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:

Enzymes are bio catalysts in living cells responsible for aiding biochemical reactions.

Their active sites are primary protein  3D structure  for binding with substrates during   reactions.

Enzymes reactions are usually affected by inhibitors(e.g methyl alcohol) which compete with the active site with the actual substrate(ethyl alcohol) ,

This is an example of competitive inhibition. The more ethyl alcohol available, the more active sites  occupied, therefore denying the inhibitor methylalchol  chances to the active site, thus preventing the formation of toxic formaldehyde, and more of non toxic substances from formation of more ADH-METHYL ALCOHOL COMPLEXES at the active sites