Biomolecules

• RNA is typically single standard but it contains uracil, not thymine.

• DNA is secondary structure has a double strand helix consisting of two polynucleotide chains. Its four nitrogenous bases are adenine (A), guanine (G), cytosine (C) and thymine (T).

In $\beta$-pleated sheet structure, peptide chains are held together by intermolecular hydrogen bonds.

This is only correct statement.

Remaining (1), (2) and (3) are incorrect statements.

In case both nitrogen and sulphur are present in an organic compound, it gives blood red colour with sodium nitroprusside.

D-Fructose is found in honey and is a keto sugar.

$\text { D-(-) - Fructose }$

Let's analyze the statements one by one:

Statement I: Glycogen is similar to amylose in its structure.

This statement is not correct. Glycogen and amylose are both polysaccharides, but they have different structures. Amylose is a linear polymer of glucose units linked by α-1,4-glycosidic bonds, which makes it a component of starch. Glycogen, on the other hand, is a highly branched polymer of glucose with α-1,4-glycosidic bonds in the linear chains and α-1,6-glycosidic bonds at the branching points. This high level of branching makes glycogen structurally distinct from amylose.

Statement II: Glycogen is found in yeast and fungi also.

This statement is correct. Glycogen is a form of energy storage in many organisms, including animals, plants, bacteria, and fungi. In fungi and yeasts, glycogen serves as a reserve of glucose which can be mobilized when needed.

Therefore, the correct option is:

Option B

Statement I is false but Statement II is true.

Let's analyze each reagent and its reaction with glucose to determine which ones do not react accordingly.

Glucose is an aldohexose, meaning it has an aldehyde group in its straight-chain form and can form a cyclic hemiacetal structure. This structure is predominant under physiological conditions.

A. Tollen's Reagent: Tollen's reagent is a solution of silver nitrate in ammonia, used to detect reducing sugars. Glucose, having a free aldehyde group or an easily openable hemiacetal in its cyclic form, can reduce Tollen's reagent to metallic silver, showing a positive test. Thus, glucose reacts with Tollen's reagent.

B. Schiff's Reagent: Schiff's reagent is used to detect aldehydic groups. It is sensitive to free aldehydes, turning pink when aldehydes are present. Since the aldehyde form of glucose is present in very minute amounts at equilibrium (most glucose is in the cyclic form), it generally does not react strongly with Schiff's reagent, leading to a negative test.

C. HCN: Hydrogen cyanide (HCN) can react with carbonyl groups to form cyanohydrins. In the case of glucose, the reactive aldehyde group can react with HCN to form glucosyl cyanohydrin. Thus, glucose does react with HCN.

D. $\mathrm{NH}_2\mathrm{OH}$ (Hydroxylamine): Hydroxylamine reacts with carbonyl compounds to form oximes. Glucose, containing an aldehyde group, can react with hydroxylamine to form glucose oxime. Therefore, glucose reacts with hydroxylamine.

E. $\mathrm{NaHSO}_3$ (Sodium Bisulfite): Sodium bisulfite reacts with aldehydes to form bisulfite adducts. However, because glucose in solution predominantly exists in its cyclic form rather than the open-chain form, it does not readily react with sodium bisulfite. Hence, glucose generally does not react with sodium bisulfite, unless placed under certain conditions that favor the open-chain form.

In conclusion, the reagents with which glucose does not react to give the corresponding tests/products are:

B. Schiff's Reagent and

E. $\mathrm{NaHSO}_3$

Therefore, the correct answer is Option C: B and E.

Fehling's solution is a chemical reagent used to differentiate between water-soluble carbohydrates and ketone-functional groups, and as a test for monosaccharides. It is used specially in the Fehling's test for reducing sugars. The test involves two solutions, generally known as Fehling's A and Fehling's B, which are mixed together and added to perform the test.

Fehling's solution 'A' is an aqueous solution of copper(II) sulfate. Thus, the correct option is:

Option A: aqueous copper sulphate

The function of Fehling's A is to provide copper(II) ions, $ Cu^{2+} $, which act as the oxidizing agent in the reaction with the reducing sugar. When Fehling's Solution A and B are mixed and heated with a reducing sugar, the copper(II) ions are reduced to copper(I) oxide, which precipitates as a red solid, indicating a positive result.

Fehling's Solution B, on the other hand, contains alkaline sodium potassium tartrate, which helps to maintain the solution in an alkaline condition and keeps the copper(II) ions in solution.

Thus, in summary, Fehling's A consists of aqueous copper sulfate, which directly matches with Option A.

Base link with 1' position of sugar in nucleoside so statement I is correct.

$\to$ When nucleoside is linked to phosphoric acid at 5' position of sugar moiety we get a nucleotide

Statement II is incorrect because not link with phosphorous acid.

$\bullet$ Proteins are polymers of $\alpha$-amino acids and they are connected to each other by peptide bond, but this is not denaturation process.

$\bullet$ Due to denaturation globules unfold and helix get uncoiled and protein loses its biological activity.

Denaturation can be caused if protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH

Sucrose is a disaccharide composed of two monosaccharide units, glucose and fructose, joined together via a glycosidic bond. Upon hydrolysis, this bond is broken, and the two constituent monosaccharides are released. Hydrolysis of sucrose results in the formation of one molecule of glucose and one molecule of fructose. However, it is important to specify the configurations of these monosaccharides as they exist in specific forms.

Glucose in sucrose is in the $\alpha$-D form while fructose is in the $\beta$-D configuration. Thus, the correct answer reflecting the products of sucrose hydrolysis is:

Option B: $\alpha$-D-Glucose + $\beta$-D-Fructose

This reflects the correct stereochemistry and identity of the monosaccharides produced from the hydrolysis of sucrose.

Sucrose on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-)-fructose. These two monosaccharides are held together by a glycosidic linkage between $\mathrm{C}$-1 of $\alpha$-glucose and $\mathrm{C}$-$2$ of $\beta$-fructose. Both, assertion and reason are true and reason is the correct explanation of the given assertion. Hence, option (a) is the correct.

In nature glucose exists in cyclic hemiacetal form and has no free $-$CHO group, thus does not give 2, 4-DNP test.

Therefore, assertion and reason both are correct and reason is the correct explanation of the given assertion.

Hence, option (a) is the correct.

The amino acids that are not synthesised in human body and must be obtained from our diet are essential amino acids. Tryptophan is an example of essential amino acids.

Therefore, both assertion and reason are incorrect.

On hydrolysis with dilute aqueous sulphuric acid, sucrose forms a equimolar mixture of $\mathrm{D}-(+)$-glucose and $\mathrm{D}$ - $(-)$-fructose. The reaction for the hydrolysis of sucrose is as follow

$\underset{\text { Sucrose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\mathrm{H}_2 \mathrm{O}$ $\xrightarrow{\mathrm{H}_2 \mathrm{SO}_4}$ $\underset{\text { D-(+)-glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}$ + $\underset{\text { D-(-)-glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}$

Solution of sucrose is dextrorotatory (specific rotation $=+66.1^{\circ}$) but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. The solution of formed product is found to be laevorotatory (specific rotation $=-20.0^{\circ}$)