Amino Acids and Proteins Questions in English

(B) All naturally occurring amino acids found in proteins,except for glycine,are chiral.
These amino acids exist in the $L$-configuration at their $\alpha$-carbon atom.
Glycine is the simplest amino acid,$NH_2-CH_2-COOH$,and is achiral because it lacks a chiral center.
General structure of an $L$-amino acid: $NH_2-CH(R)-COOH$ where the amino group is on the left in the Fischer projection.

(A) The isoelectric point $(pI)$ of an amino acid is the $pH$ at which it exists primarily as a zwitterion,$R-CH(NH_3^+)-COO^-$.
When the $pH$ of the solution is lower than the $pI$ $(pH < pI)$,the concentration of $H^+$ ions is high.
The carboxylate group $(-COO^-)$ of the zwitterion acts as a base and accepts a proton $(H^+)$ to form a carboxylic acid group $(-COOH)$.
Therefore,at $pH$ $1.0$,the species $R-CH(NH_3^+)-COO^-$ gains a proton to become $R-CH(NH_3^+)-COOH$.

(B) Glutamic acid is an acidic amino acid with three ionizable groups: two carboxylic acid groups $(-COOH)$ and one amino group $(-NH_3^+)$.
The $pK_a$ values are given as $2.2$ (for the $\alpha-COOH$),$4.3$ (for the side chain $-COOH$),and $9.7$ (for the $-NH_3^+$ group).
The isoelectric point $(pI)$ of an acidic amino acid is calculated by taking the average of the two lowest $pK_a$ values:
$pI = \frac{pK_{a1} + pK_{a2}}{2} = \frac{2.2 + 4.3}{2} = \frac{6.5}{2} = 3.25$.

(B) To synthesize the dipeptide $Phe-Ala$,we need to couple the carboxyl group of $Phe$ with the amino group of $Ala$.
$1$. The $N$-terminus of $Phe$ must be protected,and its $C$-terminus must be free. Thus,we use $ZNH-CH(CH_2C_6H_5)-COOH$ (reagent $(3)$).
$2$. The $C$-terminus of $Ala$ must be protected,and its $N$-terminus must be free. Thus,we use $H_2N-CH(CH_3)-COOCH_2C_6H_5$ (reagent $(2)$).
$3$. Coupling these two reagents followed by deprotection yields $Phe-Ala$.

(C) The isoelectric point $(pI)$ is defined as the $pH$ at which a molecule carries no net electrical charge.
At the isoelectric point,the concentration of the zwitterionic form of an amino acid is at its maximum.
Since the isoelectric point of alanine is $6$,the maximum concentration of its zwitterion occurs at $pH = 6$.

(C) The given compound is a dipeptide: $NH_2-CH(CH_3)-CO-NH-CH_2-COOH$.
Upon hydrolysis,the peptide bond $(-CO-NH-)$ breaks to yield the constituent amino acids.
The products are $NH_2-CH(CH_3)-COOH$ (Alanine) and $NH_2-CH_2-COOH$ (Glycine).
Therefore,both $(A)$ and $(B)$ are obtained.

(D) The correct answer is $ (D) $.
An isoelectric point is defined as the $ pH $ at which a particular molecule carries no net electric charge.
Amino acids contain both an amino group (which can form a cation) and a carboxylic group (which can form an anion).
At the isoelectric $ pH $:
$ 1. $ The net charge on the molecule is zero because the cationic and anionic charges are equal.
$ 2. $ The concentration of the cationic form and the anionic form are equal.
$ 3. $ The molecule exists primarily in its di-polar form,known as the Zwitter ion,which reaches its maximum concentration at this $ pH $.
Therefore,all the given statements are correct.

(C) The iso-electric point $(pI)$ of an amino acid depends on its side chain ($R$-group).
Acidic amino acids,which contain an extra carboxylic acid group in their side chain,have lower $pI$ values.
Aspartic acid is an acidic amino acid with two carboxylic acid groups and one amino group,resulting in a $pI$ of approximately $2.77$.
Glycine and Alanine are neutral amino acids with $pI$ values around $6.0$.
Lysine is a basic amino acid with an extra amino group,resulting in a high $pI$ value of approximately $9.74$.
Therefore,Aspartic acid has the lowest $pI$.

