You have a suspension of homogenized liver cells supplied with ample phosphate,

Question

You have a suspension of homogenized liver cells supplied with ample phosphate, ADP, GDP, O_2 and a trace of CO_2. It is in a sealed vessel so that no gases enter or exit. Its mitochondria are intact and fully functional. To it you add 1.0 mu mole of oxaloacetate (the "input"). No other added carbon source is available. When action ceases, w hat compound(s) are the "output"? How many mu moles of each will be present? Show all reasoning. Remember: added oxaloacetate is outside the mitochondria. How many mu moles of ATP will be produced? (assume that 2.5 ATP/NADH, 1.5 ATP/FADH_2) How many mu moles of O_2 (molecules!) will be consumed? if a high concentration of malonate is also present (other conditions remain the same). What molecules are now the "output", and how many mu moles of each? Be meticulous; show' all reasoning (circle all pertinent consumptions as well as productions of molecules of interest). How many mu moles of ATP will be produced if malonate is present? How many mu moles of O_2 molecules will be consumed if malonate is present?

Explanation / Answer

For Krebs’ cycle to occur, oxaloacetate should be there in the mitochondrial matrix. But OAA cannot cross the inner mitochondrial membrane. So, Krebs’ cycle is not possible.

Outside mitochondria, oxaloacetate converts to malate, and then via a shuttle system, it moves into the mitochondrial matrix. Here, malate is again converted to OAA.

If Acetyl CoA is there, then only Krebs’ cycle will occur.

Acetyl CoA combines with OAA to accomplish citric acid cycle. In this cycle 3 NADH+H, 1FADH2 and one GTP will be produced (Answer ‘a’).

So, total ATP generated = (3*2.5) + (1*1.5) = 9.0 ATP (Answer ‘b’).

These compounds (of Answer ‘a’) will undergo oxidative phosphorylation in the inner mitochondrial membrane to produce ATP. The entire process requires consumption of oxygen. 6 oxygen molecules will be consumed i.e. 36ATP generated from one glucose (six carbons) divided by 6 = 6 ATP are generated from one oxygen utilization. Here number of oxygen utilized = 1.5 (Answer ‘c’).

Malonate is competitive inhibitor of Complex II in oxidative phosphorylation. So, it will not allow FADH2 to generate ATP. The output with excess malonate will be NADH+H, and GTP (Answer ‘d’). So, total number of ATP generated now will be (3*2.5) only i.e. 7.5 ATP (Answer ‘e’).

In producing 7.5 ATP, number of oxygen utilized = 1.25.

** Hope it helps.

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