WBCHSE Class 12 Chemistry Unit 14 Biomolecules Notes

Class 12 Chemistry Unit 14 Biomolecules Introduction

Organic Chemistry Biomolecules :  A characteristic feature of living organisms is that they undergo growth and reproduction, though they are composed of atoms and molecules.

  • The branch of chemistry that deals with the chemical composition of living organisms and the biochemical processes brought about by those chemical compounds is called biochemistry.
  • The complex organic compounds which are essential components of living beings are regarded as biomolecules. They form the structural and functional basis of life.
  • Some important biomolecules are carbohydrates, proteins, enzymes, lipids, nucleic acid, hormones and vitamins.

Carbohydrates

General Discussion On Carbohydrates

Carbohydrates form an important class of biomolecules and are the main source of energy in the living body.

General Discussion On Carbohydrates Example:

  1. Most of the staple food in our diet consists mainly of starch.
  2. Cotton, linen and rayon fabrics are derived from the cellulose of plant cell walls.
  3. Wood is the dead cells formed by the deposition of cellulose and lignin. Thus starch, cellulose, and lignin are all carbohydrates.

Plants: Primary Source Of Carbohydrates

All green plants photosynthesize to produce carbohydrates. The raw materials used are—

  1. H2O absorbed from soil,
  2. Atmospheric CO2,
  3. Solar energy and
  4. Chlorophyll pigment.

Biomolecules Primary Source Of Carbohydrates

Definition Of Carbohydrates: The name carbohydrate was originally given to the class of compounds (containing carbon, hydrogen and oxygen) having the general formula Cx(H2O)y. Since these compounds contain carbon and hydrogen in the same ratio as in water, they were considered as the hydrates of carbon. But this definition did not survive long for the following reasons—

Organic Chemistry Biomolecules

  1. There are several compounds which are known to be carbohydrates by their chemical behaviour but do not possess the general formula Cx(H2O)y for example., rhamnose (C6H12O5) and deoxyribose (C5H10O4).
  2. Again there are some compounds such as formaldehyde (HCHO or CH2O), acetic acid [CH3COOH or C2(H2O)2 ] etc., which do not behave like carbohydrates but can be represented by the general formula Cx(H2O)y
  3. Carbon is not known to form any hydrate.
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Hence the above definition has been modified as given below:

Modern Definition: Carbohydrates are now defined as optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds which give these on hydrolysis.

There is no free aldehydic (—CHO) or keto (C=O) group in carbohydrates. They exist as hemiacetal or hemiketal formed by the condensation of the aldehydic or ketonic group with an alcoholic  —OH group present in that molecule.

WBCHSE Class 12 Chemistry Unit 14 Biomolecules Notes

Classification Of Carbohydrates

Classification Based On Taste

Classification Based On Taste Sugars: Water-soluble crystalline carbohydrates having a sweet taste are regarded as sugars. All monosaccharides and oligosaccharides fall under this category.

Classification Based On Taste Sugars Example: Glucose, fructose, sucrose, lactose, etc.

Relative Sweetness Of Some Sugars:

Biomolecules Sugars And Relative Swetness

Classification Based On Taste Non-sugars: Amorphous, tasteless carbohydrates which are slightly soluble or completely insoluble in water are called non-sugars. All polysaccharides fall under this category.

Classification Based On Taste Non-sugars Example: Starch, cellulose, etc.

Classification Based On Hydrolysis

Monosaccharides

Classification Based On Hydrolysis Monosaccharides Definition: The carbohydrates which cannot be hydrolysed to produce any simpler carbohydrates are termed monosaccharides.

Classification Based On Hydrolysis Monosaccharides Example: Arabinose (C5H10O5), ribose (C5H10O5), glucose (C6H12O6), fructose (C6H12O6), etc.

Oligosaccharides Definition: The carbohydrates which on hydrolysis, produce 2-10 monosaccharide molecules are called oligosaccharides.

Depending on the number of monosaccharide units produced, oligosaccharides can be further classified into disaccharides, trisaccharides and tetrasaccharides.

Disaccharides: Carbohydrates whose hydrolytic products are two similar or different monosaccharides are called disaccharides.

Disaccharides Example: Sucrose, lactose, maltose, etc.

Biomolecules Disaccharides

Trisaccharides: Carbohydrates whose hydrolytic products are three similar or different monosaccharides are called trisaccharides.

Trisaccharides Example: Raffinose.

Biomolecules Raffinose

Tetrasaccharides: Carbohydrates whose hydrolytic products are four similar or different monosaccharides are called tetrasaccharides.

Tetrasaccharides Example: Stachyose, hychoose

Biomolecules Tetrasaccharides

Polysaccharides Definition: Carbohydrates which, on hydrolysis, produce several monosaccharide units are called polysaccharides. The general formula of most polysaccharides is (C6H10O5)n.

Polysaccharides Example: Starch, cellulose, glycogen, etc.,

Biomolecules Polysaccharides

  1. Homopolysaccharides (homoglycans) i.e., composed of the same monosaccharide molecules such as starch, cellulose and glycogen.
  2. Heteropolysaccharides (heteroglycans), i.e.„ composed of different monosaccharide units such as insulin, heparin and hyaluronic acid. Polysaccharides are devoid of sweetness and form colloids in boiling water.

Classification Based On Reducing Property

Reducing Sugars: Carbohydrates which can reduce Fehling’s solution or Tollen’s reagent are called reducing sugars.

  • All monosaccharides (aldoses or ketoses) are reducing sugars. Again, disaccharides in which the two monosaccharide units are linked by an aldehydic or ketonic group (reducing centre) also behave as reducing sugars.
  • The other aldehydic or ketonic group remains as a hemiacetal or hemiketal.

Reducing Sugars Example: Monosaccharides: Arabinose, ribose, glucose, mannose, fructose, etc. Disaccharides: maltose, lactose, etc.

Non-Reducing Sugars Definition: Carbohydrates which cannot reduce Fehling’s solution or Tollen’s reagent are called non-reducing sugars.

Disaccharides whose monosaccharide units are linked by aldehydic or ketonic groups only are considered as non¬ reducing sugars.

Non-Reducing Sugars Example: Disaccharide: Sucrose. Polysaccharide: All polysaccharides (starch, cellulose, glycogen), etc.

Classification Of Carbohydrates:

Biomolecules Classification Of Carbohydrates

  • Identification of carbohydrates (Molisch’s test): In the aqueous solution of a carbohydrate, Molisch’s reagent (1% alcoholic solution of α-naphthol) is added, followed by the addition of concentrated H2SO4 along the side of the sloping test tube.
  • A red-violet ring is produced at the junction of the two liquids (acid and the aqueous solution of carbohydrates). All carbohydrates respond to this test.

Organic Chemistry Biomolecules

Monosaccharides

Monosaccharides are the simplest carbohydrates. They possess 3-7 carbon atoms, generally represented by Cn(H2O)n where n = 3-7. There are 20 monosaccharides found in nature. The nomenclature depends upon the nature of the carbonyl group present in the monosaccharide molecule.

Classification Of Monosaccharides

Classification Based On The Nature Of Carbonyl Group Aldose: Monosaccharides in which an aldehyde (— CHO) group is present are called aldoses. Hence, all the polyhydroxy aldehydes are considered aldoses.  The monovalent aldehyde group is found at the C1-terminal position of the aldose chain.

Classification on the basis of the nature of carbonyl group Ketose: Monosaccharides in which a ketoBiomolecules Monosaccharides Keto Groupgroup is present are termed ketoses. Hence, all polyhydroxy ketones are considered as ketoses. The bivalent keto Biomolecules The Bivalent Keto Group group can be found at any position other than the terminal carbon. But, naturally occurring ketoses contain a keto group at the second carbon atom.

Biomolecules Aldose And Ketose

Organic Chemistry Biomolecules

Classification Based On Number Of Carbon Atoms

Depending upon the number of the C-atoms, mono-saccharides can be of different types—

  1. Triose (containing three C-atoms),
  2. Tetrose (containing four C-atoms),
  3. Pentose (containing five C-atoms),
  4. Hexose (containing six C-atoms) and
  5. Heptose (containing seven C-atoms).

Classification Of Monosaccharides Based On The Number Of Carbon Atoms:

Biomolecules Classification Of Monosaccharides On The Basis Of The Number Of Carbon Atoms

Nomenclature Of Monosaccharides: Monosaccharides are named according to the number of carbon atoms and the nature of the carbonyl group.

  • The numeral prefixes indicating the number of carbon atoms, such as— tri- (3), tetra-(4), pent-(5), hex-(6) are succeeded by the suffix-ose.
  • The terms aldo & keto are used before the numeral prefixes of the carbon atoms to indicate the presence of aldehyde and ketone groups.

Nomenclature Of Monosaccharides Example: (aldo-)+(hex-)+(-ose)=aldohexose [polyhydroxy aldehydes containing 6 C-atoms] (keto-)+(hex-)+(-ose)=ketohexose [polyhydroxy ketones containing 6 C-atoms]

Organic Chemistry Biomolecules

Some Important Aspects Regarding Monosaccharides

Numbering The C-Chain With Consecutive Rank: The C-atom present in the aldehyde group of aldoses and the terminal C-atom nearest to the keto group of the ketoses are numbered as the first carbon or C1.

D- and L- Configuration:

  1. The D- and L- symbols refer to the configuration of monosaccharides. These do not indicate the dextro-rotatory and laevorotatory-specific rotations.
  2. In the case of a monosaccharide, if the asymmetric C-atom of the highest rank is similar to the asymmetric carbon of D-glyceraldehyde (H-atom on the left, —OH group on the right) then the monosaccharide is of D-configuration. If the H-atom is on the right and the —OH group is on the left, then the monosaccharide is of L-configuration.
  3. As per convention, if in a monosaccharide unit (expressed by the Fischer projection formula), the (— OH) group falls on the right of the highest ranked asymmetric carbon, then the monosaccharide has D -configuration and if the hydroxyl group falls on the left, it has L-configuration.

Biomolecules Apair Of Enantiomers

Organic Chemistry Biomolecules

Use Of + And -Sign: D- and L-symbols are followed by ‘+’ and ‘-‘ in brackets. These represent the dextro- and laevo-specific rotation. There is no relation between optical activity and the nature of rotation.

Use Of + And -Sign Example: Both glucose and fructose have D-configuration, but their optical activities are positive (dextrorotatory) and negative (laevorotatory) respectively.

Epimers: If two optical isomers containing 3 or more asymmetric C-atoms differ in the configuration of only one single C-atom, then the isomers are regarded as epimers. For example, D-( + ) -glucose and D-(+)-galactose are epimers of each other. [Follow ‘ D -class of aldehydes’]

Number Of Optical Isomers Of Monosaccharides: All monosaccharides, except dihydroxyacetone, contain at least one asymmetric C -atom. The number of optical isomers of a monosaccharide containing n -number of non-identical asymmetric C -atoms = 2n, for example., there are 4 non-identical asymmetric C -atoms in an aldohexose and therefore the number of optical isomers of aldohexose = 24 = 16.

D-Aldose Series:

Biomolecules D Aldose Series

Organic Chemistry Biomolecules

D-Ketose Series:

Biomolecules D Ketose Series

Every isomer forms two diastereoisomers due to an increase of one C-atom. Hence, two 4-C, four 5-C and eight 6-C diastereoisomers are found

Pentoses

Monosaccharides containing 5 C-atoms, expressed by the formula C5H10O5 are called pentoses.

Aldopentoses: There are four hydroxyl groups and one aldehyde group in an aldopentose molecule, i.e., aldopentose is a tetrahydroxyaldehyde. Its structural formula is:

Biomolecules Ketopentose Molecules

  • The number of optical isomers of an aldopentose = 23 = 8. These are ribose, arabinose, xylose and lyxose of D and L -configurations.
  • The most important aldopentose is D-(-)-ribose which is a component of ribonucleic acid (RNA).
  • All aldopentoses do not take part in fermentation but aldohexoses (like glucose) do take part.
  • The removal of the O-atom from the —OH group of the second C- atom of the aldopentose ribose, forms 2 -deoxyribose (CH2OH-CHOH-CHOH-CH2-CHO). It is an essential component of DNA.

Ketopentoses: There are four hydroxyl groups and one keto group in ketopentose molecules, ketopentoses are tetrahydroxy ketones with the structural formula:

Biomolecules Ketopentose Molecules

The number of optical isomers of ketopentoses due to the presence of two different asymmetric C-atoms = 22 = 4. These are ribulose and xylulose of D- and L-configuration.

Organic Chemistry Biomolecules

Hexoses

Monosaccharides containing six C -atoms and expressed by the formula C6H12O6 are called hexoses.

Aldohexoses: There are five hydroxyl groups and one aldehyde group in an aldohexose i.e., aldohexoses are pentahydroxy aldehydes.

Biomolecules Aldohexoses

  • The number of optical isomers of an aldohexose due to the presence of four different asymmetric C – atoms = 24 = 16. These are allose, altrose, glucose, mannose, gulose, iodose, galactose and tallose of D-and L-configurations. The D- and L-configurations of a particular aldohexose are mirror images of each other.
  • The most important aldohexose is glucose.

D-(+)-Glucose Or Dextrose (C6H12O6)

  1. It is an important aldohexose. The term ‘glucose’ originated from the Greek word ‘glucose’ meaning sweet. It is found in ripe grapes, honey, most fruits, blood and urine of diabetic patients. Its sweetness is comparatively less than sugar.
  2. Glucose is commonly found in nature as D-glucose. The asymmetric C-atom of the highest rank (C5) has the same configuration as that of D-glyceraldehyde and hence, it is called D-glucose.
  3. The aqueous solution of naturally occurring glucose rotates the plane of polarised light in the clockwise direction and hence, is termed dextrorotatory (+). Therefore, naturally occurring glucose is known as D-(+)- glucose or dextrose.

Biomolecules D Configuration

Preparation Of Glucose

Laboratory Method: The hydrolysis of an alcoholic solution of sucrose (cane sugar) by dilute HCl at 50-60°C temperature produces an equimolecular mixture of glucose and fructose (The hydrolysis of sucrose is called inversion of sucrose).

Biomolecules Laboratory Method

  • Crystalline Glucose precipitates on cooling the above mixture while fructose remains dissolved in the solution (as the solubility of glucose in an alcoholic solvent is less than that of fructose).
  • Crystallised glucose is then separated by filtration.

Industrial Method: Glucose Is Prepared From Starch In Industries

Starch is hydrolysed by heating it with dilute H2SO4 under pressure at 120°C to obtain glucose.

Biomolecules Industrial Method

Organic Chemistry Biomolecules

Open-Chain Structure Of Glucose: One aldehyde (—CHO) group, one primary alcoholic ( —CH2OH) group and four secondary alcoholic (—CHOH) groups are present in the open-chain structure of glucose. CHO-(CHOH)4-CH2OH (Glucose)

Detection Of Open Chain Structure Of Glucose By Experimental Observations

  1. Molecular Formula: It can be proved that C6H12O6 is the molecular formula of glucose by the analysis of its components and determination of its molecular weight.
  2. Unbranched Chain: A hexahydric alcohol, sorbitol is obtained on reduction of glucose by Na -amalgam.

Biomolecules Glucose And Sorbitol

N-hexane and 2 -iodohexane are obtained on the reduction of glucose by HI and red phosphorus at 373K.

Biomolecules N Hexane And 2 Iodohexane

The above observations prove that six C -atoms of glucose are linked by an unbranched chain (C—C— C—C—C—C).

Presence Of Aldehyde Group: Glucose reacts with hydroxylamine (NH2OH) and hydrogen cyanide, (HCN) to produce monoxide and cyanohydrin respectively.

Biomolecules Monooxime And Cyanohydrin

  • It proves the presence of one carbonyl group in glucose.
  • Weak oxidant bromine-water converts glucose into gluconic acid containing the same number of C-atoms which proves the presence of the —CHO group at the terminal position (since the —CHO group is monovalent) of glucose.

Organic Chemistry Biomolecules

Biomolecules Weak Oxidant Bromine Water

Presence Of Five Hydroxyl Groups: Glucose reacts with acetic anhydride to form a pentaacetate. Hence, a glucose molecule has five hydroxyl groups.

  • An organic compound containing two —OH groups at the same carbon is very unstable and readily converts into a carbonyl compound by giving off one water molecule.
  • Glucose is a stable compound as its water molecules cannot be removed even by heating. Thus, it can be concluded that no single C-atom of glucose has two or more hydroxyl groups attached to it or the five hydroxyl groups are linked to five different C-atoms.

Biomolecules Glucose Pentaacetate

Presence Of A 1° Alcoholic Group: Both glucose and gluconic acid are oxidised by HNO3 to produce saccharic acid or glucaric acid. This proves the presence of a 1° alcoholic group at the terminal position of the glucose chain.

Biomolecules Presence Of A 1Degree Alcoholic Group

Open-Chain Structure Of Glucose: The above discussion shows the presence of one aldehyde (—CHO) and primary alcoholic (—CH2OH) group in the glucose molecule. There are five hydroxyl groups, of which one is CH2OH and the rest four are secondary hydroxyl groups (—CHOH—) i.e., four —CHOH groups are present between —CHO and —CH2OH. So, the open-chain structure of glucose is:

Biomolecules Open Chain Structure Of Glucose

There are four different asymmetric C -atoms (starred) in a glucose molecule. Therefore, the number of stereoisomers of glucose is 24 = 16. These sixteen stereoisomers form eight pairs of enantiomers, for example., naturally occurring D-glucose and its enantiomer L-glucose.

Configuration Of Natural D-(+)- Glucose: Emil Fischer (1891) determined the exact configuration of four chiral carbons of D-(+) -glucose. It can be expressed by the Fischer projection formula.

Biomolecules D Glucose

Cyclic Structure Of D-(+)- Glucose

Limitations Of Open-Chain Structure Of Glucose

Glucose does not show all the characteristic reactions of aldehydes though it has —CHO group. For example, glucose does not form addition compounds with ammonia or sodium bisulphite and does not turn Schiff’s reagent violet in cold conditions.

Though glucose reacts with hydroxylamine to form oxime, it does not react with pentaacetate hydroxylamine.

D-(+) -glucose possess two stereoisomers- α-D-(+) glucose and β-D-(+) -glucose. D-glucose, when crystallised from water at 303K temperature, forms or-D-(-t-) – glucose.

  • Its melting point is 419K and the specific rotation of its aqueous solution is [α]D = +112°. D -glucose forms β-D-(+) -glucose on crystallisation from water at 371K. Its melting point is 423K and specific rotation of its aqueous solution [a]D = +19°.

Both α-and β-D -glucose exhibit mutarotation. If any form is dissolved in water, the specific rotation in case of or-D -glucose decreases from +112° and that of β-D – glucose increases from +19° to attain an intermediate value of +52.7°.

  • This phenomenon is called mutarotation. It refers to the change in the values of optical rotation, α- and β-D -glucose attains a dynamic equilibrium in an aqueous solution.
  • 36% of α-D -glucose and 64% of β-D -glucose are found in an equilibrium mixture at room temperature. Hence, specific rotation decreases in the case of α-D -glucose and increases in the case of β-D -glucose.

Biomolecules D Glucose Exhibit Mutarotation

Cyclic Structure Of D-(+)- Glucose Chain Structure Of Glucose Mutarotation: If the specific rotation of an optically active substance changes on its dissolution in a solvent and finally becomes fixed at a particular value, the phenomenon is called mutarotation.

  • D(+) -glucose forms two isomeric methyl glucosides. One molecule of aldehyde reacts with two molecules of alcohol to produce an acetal.
  • But, one molecule of glucose reacts with one molecule of methanol in the presence of dry HCl gas to produce a mixture of methyl α-D -glucoside (m.p. 438K; specific rotation =+158°) and methyl β-D- glucoside (m.p. 380K; specific rotation =-33°)

Biomolecules D Glucose Forms Two Isomeric Methyl Glucosides

  • Methyl glucosides do not behave like acetals. Though these are formed in anhydrous condition, they are hydrolysed in glucose and methanol acidified with dilute HCl.
  • These glucosides do not reduce Tollen’s reagent or Fehling’s solution and do not react with HCN or NH2OH. Hence, it proves the absence of free —CHO group in these glucosides.

