Does Km and Vmax depends on enzyme concentration?

[gobbler]

I know there are similar threads but I just want to bring up the topic again because I find it controversial

If Vmax is dependent on enzyme concentration and Km is the substrate concentration = 1/2Vmax. If you increase the enzyme concentration, Vmax increases, then Km must also increase to fufill this 1/2 Vmax requirement no?

I found a book that support my theory, it says that Km and Vmax are constants for a given enzyme concentration.

http://books.google.ca/books?id=FSL...v=onepage&q=enzyme concentration vmax&f=false

---

Members don't see this ad.

 

[Isoprop]

I believe you are correct, but it's been a while since I looked at biochem.

Also, Km is derived from the reaction rates of formation of enzyme, product, substrate, and enzyme-substrate complex. Reaction rates are dependent on concentration. So I'm fairly certain you are right.

 

[Charles_Carmichael]

I know there are similar threads but I just want to bring up the topic again because I find it controversial

If Vmax is dependent on enzyme concentration and Km is the substrate concentration = 1/2Vmax. If you increase the enzyme concentration, Vmax increases, then Km must also increase to fufill this 1/2 Vmax requirement no?

I found a book that support my theory, it says that Km and Vmax are constants for a given enzyme concentration.

http://books.google.ca/books?id=FSLLgHigXSoC&pg=PA186&lpg=PA186&dq=enzyme+concentration+vmax&source=bl&ots=0EktDUVE--&sig=SX7TIiuUj_-rOOA5n5Cf_0VXYUQ&hl=en#v=onepage&q=enzyme%20concentration%20vmax&f=false

The Km should not be dependent on the Vmax:

Vo = (Vmax*)/(Km + )
1/2 Vmax = (Vmax*)/(Km + )
Km + = 2
Km =

Both of the Vmax variables get crossed out when you solve the Michaelis-Menten equation, so it shouldn't matter what the Vmax is.

 

[gobbler]

Hey Kaushik, I read your previous post on this subject and I agree, if you do the whole equation thingie, it does show that Km is independent of Vmax. However, this does not explain it conceptually. I asked my biochem prof today about it and he said Km is independent of enzyme concentration and gave some explaination about the substrate concentration which I didn't catch.

 

[TieuBachHo]

I just give you my 2 cents here:

1- You have to understand the nature of enzymes is that they do not get consumed in the rxn, and thus their concentration remains CONSTANT throughout the rxn process. (It's very important in physiological processes).
2- What alters here is only the substrate concentration and we play around with diff. concentrations.

Therefore, from (1) and (2) we can conclude that Vmax and Km are only dependent on diff. substrate concentrations.

Your book says that "Km and Vmax are constants for a given enzyme concentration." This is correct because once all the active sites of the enzymes are occupied, you can not increase that rate of the rxn (the formation of products) despite how many more substrates are added to the rxn. Thus, Vmax and Km are constants.
However, if you increase the enzyme conc. you violate the condition above. Hence, there is no controversial here.

 

[wanderer]

Hey Kaushik, I read your previous post on this subject and I agree, if you do the whole equation thingie, it does show that Km is independent of Vmax. However, this does not explain it conceptually. I asked my biochem prof today about it and he said Km is independent of enzyme concentration and gave some explaination about the substrate concentration which I didn't catch.

Km is a measure of the enzyme's affinity for a particular substrate. Think of it as the probablity that an enzyme will "stick" to its substrate if it happens to bump into it. This "stickiness" doesn't have anything to do with concentration.