In a metabolic pathway, succinate dehydrogenase catalyzes the conversion of succinate to fumarate. The reaction is inhibited by malonic acid, a substance that resembles succinate but cannot be acted upon by succinate dehydrogenase. Increasing the amount of succinate molecules to those of malonic acid reduces the inhibitory effect of malonic acid. Which of the following statements correctly describes the role played by molecules described in the reaction?

A. Fumarate is the product, and malonic acid is a noncompetitive inhibitor in the reaction
B. Succinate is the substrate, and fumarate is the product in the reaction.
C. Succinate dehydrogenase is the enzyme, and fumarate is the substrate in the reaction.
D. Succinate dehydrogenase is the enzyme, and malonic acid is the substrate in the reaction.

B. Succinate is the substrate, and fumarate is the product in the reaction.

Explanation:

The substrate is the reactant, i.e. the compound that is being acted upon in the reaction. In this case, the substrate is succinate.

The enzyme is the compound that is doing the catalysis, in this case it is succinate dehydrogenase.

The product is what is made by the reaction, in this case it is fumarate

A non-competitive inhibitor is one that reduces the activity of the enzyme regardless of whether or not it has already bound the substrate. This is not true for this reaction, as the inhibitor - malonic acid - acts as a competitive inhibitor, in the presence of more succinate acid, its effects are reduced.

Therefore, the correct answer is B.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.