Organic Chemistry Biomolecules

6-Membered Ring Structure Of D-(+)-Glucose: As the open-chain structure of glucose cannot explain the above observations, scientist W. N. Haworth proposed the six-membered ring structure of glucose.

  1. There is no free aldehyde (—CHO) group in this structure. The —CHO group of the glucose molecule bonds with the —OH group of C-5 of the same molecule by intramolecular reaction to produce a hemiacetal. This hemiacetal is a heterocyclic ring formed of one oxygen atom and five carbon atoms.
  2. The anomeric carbon (C-1) turns into a chiral carbon (C) after the formation of hemiacetal. The newly formed C-atom can link with the H and —OH groups in two patterns, i.e., D -glucose is found in two stereoisomeric forms of α-D-(+) -glucose and β-D-(+) -glucose.
  3. Conventionally, if the —OH group is linked to C-1, the —OH group is linked to C-5 (determines the prefix D) and the O-atom of the ring in the Fischer projection formula is on the same side, the structure is referred to as α-D-(+) glucose and vice versa in case of β-D-(+) -glucose.
  4. The closed ring structure of glucose is similar to the pyranBiomolecules Pyran Ringring and hence, is also known as the pyranose ring. Thus, the six-membered closed-ring structure of glucose is called glucopyranose. α-and β-D-(+) -glucose are known as α- and β-D-(+) -glucopyranose respectively.

Biomolecules Anomeric Carbon.

The stereoisomers α-and β-D-(+) -glucopyranose having different C-1 configurations are referred to as anomers and C-1 of glucose is called anomeric carbon.

Anomeric Carbon: In the case of any monosaccharide, the carbon atom which takes part in the process of intramolecular hemiacetal formation is called the anomeric carbon of the monosaccharide, for example., in the case of glucose, the carbon atom of the aldehyde group participates in the intramolecular reaction to form hemiacetal.

  • So, the number-1 carbon atom, i.e., C1 -atom is called anomeric carbon. Similarly, in the case of fructose, the carbon atom of the keto group, i.e., the C2-atom takes part in the process of hemiacetal formation.
  • Therefore, the number-2 carbon atom, i.e., the C2-atom of fructose is called anomeric carbon.

Anomer: Any pair of cyclic stereoisomers which are produced from monosaccharides through intramolecular hemiacetal formation are known as anomers, for example., α-D-(+) -glucopyranose and β-D-(+) -glucopyranose form such a pair of anomers.

Explanation Of Limitations Of Open-Chain Structure With The Help Of Closed Ring Structure

  • Solid D-glucose is found as a closed ring as well as α- and β-anomers. When any anomer (either the α- or β- anomer) is dissolved in water, it gradually converts into the other anomeric form, resulting in a dynamic equilibrium between the two.
  • The equilibrium mixtures contain the anomers in a specific ratio and hence, their specific rotation attains a fixed value.
  • In closed-chain structures (36% a-anomer and 64% β-anomer) traces of open-chain structure are also found at room temperature. The optical rotation of the equilibrium mixture is +52.7°.

Biomolecules Open Chain Structure With The Help Of Closed Ring Structure

Some Important Points Regarding Mutarotation:

  1. All monosaccharides and some disaccharides like maltose, and lactose exhibit mutarotation.
  2. The addition of an acidic or basic catalyst to the aqueous solution of glucose or other monosaccharides increases the rate of mutarotation.
  3. Acidic or basic solutions do not exhibit mutarotation, for example., glucose dissolved in cresol (acidic) or pyridine (basic) do not show mutarotation. But, when it is dissolved in a mixture of cresol and pyridine, it shows mutarotation. This proves that both acidic and basic solvents are required for mutarotation.
  4. Water is an amphoteric solvent. So, glucose or other monosaccharides show mutarotation in aqueous solution.
  5. A rise in temperature increases the rate of mutarotation.
  6. It gives an idea about the closed ring structure of monosaccharides.
  7. Methyl α-D-(+)- glucoside and derivatives of other monosaccharides do not show mutarotation.
  8. Ring-chain tautomerism i.e., the transformation of the closed ring to an open chain and vice-versa causes mutarotation.

The open-chain structure can be used to explain the formation of cyanohydrin, oxime and osazone and the reduction of Fehling’s solution by the aldehyde groups.

  • These are irreversible reactions. Firstly, the open chain structure present in traces undergoes a chemical reaction. Its decrease in concentration disturbs the equilibrium and therefore the closed-ring structures change into open-chain forms.
  • These forms of glucose produce the aforesaid compounds. On the other hand, the reactions between D-glucose and Schiff’s reagent, sodium bisulphite, 2, 4- DNP and NH3 are reversible.
  • So, a sufficient concentration of open-chain structures remains in equilibrium with the ring forms. So, it can be said that the weak reagents do not disturb the equilibrium to produce more open-chain structures. Hence, these reagents do not react with D-(+) -glucose.
  • The structures of two anomeric methyl glucosides can be explained by the closed ring forms. The two anomers on separately reacting with methanol in the presence of dry HCl can produce methyl α-D -glucoside and methyl β-D- glucoside respectively.

Biomolecules Structures Of Two Anomeric Methyl Glucosides

  • Glucosides are acetal compounds that are produced during reactions between hemiacetal and methanol. These are hydrolysed by dilute acids only to form the corresponding hemiacetal which further gives glucose and methanol as products.
  • But, acetals are stable in an alkaline medium similar to ether i.e., no hydrolysis takes place. So, glucosides do not reduce Fehling’s solution or Tollens’ reagent.
  • Glucose pentaacetate has two anomeric forms similar to methyl glucosides.

Biomolecules Glucose Pentaacetate Has Two Anomeric Forms Similar To Methyl Glucosides

These pentaacetates do not hydrolyse easily in aqueous medium to form hemiacetal, i.e., open-chain forms. So, these compounds cannot react with hydroxylamine (NH2OH) to form oximes like glucose.

Structures Of Glucose And Other Monosaccharides Haworth Projection Formula

  1. As per the Haworth projection formula, α-D-(+) and β-D-(+) -glucopyranose are considered to be equiplanar hexagons. The O-atom is placed on the right corner of the hexagonal part, which is farthest from the reader.
  2. The part closest to the reader is made bold. A perpendicular vertical straight line shows that the H-atom and —OH groups are linked to the C-atom.
  3. The H-atoms and —OH groups present on the right-hand side in the Fischer projection formula are drawn below the hexagonal plane and those on the left-hand side are drawn above the hexagonal plane in the Haworth projection formula.
  4. The —CH2OH group and H-atom linked to C-5 are drawn above and below the hexagonal plane respectively.
  5. The —OH group joined to the anomeric carbon C-1 of α-anomer lies below the hexagonal plane and the —OH group of C-1 of β-anomer lies above. Similarly, the structures of other monosaccharides can be drawn using the Haworth projection formula.

Biomolecules Alpha And Beta D Glucopyranose

Organic Chemistry Biomolecules

Fischer projection Formula Of A- And P- Anomers Of D-And L-Sugars:

Biomolecules Anomers of D And L Sugars

Reactions Of D-(+)- Glucose:

Biomolecules Reactions Of D Glucose

  • Glucose reacts with excess phenylhydrazine (in the 1:3 molar ratio to give a yellow crystalline solid compound, glucosamine.
  • Osazone of monosaccharides is a crystalline solid having well well-defined melting point. For example, glucosamine has a melting point of 204° C.
  • Thus, a monosaccharide may be identified by preparing its osazone and subsequently determining its melting point-

D-(+)-glucose reacts with the cone. HCl to form 5-hydroxymethyl furfural:

Biomolecules 5 Hydroxymethyi Furfural

D-(+) -glucose undergoes a rearrangement (known as Lobry de Bruyn-van Ekenstein rearrangement) on treatment with aqueous NaOH to form an equilibrium mixture of D-mannose. D-fructose along with the starting material:

Biomolecules Lobry De Bruyn Van Ekenstein Rearrangement

Glucose undergoes fermentation by die action of zymase from yeast to give ethyl alcohol with the liberation of CO2.

Biomolecules Zymase

Organic Chemistry Biomolecules

The reaction is anaerobic. However in the presence of air (oxygen), the alcohol may be oxidised further to acetic acid or it may produce more CO2.

D-(-) – Fructose (C6H12O6)

Characteristics Of D-(-)- Fructose

  1. D-(-)-fructose is the most essential naturally occurring ketohexose. It is found in fruits and honey in free form and hence, is called fruit sugar.
  2. D-(-) -fructose remains bonded with glucose in the sucrose and hydrolysis of sucrose yields fructose. Inulin is a naturally occurring polysaccharide formed of fructose molecules.
  3. Fructose is commercially prepared by the hydrolysis of Inulin. The aqueous solution of naturally occurring fructose is optically active and laevorotatory (specific rotation is -92°). Thus, it is known as laevulose.
  4. The chiral carbon atom (C-5) of the highest rank in fructose has a three-dimensional configuration similar to that of D -glyceraldehyde. So, it is placed under the D – class of ketohexoses.
  5. Fructose also forms intramolecular acetal and ketal just like glucose. It exhibits mutarotation.
  6. Fructose has a reducing properties.

The open-chain structure of D-(-) -fructose is depicted here.

Biomolecules Anomeric Carbon

Organic Chemistry Biomolecules

  • The O-atom of the —OH group linked to C-6 of fructose bonds with the keto group of C-2. Hence, C-2 becomes an asymmetric carbon.
  • There are two possible arrangements of the —CH, OH and —OH groups. So, D-fructose can exist in two stereoisomeric forms.
  • These are α-D -fructopyranose C[α]D = -20°) and β-D-fructopyranose ([α]D = -133°). Fructose exhibits mutarotation.
  • The specific rotation of aqueous solution of naturally occurring fructose in equilibrium is -92°. Fructose is a reducing sugar.

Haworth Projection Formula Of D-(-)- Fructopyranose And D-(-)- Froctofuranose

Fructose exists as a five-membered ring when bonded to glucose in sucrose. This ring structure, being similar to a furanBiomolecules Furan Ring ring is also called a furanose ring. Similarly, the five-membered ring of fructose is called fructofuranose.

Biomolecules Anomeric CarbonIn Each Structure

Biomolecules Anomers

Reactions Of D-(-)-Fructose

Biomolecules Reactions Of D Fructose

Organic Chemistry Biomolecules

Comparison Between Glucose And Fructose Biomolecules Comparison Between Glucose And Fructose

Osazone formation by monosaccharides involves only C-l and C-2; Thus the configurations of the other carbon atoms of the starting monosaccharide remain unchanged the configuration of C-3, C-4 and C-5 of glucose, mannose and fructose are identical because they form the same osazone on treatment with excess of phenylhydrazine.

Biomolecules Osazone Formation By Monosaccharides

Disaccharides

Carbohydrates whose hydrolytic products are two similar or different monosaccharides are called disaccharides. Their general formula is C12H22O11. Three essential disaccharides are sucrose, maltose and lactose. These are hydrolysed by acids or enzymes to form 2 similar or different monosaccharides.

Biomolecules Glycosidic Linkage

Two monosaccharides are bonded by ether or oxide linkage by the removal of one water molecule to form a disaccharide. This bond is known as glycosidic linkage.

Classification Of Disaccharides: Depending upon the type of linkage, disaccharides are of two types—

  1. Reducing and
  2. Non-reducing.

If two monosaccharides are joined by a —CO group, it forms a nonreducing disaccharide as monosaccharide units in the form of acetal (or ketal) ring cannot open up to form —CHO (or —COCH2OH) groups (responsible for the reduction of Fehling’s solution or Tollen’s reagent).

If there is a free —CO group in a monosaccharide unit, it will act as a reducing agent as it remains in the form of hemiacetal or hemiketal which produces —CHO or —COCH2OH.

Classification Of Disaccharides Example:

  1. Reducing: Maltose and lactose.
  2. Non-reducing: Sucrose, Trehalose

Sucrose Or Cane Sugar (C12H22O11): It is a common disaccharide present in sugarcane, beet, palms, etc. A major source of sucrose is sugarcane, hence, it is called cane sugar.

It is a colourless, water-soluble, sweet, crystalline substance’ and is used in the preparation of sweetened food. Dilute acids or enzymes hydrolyse it to produce an equimolar mixture of D-(+) -glucose & D-(-)- fructose.

Biomolecules Sucrose Or Cane Sugar

Disaccharides Inversion Of Cane Sugar: The specific rotation of sucrose in an aqueous solution is +66.5°. Its hydrolysis produces an equimolar mixture of D-glucose (specific rotation =+52.5°) and D-fructose (specific rotation = -92°).

Organic Chemistry Biomolecules

  • As the specific rotation of laevorotatory fructose is higher than that of dextrorotatory glucose, the specific rotation of the hydrolysed solution is laevorotatory with the value being 1/2(+52.5°-92°)=-19.75°.
  • Since, the direction of optical rotation of sucrose changes (+66.5° → -19.75°), the phenomenon is called inversion of cane sugar. The mixture thus obtained is called inverted sugar. Honey is a very common example of inverted sugar.

Disaccharides Structure Of Sucrose: Sucrose is a non-reducing sugar and hence, its two monosaccharide units, glucose and fructose are linked by a —CO group.

  • Structural analysis of the ring shows that glucose exists in the form of pyranose and fructose in the form of furanose.
  • As the enzymes maltase (hydrolyses α-D-glucoside) and invertase (hydrolyses β-D-fructoside) hydrolyse sucrose, glucose is found in the form of α-glycoside and fructose as β-fructoside in the sucrose molecule.
  • The C-1 of α-D-glucose is linked to C-2 of β-D-fructose. Therefore, the structure of sucrose is given below, which explains all its properties.

Biomolecules Haworth Projection Formula Of Sucrose

Maltose Or Malt Sugar (C12H22O11): The enzyme diastase present in germinating barley or malt hydrolyses starch partially to produce maltose.

Biomolecules Hydrolyses Starch Partially To Produce Maltose

  • 1 mol maltose, on hydrolysis, produces 2 mol of D- glucose. Maltose is a reducing sugar as it’s one glucose unit remains as hemiacetal and produces osazone on reaction with excess hydrazine.
  • It shows mutarotation and reduces both Tollens’ reagent and Fehling’s solution.

Structure Of Maltose: Several experiments have proved that maltose contains two units of glucose as pyranose of which one is reducing and the other is non-reducing. The C-1 of non-reducing glucose is linked to the C-4 of reducing glucose.

Again, as maltose is hydrolysed by the enzyme maltase (that hydrolyses α-glucoside), the non-reducing glucose unit is found in the form of a-glucoside. i.e., C-1 of non-reducing α-D-glucose is linked to C-4 of the reducing glucose unit.

Biomolecules Haworth Projection Formula Of Maltose

Lactose Or Milk Sugar (C12H22O11): Since lactose is found in milk, it is called milk sugar. Dilute acids hydrolyse lactose to form an equimolar mixture of D-glucose and D-galactose.

It is a non-reducing sugar as its glucose unit is a hemiacetal and reacts with excess phenylhydrazine to form osazone. It shows mutarotation and reduces both Tollens’ reagent and Fehling’s solution.

Organic Chemistry Biomolecules

Structure Of Lactose: In lactose, glucose (reducing form) and galactose (non-reducing form) remain as pyranose and C-4 of the glucose unit is linked to C-1 of the galactose unit.

Hydrolysis of lactose by emulsion (hydrolyses fi glycoside) indicates that non-reducing galactose is present in the form of β-galactoside i. e., C-1 of non-reducing galactose is linked to C-4 of reducing glucose unit.

Biomolecules Haworth Projection Formula Of Lactose

Polysaccharides

Several monosaccharides by the removal of H2O molecules get linked together and form a macromolecule of polysaccharide. So, polysaccharides are naturally occurring condensation polymers in which monomeric monosaccharides are joined by glycosidic linkages. Some important polysaccharides are:

  1. Starch,
  2. Cellulose,
  3. Glycogen and
  4. Dextrin. Starch and cellulose are formed of only D-glucose units.

Polysaccharides Starch, Amylum (C6H10O5)n: Starch is stored as a reserve food in tuber, root and seeds. The value of n may vary from 200-1000 depending on the source.

Polysaccharides Properties of starch

  1. Starch is an amorphous granular solid.
  2. It is insoluble in cold water. If starch is heated as a suspension in water, it condenses to a viscous milk-like solution, which when cooled thickens into a jelly-like substance.
  3. It breaks into small molecules on hydrolysis. Later, it forms maltose which ultimately converts into D-glucose.

Biomolecules Ultimately Converts Into D Glucose.

It is a non-reducing polysaccharide. It does not reduce Tollen’s reagent or Fehling’s solution and does not form osazones. This proves that the hemiacetal —OH groups of C-1 of glucose are joined by glycosidic linkage.

Organic Chemistry Biomolecules

Polysaccharides Composition Of Starch: Starch is a mixture of amylose (15-20%) and amylopectin (80-85%).

  • It reacts with I2 to produce a violet-blue colour as amylose forms an inclusion complex with I2. When heated, the blue colour fades away but reappears on cooling.
  • Amylopectin does not produce a blue colour with I2. Both amylose and amylopectin are polymers of α-D glucopyranose.
  • One amylose molecule contains about 200- 1000 glucose units, while one amylopectin molecule contains 2000 -3000 glucose units.

Biomolecules Amylose Iodine Inclusion Complex

Structure Of Amylose: Amylose is a linear polymer of α-D-glucose in which C-1 of one glucose unit is linked to C-4 of another by a-glycosidic linkage.

Structure Of Amylopectin: Amylopectin is a complex polymer consisting of 10-200 unbranched linear chains.

  • Each chain is made up of 25-30 α-D-glucopyranose linked serially by or- 1, 4-glycosidic bonds.
  • Unbranched chains are linked by α-1, 6 -the glycosidic bond between the anomeric C-1 of the glucose residue at the reducing end of one chain and the C-6 of any glucose residue of another chain, which gives rise to an unbranched molecule of amylopectin.

Biomolecules Structure Of Amylose

Biomolecules Structure Of Amylopectin.

Distinction Between Glucose And Starch:

Biomolecules Property Glucose Starch

Cellulose (C6H10O5)n: The molecular mass of cellulose is 50,000 to 5,00,000. It is formed by 300-3000 β-D-glucose units. It is water-insoluble but dissolves in an ammoniacal solution of cupric hydroxide (Schweizer’s reagent).

Organic Chemistry Biomolecules

  • The addition of alcohol, acid or salt to this solution precipitates cellulose.
  • Its pure form is most abundant in plants. It is the major component of the plant cell walls and mainly brightens the cotton fabric. This process is called mercerisation and the cotton thus obtained is called mercerised cotton.

Properties Of Cellulose: Cellulose is a non-reducing sugar like starch. It does not reduce Tollen’s reagent or Fehling’s solution and does not form osazones.

  • It is not fermented by yeast. It cannot be hydrolysed easily like starch. But, when cellulose is heated with dilute H2SO4 (under pressure), it hydrolyses to produce D-glucose.
  • The enzyme emulsion or cellulase (hydrolyses β-glycosidic linkage) can hydrolyse it. In addition to concentrated NaOH, a translucent jelly-like substance is formed that brightens the cotton fabric.
  • This process is called mercerisation and the cotton thus obtained is called mercerised cotton.

Structure Of Cellulose: Cellulose is a linear polymer of serially arranged α-D- glucose units linked by β-1,4-glycosidic bonds.

The CH2OH— group of one chain links to the —OH group of C-2 of another by H -bonding to form a bundle of fibres. This linear arrangement helps to form cellulose threads.

