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Biomolecules Set-5 MCQ Online Practice Test | Class 12 | Quiz Tweek

Biomolecules Set-5

Practice Important MCQs for Biomolecules — Chemistry, Class 12.

Biomolecules is a high-yield chapter because it links structure to function—covering amino acids, peptides, proteins, enzymes, carbohydrates, nucleic acids, and biomolecular interactions. Board and competitive exams repeatedly test concepts like acid-base behavior (pI, net charge), molecular stability, tautomerism/mutations, and kinetics/biological rationale behind stability and reactivity.

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Q1. Calculate the isoelectric point ( $pI$ ) of alanine given $pK_a(-\mathrm{COOH}) = 2.34$ and $pK_a(-\mathrm{NH_3^+}) = 9.69$ .

(A)

$5.50$

(B)

$6.78$

(C)

$6.02$

(D)

$7.02$

Q2. A mixture of lysine, histidine and glutamic acid is subjected to electrophoresis at $pH = 7.0$ . Given $pK_a(\alpha\text{-COOH}) = 2.34$ , $pK_a(\alpha\text{-NH_3^+}) = 9.69$ , and side‑chain $pK_a$ values: lysine $= 10.54$ , histidine $= 6.04$ , glutamic acid $= 4.25$ . Which amino acid will migrate towards the anode (positive electrode) at $pH = 7.0$ ?

(A)

Glutamic acid only

(B)

Lysine only

(C)

Histidine only

(D)

Lysine and histidine

Q3. Assertion (A): Glycogen is mobilized more rapidly than amylopectin in animals. Reason (R): Glycogen has a higher frequency of $\alpha(1\rightarrow6)$ branch points and therefore presents a larger number of non‑reducing ends per unit mass where glycogen phosphorylase acts.

(A)

Both A and R are true but R is NOT the correct explanation of A

(B)

A is true but R is false

(C)

A is false but R is true

(D)

Both A and R are true and R is the correct explanation of A

Q4. Calculate the isoelectric point ( $pI$ ) of the dipeptide Gly–Glu (N‑terminal Glycine, C‑terminal Glutamic acid). Use $pK_a(\text{N‑terminus }\mathrm{NH_3^+}) = 9.60$ , $pK_a(\text{C‑terminus }\mathrm{COOH}) = 2.34$ , $pK_a(\text{Glu side chain }\mathrm{COOH}) = 4.25$ .

(A)

$6.95$

(B)

$3.30$

(C)

$5.78$

(D)

$4.92$

Q5. Assertion (A): RNA is chemically less stable than DNA under alkaline conditions. Reason (R): The $2'$ ‑OH group in ribose of RNA can be deprotonated under basic conditions and act as an intramolecular nucleophile to form a $2',3'$ ‑cyclic phosphate intermediate, causing backbone cleavage; DNA lacks the $2'$ ‑OH.

(A)

Both A and R are true and R is the correct explanation of A

(B)

Both A and R are true but R is NOT the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Q6. Calculate the isoelectric point ( $pI$ ) of alanine given $pK_a(\text{COOH})=2.35$ and $pK_a(\text{NH}_3^+)=9.87$ (side chain non‑ionisable).

(A)

$5.11$

(B)

$8.11$

(C)

$6.11$

(D)

$4.97$

Q7. An enzyme follows Michaelis–Menten kinetics with $V_{\max}=120\ \mu\text{mol min}^{-1}$ and $K_m=3.0\ \text{mM}$ . Using $v=\dfrac{V_{\max}[S]}{K_m+[S]}$ , calculate the initial rate $v$ when $[S]=1.0\ \text{mM}$ .

(A)

$30.0\ \mu\text{mol min}^{-1}$

(B)

$40.0\ \mu\text{mol min}^{-1}$

(C)

$16.7\ \mu\text{mol min}^{-1}$

(D)

$60.0\ \mu\text{mol min}^{-1}$

Q8. For the dipeptide Gly–Ala (glycine at N‑terminus, alanine at C‑terminus) with terminal $pK_a(\text{COOH})=2.30$ and $pK_a(\text{NH}_3^+)=9.60$ , determine the net charge at pH $1$ , pH $7$ , and pH $12$ respectively (assume no ionisable side chains).

