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  1. Problems
  2. /For one glucose, the link reaction and the Krebs cycle send 8 NADH and 2 FADH₂ to the electron…
HardBiologyHigh School (11–12)Plant Biology › Respiration in PlantsNumerical

For one glucose, the link reaction and the Krebs cycle send 8 NADH and 2 FADH₂ to the electron…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

For one glucose, the link reaction and the Krebs cycle send 888 NADH and 222 FADH2_22​ to the electron transport system. The Krebs cycle also makes 222 ATP directly. One mole of ATP holds 30.530.530.5 kJ, and one mole of glucose holds 287028702870 kJ.

  1. How many ATP does the electron transport system make from these carriers?
  1. How many ATP come from the link reaction and the Krebs cycle in all?
  1. How much energy do they keep, for one mole of glucose?
  1. What per cent of the energy of the glucose is this?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

kJ
%
Need a reminder? The key ideaShowHide

Electrons in the chain pump protons into the intermembrane space. The protons flow back through ATP synthase, whose F1_11​ head makes ATP and whose F0_00​ part is the channel. Two protons pass for each ATP. Without oxygen the chain stops. Substrate level phosphorylation needs no membrane, and photophosphorylation is driven by light.

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.

Given. NADH: 888. FADH2_22​: 222. ATP made directly: 222.

Plan. Add the ATP, then turn it into energy.

Step 1. From the carriers. 8×3+2×2=24+4=288 \times 3 + 2 \times 2 = 24 + 4 = 288×3+2×2=24+4=28.

Step 2. In all. 28+2=3028 + 2 = 3028+2=30.

Step 3. Energy. 30 mol×30.5 kJ/mol=91530\ \text{mol} \times 30.5\ \text{kJ/mol} = 91530 mol×30.5 kJ/mol=915 kJ.

Step 4. Share. 9152870×100=31.9%\dfrac{915}{2870} \times 100 = 31.9\%2870915​×100=31.9%.

Answer: 282828; 303030; 915915915 kJ; 31.9%31.9\%31.9%.

Check. The glycolysis part is not counted here. With it, the share is a little higher still.

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  1. Oxidative phosphorylationThe Krebs Cycle and the Electron Transport System

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