Biomolecules Structure Of Cellulose

Uses Of Cellulose: It is used in the manufacture of clothes, cameras and papers. Cellulose also forms many useful compounds by treatment with suitable chemical reagents. Some examples are—

  1. Celluloid, which is used in the manufacture of toys, decorative articles and photographic films.
  2. Guncotton, which is an explosive.
  3. Cellulose acetate is used in the manufacture of rayon (artificial silk), plastics and nail polishes.
  4. Methylcellulose is used as a drug for the treatment of constipation and for the manufacture of cosmetics.
  5. Ethyl cellulose is used as a thin film coating material and also as a food additive as an emulsifier.

Can cellulose be considered as food?

  • The human intestine does not synthesise the enzyme cellulase which can hydrolyse cellulose into glucose. So, human beings cannot digest cellulose.
  • On the other hand, herbivores like cows, goats, and deer have bacterially produced cellulase enzymes in their intestines. So, they can feed on grasses and plants.

Importance Of Carbohydrates

In Living Systems

Biofuels: Carbohydrates are very essential for the survival of plants and animals. It forms a major portion of our diet. From an early time, ayurvedic medicine has considered honey as an instant source of energy.

  • Hence, carbohydrates provide the required energy for metabolism occurring inside living beings, i.e., act as biofuels.
  • C6H12O6(glueose) + 6O2 → 6CO2 + 6H2O + 2880 kj.mol-1 Polysaccharides like starch and glycogen are hydrolysed enzymatically into glucose inside living cells.
  • The produced glucose is then transported to all the cells via blood. After that oxidation of glucose into CO2 and H2O during enzyme catalysed reactions yields energy.

Reserve Foods: The major reserve food of plants is the polysaccharide starch.

  • It is stored in seeds as food for the germinating plant until it starts to photosynthesize.
  • It is also stored in the form of glycogen in animal liver and muscles. During starvation or illness, this glycogen is rapidly hydrolysed to glucose which in turn, gets oxidised to produce energy.

Constituents Of Biomolecules: Two aldopentoses, D -ribose and 2 -deoxyribose are essential constituents of RNA and DNA respectively. These nucleic acids carry hereditary traits and also participate in protein synthesis.

Organic Chemistry Biomolecules

  • The monosaccharide ribose is an essential component of adenosine triphosphate (ATP).
  • Energy obtained by oxidation of biomolecules such as carbohydrates, lipids etc., is stored in the cells in the form of ATP, which, in turn, helps to perform all the cellular functions (metabolism).
  • This is the reason why ATP is also called the energy currency of the cells. Carbohydrates also exist in biosystems in combination with many proteins and lipids.

As Structural Components: The major structural component of plant and bacterial cell walls is cellulose. It is used in the textile industry (as cotton) and in the furniture industry (as wood).

As Industrial Material: Carbohydrate is used as a raw material in different industries such as textiles, paper, liquor, etc.

Proteins

Chemically, proteins are condensation polymers or polyamides of high molecular mass (>10000) whose monomers are α-amino acids. Proteins contain C, H, N, O, S, and sometimes P, I, and traces of metallic elements (Fe, Cu, Zn, Mn). Partial hydrolysis of proteins yields peptides of varying molecular mass and complete hydrolysis produces a mixture of α-amino acids of L-configuration.

α-Amino Acids

Amino acids have amino (—NH2) and carboxyl ( —COOH) groups. Depending on the relative position of these groups, they are categorised as α, β, γ, δ.

  • The most important of all is a -amino acid as all biological proteins yield a -amino acids on their hydrolysis.
  • Hence, or-amino acids are the building blocks of proteins. These are linked together by — CO—NH— bonds in proteins. About 23 amino acids have been isolated till now.
  • Amino acids differ due to the presence of different kinds of R (alkyl) groups.

Biomolecules Amino Acids

Nomenclature Of Amino Acids: Amino acids can be named as per the IUPAC convention, but they are generally known by their common names.

Nomenclature Of Amino Acids Example: The amino acid H2NCH2COOH is known as glycine instead of 2-amino ethanoic acid (IUPAC).

The common names are assigned depending on their sources and properties.

Nomenclature Of Amino Acids Example: Glycine is named so, as it tastes sweet (Greek word ‘glykos’ means sweet); tyrosine was first isolated from cheese (Greek word ‘tyros’ means cheese).

The abbreviations are formed from the first three letters of the common name given to an amino acid.

Organic Chemistry Biomolecules

Nomenclature Of Amino Acids Example: ‘Gly’ stands for glycine, and ‘Ala’ stands for alanine. Sometimes, a single letter is also used as a symbol.

Classification Of α-Amino Acids

Classification Based On The Number Of —NH2 And —COOH Groups

  • Neutral Amino Acids: Amino acids containing the same number of — NH2 and —COOH groups are known as neutral amino acids.
  • Neutral Amino Acids Example: Glycine, alanine, valine, etc.
  • Acidic Amino Acids: Amino acids containing one — NH2 group and several —COOH groups are acidic amino acids.
  • Acidic Amino Acids Example: Aspartic acid, glutamic acid, etc.
  • Basic Amino Acids: Amino acids containing one — COOH group and several — NH2 groups are basic amino acids.
  • Basic Amino Acids Example: Lysine, histidine, etc.

Classification Based On Nutritional Significance

  • Essential Amino Acids: The human body cannot synthesise 10 amino acids obtained from protein. So, these should be taken through diet. A deficiency of these amino acids retards the normal growth and development of the body. These 10 amino acids are called essential amino acids.
  • Essential Amino Acids Example: Arginine, valine, leucine, isoleucine, phenylalanine, methionine, tryptophan, threonine, histidine and lysine.
  • Non-essential Amino Acids: The amino acids which are synthesised inside the human body are called non-essential amino acids. All amino acids except the ten essential amino acids are non-essential amino acids.
  • Non-essential Amino Acids Example: Glycine, alanine, serine, cysteine, cystine, tyrosine, proline, hydroxyproline, asparagine, aspartic acid, glutamine, glutamic acid, hydroxylysine.
  • Complete Proteins: Proteins providing all essential amino acids in the correct proportion for the nourishment of the body are called complete proteins, for example., proteins in fish, meat, milk and eggs.
  • Incomplete Proteins: Proteins which cannot provide one or more essential amino acids to the human body are called incomplete proteins for example., lysine is absent in proteins found in rice, wheat and other cereals.

Natural Amino Acids:

Biomolecules Natural Amino Acids

Biomolecules Natural Amino Acids

Optical Activity Of Amino Acids: All amino acids (except glycine) have one chiral carbon atom. So, each amino acid has two possible optically active isomers of D- and L- configuration.

All-natural amino acids are of L-configuration. The L-amino acid has a —NH2 group on the left and a —H group on the right according to the Haworth projection formula.

Biomolecules Chiral Carbon

Properties Of α-Amino Acids

Physical State, Solubility And Polarity

  1. Amino acids are colourless, non-volatile, crystalline solids. They melt by decomposing at high temperatures.
  2. Amino acids are water-soluble but insoluble in petroleum ether, benzene and ether
  3. Their aqueous solutions behave like solutions of substances having high dipole moment

Zwitterion Formation

  1. The presence of both acidic (—COOH) and basic (—NH2) groups in an amino acid molecule allows the proton (H+) to dissociate from the —COOH group and combine with the — NH2 group of the same molecule. Thus, the —COOH group becomes —COO and the —NH2 group converts into —N+H3. Thus, an amino acid behaves as an ampholyte and is regarded as a dipolar ion or zwitterion
  2. Due to zwitterion formation, amino acids have high melting points and dipole moments These are insoluble in non-polar solvents but soluble in water Hence, amino acids should be expressed in the zwitterion form as
    H3N—CHR—COO.

Organic Chemistry Biomolecules

  1. —COO group acts as a proton acceptor and —N+H3 acts as a proton donor, i.e., —COO group impart* basicity and — NH3 group imparts acid character Amino acids may exist as zwitterions in solid or polar solvents but not in a gaseous medium.

Biomolecules Zwitterion

Isoelectric Point: The proton transfer of zwitterions in alkaline and acidic mediums takes place in the following way:

Biomolecules Isoelectric Point

  • Amino acids behave as cations in an acid medium and as anions in an alkaline medium. If two electrodes are dipped in the acidic solution of an amino acid and then an electric current is passed through the solution, the amino acid will move towards the cathode.
  • Similarly, if current is passed through the basic solution of an amino acid, the amino acid will move towards the anode. Hence, the movement of die amino add molecules towards a particular electrode depends on the pH of the solution.
  • It can be shown that an amino acid at a particular pH does not move towards any’ electrodes. Amino acids exist as neutrally charged zwitterion at that pH, called the iso-electric point or pi. Each amino acid has its specific pl.

Isoelectric Point Definition: The particular value of pH at which an amino acid exists as a neutrally charged zwitterion and does not move towards any electrodes on passing an electric current through it, is called the isoelectric point of that amino acid.

  • The value of pl of neutral amino adds is slightly less than 7 [pi of glycine = 6.0); aridic amino adds is 3-5.4 [pi of aspartic add =3.0) and basic amino adds is 7.6-10.8 [pl of lysine = 9.5).
  • The solubility of amino added in water is the least at the isoelectric point So, this property’ is used to separate different amino adds obtained during protein hydrolysis.

Acidity And Basicity Constants (Ka and Kb): The acidity and basicity constants of an amino acid are very low, for example., Ka of glycine = 1.6 × 10-10 and Kb = 2.5 × 10-12. Most carboxylic acids have a value of 10-5 and most aliphatic amines have a Kb value of 10-4. The reason is the presence of the — NH3 group as a weak acid and the —COO group as a weak base in a-amino acids.

Organic Chemistry Biomolecules

Identification Of α-amino Acids: The addition of ninhydrin to the aqueous solution of a -amino acid in the presence of lithium acetate (LiOAc) produces a deep bluish-violet colour. This test, known as the Ninhydrin reaction is used to detect α-amino acids.

Biomolecules Bluish Violet Coloured Compound

Peptides And Their Classification

Amino acids can bond with each other to form a long molecular chain. The (—COOH) group of one amino acid links with the (—NH2) group of another by — CO —NH— bonds when two similar or different amino acids combine. The — CO—NH— bond formed by the elimination of 1 molecule of water is called a peptide bond or peptide linkage and the compound thus obtained is called a peptide.

Biomolecules Peptide Bond Or Peptide Linkage

Classification Of Peptide: The peptide produced in the reaction between two amino acids is called a dipeptide. Similarly, the peptides which are formed in the reaction between 3, 4, and 5 amino acid molecules are termed tripeptide, tetrapeptide, pentapeptide and so on.

Organic Chemistry Biomolecules

Classification Of Peptide Example: The —COOH group of glycine reacts with the —NH2 group of alanine to form glycylalanine (dipeptide). Similarly, the condensation of glycine, alanine and valine produces glycylalanylvaline (tripeptide).

Biomolecules Glycylalanylvaline

Similarly, several amino acids are linked together by peptide bonds to form a macromolecular chain or polymer. These polymers are called polypeptides. Conventionally, the peptide chain is expressed keeping the N-terminal amino acid or residue at the left-hand side and the C-terminal amino acid or residue at the right-hand side.

Biomolecules C Terminal Amino Acid

Nomenclature Of Peptide: The peptide is named according to the order of amino acids present in the peptide chain starting from the N-terminal residue.

  • The ‘ine’ suffix is eliminated from the amino acids other than the C-terminal one and is replaced by the suffix ‘yl’ and the C-terminal residue is written without any change.
  • There are no spaces between the names of the two amino acids.

Biomolecules Alanylglycylphenylalanine

The abbreviation of the amino acids is generally used for the nomenclature of peptides.

Nomenclature Of Peptide Example: Glycylalanine can be represented as Gly-Ala or G-A, alanyl glycyl phenylalanine as Ala-Gly-Phe or A-G-F.

  • The geometry of the peptide bond: The lone electron pairs of I N-atom present in a peptide bond ( — CO—NH— ) resonates with the C—O group, for example., delocalisation of the electron pair occurs via the C=0 group. Hence, the C=N bond gains a double bond character and geometrical isomerism is observed in the peptide bonds.
  • The steric hindrance between the two R—groups on the same side of the cis-isomer makes it less stable than the trans-isomer. Hence, the — CONH group is more stable in the trans-form i.e., O and H atoms lie on the opposite sides.
  • Carbon, nitrogen and the atoms (O, H and two C) linked to them lie on the same plane.

Biomolecules Resonance In Peptide Bond

Oligopeptides, Polypeptides And Proteins: Small-chain peptides of 2-10 amino acids are known as oligopeptides and long-chain peptides consisting of more than 10 amino acids are called polypeptides.

  • Polypeptides having 100 or more a-amino acids and an overall molecular mass of more than 10,000 are known as proteins.
  • The difference between a polypeptide and a protein is not distinct. Polypeptides of a relatively smaller number of a-amino acids bearing conformation similar to that of proteins are also regarded as proteins, for example., insulin is a protein containing 51 amino acids.
  • Proteins are amphoteric. So, these can neutralise acids or bases and possess an isoelectric point like α-amino acids. The solubility of polypeptides is the least at its isoelectric point and they are separated based on this property.

Classification Of Proteins

Classification Based On Molecular Structure: Structurally, proteins are of two types:

  1. Fibrous proteins and
  2. Globular proteins.

Fibrous Proteins: These proteins are slender, thread-like structures and water-insoluble and lie parallel to each other like fibres.

  • H-bonds and disulphide bonds join the peptide chains together.
  • The strong intermolecular force makes them water-insoluble. These proteins are components of connective tissues in living systems.

Fibrous Proteins Example: It is an essential component of skin, nails, hair, etc. Collagen, α-keratin, and myosin are different examples of this class of protein. These proteins remain stable at slight variations in temperature and pH.

Globular Proteins: These proteins intertwine among themselves to form a small sphere.

  • H -bonds, disulphide bonds, van der Waals force of attraction and dipolar interaction act between the polypeptide chains.
  • The hydrophobic R-groups lie on the inner side and the hydrophilic polar groups on the outer side of the bundle.
  • So, these are readily soluble in water. These proteins participate in different activities of plant and animal cells.

Globular Proteins Example: Enzyme (like trypsin), hormone (like insulin), transport protein (like haemoglobin), protective protein (like antibody). These proteins are more sensitive to changes in pH and temperature than the fibrous proteins.

Classification Based On The Nature Of Hydrolytic Product

  1. Simple Proteins: The proteins whose hydrolytic products are α-amino acids are called simple proteins.
    1. Simple Proteins Example: Albumin in egg white, glutinine in wheat, oxygenin in rice, keratin in hair and nails.
  2. Conjugated Proteins: Proteins whose hydrolytic products consist of α-amino acids and non-proteinaceous compounds are called conjugated proteins.
    1. The nonprotein part is called the prosthetic group and the remaining part excluding the prosthetic group is called the apoprotein.
    2. The main role of the prosthetic group is to control the biochemical activities of the protein. There are different conjugated proteins based on the nature of the prosthetic group.

Biomolecules Classification On The Basis Of Nature Of Hydrolytic Product

Organic Chemistry Biomolecules

Derived Proteins: The compounds produced on partial hydrolysis of simple or conjugated proteins by acid, base or enzyme are called derived proteins.

Derived Proteins Example: Proteose, peptones and polypeptides.

Structures Of Proteins

Several a -amino acids combine by peptide bonds to produce organic polymers of 3-D configuration. The structure of a protein can be categorised into any one of the four structural tiers.

Primary Structure Of Proteins: The arrangement or sequence of amino acids in a protein molecule refers to its primary structure. The modification of a single amino acid will change the sequence, which will form a different protein. The characteristic chemical and biological properties of different proteins are due to the differences in their primary structure.

Secondary Structure Of Proteins: The conformation of molecular chains in a protein refers to its secondary structure. Experimentally, it has been observed that a protein molecule has two conformations:

  1. Right-handed α-helix and
  2. β-pleated sheet.

α-Helix Conformation: A protein chain with a large R-group exists in the form of a right-handed α-helix. R-groups are extended away from the axis of the helix. This conformation arises due to the intramolecular hydrogen bond (between C=O of an amino acid residue and N—H of a fourth amino acid residue).

β-Pleated Sheet Conformation: The protein chains acquire a coiled structure to attain the β-pleated sheet conformation due to the presence of moderately sized R-groups.

  • The two adjacent protein chains are linked by H-bonds. The alternate R-groups linked to the a -carbon lie on the same side. Several chains together form a sheet—either parallel or antiparallel.
  • The N-terminals lie on the same side in parallel conformation. In anti-parallel conformation, the N-terminal of one chain and the C-terminal of another chain lie on the same side.

Biomolecules Alpha Helix Structure Of Protein

Tertiary Structure Of Proteins: Although there are only two types of conformation in a protein molecule, large protein molecules fold to form 3-D structures.

  • The 3-D structure formed due to the folding of protein molecules keeping the original conformation (α-helix or β- pleated sheet) intact, is called tertiary structure.
  • It is specific for each protein and is rigid and stable. The two major tertiary structures are discussed below.

Tertiary Structure Of Fibrous Proteins: An identical secondary structure is present in the complete chains of fibrous proteins, for example., α-keratin (a major protein in hair and wool) always exists as α-helix.

  • Several α-helices intertwine with each other to form a rope or rod-like tertiary structure.
  • The triple helical structure of collagen is shown below. α-helix is also known as 3.613 helix as 3.6 amino acids are found in one turn and one 13-membered ring is formed by H-bonds. The two adjacent turns are 54 pm away from each other.

Biomolecules Tertiary Structure Of Fibrous Protein

Tertiary Structure Of Globular Proteins: Globular proteins do not possess identical secondary structures of protein chains, like fibrous proteins. Some part of the molecule has α-helix, some parts are in the form of β- pleated sheet, and some parts do not have any secondary structure.

  • The latter part with no secondary structure is known as random coils. All these different parts are convoluted together to form a globular structure.
  • It is possible due to the presence of different attractive forces acting among the different R-groups in the side chains. These forces are hydrogen bonds, ionic or salt bridges, disulphide bonds and van der Waals’ force of attraction.

Hydrogen Bonds: The major radicals present in the side chains of amino acids of protein molecules are —OH, — NH2 and — COOH which exhibit H-bonds, for example., tyrosine and aspartic acid are linked by H-bonds.

Biomolecules Hydrogen Bonds.

Attractive Forces Due To Salt Bridge Formation: The internal neutralisation occurring between the — COOH group and — NH2 group of side chains of amino acids a protein molecule gives rise to —COO and — NH3 groups. A salt bridge is formed generating an attractive force between these two groups.

Biomolecules Tertiary Salt Bridge

Van Der Waals’ Forces Of Attraction: This force acts between the non-polar (hydrocarbon) side chains of amino acids of the protein.

Biomolecules Van Der Waals Forces Of Attraction

Disulphide Bridge: Two — SH groups of two cysteine amino acids of a protein molecule are oxidised to form a disulphide bridge.

Biomolecules Disulphide Bridge

Differences Between Globular And Fibrous Proteins:

Biomolecules Differences Between Globular And Fibrous Proteins

Quaternary Structure Of Proteins: Some complex proteins are formed of two or more molecular chains of polypeptides which are called subunits or protomers.

  • The complex three-dimensional structure formed by the aggregation of different protomers by H-bonds, electrostatic force, and van der Waals force of attraction is called a quaternary structure.
  • Haemoglobin bears a quaternary structure of spherical-shaped tetramer consisting of four polypeptide chains (two identical α-chains of 141 amino acids each and two identical β-chains of 146 amino acids each). Each polypeptide chain bears a terminal heme group (iron- protoporphyrin complex).

Biomolecules Quaternary Structure Of Haemoglobin

Denaturation Of Proteins: Every protein in biological systems possesses a particular 3-D configuration and a specific biological activity.

