Biomolecules

Structure (1) given is of sucrose which is non reducing.

For non reducing sugar compound should have acetal linkage not hemi acetal linkage.

Gly ala val

Gly val ala

Val gly ala

Val ala gly

Ala val gly

Ala gly val

Total tri peptides $=6$

DNA and RNA show chirality because they contain a chiral sugar in their structure.

In DNA, the sugar present is chiral 2-deoxy-Ribose.

In RNA, the sugar present is chiral Ribose.

$ \begin{aligned} & {[\alpha]_{D \text {-sucrose }}=+66.5^{\circ}, \quad[\alpha]_{D \text {-Glucose }}=+52.5^{\circ}} \\ & {[\alpha]_{D \text {-Fructose }}=-92.4^{\circ}} \end{aligned} $

⇒ Sucrose is dextrorotatory and hydrolysed product is laevorotatory.

A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis.

• Enzymes are biological catalysts that increase the rate of reactions by lowering their activation energy.

• Acid catalysis also provides an alternative reaction pathway, but enzymes are typically much more efficient at lowering activation energy under physiological conditions.

• For the hydrolysis of sucrose, the enzyme sucrase (invertase) significantly reduces the activation energy compared to acid-catalyzed hydrolysis, making the reaction much faster.

• Therefore, statement A is correct.

B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact.

• Denaturation is the process by which proteins lose their native three-dimensional structure (secondary, tertiary, and if present, quaternary structures) due to disruptions of weak interactions (like hydrogen bonds, hydrophobic interactions, ionic bonds).

• The primary structure, which is the sequence of amino acids linked by strong covalent peptide bonds, is generally not broken during denaturation. Breaking peptide bonds would require harsh conditions like strong acid/base hydrolysis or specific proteases.

• Therefore, statement B is correct.

C. Nucleotides are joined together by glycosidic linkage between C1 and C4 carbons of the pentose sugar.

• Nucleotides are linked together in nucleic acids (DNA and RNA) by phosphodiester bonds.

• A phosphodiester bond forms between the phosphate group attached to the 5' carbon of one nucleotide and the hydroxyl group on the 3' carbon of the pentose sugar of the next nucleotide.

• A glycosidic linkage (specifically, an N-glycosidic bond) connects the nitrogenous base to the C1 carbon of the pentose sugar within a single nucleotide. It does not link nucleotides together, nor does it involve C1 and C4 carbons between different pentose sugars.

• Therefore, statement C is incorrect.

D. Quaternary structure of proteins represents overall folding of the polypeptide chain.

• The tertiary structure describes the overall three-dimensional folding of a single polypeptide chain.

• The quaternary structure refers to the arrangement of multiple polypeptide subunits (each having its own tertiary structure) to form a functional protein complex. It describes how these individual chains associate with each other.

• Therefore, the statement describes tertiary structure, not quaternary structure.

• Therefore, statement D is incorrect.

Conclusion:

Statements A and B are correct. Statements C and D are incorrect.

The correct option should include only A and B.

The final answer is $\boxed{\text{A and B Only}}$

Arginine and Tryptophan are essential amino acids.

Option (A)

Glucose, when oxidized at the aldehyde group (–CHO), forms gluconic acid. When it is oxidized more strongly at both ends (–CHO and –CH₂OH), it forms saccharic acid.

Option (B)

In a basic medium, fructose changes (tautomerises) into its isomers, D-glucose and D-mannose.

Since glucose and mannose are reducing sugars, this explains why fructose also gives a positive test with Fehling’s solution. This solution contains a weak base (dilute NaOH), which allows the reaction to occur.

Option (C)

When sucrose is hydrolysed using an acid, it breaks down into D-glucose and D-fructose in equal amounts (1:1). This mixture is called invert sugar.

Option (D)

The specific rotation of these sugars is as follows:

For D-fructose: $-92.5^{\circ}$

For D-glucose: $+52.5^{\circ}$

Therefore, the specific rotation of the invert sugar mixture is the average of the two:

$\text{Specific rotation of invert sugar} = -\frac{40}{2} = -20^{\circ}$

When oligopeptides undergo hydrolysis, they yield a small number of α-amino acids. In contrast, proteins do not produce β-amino acids upon hydrolysis. Thus, Statement I is incorrect.

Furthermore, natural proteins, when subjected to denaturation by acids, do not convert fibrous proteins from a water-soluble to a water-insoluble form, as fibrous proteins are inherently not water-soluble. Therefore, Statement II is also incorrect.

In conclusion, both Statement I and Statement II are incorrect.

