WBBSE Solutions For Class 10 Life Science And Environment Chapter 3 Mendel’s Laws And Their Deviation

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.
• 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.

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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.

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

1.

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

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