlargement of growth?

Answer:

The phase of cell enlargement of growth:

The phase of growth, in which the newly divided cells absorb water, and synthesise various proteins, nucleic acids and other organic matters to increase the protoplasmic volume, is known as the phase of cell enlargement of growth. By this process, the overall size of a cell increases.

Question 11. What is meant by the phase of cell differentiation growth?

Answer:

The phase of cell differentiation growth:

After the phase of cell division and cell enlargement, cells undergo structural and functional transformations. In this process, cells develop specialised features and take part in the formation of different tissues and organs. This phase of growth phase of is called of growth cell differentiation.

Question 12. What is meant by continuous growth?

Answer:

Continuous growth:

The type of growth, in which the growing process starts since birth and continues until the attainment of ultimate size and maturity, is treated as continuous growth.

Example-All higher animals including humans – show continuous growth.

Wbbse Class 10 Life Science Solutions

Question 13. What is meant by discontinuous growth?

Answer:

Discontinuous growth:

In certain organisms, the growing process stops intermittently, after every quick period of growth. This type of growth is known as discontinuous growth.

Example: In insects and other arthropods, growth stops before moulting and it starts again after completion of it.

Question 14. How does the vegetative growth of plants occur?

Answer:

Vegetative growth of plants occurs by cell division of primary and secondary meristems. By this process, leaves, branches and roots of plants grow significantly.

Wbbse Class X Life Science

Question 15. What is meant by prenatal phase?

Answer:

Prenatal phase:

The phase of life, starting after the formation of the zygote and ending with the birth of a baby, is known as the prenatal phase. In humans, the span of the prenatal phase is 280 days.

Question 16. What is meant by the neonatal phase?

Answer:

Neonatal phase:

The span of life, starting from the birth of a baby till the completion of one month, is known as the neonatal phase. During this phase, a human baby remains completely dependent upon others.

Question 17. Write down the characteristics of neonates.

Answer:

A few mention-worthy characteristics of neonates are

1. The child can recognise faces, some sounds, and lights.
2. They express themselves only by crying.
3. Their weight increase rapidly.

Question 18. What is meant by childhood?

Answer:

Childhood:

The phase of life, starting from the beginning of the second month until the end of ten years, is known as childhood. During this period, physical, mental and emotional development occurs at a faster rate along with various motor activities.

Question 19. What is meant by adolescence?

Answer:

Adolescence:

The phase of human life, in which secondary sexual characters appear in males and females and the respective reproductive organs start producing gametes and secrete sex hormones, is known as adolescence. This span usually ranges between 10-19 years of age.

Question 20. What are the secondary sexual characters?

Answer:

The secondary sexual characters:

Secondary sexual characteristics are the features that appear during puberty in human beings. These visibe features include maturation of gonads, the emergence of public hair in both males and females, enlarged breasts, the onset of menstruation and widened hips in females and cracked voices, growth of facial hair as well as protrusion of Adam’s apple in males.

Human Growth Stages

Question 21. Write down some characteristics and features of the adolescent phase.

Answer:

Some characteristic features of the adolescent phase are as follows-

1. The rapid growth of the body takes place.
2. Secondary sexual characters develop on the body.
3. Attraction towards the opposite sex develops.
4. The rapid growth of thinking, analytical power and memory takes place.

Question 22. What do you mean by matured phase in human development?

Answer:

Matured phase in human development:

In the case of human development, the phase or period of life from the age of 19 to 60 years is treated as a matured phase. During this phase, physical growth declines and finally stops. Mental maturity, responsibility and decision-making ability reach their peak.

Wbbse Class 10 Life Science Solutions Chapter 2 Question 23.

Question 23. Mention the internal factors responsible for mental and physical changes in an adolescent boy.

Answer:

The internal factors responsible for mental and physical changes in an adolescent boy

In an adolescent boy, different hormones play an important role in changing the mental and physical state. These hormones include GnRH- from the hypothalamus, ICSH from the pituitary and testosterone from the testis.

Question 24. Mention the internal factors responsible for mental and physical changes in an adolescent girl.

Answer:

In an adolescent girl, different hormones play an important role in changing her mental and physical state. These hormones include GnRH from the hypothalamus, FSH and LH from the pituitary and oestrogen and progesterone from the ovary.

Question 25. Mention the signs of old age in a human being.

Answer:

As a person reaches the old age, several signs of old age or senescence appear.

These are as follows

1. Bone and joints become weak.
2. The power of vision and hearing weakens.
3. Loss of memory, fatigue and lack of confidence are manifested.
4. Hairs become thin and grey.

Question 26. What is meant by senescence?

