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gtnhenbr [62]
3 years ago
12

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]3 years ago
5 0

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.

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How many moles of liquid water must freeze to remove 100 kJ of heat? (ΔHf = –334 J/<br> g.
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First figure out how many grams must freeze and then convert the grams to moles. 
<span>Hf = -334 J/g. Convert this to KJ/g by dividing by 1000. (There are 1000 Joules in a kJ). </span>
<span>Hf = -334 J/g ÷ 1000 J/kj = -0.334 kJ/g </span>
<span>Now, divide 100 kJ by -0.334 kJ/g (see how the units are lining up?) </span>
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