In Statement I, the reaction shown is incorrect. During the hydrolysis of sucrose, D-(+)-glucose and D-(-)-fructose are produced. However, the statement incorrectly specifies D-(+)-fructose, making Statement I inaccurate.

Statement II is accurate. When sucrose undergoes hydrolysis, it results in the formation of invert sugar, a mixture of glucose and fructose. This change involves the inversion of optical rotation, which is where the term "invert sugar" comes from.

Fat-soluble vitamins are A, D, E and K. From the list:

A: Vitamin B₁ – water-soluble

B: Vitamin C – water-soluble

C: Vitamin E – fat-soluble

D: Vitamin B₁₂ – water-soluble

E: Vitamin K – fat-soluble

So the fat-soluble ones are C (E) and E (K), which is Option C.

To determine the number of tetrapeptide sequences possible using the amino acids glycine (Gly), alanine (Ala), valine (Val), and leucine (Leu), we consider the concept of permutations. Since each tetrapeptide sequence consists of four different amino acids, we need to calculate the number of ways to arrange these four different elements.

The formula for the number of permutations of a set of $ n $ distinct elements is given by $ n! $ (n factorial), where $ n! = n \times (n-1) \times (n-2) \times \cdots \times 1 $.

For our case:

$ n = 4 $

Hence, the number of possible sequences (or permutations) is:

$ 4! = 4 \times 3 \times 2 \times 1 = 24 $

Thus, there are 24 different tetrapeptide sequences possible using each of the four amino acids exactly once.

To clarify the statements regarding amino acids:

Chirality in Amino Acids: Most naturally occurring amino acids contain at least one chiral center, but there are exceptions. For example, Isoleucine has two chiral centers, and Glycine is an exception as it is optically inactive due to lacking a chiral center.

Cysteine and Dimerization: The amino acid Cysteine can easily undergo dimerization. This is due to its free sulfhydryl (SH) group, which can form disulfide bonds, leading to the formation of a dimer.

Carboxyl Groups in Amino Acids: Glutamic acid is not the only amino acid with a carboxyl group in its side chain. Aspartic acid also contains a carboxyl group in its side chain.

Optical Activity: While many naturally occurring amino acids are indeed optically active, Glycine is an exception as it does not possess a chiral center and is hence optically inactive.

In summary:

Not all naturally occurring amino acids are optically active (e.g., Glycine).

Not all naturally occurring amino acids contain only one chiral center (e.g., Isoleucine has two).

Both Glutamic acid and Aspartic acid contain side chain carboxyl groups.

Cysteine can dimerize due to its SH group.

The nine essential amino acids that must be obtained from the diet are:

Histidine

Isoleucine

Leucine

Lysine

Methionine

Phenylalanine

Threonine

Tryptophan

Valine

From the provided list:

Valine (A) is essential.

Proline (B) is non-essential.

Lysine (C) is essential.

Threonine (D) is essential.

Tyrosine (E) is non-essential (it is considered conditionally essential, but under normal conditions it is non-essential).

Thus, the essential amino acids among those listed are Valine, Lysine, and Threonine.

So the correct answer is:

$\textbf{Option A: (A), (C) and (D) only}$

A) Amylose - Polysaccharide of glucose units with glycosidic linkage $\alpha-1,4$.

The source is plant.

So, the linkage is $\alpha-c_1-C_4$.

Glucose units in amylose are linked by alpha bonds between the $1^{\text {st }}$ and $4^{\text {th }}$ carbon atoms.

Amylose match with $\alpha-c_1-c_4$, plant

(A) with (IV)

B) Cellulose - The glycosidic linkage in cellulose is $\beta-1,4$. Glucose monomers are joined by linkage between the $c_1$ of the one glucose ling and the $C_4$ of the next ring.

Source is plant

Cellulose match with $\beta-C_1-C_4$, plant.

(B) with (I)

c) Glycogen - Glycogen is a polymer of glucose that is stored in animals as an energy reserve. It is used as carbohydrate storage in animals. It is a polymer of $\alpha$-glucose with $\alpha-1,4$-glycosidic and $\alpha-1,6$-glycosidic links.

$\alpha-1,4$-linkage forms during linear polymer.

$\alpha-1,6$ - linkage forms during the formation of branching.

Glycogen match with $\alpha-C_1-C_4$, animal

(C) with (II)

D) Amylopectin - It is the polymer of glucose units linked by $\alpha-1,4$ and $\alpha-1,6$ glycosidic bonds.

$x-1,4$ - glucose units linked in a linear way.

