Amino Acids and Proteins Questions in English

(A) tripeptide is formed by the combination of $3$ amino acids.
Since there are $20$ different types of naturally occurring amino acids and each position in the tripeptide can be occupied by any of these $20$ amino acids,the total number of possible tripeptides is calculated as $20 \times 20 \times 20 = 20^{3}$.
Therefore,the total number of tripeptides is $8000$.

(D) An amino acid is optically active if it contains at least one chiral carbon atom (an asymmetric carbon atom bonded to four different groups).
Glycine has the structure $NH_2-CH_2-COOH$.
In glycine,the central carbon atom is bonded to two identical hydrogen atoms,meaning it does not have a chiral center.
Therefore,glycine is the only naturally occurring amino acid that is not optically active.

(C)
Fibrous proteins are generally insoluble in water due to the presence of strong intermolecular hydrogen bonds and disulfide linkages that hold the polypeptide chains together.

(C) Thyroxine,also known as tetraiodothyronine or $T_4$,is an iodinated derivative of the amino acid tyrosine.
It is a hormone produced by the thyroid gland and plays a crucial role in regulating the metabolism and growth in the body.

(C) Insulin is composed of two peptide chains referred to as the $A$ chain and $B$ chain.
The $A$ chain consists of $21$ amino acid residues and the $B$ chain consists of $30$ amino acid residues.
These two chains are linked together by two inter-chain disulphide bridges.
Additionally,there is one intra-chain disulphide bridge within the $A$ chain.
Therefore,the total number of disulphide linkages in insulin is $3$.

(B) $Proteins$ are the condensation polymers of $\alpha$-amino acids.
$Proteins$ contain peptide bonds,which are represented by the linkage:
$(-CO-NH-)$.

(B) $1$. Analyze the amino acids provided:
$a)$ Lysine: Essential,Basic
$b)$ Alanine: Non-essential,Neutral
$c)$ Serine: Non-essential,Neutral
$d)$ Arginine: Essential,Basic
$e)$ Tyrosine: Non-essential,Neutral
$2$. $X$ represents essential and basic amino acids: Lysine $(a)$ and Arginine $(d)$.
$3$. $Y$ represents non-essential and neutral amino acids: Alanine $(b)$,Serine $(c)$,and Tyrosine $(e)$.
$4$. Therefore,$X = a, d$ and $Y = b, c, e$.

(C) Statement-$I$ is correct: Lysine and arginine are indeed basic amino acids (due to the presence of extra amino groups) and they are essential amino acids (cannot be synthesized by the human body).
Statement-$II$ is incorrect: While leucine and phenylalanine are neutral amino acids,they are both essential amino acids,not non-essential.
Therefore,statement-$I$ is correct,but statement-$II$ is not correct.

(D) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
Among the given options:
$1$) Leucine is an essential amino acid.
$2$) Histidine is an essential amino acid (often classified as semi-essential in adults but essential in infants).
$3$) Tyrosine and Cysteine are non-essential amino acids as they can be synthesized by the body.
Therefore,$A$ (Leucine) and $D$ (Histidine) are the essential amino acids.
The correct option is $D$.

(C) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through diet.
Among the given options,Leucine is an essential amino acid.

Essential Amino Acid	Non-essential Amino Acid
Arginine	Alanine
Histidine	Asparagine
Isoleucine	Aspartate
Leucine	Glutamate
Lysine	Glycine
Methionine	Serine
Phenylalanine	Tyrosine
Threonine
Tryptophan
Valine

(C) The structure of Asparagine is $H_2N-CO-CH_2-CH(NH_2)-COOH$.
It contains an amino group $(-NH_2)$,a carboxylic acid group $(-COOH)$,and an amide group $(-CONH_2)$ in its side chain.
Therefore,the correct option is $C$.