(A)

$(+2,\; +1,\; 0)$

(B)

$(+1,\; +1,\; 0)$

(C)

$(+1,\; -1,\; -2)$

(D)

$(+1,\; 0,\; -1)$

Q9. Assertion (I): Replacing an $A\text{-}T$ base pair by a $G\text{-}C$ base pair in the central region of a DNA duplex increases the melting temperature ( $T_m$ ). Reason (II): The increase in $T_m$ is solely due to the extra hydrogen bond in a $G\text{-}C$ pair compared to an $A\text{-}T$ pair.

(A)

Both (I) and (II) are true, and (II) is the correct explanation of (I).

(B)

Both (I) and (II) are true, but (II) is not the correct explanation of (I).

(C)

(I) is true but (II) is false.

(D)

(I) is false but (II) is true.

Q10. Protein X contains two cysteine residues that form an intramolecular disulfide bond ( $\mathrm{S}\!-\!\mathrm{S}$ ) in the oxidizing extracellular environment. Mutant Y replaces those cysteines with serines. Predict their relative conformational (thermal) stabilities in (i) the extracellular non‑reducing environment and (ii) the cytosolic reducing environment.

(A)

X is more stable than Y in both (i) and (ii).

(B)

X is more stable than Y in (i); stabilities are similar in (ii).

(C)

Y is more stable than X in (i); X is more stable in (ii).

(D)

X and Y have identical stabilities in both (i) and (ii).

Q11. Calculate the isoelectric point $pI$ of glycine given $pK_a(\mathrm{COOH})=2.34$ and $pK_a(\mathrm{NH_3^+})=9.60$ . (Use $pI=\dfrac{pK_a(\mathrm{COOH})+pK_a(\mathrm{NH_3^+})}{2}$ .)

(A)

(A) $7.00$

(B)

(B) $5.97$

(C)

(D)

(D) $3.97$

Q12. Consider the tripeptide H $_2$ N–Gly–Lys–Glu–COOH. Given $pK_a(\mathrm{N\text{-}term})=9.60$ , $pK_a(\mathrm{C\text{-}term})=2.20$ , $pK_a(\mathrm{Lys\ side\ chain})=10.50$ , $pK_a(\mathrm{Glu\ side\ chain})=4.25$ . What is the net charge of this tripeptide at pH $7.0$ ?

(A)

(A) $+2$

(B)

(B) $+1$

(C)

(D)

(D) $0$

Q13. An enzyme obeys Michaelis–Menten kinetics with $K_m=2.0\ \mathrm{mM}$ and $V_{max}=100\ \mu\mathrm{mol\,min^{-1}}$ . Using $v=\dfrac{V_{max}[S]}{K_m+[S]}$ , at what substrate concentration $[S]$ will $v=75\ \mu\mathrm{mol\,min^{-1}}$ ?

(A)

(A) $6.0\ \mathrm{mM}$

(B)

(B) $1.5\ \mathrm{mM}$

(C)

(D)

(D) $2.0\ \mathrm{mM}$

Q14. Aspartic acid has $pK_a(\alpha\text{-COOH})=2.09$ , $pK_a(\text{side-chain COOH})=3.86$ and $pK_a(\alpha\text{-NH}_3^+)=9.82$ . Estimate the net charge of aspartic acid at pH $3.0$ (use Henderson–Hasselbalch qualitatively to account for fractional ionization).

(A)

(A) $+1.00$

(B)

(B) $-1.00$

(C)

(D)

(D) $+0.50$

Q15. Which tautomeric shift and mispairing is most likely to cause an $A\!:\!T \to G\!:\!C$ transition mutation after DNA replication?

(A)

(A) $\mathrm{G}$ in enol form pairing with T so that A→G results

(B)

(B) $\mathrm{A}$ in imino form pairing with C during replication; in the next round the original $A\!:\!T$ can become $G\!:\!C$

(C)

(D)

(D) $\mathrm{T}$ in keto form pairing with G causing the transition

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