$\alpha-1,6$ - These bonds branch out from the linear chains of glucose units.

They are stored in large complexes called starch granules in plants.

Amylopectin match with $\alpha-c_1, c_4, \alpha-c_1-c_6$, plant (D) with (III).

Answer: $(A)-($ IV $),(B)-(I),(C)-($ II $),(D)-($ III $)$. Option (3).

(A) Sucrose $\rightarrow \alpha_1-\beta_2$ Glycosidic linkage

(B) Maltose $\rightarrow \alpha$ 1-4 Glycosidic linkage

(C) Lactose $\rightarrow \beta$ 1-4 Glycosidic linkage

(D) Amylopectin $\rightarrow \alpha$ 1-4 and $\alpha$ 1-6 Glycosidic linkage

$\mathrm{A}-\mathrm{III}, \mathrm{~B}-\mathrm{I}, \mathrm{C}-\mathrm{IV}, \mathrm{D}-\mathrm{II}$

(A) Starch $\xrightarrow{\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}$ Glucose

(B) $\mathrm{\mathop {Cane\,sugar}\limits_{(Sucrose)} \buildrel {{H^ + }/{H_2}O} \over \longrightarrow \mathop {glu\cos e}\limits_{50\% } \, + \mathop {fructose}\limits_{50\% }}$

(C) $\mathrm{\mathop {Milk\,sugar}\limits_{(Lactose)} \buildrel {{H^ + }/{H_2}O} \over \longrightarrow glu\cos e\, + \,galactose}$

(D) Amylopectin $\xrightarrow{\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}$ Glucose

(E) Amylose $\xrightarrow{\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}$ Glucose

So, correct options are B, C and E only

Statement I: D-glucose pentaacetate does not react with 2,4-dinitrophenylhydrazine (DNPH). This is because D-glucose pentaacetate is a derivative where all the hydroxyl groups of D-glucose have been acetylated, eliminating the presence of a free aldehyde group. Since DNPH reacts specifically with free aldehyde or ketone groups to form hydrazones, D-glucose pentaacetate does not participate in this reaction.

Statement II: Starch, when heated with concentrated sulfuric acid at 100°C and under a pressure of 2-3 atmospheres, undergoes hydrolysis. This process breaks the glycosidic linkages in starch, resulting in the production of glucose. This statement is accurate as starch is a polysaccharide composed of glucose units and can be broken down into glucose under these conditions.

The correct information based on the statements is:

Statement I is false, as D-glucose pentaacetate does not react with 2,4-dinitrophenylhydrazine due to the absence of a free aldehyde group.

Statement II is true, as the described method effectively converts starch into glucose.

In Ribose carbohydrate present in DNA is $\beta-2-$ Deoxy-D-Ribose whose structure is

which is a reducing $D$-sugar in $\beta$ anomeric form & it is a pentose sugar.

$\alpha$-helix and $\beta$-pleated sheet belongs to secondary structure of protein, which have hydrogen bonds.

In $\mathrm{A}, \mathrm{B}, \mathrm{D}-\mathrm{OH}$ group in right hand side then D-configuration is assigned.

Correct Answer: Ascorbic Acid

Ascorbic acid is commonly known as Vitamin C. It is a water-soluble vitamin essential for various bodily functions, including:

$\textbf{Antioxidant Function:}$ It helps protect cells from damage caused by free radicals.

$\textbf{Collagen Synthesis:}$ It is crucial for the formation of collagen, a protein necessary for skin, cartilage, tendons, ligaments, and blood vessels.

$\textbf{Immune Support:}$ It supports the immune system by enhancing the body's natural defenses.

$\textbf{Iron Absorption:}$ It improves the absorption of non-heme iron from plant-based foods.

Among the given options, only ascorbic acid qualifies as a vitamin.

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.

Sucrose and cellulose does not reduces Tollen's reagent. This is because they lack free aldehyde or ketone group necessary for the reaction.

Statement-I: Lysine, arginine are essential and basic amino acids.

• Essentiality: Lysine is an essential amino acid, meaning the human body cannot synthesize it and it must be obtained from the diet. Arginine is often considered a conditionally essential amino acid (essential during growth, pregnancy, or illness) but is frequently grouped with essential amino acids in general contexts.

• Basicity: Basic amino acids have an extra amino group or other basic group in their side chain, which is protonated and positively charged at physiological pH.

  • Lysine has a primary amino group (-NH2) at the end of its side chain.

  • Arginine has a guanidinium group in its side chain.

  Both are indeed basic amino acids.