Answer:

Senescence:

Beyond the age of 60 years, certain decelerating changes start to occur in the human body, which continues until death. In the course of time, these gradual changes make a person older. This process of ageing is known as senescence.

Question 27. Mention two changes related to vision and bones at late adulthood or the senescence phase of human development.

Answer:

During adulthood or the senescence phase of human development, some noticeable changes are-

1. Vision: Due to the weakening of eyesight, symptoms of presbyopia can be seen.
2. Bones: Due to the decaying of bones and joints, symptoms of osteoarthritis and osteoporosis can be noticed.

Question 28. What are the measures should be obeyed by people of old age to stay healthy?

Answer:

The measures should be obeyed by people of old age to stay healthy:

To stay healthy people at old age must follow some rules, like

1. Eating adequate but healthy food as the metabolism rate decreases during this age.
2. Easy but regular exercising to stay physically active.
3. Starting a routine to enhance mobility, flexibility balance, to improve sleep and to boost mood and self-confidence.
4. Remaining socially connected with friends and families to maintain physical and mental health.

 

Chapter 2 Growth And Development Very Short Answer Type Questions Multiple Choice Questions And Answers [Mcq]

Question 1. The phase of growth from start to end, when the process of growth occurs at a fast pace, is called—

1. Grand period
2. Lag period
3. Steady period
4. Stagnant period

Answer: 1. Grand period

Question 2. The branch of biological science that deals with the study of senescence and ageing is called—

1. Anthropology
2. Psychology
3. Physiology
4. Gerontology

Answer: 4. Gerontology

Question 3. A permanent increase of dry weight is called—

1. Development
2. Differentiation
3. Division
4. Growth

Answer: 4. Growth

Question 4. The number of phases of growth of organisms is—

1. Four
2. Five
3. Three
4. Two

Answer: 3. Three

Wbbse Class 10 Life Science Solutions Chapter 2

Question 5. The growth of a plant is measured with—

1. Potometer
2. Arc indicator
3. Barometer
4. All of these

Answer: 2. Arc indicator

Question 6. Which of the following is an example of an external factor of growth?

1. Gene
2. Sunlight
3. Hormone
4. All of these

Answer: 3. Hormone

Question 7. Which of the following occurs first during the growth of a developing organism?

1. Cell maturation
2. Cell transformation
3. Cell development
4. Cell division

Answer: 4. Cell division

Question 8. The growth in width of a plant is treated as—

1. Vegetative growth
2. Primary growth
3. Secondary growth
4. Reproductive growth

Answer: 3. Secondary growth

Question 9. The metabolic process that involves the increase in dry weight of the body is denoted as—

1. Anabolic metabolism
2. Catabolic metabolism
3. Differential metabolism
4. All of these

Answer: 1. Anabolic metabolism

Question 10. The metabolic process in which the dry weight of the body decreases is called—

1. Anabolic metabolism
2. Catabolic metabolism
3. Differential metabolism
4. All of these

Answer: 2. Catabolic metabolism

Question 11. In plants, the growth that occurs due to cell division of secondary meristem is called—

1. Reproductive growth
2. Secondary growth
3. Vegetative growth
4. Primary growth

Answer: 2. Secondary growth

Question 12. If catabolism takes over anabolism, the dry weight of the body decreases. This is called—

1. Negative growth
2. Regular growth
3. Irregular growth
4. Positive growth

Answer: 1. Negative growth

Question 13. The second phase of growth is—

1. Cell enlargement phase
2. Cell division phase
3. Cell differentiation phase
4. Late phase

Answer: 1. Cell enlargement phase

Wbbse Class 10 Life Science Solutions Chapter 2

Question 14. The last phase of growth is—

1. The phase of cell enlargement
2. The phase of cell division
3. The phase of cell differentiation
4. Late phase

Answer: 3. Phase of cell differentiation

Question 15. The phrase, in which the origin of different types of blood cells occurs from stem cells, is treated as—

1. The phase of cell division
2. The phase of cell enlargement
3. The phase of cell differentiation
4. The phase of cell ageing

Answer: 3. Phase of cell differentiation

Question 16. Increase in the number of cells in meristem can be treated as—

1. The phase of cell enlargement
2. The phase of cell division
3. The phase of cell differentiation
4. The phase of cell elongation

Answer: 2. Phase of cell division

Question 17. Transformation of meristematic tissue into permanent tissue occurs in—

1. The phase of cell differentiation
2. The phase of cell enlargement
3. The phase of cell division
4. All of these

Answer: 1. Phase of cell differentiation

Question 18. Which phase of human development is associated with the maturity of reproductive organs and reproductive glands?

