Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
It is desired to cultivate 150 mg (dry weight) microbial cells with an empirical
formula of C 5H7O2N. If the phosphorus requirement for the cells is 20% of the
nitrogen requirement (on a weight/weight basis), the minimum amount of
phosphorous required to be added to the cultivation medium is ________ mg.
(rounded off to one decimal place)
0
1
2
3
4
0 1 2 3
x
Official Answer Key:
3.6 to 3.8
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2026 Q.62)**
**1. Governing Concept & Definition**:
- **Topic**: Microbial Stoichiometry & Nutrient Requirements
- From the empirical formula of biomass $\text{C}_5\text{H}_7\text{O}_2\text{N}$, the mass fraction of nitrogen is determined, and phosphorus requirement is given as $20\%$ of nitrogen requirement by weight.
**2. Detailed Step-by-Step Calculation**:
- **Molecular Weight of Biomass Formula $\text{C}_5\text{H}_7\text{O}_2\text{N}$**:
$\text{MW} = 5(12) + 7(1) + 2(16) + 14 = 60 + 7 + 32 + 14 = 113\text{ g mol}^{-1}$
- **Nitrogen Mass Fraction**:
$\omega_N = \frac{14}{113} \approx 0.12389$
- **Step 1: Nitrogen required for $150\text{ mg}$ cells**:
$m_N = 150\text{ mg} \times \frac{14}{113} \approx 18.584\text{ mg}$
- **Step 2: Phosphorus required ($20\%$ of nitrogen)**:
$m_P = 0.20 \times m_N = 0.20 \times 18.584\text{ mg} \approx 3.717\text{ mg} \approx 3.7\text{ mg}$
**3. Final Answer**:
- **Calculated Value**: **3.7 mg**
- **Official GATE Accepted Range**: **3.6 to 3.8**
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