Question

Alcohol dehydrogenase (ADH) is an enzyme that aids in the decomposition of ethyl alcohol (CH3OH) 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? A. There will be an increase in formaldehyde because ADH activity increases. B. Competitive inhibition will be terminated because ethyl alcohol will bind to methyl alcohol and decrease ADH activity. C. The peptide bonds in the active site of the enzyme will be denatured, inhibiting the enzyme. D.Competitive inhibition will decrease because the proportion of the active sites occupied by substrate will increase.

Answer 1

Answer:

D.Competitive inhibition will decrease because the proportion of the active sites occupied by substrate will increase.

Explanation:

Methyl alcohol is a competitive inhibitor of ethyl alcohol. This means that both of them compete with each other for the same enzyme (ADH). Methyl alcohol leads to poisoning because it is converted to toxic formaldehyde in body. It can be treated by increasing the concentration of ethyl alcohol.

More ethyl alcohol (substrate) will reduce the competitive inhibition by methyl alcohol (inhibitor). This will happen because a substrate needs active sites on an enzyme to attach to it. If ethyl alcohol is increased in concentration, it will occupy more active sites and their availability to methyl alcohol will be reduced. Thus, the enzyme will not be able to act on methyl alcohol.