[orgo814]

a) What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5 x 10^-5 M if Km = 2.9 x 10^-4 M and Ki = 2.0 x 10^-5 M. b) To what concentration must the substrate be increased to reestablish the velocity at the original unhibited value?

The equation I would use for this (usually) would be Vi = [ S]Vmax/Km (1 + [ I]/Ki) + [ S]

However, I do not have Vmax or a velocity so I am not sure how to solve for [ I]

[Babcock_Hall]

Yes, but you have relative velocities, with and without the inhibitor.

[orgo814]

I don't understand. What do you mean?

[Babcock_Hall]

Suppose we let v₁ be the velocity with no inhibitor and we let v₂ be the velocity with an inhibitor.  Can you write an equation that relates the two velocities?

[orgo814]

Only thing I can think of is the equation above which includes inhibition (1 + [ I]/Ki) and then without just doesn't include that term but don't I still need to include Vmax etc. I'm not best at deriving these sorts of things

[Babcock_Hall]

I would start with v₂/v₁ = 0.25 and see how it turns out.  It looks messy algebraically, but solvable.

[orgo814]

I got it. Thanks. Basically just canceling out that Vmax term and what not

[orgo814]

Ok I actually understand solving for the concentration of substrate. However, I'm having issues with the second part... Basically cuz I don't have a velocity just a ratio

[Borek]

Second part is about ratio as well.