GATE 2025 BT Q.28: solution and answer

GATE 2025 · BTQ.28 · 1 mark · NAT

If the doubling time of a bacterial population is 3 hours, then its average specific growth rate during this period is ______ $h^{- 1}$ .

(Round off to two decimal places)

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Answer and solution

ANSWER

The average specific growth rate is $μ = \frac{ln 2}{t _{d}} = \frac{0.6931}{3} \approx 0.23 h^{- 1}$ .

Official answer: 0.20 to 0.25Source: Official answer key, GATE 2025 BT (IIT Roorkee)

In exponential growth, $\frac{d X}{d t} = μ X$ , where $X$ is the cell concentration and $μ$ is the specific growth rate ( $h^{- 1}$ ). Integrating from $X_{0}$ at $t = 0$ gives $X = X_{0} e^{μ t}$ .
The doubling time $t_{d}$ is the time for $X$ to reach $2 X_{0}$ . So $2 X_{0} = X_{0} e^{μ t_{d}}$ , which gives $e^{μ t_{d}} = 2$ and $μ = \frac{ln 2}{t _{d}}$ . Because the population exactly doubles over $t_{d}$ , this $μ$ is also the average specific growth rate over that interval, $\overset{μ}{ˉ} = \frac{ln ( X / X _{0} )}{t}$ .
Substitute $t_{d} = 3 h$ : $μ = \frac{ln 2}{3 h} = \frac{0.6931}{3} h^{- 1} = 0.2310 h^{- 1}$
Rounded to two decimal places, $μ \approx 0.23 h^{- 1}$ .

KEY IDEA

In exponential growth $X = X_{0} e^{μ t}$ , so the specific growth rate and the doubling time are linked by $μ t_{d} = ln 2$ . That gives $μ = 0.693/ t_{d}$ .

WATCH OUT FOR

Taking $μ = 1/ t_{d} = 0.33 h^{- 1}$ , which mixes up the specific growth rate with the number of doublings per hour. Using $lo g_{10} 2$ instead of $ln 2$ , which gives $0.10 h^{- 1}$ . Both answers fall outside the key. The natural log is needed because $μ$ is defined through $e^{μ t}$ .

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GATE 2025 BT Q.28: solution and answer

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