  • These proteins are called native proteins. If the native proteins are subjected to physical changes like temperature variation or chemical changes like pH variation or brought in contact with UV rays, the globular or helical coils are opened up and water-soluble globular proteins are coagulated into water-insoluble fibrous protein.
  • This coagulation disintegrates the original structure of protein and its biological reactivity is lost. This phenomenon is called protein denaturation and the coagulated protein is called denatured protein.
  • As a result of denaturation, there is no modification in primary structure but secondary and tertiary structure undergo modification. The albumin of egg white is converted into water-insoluble fibrous protein.
  • Similarly, the conversion of milk into curd is also an example of protein denaturation where lactic acid secreted by bacteria is responsible for this process.
  • During the preparation of cottage cheese by heating milk with lemon juice or lactic acid, globular milk protein lactalbumin is converted into fibrous protein. Here also the protein gets denatured.

Biomolecules Denaturation Of Proteins

  • Protein denaturation may be reversible or irreversible.
  • Coagulation of egg white on heating is an irreversible process.
  • Some denatured proteins return to their original secondary and tertiary structures when provided with previous pH and temperature and hence, become metabolically active.
  • This phenomenon is observed among enzymes and is called renaturation (the opposite process of denaturation).

Biological Functions Of Proteins

  1. Structural Materials: The primary building block of any tissue is protein. Half of the total proteins in the human body act as structural components, for example., skin, hair, nails, wool, and feathers containing keratin (fibrous protein), collagen in tendons (a group of muscle fibres), myosin in muscles, fibroin in silk, etc.
  2. Transport Agent: Some proteins act as transport agents or carriers which help in the transport of oxygen, metals, fatty acids and hormones, for example., haemoglobin in human blood carries oxygen from the lungs to different parts of the body. JMJ Enzymes: Some proteins act as enzymes and catalyse biochemical reactions.
  3. Metabolic Regulators: Some globular proteins control the metabolism of a body, for example., insulin secreted from the pancreas controls blood glucose level, thyroglobulin, a glycoprotein secreted from the thyroid gland synthesises the amine hormone thyroxine, fibrinogen in the blood plasma converts into insoluble protein fibrin to clot blood.
  4. Antibodies: It defends the body against pathogens. Certain toxins called antigens are secreted by the invading infectious viruses, bacteria or external proteins.

Antibodies are synthesised to destroy these antigens and thus prevent the body from diseases. Gamma-globulin is an antibody present in blood plasma.

Enzymes And Hormones

Enzymes

Enzymes are biological catalysts. The activation energy of a biochemical reaction is decreased in the presence of an enzyme and so the reaction occurs rapidly and easily. All enzymes are chemically protein in nature and are synthesised in living systems.

Nomenclature Of Enzymes

  1. The name of an enzyme is derived by adding the suffix ‘ase’ to the name of the substrate on which that particular enzyme acts.
    1. Nomenclature Of Enzymes Example: The hydrolytic enzyme of maltose is maltase, that of arginine is arginase, cellulose is cellulose.
  2. As per the IUPAC convention, the enzyme can be named depending on the nature of the chemical reaction it catalyses.
    1. Nomenclature Of Enzymes Example: Oxidoreductase for redox reactions, hydrolase for hydrolytic reactions, transferase for reactions involving functional group transfer, and isomerase in case of isomerisation reactions to exhibit catalytic activity.

Organic Chemistry Biomolecules

Composition Of Enzymes

Enzymes Are Globular Proteins: Some enzymes act as catalysts while some others combine with a non-protein component called a co-factor to exhibit catalytic activity. Co-factors are of 2 types:

Inorganic ions: Zn2+, Mg2+, Mn2+, Fe2+, Cu2+, Co2+ Mo3+, Na+, K+, etc.

Composition Of Enzymes Organic Molecules

  1. Some organic molecules, called coenzymes remain loosely attached to the enzyme and can be easily separated by dialysis. For example, pyridoxal phosphate (PLP) is a coenzyme originated from pyridoxin or vitamin B6.
  2. Some organic molecules called prosthetic groups remain strongly attached to the enzyme by covalent bonds and can be separated only by hydrolysis. The enzyme-cofactor complex is called holoenzyme and the remaining inactive enzyme after the cofactor is excluded is called apoenzyme.

Important Characteristics Of Enzymes

  1. Enzymes accelerate the rate of a biochemical reaction. It has been observed that biochemical reactions in the presence of enzymes occur approximately 106 times faster than the usual rate (in the absence of enzymes).
  2. A small amount (~ 10-5 mol) of enzyme is required for catalysis.
  3. A specific enzyme can catalyse a specific reaction only.
  4. Enzymes remain active at an optimum temperature of 37 °C and an optimum pH of 7.0
  5. Certain organic and inorganic molecules can act as enzyme inhibitors and retard enzyme activity.
  6. Several enzymes show stereospecificity. They react with or produce a particular stereoisomer only, for example., stearoyl-CoA 9 -desaturase converts a derivative of stearic acid into the cis-isomer of a derivative of oleic acid (unsaturated fatty acid).

Some Important Enzymes And Their Catalytic Behaviour:

Biomolecules Some Important Enzymes And Their Catalytic Behaviour

Applications Of Enzymes

Production Of Essential Substances:

  1. Invertase, zymase and maltase are used in the alcoholic beverage industry.
  2. The enzyme is used in making paneer.
  3. α-amylase is used in the preparation of sweet syrup from starch.
  4. Invertase enzyme is used to prepare invert sugar.

Cardiological Treatment: The enzyme streptokinase can decoagulate clotted blood. This enzyme can be used to minimise heart attacks caused due to blood clotting in the coronary artery.

Diseases Related To Enzyme Deficiency: Deficiency of enzymes leads to certain diseases.

  1. Lack of phenylalanine hydroxylase enzyme (that converts phenylalanine into tyrosine) causes phenylketonuria (PKU). In the absence of this enzyme, phenylalanine is converted into phenylpyruvate by another enzyme. The presence of both phenylalanine and phenylpyruvate damages the brain severely and causes mental retardation.
  2. The disease albinism occurs due to a deficiency of the enzyme tyrosinase. Its absence causes less amount of melanin (pigment giving colour to skin and hair) secretion. The skin and hair colour of the affected human beings and animals, as a result, turn white.

Hormones

Hormones are secreted from endocrine glands, which are transported to different organs and tissues by blood to control metabolic reactions. These are also called chemical messengers because they send messages from one cell to another.

Classification Of Hormones:

Biomolecules Classification Of Hormones

Organic Chemistry Biomolecules

Based on chemical structures, hormones can be divided into three classes.

  1. Steroids,
  2. Proteins or polypeptides and
  3. Amines. Steroid hormones can be further subdivided into two sub-classes
  1. Sex hormones and
  2. Adrenocortical hormones. Sex hormones are of two types—male sex hormones and female sex hormones

Structures Of Few Important Hormones

Biomolecules Structures Of Few Important Hormones

Sources, Chemical Nature And Functions Of Some Important Hormones:

Biomolecules Sources Chemical Nature And Functions Of Some Important Hormones

Lipids And Vitamins

Lipids are the compounds mainly formed of carbon, hydrogen and oxygen. Sometimes sulphur, phosphorus and nitrogen are also present in the lipids. A large part of a lipid consists of hydrocarbons and hence, lipids are soluble in non-polar organic solvents.

Organic Chemistry Biomolecules

Lipids And Its Classification

Lipids And Its Classification Definition: Certain organic compounds which are water-insoluble (like oils, fats, steroids, and terpenes) but soluble in non-polar organic solvents like chloroform, ether, carbon tetrachloride and benzene are called lipids.

Lipids And Its Classification Lipids are of two types:

  1. Hydrolysable lipid and
  2. Nonhydrolysable lipid.

Biomolecules Hydrolysable Lipid And Non Hydrolysable Lipid

Hydrolysable Lipids: The lipids which can be hydrolysed into smaller molecules are called hydrolysable lipids. These lipids mostly contain an ester group. They are subdivided into three classes:

  1. Waxes,
  2. Fats or oils or triglycerides and
  3. Phospholipids. Hydrolysable lipids are also called complex lipids.

Hydrolysable Lipids Waxes: These are the simplest hydrolysable lipids. Esters (RCOOR7) produced from alcohol (R’H) and fatty acid (RCOOH) of high molecular mass are called waxes.

Hydrolysable Lipids Importance Of Wax: The long hydrocarbon chains make waxes hydrophobic.

  • They form a protective coating over the feathers of birds and prevent them from getting wet.
  • The waxy layer over the leaves of trees prevents excess water loss or absorption. Spermaceti (a wax) present on the head of whales helps to control buoyancy during deep sea diving.

Hydrolysable Lipids Example: Spermaceti [CH3(CH2)15OCO(CH2)14CH3] and cetyl palmitate are wax. Bee wax is myristyl palmitate, [CH3(CH2)14COO(CH2)29CH3].

Hydrolysable Lipids Triglycerides: The most common lipids are triglycerols or triglycerides. These are a kind of triesters whose hydrolysis yields one mole of glycerol and three moles of fatty acids.

Biomolecules Triglycerides

Organic Chemistry Biomolecules

Oils and fats are triglycerols. Triglycerols which are solids at room temperature are known as fats and those which remain liquid at room temperature are called oils.

Triglycerides Example:

Biomolecules Glyceryl Tristearate

Importance Of Triglycerides: The principal role of triglycerides in cells is to store energy. Complete catabolism of triglycerols into CO2 and H2O produces a huge amount of energy.

  • Carbohydrates yield energy for a short period, while triglycerols provide energy for a longer period.
  • Triglycerols store ~9 kcal-g-1 energy while carbohydrates store only -4 kcal-g-1 energy.
  • Theoretically, triglycerols can be used as fuel in automobiles because their combustion produces a lot of heat. Nowadays, plant-based oils are mixed with diesel to use as fuels.

Phospholipids: Hydrolysable lipids containing phosphate groups are called phospholipids. The common phospholipids are phosphoryl glycerol or phosphoglycerides.

  • Substitution of a fatty acid group in a triglycerol by a phosphate group produces a phospholipid.
  • The simplest class of phosphoglycerides is phosphatidic acid. In some phospholipids, an extra alcoholic group may form an ester with phosphoric acid. Cephalins and lecithin belong to this class.

Biomolecules Phosphatidic Acid

Triglycerides Importance Of Phospholipids: Phospholipids possess a polar head and two non-polar tails. These form a lipid bilayer in aqueous solutions as found in cell membranes.

  • The polar groups lie towards the inner face (cytoplasm) and outer face (extracellular fluid) and the non-polar groups lie in between the two heads.
  • These non-polar tails prevent the watery part of the cytoplasm from leaking out and thus, protect the cell.

Non-Hydrolysable Lipids

The lipids which cannot be hydrolysed into smaller molecules are called non-hydrolysable lipids. These can be subdivided into four classes:

  1. Lipid-soluble vitamins,
  2. Eicosanoids,
  3. Terpenes and
  4. Steroids. Non- hydrolysable lipids are also called simple lipids.

Fat-Soluble Vitamins: Vitamins A, D, E and K are fat-soluble vitamins. Vitamin A present in the human body is converted into light-sensitive compound 11-cis-retanol, which helps in eyesight.

  • Vitamin D controls the metabolism of phosphorus and calcium. Vitamin E, being an antioxidant, prevents oxidation of unsaturated fatty acid chains.
  • Vitamin K controls the synthesis of prothrombin and certain proteins needed for blood clotting.

Non-Hydrolysable Lipids Eicosanoids: These are a group of organic molecules formed of 20 carbon atoms and are derived from arachidonic acid. Eicosanoids are of four types—

  1. Prostaglandins,
  2. Leukotrienes,
  3. Thromboxanes and
  4. Prostacyclins.

Eicosanoids Importance Of Eicosanoids: These are present in low concentration in cells and are localised in activity, i.e., act near the site of synthesis.

  • Each eicosanoid has its specific biological activity. Prostaglandins reduce blood pressure, prevent blood platelet coagulation, control inflammation and decrease digestive juice secretion.
  • Prostacyclins dilate blood vessels and prevent blood platelet coagulation. Thromboxanes constrict blood vessels and accelerate blood coagulation and leukotrienes contract smooth muscles of lungs.

Non-Hydrolysable Lipids Terpenes: Terpenes are a group of lipids which are formed of 5-C isoprene units.

Biomolecules Isoprene Unit

These are closed or open-chain hydrocarbons, alcohols, aldehydes or ketones. Oxygen-containing terpenes are called terpenoids. Terpene is a major component of essential oils extracted by distillation of plant extracts.

Terpenes Examples: Myrcene and menthol.

Biomolecules Myrcene And Menthol

Non-Hydrolysable Lipids Importance Of Terpenes: Triterpene (six isoprene units) and tetraterpene (eight isoprene units) play important biological roles, for example., triterpene squalene is a precursor of steroid molecules.

Tetraterpenes are found in the form of carotenoids such as lycopene in tomatoes and β-carotene in carrots. Vitamin A is synthesised inside the human body from β-carotene.

Non-Hydrolysable Lipids Steroids: Biomolecules with perhydro-1, 2-cyclopentanophenan threne system are known as steroids.

Biomolecules Steroids

Steroids Example: Cholesterol, aldosterone, testosterone.

Non-Hydrolysable Lipids Importance Of Steroids: The steroid cholesterol is found abundantly in the human body. It is an essential component of the cell membrane.

  • It is also a precursor of Vitamin D and other steroids. Gall gallstone formed in the gall bladder is primarily cholesterol.
  • Testosterone and estradiol are the most effective natural male and female sex hormones respectively.

Vitamins

Vitamins do not provide energy or act as building blocks of tissues but play a major role in the maintenance of good health. Their deficiency leads to diseases related to nutritional deficiency.

Organic Chemistry Biomolecules

The human body can synthesise vitamin A, some members of vitamin B and vitamin K. Plants can synthesise all vitamins.

Vitamins Definition: A special class of biomolecules other than carbohydrates, proteins and lipids, of which most are not synthesised inside the human body and are therefore, taken through diet in small quantities for proper metabolism, growth and nourishment are called vitamins.

Vitamins Sources: The foods are the main source of vitamins— milk, butter, paneer, fruits, green vegetables, meat, fish, eggs, etc. Vitamins can be synthesised in the laboratory and are available in the form of tablets or capsules in the market.

Classification Of Vitamins: Vitamins are complex organic molecules. Different vitamins possess different structures. These are designated by English alphabets like A, B, C, D, E, and K. Different sub-groups of some vitamins are also designated as B1 B2, B6 and B12. Around 25 vitamins are known to date. These can be classified into two classes as follows:

  1. Water-soluble Vitamins: These are vitamin B-complex (B1, B2, B5, i.e., nicotinic acid, B6, B12, pantothenic acid) and vitamin C. These water-soluble vitamins should be taken through our diet daily as these are excreted out in the form of urine and are not stored inside the body (except vitamin B12).
  2. Fat-soluble Vitamins: These are oily substances which are soluble in lipids but not in water, for example., vitamins A, D, E, and K. Excess intake of vitamins A and D is harmful and may cause hypervitaminosis. Biotin (vitamin H) is insoluble in both water and lipids. Deficiency of one or more vitamins leads to avitaminosis. It is often noticed in human beings.

Some Essential Vitamins, Their Features, Sources And Deficiency Diseases:

Biomolecules Some Essential Vitamins Their Features Sources And Deficiency Diseases

Differences Between Hormones And Vitamins:

Biomolecules Difference Between Homones And Vitamins

Nucleic Acids

Biomolecules present in chromosomes of the cell nucleus other than protein is called nucleic acid.

Nucleic Acid Definition: The complex organic polymers formed of nucleotide monomers located in the nucleus and cytoplasm which carry the essential hereditary traits from one generation to another and take part in protein synthesis are called nucleic acid.

Types Of Nucleic Acids And Their Components

Nucleic acids are of two types:

  1. Deoxyribonucleic acid (DNA) and
  2. Ribonucleic acid (RNA). Both DNA and RNA are organic polymers. These are the polymers of nucleotides.

Biomolecules Deoxyribonucleic Acid

Structural Components Of Nucleotides

Three types of compounds combine to form one nucleotide—

  1. A heterocyclic nitrogenous base,
  2. A pentose sugar and
  3. A phosphate group.

Heterocyclic Nitrogenous Base

These are of two types—

  1. Purine derivatives and
  2. Pyrimidine derivatives.

Biomolecules Purine Derivatives And Pyrimidine Derivatives

  1. Bases of DNA: Adenine, Guanine, Cytosine and Thymine
  2. Base of RNA: Adenine, Guanine, Cytosine and Uracil

Heterocyclic Nitrogenous Base Pentose Sugar: The sugar unit of DNA is deoxyribose and that of RNA is ribose. C-atoms ofpentose sugar are distinguished from the C-atoms of nitrogenous bases by marking them with numbers like 1 2′, and 3′ (read as one-prime, two-prime and so on).

Biomolecules The Sugarunit Of DNA Is Deoxyribose And That Of RNA Is Ribose

Heterocyclic Nitrogenous Base Phosphate Group: Both DNA and RNA contain a phosphate group. O -atom of a hydroxyl group of pentose sugar is linked to a phosphate group.

Nucleoside And Nucleotide

Nucleoside And Nucleotide Nucleotide: A nitrogenous heterocyclic base (A, G, C, U, T) and a pentose sugar (ribose or deoxyribose) combine to form a compound called nucleoside.

  • Addition of a phosphate group to a nucleoside produces a nucleotide.
  • Nucleosides containing ribose and deoxyribose sugars are called ribonucleoside and deoxyribonucleoside respectively. C-1′ of sugar is bonded to the N-atom of the base in both nucleosides.

Biomolecules Ribonucleoside And Deoxyribonucleoside Respectively

Nucleosides In RNA:

Biomolecules Nucleosides In RNA

Organic Chemistry Biomolecules

Nucleosides In DNA:

Biomolecules Nucleosides In DNA

Nucleoside And Nucleotide Nucleotide: A compound containing a heterocyclic nitrogenous base (A, G, C, U, T), a pentose sugar (ribose or deoxyribose) and a phosphate group is called nucleotide. The phosphate group is linked to the C-5′ of sugar.

Nucleotide

= 1 pentose sugar + 1 heterocyclic base + 1 phosphate group

= 1 nucleoside +1 phosphate group

Biomolecules Deoxyribonucleotide And Ribonomeric Unit Of RNA

Di, tri-, tetra- and polynucleotides have 2, 3, 4, or several nucleotides attached. A phosphodiester bond is present between 5′-C of a sugar and 3′-C of another sugar.

Biomolecules Formation Of Dinucleotide

Biomolecules A Simplified Version Of Nucleic Acid Chain Is Shown Below

Organic Chemistry Biomolecules

  • Every nucleotide contains C-1′ of pentose sugar Jinked to a nitrogenous base and C-3’to a phosphate group. C-S’of pentose sugar is linked to a —OH group which reacts with the phosphate group of the next nucleotide to form a phosphodiester bond. This bond joins two nucleotides.
  • Hence, a free phosphate group at 5′ C of one nucleotide terminus and a free —OH group at 3’C of the other nucleotide terminus is found in a chain of DNA or RNA.
  • In every polynucleotide chain, a 5′ -terminal bears a free phosphate radical & a 3′ – terminal bears a free —OH group. So, a polynucleotide chain is marked as 5′- 3′.

RNA [Ribonucleotide]:

Biomolecules RNA Ribonucleotide

DNA [Deoxyribonucleotide]:

Biomolecules DNA Deoxyribonucleotide

Hydrolytic Products Of DNA And RNA:

Biomolecules Hydrolytic Products Of DNA And RNA

Organic Chemistry Biomolecules

Differences Between Nucleotide And Nucleoside:

Biomolecules Differences Between Nucleotide And Nucleoside

Biomolecules Structures Of ATP And ADP And AMP

Structures Of Nucleic Acids

The structure of nucleic acids can be discussed under the following divisions of

  1. Primary structure and
  2. Secondary structure.

Structure Of Nucleic Acid Primary Structure: The nucleotide sequence in a nucleic acid molecule is the primary structure of that nucleic acid.

