Biomolecules

Vitamin deficiencies can lead to various health issues. Here's a brief explanation of two specific vitamin deficiencies:

Vitamin B₁ (Thiamine) Deficiency: A lack of vitamin B₁ can lead to a condition known as beriberi. This condition affects the cardiovascular and nervous systems, causing symptoms such as weakness, confusion, and difficulty breathing.

Vitamin B₆ (Pyridoxine) Deficiency: Insufficient vitamin B₆ can result in convulsions or seizures. This vitamin is crucial for neurotransmitter synthesis and proper brain function.

Thus, the deficiency of vitamin B₁ causes beriberi, and the deficiency of vitamin B₆ causes convulsions.

Five -OH group are present in 152 both open and closed (ring) structure of D-glucose.

A. Hydrolysis of starch to maltose: This reaction is catalyzed by the enzyme Diastase. So, A matches with I.

B. [This entry appears to be unrelated to enzymes and reactions. It seems to be mistakenly included. Therefore, we will skip it.]

C. Hydrolysis of sucrose to glucose and fructose: This process is facilitated by the enzyme Invertase. Thus, C corresponds to III.

D. Glucose to ethanol: The enzyme responsible for this conversion is Zymase. Hence, D matches with IV.

Based on these matches:

A - I

C - III

D - IV

Therefore, the correct answer is Option D: A-I, C-III, D-IV.

The correct answer is Option C: Glutamine.

Essential amino acids are those that the body cannot produce in sufficient quantities on its own, so they must be obtained from the diet. Here's a quick breakdown of the options:

$\textbf{Lysine}$: An essential amino acid.

$\textbf{Histidine}$: Generally considered essential, especially in growing individuals.

$\textbf{Methionine}$: An essential amino acid.

$\textbf{Glutamine}$: Although very important for many functions (e.g., muscle recovery, immune function), it is non-essential because the human body can synthesize it.

Thus, Option C, Glutamine, is not an essential amino acid.

Fructose is not a disaccharide.

A disaccharide is a type of sugar that forms when two monosaccharides join together through a glycosidic linkage. Fructose, on the other hand, is a monosaccharide, meaning it is a single sugar molecule and not composed of two units like disaccharides are.

$ \begin{array}{ll} \hline \text { Name } & \text { Chemical formula } \\ \hline \text { Morphine } & \mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_3 \\ \hline \text { Heroin } & \mathrm{C}_{21} \mathrm{H}_{23} \mathrm{NO}_5 \\ \hline \text { Penicillin } & \mathrm{C}_{16} \mathrm{H}_{18} \mathrm{~N}_2 \mathrm{O}_4 \mathrm{~S} \\ \hline \text { Terpineol } & \mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O} \\ \hline \text { Cimetidine } & \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_6 \mathrm{~S} \\ \hline \end{array} $

Let's analyze each option:

Option A: Vitamin A is a fat-soluble vitamin, not water-soluble. Its deficiency can lead to xerophthalmia, which is correct, but the solubility is mismatched.

Option B: Vitamin B₆ is water-soluble, but its deficiency does not cause scurvy. Scurvy is actually due to a deficiency of Vitamin C.

Option C: Vitamin D is a fat-soluble vitamin, and its deficiency can lead to rickets. This set is correctly matched.

Option D: Vitamin C is water-soluble, not fat-soluble, and its deficiency leads to scurvy. Also, convulsions are not typically associated with Vitamin C deficiency.

Thus, the correctly matched set is:

$\textbf{Option C: Vitamin D - fat soluble - rickets.}$

For statement I: In double-stranded DNA, according to Chargaff’s rules, the number of cytosine (C) bases is equal to the number of guanine (G) bases. This is because C pairs with G. Therefore, when DNA is hydrolyzed, cytosine and guanine are released in equal quantities.

For statement II: RNA is typically single-stranded and does not follow the strict base pairing rules found in DNA. Thus, the relationship between adenine (A) and uracil (U) in RNA is not necessarily one of equal quantities.

