Mock test 4

Mock Test 04: Numerical Biochemistry

20:00

1. An enzyme follows Michaelis-Menten kinetics. If the velocity of the reaction is 80% of Vmax, the substrate concentration [S] in terms of Km is: A) 2 Km B) 4 Km C) 8 Km D) 10 Km

Solution: B. Using $v = \frac{V_{max}[S]}{K_m + [S]}$, set $v = 0.8 V_{max}$. Then $0.8(K_m + [S]) = [S]$. Solving gives $[S] = 4 K_m$.

2. Calculate the pI (Isoelectric point) of Aspartic acid. (pKa values: pK1=2.09, pK2=9.82, pKR=3.86) A) 5.95 B) 2.97 C) 6.84 D) 3.55

Solution: B. For acidic amino acids, $pI = \frac{pK_1 + pK_R}{2}$. $pI = \frac{2.09 + 3.86}{2} = 2.97$.

3. In a 500 bp double stranded DNA, if there are 150 Adenine bases, what is the total number of Hydrogen bonds? A) 1150 B) 1350 C) 1500 D) 1000

Solution: B. A=T=150. Remaining bases = 1000 - 300 = 700. So G=C=350 pairs. Bonds = (150 x 2) + (350 x 3) = 300 + 1050 = 1350.

4. The $\Delta G^\circ'$ for the hydrolysis of ATP is -30.5 kJ/mol. What is the Equilibrium Constant (Keq) at 25°C? A) 2.2 x 10^5 B) 1.5 x 10^4 C) 4.5 x 10^6 D) 1.2 x 10^3

Solution: A. Use $\Delta G^\circ' = -RT \ln K_{eq}$. $K_{eq} = e^{-\Delta G^\circ' / RT}$. Converting units ($30500 / (8.314 \times 298)$), we get approx $2.2 \times 10^5$.

5. For a reaction A → B, $\Delta H$ is -20 kJ/mol and $\Delta S$ is -50 J/mol·K. At what temperature (K) will the reaction be at equilibrium? A) 200 K B) 400 K C) 500 K D) 300 K

Solution: B. At equilibrium, $\Delta G = 0$, so $T = \Delta H / \Delta S$. $T = -20000 / -50 = 400 K$.

6. How many rounds of Beta-oxidation are required to fully oxidize a 16-carbon Palmitic acid? A) 8 B) 7 C) 16 D) 15

Solution: B. Rounds = $(n/2) - 1$. For $n=16$, rounds = $8 - 1 = 7$.

7. A protein solution has an absorbance of 1.0 at 280 nm. If the path length is 1 cm and molar extinction coefficient is 50,000 M⁻¹cm⁻¹, the concentration is: A) 20 μM B) 50 μM C) 10 μM D) 100 μM

Solution: A. $A = \epsilon cl$. $c = A / \epsilon = 1 / 50000 = 2 \times 10^{-5} M = 20 \mu M$.

8. If the turnover number (kcat) of an enzyme is 100 s⁻¹ and total enzyme concentration is 2 μM, the Vmax is: A) 200 μM/s B) 50 μM/s C) 0.02 μM/s D) 20 μM/s

Solution: A. $V_{max} = k_{cat} \times [E_t] = 100 \times 2 \mu M = 200 \mu M/s$.

9. A DNA molecule is 3.4 micrometers long. How many base pairs does it contain? (Assuming B-DNA) A) 1000 B) 10,000 C) 3400 D) 100,000

Solution: B. Rise per bp = 0.34 nm. Length = $3.4 \mu m = 3400 nm$. $BP = 3400 / 0.34 = 10,000$.

10. Calculate the number of ATPs produced from the complete oxidation of 1 molecule of Acetyl-CoA in the TCA cycle. A) 12 B) 10 C) 24 D) 38

Solution: B. 1 Acetyl-CoA yields: 3 NADH (7.5 ATP), 1 FADH2 (1.5 ATP), and 1 GTP (1 ATP). Total = 10 ATP. (Note: Using modern 2.5/1.5 ratios).