Question

If a small molecule inhibitor of complex I were toblock the transfer of electrons from the Fe-S chain in complex I to coenzyme Q predict the following (assume oxygen is in ample supply):

a. Redox state (you can simply state “oxidized” or “reduced”) of:

i. FMN

ii. Complex II Fe-S centers

iii. Cyt c₁

iv. Cu_b

b. Flux through the citric acid cycle (you can simply state “increased” or “decreased”).

c. Explain the rationale for your answer to “ii.” (redox state of complex II Fe-S centers) and “b.” (Flux of the citric acid cycle)

Answer 1

a.

i. FMN - reduced

ii. Complex II Fe-S center - oxidized

iii. cyt c1 - oxidized

iv. Cub - oxidized

b. Flux through the citric acid cycle - decreased.

c. The redox status of complex Fe-S centers depends on the availability of FADH₂ which transfers electron to coenzyme Q. In case of inhibition of electron transfer from NADPH to coenzyme Q via complex 1, FADH₂ is the only reducing cofactor available to operate the electronic transport chain. FADH₂ is formed in citric acid cycle as well as fatty acid oxidation, and remains available to the mitochondrial electron transport chain. Therfore, FADH₂ will reduce the Fe-S of complex II, followed by electron transfer ultimately to oxygen. After consumption of FADH₂, Fe-S become oxidized again ready to accept electron.

The flux through citric acid cycle will decrease bacause NADPH will not be utilized in mitochondrial electron transport chain due to presence of inhibitor. However, FADH₂ will be utilized via complex ii. Therefore to avoid further formation of NADPH in citric acid cycle, alternative pathways like fatty acid oxidation will be activated to produce more FADH₂, lowering the citric acid cycle flux.