(B) The isoelectric point $(pI)$ is calculated using the formula: $pI = \frac{pK_{a1} + pK_{a2}}{2}$.
Given $pK_{a1} = 2.2$ (for the carboxyl group).
Given $pK_b = 4.4$ for the amino group,we find $pK_{a2}$ using the relation $pK_{a2} = 14.0 - pK_b = 14.0 - 4.4 = 9.6$.
Substituting these values into the formula: $pI = \frac{2.2 + 9.6}{2} = \frac{11.8}{2} = 5.9$.

(C) The given amino acid is glutamic acid,which has two carboxylic acid groups and one amino group.
For an acidic amino acid,the isoelectric point $(pI)$ is calculated as the average of the two lowest $pK_a$ values.
$pI = \frac{pK_{a1} + pK_{a2}}{2}$
$pI = \frac{2 + 4}{2} = \frac{6}{2} = 3.0$
Therefore,the correct option is $C$.

(B) The reaction proceeds as follows:
$1$. Acetylene $(HC\equiv CH)$ undergoes hydration in the presence of $HgSO_4$ and $H_2SO_4$ to form acetaldehyde $(CH_3CHO)$ as product $(A)$.
$2$. Acetaldehyde $(CH_3CHO)$ reacts with $NH_3$ and $HCN$ (Strecker synthesis) followed by hydrolysis $(H_3O^{\oplus})$ to form the amino acid alanine $(CH_3-CH(NH_2)-COOH)$ as product $(B)$.

(C) chiral centre is a carbon atom bonded to four different groups.
In $Glycine$ $(NH_2-CH_2-COOH)$,the $\alpha$-carbon is bonded to two hydrogen atoms,which are identical.
Therefore,$Glycine$ is the only achiral amino acid among the proteinogenic amino acids.

(A) Sanger's reagent is used for the identification of the $N$-terminal amino acid in a peptide chain.
It reacts with the free amino group of the $N$-terminal amino acid to form a yellow-colored dinitrophenyl $(DNP)$ derivative.
The chemical name of Sanger's reagent is $2,4-$Dinitrofluorobenzene $(DNFB)$.
Therefore,the correct option is $(A)$.

(A) The abbreviation $Phe-Val-Ala$ represents a tripeptide consisting of Phenylalanine $(Phe)$,Valine $(Val)$,and Alanine $(Ala)$ in that specific order.
$1$. $Phe$ (Phenylalanine) has the side chain $-CH_2C_6H_5$.
$2$. $Val$ (Valine) has the side chain $-CH(CH_3)_2$.
$3$. $Ala$ (Alanine) has the side chain $-CH_3$.
In the structure of a peptide,the amino acids are linked by peptide bonds $(-CONH-)$. The structure starts with the $N$-terminal amino acid $(Phe)$ and ends with the $C$-terminal amino acid $(Ala)$.
Looking at the options:
- Option $A$ shows the sequence $Phe-Val-Ala$ correctly: the first amino acid has a benzyl side chain $(Phe)$,the second has an isopropyl side chain $(Val)$,and the third has a methyl side chain $(Ala)$.
Therefore,the correct structure is represented by option $A$.

(B) peptide bond is a chemical bond formed between two molecules when a carboxylic group $(-COOH)$ of one molecule reacts with an amino group $(-NH_2)$ of the other,releasing a water molecule $(H_2O)$.
The $-CO-NH-$ linkage is known as a peptide bond.
In the given structure,there are $2$ such $-CO-NH-$ linkages,which are marked with red lines in the provided image. Therefore,the total number of peptide bonds is $2$.

