GATE 2023 BT Q.54: solution and answer

GATE 2023 · BTQ.54 · 2 marks · NAT

An enzyme (E) catalyzes the biochemical reaction $A \to B$ with $k_{c a t}$ equal to $500 s^{- 1}$ . If the initial reaction velocity ( $V_{0}$ ) is $10 μ M.s^{- 1}$ at the total enzyme concentration $[E_{t}]$ of $30 nM$ and substrate concentration [A] of $40 μ M$ , the value of $K_{m}$ (in $μ M$ ) is ________.

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Answer and solution

ANSWER

$K_{m} = 20 μ M$

Official answer: 20Source: Official answer key, GATE 2023 BT (IIT Kanpur)

Symbols: $k_{c a t}$ is the turnover number, $[E_{t}]$ the total enzyme concentration, $V_{ma x}$ the maximum rate, $V_{0}$ the initial rate and $K_{m}$ the Michaelis constant.
Maximum rate: $V_{ma x} = k_{c a t} [E_{t}]$ . Put the concentrations in the same units first: $[E_{t}] = 30 nM = 0.030 μ M$ .
$V_{ma x} = 500 s^{- 1} \times 0.030 μ M = 15 μ M s^{- 1}$
Michaelis-Menten: $V_{0} = \frac{V _{ma x} [ A ]}{K _{m} + [ A ]}$ . Rearranging gives $K_{m} = [A] (\frac{V _{ma x}}{V _{0}} - 1)$ .
$K_{m} = 40 (\frac{15}{10} - 1) = 40 \times 0.5 = 20 μ M$
Check: $V_{0} = \frac{15 \times 40}{20 + 40} = 10 μ M s^{- 1}$ , which matches the given rate. It also fits the rule that $V_{0} = \frac{2}{3} V_{ma x}$ when $[A] = 2 K_{m}$ .

KEY IDEA

$V_{ma x} = k_{c a t} [E_{t}]$ , then solve Michaelis-Menten for $K_{m}$ : $K_{m} = [A] (\frac{V _{ma x}}{V _{0}} - 1)$ .

WATCH OUT FOR

Forgetting to convert $[E_{t}]$ from nM to $μ$ M. Using $30$ directly gives $V_{ma x} = 15000 μ M s^{- 1}$ and a $K_{m}$ that is wrong by orders of magnitude. The other slip is treating $V_{0}$ as $V_{ma x}$ .

From the GATE 2023 BT paper. Every question of this paper · Download the full paper · All 9 Enzyme kinetics questions, with notes

GATE 2023 BT Q.54: solution and answer

Answer and solution

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