Class 12 chemistry chapter 14 Biomolecules solutions

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Class 12 chemistry textbook Solutions for Class 12, Chemistry  Chapter 14 Biomolecules maharashtra tate board are provided here with simple step-by-step  detailed explanations. These solutions for Biomolecules  are very popular among Class 12 students for chemistry chapter 14 Biomolecules  Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the chemistry textbook Solutions Book of Class 12 chemistry Chapter 14 are provided here for you for free. You will also love the experience on ybstudy class 12 Solutions. All chemistry textbook Solutions. Solutions for class 12, These chemistry textbook solutions are prepared by  Chemistry experts and are 100% accurate.

1. Select the most correct choice.
i. CH2OH-CO-(CHOH)4CH2OH is an 
example of
a. Aldohexose
b. Aldoheptose
c. Ketotetrose
d. Ketoheptose

ii. Open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group

iii. Which of the following does not apply 
to CH2NH2- COOH
a. Neutral amino acid
b. L – amino acid
c. Exists as zwitter ion
d. Natural amino acid

iv. Tryptophan is called essential amino 
acid because
a. It contains aromatic nucleus.
b. It is present in all the human
proteins.
c. It cannot be synthesised by human 
body.
d. It is essential constituent of
enzymes.

v. A disulfide link gives rise to the following structure of protein.
a. Primary
b. Secondary
c. Tertiary
d. Quaternary

vi. RNA has
a. A – U base pairing
b. P-S-P-S backbone
c. double helix
d. G – C base pairing

2. Give scientific reasons :
i. The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
Answer :

ii. On complete hydrolysis DNA gives 
equimolar quantities of adenine and 
thymine.
Answer : A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine and similarly, the quantity of cytosine is equal to that of guanine.

iii. α – Amino acids have high melting points compared to the corresponding 
amines or carboxylic acids of comparable molecular mass.
Answer : Amino acids have two functional groups −NH2 and −COOH. Hence the amino acids exist as a dipolar ion (or) zwitterion. As a result, strong electrostatic forces exist between zwitterions. Hence amino acids have high melting points than the corresponding halo-acids.

iv. Hydrolysis of sucrose is called 
inversion.
Answer : Sucrose (C12H22O11) is
dextrorotatory (+66.50). On hydrolysis with
dilute acid or an enzyme called invertase
sucrose gives equimolar mixture of D-(+)
glucose and D-(-) fructose.

Since the laevoratotion of fructose (-92.40)
is larger than the dextrorotation of glucose
(+52.70), the hydrolysis product has net
laevorotation. Hence hydrolysis of sucrose
is also called inversion of sucrose, and the
product is called invert sugar.

v. On boiling egg albumin becomes 
opaque white.
Answer : As an egg cooks, the heat causes the bonds within the proteins to break, a process called denaturation. As these proteins chains unfold and entangle with other proteins, new bonds form between these amino acids and the amino acids of neighboring proteins, causing the texture to change and it turns white.

3. Answer the following
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. (……… )
Answer : DNA

b. The polynucleotide contains uracil. 
(……… )
Answer : RNA

c. The polynucleotide contains D-ribose (……… ).
Answer : DNA

d. The polynucleotide contains Guanine (……… ).
Answer : Both DNA & RNA

ii. Write the sequence of the 
complementary strand for the 
following segments of a DNA 
molecule. 
a. 5′ – CGTTTAAG – 3′
Answer :
5′ – CGTTTAAG – 3′
3′ – GCAAATTG – 3′

b. 5′ -CCGGTTAATACGGC – 3′
Answer :
5′ -CCGGTTAATACGGC – 3′
3′ – GGCCAATTA+GCCG – 5′

iii. Write the names and schematic representations of all the possible 
dipeptides formed from alanine, 
glycine and tyrosine.
Answer :

iv. Give two evidences for presence of
formyl group in glucose.
Answer : To prove the presence of −CHO group: When glucose reacts with hydroxylamine, oxime is formed and cyanohydrins are formed on the addition of hydrogen cyanide to it. This reaction can confirm the presence of carbonyl group in glucose.

4. Draw a neat diagram for the following:
i. Haworth formula of glucopyranose

ii. Zwitter ion

iii. Haworth formula of maltose

iv. Secondary structure of protein

vii. One purine base from nucleic acid

viii. Enzyme catalysis