(A) The structure of serine is $HOCH_2-CH(NH_2)-COOH$.
In a Fischer projection for an $L$-amino acid,the $-COOH$ group is placed at the top,the $-R$ group (side chain) at the bottom,and the $-NH_2$ group is on the left side.
For serine,the side chain $-R$ is $-CH_2OH$.
Thus,placing $-COOH$ at the top,$-CH_2OH$ at the bottom,$-NH_2$ on the left,and $-H$ on the right gives the correct $L$-serine configuration.
This corresponds to the structure shown in option $A$.

(B) zwitterion is a molecule that has both a positive and a negative charge,allowing it to exist as a dipolar ion. This requires the presence of both an acidic group (like $-COOH$ or $-SO_3H$) and a basic group (like $-NH_2$) in the same molecule.
In $N$-acetylalanine,the amino group $(-NH_2)$ is converted into an amide group $(-NH-CO-CH_3)$. The nitrogen atom in an amide is not basic because its lone pair of electrons is involved in resonance with the carbonyl group $(C=O)$.
Therefore,$N$-acetylalanine cannot accept a proton to form a positive charge,and thus it cannot exist in a zwitterionic form at $pH = 7$.

(D) The dipeptide $Gln-Gly$ consists of Glutamine $(Gln)$ at the $N$-terminus and Glycine $(Gly)$ at the $C$-terminus. The structure of $Gln-Gly$ is $H_2N-CH(CH_2CH_2CONH_2)-CONH-CH_2-COOH$.
When treated with $CH_3COCl$ (acetyl chloride),the primary amino group $(-NH_2)$ at the $N$-terminus is acetylated to form an acetamide group $(-NHCOCH_3)$.
The amide group $(-CONH_2)$ present in the side chain of Glutamine is significantly less nucleophilic due to resonance stabilization and is generally not acetylated under these conditions.
Therefore,the product is $CH_3CONH-CH(CH_2CH_2CONH_2)-CONH-CH_2-COOH$,which corresponds to the structure shown in option $D$.

(B) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet. $Valine$ is an essential amino acid. Other examples include $Histidine$,$Isoleucine$,$Leucine$,$Lysine$,$Methionine$,$Phenylalanine$,$Threonine$,and $Tryptophan$.

(B) For an acidic amino acid like aspartic acid,the isoelectric point $(pI)$ is calculated as the average of the two $pK_a$ values that flank the zwitterionic form.
The zwitterionic form of aspartic acid exists between the first deprotonation of the $\alpha$-carboxyl group $(pK_1 = 1.88)$ and the deprotonation of the side-chain carboxyl group $(pK_R = 3.65)$.
$pI = \frac{pK_1 + pK_R}{2}$
$pI = \frac{1.88 + 3.65}{2} = \frac{5.53}{2} = 2.765 \approx 2.77$

(A) The given compound is purine.
In purine,the lone pair of electrons on the $N-9$ atom is involved in the aromatic sextet of the imidazole ring.
Therefore,this lone pair is not available for protonation,and $N-9$ does not contribute to the basicity of the compound.
Other nitrogen atoms $(N-1, N-3, N-7)$ have lone pairs in $sp^2$ hybridized orbitals that are not involved in aromaticity,making them available for protonation.

(C) Ninhydrin is often used to detect $\alpha$-amino acids and also free amino and carboxylic acid groups on proteins and peptides.
When about $0.5 \ mL$ of a $0.1\%$ solution of ninhydrin is boiled for one or two minutes with a few $mL$ of dilute amino acid or protein solution,a blue color develops.
Ninhydrin degrades amino acids into aldehydes,ammonia,and $CO_2$ through a series of reactions; the net result is ninhydrin in a partially reduced form,hydrindantin.
Ninhydrin then condenses with ammonia and hydrindantin to produce an intensely blue or purple pigment,sometimes called Ruhemann's purple.

(D) Statement $(D)$ is incorrect.
Proteins are polymers of amino acids and,in addition to $C, H, O$ and $N$,many proteins also contain sulfur $(S)$ and sometimes phosphorus $(P)$ or other elements.

