Alcohol dehydrogenase (ADH) is an enzyme that aids in the decomposition of ethyl alcohol (C2H5OH) into nontoxic substances. Methyl alcohol acts as a competitive inhibitor of ethyl alcohol by competing for the same active site on ADH. When attached to ADH, methyl alcohol is converted to formaldehyde, which is toxic in the body.

Which of the following statements best predicts the effect of increasing the concentration of substrate (ethyl alcohol), while keeping the concentration of the inhibitor (methyl alcohol) constant?

Question

Alcohol dehydrogenase (ADH) is an enzyme that aids in the decomposition of ethyl alcohol (C2H5OH) into nontoxic substances. Methyl alcohol acts as a competitive inhibitor of ethyl alcohol by competing for the same active site on ADH. When attached to ADH, methyl alcohol is converted to formaldehyde, which is toxic in the body.

Which of the following statements best predicts the effect of increasing the concentration of substrate (ethyl alcohol), while keeping the concentration of the inhibitor (methyl alcohol) constant?

BlackTom AI · Accepted Answer

In competitive inhibition the substrate and inhibitor vie for the same active site on the enzyme, so the relative concentrations of substrate [S] and inhibitor [I] determine which ligand occupies the active site more often.

1) Relevant concepts:
- Competitive inhibition: inhibitor binds reversibly at the active site, blocking substrate binding. 
- Michaelis–Menten consequences: competitive inhibitors increase the apparent Km (lower apparent affinity) but do not change Vmax. 
- Quantitatively: v = (Vmax [S]) / (αKm + [S]), where α = 1 + [I]/Ki (Ki = inhibitor dissociation constant).

2) Step-by-step reasoning:
- With a fixed [I], α is constant (α = 1 + [I]/Ki).
- As [S] increases, the denominator αKm + [S] becomes dominated by [S]. 
- Taking the limit [S] >> αKm gives v ≈ Vmax, so the enzyme can reach its maximum rate despite the inhibitor being present.
- Thus, increasing [S] increases the fraction of active sites occupied by ethyl alcohol relative to methyl alcohol, reducing the effective inhibition.

3) Intermediate steps shown:
- α = 1 + [I]/Ki (constant for fixed [I]).
- v = Vmax[S] / (αKm + [S]).
- If [S] → large, αKm/[S] → 0, so v → Vmax. This mathematically demonstrates that the inhibitory effect is overcome at high [S].

4) Why the correct answer is correct:
- Because methyl alcohol is a competitive inhibitor, raising ethyl alcohol concentration increases the probability that ethyl alcohol, not methyl alcohol, occupies the active site. The competitive inhibition effect therefore decreases.

5) Why the other choices are incorrect:
- "Competitive inhibition will be terminated because ethyl alcohol will bind to methyl alcohol…" is mechanistically incorrect; substrates do not bind inhibitors to negate inhibition. 
- "There will be an increase in formaldehyde because ADH activity increases." Although overall ADH turnover toward ethanol could rise, methyl alcohol must bind ADH to form formaldehyde; increased ethanol actually reduces methyl alcohol binding and thus would not necessarily increase formaldehyde production. The option is misleading/wrong. 
- "The peptide bonds in the active site will be denatured…" Denaturation is not caused by increasing substrate concentration; this describes irreversible structural damage, not competitive inhibition dynamics.

6) Conclusion:
Therefore, increasing ethyl alcohol concentration (with methyl alcohol constant) reduces the degree of competitive inhibition because a larger proportion of active sites will be occupied by substrate. This confirms the correct answer: "Competitive inhibition will decrease because the proportion of the active sites occupied by substrate will increase."

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类似问题

How does a competitive inhibitor affect Km and Vmax ?

Which type of enzyme inhibitor does not affect the maximal velocity of a reaction?

Which of the following sentences best describes what happens during competitive inhibition of an enzyme?

A competitive inhibitor of an enzyme:

A competitive inhibitor of an enzyme:

This type of inhibition is _______________. We know because __________________. In this type of inhibition, ___________ is forbidden.

9. This figure describes inhibition of an enzyme by ____________________.

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希望你的学习变得更简单

This type of inhibition is ___. We know because . In this type of inhibition, _ is forbidden.