(A) The given amino acids are: $Val$ (Valine),$Gly$ (Glycine),$Leu$ (Leucine),$Lys$ (Lysine),$Pro$ (Proline),and $Ser$ (Serine).
Essential amino acids are those that cannot be synthesized by the body and must be obtained through the diet. Among the given list,$Val$,$Leu$,and $Lys$ are essential amino acids.
Non-essential amino acids are those that can be synthesized by the body. Among the given list,$Gly$,$Pro$,and $Ser$ are non-essential amino acids.
Thus,there are $3$ essential and $3$ non-essential amino acids.
Therefore,the correct option is $A$.

(D) The structure of Threonine is $CH_3-CH(OH)-CH(NH_2)-COOH$.
In this structure,the carbon atom attached to the $-NH_2$ group is chiral because it is bonded to four different groups: $-H$,$-NH_2$,$-COOH$,and the $-CH(OH)CH_3$ group.
The carbon atom attached to the $-OH$ group is also chiral because it is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and the $-CH(NH_2)COOH$ group.
Therefore,Threonine has $2$ chiral centres.

(C) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet. Non-essential amino acids are those that the body can synthesize on its own.

Essential Amino Acids	Non-essential Amino Acids
$Arginine$ $(Arg)$	$Alanine$ $(Ala)$
$Histidine$ $(His)$	$Asparagine$ $(Asn)$
$Isoleucine$ $(Ile)$	$Aspartic$ $acid$ $(Asp)$
$Leucine$ $(Leu)$	$Glutamic$ $acid$ $(Glu)$
$Lysine$ $(Lys)$	$Glutamine$ $(Gln)$
$Methionine$ $(Met)$	$Glycine$ $(Gly)$
$Phenylalanine$ $(Phe)$	$Proline$ $(Pro)$
$Threonine$ $(Thr)$	$Serine$ $(Ser)$
$Tryptophan$ $(Trp)$	$Cysteine$ $(Cys)$
$Valine$ $(Val)$	$Tyrosine$ $(Tyr)$

Based on the table,$Glutamine$ is a non-essential amino acid.

(B) dipeptide is formed by the condensation reaction between the carboxyl group $(-COOH)$ of one amino acid and the amino group $(-NH_2)$ of another amino acid.
In the dipeptide $Gly-Ala$,$Gly$ (Glycine) is the $N$-terminal amino acid and $Ala$ (Alanine) is the $C$-terminal amino acid.
Glycine structure: $H_2N-CH_2-COOH$
Alanine structure: $H_2N-CH(CH_3)-COOH$
When $Gly$ and $Ala$ combine,the $-COOH$ group of $Gly$ reacts with the $-NH_2$ group of $Ala$ to form a peptide bond $(-CO-NH-)$.
The resulting structure is $H_2N-CH_2-CO-NH-CH(CH_3)-COOH$.

(D) $S$-containing amino acids are Cysteine and Methionine.
Serine is an $OH$-containing amino acid.
Lysine is a basic amino acid containing an extra $-NH_2$ group.
Therefore,the correct options are $(B)$ and $(D)$.

(C) To determine which amino acids contain an $-OH$ (hydroxyl) group in their side chain,let us examine the structures:
$1$. Lysine: Contains an amino group $(-NH_2)$ in its side chain.
$2$. Serine: Contains a hydroxymethyl group $(-CH_2OH)$ in its side chain,which includes an $-OH$ group.
$3$. Tyrosine: Contains a phenolic group $(-C_6H_4OH)$ in its side chain,which includes an $-OH$ group.
$4$. Valine: Contains an isopropyl group $(-CH(CH_3)_2)$ in its side chain,which does not contain an $-OH$ group.
Therefore,both Serine and Tyrosine contain an $-OH$ group. The correct option is $(C)$.

(D) Essential amino acids cannot be synthesized in our body,so we need to take them in our diet.
Amino acids are soluble in water but generally insoluble in organic solvents like ether.
Due to their crystalline nature,they are high melting solids because of strong dipole-dipole interactions.
In aqueous solutions,they exist as dipolar zwitter ions,as shown below:
$H_3N^+-CH(R)-COO^-$

(A) Serine is an amino acid with the chemical structure $HOCH_2-CH(NH_2)-COOH$.
Apart from the amino group $(-NH_2)$ and the carboxylic acid group $(-COOH)$,it contains a hydroxyl group $(-OH)$ in its side chain.