(C) The $pI$ (isoelectric point) values for the given amino acids are as follows:
$Asp$ $(pI = 3.0)$,
$Gly$ $(pI = 6.0)$,
$Lys$ $(pI = 9.8)$,
$Arg$ $(pI = 10.8)$.
The $pKa$ values of the side chains or the overall acidity of amino acids correlate with their $pI$ values.
Acidic amino acids like $Asp$ have lower $pI$ values,while basic amino acids like $Lys$ and $Arg$ have higher $pI$ values.
Therefore,the increasing order of $pKa$ values is $Asp < Gly < Lys < Arg$.

(A) The given tripeptide is hydrolyzed to obtain three amino acids: $Valine$ $(Val)$,$Serine$ $(Ser)$,and $Threonine$ $(Thr)$.
By observing the structure from the $N$-terminal (left side) to the $C$-terminal (right side),the sequence of amino acids is $Val-Ser-Thr$.

(D) The reaction involves the treatment of the dipeptide $Asn-Ser$ with excess acetic anhydride $((CH_3CO)_2O)$ in the presence of a base $(NEt_3)$.
Acetic anhydride is an acetylating agent that reacts with nucleophilic functional groups.
In the dipeptide $Asn-Ser$:
$1$. The $N$-terminal amino group $(-NH_2)$ of $Asn$ is nucleophilic and will be acetylated to form an amide $(-NHCOCH_3)$.
$2$. The side-chain amide group $(-CONH_2)$ of $Asn$ is also nucleophilic and will be acetylated to form an imide $(-CONHCOCH_3)$.
$3$. The side-chain hydroxyl group $(-OH)$ of $Ser$ is nucleophilic and will be acetylated to form an ester $(-OCOCH_3)$.
Therefore,all three nucleophilic sites are acetylated,resulting in the structure shown in option $D$.

(B) The $Biuret$ test,$Ninhydrin$ test,and $Xanthoproteic$ test are standard chemical tests used to identify amino acids or proteins.
$Barfoed$ test is specifically used to detect the presence of monosaccharides,not amino acids.

(B) Protonation occurs at the site where the lone pair of electrons is most available for donation to a proton $(H^+)$.
In adenine,the nitrogen atoms at positions $b, c,$ and $d$ have lone pairs that are not involved in the aromatic sextet of the rings.
Specifically,the nitrogen at position $d$ is the most basic site because its lone pair is in an $sp^2$ orbital and is not involved in the aromatic system,making it highly available for protonation.
Protonation at $b, c,$ or $d$ results in a conjugate acid that is stabilized by resonance.
The exocyclic $-NH_2$ group $(a)$ has its lone pair involved in resonance with the ring,and the $-NH-$ group $(e)$ has its lone pair involved in the aromaticity of the imidazole ring,making them less basic.
Thus,the favourable sites for protonation are $b, c,$ and $d$.

(C) The identification of amino acids based on functional group tests is as follows:
$1$. Ester test: Serine $(Ser)$ contains a hydroxyl $(-OH)$ group,which can undergo esterification. Thus,$A-R$.
$2$. Carbylamine test: Lysine $(Lys)$ contains a primary amine $(-NH_2)$ group,which gives a positive carbylamine test. Thus,$B-S$.
$3$. Phthalein dye test: Tyrosine $(Tyr)$ contains a phenolic group,which reacts with phthalic anhydride to form a phthalein dye. Thus,$C-P$.
Therefore,the correct matching is $A-R, B-S, C-P$.

(B) The basicity of an amino acid is determined by the number of amino groups relative to the number of carboxylic acid groups in its side chain.
Lysine contains two amino groups and one carboxylic acid group,making it a basic amino acid.
Histidine is also basic,but Lysine is more basic due to the presence of an aliphatic amino group in its side chain.
Comparing their $pI$ values:

Compound	$pI$ value
$A$. Histidine	$7.6$
$B$. Serine	$5.7$
$C$. Lysine	$9.8$
$D$. Asparagine	$5.4$

$A$ higher $pI$ value indicates a more basic character. Therefore,Lysine is the most basic among the given options.