1. Childhood
2. Adolescence
3. Late adulthood
4. Infancy

Answer: 2. Adolescence

Question 19. The gradual transformation occurring in a plant from germination of seed to the onset of ageing, is called—

1. Senescence
2. Metabolism
3. Growth and development
4. Assimilation

Answer: 3. Growth and development

Question 20. The gradual complexity, that appears in an organism during its life cycle, by means of growth and following changes, is known as

1. Growth
2. Development
3. Division
4. Degeneration

Answer: 2. Development

Question 21. The total number of developmental phases in human life is—

1. Three
2. Four
3. Five
4. Six

Answer: 3. Five

Question 22. If anabolism takes over catabolism, the dry weight of the body increases. This is called

1. Negative growth
2. Regular growth
3. Irregular growth
4. Positive growth

Answer: 4. Positive growth

Life Science Class 10 Wbbse

Question 23. The organ which is responsible for husky or cracked voice in males is—

1. Chest muscle
2. Larynx
3. Bronchiole
4. Tongue

Answer: 2. Larynx

Question 24. The developmental phase of human life, in which osteoporosis and osteoarthritis appear, I is called the—

1. Childhood
2. Adolescence
3. Matured age
4. Old age

Answer: 4. Old age

Question 25. The developmental phase of human life, in which motor activities grow more, is the—

1. Childhood
2. Neonatal stage
3. Adolescence
4. Matured age

Answer: 1. Childhood

Question 26. The human developmental phase, in which sexual urge starts to develop, is called the—

1. Childhood
2. Matured stage
3. Adolescence
4. Neonatal stage

Answer: 3. Adolescence

Question 27. The ultimate stage of human development is—

1. Development of motor activity
2. The quick development of intelligence
3. Development of abstract idea
4. Degeneration of bone joints

Answer: 4. Degeneration of bone joints

Question 28. Which of the following is not a character of senescence?

1. Wrinkled skin
2. Fatigue
3. Development of bone
4. Degeneration of bones

Answer: 3. Development of bone

Life Science Class 10 Wbbse

Question 29. The stage of human development, in which the ability for detecting light and sound develops is—

1. Neonatal stage
2. Childhood
3. Adolescence
4. Matured stage

Answer: 1. Neonatal stage

Wbbse Class X Life Science

Question 30. Among vertebrates, indirect development occurs in—

1. Frogs
2. Birds
3. Reptiles
4. Mammals

Answer: 1. Frogs

Question 31. The phase of cell enlargement is controlled by—

1. Thyroxin
2. Somatotropic hormone
3. Insulin
4. ADH

Answer: 2. Somatotropic hormone

Question 32. Development of an organism starts from—

1. Egg cell
2. Male gametes
3. Zygote
4. Ovary

Answer: 3. Zygote

Question 33. The glands which become hyperactive during adolescence are—

1. Sebaceous glands
2. Sweat glands
3. Reproductive glands
4. All of these

Answer: 4. All of these

Question 34. Secondary sexual characters appear during—

1. Childhood
2. Matured age
3. Adolescence
4. Neonatal stage

Answer: 3. Adolescence

Question 35. The quick loss of calcium occurs in the bones of human females—

1. Before puberty
2. Predelivery
3. Postdelivery
4. After puberty

Answer: 3. Postdelivery

Cell Division and Its Role in Growth

Chapter 2 Growth And Development Answer In A Single Word Or Sentence

Question 1. By which life process, the size, shape and dry weight of an organism increase permanently?
Answer: Growth

Question 2. Which type of growth occurs among woody perennial plants?
Answer: Indefinite growth

Question 3. Which type of growth occurs in plants that grow with time, produce flowers and fruits and then die?
Answer: Definite growth

Question 4. Name two plants showing indefinite growth.
Answer: Banyan and mango

Question 5. Name two plants showing definite growth.
Answer: Paddy and wheat

Life Science Class 10 Wbbse

Question 6. How do you denote the growth of an organism in unit time?
Answer: Growth rate

Question 7. Mention two internal factors of growth regulation.
Answer: Genes and hormones

Question 8. What is produced after the division of a zygote?
Answer: An embryo

Question 9. Which type of cell division leads to the development of an embryo from a fertilised ovum?
Answer: Mitotic cell division

Question 10. In which phase of growth are the tissues formed?
Answer: Phase of cell differentiation

Question 11. What happens to a cell with the synthesis of new organic matter within it?
Answer: With the synthesis of new organic matters within a cell, the size of the cell increases gradually.

Question 12. Why is the growth of plants treated as indefinite growth?
Answer: Since there is no fixed span of growth among plants, the growth of plants is treated as indefinite growth.