  • Nucleic acid refers to a polynucleotide chain which is formed of several pentose sugar units joining 5′-C and 3′-C by phosphodiester bonds. Hence, the sugar and phosphate units are arranged sequentially in a nucleic acid.
  • Any four nitrogenous bases (A, G, C and T or A, G, C and U) remain linked consecutively with four sugar units.
  • The particular sequence of four nitrogenous bases in the sugar—phosphate framework of a nucleic acid refers to the primary structure of that nucleic acid.

Biomolecules A Part Of DNA Polynucleotide Strand

Biomolecules Structure Of DNA.

Secondary Structure Of DNA: Eminent scientists James Watson and Francis Crick (1953) proposed the Watson-Crick model on the physical structure of deoxyribonucleic acid (DNA) developed by X-ray diffraction analysis. From this analysis, it is found that—

  1. Every DNA molecule is formed of two polynucleotide chains.
  2. The two chains rotate clockwise to form a double helix.
  3. The two polynucleotide chains are joined anti-parallelly as one chain runs in the 5′-3′ direction while the other in 3 – 5.
  4. A polynucleotide chain has a sugar-phosphate backbone whose nitrogenous bases are projected inwards perpendicularly in a particular sequence.
  5. The pyrimidine base of one chain is linked to the purine base of another chain by hydrogen bonding. The distance between two consecutive nitrogenous bases in a DNA strand is 3.4 A.
  6. The diameter of the double-stranded DNA helix is 20A. A complete spiral strand of a DNA helix consists of 10 complementary base-pairs and it is 34A long.
  7. In the double-helical DNA molecule, the nitrogenous base sequence in one chain gives a specific idea about the complementary base sequence in the other chain. The base A is linked by two hydrogen bonds to T and G is linked by three hydrogen bonds to C.

As per Chargaff’s rule, the sum of the purines is equal to the sum of pyrimidines, i.e., A + G = T + C is applicable to all polynucleotide strands.

Biomolecules Polynucleotide Strands

Secondary Structure Of DNA Functions Of DNA

  1. It is the chemical component of genes that carries biological information. It controls the transfer of hereditary traits between generations and its expression in an individual.
  2. DNA transfers the characteristics from parent to daughter cell by replication.
  3. The genetic information stored is expressed by RNA synthesis and polypeptide synthesis.

Different Types Of RNA

RNA is a single-stranded linear or helical polymer of ribo¬nucleotide. The pentose sugar is ribose & nitrogenous bases are adenine, guanine, cytosine and uracil. RNA are of 3 types:

  1. mRNA (messenger RNA): It is synthesised from DNA. It carries genetic information from DNA to cytoplasm and takes part in protein synthesis.
  2. rRNA (ribosomal RNA): It helps in the attachment of m-RNA to ribosomes, thus forming a polysome. This polysome acts as a template for the synthesis of different proteins.
  3. tRNA (transfer RNA): It is a relatively small molecule. f-RNA sends a particular amino acid to the specified location to construct the correct sequence of amino acids during protein synthesis.

Organic Chemistry Biomolecules

Biological Role Of Nucleic Acids

Two important roles of nucleic adds are—

  1. Replication and
  2. Protein synthesis

Replication: The process of formation of two identical daughter DNA molecules from parent DNA molecule using enzymes and proteins is called replication.

Procedure: At first, uncoiling of supercoiled DNA occurs with the help of suitable enzymes.

  • The H-bonds between the DNA strands are broken, and each strand acts as a template strand.
  • By following this template DNA and with the help of base-pairing, complementary deoxyribonucleotides are added to each strand.
  • In this way, two daughter strands are formed.
  • Each DNA molecule consists of a parent strand (original strand) and its complementary daughter strand (newly synthesised strand).
  • So, this process is called semi-conservative replication. It takes place in 5,-3/ direction only

Synthesis Of Protein: Different kinds of RNA molecules (m-RNA, r-RNA and t-RNA) take part in protein synthesis. The central dogma of molecular biology describes the protein synthesis in two steps—

  1. Transcription and
  2. Translation by which the information in genes flows into proteins ((DNA—>RNA—^Protein). The biological information required for the synthesis of a particular protein is obtained from a DNA molecule only.

Biomolecules Central Dogma Synthesis Of Proteins

Differences Between DNA and RNA:

Biomolecules Differences Between DNA And RNA

DNA Fingerprint

The fingerprint of every human being is unique, i.e., no two individuals bear similar fingerprints. Thus, it is in vogue to identify a person by his or her fingerprint. However, there is a possibility of modifying the skin of the fingertip by surgery.

Organic Chemistry Biomolecules

  • Hence, nowadays DNA testing is a much more surer process for the identification of a person.
  • DNA fingerprint refers to the specific nucleotide sequence present in the double-stranded DNA of an individual. Every cell of a particular individual has an identical nucleotide sequence in its DNA. One cannot modify this sequence voluntarily.

Uses of DNA-fingerprint

  1. To identify criminals by forensic testing.
  2. Paternity test, i.e., to identify the father of an individual.
  3. To identify a dead person by comparing his or her DNA with his or her parents or children.
  4. To determine a racial group.

Class 12 Chemistry Unit 14 Biomolecules Very Short Questions And Answers

Question 1. Which polysaccharide is stored in animal liver?
Answer: Glycogen,

Question 2. What kind of ring is formed by fructose in sucrose?
Answer: In the form of a 5-membered fructofuranose ring.

Question 3. State the configuration of natural a -amino acid.
Answer: L-configuration.

Question 4. Name two essential amino acids.
Answer: Valine and phenylalanine.

Question 5. State nature of glycosidic bonds in starch and cellulose.
Answer: α-glycosidic bond in starch and β-glycosidic bond in cellulose.

Question 6. Which enzyme is used in the treatment of heart disease?
Answer: Streptokinase

Question 7. Which solvent allows smooth mutarotation of glucose?

  1. Cresol
  2. Pyridine
  3. Cresol + pyridine

Answer: Cresol + pyridine.

Question 8. Which bond helps to stabilise a -helix?
Answer: Intermolecular hydrogen bond.

Question 9. State the pH at which the solubility of amino acids is the least.
Answer: At an isoelectric point, the solubility of amino acids is the least

Question 10. Which phenomenon does the coagulation of egg white refer to?
Answer: Denaturation of protein.

Question 11. State the possible formula of a tripeptide which on hydrolysis produces glycine alanine and valine.
Answer: The tripeptide may be either of the following: Gly-Ala-Val, Gly-Val-Ala, Ala-Gly-Val, Ala-Val-Gly, Val-Ala-Gly, Val-Gly-Ala.

Question 12. State the name and structural formula of bee wax.
Answer: Myrisyl palmitate [CH3(Ch2)14COO(CH2)29CHg3]

Question 13. How many 5-membered rings and 6-membered rings are found in steroids?
Answer: One 5 -membered ring and three 6 -membered rings.

Question 14. Which steroid is abundant in living bodies?
Answer: Cholesterol.

Question 15. Name the components of essential oil.
Answer: Monoterpene (10-C terpene), sesquiterpene (15-C terpene).

Organic Chemistry Biomolecules

Question 16. Name the two classes of nitrogenous bases in nucleic acid.
Answer: Purine and pyrimidine.

Question 17. Name the two pentose sugars present in DNA and RNA.
Answer: β-D-2-deoxyribose in DNA and β-D-ribose in RNA.

Question 18. Define invert sugar.
Answer: The equimolar mixture of glucose and fructose was obtained due to sucrose hydrolysis.

Question 19. Give an example of an amino acid containing 2′ amine.
Answer: ProlineBiomolecules Proline

Question 20. Why do enzymes accelerate biochemical reactions?
Answer: The activation energy of enzyme-catalysed biochemical reactions is comparatively smaller than that of reactions taking place without enzymes.

Question 21. How many isoprene units are present in β-carotene? What kind of terpene is it?
Answer: β-carotene has 8 isoprene units. It is a tetraterpene.

Organic Chemistry Biomolecules

Question 22. What is the structural feature characterising reducing sugars?
Answer: Reducing sugars contain free aldehydic or ketonic groups.

Question 23. Mention the names of two carbohydrates which act as biofuels.
Answer: Glycogen in animals and starch in plants act as biofuels.

Question 24. Name the enzymes present in the saliva of human beings and mention their role.
Answer: The saliva of human beings contains amylase as the enzyme. Amylase first hydrolyses starch to maltose and then to glucose.

Question 25. Why is a -helix named as 3.613 helix?
Answer: Since each turn of the α-helix has approximately 3.6 amino acids and the hydrogen bonding leads to the formation of a 13-membered ring, the α-helix is termed as 3.613 helix.

Question 26. Mention the amino acids which exhibit aromatic properties.
Answer: Phenylalanine, tryptophan, tyrosine and histidine exhibit aromatic properties.

Question 27. Indicate the total number of basic groups in the following form of lysine:

Biomolecules Lysine Contains Two Basic Groups

Answer: Lysine contains two basic groups. These are —COO and —NH2

Question 28. Name two α-amino acids which form a dipeptide, whose methyl ester is 100 times more sweet than cane sugar.
Answer: Aspartic acid and phenylalanine, [aspartame is an artificial sweetener 100 times sweeter than cane sugar. It is the methyl ester of the dipeptide derived from aspartic acid and phenylalanine]

Organic Chemistry Biomolecules

Question 29. Define native state concerning protein.
Answer: The native state of a protein is its properly folded and/or assembled form, which is operative and functional. They are unaltered by denaturating agents such as heat, the action of enzymes or chemicals.

Question 30. Name the enzyme that breaks large proteins into small peptides.
Answer: The enzyme pepsin plays an important role in the digestion of proteins by causing the fission of the large amino acid chains to peptides, which are short chains of 4-9 amino acids.

Question 31. What happens when L-glucose is treated with Tollens’ reagent?
Answer: L-glucose is oxidised to the corresponding gluconate ion.

Biomolecules L Glucose Is Oxidised To The Corresponding Gluconate Ion

Question 32. What change in free energy occurs during the conversion of ATP into ADP? Which bonds link the phosphoric acid molecules together in ATP?
Answer:

  1. There is a decrease in free energy during the conversion of ATP into ADP (AG = -ve)
  2. Phosphoric anhydride bonds Biomolecules Phosphoric Anhydride Bonds link the phosphoric acid molecules in ATP.

Question 33. Why cannot vitamin C be stored in our bodies?
Answer: As vitamin C is water-soluble, therefore, it is readily excreted through urine and hence, cannot be stored in our body.

Organic Chemistry Biomolecules

Question 34. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer: The hydrolysis products are: thymine, 2-deoxy-D-ribose and phosphoric acid.

Question 35. Amino acids show amphoteric behaviour. Why?
Answer: Amino acids are amphoteric as basic (—NH2) and acidic (—COOH) both functional groups are present,

Question 36. Which disaccharides do not exhibit mutorotation? give an example.
Answer: Disaccharides, which contain glycosidic bonds between the anomeric carbon atoms of the constituent monosaccharide units, do not exhibit mutarotation. Example: Sucrose.

Question 37. Write the products of hydrolysis of lactose.
Answer: D-glucose and D-galactose.

Question 38. What type of bonding helps stabilise the α-helix structure of proteins?
Answer: The α-helix structure of proteins is stabilised by intramolecular H-bonding between C—O of one amino acid residue and the N—H of the fourth amino acid residue within a single chain.

Organic Chemistry Biomolecules

Question 39. Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called?
Answer: Lactose is the sugar present in milk. It contains two monosaccharide units. Such oligosaccharides are known as disaccharides;

Question 40. In nucleoside, a base is attached at 1′ position of the sugar moiety. Nucleotide is formed by linking of phosphoric acid unit to the sugar unit of nucleoside. At which position of the sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide?
Answer: Phosphoric acid unit is linked to C5 of sugar unit of nucleoside to give a nucleotide;

Question 41. Name the linkage connecting monosaccharide units in polysaccharides.
Answer: Glycosidic linkage;

Question 42. Under what conditions glucose is converted to gluconic and saccharic acid?
Answer: Glucose is oxidised to gluconic acid by bromine water (Br2/H2O) and to saccharic acid by the cone. HNO3;

Organic Chemistry Biomolecules

Question 43. Monosaccharides contain a carbonyl group and hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule is also considered for classification. In which class of monosaccharide will you place fructose?
Answer: Fructose is a monosaccharide containing six carbon atoms including the keto group. So it is called ketohexose;

Question 44. Vitamin B complex is a combination of several other vitamins. Name the constituent vitamins.
Answer: Vitamin B is a complex vitamin and consists of vitamins B1, B2, B6, B12, biotin, folic acid, pantothenic acid and nicotinic acid.

Question 45. How do enzymes effectively help a substrate to be attacked by the reagent?
Answer: The active sites of enzymes hold the substrate molecule in a suitable position so that the substrate can be attacked by the reagent effectively;

Question 46. Why are naturally occurring glucose and fructose called dextrose and laevulose respectively?
Answer: The aqueous solution of naturally occurring glucose is dextrorotatory while that of fructose is laevorotatory. Hence, glucose and fructose are known as dextrose and laevulose respectively;

Question 47. Explain the role of osazone structure in identifying monosaccharides.
Answer: The Osazone product of a particular monosaccharide has a well-defined melting point which is useful for identifying monosaccharides;

Question 48. Name the enantiomer of α-D-(+) -glucose.
Answer: α-L(-)- glucose

Organic Chemistry Biomolecules

Question 49. Which amino acid is not optically active?
Answer: Glycine

Question 50. What idea can be obtained about the structure of monosaccharides from mutarotation?
Answer: The idea of a closed ring structure of monosaccharides is obtained from mutarotation;

Question 51. Which two compounds combine to produce inverted sugar?
Answer: Glucose and fructose

Question 52. Name the disaccharide present in milk.
Answer: Lactose;

Question 53. How many molecules of phenylhydrazine react with glucose to produce osazone?
Answer: 3

Question 54. Name the closed ring structure in glucose.
Answer: Pyranose

Organic Chemistry Biomolecules

Question 55. Name a sulphur-containing amino acid.
Answer: Cystine

Question 56. Name the hydrolytic product of cellulose.
Answer: D-glucose;

Question 57. Why human beings cannot digest cellulose?
Answer: The human intestine cannot synthesise cellulase which can hydrolyse cellulose into glucose;

Question 58. Which vitamin deficiency causes pernicious anaemia?
Answer: B12

Question 59. Which enzyme decoagulates clotted blood?
Answer: Streptokinase;

Question 60. How many a-amino acid molecules are found in one insulin molecule?
Answer: 51

Question 61. State the role of insulin in the human body.
Answer: Decreases blood glucose level

Question 62. Which enzyme converts glucose into ethanol?
Answer: Zymase;

Class 12 Chemistry Unit 14 Biomolecules Short Questions And Answers

Question 1. α-amino monocarboxylic acids have two pka values. Explain.
Answer: One pka value arises from the dissociation of R—CH(NH3)COOH and the second value arises from the dissociation of R—CH(NH2)—COOH.

Biomolecules Alpha Amino Monocarboxylic Acids

Organic Chemistry Biomolecules

Question 2. What is the reason for the specific action of enzymes?
Answer: Each enzyme is specific for a particular reaction.

  • This is because enzymes contain active sites of definite shape and size on their surfaces so that only specific substrates can fit into these active sites (just like a key fits into a lock).
  • Such specific binding leads to the formation of enzyme-substrate complex which accounts for specific action of enzymes.

Question 3. Indicate the nature of changes that occur when egg protein is boiled.
Answer: Egg contains globular proteins. On boiling such proteins undergo coagulation irreversibly to form fibrous proteins.

  • In other words, proteins lose their biological activity and thus get denatured.
  • During denaturation, the secondary and tertiary structures of proteins are destroyed without affecting the primary structure.

Question 4. How will you prove that— The carbonyl group is present at the C-2 atom of a fructose molecule?

Answer: Hydrolysis of fructose cyanohydrin (which is formed due to the reaction between fructose and HCN) yields an acid. The acid is reduced by HI/P to form 2-methylhexanoic acid. This proves the presence of the carbonyl group at the C-2 atom of fructose.

Biomolecules Fructose And 2 Methylhexanoic Acid

Question 5. What is the monomer unit of protein? give one example of such a monomer which contains sulphur. Write its zwitterionic form.
Answer: The monomer unit of proteins is α-amino acid. An α-amino acid containing sulphur is cysteine.

Biomolecules Cysteine And Zwitterionic

Question 6. Identify:

  1. The vitamin responsible for blood coagulation,
  2. The vitamin is not stored inside the human body.
  3. The vitamin whose deficiency causes scurvy
  4. The vitamin whose deficiency causes beri¬beri;
  5. The vitamin included under lipid;
  6. The enzyme that causes hydrolysis of cellulose
  7. The enzyme that can reduce chances of heart attack;
  8. The enzyme used in the preparation of inverted sugar;
  9. The group B vitamin is synthesised in the human body.

Answer:

  1. Vitamin K;
  2. Vitamin C;
  3. Vitamin C;
  4. Vitamin B1
  5. Vitamin A, D, E and K;
  6. Cellulase;
  7. Streptokinase;
  8. Invertase;
  9. Vitamin B12

Question 7. If the base sequence of one strand of DNA double- • stranded molecule is ‘ATCGTCCA’, state the complementary base sequence.
Answer: Adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C) by hydrogen bonds in a DNA molecule. Therefore, the complementary base sequence will be ‘TAGCAGGT’ corresponding to ‘ATCGTCCA’.

Organic Chemistry Biomolecules

Question 8. Indicate the nature of linkages responsible for the formation of

  1. Cross-linking of polypeptide chains
  2. β-sheet structure.

Answer:

  1. Cross-linking of polypeptide chains is due to hydrogen bonds or disulphide bonds.
  2. β-sheet structure is due to hydrogen bonds betweenBiomolecules Beta Sheet Structure Is Due To Hydrogen Bonds groups belonging to different
    polypeptide chains.

Question 9. What is essentially the difference between α-form and β-form of D-glucose?
Answer: α-and β-forms of D-(+)-glucose differ only in the configuration of Cj (Le„ the configuration of anomeric carbon). In the α-form, C1 —OH and C2—OH are on the same side, while in the β-form, C1 —OH and C2—OH are on the opposite sides.

Question 10. Fructose contains a keto group but still reduces Tollens’ reagent. Explain.
Answer: Under the basic conditions of Tollens’ reagent (ammoniacal silver nitrate solution), fructose undergoes Lobry de Bruyn van Ekenstein rearrangement to form an equilibrium mixture of fructose, glucose and mannose. Since both glucose and mannose contain the free—CHO group, they reduce Tollens’ reagent.

Question 11. Glycine exists as a zwitterion but o-and paminobenzoic acids do not— Give reason.
Answer: α-and β-aminobenzoic acids are resonance hybrids. In such resonance, the lone pair on the amino nitrogen is involved in conjugation with the aromatic ring.

  • This makes the —NH2 group poorly basic and the —COOH group poorly acidic.
  • Therefore the weakly acidic —COOH group cannot transfer a proton to the weakly basic —NH2 group, obviously o-and p-aminobenzoic acids do not exist as zwitterions.

Biomolecules Conjugation With The Aromatic Ring

On the other hand, in glycine, no such benzene ring is present. Thus, the NH2 group is sufficiently basic to accept a proton from the acidic —COOH group to form a zwitterion.

