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Follow the carbon atoms in one turn of the Krebs cycle…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Follow the carbon atoms in one turn of the Krebs cycle.
- How many carbon atoms are there in citric acid?
- How many carbon dioxide molecules are lost in one turn?
- How many carbon atoms are there in the oxaloacetic acid at the end of the turn?
- Why can the cycle go on turning with only a small stock of oxaloacetic acid?
Need a reminder? The key ideaShow
Pyruvic acid enters the matrix. The link reaction gives acetyl CoA, carbon dioxide and NADH. In the Krebs cycle, acetyl CoA (2C) joins OAA (4C) to give citric acid (6C). Each turn releases 2 carbon dioxide and forms 3 NADH, 1 FADH and 1 ATP. One glucose gives 2 pyruvic acid and so 2 turns.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. (a) Acetyl CoA brings carbon atoms and oxaloacetic acid has . Citric acid has .
Step 2. (b) Two decarboxylations follow, so carbon dioxide molecules are lost.
Step 3. (c) carbon atoms remain, and oxaloacetic acid is made again.
Step 4. (d) The cycle makes its own oxaloacetic acid at the end of each turn, so a small stock is used again and again.
Answer: ; ; ; the cycle makes oxaloacetic acid again.
Check. The 2 carbon atoms that were lost equal the 2 that came in with acetyl CoA, so the cycle stays in balance.
Watch out. The carbon dioxide of one turn does not come from the acetyl group of that same turn. The balance is what matters.
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