Biomolecules

The characteristic feature of given amino acids are

$ \text { Ser } \mathrm{HO}-\mathrm{CH}_2- $

Cys $\mathrm{HS}-\mathrm{CH}_2-$

Lys $\mathrm{H}_2 \mathrm{~N}-\left(\mathrm{CH}_2\right)_4-$

Thr $\mathrm{H}_3 \mathrm{C}-\mathrm{CHOH}-$

$ \text { Tyr } \quad(p) \mathrm{HO}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{CH}_2- $

Thus $X$ and $Y$ respectively are Tyr and Gln as they contain phenolic hydroxy group and amide group respectively.

$ \text { The given general formula is } $

Then, $R=-\mathrm{CH}_2-\mathrm{C}_6 \mathrm{H}_4 \mathrm{OH}$

(para position) gives tyrosine.

When $\mathrm{R}=\mathrm{CH}_2-\mathrm{SH} \longrightarrow$ Cysteine

When $R=\mathrm{CH}_2-\mathrm{CONH}_2 \longrightarrow$

Asparagine.

When

$ R=-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{S}-\mathrm{CH}_3 \longrightarrow $

Methionine not serine.

As serine has the $R$ group : $-\mathrm{CH}_2-\mathrm{OH}$

The incorrect statement about monosaccharide units formed by hydrolysis of maltose is given in option (b). Its correct form is already given in option (a).

Malt contain enzyme that converts starch into maltose specifically the enzyme diastase is responsible for breaking down starch into maltose

$ \underset{\substack{\text { Starch } \\(X)}}{\left(\mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_5\right)_n} \xrightarrow[\text { Amylose }]{\text { Diastase/ }} n \times \underset{\substack{\text { Maltose } \\(Y)}}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}} $

$ \text { The reaction involved is as follows, } $

The correct answer is:

✅ Option B: Insulin

Explanation:

• Polypeptide hormones are made up of chains of amino acids.

• Insulin is a classic example — it’s composed of two peptide chains (A and B chains) linked by disulfide bonds.

• Epinephrine is a catecholamine (derived from the amino acid tyrosine).

• Estrogen and androgen are steroid hormones derived from cholesterol.

Hence, Insulin is the polypeptide hormone among the options.

The reaction will takes place as follows

Statement $ I $ is incorrect but statement II is correct. The correct form of statement $ I $ is,

Lactose commonly known as milk sugar is made up of $ \beta $-D-galactose $ (A) $ and $ \beta $-D-glucose $ (B) $.

Vitamin $\mathrm{B}_6$ is also called as pyridoxine.

Vitamin $\mathrm{B}_{12}$ is also called as cyanocobalamin.

Vitamin $B_2$ is also called as Riboflavin.

Vitamin $B_1$ is also called as Thiamine.

Cellulose is a polysaccharide and it is made of $\beta$-D glucose units joined through 1, 4-glycosidic linkage.

Starch is a condensation polymer of hundreds of $\alpha-\mathrm{D}(+)$-glucose monomer. Thus, statement I is correct.

Amylose is solube in water and can be hydrolysed into glucose units by enzymes $\alpha$-amylase. Amylase catalyses the hydrolysis of starch into sugar. So, statement (II) is incorrect. Amylose is an unbranched linear polymer composed of $\alpha$-D-glucose units linked by $\alpha-(1,4)$ linkages. So, statement (III) is incorrect.

Cellulose is straight chain polymer compound of only D-glucose units joined by $\beta$-glycosidic linkage. So, statement (IV) is correct.

Deficiency of vitamin $B_{12}$ causes pernicious anaemia. (A decrease in red blood cells).

Vitamin D is soluble in fat and its sources are egg yolk, exposure to sunlight, fish etc. Deficiency of vitamin D causes rickets disease.

The given Haworth's projection is of $\alpha$-D-(+) glucopyranose.

It has 6-membered pyranose ring having 5 -carbon atoms and one oxygen-atom. It is alpha because at $\mathrm{C}-1$ atom, -OH group is present downwards.

The asparagine amino acid consists of an alpha-amino group ( $-\mathrm{NH}_2$ ), an alpha carboxylic group $(-\mathrm{COOH})$ and a side chain of

The incorrect statement about maltose is given in option (c) because it is composed of 2 molecules of $\alpha$-D-glucose.

Both (A) and (R) are true but (R) is not correct explanation for (A), as the amino acids exist in dipolar form due to presence of positive $\left(-\stackrel{+}{\mathrm{N}} \mathrm{H}_3\right)$ group and negative ( $-\mathrm{COO}^{-}$) group. The dipolar form is called as Zwitter ion form.

Fructose possess molecular formula $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$. Structure of $D$ and $L$ fructose are

$D$ and $L$ configurations are relative configurations. Glucose and fructose

both have -OH group on the bottom carbon to the right side. They have same $D$-configuration. Hence, they are called $D$-glucose and $D$-fructose. So, assertion (A) is true. Glucose rotates plane polarised light to the right side or clockwise direction. So, it is dextrorotatory.

Whereas fructose rotates plane polarised light to left side or anti-clockwise. So, fructose is laevorotatory. So, given reason (R) is false.

1 mole of sucrose on acid hydrolysis produces 1 mole of glucose and 1 mole of fructose.

$ \underset{\text { Sucrose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}} \xrightarrow{\text { Acid hydrolysis }} \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6} $

Molecular mass of sucrose

$ \begin{aligned} & =(12 \times \text { atomic mass of } \mathrm{C})+(22 \times \text { mass of } \\ & \mathrm{H})+(11 \times \text { Atomic mass of } \mathrm{O}) \\ & =(12 \times 12)+(22 \times 1)+(11 \times 16)=342 \mathrm{~g} \end{aligned} $

Therefore, amount of sucrose needed to produce 1 mole of glucose using acid hydrolysis is 342 g .

The structure of proteins is destroyed by the application of heat or chemical agents and the process i. called denaturation. It involves the disruption of both the secondary and tertiary structures of a protein, but the primary structure, which involves a sequence of amino acids does not get disturbed and remains intact. This results in the loss of the biological activity of the proteins.

The protein in milk is normally suspended in a colloidal solution, which means that the small protein molecules float around freely and independently. On adding lime juice to milk the pH decreases and the small protein molecules gets coagulated and forms large chain. This is denaturing of protein and causes curdling.

Therefore the (A) is false but (R) is true.

Xerophthalmia is a medical condition in which eye fails to produce tears. It may be caused by deficiency of vitamin (A). Hence, option (d) is the correct answer.

(ii) Anomeric crystalline form

(i) do not give Schiff's test, because, - CHO group is not free.

(iv) Pentaacetate do not react with$ \text { hydroxylamine, because } $

is not available.

So, statements (i), (ii) and (iv) support the cyclic structure of glucose.

$ \text { Zwitter ionic form of aspartic acid is } $

As, the species bear negatively charge over $\mathrm{COO}^{\ominus}$ and positivity charge over $\mathrm{H}_3 \stackrel{\oplus}{\mathrm{~N}}$-; due to migration of H -atom from -COOH to $-\mathrm{NH}_2$.

Hence, option (b) is the correct answer.

$ \text { (i) Histidine } $

$ \text { (ii) Valine } $

$ \text { (iii) Arginine } $

$ \text { (iv) Proline } $