(A) Histidine is an amino acid with three ionizable groups: the $\alpha$-amino group,the $\alpha$-carboxyl group,and the imidazole side chain.
In a strongly acidic solution $(pH = 2)$,the concentration of $H^{+}$ ions is very high.
At this $pH$,all basic groups will be protonated:
$1$. The carboxyl group exists as $-COOH$ (since $pH < pK_a$ of $-COO^-$).
$2$. The $\alpha$-amino group exists as $-NH_3^+$.
$3$. The imidazole side chain also gets protonated to exist as an imidazolium ion $(-NH^+)$.
Therefore,the structure will have a net charge of $+2$. This corresponds to the structure shown in option $A$.

(B) The ceric ammonium nitrate test is used to detect the presence of an alcoholic $-OH$ group. Among the given amino acids,$Serine$ $(Ser)$ contains an alcoholic $-OH$ group.
The carbylamine test is used to detect the presence of a primary amine $(-NH_2)$ group. Among the given amino acids,$Lysine$ $(Lys)$ contains a primary amine group in its side chain.
Therefore,the peptide $Ser - Lys$ will give a positive result for both tests.

(C) $1$. At the isoelectric point $(pI)$,the amino acid exists as a zwitter ion,meaning it has a net charge of zero. This statement is correct.
$2$. When $pH > pI$,the amino acid exists as an anion (negatively charged) and will move towards the anode (positive electrode) in an electric field. This statement is correct.
$3$. Denaturation of proteins involves the loss of secondary and tertiary structures due to physical changes like heating or chemical changes. However,the primary structure (the sequence of amino acids) remains intact. Therefore,the statement that primary structures are denatured by heating is incorrect.
$4$. Glycine is the simplest $\alpha$-amino acid and is optically inactive. Alanine is the next simplest and is the first optically active $\alpha$-amino acid. This statement is correct.

(A) Insulin is a peptide hormone produced by the beta cells of the pancreatic islets.
It consists of two polypeptide chains,chain $A$ and chain $B$,which are linked together by disulfide bonds.
Chain $A$ contains $21$ amino acids and chain $B$ contains $30$ amino acids.
Therefore,the total number of amino acids in insulin is $21 + 30 = 51$.

(D) $1$. Assign priority to the groups attached to the chiral center using Cahn-Ingold-Prelog $(CIP)$ rules: $-NH_2$ $(1)$,$-COOH$ $(2)$,$-CH_3$ $(3)$,$-H$ $(4)$.
$2$. In the given structure,the lowest priority group $(-H)$ is on a wedge (pointing towards the viewer).
$3$. Looking at the arrangement of the remaining groups ($1$,$2$,$3$),the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ appears clockwise,which corresponds to $R$ configuration. Since the lowest priority group is on a wedge,we reverse the result,making the configuration $S$.
$4$. For $D/L$ nomenclature,we look at the Fischer projection or the orientation of the amino group in amino acids. In the given structure,the $-NH_2$ group is on the left side when the carbon chain is vertical with the most oxidized group at the top,which corresponds to the $L$ configuration.
$5$. Therefore,the compound has both $L$ and $S$ configurations.

(A, D) Amino acids are the building blocks of proteins.
Among the given options,$Methionine$ and $Cysteine$ are the two standard amino acids that contain sulphur in their side chains.
$Methionine$ contains a thioether group,while $Cysteine$ contains a thiol $(-SH)$ group.
Therefore,both $a$ and $d$ are correct.

(C) At low $pH$ (highly acidic medium),the amino acid exists in its cationic form.
In this state,the carboxylate group is protonated $(COO^- \rightarrow COOH)$ while the amino group remains protonated $(NH_3^+)$.
Among the given options,$pH = 2$ represents the acidic condition required for this structure to exist.

(B) In the imidazole ring of histidine,there are two nitrogen atoms: the pyrrole-like nitrogen $(\alpha-N)$ and the pyridine-like nitrogen $(\beta-N)$.
The lone pair on the $\alpha-N$ is involved in the aromatic sextet of the imidazole ring.
The lone pair on the $\beta-N$ is in an $sp^2$ orbital perpendicular to the aromatic ring and is available for protonation.
Protonation at the $\beta-N$ leads to the formation of an imidazolium ion,which is stabilized by two equivalent resonance structures,making it the more basic site.