  Therefore, Statement-I is correct.

Statement-II: Leucine, phenylalanine are non essential and neutral amino acids.

• Essentiality:

  • Leucine is one of the nine essential amino acids.

  • Phenylalanine is also an essential amino acid.

  Thus, the claim that they are "non-essential" is incorrect.

• Neutrality:

  • Leucine has an aliphatic, nonpolar side chain, making it a neutral amino acid.

  • Phenylalanine has an aromatic, nonpolar side chain, making it a neutral amino acid.

  Thus, the claim that they are "neutral amino acids" is correct.

Since the first part of the statement (non-essential) is incorrect, Statement-II as a whole is not correct.

Conclusion:

Statement-I is correct.

Statement-II is not correct.

This corresponds to Option C.

The final answer is $\boxed{C}$

The correct structure of $\beta$-D-(-)-fructofuranose is

The reaction requires a free aldehyde group which is not readily available in cyclic structure.

Enzymetic reaction given in option (b) is incorrect for conversion of starch to maltose. The correct form is,

$ \text { Starch } \xrightarrow{\text { Amylase }} \text { Maltose } $

Essential amino acid : Leucine and Histidine

Non essential amino acid : Tyrosine and cysteine.

Essential are those which human body does not produce and vice-versa.

A nucleoside of RNA is composed of a nitrogenous base (like uracil adenine, cytosine or guanine) attached to a ribose sugar molecules.

The correct structure is

1. Amylose:

   • Amylose is a linear polymer made up of D-glucose units.

   • The glucose units are linked together by α-1,4-glycosidic bonds.

2. Amylopectin:

   • Amylopectin is a branched polymer of D-glucose units.

   • The linear chains are formed by α-1,4-glycosidic bonds.

   • Branching occurs due to α-1,6-glycosidic bonds at intervals.

3. Cellulose:

   • Cellulose is a linear polymer of D-glucose units.

   • The glucose units are linked by β-1,4-glycosidic bonds. This type of linkage gives cellulose its rigid, structural properties.

Now, let's match List-I (Glycosidic linkage) with List-II (Polysaccharide):

• (A) α-1,4: This linkage is characteristic of the linear chains in both amylose and amylopectin. Since amylose only has α-1,4 linkages for its main structure, it's the most direct match here. So, A - (I) Amylose.

• (B) β-1,4: This linkage is unique to (III) Cellulose among the given options. So, B - (III) Cellulose.

• (C) α-1,4, α-1,6: The presence of both α-1,4 and α-1,6 linkages indicates a branched polysaccharide with alpha linkages, which is characteristic of (II) Amylopectin. So, C - (II) Amylopectin.

Therefore, the correct matching is:

• A - I

• B - III

• C - II

The aminoacids, which can be synthesised in the body are called non-essential amino acid. While those amino acids which cannot be synthesised in the body must be obtained through diet. Thus, among the given amino acids.

Essential amino acid is $X=$ Lysine, Arginine, Non-essential amino acid is $Y=$ Alanine, Serine, Tyrosine

Statement–I:

Cane sugar is a disaccharide of α-D-glucose and β-D-fructose.

• Cane sugar is sucrose.

• The structure of sucrose consists of α-D-glucose and β-D-fructose units joined by an α,β‑1→2 glycosidic linkage.

✅ Statement-I is correct.

Statement–II:

Milk sugar is a disaccharide of α-D-glucose and β-D-galactose.

• Milk sugar is lactose.

• Lactose is composed of β-D-galactose and β-D-glucose, linked by a β‑1→4 glycosidic bond.

• The glucose part in lactose is β-D-glucose, not α-D-glucose.

❌ Statement-II is not correct.

✅ Final Answer:

Option C:

Statement-I is correct, but Statement-II is not correct.

• Convulsions occur when there is an issue with the normal functioning of the nervous system.

• Vitamin that prevents convulsions: Pyridoxine (Vitamin B₆) is required for the synthesis of neurotransmitters such as GABA.

  Deficiency of Vitamin B₆ → decreased GABA formation → neuronal hyperexcitability → convulsions.

Now let’s check the source:

• Vitamin B₆ (Pyridoxine) is found in cereals, pulses, nuts, and non-vegetarian foods.

✅ Correct answer:

X = Pyridoxine (Vitamin B₆)

Y = Cereals

Answer: Option D — Pyridoxine, cereals

Statement I is correct, but statement II is not correct. The correct form is, $\alpha$-Helix and $\beta$-pleated sheet structure of protein represent secondary structure of it.