Biomolecules Glycine And Zwitterion

Organic Chemistry Biomolecules

Question 12. A decapeptide (M-mass =796) on complete hydrolysis gives glycine, alanine and phenylalanine. Glycine contributes 47% to the total mass of the hydrolysed products. Calculate the number of glycine units present in the decapeptide.
Answer:

The molecular mass of decapeptide = 796

The molar mass of glycine (H2NCH2COOH) = 75

The number of H2O molecules that will be required for the hydrolysis of the decapeptide must be 9.

Decapeptide + 9H2O → Glycine + Alanine + Phenylalanine

Now, the total mass of all the amino acids obtained after the addition of 9 molecules of H2O = 796 + (9 ×18) = 958

∴ The total mass of glycine in the hydrolysed products

⇒ \(=\frac{958 \times 47}{100} \approx 450\)

But the molecular mass of glycine = 75

∴ No. of glycine units in the decapeptide \(=\frac{450}{75}=6\)

Question 13. Which of the following reagents converts glucose to gluconic acid—

  1. Na-Hg/H2O
  2. Br2/H2O
  3. HNO3
  4. NaBH4

Answer: 2. Br2/H2O

Organic Chemistry Biomolecules

Question 14. Write down the structure of D-glucose. How it differ from the structure of D-fructose?
Answer:

Biomolecules D Glucose And D Fructose

Question 15. In which of the following peptide bonds is present—

  1. CH3CH2CON(CH3)2
  2. H2NCH2CH2CO2C2H5
  3. C6H5CONHOC2H5
  4. H2NCH2CONHCHCO2H CH3

Answer: 4. H2NCH2CONHCHCO2H CH3

Question 16. Which of the following bases is not present in DNA—

  1. Uracil
  2. Thymine
  3. Guanine
  4. Cytosine

Answer: 1. Uracil

Question 17. After watching a programme on TV about the presence of carcinogens (cancer-causing agents) potassium bromate and potassium iodate in bread and other bakery products, Ritu a class XII student decided to be aware others about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct the canteen contractor to stop selling sandwiches, pizzas, burgers, and other bakery products to the students. The principal took immediate action and instructed the canteen contractor to replace the bakery products with some protein and vitamin-rich food like fruits, salads, sprouts etc. The decision was welcomed by the parents and students. After reading the above passage, answer the following questions.

  1. What are the values (at least two) displayed by Ritu?
  2. Which polysaccharide component of carbohydrates is commonly present in bread?
  3. Write the two types of secondary structure of proteins.
  4. Give two examples of water-soluble vitamins.

Answer:

  1. Concern for students’ health
    1. Caring in nature
    2. Socially alert.
    3. Starch.
  2. α-helix and β-pleated structures.
  3. Vitamin B1, B2, B6 and C.

Organic Chemistry Biomolecules

Question 18. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer: Molecules of glucose and sucrose contain several —OH groups and hence, are capable of forming extensive H -bonds with water molecules. On the other hand, cyclohexane and benzene are hydrocarbons and hence, their molecules can not form H-bonds with H20 molecules. Thus, glucose and sucrose are soluble in water, while cyclohexane and benzene are not.

Question 19. What are the expected products of the hydrolysis of lactose?
Answer: Lactose (a disaccharide), on hydrolysis, yields a 1:1 mixture of the monosaccharides—D-(+) -glucose and D-(+) – galactose.

Biomolecules Hydrolysis Of Lactose

Question 20. How do you explain the absence of an aldehyde group in the pentaacetate of D-glucose?
Answer: In its cyclic structure, glucose does not contain any free aldehyde group because it exists in the hemiacetal form.

  • When, however, it is dissolved in water, it may undergo ring opening to generate the open chain aldehydic form, thereby indicating the presence of an aldehyde group in glucose.
  • When glucose (hemiacetal) is treated with acetic anhydride, the —OH group at C-1, along with four other — OH groups at C-2, C-3, C-4 and C-6 are involved in the formation of acetyl derivative.
  • The resulting pentaacetate (having a cyclic structure) does not contain a free — OH group at C-1, so it can not produce an open-chain aldehydic form. This indicates that glucose pentaacetate does not contain
    an aldehyde group.

Biomolecules Pentaacetate Of D Glucose

Organic Chemistry Biomolecules

Question 21. The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.
Answer: Due to the presence of both acidic and basic groups in the molecule, amino acids exist as Zwitterions having dipolar structures with salt-like character.

Biomolecules Amino Acid And Zwitterion

  • Because of such a Zwitterionic structure, molecules are held together by strong electrostatic forces of attraction. Thus, amino acids have high melting points like ionic compounds.
  • Furthermore, due to their dipolar nature, they interact strongly with H2O molecules and hence, amino acids are highly soluble in water.
  • On the other hand, corresponding halo acids cannot exist in the Zwitterionic form and hence they behave like simple carboxylic acids having relatively lower melting points and lower solubility in water.

Question 22. Where does the water present in the egg go after boiling the egg?
Answer: When an egg is boiled, the soluble globular protein albumin present in the egg white first undergoes denaturation and then coagulation to give fibrous protein. The water present in the egg gets attached to the fibrous protein molecules through hydrogen bonds.

Question 23. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer: Since there is no relationship among the quantities of different bases obtained by hydrolysis of RNA, it suggests that the base-pairing principle, i.e., adenine (A) pairs with uracil (U) and cytosine (C) pairs with guanine (G) is not followed. So RNA does not exist in the form of a double strand but it has a single strand.

Question 24. What are monosaccharides?
Answer:

  • Monosaccharides are simple carbohydrates which can not be hydrolysed to still simpler carbohydrates. Usually, they have the general formula (CH2O)n, where n = 3-7.
  • These are of two types: aldoses (which contain an aldehyde group) and ketoses (which contain a keto group). Thus, glucose (C6H12O6) is an aldose while fructose (C6H12O6) is a ketose.

Question 25. What are reducing sugars?
Answer: The carbohydrates which can reduce Fehling’s solution to give a red precipitate of Cu2O or reduce Tollens’ reagent to produce metallic silver are classified as reducing sugars.

  • All monosaccharides (both aldoses and ketoses) and disaccharides such as maltose and lactose are reducing sugars.
  • Reducing sugars contain free aldehydic or ketonic groups.

Question 26. Write two main functions of carbohydrates in plants.
Answer:

  1. Structural Material For Plant Cell Walls: Cellulose (a polysaccharide) acts as the main structural material of the plant cell walls.
  2. Reserve Food Material: Starch (a polysaccharide) is the chief reserve food material in the plants. It is stored in seeds and acts as the reserve food material for sapling till it is capable of preparing its food by photosynthesis.

Organic Chemistry Biomolecules

Question 27. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:

  1. Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose.
  2. Disaccharides: Maltose and lactose.

Question 28. What do you understand by the term glycosidic linkage?
Answer:

  • In oligosaccharides and polysaccharides, any two adjacent monosaccharide units are joined together by an ether linkage or oxide linkage formed by the loss of a molecule of water.
  • Such a linkage between two monosaccharide units through an oxygen atom is called glycosidic linkage. For example, see the structure of sucrose or maltose in the text part.

Question 29. What is glycogen? How is it different from starch?
Answer:

  • Glycogen is a branched polymer of α-D-glucose. Just as glucose is stored in plants as starch, it (glucose) is stored as glycogen in the liver and muscles of human beings.
  • When the body needs glucose during fasting or hard work, glycogen undergoes hydrolysis by the action of enzymes to provide glucose. However, starch is not a single compound, but is a mixture of two components:
  • A water-soluble component called amylose (which is a linear polymer of α-D-glucose) and
  • A water-insoluble component called amylopectin (a branched polymer of α-D-glucose). Glycogen differs from amylopectin in that the former is a more highly branched polymer of α-D-glucose.

Question 30. What are the hydrolysis products of sucrose and lactose?
Answer: Sucrose (C12H22O11, a disaccharide) undergoes hydrolysis to give a 1: 1 molar mixture of D-(+) -glucose (C6H12O6) and D-(-) -fructose (C6H12O6). On the other hand, lactose a disaccharide) undergoes hydrolysis to give a 1: 1 molar mixture of D-(+) -glucose and D-(+) -galactose.

Question 31. What is the basic structural difference between starch and cellulose?
Answer: Starch is not a single compound but is a mixture of two components:

  • Amylose (a linear polymer of α-D-glucose and
  • Amylopectin (a branched polymer of a-D-glucose ). In both these components, D-glucose units are linked through the α-glycosidic linkage between C1 of one glucose unit with C4 of the next glucose unit.
  • Cellulose is however a single compound. It is a linear polymer of β-D-glucose. C-1 of one glucose unit is connected to C-4 of another through β-glycosidic linkage.

Organic Chemistry Biomolecules

Question 32. What happens when D-glucose is treated with the following reagents?

  1. HI
  2. Bromine water
  3. HNO3

Answer: D-glucose is reduced by HI to give a mixture of n-hexane and 2-iodohexane.

Biomolecules N Hexane And 2 Iodohexane

Question 33. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:

  1. D-glucose does not form NaHSO3 addition product, aldehyde ammonia adduct, 2, 4-DNP derivative and does not respond to Schiff’s reagent test.
  2. When dissolved in an aqueous solution, both α-D-glucose and β-D-glucose exhibit the phenomenon of mutarotation.
  3. Glucose reacts with NH2OH to form an oxime but the pentaacetyl derivative of glucose does not react with NH2OH.
  4. D-glucose reacts with methanol in the presence of dry HCl gas to form two isomeric methyl-D-glucosides

Question 34. How do you explain the amphoteric behaviour of amino acids?
Answer: Amino acids exist as zwitterion. In the acid medium —COO group of the zwitterion accepts a proton to form a cationic species while in the basic medium, the —N+H3 group loses a proton to form an anionic species.

Biomolecules Amino Acids Exist As Zwitterion

Thus —N+H3 group acts as the acidic group, while the —COO group acts as the basic group thereby justifying that amino acids exhibit amphoteric behaviour.

Question 35. What is the effect of denaturation on the structure of; proteins?
Answer:

  • As a result of denaturation, secondary and tertiary structures of proteins are destroyed but primary structure remains unchanged.
  • During denaturation, H-bonds are broken, globules unfold and helices get uncoiled to form thread-like molecules. In other words, globular proteins (which are soluble in water) undergo coagulation to give fibrous proteins (which are insoluble in water).

Question 36. Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer:

  • Vitamin A is essential to our body as its deficiency causes xerophthalmia (hardening of the cornea), night blindness and xerosis. Sources: Fish liver oil, carrot, butter and milk.
  • Vitamin C is essential to us as its deficiency causes scurvy (bleeding gums) and pyorrhea. Sources: Citrous fruits [for example., oranges, lemon, amla, tomatoes) and green leafy vegetables.

Organic Chemistry Biomolecules

Question 37. The two strands in DNA are not identical but are complementary. Explain.
Answer: DNA has a double helix structure in which the two strands are held together by H-bonds between the purine base of one strand and the pyrimidine base of the other and vice-versa.

  • Thymine (T) pairs with adenine (A) through two H-bonds and cytosine (C) pairs with guanine (G) through three H-bonds.
  • So, opposite of each adenine (A) on one strand there must be a thymine (T) on the other strand and opposite of guanine (G), there must be cytosine (C).
  • In other words, the two strands of DNA are complementary and not identical.

Question 38. Write the important structural and functional differences between DNA and RNA.
Answer:

Structural Differences:

Biomolecules Structural Differences Feature and DNA And RNA

Functional Differences:

Biomolecules Functional Differences DNA And RNA

Question 39. Write the structure of serine at pH = 1 and pH = 11
Answer: Serine exists as zwitterion at neutral pH. In pH = 1, —COO accepts a proton to give cation and at pH = 11, —NH3 donating a proton to the base exists as an anion.

Biomolecules Serine Exists As Zwitterion At Neutral pH

Question 40. Draw the Fischer projection formula of the enantiomer of α-D-(+)-glucopyranose and write its name.
Answer: It is α, as oxygen at C-1 and C-5 are on the same side. It is L as C-5 oxygen is oriented towards the right. It is (-), Le., laevorotatory (enantiomer of dextrorotatory glucopyranose).

Biomolecules Enantiomer Of Dextrorotatory Glucopyranose

Question 41. How many chiral carbons are found in an aldohexose? How many D and L-isomers are possible in each? State the relationship between D and L-isomers of an aldohexose. Give example.
Answer: The structural formula of aldohexose is:

Biomolecules The Structural Formula Of AldohexoseNumber of chiral carbons in aldohexose (n) = 4; the number of 3-D isomers =2n = 24 = 16.

Out of these 16 isomers, 8 are of D- and 8 are of L-configuration. The D-and L-isomers of the same aldohexose are enantiomers, for example., D-(+)-glucose and L-(-)-glucose.

Question 42. Name the lipids having perhydro-1, 2-cydopentano- phenanthrene system. Name an important steroid formed in the gall bladder and state its function.
Answer: Lipids having a perhydro-1, 2-cyclopentane-phenanthrene system are called steroids. The important steroid formed in the gall bladder is cholesterol. It is an essential component of the cell membrane. It acts as a precursor during vitamin D and steroid synthesis.

Question 43. What happens when glucose is treated with dilute NaOH?
Answer: When glucose is treated with dilute NaOH, it forms a mixture of D-glucose, D-fructose and D-mannose due to reversible isomerisation. This reaction is generally known as Lobry de Bruyn-van Ekenstein.

Question 44. Which aldohexose other than D-glucose can produce the same dicarboxylic acid obtained due to oxidation of D-glucose by HNO3?
Answer: D-glucose and L-gulose both yield the same dicarboxylic acid on oxidation by HNO3.

Biomolecules D Glucose And L Gulose

Question 45. Different values of specific rotation of optically active amino acids are obtained at different pH. Explain.
Answer:

Biomolecules The Values Of Specific Rotation Of Dipolar Zwitterion

The values of specific rotation of dipolar zwitterion, cation (conjugate acid) and anion (conjugate base) are different as the groups linked to the chiral carbon in each are different. A specific form exists at a particular pH. So, the specific rotation varies according to the changes in the pH of amino acids.

Organic Chemistry Biomolecules

Question 46. A tripeptide on complete hydrolysis produces glycine, alanine and phenylalanine In 1: 1: 1 molar ratio. What are the probable structures of the tripeptide?
Answer: These three amino acids can be obtained from six tripeptides and these are Gly-Ala-Phe, Gly-Phe-Ala, Ala-GlyPhe, Ala-Phe-Gly, Phe-Ala-Gly and Phe-Gly-Ala.

Question 47. Why disaccharides obtained from monosaccharides like glucose (C6H12O6) have the formula C12H22On instead of C12H24O12?
Answer: Two glucose (C6H12O6) units are joined along with the elimination of a water molecule (H2O) to form a disaccharide. Hence, the resulting disaccharide has the formula C12H22O11 rather than C12H24O12.

Question 48. Write down two differences between amylopectin and cellulose based on their structure.
Answer: Two differences between amylopectin and cellulose are—

Biomolecules Two Differences Between Amylopectin And Cellulose

Question 49. A DNA molecule has a higher melting point having more GC pairs than another DNA molecule which has a greater number of AT pairs. What conclusion can be drawn from this fact?
Answer: The DNA molecule with a higher number of GC pairs will have a higher melting point than the DNA molecule with a higher number of AT pairs (i.e., a lower number of GC pairs).

  • This is due to the presence of three H-bonds in the GC base pair as compared to two H-bonds in the AT base pair.
  • Hence, stronger hydrogen bonding results in a higher melting point of the DNA molecule which has more GC pairs.

Question 50. Why do glucose and fructose produce the same osazone?
Answer: During the reaction of glucose and fructose with excess phenylhydrazine to form osazone, only the C-1 and C-2 atoms of glucose and fructose participate in the reaction. The rest of the molecule remains intact. Hence, glucose and fructose produce the same osazone.

Question 51. 2 DNA samples, X and Y have m.p 340 and 350K respectively. Which sample has a higher CG base pair & why?
Answer: CG base pairing occurs through 3 H-bonds while AT base pairing occurs through 2 H-bonds. Thus, CG base pairing is stronger than AT base pairing. Hence, a sample containing more CG base pairs is more compact and has a higher melting point. So, the sample Y has higher CG content.

Question 52. The structure of aspartame (a peptide and an artificial sweetener) is given below:

Biomolecules Structure Of Aspartame

  1. Name the functional groups present in aspartame.
  2. Give the zwitterionic structure.
  3. Name the amino acids obtained from hydrolysis of aspartame

Answer: The functional groups present in aspartame are amine (—NH2), carboxylic acid (—COOH), amide

Biomolecules The Functional Groups Present In Aspartame Are Amine Carboxylic Acid

The amino acids obtained on hydrolysis of aspartame are aspartic acid and phenylalanine.

Question 53. Name the following:

  1. Acidic group in DNA.
  2. A vitamin which is neither fat-soluble nor water-soluble.
  3. Disease caused due to deficiency of tyrosinase.
  4. An or- amino acid with no chiral carbon atom.
  5. A vitamin which prevents hair loss.
  6. The nucleic acid base with two possible binding sites.

Organic Chemistry Biomolecules

Answer:

  1. The phosphate group.
  2. Vitamin H.
  3. Albinism
  4. Glycine.
  5. Vitamin H (Biotin).
  6. Thymine.

Question 54. Explain why the ka and kb values of α-amino acids are very low.
Answer: The ka values of α-amino acids are very low because, in α-amino acids, the acidic group is —NH3 instead of —COOH group in carboxylic acids. Similarly, the kb values of α-amino acids are very low because the basic group is e —COO instead of —NH2 group in aliphatic amines.

Question 55. Distinguish between anomer and epimer with suitable examples.?
Answer: Carbohydrate molecules which differ in configuration at the glycosidic or anomeric carbon are called anomers while those that differ in configuration at any carbon other than anomeric carbon are called epimers.

  • For example, α-D-glucose and β-D-glucose are anomers of each other as they differ in configuration at C-1 (glycosidic or anomeric carbon).
  • Glucose and galactose are epimers of each other since they differ in configuration at C-4 (other than the glycosidic carbon C-1).

Biomolecules Anomers And Epimers

Question 56. An optically active amino acid (A) having molecular formula C5H11NO2 can exist in three forms depending upon the pH of the medium. Write the structure of (A) in an aqueous medium. What are such ions called? In which medium do the cationic and the anionic forms of the amino acid (A) exist and towards which electrode do they migrate in an electric field?
Answer: The optically active amino acid (A) is valine Me2CH—CH(NH2)—COOH. Depending upon the pH of the medium, valine can exist in the following three forms:

Biomolecules The Optically Active Amino Acid

In aqueous medium, valine exists as a zwitterion. In an acidic medium, it will exist; in an electric field, it will migrate towards the cathode. In a basic medium, it will exist; in an electric field, it will migrate towards the anode.

Question 57. A tetrapeptide on partial hydrolysis produces three dipeptides such as Ser-Thr, Thr-Hyp and Pro-Ser. Identify the tetrapeptide. Write its structure.
Answer: Tetrapeptlde can be obtained by adding 3 dipeptides:

Biomolecules Tetrapeptlde Can Be Obtained By Adding 3 Dipeptides

Therefore, the tetrapeptide is (Pro—Ser—Thr—Hyp) prolylserylthreonylhydroxyproline. Its structure is:

Biomolecules Prolylserylthreonylhydroxyproline

Organic Chemistry Biomolecules

Question 58. The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicate the correlation of the configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has a ‘D’ or ‘L’ configuration.