(C) Human insulin is a peptide hormone composed of two polypeptide chains,chain $A$ and chain $B$,linked by disulfide bridges.
Chain $A$ consists of $21$ amino acids and chain $B$ consists of $30$ amino acids.
Therefore,the total number of amino acids in insulin is $21 + 30 = 51$.

(A) An amino acid is optically active if it contains at least one chiral (asymmetrical) carbon atom.
Glycine has the structure $H_2N-CH_2-COOH$.
In glycine,the central carbon atom is bonded to two identical hydrogen atoms,meaning it does not have a chiral center.
Therefore,glycine is the only naturally occurring amino acid that is optically inactive.

(C) Amino acids contain both acidic $(-COOH)$ and basic $(-NH_2)$ groups.
In an acidic medium (low $pH$),the $-NH_2$ group accepts a proton to form a cation $(R-CH(NH_3^+)-COOH)$.
In a basic medium (high $pH$),the $-COOH$ group loses a proton to form an anion $(R-CH(NH_2)-COO^-)$.
At the isoelectric point,they exist as a zwitterion $(R-CH(NH_3^+)-COO^-)$.
Thus,the assertion is correct.
However,the reason states that they exist as an anionic form in acidic medium and a cationic form in basic medium,which is the reverse of the actual chemical behavior.
Therefore,the reason is wrong.

(A) The general structure of an $\alpha$-amino acid is $NH_2-CH(R)-COOH$.
In tryptophan,the $R$ group is an indolyl-methyl group (specifically,a $3$-indolylmethyl group).
In histidine,the $R$ group is an imidazolyl-methyl group (specifically,a $4$-imidazolylmethyl group).
Therefore,the correct representation for $R$ in tryptophan $(X)$ and histidine $(Y)$ is the indolyl-methyl group and the imidazolyl-methyl group respectively.

(B) The general formula of an $\alpha$-amino acid is $H_2N-CH(R)-COOH$.
For the amino acid serine,the side chain $-R$ is a hydroxymethyl group,which is represented as $-CH_2OH$.
Therefore,the correct option is $B$.

(A) The amino acid $Tryptophan$ contains an indole ring in its side chain.
An indole ring is a bicyclic structure consisting of a six-membered benzene ring fused to a five-membered nitrogen-containing pyrrole ring.
Among the given options,$Tryptophan$ is the only amino acid that possesses this specific structural feature.

(C) Statement-$I$ is correct because the primary structure of a protein refers to the specific sequence of amino acids linked by peptide bonds.
Statement-$II$ is incorrect because $\alpha$-helix and $\beta$-pleated sheet structures are examples of the secondary structure of proteins,not the tertiary structure. The tertiary structure refers to the overall three-dimensional folding of the polypeptide chain.

(C) The $Primary$ structure of a protein refers to the specific sequence of amino acids in a polypeptide chain,which defines its constitution.
$Secondary$ structure refers to the local folding of the polypeptide chain into structures like $\alpha$-helices and $\beta$-pleated sheets,stabilized by hydrogen bonding.
$Tertiary$ structure refers to the overall three-dimensional shape of a single polypeptide chain,determined by interactions between side chains.
$Quaternary$ structure refers to the arrangement of multiple polypeptide chains in a multi-subunit protein.

(B) Fibrous proteins are long,thread-like structures that are insoluble in water. Examples include $Keratin$ (found in hair,wool,skin) and $Myosin$ (found in muscles).
Globular proteins have a spherical shape and are generally soluble in water. Examples include $Insulin$ and $Albumin$.
Therefore,$Keratin$ $(A)$ and $Myosin$ $(C)$ are fibrous proteins.

(A) Chemically,a peptide linkage is an amide bond formed between the $-COOH$ group of one amino acid and the $-NH_2$ group of another amino acid.
During this condensation reaction,the hydroxyl group $(-OH)$ from the carboxyl group and a hydrogen atom $(-H)$ from the amino group combine to form a water molecule $(H_2O)$,which is eliminated.
This results in the formation of a peptide bond $(-CO-NH-)$.
Therefore,the molecule eliminated during peptide bond formation is $H_2O$.

