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:

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

The monosaccharide units obtained from hydrolysis of oligosaccharides may be or may notbe same.

Example :

Let us recall the vitamin deficiencies and the corresponding diseases:

1. Scurvy – caused due to deficiency of Vitamin C (Ascorbic acid).

   → So, $ A \rightarrow III $

2. Convulsions – occur due to deficiency of Vitamin B$_6$ (Pyridoxine).

   → So, $ B \rightarrow I $

3. Cheilosis – characterized by cracks at the corners of the mouth; caused by deficiency of Vitamin B$_2$ (Riboflavin).

   → So, $ C \rightarrow IV $

4. Xerophthalmia – dryness of the cornea and conjunctiva; caused by deficiency of Vitamin A.

   → So, $ D \rightarrow II $

Hence, the correct matching is:

$ A - III,\; B - I,\; C - IV,\; D - II $

Therefore, the correct answer is Option C.

When the pH changes, the tertiary structure of a protein gets disturbed or broken. However, the amino acids in the protein remain connected to each other through peptide bonds.

Statement I: The two given cyclic structures are the $\alpha$ and $\beta$ forms (anomers) of D-(+)-glucose.

They are called anomers because they differ only in the position (orientation) of the $-OH$ group at the anomeric carbon, i.e. $C_1$ in glucose.

Statement II:

In the open-chain forms of D-glucose and D-fructose, the chiral carbons $C_3$, $C_4$, and $C_5$ have the same arrangement of groups (same configuration) in both molecules.

$ \text { -OH position is same at } C_3, C_4 \text { and } C_5 \text {. } $

Statement A: Glucose exists in two anomeric forms.

This statement is correct. Glucose exists in two different crystalline forms, which are named as the $\alpha$-form and the $\beta$-form. These are known as anomers.

Statement B: Anomers of glucose differ in configuration at $\mathrm{C}-1$ in cyclic hemiacetal structure.

This statement is correct. The $\alpha$ and $\beta$ forms of glucose are cyclic hemiacetals that differ only in the spatial arrangement of the hydroxyl group (-OH) at the first carbon atom ($\mathrm{C}-1$). This specific carbon is referred to as the anomeric carbon.

Statement C: Melting point of $\alpha$ - anomer of glucose is greater than $\beta$ - anomer.

This statement is incorrect. According to experimental data, the melting point of the $\alpha$-form of glucose is $419 \text{ K}$, whereas the melting point of the $\beta$-form is slightly higher, at $423 \text{ K}$. Therefore, the melting point of the $\beta$-anomer is greater than that of the $\alpha$-anomer.

Statement D: Specific rotation of $\alpha$ - anomer is $+19^{\circ}$ while for $\beta$ - anomer is $+112^{\circ}$

This statement is incorrect because the values are interchanged. The specific rotation of the pure $\alpha$-anomer is $+112.2^{\circ}$, and the specific rotation of the pure $\beta$-anomer is $+19.2^{\circ}$. Over time, both undergo mutarotation to reach an equilibrium specific rotation of $+52.5^{\circ}$.

Statement E: $\alpha$ and $\beta$ - anomers of glucose are prepared by crystallization of saturated glucose solution at 303 K and 371 K respectively.

This statement is correct. The $\alpha$-form is typically obtained by the crystallization from a concentrated aqueous solution of glucose at room temperature, around $303 \text{ K}$. On the other hand, the $\beta$-form is obtained by crystallizing it from a hot and saturated aqueous solution at $371 \text{ K}$.

From the analysis above, the mathematically and factually correct statements are A, B, and E.

Therefore, the correct choice is Option D.

Maltose and lactose are both disaccharides.

In each of these structures, one anomeric carbon is used to form the glycosidic linkage, but the other anomeric carbon remains free as a hemiacetal (–OH at anomeric carbon).

If a disaccharide has a free hemiacetal group, it can open up to the open-chain form and can reduce mild oxidising agents (like Tollens’ reagent or Fehling’s solution).

Therefore, both maltose and lactose are reducing sugars because both contain a hemiacetal linkage (free anomeric carbon).

Match each amino acid with its correct one-letter symbol and whether it is essential or non-essential:

• Arginine $\rightarrow$ one-letter symbol is $R$

  In standard school-level classification here, it is matched as Essential

  So, $A \rightarrow II$

• Aspartic acid $\rightarrow$ one-letter symbol is $D$

  It is Non-essential

  So, $B \rightarrow I$

• Lysine $\rightarrow$ one-letter symbol is $K$

  It is Essential

  So, $C \rightarrow IV$

• Glutamic acid $\rightarrow$ one-letter symbol is $E$

  It is Non-essential

  So, $D \rightarrow III$

Thus the correct matching is:

$ A-II,\; B-I,\; C-IV,\; D-III $

So, the correct option is Option A.

The correct pair is:

$\text{Amino acid (A) = Y, Hormone (B) = Thyroxine}$

So, the correct option is:

$\boxed{\text{Option C}}$

Reason:

• The one-letter code $Y$ stands for tyrosine.

• Thyroxine is an iodinated derivative of tyrosine.

• Insulin is a protein hormone, not an iodinated derivative of an amino acid.

Hence,

$\boxed{Y \rightarrow \text{Thyroxine}}$

1. Vitamin B$_1$: The chemical name for Vitamin B$_1$ is Thiamine. Its deficiency in the human body can lead to a disease known as Beri-Beri.

Therefore, A matches with III.

2. Vitamin B$_2$: The chemical name for Vitamin B$_2$ is Riboflavin. A deficiency in this vitamin leads to cheilosis, which is the fissuring at the corners of the mouth and lips.

Therefore, B matches with IV.

3. Vitamin B$_6$: The chemical name for Vitamin B$_6$ is Pyridoxine. Its deficiency can cause convulsions.

Therefore, C matches with I.

4. Vitamin C: The chemical name for Vitamin C is Ascorbic acid. It is a water-soluble vitamin found in citrus fruits, and its deficiency causes Scurvy (bleeding gums).

Therefore, D matches with II.

Now, let us combine our findings:

A $\rightarrow$ III

B $\rightarrow$ IV

C $\rightarrow$ I

D $\rightarrow$ II

Looking at the options provided, the sequence A-III, B-IV, C-I, D-II perfectly matches Option C.

Option C is the correct answer.

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.