Biomolecules The Relation With One Of The Isomers Of Glyceraldehyde

Answer: The highest numbered asymmetric centre [i.e., C5 in this case) of the given compound contains a hydroxyl group on the left side, therefore, the compound has Lconfiguration;

Question 59. Aldopentoses named ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration?
Answer: In the Fischer projection formulae of ribose and 2-deoxyribose, the OH group is on the right side of the highest numbered asymmetric centre [i.e., C-4) and hence both the compounds have D-configuration;

Question 60. Which sugar is called inverted sugar? Why is it called so?
Answer:

  • An equimolar mixture of D-(+) -glucose and D-(-) -fructose (which are obtained by hydrolysis of sucrose) is called inverted sugar. Sucrose is dextrorotatory with a specific rotation of +66.5°.
  • On hydrolysis it produces a 1: 1 mixture of glucose (+52.5°) and fructose (-92°) and this mixture has a specific rotation of -19.75° i.e., the mixture is laevorotatory.
  • Because of such change in the sign of optical rotation, the hydrolysis of sucrose is called inversion of sucrose and the mixture of glucose and fructose obtained by hydrolysis is called inverted sugar;

Question 61. Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of the amino group concerning the carboxyl group. Which type of amino acids form polypeptide chains in proteins?
Answer: α-Amino acids form polypeptide chains in proteins;

Question 62. α-Helix is a secondary structure of proteins formed by twisting the polypeptide chain into right-handed screw-like structures. Which type of interactions are responsible for making the α-helix structure stable?
Answer: In α-helix structure, a polypeptide chain is stabilised by the formation of H-bonds betweenBiomolecules Formation Of H Bonds the group of one amino acid residue and the Biomolecules Amino Acid Residue Formation Of H Bonds group of the fourth amino acid residue in the chain;

Question 63. Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with the simultaneous reduction of another substrate?
Answer: Enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate are called oxidoreductase;

Question 64. During the curdling of milk, what happens to the sugar present in it?
Answer: During the curdling of milk, the milk sugar (i.e lactose) is converted to lactic acid by the bacteria present in milk;

Organic Chemistry Biomolecules

Question 65. How do you explain the presence of five—OH groups in glucose molecules?

Biomolecules OH Groups In Glucose Molecule

Answer: Glucose reacts with an excess of acetic anhydride to form a pentadactyl derivative thereby showing the presence of five hydroxyl groups in the molecule;

Question 66. Why does compound (A) given below not form an oxime?
Answer: Glucose pentaacetate (structure A) does not have a free hydroxyl ( —OH) group at the anomeric carbon C-1 and hence, it cannot be converted to the open-chain form having a free aldehyde ( — CHO) group. Thus, it does not form an oxime;

Question 67. Why must vitamin C be supplied regularly in the diet?
Answer: Vitamin C is a water-soluble substance so excess of it is readily excreted in urine and cannot be stored in our body. Consequently, it should be supplied regularly in the diet;

Question 68. Sucrose is dextrorotatory but the mixture obtained after hydrolysis is laevorotatory. Explain.
Answer: On hydrolysis, sucrose (dextrorotatory) gives a 1:1 mixture of glucose (dextrorotatory, +52.5° ) and fructose (laevorotatory, -92°). Since the laevorotation of fructose is greater than the dextrorotation of glucose, the mixture is laevorotatory;

Question 69. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.
Answer: In an aqueous solution, the carboxyl ( — COOH) group loses a proton and the amino ( — NH2) group accepts a proton to form a zwitterion. Thus amino acids behave like salts;

Biomolecules Zwitterion

Question 70. Structures of glycine and alanine are given below. Show the peptide linkage in glycyl alanine.

Biomolecules Glycine And Alanine

Answer: In the formation of glycyialaninc, the carboxyl group of glycine interacts with an amino group of alanine to form a peptide bond.

Biomolecules Glycylalanine

Question 71. A protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to a physical change like a temperature change or a chemical change like, a change in pH, denaturation of protein takes place. Explain the cause.
Answer:

  • Due to a physical or chemical change, hydrogen bonds in proteins are disturbed, globules unfold and helices get uncoiled to form a thread-like structure.
  • Consequently, secondary and tertiary structures are destroyed and the protein loses its biological activity. This is called the denaturation of proteins;

Question 72. The activation energy for the acid-catalysed hydrolysis of sucrose is 6.22kJ.mol-1, while the activation energy is only 2.15kJ-mol-1 when hydrolysis is catalysed by the enzyme sucrase. Explain.
Answer:

  • Enzymes are biocatalysts. They reduce the magnitude of activation energy by providing an alternative path. In the hydrolysis of sucrose the enzyme sucrase reduces the activation energy from 6.22 kj- mol-1 to 2.15 kj mol-1.
  • Thus enzyme catalysed reactions proceed at a much faster rate than ordinary chemical reactions using conventional catalysts;

Organic Chemistry Biomolecules

Question 73. How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:

  • Glucose reacts with hydroxylamine (NH2OH) to form a monoxime and adds one molecule of hydrogen cyanide (HCN) to give a cyanohydrin. Thus, glucose contains a carbonyl group, which may be aldehydic or ketonic.
  • However, glucose (C6H12O6) undergoes oxidation by bromine-water (a mild oxidising agent) to give gluconic acid (C6H12O7) with the same number of carbon atoms.
  • This indicates that the carbonyl group present in glucose is an aldehydic group;

Question 74. Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
Answer:

  • In the formation of dinucleotides, the 3′ — OH group of the pentose sugar of one nucleotide unit and the 5′ — OH group of the pentose sugar of the other nucleotide unit are involved in generating the phosphodiester linkage.
  • Phosphoric acid is involved in the formation of phosphodiester linkage.
  • The word ‘diester’ in this linkage indicates that two —OH groups of phosphoric acid are involved in the formation of two ester linkages,

Question 75. What are glycosidic linkages? In which type of biomolecules are they present?
Answer:

  • In oligosaccharides and polysaccharides, any two adjacent monosaccharide units are joined together by an ether linkage or oxide linkage formed by the loss of a molecule of water.
  • Such a linkage between two monosaccharide units through an oxygen atom is called glycosidic linkage.

Question 76. Which monosaccharide units are present in starch, cellulose and glucose and which linkages link these units?
Answer:

  • All these polysaccharides (i.e., starch, cellulose and glycogen) are composed of glucose units. In starch and glycogen, α-glycosidic linkages are present.
  • On the other hand in cellulose, β-glycosidic linkages are present between glucose units;

Question 77. Describe the term D- and L- configuration used for amino acids with examples.
Answer: Consider the following configurations of the a -amino acid,

  • R—CH(NH2)—COOH, in which the main carbon chain is shown vertically keeping the —COOH group at the top. If the — NH2 group is on the right side of the α-carbon (as in str. 1), it is referred to as D-amino acid.
  • On the other hand, if the — NH2 group is on the left side of the α-carbon (as in str. 2), it is referred to as L-amino acid;

Biomolecules D Amino Acid And L Amino Acid

Question 78. Coagulation of egg white on boiling is an example of denaturation of protein. Explain it in terms of structural changes.
Answer:

  • When egg white is boiled, the soluble globular protein albumin present in it is denatured resulting in the formation of insoluble fibrous protein.
  • During this process, secondary and tertiary structures of the protein (albumin) are destroyed but the primary structure (i.e„ the sequence of α-amino acids) remains unchanged;

Question 79. Carbohydrates are essential for life in both plants and animals. Name the carbohydrates used as storage molecules in plants and animals, and also name the carbohydrate present in wood or fibre of cotton cloth.
Answer:

Starch’ is the carbohydrate which is used as a storage molecule in plants, while ‘glycogen’ is the carbohydrate which is used as storage material in animals. Cellulose is present in wood or the fibre of cotton cloth;

Class 12 Chemistry Unit 14 Biomolecules Multiple Choice Questions And Answers

Question 1. In an aqueous solution, glucose remains as—

  1. Only in the open-chain form
  2. Only in the pyranose form
  3. Only in furanose form
  4. In all three forms of equilibrium

Answer: 4. In all three forms of equilibrium

Biomolecules Pyranose Form And Furanose Form

The six-membered cyclic form is generally referred to as the ‘pyranose’ form, and the five-membered cyclic form is called the ‘furanose’ form.

Closure of the ring creates a chiral centre at C- 1, resulting in two diastereomers (sometimes called “anomers”)-the alpha (α) and beta (β) forms.

Question 2. Which one is not a constituent of nucleic acid—

  1. Uracil
  2. Guanidine
  3. Phosphoric acid
  4. Ribose sugar

Answer: 2. Guanidine

Organic Chemistry Biomolecules

Nucleic acid consists of bases, sugars and phosphate groups. Thus, guanidine is not a constituent of nucleic acid.

Question 3. The correct structure of the dipeptide gly-ala is—

Biomolecules The Correct Structure Of The Dipeptide Gly Ala

Answer: 3

Biomolecules Structure Of The Dipeptide Gly Ala.

Question 4. Ribose and 2-deoxyribose can be differentiated by—

  1. Fehling’s reagent
  2. Tollens’ reagent
  3. Barfoed’s reagent
  4. Osazone formation

Answer: 4. Osazone formation

2-deoxyribose has no ketonic or oxidisable hydroxy group and hence, it does not respond in osazone reaction.

Question 5. The number of amino acids and number of peptide bonds in a linear tetrapeptide (made of different amino acids) are respectively—

  1. 4 and 4
  2. 5 and 5
  3. 5 and 4
  4. 4 and 3

Answer: 4. 4 and 3

For tetrapeptide, there are four amino acids linked together by (4- 1) = 3 amide or peptide linkages.

Question 6. Among the following statements about the molecules X and Y, the one(s) which is (are) correct is (are)—

Biomolecules Statements About The Molecules X And Y

Organic Chemistry Biomolecules

  1. X and Y are diastereoisomers
  2. X and Y are enantiomers
  3. X and Y are aldohexoses
  4. X is a D-sugar and Y is a L-sugar

Answer: 2,3 and 4

  • X and Y are mirror images of each other. Thus, these are enantiomeric pairs having four stereogenic centres.
  • In both, X and Y, the aldehyde functional group (—CHO) is present having a total no. of 6 carbon in each of them.
  • Thus, both of them are aldohexose. X is D-glucose thus its mirror image, Y is L-glucose.

Question 7. Within the list shown below, the correct pair of structures of alanine in the pH range 2-4 and 9-11 is—

  1. H3N-CH(CH3)COOH
  2. H2NCH(CH3)COO
  3. H3N—CH(CH3)COO
  4. H2NCH(CH3)COOH
  1. 1, 2
  2. 1, 3
  3. 2, 3
  4. 3, 4

Answer: 1. H3N-CH(CH3)COOH

Biomolecules Alanine In Zwitterionic Form Alaninein Acid Medium

Question 8. ADP and ATP differ in the number of—

  1. Phosphate units
  2. Ribose units
  3. Adenine base
  4. Nitrogen atom

Answer: 1. Phosphate units

ADP is adenosine diphosphate, thus it contains 2 phosphates. On the other hand, ATP is adenosine triphosphate, thus, ADP and ATP differ in the no. of phosphate units.

Question 9. In photosynthesis, the synthesis of each glucose molecule is related to—

  1. 8 molecules of ATP
  2. 6 molecules of ATP
  3. 18 molecules of ATP
  4. 10 molecules of ATP

Answer: 3. 18 molecules of ATP

Question 10. Thiol group is present in—

  1. Cytosine
  2. Cystine
  3. Cysteine
  4. Methionine

Answer: 3. Cysteine

Organic Chemistry Biomolecules

Question 11. Which of the following compounds will behave as reducing sugar in an aqueous KOH solution—

Biomolecules Sugar In An Aqueous KOH Solution

Answer: 3. The —OCOCH3 group, attached to the anomeric carbon in the compound, on hydrolysis in basic medium (aq. KOH), gets converted into the —OH group. The —OH group, present in the compound shows its reducing property.

Biomolecules Ketohexose

Question 12. The predominant form of histamine present in human blood is (pKa, Histidine = 6.0 )—

Biomolecules The Predominant Form Of Histamine Present In Human Blood

Answer: 4

Biomolecules Aliphatic Amine Most Basic In Nature

Organic Chemistry Biomolecules

Question 13. Glucose on prolonged heating with HI gives—

  1. N-hexane
  2. 1 -hexene
  3. Hexanoic acid
  4. 6-iodohexanal

Answer: 1. N-hexane

Biomolecules Glucose And N Hexane

Question 14. The difference between amylose and amylopectin is—

  1. Amylopectin has 1→ 4 α-linkage and 1 → 6 α-linkage
  2. Amylose has 1→ 4 α-linkage and 1→ 6 β-linkage
  3. Amylopectin has 1 → 4 α-linkage and 1 → 6 β-linkage
  4. Amylose is made up of glucose and galactose

Answer: 1. Amylopectin has 1→ 4 α-linkage and 1 → 6 α-linkage

Amylose and amylopectin both are polymers of α-D-(+) glucose. In amylose, the glycosidic linkage is formed between C-l of a glucose unit and C-4 of another glucose unit. Again, amylopectin is a branched polymer where glycosidic linkages are present between C-1-C-4 and C-1-C-6.

Question 15. Find the hydrolysis product of maltose—

  1. α-D-glucose + α-D-glucose
  2. α-D-glucose + α-D-fructose
  3. α-D-glucose + α-D-galactose
  4. α-D-glucose + α-D-glucose

Answer: 1. α-D-glucose + α-D-glucose

Biomolecules Maltose And Maltase Glucose

Question 16. Lysine is least soluble in water in the pH range—

  1. 3 to 4
  2. 5 to 6
  3. 6 to 7
  4. 8 to 9

Answer: 4. 8 to 9

Any amino acid has its lowest solubility at its isoelectric point.

Question 17. Thymine is—

  1. 5-methyl uracil
  2. 4-methyIuracil
  3. 3-methyluradl
  4. 1-methyl uracil

Answer: 1. 5-methyl uracil

Organic Chemistry Biomolecules

Thymine is also known as 5-methyluracil.

Question 18. Biomolecules Heptanoic Acid

  1. Heptanoic acid
  2. 2-isohexane
  3. Heptane.
  4. Heptanol

Answer: 1. Heptanoic acid

Biomolecules Glucose Cyanohydrin

Question 19. Biomolecules Increasing Acidic StrengthsArrange X, Y and Z in order of increasing acidic strengths—

  1. X > Z > Y
  2. Z < X > Y
  3. X > Y > Z
  4. Z > X > Y

Answer: 1. X > Z > Y

  • Carboxylic acid is a stronger acid than NH3, therefore, X is the strongest acid.
  • Since —COOH has a -I effect which decreases with distance, therefore, -I effect is more pronounced on Z than on Y.
  • As a result, Z is more acidic than Y, Thus, the overall order of decreasing acidic strength is X > Z > Y.

Question 20. Glycogen is a branched chain polymer of α-D-glucose units in which the chain is formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of Cl—C6 glycosidic linkage. The structure of glycogen is similar to.

  1. Amylose
  2. Amylopectin
  3. Cellulose
  4. Glucose

Answer: 2. Amylopectin

Question 21. Which of the following polymers is stored in the Uver of animals—

  1. Amylose
  2. Cellulose
  3. Amylopectin
  4. Glycogen

Answer: 4. Glycogen

Organic Chemistry Biomolecules

Question 22. Which of the following pairs represents anomers-

Biomolecules Which Of The Following Pairs Represents Anomers

Answer: 3

Question 23. Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet structure, α-helix structure of protein is stabilised by—

  1. Peptide bonds
  2. Van der Waals forces
  3. Hydrogen bonds
  4. Dipole-dipole interactions

Answer: 3. Hydrogen bonds

Organic Chemistry Biomolecules

Question 24. In disaccharides, if the reducing groups of monosaccharides, i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar—
Answer:

Biomolecules Disaccharide Is A Non Reducing Sugar.

Answer: 2

Question 25. Which of the following acids is a vitamin—

  1. Aspartic acid
  2. Ascorbic acid
  3. Adipic acid
  4. Saccharic acid

Answer: 2. Ascorbic acid

Question 26. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present—

  1. 5′ and 3′
  2. 1′ and 5′
  3. 5′ and 5′
  4. 3′ and 3′

Answer: 1. 5′ and 3′

Question 27. Nucleic acids are the polymers of

  1. Nucleosides
  2. Nucleotides
  3. Bases
  4. Sugars

Answer: 2. Nucleotides

Organic Chemistry Biomolecules

Question 28. Which of the following statements is not true about glucose—

  1. It is an aldohexose
  2. On heating with HI, it forms n-hexane
  3. It is present in furanose form
  4. It does not give a 2,4-DNP test

Answer: 3. It is present in furanose form

Question 29. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be

  1. Primary structure of proteins
  2. Secondary structure of proteins
  3. Tertiary structure of proteins
  4. Quaternary structure of proteins

Answer: 1. Primary structure of proteins

Question 30. DNA and RNA contain four bases each. Which of the following bases is not present in RNA—

  1. Adenine
  2. Uracil
  3. Thymine
  4. Cytosine

Answer: 3. Thymine

Question 31. Which of the following B group vitamins can be stored in our body—

  1. Vitamin B1
  2. Vitamin B2
  3. Vitamin B6
  4. Vitamin B12

Answer: 4. Vitamin B12

Organic Chemistry Biomolecules

Question 32. Which of the following bases is not present in DNA—

  1. Adenine
  2. Thymine
  3. Cytosine
  4. Uracil

Answer: 4. Uracil

Question 33. Three cyclic structures of monosaccharides are given below. Which of these are anomers—

Biomolecules Cyclic Structures Of Monosaccharides

Answer: 1

Question 34. Which of the following reactions of glucose can be explained only by its cyclic structure—

  1. Glucose forms pentaacetate
  2. Glucose reacts with hydroxylamine to form an oxime
  3. Pentaacetate of glucose does not react with hydroxylamine
  4. Glucose is oxidised by nitric acid to gluconic acid

Answer: 3. Pentaacetate of glucose does not react with hydroxylamine

Organic Chemistry Biomolecules

Question 35. Optical rotations of some compounds along with their structures are given below. Which of them have D- configuration—

Biomolecules Optical Rotations Of Some Compounds

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

Answer: 1. 1,2,3

Question 36. The structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units—

Biomolecules Anomeric Carbon Atomsin Monosaccharide Units

  1. ‘a’ carbon of glucose and ‘a’ carbon of fructose
  2. ‘a’ carbon of glucose and V carbon of fructose
  3. ‘a’ carbon of glucose and ‘b’ carbon of fructose
  4. ‘f’ carbon of glucose and ‘/ carbon of fructose

Answer: 3. ‘a’ carbon of glucose and ‘b’ carbon of fructose

Organic Chemistry Biomolecules

Question 37. Three structures are given below in which two glucose units are linked. Which of these linkages between glucose units are between C-1 and C-4 and which linkages are between C-1 and C-6—

Biomolecules Three Structures Are Given Below In Which Two Glucose

  1. (A) is between C-1 and C-4, (B) and (C) is between C-1 and C-6
  2. (A) and (B) are between C-1 and C-4, (C) is between C-1 and C-6
  3. (A) and (C) are between C-1 and C-4, (B) is between C-1 and C-6
  4. (A) and (C) are between C-1 and C-6, (B) is between C-1 and C-4

Answer: 3. (A) and (C) are between C-1 and C-4, (B) is between C-1 and C-6

Question 38. Carbohydrates are classified based on their behaviour on hydrolysis and also as reducing or nonreducing sugar. Sucrose is a.

  1. Monosaccharide
  2. Disaccharide
  3. Reducing sugar
  4. Non-reducing sugar

Answer: 2 and 4

Question 39. Proteins can be classified into two types based on their molecular shape, i.e., fibrous proteins and globular proteins. Examples of globular proteins are—

  1. Insulin
  2. Keratin
  3. Albumin
  4. Myosin

Answer: 1 and 3

Question 40. Which of the following carbohydrates are branched polymers of glucose—

  1. Amylose
  2. Amylopectin
  3. Cellulose
  4. Glycogen

Answer: 2 and 4

Organic Chemistry Biomolecules

Question 41. Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following are acidic—

Biomolecules Amino And Carboxyl Groups In Their Molecule

Answer: 2 and 4

Question 42. Lysine, H2N—(CH2)2—CH—COOH is.