(A) Glycylalanine is a dipeptide formed by the condensation of glycine $(H_2NCH_2COOH)$ and alanine $(H_2NCH(CH_3)COOH)$.
During the formation of the peptide bond,a water molecule is eliminated between the carboxyl group of glycine and the amino group of alanine,resulting in the dipeptide glycylalanine.

(B) $Keratin$ is a fibrous structural protein found in the outer protective layers of animals,such as hair,wool,nails,and skin.
$Silk$ is composed of a different fibrous protein called $Fibroin$.
$Muscles$ are primarily composed of contractile proteins like $Actin$ and $Myosin$.
Therefore,$Keratin$ is absent in $Muscles$ and $Silk$.

(C) During the denaturation of proteins,the $1^{\circ}$ (primary) structure remains intact because the peptide bonds are not broken.
However,the $2^{\circ}$ (secondary) and $3^{\circ}$ (tertiary) structures are destroyed due to the disruption of hydrogen bonds and other non-covalent interactions.
As a result,the protein loses its specific shape and its biological activity is lost.

(D) Proteins are polymers of amino acids linked by peptide bonds.
Upon complete hydrolysis,proteins yield only $\alpha$-amino acids.
An $\alpha$-amino acid has the general structure $R-CH(NH_2)-COOH$,where the amino group is attached to the $\alpha$-carbon atom.

(C) nucleotide is the basic structural unit of nucleic acids ($DNA$ and $RNA$). It consists of three main components:
$1$. $A$ nitrogenous base (purine or pyrimidine) attached to the $1'$ carbon of the sugar.
$2$. $A$ pentose sugar (ribose in $RNA$ or $2'$-deoxyribose in $DNA$).
$3$. $A$ phosphate group attached to the $5'$ carbon of the sugar.
Looking at the provided structures:
- The structure must show the phosphate group attached to the $5'-CH_2$ group.
- The nitrogenous base must be attached to the $1'$ carbon.
- The $3'$ carbon must have a hydroxyl $(-OH)$ group.
Based on the standard structural representation of a nucleotide,the correct structure is shown in option $(C)$.

(A) Fibrous proteins are long,thread-like molecules held together by strong hydrogen bonds and disulphide linkages. They are generally insoluble in water.
Globular proteins have a compact,spherical shape where the polypeptide chain twists around itself. The polar groups are on the surface,making them soluble in water.
Therefore,the correct matches are:
Fibrous $(A)$: $1$ (Hydrogen bonding) and $4$ (Disulphide linkage).
Globular $(B)$: $2$ (Water soluble) and $3$ (Spherical shape).
Thus,the correct option is $A-1, 4; B-2, 3$.

(C) The isoelectric point $(pI)$ of an amino acid is the $pH$ at which the molecule carries no net electric charge.
For a simple amino acid,the isoelectric point is calculated as the average of the two $pK_a$ values:
$pI = \frac{pK_{a_1} + pK_{a_2}}{2}$
Given $pK_{a_1} = 2.56$ and $pK_{a_2} = 9.38$,
$pI = \frac{2.56 + 9.38}{2} = \frac{11.94}{2} = 5.97$.

(C) The general structure of an $\alpha$-amino acid is $H_2N-CH(R)-COOH$.
Evaluating the options:
$A$. Tyrosine has $R = -CH_2-C_6H_4-OH(p)$,which is correct.
$B$. Cysteine has $R = -CH_2-SH$,which is correct.
$C$. Serine has $R = -CH_2-OH$. The structure given,$R = -CH_2-CH_2-S-CH_3$,corresponds to Methionine,not Serine. Thus,this is incorrectly matched.
$D$. Asparagine has $R = -CH_2-CONH_2$,which is correct.

