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Biomolecules Set-2 MCQ Online Practice Test | Class 12 | Quiz Tweek

Biomolecules Set-2

Practice Important MCQs for Biomolecules — Chemistry, Class 12.

Biomolecules is a high-yield Class 12 Chemistry chapter because it builds the chemical reasoning behind the structure, properties, and reactions of amino acids, peptides, enzymes, DNA/RNA, and biologically relevant equilibria. In both board and competitive exams, questions commonly test correct application of $pK_a/pI$ , charge calculations, peptide ionization concepts, and enzyme kinetics (Michaelis–Menten), along with mechanistic understanding of nucleic acid hydrolysis and protein folding.

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Q1. Glycine has $pK_a(\mathrm{COOH})=2.34$ and $pK_a(\mathrm{NH_3^{+}})=9.60$ . Calculate its isoelectric point ( $pI$ ).

(A)

$6.00$

(B)

$7.95$

(C)

$5.97$

(D)

$3.63$

Q2. Consider the tetrapeptide $H\text{-}\mathrm{Ala}\text{-}\mathrm{Lys}\text{-}\mathrm{Asp}\text{-}\mathrm{Gly}\text{-}\mathrm{OH}$ (free N‑ and C‑termini). Assume $pK_a(\mathrm{COOH,\,C\text{-}term})=2.34$ , $pK_a(\mathrm{NH_3^{+},\,N\text{-}term})=9.60$ , $pK_a(\mathrm{Lys\ side\ chain})=10.53$ and $pK_a(\mathrm{Asp\ side\ chain})=3.86$ . What is the net charge of this peptide at $pH=7.0$ ?

(A)

$0$

(B)

$+1$

(C)

$-1$

(D)

$+2$

Q3. Two double‑stranded DNA fragments X and Y are each $20$ base pairs long. Fragment X has $40\%$ GC content and fragment Y has $60\%$ GC content. Using the approximation

T_m\approx 2^\circ\mathrm{C}\times(\#\text{ A–T pairs})+4^\circ\mathrm{C}\times(\#\text{ G–C pairs}),

estimate $T_m$ for X and Y and state which has the higher $T_m$ and by how much.

(A)

X: $64^\circ\mathrm{C}$ , Y: $56^\circ\mathrm{C}$ (X higher by $8^\circ\mathrm{C}$ )

(B)

X: $52^\circ\mathrm{C}$ , Y: $68^\circ\mathrm{C}$ (Y higher by $16^\circ\mathrm{C}$ )

(C)

X: $56^\circ\mathrm{C}$ , Y: $60^\circ\mathrm{C}$ (Y higher by $4^\circ\mathrm{C}$ )

(D)

X: $56^\circ\mathrm{C}$ , Y: $64^\circ\mathrm{C}$ (Y higher by $8^\circ\mathrm{C}$ )

Q4. An enzyme sample (sample I) loses tertiary structure when treated with $8\ \mathrm{M}$ urea but regains full activity after dialysis to remove urea. A second sample (sample II) treated with $8\ \mathrm{M}$ urea plus $0.2\ \mathrm{M}$ $\beta$ ‑mercaptoethanol ( $\mathrm{HSCH_2CH_2OH}$ ) does not regain activity after dialysis. Which explanation best accounts for these observations?

(A)

Both urea and $\beta$ ‑mercaptoethanol hydrolyse peptide bonds irreversibly; sample I survived by chance while sample II was cleaved.

(B)

Urea disrupts noncovalent interactions (H‑bonds, hydrophobic packing) that can reform on removal, whereas $\beta$ ‑mercaptoethanol reduces disulfide bonds ( $\mathrm{S\!-\!S}\rightarrow 2\,\mathrm{SH}$ ), breaking covalent cross‑links required for correct folding so native structure cannot be restored by simple dialysis.

(C)

Urea permanently oxidises disulfide bonds preventing refolding, while $\beta$ ‑mercaptoethanol forms stable thioether adducts that block the active site.

(D)

Urea racemises $\alpha$ ‑chiral centres rendering the protein inactive, and $\beta$ ‑mercaptoethanol converts essential lysine residues into unreactive derivatives.

Q5. Histidine has $pK_a(\mathrm{COOH})=1.8$ , $pK_a(\mathrm{imidazole\ side\ chain})=6.0$ and $pK_a(\mathrm{NH_3^{+}})=9.2$ . For a solution of histidine at $pH=6.0$ , calculate the average net electric charge per histidine molecule (assume independent ionizations; give answer to two decimal places).

(A)

$+0.50$

(B)

$+1.00$

(C)

$0.00$

(D)

$-0.50$

Q6. The amino acid glycine has $pK_a(\mathrm{COOH})=2.34$ and $pK_a(\mathrm{NH_3^+})=9.60$ . Using $pI=\dfrac{pK_{a1}+pK_{a2}}{2}$ , the isoelectric point of glycine (approx.) is:

(A)

$2.34$

(B)

$5.97$

(C)

$9.60$

(D)

$6.60$

Q7. Consider the tripeptide $\mathrm{H_2N\!-\!Lys\!-\!Ala\!-\!Glu\!-\!COOH}$ . Given $pK_a(\mathrm{terminal\ COOH})=2.0,\; pK_a(\alpha\mathrm{-NH_3^+})=9.0,\; pK_a(\mathrm{Lys\ side\ chain})=10.5,\; pK_a(\mathrm{Glu\ side\ chain})=4.25$ . At pH $7.0$ the net charge on the peptide is:

(A)

$+2$

(B)

$+1$

(C)

$0$

(D)

$-1$

Q8. An enzyme follows Michaelis–Menten kinetics with $V_{\max}=120\ \mu\mathrm{M\,min^{-1}}$ and $K_m=6\ \mu\mathrm{M}$ . At $[S]=6\ \mu\mathrm{M}$ (i.e. $[S]=K_m$ ) the initial velocity is $60\ \mu\mathrm{M\,min^{-1}}$ . In presence of a competitive inhibitor the apparent $K_m$ increases to $18\ \mu\mathrm{M}$ while $V_{\max}$ is unchanged. At $[S]=6\ \mu\mathrm{M}$ the new initial velocity is:

(A)

$30\ \mu\mathrm{M\,min^{-1}}$

(B)

$60\ \mu\mathrm{M\,min^{-1}}$

(C)

$15\ \mu\mathrm{M\,min^{-1}}$

(D)

$90\ \mu\mathrm{M\,min^{-1}}$

Q9. Assertion (A): DNA is more susceptible to alkaline hydrolysis than RNA.
Reason (R): The ribose in RNA has a $2'$ –OH that can attack the adjacent phosphate to form a $2',3'$ –cyclic phosphate, causing strand cleavage; deoxyribose lacks the $2'$ –OH.
Choose the correct option:

(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.

Q10. For an enzyme obeying Michaelis–Menten kinetics the rate equation is $v=\dfrac{V_{\max}[S]}{K_m+[S]}$ . For which substrate concentration $[S]$ (expressed in terms of $K_m$ ) will $v=0.75\,V_{\max}$ ?

(A)

$0.75\,K_m$

(B)

$3\,K_m$

(C)

$\dfrac{K_m}{3}$

(D)

$\dfrac{4}{3}\,K_m$

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