Wbbse Class X Life Science

Question 13. Why is the growth of animals treated as definite growth?
Answer: Since the growth of animals continues for a specific period, animal growth is treated as definite growth.

Question 14. Which phase of human growth is treated as a period of turmoil?
Answer: Adolescent period

Question 15. How do you denote the span of human life from the age of two months up to ten years?
Answer: Childhood

Question 16. How do you denote the phase between childhood and adulthood?
Answer: Adolescence

Question 17. At which phase, a maximum level of physical and mental changes occur in human life?
Answer: Adolescence

Question 18. How do you denote the period of human life between the age of 19 to 60 years?
Answer: Adulthood or matured phase

Question 19. How do you denote the growth of an organism, in which all parts of the body grow at an even rate?
Answer: The growth of certain organisms, like fishes, where all parts of the body grow at an even rate, is called isometric growth.

Question 20. How do you denote the growth of an organism, in which different parts of the body grow at uneven rates?
Answer: The growth of certain organisms, where different parts of the body grow at an uneven rate, is called allometric growth.

Question 21. Which part of the human body shows the slowest rate of growth?
Answer: Human head

Question 22. Which part of the human body shows the highest rate of growth?
Answer: Human legs

Life Science Class 10 Wbbse

Question 23. At which age of human life, does height increase at the highest rate?
Answer: During the first three years of human life, height’ increases at the highest rate.

Question 24. At which growth phase, do thickening and ornamentation of cell wall occur in plant cells?
Answer: Phase of cell differentiation

Question 25. Which hormone becomes highly active in human adolescent males?
Answer: Testosterone

Wbbse Class X Life Science

Question 26. In which type of growth, all the organs do not increase in the same proportion?
Answer: Allometric growth

Question 27. Give an example of an organism with allometric growth.
Answer: Human

Question 28. What is a toddler?
Answer:

Toddler

A toddler is a child who learns to walk with a great cognitive, emotional and social development of 1-3 years of age.

Importance of Nutrition in Growth

Question 29. Which age of a human is called the adolescent stage.
Answer: 10-19 years

Question 30. In which phase of growth of human metabolism decreases?
Answer: Old age

Question 31. Give one example of isometric growth.
Answer: Isometric growth can be seen in fish.

Question 32. Which is the first site of growth?
Answer: Embryo

Question 33. Which type of metabolism is growth?
Answer: Anabolism

 

Chapter 2 Growth And Development Fill In The Blanks

Question 1. Due to _________, structural complexity increases in a living body.
Answer: Development

Question 2. An organism attains _________ by the process of growth and development.
Answer: Maturity

Question 3. Growth in animals occurs for a _________ period of time.
Answer: Limited

Question 4. After fertilisation, a unicellular _________ is formed.
Answer: Zygote

Question 5. During cellular differentiation, _________ cell wall is formed.
Answer: 5Secondary

Question 6. The period of human life from birth to the end of the first month is called the _________ phase.
Answer: Neonatal

Life Science Class 10 Wbbse

Question 7. Attaining sexual maturity is denoted as the onset of _________.
Answer: Puberty

Question 8. During the cell enlargement phase, a young cell absorbs _________ to increase its volume.
Answer: Water

Question 9. In plants, the cell layer below _________ shows cellular enlargement.
Answer: Cambium

Question 10. During the _________ period, attraction towards the opposite sex increases significantly.
Answer: 1Adolescent

Question 11. During _________a human baby learns to utter meaningful words.
Answer: Childhood

Question 12. In human females, two hormones become highly active during puberty, these are _________ and progesterone.
Answer: Oestrogen

Question 13. The age 10-19 of humans is called _________ phase.
Answer: Adolescent

Question 14. The age of 60 or more is included in _________ phase.
Answer: Old age

Question 15. The motor activities of human is developed in _________ phase.
Answer: Childhood

 

Chapter 2 Growth And Development State True Or False

Question 1. The size of all organisms changes permanently during growth.
Answer: True

Question 2. Hormones play an important role in the growth of an organism.
Answer: True

Wbbse Class X Life Science

Question 3. Human development is divided into four phases.
Answer: False

Question 4. Cell enlargement occurs by continuous intake of water and synthesis of organic matter within the cell.
Answer: True

Question 5. The number of cells increases by mitosis.
Answer: True

Question 6. The growth of any organism occurs with an increase in dry weight of the cytoplasm of all its cells.
Answer: True

Life Science Class 10 Wbbse

Question 7. The age between 20 to 40 years denotes adulthood.
Answer: False

Question 8. In human life, 10 to 15 years of age indicates childhood.
Answer: False

Question 9. Germ cell production in gonads starts during adolescence.
Answer: True

Question 10. Gradual loss of memory occurs during old age.
Answer: True

Question 11. The study of human development is known as gerontology.
Answer: False

 

Chapter 2 Growth And Development Match The Columns

Question 1.

