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Maharashtra State Board Class 12
Chemistry Solutions Chapter 14
Biomolecules
- Select the most correct choice. Interactive textbooks for Maharashtra board students Online learning platforms Question i. CH 2 OH-CO-(CHOH) 4 -CH 2 OH is an example of a. Aldohexose b. Aldoheptose c. Ketotetrose d. Ketoheptose Answer: (d) Ketoheptose Question ii. The open chain formula of glucose does not contain a. Formyl group b. Anomeric hydroxyl group c. Primary hydroxyl group d. Secondary hydroxyl group Answer: (b) Anomeric hydroxyl group Question iii. Which of the following does not apply to CH 2 NH 2 – COOH a. Neutral amino acid b. L – amino acid c. Exists as zwitterion d. Natural amino acid Answer: (d) Natural amino acid
Question iv. Tryptophan is called essential amino acid because a. It contains an aromatic nucleus. b. It is present in all the human proteins. c. It cannot be synthesized by the human body. d. It is an essential constituent of enzymes. Answer: (c) It cannot be synthesised by human body. Question v. A disulfide link gives rise to the following structure of protein. a. Primary b. Secondary c. Tertiary d. Quaternary Answer: (c) Tertiary Question vi. RNA has a. A – U base pairing b. P – S – P – S backbone c. double helix d. G – C base pairing Answer: (a) A – U base pairing
- Give scientific reasons : Question i. The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test. Answer: (1) In disaccharide sucrose, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a nonreducing sugar. As there is no free aldehyde group, it does not reduce Tollen’s reagent to metallic silver. Hence, sucrose gives negative Tollen’s test.
forms a salt, which is a dipolar ion called zwitter ion. Question iv. Hydrolysis of sucrose is called inversion. Answer: Sucrose is dextro rotatory. On hydrolysis it gives equimolar mixture of D – ( + ) glucose and D – ( -) fructose. Since the laevorotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.7°), the hydrolysis product has net laevorotation. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro ( + ) to laevo (-) and the product is called as invert sugar and so the hydrolysis of sucrose is called inversion. Question v. On boiling, egg albumin becomes opaque white. Answer: Upon boiling the egg, denaturation αcurs. During denaturation, secondary and tertiary structures are destroyed, but primary structure remains intact. Egg contains soluble globular proteins, which forms insoluble fibrous proteins (opque) on boiling egg.
- Answer the following Question i. Some of the following statements apply to DNA only, some to RNA only and some to both. Lable them accordingly. a. The polynucleotide is double stranded. ( …………… ) b. The polynucleotide contains uracil. ( …………… )
c. The polynucleotide contains D-ribose ( …………… ). d. The polynucleotide contains Guanine ( …………… ). Answer: (1) The polynucleotide is double stranded. (DNA) (2) The polynucleotide contains uracil. (RNA) (3) The polynucleotide contain D-ribose (RNA) (4) Thc polynucleotide contains Guanine (DNA, RNA) Question ii. Write the sequence of the complementary strand for the following segments of a DNA molecule. a. 5′ – CGTTTAAG – 3′ b. 5′ – CCGGTTAATACGGC – 3′ Answer: (1) DNA molecule : 5′ – CGTTTAAG – 3′ The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G. (2) DNA molecule : 5′ – CCGGTTAATACGGC – 3′ The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G. Question iii. Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine. Answer: (1) Dipeptide formed from alanine and glycine
(formyl group) group.
- Draw a neat diagram for the following: Question i. Haworth formula of glucopyranose Answer: Question ii. Zwitter ion
Answer: Question iii. Haworth formula of maltose Answer: Question iv. Secondary structure of the protein Answer: The structure of proteins can be studied at four different levels i.e. primary, secondary, tertiary and quaternary levels. Each level is more complex than the previous one. (1) Primary structure of proteins : (a) Representation by structural formula (b) Representation with amino acid symbols
(1) Each turn of the helix has 3.6 amino acids. (2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain. (3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core. Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure. β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are
The C = O and N – H bonds lie in the planes of the sheet. Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains. The R groups are oriented above and below the plane of the sheet. The β-pleated sheet arrangement is favoured by amino acids with small R groups.
Haemoglobin can do its function of oxygen transport only when all the four subunits are together. Question v. AMP Answer:
Question vi. dAMP Answer: Question vii. One purine base from nucleic acid Answer: Question viii. Enzyme catalysis
respectively? (2) Which other functional groups are present in these three compounds? Answer: (1) Glucose contains five hydroxyl (- OH) groups. Fructose contains five hydroxyl ( – OH) groups. Ribose contains four hydroxyl ( – OH) groups. (2) Glucose contains aldehyde ( – CHO) as other functional group. Fructose contains ketonic group as other functional group. Ribose contains aldehyde ( – CHO) as other functional group. Use your brain power! (Textbook Page No 299) Question 1. Give IUPAC names to the following monosaccharides. Answer: (1) Aldotriose (2) Aldopentose (3) Ketoheptose Problem 14.1 : (Textbook Page No 300) Question 1. An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (- OH) groups must be present in the original compound? Solution :
In acetylation reaction H atom of an (- OH) group is replaced by an acetyl group (- COH 3 ). This results in an increase in molecular mass by [(12 + 16 + 12 + 3 x 1) – 1] that has, 42 u. In the given alcohol, increase in molecular mass = 174 u – 90 u = 84 u ∴ Number of – OH groups=\frac{84 \mathrm{u}}{42 \mathrm{u}}= Use your brain power! (Textbook Page No 301) (1) Write structural formula of glucose showing all the bonds in the molecule. (2) Number all the carbons in the molecules giving number 1 to the ( – CHO) carbon. (3) Mark the chiral carbons in the molecule with asterisk (*). (4) How many chiral carbons are present in glucose? Answer: Refer structural formula of glucose for (1) (2) and (3). (4) There are 4 chiral carbon atoms present in glucose. Use your brain power! (Textbook Page No 306) Question 1. (1) Is galactose an aldohexose or a ketohexose? (2) Which carbon in galactose has different configuration compared to glucose?
Answer: In tryptophan, nitrogen atom present in cyclic structure cannot donate pair of electrons as it is stabilized by resonance. The other amino group and carboxylic group present in the side chain neutralize each other. Tryptophan has equal number of amino and carboxylic groups. Hence, tryptophan is a neutral amino acid. In histidine, amino groups are more in number than carboxyl groups therefore histidine ¡s basic in nature. Can you think? (Textbook Page No 309) Question 1. Compare the molecular masses of the following compounds and explain the observed melting points. Answer: Above compounds have same molecular masses but they have different melting points, a-amino acids have higher melting points compared to the corresponding amines or carboxylic acids of
comparable masses. This property is due to the presence of both carboxylic group (acidic) and amino group (basic) in the molecule. In aqueous solution, protons transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called – Zwitter ion. Use your brain power! (Textbook Page No 310) Question 1. (1) Write the structural formula of dipeptide formed by combination of carboxyl group of alanine and amino group of glycine. (2) Name the resulting dipeptide. (3) Is this dipeptide same as glycyalanine or its structural isomer? Answer: (2) ala-glycine. OR ala-gly (3) It is a structural isomer. Question 54. Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine. Problem 14.3 : (Textbook Page No 311) Question 1. Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine,
