GATE 2022 BT Q.50: solution and answer

GATE 2022 · BTQ.50 · 2 marks · MSQ

Which of the following plot(s) is(are) CORRECT for an enzyme that obeys Michaelis-Menten kinetics, assuming $[S] ≪ K_{m}$ ?

$[S]$ is the concentration of the substrate, $K_{m}$ is the Michaelis constant, and $v_{0}$ is the initial reaction velocity.

GATE 2022 BT question 50: Each option is a sketch graph with no numbers on the axes. — Figure from the GATE 2022 BT paper. It is part of the question.

(A)
Figure — $v_{0}$ (vertical) against $[S]$ (horizontal): a straight line through the origin with a constant positive slope.
(B)
Figure — $v_{0}$ (vertical) against $[S]$ (horizontal): a sigmoidal curve, flat at first, rising steeply, then levelling off to a plateau.
(C)
Figure — $v_{0}$ (vertical) against $[S]$ (horizontal): a concave-up curve from near the origin that gets steeper as $[S]$ increases.
(D)
Figure — $[S]$ (vertical) against time (horizontal): a curve starting at a nonzero $[S]$ at time zero and decaying, steeply at first and then more slowly, toward the time axis.

Answer and solution

ANSWER

A and D. When $[S] ≪ K_{m}$ , a Michaelis-Menten enzyme follows first-order kinetics, $v_{0} \approx (V_{ma x} / K_{m}) [S]$ . That makes $v_{0}$ against $[S]$ a straight line through the origin (A), and makes $[S]$ fall exponentially with time (D).

Official answer: A, DSource: Official answer key, GATE 2022 BT (IIT Kharagpur)

Michaelis-Menten equation: $v_{0} = \frac{V _{ma x} [ S ]}{K _{m} + [ S ]}$ where $V_{ma x}$ is the maximum velocity (reached when the enzyme is saturated) and $K_{m}$ is the Michaelis constant, the substrate concentration at which $v_{0} = V_{ma x} /2$ .
Limit $[S] ≪ K_{m}$ : the denominator is $K_{m} + [S] \approx K_{m}$ , so $v_{0} \approx \frac{V _{ma x}}{K _{m}} [S]$ The rate is first order in substrate, with first-order rate constant $k = V_{ma x} / K_{m}$ (units of $time^{- 1}$ ).
(A) $v_{0} = k [S]$ is a straight line through the origin with slope $V_{ma x} / K_{m}$ . A is CORRECT.
(B) A sigmoidal $v_{0}$ against $[S]$ curve is the signature of cooperative (allosteric, Hill-type) enzymes. A Michaelis-Menten enzyme gives a rectangular hyperbola, which has no inflection point, and it only approaches its plateau $V_{ma x}$ when $[S] ≫ K_{m}$ , which is outside the stated condition. B is wrong.
(C) A concave-up curve would mean the rate rises faster than in proportion to $[S]$ (order above 1). In this limit the order is exactly 1. Over the full range the hyperbola is concave down everywhere: $\frac{d ^{2} v _{0}}{d [ S ] ^{2}} = - \frac{2 V _{ma x} K _{m}}{( K _{m} + [ S ] ) ^{3}} < 0$ C is wrong.
(D) Substrate depletion over time. The same rate law gives the instantaneous rate at every moment, and because $[S]$ only falls, the condition $[S] ≪ K_{m}$ keeps holding throughout. So $- \frac{d [ S ]}{d t} = \frac{V _{ma x}}{K _{m}} [S]$ Separating variables and integrating from $[S]_{0}$ at $t = 0$ : $[S] = [S]_{0} e^{- (V_{ma x} / K_{m}) t}$ This is an exponential decay: it starts at $[S]_{0}$ , falls steeply and then flattens toward zero, which is the shape shown. D is CORRECT.
Correct options: A and D.

KEY IDEA

At low substrate ( $[S] ≪ K_{m}$ ) the Michaelis-Menten equation reduces to first-order kinetics, $v_{0} \approx (V_{ma x} / K_{m}) [S]$ . A first-order rate law gives two shapes: rate linear in $[S]$ through the origin, and substrate decaying exponentially in time, $[S] = [S]_{0} e^{- (V_{ma x} / K_{m}) t}$ .

WATCH OUT FOR

Picking B because Michaelis-Menten kinetics 'saturates'. The hyperbolic plateau only appears at $[S] ≫ K_{m}$ , and a sigmoid (with an inflection point) belongs to cooperative enzymes, not Michaelis-Menten ones. Another trap is rejecting D because $v_{0}$ is defined as an initial velocity. The rate law holds at every instant, and $[S]$ only decreases, so the low-substrate condition stays true and first-order decay follows.

From the GATE 2022 BT paper. Every question of this paper · Download the full paper · All 9 Enzyme kinetics questions, with notes

GATE 2022 BT Q.50: solution and answer

Answer and solution

More Enzyme kinetics questions