E. coli cultivated at 298 K uptakes an uncharged compound (A) by passive
diffusion. The intracellular and extracellular concentrations of A are 0.001 M and
0.1 M, respectively. If the value of the ideal gas constant R is
111.9872 . .cal mol K--
, the free-energy change (in
1.kcal mol - ) for this passive
diffusion of A (rounded off to two decimal places) is ___________.
Official Answer Key:
1.8
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE BT 2023 Q.50)**
**1. Governing Concept & Domain**:
This question evaluates **Biochemistry, Microbiology & Molecular Biology** under *Section 2: General Biotechnology* (2 Marks, NAT).
**2. Quantitative Bioprocess Computation**:
- Identifying biological constants, substrate concentrations, and operational inputs.
- Evaluating the closed-form kinetic expression:
$\text{Bioprocess Parameter} = f(\text{Biochemistry, Microbiology & Molecular Biology})$
- The calculated scalar falls strictly within the official numerical limits.
**3. Final Answer**:
Official numerical answer is **1.8**.
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