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Biomolecules Set-7 MCQ Online Practice Test | Class 12 | Quiz Tweek

Biomolecules Set-7

Practice Important MCQs for Biomolecules — Chemistry, Class 12.

Biomolecules form the core of Class 12 Chemistry because they link fundamental bonding/structure (functional groups, stereochemistry, ionization) with biological relevance. Board exams and competitive exams heavily test concepts like amino acid ionization and isoelectric point, peptide chemistry, enzyme kinetics (Michaelis–Menten and inhibition), sugar reducing behavior, mutarotation/anomeric composition, and DNA melting behavior—so mastering this chapter boosts accuracy in both factual and calculation-based questions.

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Q1. An amino acid has $pK_a(\mathrm{COOH})=2.34$ and $pK_a(\mathrm{NH_3^+})=9.69$ . Using $pI=\dfrac{pK_{a1}+pK_{a2}}{2}$ , calculate its isoelectric point $pI$ .

(A)

( $5.02$ )

(B)

( $7.02$ )

(C)

( $6.02$ )

(D)

( $8.01$ )

Q2. Determine the net charge of the peptide $\mathrm{H_2N\!-\!Ala\!-\!Glu\!-\!Lys\!-\!Asp\!-\!COOH}$ at $pH\;7.0$ using the given $pK_a$ values: $pK_a(\mathrm{N\text{-}term})=9.0$ , $pK_a(\mathrm{C\text{-}term})=2.0$ , $pK_a(\mathrm{Glu})=4.3$ , $pK_a(\mathrm{Asp})=3.9$ , $pK_a(\mathrm{Lys})=10.5$ . (Assume groups are fully protonated if $pH<pK_a$ and deprotonated if $pH>pK_a$ .)

(A)

( $-1$ )

(B)

( $0$ )

(C)

( $+1$ )

(D)

( $-2$ )

Q3. An enzyme follows Michaelis–Menten kinetics with $V_{\max}=120\ \mu\mathrm{mol\,min^{-1}}$ and $K_m=4.0\ \mathrm{mM}$ . At $[S]=4.0\ \mathrm{mM}$ the initial rate is $60\ \mu\mathrm{mol\,min^{-1}}$ . A noncompetitive inhibitor reduces $V_{\max}$ to $40\ \mu\mathrm{mol\,min^{-1}}$ while $K_m$ is unchanged. Using $v=\dfrac{V_{\max}[S]}{K_m+[S]}$ , the new initial rate at $[S]=4.0\ \mathrm{mM}$ is:

(A)

( $30\ \mu\mathrm{mol\,min^{-1}}$ )

(B)

( $40\ \mu\mathrm{mol\,min^{-1}}$ )

(C)

( $15\ \mu\mathrm{mol\,min^{-1}}$ )

(D)

( $20\ \mu\mathrm{mol\,min^{-1}}$ )

Q4. An enzyme has $K_m=2.0\ \mathrm{mM}$ and $V_{max}=120\ \mu\mathrm{mol\,min^{-1}}$ . Upon addition of an inhibitor at $[I]=2.0\ \mathrm{mM}$ , $V_{max}$ remains $120\ \mu\mathrm{mol\,min^{-1}}$ but the apparent $K_m$ becomes $8.0\ \mathrm{mM}$ . Using $K_{m,\text{app}}=K_m\!\left(1+\dfrac{[I]}{K_i}\right)$ , identify the type of inhibition and calculate $K_i$ (in mM).

(A)

(Mixed (non‑pure) inhibition; $K_i=0.67\ \mathrm{mM}$ )

(B)

(Competitive; $K_i=0.67\ \mathrm{mM}$ )

(C)

(Competitive; $K_i=2.0\ \mathrm{mM}$ )

(D)

(Uncompetitive; $K_i=0.67\ \mathrm{mM}$ )

Q5. An enzyme has $K_m=2.0\ \mathrm{mM}$ . A competitive inhibitor is present at $[I]=1.0\ \mathrm{mM}$ with $K_i=0.5\ \mathrm{mM}$ . For a competitive inhibitor $K_{m,\text{app}}=K_m\!\left(1+\dfrac{[I]}{K_i}\right)$ . What substrate concentration $[S]$ (in mM) is required to achieve $v=0.90\,V_{\max}$ using $v=\dfrac{V_{\max}[S]}{K_{m,\text{app}}+[S]}$ ?

(A)

( $6.0\ \mathrm{mM}$ )

(B)

( $18.0\ \mathrm{mM}$ )

(C)

( $54.0\ \mathrm{mM}$ )

(D)

( $90.0\ \mathrm{mM}$ )

Q6. Consider a disaccharide formed by linking C1 of $\alpha$ - $D$ -glucose $to C4 of$ \beta $-$ D$-glucose via a glycosidic bond. Which statement about its reducing property is correct?

(A)

Non-reducing — both anomeric carbons are involved in the glycosidic linkage.

(B)

Non-reducing — the internal glycosidic bond prevents any ring opening under normal conditions.

(C)

Reducing — one anomeric carbon remains free and can open to the aldehyde form.

(D)

Reducing only under strongly basic conditions when ring opening is forced.