Based on this analysis:

Statement I is correct.

Statement II is incorrect.

So, the correct answer is Option C: Statement I is correct but Statement II is incorrect.

The constitution of a protein refers to its basic makeup—the specific sequence of amino acids that form the protein. This level of organization is known as the $\textbf{primary structure}$ of the protein.

Here's a quick breakdown:

$\textbf{Primary Structure:}$ The linear sequence of amino acids held together by peptide bonds.

$\textbf{Secondary Structure:}$ Local arrangements such as alpha helices and beta sheets formed by hydrogen bonding.

$\textbf{Tertiary Structure:}$ The overall three-dimensional folding of a single polypeptide.

$\textbf{Quaternary Structure:}$ The arrangement of multiple polypeptide subunits in a multi-subunit protein.

Since "constitution" is about the fundamental composition, the correct answer is:

Option C: Primary structure.

Carrots are a source of vitamin A, which is a fat-soluble vitamin essential for maintaining good vision and supporting the immune system. On the other hand, curd provides vitamin B₁₂, a water-soluble vitamin that is important for blood cell formation.

A nucleoside consists solely of a sugar and a nitrogenous base. When a phosphate group is added, it becomes a nucleotide. Therefore, the simplified version of a nucleoside is represented by:

$\text{Nucleoside} = \text{Sugar} + \text{Base}$

So, the correct answer is Option B: Sugar - base.

Essential amino acids are those amino acids, which the body cannot produce by themselves.

While non-essential amino acids are produced by body itself.

$ \begin{array}{ll} \hline \text { Essential } & \text { Non-essential } \\ \hline \text { Amino acid } & \text { Amino acid } \\ \hline \text { Valine (Val) } & \text { Glycine (Gly) } \\ \hline \text { Leucine (Leu) } & \text { Proline (Pro) } \\ \hline \text { Lysine (Lys) } & \text { Serine (Ser) } \\ \hline \end{array} $

Tollens' reagent, which is an ammoniacal silver nitrate solution $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2 \mathrm{OH}\right]$, is used to test for reducing sugars. Reducing sugars have a free aldehyde group that reacts with Tollens' reagent.

In the case of sucrose (C₁₂H₂₂O₁₁), there is no free anomeric carbon available to act as a reducing agent. Therefore, sucrose does not react with Tollens' reagent.

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

Among the given structure valine is an essential amino acid.

Other essential amino acids are : Leucine, isoleucine, lysine etc. Essential amino acid are not synthesised in human body. Hence, it is obtained from other food sources.

The correct answer is Option B: Progesterone.

Here's why:

After ovulation, the corpus luteum releases progesterone.

Progesterone prepares the endometrium (the lining of the uterus) by thickening it, making it suitable for the implantation of a fertilized egg.

Although estradiol (Option A) plays a key role in the initial growth of the uterine lining, it is progesterone that ensures the lining is receptive.

Testosterone (Option C) and thyroxin (Option D) do not have a direct role in preparing the uterus for implantation.

Thus, progesterone is essential for a successful implantation process.

Deficiency of riboflavin (vitamin $ B_2 $) can lead to conditions such as cheilosis, which is characterized by the cracking or scaling of the lips, as well as digestive disorders.

A tetrapeptide consists of four amino acids: alanine, serine, glycine, and valine. Given that alanine is the C-terminal amino acid and the N-terminal amino acid must be chiral, we cannot use glycine at the N-terminal position because it is achiral. This means the N-terminal amino acid can be either valine or serine, as both are chiral.

Considering these constraints, we have the following possible sequences for the tetrapeptide:

Val-Gly-Ser-Ala

Val-Ser-Gly-Ala

Ser-Gly-Val-Ala

Ser-Val-Gly-Ala

These sequences respect the condition of having a chiral N-terminal amino acid, with alanine fixed as the C-terminal amino acid.

