Q.44 Maximum specific growth rate (μmax) of a microorganism is calculated by taking the (ln = loge, X = biomass, t = time) (A) slope of ln X vs t of the growth cycle (B) slope of ln X vs t during the exponential growth phase (C) slope of X vs t (D) slope of X vs t during the exponential phase of growth

Q.44 Maximum specific growth rate (μmax) of a microorganism is calculated by taking the (ln = loge, X = biomass, t = time)

(A) slope of ln X vs t of the growth cycle

(B) slope of ln X vs t during the exponential growth phase

(C) slope of X vs t

(D) slope of X vs t during the exponential phase of growth

Correct Answer: (B) slope of ln X vs t during the exponential growth phase

Maximum specific growth rate (μ_max) derives from dX/dt = μX during exponential phase where μ remains constant, yielding ln X = ln X₀ + μt. The slope of this linear region equals μ_max (h⁻¹).

Option Analysis

  • (A): Incorrect. Entire growth cycle includes lag (μ≈0), exponential (μ=μ_max), stationary (μ=0), death (μ<0) phases; slope varies, doesn’t yield μ_max.

  • (B): Correct. Exponential phase: X = X₀e^(μt) → ln X vs t = straight line (R²>0.99); slope = μ_max.

  • (C): Incorrect. X vs t yields absolute growth rate (dX/dt), not specific rate (μ = (1/X)(dX/dt)).

  • (D): Incorrect. X vs t during exponential phase curves upward (accelerating growth); slope increases continuously.

Maximum specific growth rate μmax calculation via ln X vs t slope during exponential phase defines Q.44 core concept essential for biochemical engineering bioreactor optimization and microbial kinetics analysis.

Growth Kinetics Mathematics

Exponential Phase: dX/dt = μX → ∫(dX/X) = ∫μ dt → ln(X/X₀) = μt → μ_max = slope(ln X vs t)

text
t (h) | X (g/L) | ln X
0.0 | 0.10 | -2.30
1.0 | 0.27 | -1.31
2.0 | 0.74 | -0.30
3.0 | 2.01 | 0.70
μ_max = Δ(ln X)/Δt = [0.70 - (-2.30)]/3.0 = 1.0 h⁻¹

Doubling Time: t_d = ln(2)/μ_max = 0.693/μ_max

Phase-Specific Analysis

  • Lag: μ≈0 (adaptation); ln X ≈ constant

  • Exponential: μ=μ_max (nutrient excess, no inhibition); linear ln X vs t

  • Stationary: μ=0 (nutrient limitation, toxins); ln X plateau

  • Death: μ<0 (starvation); ln X declines linearly

Bioprocess Applications

Chemostat Design: D = μ_max for washout avoidance (D < μ_max)
Batch Optimization: Harvest at μ_max peak (2-3 doublings post-lag)
Fed-batch: μ = μ_max(S)/(K_S + S) → exponential feed maintains μ_max

GATE Biotechnology Integration

Q.44 culminates Q.35-43 progression: μ_max → synchronous culture (Q.42) → Pol III replication timing (Q.43) → IPTG induction (Q.41) → pathway localization (Q.35). Essential for Y_X/S = ΔX/ΔS yield calculations and Monod kinetics (μ = μ_max S/(K_S + S)).

Exam Tip: Linearize via semilog plot; R²>0.98 confirms exponential phase. Exclude lag/stationary data points.

 

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