(C) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs around each carbon atom.
$C_1$ is bonded to one nitrogen atom,two hydrogen atoms,and one carbon atom $(C_2)$. It forms four sigma bonds and has no lone pairs. Therefore,the steric number is $4$,which corresponds to $sp^3$ hybridisation.
$C_2$ is bonded to one carbon atom $(C_1)$,one oxygen atom (via a double bond),and one oxygen atom (via a single bond). It forms three sigma bonds and has no lone pairs. Therefore,the steric number is $3$,which corresponds to $sp^2$ hybridisation.
Thus,the hybridisation of $C_1$ is $sp^3$ and $C_2$ is $sp^2$.

(A) Nitrogen is an essential component of amino acids,which are the building blocks of proteins. Therefore,nitrogen is an essential constituent of proteins.

(B) The helical structure of proteins is stabilized by hydrogen bonds.
In the helical structure,the carbonyl oxygen atoms in $C=O$ point in one direction,towards the amide hydrogen atoms in $NH$ groups,$4$ residues away.
Together these groups form a hydrogen bond,which is one of the main forces of secondary structure stabilization in proteins.
Hydrogen bonds are shown by dashed lines in the structure.

(C) Amino acids are classified as basic if they contain an extra amino group in their side chain.
$Lysine$,$Arginine$,and $Histidine$ are the three basic amino acids.
$Glycine$ is a neutral amino acid.
Among the basic amino acids,$Histidine$ is the least basic because its imidazole ring has a $pK_a$ value of approximately $6.0$,which is close to physiological $pH$,whereas $Lysine$ $(pK_a \approx 10.5)$ and $Arginine$ $(pK_a \approx 12.5)$ are much more basic.

(A) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
Among the given options,$Methionine$ is an essential amino acid.
$Tyrosine$,$Proline$,and $Glycine$ are non-essential amino acids as they can be synthesized by the human body.

(C) The secondary structure of proteins refers to the local folding of the polypeptide chain into regular structures such as the $\alpha$-helix and $\beta$-pleated sheets.
These structures are stabilized primarily by $Hydrogen$ bonds formed between the carbonyl oxygen $(C=O)$ and the amide hydrogen $(N-H)$ of the peptide backbone.

(D) Denaturation is a process in which the protein loses its higher-order structures (secondary and tertiary) due to physical or chemical changes like temperature or pH variation.
As a result,the protein loses its biological activity because the specific three-dimensional shape required for its function is destroyed.
The primary structure,which is the sequence of amino acids,remains intact during denaturation.

(B) The reaction of proteins with concentrated $HNO_3$ results in the nitration of aromatic amino acid residues (like tyrosine and tryptophan) present in the protein chain,forming yellow-colored nitro compounds. This specific qualitative test for proteins is known as the $Xanthoproteic$ test.

(B) The $Biuret$ test is a chemical test used for detecting the presence of peptide bonds in proteins.
In the presence of peptides,the copper$(II)$ ions in the $Biuret$ reagent form a violet-colored complex in an alkaline solution,which is characteristic of proteins.

(A) The test described is the $Biuret$ test.
In the $Biuret$ test,when a protein solution is treated with an alkaline solution of copper sulfate $(CuSO_4)$,a violet or purple color is produced.
This color is due to the formation of a coordination complex between $Cu^{2+}$ ions and the peptide bonds in the protein.
Therefore,the correct option is $A$.

(A) The isoelectric point $(pI)$ of an amino acid is the $pH$ at which the molecule carries no net electric charge.
For a neutral amino acid,the $pI$ is calculated as the average of the $pK_a$ values of the two ionizable groups:
$pI = \frac{pK_{a_1} + pK_{a_2}}{2}$
Given $pK_{a_1} = 2.34$ and $pK_{a_2} = 9.6$:
$pI = \frac{2.34 + 9.6}{2} = \frac{11.94}{2} = 5.97$
Therefore,the isoelectric point is $5.97$.