Saccharomyces cerevisiae is cultured in a chemostat (continuous fermentation) at a dilution rate of
Q.5 h-1. The feed substrate concentration is 10 g.L-1. The biomass concentration in the chemostat at
steady state will be _____________ g.L-1.
Assumptions: Feed is sterile, maintenance is negligible and maximum biomass yield with respect to
substrate is 0.4 (g biomass per g ethanol).
Microbial growth kinetics is given by
m
s
s
K
s
where µ is specific growth rate (h-1), µm = 0.7 h-1, Ks = 0.3 g.L-1 and s is substrate concentration
(g.L-1).
Official Answer Key:
3.65 - 3.75
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE BT 2016 Q.57)**
**1. Governing Concept & Syllabus Classification**:
- **Section**: BT
- **Question Type**: **NAT** (2 Marks)
- **Core Topic**: Biochemistry, Microbiology & Molecular Biology
**2. Detailed Step-by-Step Analytical & Scientific Derivation**:
- **Mathematical & Quantitative Formulation**:
- Problem Statement: *"Saccharomyces cerevisiae is cultured in a chemostat (continuous fermentation) at a dilution rate of
0.5 h-1. The feed substrate concentration is 10 g.L-1. The biomass concentration in the chemostat at
steady state will..."*
- 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.
$S = \frac{K_s D}{\mu_{max} - D}, \quad X = Y_{X/S}(S_0 - S)$
- **Official Answer Key Interval**: Evaluated numerical range is **3.65 to 3.75** (Target value: **3.65 to 3.75**).
**3. Final Answer / Verification Summary**:
- **Official GATE Answer Key**: **3.65 to 3.75**
- **Scoring Criteria**: 2 Marks (Negative marking: -0.0 marks).
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