Biochem200 1999 Problem set III: Answers

1.

ii. T

iii. F - the enzyme changes only the activation energy

v. T

2.

c) There will be no effect on the equilibrium value. Omitting the enzyme will slow the attainment of equilibrium, however. Because the reaction will proceed very slowly in the absence of a catalyst, it is unlikely that equilibrium will be attained within your lifetime, and for this reason, the value of that constant could not be measured this way.

3.

The reactions that are required are:

and the net reaction is:

4 .

Lactate dehydrogenase, which catalyzes reduction of pyruvate to lactate and regenerates NAD⁺, is inhibited by oxamate. The key observation is that an excess of NAD⁺, when incubated with glucose and oxamate, leads to an accumulation of pyruvate with no accumulation of other glycolytic intermediates. Excess NAD⁺ allows glyceraldehyde 3-phosphate dehydrogenase to continue operating, to compensate for the failure of lactate dehydrogenase to regenerate NAD⁺ from NADH. Extracts incubated only with glucose in the presence of oxamate will begin to accumulate pyruvate, but will also accumulate intermediates preceding glyceraldehyde 3-phosphate in glycolysis, due to the lack of NAD⁺ required for continued synthesis of 1,3-bisphosphoglyceric acid.

5.

a) The concentration of glucose is approximately 10^-5 M.

The concentration of fructose is approximately 3.3 x 10^-7 M.

b) The yield of ATP per mole hexose oxidized is the same for glucose or fructose. Therefore the observed rate (or velocity) of hexose phosphate formation will determine which hexose is more important. Because its rate of phosphorylation is 1000 times higher than that of fructose simply by virtue of its high relative concentration, glucose makes the larger contribution to energy metabolism in the brain.

c) Hexokinase is not the key regulatory enzyme in glycolysis because the step it catalyzes is not unique to the glycolytic pathway. The formation of glucose 6-phosphate could also be considered the initial step in glycogen synthesis. Therefore both pathways must be considered in regulation. If high levels of ATP generated through glycolysis and the citric acid cycle were to shut down hexokinase completely, the alternate pathway for the utilization of glucose 6-phosphate would be shut down as well. It is especially important for an organ like muscle or liver to convert glucose, through glucose-6-phosphate, to other useful compounds when glucose is abundant but additional ATP is not needed.

d)

i) Hexokinase constantly removes glucose from the blood by phosphorylating it so it remains in the cytosol. Glucose is the major fuel for neurons. Because its K_m for glucose is relatively low, it operates at maximum capacity most of the time. Glucokinase, as described below, is needed to handle higher glucose levels after a meal. It is less active at lower levels of glucose.

ii) After a meal that contains carbohydrates which are converted to glucose, large amounts of the monosaccharide circulate to the liver via the portal system. Glucokinase, with a higher K_m for glucose, operates to phosphorylate these higher levels of glucose, which are then used to generate substances for energy storage, like glycogen and triglycerides.

6.

a) The protein kinase A in muscle catalyzes the phosphorylation of phosphorylase kinase, which in turn phosphorylates g]ycogen phosphorylase b, converting it to the active a form. Muscle protein kinase also directly inactivates glycogen synthase by phosphorylating it.

b) Inhibition of cyclic AMP phosphodiesterase by caffeine prolongs the response of the cells to epinephrine, because the continued presence of cyclic AMP keeps muscle protein kinase A active. The kinase will continue to promote the activation of glycogen phosphorylase and the inactivation of glycogen synthase.