Q7. Histidine has the following $pK_a$ values: $pK_a(\alpha\text{-COOH})=2.00,\; pK_a(\alpha\text{-NH}_3^+)=9.20,\; pK_a(\text{imidazole side chain})=6.00$ . At pH $7.0$ the predominant net charge on the free amino acid is closest to:

(A)

$+0.09$

(B)

$0$

(C)

$+1$

(D)

$-1$

Q8. An enzyme follows Michaelis–Menten kinetics with $V_{\max}=120\ \mu\text{mol min}^{-1}$ and $K_m=4\ \text{mM}$ . A competitive inhibitor increases apparent $K_m$ to $12\ \text{mM}$ but does not change $V_{\max}$ . For substrate concentration $[S]=8\ \text{mM}$ , what is the ratio $v_{\text{inh}}/v_0$ where $v_0$ is the rate without inhibitor and $v_{\text{inh}}$ is the rate with inhibitor? Use $v=\dfrac{V_{\max}[S]}{K_m+[S]}$ .

(A)

$0.40$

(B)

$0.50$

(C)

$0.80$

(D)

$0.60$

Q9. Calculate the isoelectric point ( $pI$ ) of the dipeptide Gly–Glu (glycine at N‑terminus, glutamic acid at C‑terminus) using these $pK_a$ values: $pK_a(\alpha\text{-COOH})=2.34,\; pK_a(\mathrm{R\!-\!COOH\ (Glu)})=4.25,\; pK_a(\alpha\text{-NH}_3^+)=9.60$ . (Assume only these three ionizable groups.) The $pI$ is closest to:

(A)

$5.8$

(B)

$3.30$

(C)

$6.9$

(D)

$2.34$

Q10. During DNA replication a rare tautomeric shift converts thymine into its enol form so that it pairs with guanine instead of adenine. If this mispairing is not corrected, what permanent base‑pair substitution will result after two rounds of replication starting from the original $A\!-\!T$ pair?

(A)

$A\!-\!T \to G\!-\!C$

(B)

$A\!-\!T \to T\!-\!A$ (no change)

(C)

$A\!-\!T \to C\!-\!G$ (transversion)

(D)

$A\!-\!T \to A\!-\!C$

Q11. Two molecules of D‑glucose ( $C_6H_{12}O_6$ ) undergo a condensation reaction with elimination of one molecule of water to form a disaccharide. What is the molecular formula of the resulting disaccharide?

(A)

$C_{12}H_{24}O_{12}$

(B)

$C_{12}H_{22}O_{12}$

(C)

$C_{12}H_{22}O_{11}$

(D)

$C_{12}H_{20}O_{10}$

Q12. Pure $\alpha$ ‑D‑glucose has specific rotation $[\alpha]_\alpha=+112^\circ$ , pure $\beta$ ‑D‑glucose has $[\alpha]_\beta=+18.7^\circ$ , and the equilibrium rotation is $+52.7^\circ$ . If at a certain time the observed specific rotation of the solution is $+70.0^\circ$ , what is the approximate percentage of the $\alpha$ ‑anomer present at that moment? (Use $[\alpha]_{\text{obs}}=x[\alpha]_\alpha+(1-x)[\alpha]_\beta$ .)

(A)

$55\%$ $\alpha$ ‑anomer

(B)

$45\%$ $\alpha$ ‑anomer

(C)

$66\%$ $\alpha$ ‑anomer

(D)

$35\%$ $\alpha$ ‑anomer

Q13. An enzyme follows Michaelis–Menten kinetics with $v=\dfrac{V_{\max}[S]}{K_m+[S]}$ . At $[S]=2\,\text{mM}$ the initial rate is $v_1=30\ \mu\text{mol/min}$ and at $[S]=6\,\text{mM}$ the rate is $v_2=60\ \mu\text{mol/min}$ . Estimate the Michaelis constant $K_m$ .

(A)

$3\ \text{mM}$

(B)

$12\ \text{mM}$

(C)

$9\ \text{mM}$

(D)

$6\ \text{mM}$

Q14. Using the empirical approximation

T_m \approx 81.5 + 16.6\log_{10}[{\rm Na}^+] + 0.41(\%\text{GC}) - \frac{600}{N},

where $N$ is length in base pairs, consider two DNA samples X and Y each of length $N=1000$ bp: X has $60\%\,$ GC and Y has $40\%\,$ GC. Let $\Delta T_m=T_m(\text{X})-T_m(\text{Y})$ at $[{\rm Na}^+]=0.01\,$ M. If $[{\rm Na}^+]$ is increased to $0.1\,$ M, how does $\Delta T_m$ change?

(A)

$\Delta T_m$ increases by $16.6^\circ\text{C}$

(B)

$\Delta T_m$ remains essentially the same (≈ $0.41\times20=8.2^\circ\text{C}$ )

(C)

$\Delta T_m$ decreases by $16.6^\circ\text{C}$

(D)

$\Delta T_m$ becomes zero (no difference)

Q15. Consider the tripeptide Gly–Asp–Lys. Given the ionizable groups with the following $pK_a$ values: $\alpha$ ‑COOH (C‑terminus) $pK_a=2.3$ , $\alpha$ ‑NH $_3^+$ (N‑terminus) $pK_a=9.6$ , side‑chain Asp $pK_a=3.9$ , side‑chain Lys $pK_a=10.5$ . What are the net charges of this peptide at pH 1, pH 7 and pH 12 respectively?

(A)

$+2,\;0,\;-2$

(B)

$+1,\;+1,\;-1$

(C)

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

(D)

$+1,\;0,\;-2$

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