Answer: 1. B, 2. A, 3. D, 4. C, 5. F, 6. E

Question 2.

Answer: 1. B, 2. A, 3. D, 4. C, 5. F, 6. E

Wbbse Class 10 Life Science Solutions

Chapter 2 Growth And Development Find The Odd One Out

Question 1. Childhood, Adulthood, Old age, Birth
Answer: Birth

Question 2. The phase of cell division, Phase of cell enlargement, Phase of cell differentiation, Cell cycle
Answer: Cell cycle

Life Science Class 10 Wbbse

Question 3. Rapid growth, development of sexual character, Development of sex organs, Loss of bone density
Answer: Loss of bone density

Question 4. Neonate, Adolescent, Old age, Phase of cell differentiation.
Answer: Phase of cell differentiation

 

Chapter 2 Growth And Development Fill In The Blanks By Looking At The First Pair

Question 1. Formation of new cells: Phase of cell division:: Formation of different tissues:_____________
Answer: Phase of cell differentiation

Question 2. The phase of cell enlargement: Phase of growth:: Adolescence:_____________
Answer: Phase of development

Wbbse Class 10 Life Science Solutions

Question 3. 2 months year: Childhood:: 10-19 years: _____________
Answer: Adolescence

Question 4. The onset of activity of sex hormones: Adolescence Optimum activity of sex hormones: _____________
Answer: Adulthood

Question 5. Identification of sound: Neonate:: Decrease of hearing: _____________
Answer: Old age

 

Chapter 2 Growth And Development Among The Four Concepts Given Three Of Them Belong To One Find That

Question 1. Old age, Childhood, Adulthood, Phases of development
Answer: Phases of development

Question 2. Phase of cell division, Phase of cell enlargement, Phase of cell differentiation, Phases of growth
Answer: Phases of growth

Question 3. Formation of phloem tissue, Formation of nerve tissue, Formation of muscle tissue, Phase of cell differentiation
Answer: Phase of cell differentiation

Question 4. Loss of bone density, Old age, Loss of hearing, Loss of vision
Answer: Old age

 

Chapter 2 Growth And Development Advanced Questions and Answers

Question 1. What is fascicular cambium? What is its importance during grafting?

Answer:

Fascicular cambium:

The fascicular cambium is a meristem present between the primary xylem and phloem of vascular bundles.

Importance of Fascicular cambium:

During grafting it produces secondary xylem and secondary phloem in both stock and scion. This newly formed vascular tissues mix together through which it connect the two stems.

Question 2. How does the vegetative reproduction of plants take place through layering?

Answer:

In this process, the outer epidermis of a node portion of a branch is removed to expose the vascular tissue. This portion this then covered with soil, cow dung and cloth. Watering is done at regular intervals.

After a few days, roots develop in this portion by the activity of meristem. At this stage, the branch is removed from the main plant and is planted separately.

Wbbse Class 10 Life Science Solutions

Question 3. What is parthenogenesis?

Answer:

Parthenogenesis

The process through which an offspring is formed without fertilisation from an unfertilised egg is called parthenogenesis.

Example-Wasp, are honey bees, grape plants, etc.

Question 4. What is bulbil?

Life Science Class 10 Wbbse

Answer:

Bulbil

Bulbil is a type of vegetative reproductive structure in which a round, multicellular structure develop at the axil of leaves.

Example-Dioscorea sp.

Question 5. What is an actinomorphic flower?

Answer:

Actinomorphic flower

A flower which can be cut into two equal halves through more than one plane is called an actinomorphic or regular flower.

Example-Hibiscus.

Question 6. What is the zygomorphic flower?

Answer:

Zygomorphic flower

A flower which can be cut into two equal halves through only one plane is called a zygomorphic or irregular flower.

Example-Pea flower.

Question 7. What is the type of growth in perennial woody plants?

Answer: Indefinite growth

Question 8. The annual plants dies after growing for a certain period, flowering and fruiting. What type of growth it is?

Answer: Definite growth

Wbbse Class 10 Life Science Solutions

Question 9. What is the tanner scale or tanner stage?

Answer:

Tanner scale or tanner stage

The predefined chart used to measure the physical development of a child, adolescent and mature adult is called the tanner scale or tanner stage. It is mainly used to measure secondary sexual growth, so it is also known as Sexual Maturity Rating (SMR).