GATE 2022 BT Q.19: solution and answer

GATE 2022 · BTQ.19 · 1 mark · MCQ

Which of the following statements about reversible enzyme inhibitors are CORRECT?

P. Uncompetitive inhibitors bind only to the enzyme-substrate complex

Q. Non-competitive inhibitors bind only at a different site from the substrate

R. Competitive inhibitors bind to the same site as the substrate

(A)
P and Q only
(B)
P and R only
(C)
Q and R only
(D)
P, Q and R

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Answer and solution

ANSWER

(D) P, Q and R. Uncompetitive inhibitors bind only to ES, non-competitive inhibitors bind at a site other than the active site, and competitive inhibitors bind at the active site that the substrate uses. All three statements are correct.

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

Write the general reversible-inhibition scheme. Free enzyme $E$ binds substrate $S$ to form $E S$ , which turns over to product. $v$ is the initial rate, $V_{ma x}$ is the maximum rate at saturating $[S]$ , and $K_{m}$ is the Michaelis constant. A reversible inhibitor $I$ can form $E I$ (dissociation constant $K_{i}$ ), $E S I$ (dissociation constant $K_{i}^{'}$ ), or both. Which of these complexes can form is what separates the three classes.
P, uncompetitive: $I$ binds only to $E S$ , forming $E S I$ . It has no affinity for free $E$ , because its binding site only appears once substrate is bound. Taking $E S I$ into the enzyme balance gives $v = \frac{V _{ma x} [ S ]}{K _{m} + [ S ] ( 1 + \frac{[ I ]}{K _{i}^{'}} )}$ Dividing top and bottom by $(1 + [I] / K_{i}^{'})$ shows that the apparent $V_{ma x}$ and the apparent $K_{m}$ both fall by the same factor $(1 + [I] / K_{i}^{'})$ . This is the kinetic signature of binding to $E S$ only. P is correct.
Q, non-competitive (pure): $I$ binds at a site other than the active site (an allosteric site). So it can attach to free $E$ and to $E S$ equally, with $K_{i} = K_{i}^{'}$ : $v = \frac{V _{ma x} [ S ]}{( 1 + \frac{[ I ]}{K _{i}} ) ( K _{m} + [ S ] )}$ $I$ does not compete with $S$ for the active site, so $K_{m}$ is unchanged and $V_{ma x}$ falls to $V_{ma x} / (1 + [I] / K_{i})$ . Raising $[S]$ cannot overcome it. Its binding site is always different from the substrate's, so Q is correct.
R, competitive: $I$ resembles the substrate and occupies the active site itself. $S$ and $I$ exclude each other, so the inhibitor binds only free $E$ (forming $E I$ ): $v = \frac{V _{ma x} [ S ]}{K _{m} ( 1 + \frac{[ I ]}{K _{i}} ) + [ S ]}$ The apparent $K_{m}$ rises by the factor $(1 + [I] / K_{i})$ and $V_{ma x}$ is unchanged, because a high enough $[S]$ outcompetes $I$ for the shared site. R is correct.
All three statements are correct, so the answer is (D) P, Q and R.

KEY IDEA

Each class of reversible inhibitor is defined by what it binds. Competitive binds free $E$ at the active site, so $K_{m}$ rises and $V_{ma x}$ stays the same. Uncompetitive binds only $E S$ , so $K_{m}$ and $V_{ma x}$ fall by the same factor. Pure non-competitive binds $E$ and $E S$ equally at a separate site, so $K_{m}$ stays the same and $V_{ma x}$ falls.

WATCH OUT FOR

Mixing up uncompetitive and non-competitive because the names sound alike, and so rejecting P (thinking uncompetitive inhibitors also bind free E) or rejecting Q. Another mistake is to drop a statement because it contains the word "only". Here "only" matches the textbook definition in both P and Q.

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.19: solution and answer

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