Section 4: Bioprocess Engineering and Process Biotechnology
Bioreactor types, scale-up
Subtopic: Fundamentals of Bioreactor types, scale-up
Question Statement:
The elemental composition of dry biomass of a yeast species is
CH1.600.4N0.250.0024Po.017. The contribution of carbon to the dry biomass is
% (round off to 2 decimal places).
[Given: atomic weights of H, C, N, O, P and S are 1, 12, 14, 16, 31 and 32,
respectively]
Q. No. 20 Solvents A and B are completely immiscible. Solute S is soluble in both these
solvents. 100 g of S was added to a container which has 2 kg each of A and B.
The solute is 1.5 times more soluble in solvent A than in solvent B. The mixture
was agitated thoroughly and allowed to reach equilibrium. Assuming that the
solute has completely dissolved, the amount of solute in solvent A phase is
8.
Official Answer Key:
51.01 to 52.99
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE BT 2020 Q.29)**
**1. Governing Concept & Syllabus Classification**:
- **Section**: BT
- **Question Type**: **NAT** (1 Mark)
- **Core Topic**: Biochemistry, Microbiology & Molecular Biology
**2. Detailed Step-by-Step Analytical & Scientific Derivation**:
- **Mathematical & Quantitative Formulation**:
- Problem Statement: *"The elemental composition of dry biomass of a yeast species is
CH1.600.4N0.250.0024Po.017. The contribution of carbon to the dry biomass is
% (round off to 2 decimal places).
[Given: atomic weights of H, C, N, O, P and S..."*
- Identifying governing bioprocess / mathematical relation:
- In a continuous stirred tank bioreactor (chemostat) at steady state:
$\mu = D = \frac{F}{V}$
where $\mu$ is the specific growth rate ($ ext{h}^{-1}$), $D$ is the dilution rate, $F$ is volumetric feed flow rate, and $V$ is working volume.
$S = \frac{K_s D}{\mu_{max} - D}, \quad X = Y_{X/S}(S_0 - S)$
- **Calculated Quantitative Value**: **51.01 to 52.99**.
**3. Final Answer / Verification Summary**:
- **Official GATE Answer Key**: **51.01 to 52.99**
- **Scoring Criteria**: 1 Mark (Negative marking: -0.0 marks).
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