At high concentrations, inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate,

Question

3 years ago

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At high concentrations, inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate,

which of the following compounds would you expect to immediately increase in concentration following the addition of fluoride?
A. 2-phosphoglycerate.
B. Glucose.
C. 3-phosphoglycerate
D. Phosphoenolpyruvate
E. Pyruvate

Chemistry

1 answer:

Alenkinab [10] · Answer 1 · 2022-07-15T04:07:42+03:00

Answer:

<h2>A. 2-phosphoglycerate </h2>

Explanation:

Glycolysis is the process of breakdown of glucose into two 3-carbon molecules called pyruvate. The energy released during glycolysis is used to make ATP.

Enolase is the enzyme which plays very important role in glycolysis. In the 9th step of glycolysis, Enolase converts 2-phosphoglycerate into phosphoenolpyruvate.

This reaction of conversion of 2-phosphoglycerate to phosphoenolpyruvate is a reversible dehydration reaction.

Fluoride inhibits enolase, so when enolase is become non-functional then there is no convertion of 2-phosphoglycerate to phosphoenolpyruvate, so the concentraion of 2-phosphoglycerate is increases by the addition of fluoride.