Quantcast

KA enzyme kinetics

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
6

663697


Members don't see this ad.
Individuals heterozygous for the PKU mutation have very slightly increased levels of phenylalanine. How would you expect the phenylalanine hydroxylase reaction rate of an individual heterozygous for the PKU mutation to be different in than an individual with no mutated genes?

"The reaction rate would decrease due to a decrease in the reaction's Vmax"

Explanation: Having a heterozygous PKU mutation would mean that a portion of the expressed phenylalanine hydroxylase will be mutated. This would lower the effective [E]. A reaction’s Vmax is dependent on enzyme concentration, while an enzyme’s Km is not.

----

Just wondering why the Km wouldn't increase for this type of mutation. The passage says the recessive disorder arises from an "inactivating mutation in the gene coding for the phenylalanine hydroxylase enzyme". Given this I don't understand why we can't assume that a heterozygote would be inactivated to a lesser degree, thus reducing its affinity for substrate.
 

Nugester

Full Member
2+ Year Member
Joined
Jul 4, 2017
Messages
842
Reaction score
827
Less enzyme available means it won't take much substrate to saturate, which means Vmax is reached at lower levels. Because Vmax is decreased and Km is 1/2Vmax, it should decrease as well. Remember, PKU results from a point mutation. For a person who is heterozygous, you still have funcitonal enzyme around, but at lower levels. That's why they are asymptomatic/normal. Increasing Km means you need more substrate to reach Vmax. However there is no affinity issues between the enzyme and substrate, only LESS enzyme and it still works.
 
  • Like
Reactions: 1 user

aldol16

Full Member
5+ Year Member
Joined
Nov 1, 2015
Messages
5,435
Reaction score
4,224
Less enzyme available means it won't take much substrate to saturate, which means Vmax is reached at lower levels. Because Vmax is decreased and Km is 1/2Vmax, it should decrease as well. Remember, PKU results from a point mutation. For a person who is heterozygous, you still have funcitonal enzyme around, but at lower levels. That's why they are asymptomatic/normal. Increasing Km means you need more substrate to reach Vmax. However there is no affinity issues between the enzyme and substrate, only LESS enzyme and it still works.

You have this confused. Km is not 1/2 Vmax. It is the [substrate] at which 1/2 Vmax is reached. Therefore, it doesn't necessarily change with Vmax. You can see this with non-competitive inhibition.
 

Nugester

Full Member
2+ Year Member
Joined
Jul 4, 2017
Messages
842
Reaction score
827
You have this confused. Km is not 1/2 Vmax. It is the [substrate] at which 1/2 Vmax is reached. Therefore, it doesn't necessarily change with Vmax. You can see this with non-competitive inhibition.
Yep I realize that now ha. Time to go read back and brush up on kinetics o_O
 

aldol16

Full Member
5+ Year Member
Joined
Nov 1, 2015
Messages
5,435
Reaction score
4,224
Just wondering why the Km wouldn't increase for this type of mutation. The passage says the recessive disorder arises from an "inactivating mutation in the gene coding for the phenylalanine hydroxylase enzyme". Given this I don't understand why we can't assume that a heterozygote would be inactivated to a lesser degree, thus reducing its affinity for substrate.

Because it's not inactivated to a lesser degree. There's no gradation here. It's either inactivated or it's not - that's what an inactivating mutation does. So you essentially have two enzyme populations - good enzyme and bad enzyme. Good enzyme bind substrates normally. Bad enzymes don't react with substrates at all. So even if they bind, they're not going to give any rate (V = 0).
 

rabbott1971

Full Member
2+ Year Member
Joined
Apr 26, 2016
Messages
928
Reaction score
1,270
Km is saying how much the enzyme wants to bind to the substrate. Km is a feature of the enzyme, whereas Vmax is a feature of the reaction. Even if Vmax goes down, for whatever reason, it won't change how the enzyme feels about the substrate.
 
  • Like
Reactions: 1 user
6

663697

Because it's not inactivated to a lesser degree. There's no gradation here. It's either inactivated or it's not - that's what an inactivating mutation does. So you essentially have two enzyme populations - good enzyme and bad enzyme. Good enzyme bind substrates normally. Bad enzymes don't react with substrates at all. So even if they bind, they're not going to give any rate (V = 0).
So since it's heterozygous, there will still be some functional enzymes produced, but not enough to have the same Vmax? Would this be explained by Vmax = kcat * [E] (E lowers so Vmax lowers), or does kcat also change in this case?

I think I was getting tripped up by the part that said "a portion of the expressed phenylalanine hydroxylase will be mutated". I assumed this meant the binding site was affected thereby affecting the binding affinity as well.
 

Nugester

Full Member
2+ Year Member
Joined
Jul 4, 2017
Messages
842
Reaction score
827
So since it's heterozygous, there will still be some functional enzymes produced, but not enough to have the same Vmax? Would this be explained by Vmax = kcat * [E] (E lowers so Vmax lowers), or does kcat also change in this case?

I think I was getting tripped up by the part that said "a portion of the expressed phenylalanine hydroxylase will be mutated". I assumed this meant the binding site was affected thereby affecting the binding affinity as well.

For me, it is more intuitive to remember what is going on in the body, though it doesn't hurt to memorize formulas and relationships. Remember, the important part to remember is that this is a heterozygote. They will have abnormal and NORMAL enzyme. The normal enzyme can break down that phenylalanine with ease. Abnormal enzyme won't do anything, as a result, phenyalanine accumulates, which is deadly if untreated. Abnormal enzyme doesn't bind phenylalanine. That's why we can detect it in newborn screening from blood etc. All this means is LESS substrate needed to occupy the NORMAL enzyme. Compare that to the wild-type homozygote. That implies Vmaxhetero<Vmaxhomo. I would imagine kcat is intrinsic to the enzyme and since we have functional enzyme still, kcat would not change. Hence lower "E" gives ya lower Vmax.
 
Last edited:

aldol16

Full Member
5+ Year Member
Joined
Nov 1, 2015
Messages
5,435
Reaction score
4,224
So since it's heterozygous, there will still be some functional enzymes produced, but not enough to have the same Vmax? Would this be explained by Vmax = kcat * [E] (E lowers so Vmax lowers), or does kcat also change in this case?

I think I was getting tripped up by the part that said "a portion of the expressed phenylalanine hydroxylase will be mutated". I assumed this meant the binding site was affected thereby affecting the binding affinity as well.

kcat does not depend on enzyme concentration. kcat is basically the maximum rate of turnover per catalytic site - since it's normalized by number of catalytic sites available, it doesn't depend on enzyme concentration. Here, what's happening is that you have either 1) fully functional or 2) fully non-functional enzyme produced. So say you produce 1 mol of enzymes. What this means is that with this mutation, say you get 0.5 mol functional and 0.5 mol non-functional. There's no in between. This decreases the effective enzyme concentration because you're really only getting 0.5 mol of enzyme. That's why Vmax goes down. Km doesn't go down because the functional enzyme still binds the substrate with the same affinity. The non-functional enzyme is invisible and therefore doesn't matter at all for Km.
 
  • Like
Reactions: 1 user
Top