The stabilization of secondary (2°) and tertiary (3°) structures of proteins mainly involves interactions such as disulfide bridges (S-S linkages), van der Waals forces, hydrogen bonding, and hydrophobic interactions.

Option A: O-O linkage is not usually involved in the stabilization of protein structures. Oxygen-oxygen single bonds are generally unstable and rarely observed in nature.

Option B: S-S linkage (disulfide bridges) between cysteine residues in a protein can help stabilize the protein's structure, particularly its tertiary structure.

Option C: Van der Waals forces can contribute to the stabilization of protein structures, particularly in the interior of the protein where they can promote hydrophobic interactions.

Option D: Hydrogen bonding can play a critical role in the formation and stabilization of secondary structures such as alpha-helices and beta-sheets.

Therefore, the correct answer is Option A (-O-O- linkage).

The one-letter code for each amino acid is as follows:

Glutamic Acid (A) is represented by the letter E (III).

Glutamine (B) is represented by the letter Q (I).

Tyrosine (C) is represented by the letter Y (IV).

Tryptophan (D) is represented by the letter W (II).

These one-letter codes are part of the standard notation used in biology and biochemistry to represent the 20 standard amino acids, plus a few others. The codes are designed to be short and easy to remember, while still being distinct from each other.

In protein chemistry, each amino acid is assigned a one-letter code to simplify notation. The matching for the given amino acids to their one-letter codes is as follows:

• Arginine (A) - R (IV)
• Aspartic Acid (B) - D (I)
• Asparagine (C) - N (II)
• Alanine (D) - A (III)

So, the correct answer is Option C: (A) - IV, (B) - I, (C) - II, (D) - III.

Let's match the enzymes with the corresponding reactions:

• The enzyme that catalyzes the hydrolysis of sucrose to glucose and fructose is invertase. So, A matches with III.


• The enzyme that catalyzes the conversion of glucose to ethyl alcohol and CO(_2) (a step in fermentation) is zymase. So, B matches with I.


• The enzyme that catalyzes the conversion of starch to maltose is diastase. So, C matches with IV.


• The enzyme that breaks down proteins into amino acids is pepsin. So, D matches with II.

So, the correct option is:

Option B A-III, B-I, C-IV, D-II

The reaction of $\alpha$-halocarboxylic acid with dilute ammonia is known as the haloform reaction. In this reaction, the halogen atom of $\alpha$-halocarboxylic acid is replaced by an amino group to form $\alpha$-amino carboxylic acid. The yield of $\alpha$-amino carboxylic acid is good because the amino group is a better nucleophile than the halogen atom, and it attacks the carbonyl carbon of the carboxylic acid to form an intermediate, which is then hydrolyzed to form the product. On the other hand, when alkyl halides are reacted with ammonia, the yield of amines is very low because alkyl halides are less reactive than $\alpha$-halocarboxylic acids.

The reason (R) is also true. Amino acids exist in the zwitterion form in an aqueous medium. In this form, the amino group has a positive charge, and the carboxyl group has a negative charge. The zwitter ion form is stabilized by the interaction between the opposite charges, which is known as an electrostatic interaction. This interaction makes the zwitterion form more stable than the neutral form.

C₂ and C₃OH are cis

C₃ and C₄ are anti to each other.

The tetrapeptide code is

FLDY

F $\to$ Phenyl alanine

L $\to$ Leucine

D $\to$ Aspartic acid

Y $\to$ Tyrosine

(i) $\alpha$-D-(-)-Fructofuranose -

(ii) $\beta$-D-(-)-Fructofuranose -

(iii) $\alpha$-D-(-)-Glucopyranose -

(iv) $\beta$-D-(-)-Glucopyranose-

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.

Carbohydrates which differ in configuration at a carbon other than the anomeric carbon. Glucose and mannose are differ in configuration around $\mathrm{C}_2$ and hence these are $\mathrm{C}_2$ epimers.