GATE BT 2021 • Q35
NAT • 1 Mark
Section 2: General Biotechnology
Biomolecules, enzyme kinetics
Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
to 0.25
1
0
24
5
NAT
7 to 7
1
25
5
NAT
1.60 to 1.80
1
26
5
MCQ
2
27
5
MCO
2
28
5
MCQ
D
2
29
5
MCQ
2
30
5
MCQ
D
2
31
5
MCQ
2
32
5
MCQ
D
2
33
5
MCO
2
GATE 2021 Answer Key for Biotechnology (BT)
Q. No.
Answer
Negative
MCa/MSQ/NAT
Name
Key/Range
34
5
MCO
C
2
35
5
MSQ
A; B; D
2
36
5
MSQ
A; B; C
2
37
5
MSQ
B; C; D
2
38
5
MSQ
A; B
2
39
5
315 to 315
2
40
5
NAT
NAT
4 to 4
2
41
5
NAT
0 to 0 OR 4 to
4
2
42
5
NAT
2 to 2
2
43
5
NAT
Official Answer Key:
1.60 - 1.80
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2021 Q.35)**
**1. Governing Concept & Definition**:
- **Topic**: Enzyme Kinetics & Substrate Concentration Determination
- Michaelis-Menten rate equation:
$v_0 = \frac{V_{\max} [S]}{K_m + [S]}$
- For reaction velocity to reach a specified fraction $f$ of $V_{\max}$ ($v_0 = f \cdot V_{\max}$):
$[S] = \frac{f}{1 - f} K_m$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- $K_m = 0.005\text{ M} = 5\text{ mM}$
- Target velocity $v_0 = \frac{1}{4} V_{\max} \implies f = 0.25$
- **Calculation of $[S]$**:
$[S] = \frac{0.25}{1 - 0.25} K_m = \frac{0.25}{0.75} K_m = \frac{1}{3} K_m$
$[S] = \frac{5\text{ mM}}{3} \approx 1.667\text{ mM} \approx 1.67\text{ mM}$
**3. Final Answer**:
- **Calculated Value**: **1.67 mM**
- **Official GATE Accepted Range**: **1.60 to 1.80**
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