  1. α-amino acid
  2. Basic amino acid
  3. Amino acid synthesised in the body
  4. β-amino acid

Answer: 1 and 2

Question 43. Which of the following monosaccharides are present as a five-membered cyclic structure (furanose structure)—

  1. Ribose
  2. Glucose
  3. Fructose
  4. Galactose

Answer: 1 and 3

Question 44. In fibrous proteins, polypeptide chains are held together by

  1. Van der Waals forces
  2. Disulphide linkage
  3. Electrostatic forces of attraction
  4. Hydrogen bonds

Answer: 2 and 4

Question 45. Which of the following are purine bases—

  1. Guanin
  2. Adenine
  3. Thymine
  4. Uracil

Answer: 1 and 2

Organic Chemistry Biomolecules

Question 46. Which of the following terms are correct about enzyme—

  1. Proteins
  2. Dinucleotides
  3. Nucleic acids
  4. Biocatalysts

Answer: 1 and 4

Question 47. Alteration of which is responsible for DNA mutation—

  1. Ribose unit
  2. Nitrogenous base
  3. Phosphate
  4. None of the above

Answer: 4. None of the above

Question 48. Which one of the following is a globular protein—

  1. Collagen
  2. Myoglobin
  3. Myosin
  4. Fibroin

Answer: 2. Myoglobin

Question 49. The end product of protein metabolism is—

  1. Peptide
  2. Peptone
  3. Proton
  4. α-amino acid

Answer: 4. a -amino acid

Organic Chemistry Biomolecules

Question 50. Complete hydrolysis of cellulose yields—

  1. D-fructose
  2. D-ribose
  3. D-glucose
  4. L-glucose

Answer: 3. D-glucose

Question 51. Which one of the following is non-reducing sugar—

  1. Glucose
  2. Fructose
  3. Lactose
  4. Sucrose

Answer: 4. Sucrose

Question 52. Which compound does not exhibit mutarotation—

  1. Sucrose
  2. D-glucose
  3. L-galactose
  4. None

Answer: 1. Sucrose

Question 53. The participating group in the disulphide bond of proteins is—

  1. Thioether
  2. Thioester
  3. Thiol
  4. Thiolactose

Answer: 3. Thiol

Organic Chemistry Biomolecules

Question 54. The sequence in a nucleotide of nucleic acid is—

  1. Phosphate-base
  2. Sugar-base-phosphate
  3. Base-sugar-phosphate
  4. Base-phosphate-sugar

Answer: 2. Sugar-base-phosphate

Question 55. Which chemical compounds act as an emulsifier—

  1. Phosphoric acid
  2. Fatty acid
  3. Bile acids
  4. Mineral acids

Answer: 3. Bile acids

Question 56. The sequence in which amino acids are arranged in a protein molecule refers to its—

  1. Primary structure
  2. Secondary structure
  3. Tertiary structure
  4. Quaternary structure

Answer: 1. Primary structure

Organic Chemistry Biomolecules

Question 57. Stability is imparted to the protein helix by—

  1. Dipeptide bond
  2. Hydrogen bond
  3. Ether bond
  4. Peptide bond

Answer: 2. Hydrogen bond

Question 58. The monomeric unit of starch is—

  1. Glucose
  2. Fructose
  3. Glucose and fructose
  4. Mannose

Answer: 1. Glucose

Question 59. Which of the following responds to Molisch’s reagent—

  1. All carbohydrates
  2. Sucrose
  3. Fructose
  4. Glucose

Answer: 1. All carbohydrates

Question 60. Carboxylic acid and amino groups of an amino acid are ionised at pKa1 = 2.34 and pKa2 = 9.60. The pH at which the amino acid will attain its isoelectric point is—

  1. 5.97
  2. 2.34
  3. 9.60
  4. 6.97

Answer: 1. 5.97

Question 61. The percentage of α-D -glucose and β-D-glucose in D-glucose is—

  1. 50%
  2. 4% and 36%
  3. 36% and 64%
  4. 33% each along with open open-chain structure

Answer: 3. 36% and 64%

Question 62. Glucaric acid [HOOC(CHOH)4COOH] on the reaction with HIO4 produces—

  1. 4HCOOH+2CHO
  2. 3HCOOH + HCHO + OHC — COOH
  3. 2HC00H + 2OHC —COOH
  4. 6HCOOH

Answer: 2. 3HCOOH + HCHO + OHC — COOH

Question 63. Which amino acid does not contain any chiral carbon—

  1. Histidine
  2. Glycine
  3. α-alanine
  4. Threonine

Answer: 2. Glycine

Organic Chemistry Biomolecules

Question 64. Which responds to Benedict’s reagent but not to ninhydrin—

  1. Protein
  2. Monosaccharide
  3. Lipid
  4. Amino acid

Answer: 2. Monosaccharide

Question 65. The hydrolytic reaction of sucrose is called—

  1. Inhibition
  2. Inversion
  3. Saponification
  4. Hydration

Answer: 2. Inversion

Question 66. Which structural characteristic distinguishes proline from other amino acids—

  1. Optical inactivity
  2. Presence of aromatic group
  3. The presence of two hydroxylic group
  4. It is a secondary amine group

Answer: 4. It is a secondary amine group

Question 67. A compound (C6H12O6) in reaction with phenylhydrazine gives a yellow ppt. and with Na produces a mixture of sorbitol and mannitol. The compound is—

  1. Fructose
  2. Glucose
  3. Mannose
  4. Sucrose

Answer: 1. Fructose

Question 68. Compounds X and Y obtained in the following reaction are— 

Biomolecules Glucose And Osazone

  1. C6H5NH2 and NH2OH
  2. C6H5NH2 and NH3
  3. NH2OH and H2O
  4. C6H5NHNHOH and NH3

Answer: 2. C6H5NH2 and NH3

Organic Chemistry Biomolecules

Question 69. Which of the following hexoses form the same osazones on reaction with phenylhydrazine—

  1. D-glucose, D-galactose and D-talose
  2. D-fructose, D-mannose and D-galactose
  3. D-glucose, D-mannose and D-galactose
  4. D-glucose, D-fructose and D-mannose

Answer: 4. D-glucose, D-fructose and D-mannose

Question 70. The least solubility of amino acids in water is at—

  1. pH = 7
  2. pH > 7
  3. pH <7
  4. pi

Answer: 4. pi

Question 71.Biomolecules Amylase And Maltase P and Q are—

  1. Invertase and zymase
  2. Amylase and maltase
  3. Diastase and lipase
  4. Pepsin and trypsin

Answer: 2. Amylase and maltase

Question 72. The positions at which base and phosphate groups are linked in DNA and RNA are—

  1. C’ and C’
  2. C’ and C’
  3. C’ and C.

Answer: 3. C’ and C.

Question 73. Which one of the given is the C-2 epimer of D-glucose—

  1. D -galactose
  2. L -glucose
  3. D-mannose
  4. D-fructose

Answer: 3. D-mannose

Organic Chemistry Biomolecules

Question 74. Which one of the following is a zwitterion—

  1. Urea
  2. Glycine hydrochloride
  3. Ammonium acetate
  4. L-alanine

Answer: 4. L-alanine

Question 75. The hydrolytic product of sucrose is—

  1. Galactose
  2. Glucose
  3. Fructose
  4. Ribose

Answer: 2 and 3

Question 76. Which compound shows mutarotation—

  1. Glucose
  2. Fructose
  3. Sucrose
  4. Starch

Answer: 1 and 2

Question 77. Which amino acid is chiral—

  1. Alanine
  2. Glycine
  3. Phenylalanine
  4. Glutamine

Answer: 1 and 3

Question 78. Which compound possesses a transitional element—

  1. Vitamin B12
  2. Chlorophyll
  3. Haemoglobin
  4. DNA

Answer: 1,3 and 4

Organic Chemistry Biomolecules

Question 79. Globular protein is absent in—

  1. Blood
  2. Keratin
  3. Egg
  4. Muscles

Answer: 2 and 4

Question 80. Which of the given statements is true—

  1. An amino acid contains an amino group and a carboxylic group
  2. Amino acids are the structural components of peptide and protein
  3. Amino acids exist as zwitterions
  4. Amino acids are negatively charged in an alkaline medium

Answer: 1,2,3 and 4

Class 12 Chemistry Unit 14 Biomolecules Match The Following Questions And Answers

Question 1.

Biomolecules Vitamins And Diseases

Organic Chemistry Biomolecules

Answer: 1-C, F,  2-G, 3-A, 4-H, 5-D, 6-E, 6-B

Question 2.

Biomolecules Enzyme And Reactions

Answer: 1-D, 2-C, 3-E, 4-A, 5-B

Class 12 Chemistry Unit 14 Biomolecules Assertion-Reason Type

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.

  1. (A) and (R) both are correct statements and (R) is the correct explanation for (A).
  2. (A) and (R) both are correct statements but (R) is not a correct explanation for (A).
  3. (A) is a correct statement but (R) is a wrong statement.
  4. (A) and (R) both are incorrect statements.
  5. (A) is a wrong statement but (R) is a correct statement.

Question 1. Assertion (A): D (+) – glucose is dextrorotatory.

Reason (R): ‘D’ represents its dextrorotatory nature.

Answer: 3. (A) is the correct statement but (R) is the wrong statement.

Question 2. Assertion (A): Vitamin D can be stored in our body.

Reason (R): Vitamin D is a fat-soluble vitamin.

Answer: 1. (A) and (R) both are correct statements and (R) is a correct explanation for (A).

Question 3. Assertion (A): ft -glycosidic linkage is present in maltose

Biomolecules Beta Glycosidic Linkage Is Present In Maltose

Organic Chemistry Biomolecules

Reason (R): Maltose is composed of two glucose units in which C-l of one glucose unit is linked to C-4 of another glucose unit.

Answer: 4. (A) and (R) both are incorrect statements.

Question 4. Assertion (A): AU naturally occurring or-amino acids except glycine are optically active.

Reason (R): Most naturally occurring amino acids have L-configuration.

Answer: 5. (A) is a wrong statement but (R) is the correct statement.

Question 5. Assertion (A): Deoxyribose, C5H10O4 is not a carbohydrate.

Organic Chemistry Biomolecules

Reason (R): Carbohydrates are hydrates of carbon so compounds which follow the Cx(H2O)y formula are carbohydrates.

Answer: 2. (A) and (R) both are correct statements but (R) is not the correct explanation for (A).

Question 6. Assertion (A): Glycine must be taken through diet.

Reason (R): It is an essential amino acid.

Answer: 2. (A) and (R) both are correct statements but (R) is not a correct explanation for (A).

Question 7. Assertion (A): In the presence of an enzyme, the reagent can effectively attack the substrate molecules.

Organic Chemistry Biomolecules

Reason (R): Active sites of enzymes hold the substrate molecule in a suitable position.

Answer: 1. (A) and (R) both are correct statements and (R) is the correct explanation for (A).

Class 12 Chemistry Unit 14 Biomolecules Fill in the blanks

Question 1. Polysaccharides mix in boiling water to form___
Answer: Colloids;

Question 2. An increase in temperature____ the mutarotation rate.
Answer: Increases;

Question 3. The simplest amino acid is____
Answer: Glycine;

Question 4. Sucrose hydrolysis is called____
Answer: Inversion of sucrose

Question 5. The heterocyclic bases present in nucleic acids are called____
Answer: Purine and pyrimidine;

Organic Chemistry Biomolecules

Question 6. Polyhydroxy ketones are known as____
Answer: Ketoses

Organic Chemistry Biomolecules

Question 7. Amylopectin is a branched polymer of____
Answer: Glucose

Question 8. Phospholipids are mixed glycerides of and____
Answer: Fatty acid, phosphoric acid

Question 9. Thyroxine is a____ hormone.
Answer: Amine

Question 10. The scurvy disease occurs due to a deficiency of____
Answer: Vitamin C

Question 11. The chemical name of Vitamin B12 is____
Answer: Cyanocobalamine

Organic Chemistry Biomolecules

Question 12. Lecithin is a____
Answer: Phospholipid

Question 13. The acidic property of glycine pertains to____
Answer: —COOH

Question 14. The power-house of animal cells is called____
Answer: Mitochondria

Class 12 Chemistry Unit 14 Biomolecules Warm Up Exercise

Question 1. Why are monosaccharides except dihydroxyacetone optically active?
Answer: Monosaccharides except dihydroxyacetone contain almost one asymmetric C-atom making them optically active.

Organic Chemistry Biomolecules

Question 2. Name two disaccharides that on hydrolysis produce two similar and two different monosaccharides.
Answer: Hydrolysis of sucrose produces two different monosaccharides glucose and fructose.

Biomolecules Hydrolysis Of Sucrose Produces Two Different Monosaccharides Glucose And Fructose

Hydrolysis of maltose produces only one monosaccharide i.e., glucose.

Biomolecules Hydrolysis Of Moltose Produces Only One Monosaccharide Glucose

Question 3. Though benzene is water-insoluble, glucose and sucrose are water-soluble. Why?
Answer: Benzene cannot form an H-bond with water molecules. There are many —OH groups present in glucose and sucrose which can form H-bonds with water molecules and get dissolved in water.

Organic Chemistry Biomolecules

Question 4. What do you mean by D- and L-glucose? State the relationship between D and L signs with the nature of optical rotation.
Answer:

  • First part: D-and L-configurations are assigned based on the configuration of the glucose molecule.
  • If the — OH group on the C-5 lies towards the right-hand side, it is called D-glucose whereas if the — OH group is on the left-hand side, then it is termed as L-glucose.
  • Second part: D- and L-glucose are a pair of enantiomers. They are mirror images of each other.

Question 5. How many chiral carbons are present in an aldohexose and a ketohexose? How many 3D symmetries are possible in each?
Answer: There are four chiral carbons present in an aldohexose for which 24 = 16 no. of 3D stereoisomers are obtained. There are three chiral carbons present in a ketohexose for which 23 = 8 no. of 3D stereoisomers are obtained.

Question 6. Why does glucose not exhibit all characteristic reactions of aldehydes?
Answer: Glucose does not exhibit all characteristic reactions of aldehydes because though it has an aldehyde functional group in an open-chain structure, the aldehyde group generally exists in hemiacetal form where aldehyde functionality is absent.

Organic Chemistry Biomolecules

Question 7. Aldehyde reacts with two molecules of ethanol to form acetal, whereas glucose requires one molecule of ethanol. Why?
Answer: Since glucose exists in hemiacetal form, it requires 1 molecule of ethanol to form acetal.

Question 8. How can you differentiate between D-glucose and D- fructose using Tollens’ reagent or Fehling’s solution?
Answer: Both glucose and fructose react with Tollens’ reagent and Fehling’s solution to give the same products for which they cannot be differentiated by these two solutions.

Question 9. Sucrose is a non-reducing sugar. Explain
Answer: Since there is no free —CHO or —COCH2OH group in sucrose, it is a non-reducing sugar.

Question 10. Name an optically inactive amino acid.
Answer: Glycine.

Question 11. Hydrolysis of a tripeptide produces glycine, valine, and phenylalanine. Write down the possible sequence of the tripeptide using 3-lettered symbols.
Answer: Gly-Val-Phe, Gly-Phe-Val, Val-Gly-Phe, Val-Phe-Gly, Phe-Gly-Val, Phe-Val-Gly.

Organic Chemistry Biomolecules

Question 12. What is meant by the C-and N-terminal of a peptide?
Answer: The C-terminal is the end of an amino acid chain terminated with a free carboxyl group (—COOH) whereas the N-terminal is the end of the same amino acid with a free amine (—NH2) group.

Question 13. What kind of attractive force acts in a -helix?
Answer: Intermolecular H-bonds.

Question 14. Why are proteins optically active?
Answer: Since amino acids are the building blocks of proteins. So proteins are optically active as amino acids (except glycine) are optically active.

Question 15. Towards which electrode will an amino acid move whose pH > pi?
Answer: Amino acid will move towards the anode whose pH > pi.

Question 16. Which force will act between glutamic acid and lysine in a protein?
Answer: Attractive forces due to salt bridge formation.

Question 17. Name two amino acid residues that can be linked by disulfide bonds.
Answer: Cys-Cys

Question 18. Which enzyme helps to decoagulate clotted blood?
Answer: Streptokinase.

Question 19. Which disease occurs due to a lack of phenylalanine hydroxylase?
Answer: Phenylketonuria (PKU)

Question 20. Which enzyme deficiency causes albinism?
Answer: Tyrosinase.

Organic Chemistry Biomolecules

Question 21. Which enzyme converts sucrose into glucose and fructose?
Answer: Invertase.

Question 22. What do you mean by adrenocortical hormones?
Answer:  Adrenocortical hormones are polycyclic hormones j produced by the adrenal cortex. These hormones play j some important roles which are crucial for body j responses to stress and regulate other functions in the j body.

Question 23. Which glands synthesize hormones?
Answer:  Endocrine glands synthesize hormones.

Question 24. How many C-atoms are present in eicosanoids?
Answer: 20

Question 25. Which lipids can form micelles like soaps?
Answer: Phospholipids.

Question 26. Which lipids contain perhydro-1, 2- cyclopentanophenanthrene system?
Answer: Steroids

Question 27. Which steroid is the most abundant in the human body?
Answer: Cholesterol.

Question 28. Fresh tomato is a better source of vitamin C than canned tomatoes. Why?
Answer: In the case of canned tomatoes, ascorbic add (vitamin C) gets oxidized by aerial oxygen to dehydroascorbic add for which the amount of vitamin C is reduced. But for fresh tomatoes, there is no reduction in the amount of vitamin C. That is why fresh tomatoes are a better source of vitamin C than canned tomatoes.

Organic Chemistry Biomolecules

Question 29. Which vitamin molecule contains a cobalt atom?
Answer: Vitamin B12.

Question 30. Which vitamin is formed from p-carotene present in carrots?
Answer: Vitamin A.

Question 31. Which vitamin helps in blood clotting?
Answer: Vitamin K.

Question 32. What is cyanocobalamin? Which disease is caused due to its deficiency?
Answer:  Vitamin- B12. Pernicious anemia

Question 33. Point out the vitamins present in the following:

  1. Egg,
  2. Milk,
  3. Orange,
  4. Spinach,
  5. Tomato,
  6. Mango,
  7. Groundnut,
  8. Amla and
  9. Fish.

Answer:

  1. Vitamin A. B-complex, D, E.
  2. Vitamin A B-complex, D, E.
  3. Vitamin C.
  4. Vitamin K.
  5. Vitamin C.
  6. Vitamin A.
  7. Vitamin E.
  8. Vitamin C.
  9. Vitamin A, and B12.

Question 34. State different types of RNA.
Answer:  m-RNA, rRNA, and f-RNA

Question 35. Name the complementary bases of adenine and guanine.
Answer: Thymine and Cytosine fin DNA) or Uracil fin RNA).

Question 36. Name a pyrimidine present in DNA but not in RNA.
Answer: Thymine

Question 37. Name a pyrimidine present in RNA only.
Answer:  Uradl

Question 38. Which carbon pentose sugar is linked to the base of a nucleotide?
Answer: C-1′

Organic Chemistry Biomolecules

Question 39. Write down the full names of DNA and RNA.
Answer: DNA: Deoxyribonucleic acid; RNA: Ribonucleic acid.

Question 40. Name the common bases present in DNA and RNA.
Answer: Adenine, Guanine, Cytosine

Question 41. Name the monomeric unit of nucleic acid.
Answer: Nucleotide

Question 42. How many H-bonds are there between A and T?
Answer: 2

Question 43. How many H-bonds are there between G and C?
Answer: 3

Question 44. Which nucleic acid does not have uracil?
Answer: DNA

Question 45. Which nucleic acid does not have thymine?
Answer: RNA

Question 46. Name a nucleotide of RNA carrying adenine.
Answer: AMP

Question 47. Which nucleic acid acts as a genetic material?
Answer: DNA

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