Question

Problem 1. Mitochondria may be prepared from cells and tested in cell free experiments. If supplied with pyruvate and oxygen, isolated mitochondria are capable of aerobic metabolism.

C. You as the investigator measure the energy state of the mitochondria as the ratio [NADH]/[NAD+]

1. Describe how the ratio is high and low for high and low energy charge, respectively

2. You remove the supply of oxygen. Explain how the ratio changes in response

3. You add an inhibitor of ATP synthase. Explain how the ratio changes in response

D. You as the investigator makes three measurements at the end of a test that lasts 10 min (exact time is not important). You measure 10 nmol (10-9 moles) of pyruvate consumed (this number is important). You also measure the nmol of CO2 and citrate produced.

1. Explain how 30 nmol of CO2 were produced.

2. Explain how 0 nmol of citrate were produced.

3. Based on the measurements above, predict how many nmol of acetyl-CoA were produced. Explain.

Answer 1

1) The ratio of [NADH]/[NAD+] will be low for high energy charge and will be high for low energy charge.

In presence of oxygen and highly charged mitochondria, the electron transport chain will transfer electrons from complex I to complex III to complex IV and finally O2. During this electron transfer, NADH is oxidised into NAD+. Thus, NADH concentration decreases. While during low charged of mitochondria, this electron transfer will be slower and the NADH concentration will be higher.

2) If O2 is removed then electron transfer will be shut. So, NADH will not be oxidised to NAD+. Thus, the ratio increases if O2 is removed.

3) If an inhibitor of ATP synthase is added the inhibitor will not affect the ratio. The inhibitor doesn't inhibits the flow of electrons transfer. Thus, NADH/NAD+ decreases as normal situation.

D) One molecule of CO2 is generated during conversation of pyruvate into acetyl Co-A by pyruvate dehydrogenase. Another two molecules of CO2 are produced by enzymes isocytrate dehydrogenase and alphaketoglutarate dehydrogenase in the TCA cycle.

So, 3 molecules of CO3 are produced from one molecule of pyruvate.

Thus, if 10 nmol of pyruvate is consumed then 10*3= 30nmol of CO2 will be produced.

2) While pyruvate enters into TCA cycle as acetyl Co-A and produces CO2,NADH, FADH2,ATP as products, no extra citrate is produced because the citrate will enter another round of cycle at the end of one cycle. Thus, zero nmol of citrate is produced.

3) zero nmol of acetyl Co-A will be produced because all of the acetyl Co-A will enter into the TCA cycle.