Biochem question

[basophilic]

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Say if you have a general pathway A --> B --> C --> D

If you have an enzyme deficiency such that the reaction C --> D is not possible, this generally means C will accumulate in relative excess. But shouldn't B and subsequently A accumulate in excess as well (assuming no feedback inhibition)?

 

[MCATforsale1]

The conversion of substrates before C is not blocked. A can be converted to B and B can be converted to C, so there will not be a buildup of A and B.

 

[balopathic45]

You're assuming that the reverse reaction (C to B) is favorable. However, with enzyme-catalyzed reactions, it's much safer to assume that the reverse reaction is so unfavorable that it doesn't actually take place thus pushing the pathway in one direction.
For example, in glycolysis, hexokinase catalyzes the phosphorylation of glucose to glucose-6-phosphate. The reverse reaction is so unfavorable that you need a completely different enzyme (glucose-6-phosphatase) to make it happen. This is why only the liver (which contains glucose-6-phosphatase) can serve as a source of blood glucose while the muscles cannot (because they only have hexokinase).

 

[basophilic]

The conversion of substrates before C is not blocked. A can be converted to B and B can be converted to C, so there will not be a buildup of A and B.

I know but if you have too high C, then going from B to C will become increasingly nonspontaneous, which is why I felt B would accumulate and subsequently A will accumulate too.

 

[basophilic]

You're assuming that the reverse reaction (C to B) is favorable. However, with enzyme-catalyzed reactions, it's much safer to assume that the reverse reaction is so unfavorable that it doesn't actually take place thus pushing the pathway in one direction.
For example, in glycolysis, hexokinase catalyzes the phosphorylation of glucose to glucose-6-phosphate. The reverse reaction is so unfavorable that you need a completely different enzyme (glucose-6-phosphatase) to make it happen. This is why only the liver (which contains glucose-6-phosphatase) can serve as a source of blood glucose while the muscles cannot (because they only have hexokinase).

You're right that different enzymes are needed for rate-determining steps, and I guess the fact that they'd be under different control mechanisms makes it a bit complicated.
But the other enzymes DO catalyze reversible reactions, so that if you, say, have too much C then enzyme BC (catalyzing B-->C reaction) will work backwards, no? In FLs at least, when they give biochemical pathways, they don't always point out or imply rate-determining vs. near-equilibrium steps.

 

[MCATforsale1]

Basophillic - The quotes aren't working or too complicated for me.

You're assuming the reactions are reversible, which wasn't information provided, and you're bringing in spontaneity, which is an altogether different topic. Based on the information provided, A-> B -> C is valid for as far as we know and therefore it is logical to assume that C will build up. Equilibrium was also not mentioned, and while you bring up a valid point, there is a great deal of missing information, as I mentioned earlier. For a basic question like this, consider basic answers, which in this case is that C will build up because the previous steps are not hindered. That is the point that you should understand.

Hope that helps.

 

[balopathic45]

You're right that different enzymes are needed for rate-determining steps, and I guess the fact that they'd be under different control mechanisms makes it a bit complicated.
But the other enzymes DO catalyze reversible reactions, so that if you, say, have too much C then enzyme BC (catalyzing B-->C reaction) will work backwards, no? In FLs at least, when they give biochemical pathways, they don't always point out or imply rate-determining vs. near-equilibrium steps.

I'm confident the actual MCAT will be much more explicit and clear about the scenario. But, in general, for pathway problems I think its safe to assume they are not reversible unless stated.