(Take pKa1, pKa2 for glycine as 2.34…

(Take pKa1, pKa2 for glycine as 2.34 and 9.60; for glutamic acid as 2.19 (α-COOH), 4.25 (side-chain COOH) and 9.67; for lysine as 2.18, 8.95…

One or more options correct

Choose the correct statement(s). (Take

pK_{a1}

pK_{a2}

for glycine as

2.34

and

9.60

; for glutamic acid as

2.19

(

\alpha

-COOH),

4.25

(side-chain COOH) and

9.67

; for lysine as

2.18

8.95

(

\alpha

-NH

_3^+

) and

10.53

(side-chain NH

_3^+

).)

A. The isoelectric point of glycine is $5.97$ , and near this pH the zwitterion $\mathrm{H_3N^+CH_2COO^-}$ is the predominant species
B. In an electrophoresis experiment at $\mathrm{pH}=7.0$ , lysine migrates towards the cathode
C. In an electrophoresis experiment at $\mathrm{pH}=6.0$ , glutamic acid migrates towards the cathode
D. The tripeptide Ala–Gly–Val contains two peptide (amide) linkages and gives three amino-acid molecules on complete acid hydrolysis

Show answer and solution

Answer: A, B, D

A is correct: for a neutral amino acid

\mathrm{pI}=\tfrac12(pK_{a1}+pK_{a2})=\tfrac12(2.34+9.60)=5.97

, where the zwitterion dominates. B is correct: for lysine the neutral zwitterion lies between the two ammonium dissociations,

\mathrm{pI}=\tfrac12(8.95+10.53)=9.74

. At pH

7<\mathrm{pI}

the molecule carries a net positive charge and moves to the cathode. C is incorrect: for glutamic acid

\mathrm{pI}=\tfrac12(2.19+4.25)=3.22

(average of the two acidic constants). At pH

6>\mathrm{pI}

it carries a net negative charge and moves to the anode. D is correct: a tripeptide of three residues has

3-1=2

peptide bonds and hydrolyses to three amino acids. Hence A, B and D.

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake