GATE 2024 BT Q.34: solution and answer

GATE 2024 · BTQ.34 · 1 mark · NAT

If the rate at which E. coli divides is $0.5 h^{- 1}$ , then its doubling time is ________ h.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The doubling time is $t_{d} = 1/ k = 1/ (0.5 h^{- 1}) = 2 h$ .

Official answer: 2Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

The given quantity is a division rate $k$ , the number of divisions (generations) per hour: $k = 0.5 h^{- 1}$ . The question calls it the rate at which the cells divide, not a specific growth rate $μ$ .
Binary fission: each division doubles the cell number, so after $n$ divisions $N = N_{0} 2^{n}$ , where $n = k t$ is the number of divisions completed in time $t$ .
The doubling time (generation time) $t_{d}$ is the time taken for exactly one division, $n = 1$ : $k t_{d} = 1 \Rightarrow t_{d} = \frac{1}{k}$ .
$t_{d} = \frac{1}{0.5 h ^{- 1}} = 2 h$
Check with the exponential form: $N = N_{0} 2^{k t} = N_{0} e^{(k l n 2) t}$ , so the matching specific growth rate is $μ = k ln 2 = 0.5 \times 0.693 = 0.347 h^{- 1}$ , and $t_{d} = \frac{ln 2}{μ} = \frac{0.693}{0.347 h ^{- 1}} = 2 h$ . The two forms agree.

KEY IDEA

A division rate $k$ counts generations per unit time, and one generation doubles the population, so the doubling time is simply its reciprocal: $t_{d} = 1/ k$ . The formula $t_{d} = ln 2/ μ$ applies only when the rate given is the specific growth rate $μ$ in $d X / d t = μ X$ . The two are related by $μ = k ln 2$ .

WATCH OUT FOR

Treating $0.5 h^{- 1}$ as the specific growth rate $μ$ and writing $t_{d} = ln 2/ μ = 0.693/0.5 = 1.39 h$ . The question gives the rate at which the cells divide, which is a division rate, so $t_{d} = 1/ k = 2 h$ .

From the GATE 2024 BT paper. Every question of this paper · Download the full paper · All 24 Growth kinetics questions, with notes

GATE 2024 BT Q.34: solution and answer

Answer and solution

More Growth kinetics questions