I really need help there is two parts to this Four electron carriers, a, b, c, a
ID: 837772 • Letter: I
Question
I really need help there is two parts to this
Four electron carriers, a, b, c, and d, whose reduced a nd oxidized forms can be distinguished spectrophotometrically, are required for respiration in a bacterial electron-transport system. In the presence of substrates and oxygen, three differe nt inhibitors block respiration, yielding the patterns of oxidation states shown below. Inhibitor
O and R indicate fully oxidized and fully reduced, respect ively.
a) What is the order of the carriers in the chain f rom substrates to O 2 ? (10 points
Inhibitor a b c d 1 O O R O 2 R R R O 3 O R R OExplanation / Answer
I would say c must be first because it is always reduced. If the chain is disrupted at any point, you would expect the latter steps to be oxidized (not carrying electrons).
Similarly, I would say d should be the last step. If it is, and delivers its electrons to oxygen, then you would expect it to always be oxidized if oxygen is always present.
Thenb would b would be step two because it is reduced while a is oxidized in trial three. That indicates to me that if the chain is working, c is transferring electrons and reducing b, but that's where the chain stops. b is stuck in its reduced form.
Thus the order would be c --> b --> a --> d
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From the data we can see that inhibitor #1 inhibits the second step of our proposed chain. That is, it inhibits b (step 2) because c is still working (step 1).
Inhibitor #2 can only inhibit the last step (i.e. carrier d) because a, b, and c are all reduced and functioning.
Inhibitor #3 should act on carrier a because it remains oxidized even in the presence of carrier 2, b. That is, carrier a would be the next step in our proposed chain.
Hopefully that makes sense! Let me know if you're still confused.