Q.69 A bacterial culture (5 ×108 cells/ml) is maintained in a chemostat of working volume 10 L. If the
doubling time of the bacteria is 50 min, the required rate of flow of nutrients (in ml/min) is
_______________.
Chemostat Basics
A chemostat is a continuous culture device where fresh nutrient medium flows in at rate F (volume/time), and culture overflows at the same rate, keeping volume V constant.
At steady state, dilution rate D = F/V equals specific growth rate μ to prevent washout.
- Cell density (here 5 × 108 cells/ml) confirms steady state but doesn’t affect flow calculation
- μ depends only on growth kinetics (doubling time), not biomass concentration
⚙️ Step-by-Step Calculation
Given: td = 50 min, V = 10 L
Step 1: Convert doubling time to growth rate
μ = ln(2) / td = 0.693 / 50 ≈ 0.01386 min-1
🔑 Key Formula
D = μ = F/V
F = μ × V = 0.01386 × 10 L = 0.1386 L/min
138.6 ml/min ≈ 139 ml/min
❓ Why Cell Density is Irrelevant
The given density (5 × 108 cells/ml) confirms steady state but does not affect D or F.
- Chemostat equations balance growth rate with dilution independently of biomass X
- μ depends only on intrinsic doubling time, not cell concentration
- Cell count validates steady-state assumption but isn’t used in flow calculation
⚠️ Common Pitfalls Explained
| Mistake | Why Wrong | Correct Approach |
|---|---|---|
| Using cell density in formula | X cancels out in steady-state equation | Only need μ and V |
| Wrong time units | Using hours gives 8.3 ml/min | Keep minutes consistent |
| Forget L→ml conversion | Miss ×1000 factor | 10 L = 10,000 ml |
🎯 Final Answer & Exam Tips
✅ Required flow rate = 139 ml/min
Perfect for GATE Biotech, CSIR NET, GATE BT exams
- Washout condition: If F higher → D > μ → culture washes out
- Practice variations: Change td or V for similar problems
- Mnemonic: “Flow = Growth × Volume”