(D) $1$. The amino acid '$X$' contains a phenolic hydroxy group. Among the given options,$Tyrosine$ $(Tyr)$ contains a phenolic hydroxy group attached to the benzene ring.
$2$. The amino acid '$Y$' contains an amide group. Among the given options,$Glutamine$ $(Gln)$ contains an amide group $(-CONH_2)$ in its side chain.
$3$. Therefore,'$X$' is $Tyr$ and '$Y$' is $Gln$.

(A) The structure of asparagine is $H_2N-CO-CH_2-CH(NH_2)-COOH$.
It contains an amino group $(-NH_2)$,a carboxylic acid group $(-COOH)$,and an amide group $(-CONH_2)$ in its side chain.

(B) In aqueous solution,the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton,giving rise to a dipolar ion known as a zwitterion. Thus,Assertion $(A)$ is true.
Most naturally occurring amino acids have the $L$-configuration,where the $-NH_2$ group is on the left side in the Fischer projection. Thus,Reason $(R)$ is true.
However,the $L$-configuration of amino acids is not the reason why they exist as dipolar ions in aqueous solution. The dipolar nature is due to the acid-base properties of the functional groups present. Therefore,$(R)$ is not the correct explanation for $(A)$.

(B) In aqueous solution,the carboxyl group $(-COOH)$ can lose a proton to form a carboxylate ion $(-COO^-)$ and the amino group $(-NH_2)$ can accept a proton to form an ammonium ion $(-NH_3^+)$,giving rise to a dipolar ion known as a zwitter ion.
This zwitter ion is electrically neutral but contains both positive and negative charges.
In the zwitter ionic form,amino acids show amphoteric behavior as they react with both acids and bases.
Proline is indeed a natural amino acid that contains a secondary amine group (it is a cyclic amino acid).
Both the Assertion $(A)$ and the Reason $(R)$ are true,but the fact that proline has a secondary amino group is not the reason why amino acids exist as zwitter ions in aqueous solution. Therefore,$(R)$ is not the correct explanation for $(A)$.

(A) The structures of the given amino acids are as follows:
$1$. Histidine contains an imidazole ring,which is a heterocyclic ring.
$2$. Valine is an aliphatic amino acid with an isopropyl side chain.
$3$. Arginine contains a guanidino group in its side chain,which is not a heterocyclic ring.
$4$. Proline contains a pyrrolidine ring,which is a heterocyclic ring.
Therefore,both Histidine $(i)$ and Proline $(iv)$ contain heterocyclic rings.

(B) Aspartic acid is an acidic amino acid with the structure $HOOC-CH(NH_2)-CH_2-COOH$.
In its Zwitter ionic form,the proton from the $\alpha$-carboxyl group $(-COOH)$ migrates to the amino group $(-NH_2)$ to form $-NH_3^{\oplus}$ and $-COO^{\ominus}$.
Since aspartic acid has an additional carboxyl group in its side chain,the $\alpha$-carboxyl group is the most acidic and loses its proton first at physiological pH.
Thus,the correct Zwitter ionic form is $H_3\stackrel{\oplus}{N}-CH(COO^{\ominus})-CH_2-COOH$.

(B) At the isoelectric point,the net charge on an amino acid is zero,which results in minimum intermolecular electrostatic repulsion and minimum interaction with water molecules. Consequently,amino acids have the least solubility in water at their isoelectric point. Therefore,the statement that they have maximum solubility at their isoelectric point is incorrect.

(D) codon is a specific sequence of $3$ adjacent bases on a strand of $DNA$ or $RNA$.
It specifies a particular amino acid that is to be incorporated into a protein chain during the process of translation.
Therefore,$\underline{A} = 3$ bases,$\underline{B} = \text{amino acid}$,and $\underline{C} = \text{protein}$.

(C) Tripeptides are polymers of amino acids in which three individual amino acid units,called residues,are linked together by amide bonds (peptide bonds).
Since we have $3$ distinct amino acids (glycine,alanine,and phenylalanine),the number of possible tripeptides is given by the number of permutations of these $3$ distinct items.
The number of arrangements is $3! = 3 \times 2 \times 1 = 6$.
Therefore,there are $6$ different ways to link these amino acids.