What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5x10-3 M if Km=2.9x10-4 and Ki=2x10-5?
Well, let's calculate this in a more amusing way. Brace yourself for some enzyme comedy!
So, we have a substrate concentration of 1.5x10-3 M and a Km of 2.9x10-4. That's quite a challenge for the enzyme!
Now, the Ki value of the competitive inhibitor is 2x10-5. This inhibitor is a sneaky little antagonist trying to ruin the enzyme's day!
To calculate the concentration needed for 75% inhibition, let's use a formula. But don't worry, this won't be as complicated as an algebra equation!
The formula is:
[(Inhibitor) / (substrate)] = (1 + [I] / Ki)
We rearrange this to:
[(Inhibitor) / (substrate)] - 1 = [I] / Ki
Now, to find the concentration of the inhibitor, we multiply both sides by Ki:
[(Inhibitor) / (substrate)] - 1 x Ki = [I]
Putting in the values we have:
[(Inhibitor) / (1.5x10-3)] - 1 x 2x10-5 = [I]
Now, let's solve this equation and unmask the concentration of the competitive inhibitor! Drum roll, please...
[(Inhibitor) / (1.5x10-3)] - 1 x 2x10-5 = [I]
[(Inhibitor) / (1.5x10-3)] - 1 = [I] / 2x10-5
[(Inhibitor) / (1.5x10-3)] - 1 = 75% (because 75% inhibition is what we're after)
[(Inhibitor) / (1.5x10-3)] = 1 + 0.75
[(Inhibitor) / (1.5x10-3)] = 1.75
Now, let's isolate the concentration of the inhibitor:
(Inhibitor) = 1.75 x 1.5x10-3
(Inhibitor) = 2.625x10-3
Ta-da! The concentration of the competitive inhibitor required to yield 75% inhibition is 2.625x10-3 M!
Remember, this is a laughing matter, so don't let those numbers give you a frown. Just imagine the enzyme and the inhibitor dancing in a wild circus routine!
To calculate the concentration of the competitive inhibitor required to yield 75% inhibition, we can use the following equation:
% Inhibition = (Ki / (Ki + [I])) * 100
Where Ki is the dissociation constant for the enzyme-inhibitor complex, and [I] is the concentration of the competitive inhibitor.
Given that Km = 2.9x10-4 M, Ki = 2x10-5 M, and the desired % Inhibition is 75%, we can rearrange the equation as follows:
75 = (2x10-5 / (2x10-5 + [I])) * 100
Simplifying the equation further:
0.75 = 2x10-5 / (2x10-5 + [I])
To isolate [I], we can multiply both sides of the equation by (2x10-5 + [I]):
0.75 * (2x10-5 + [I]) = 2x10-5
Expanding the left side of the equation:
1.5x10-5 + 0.75[I] = 2x10-5
Subtracting 1.5x10-5 from both sides of the equation:
0.75[I] = 2x10-5 - 1.5x10-5
0.75[I] = 0.5x10-5
Dividing both sides of the equation by 0.75:
[I] = (0.5x10-5) / 0.75
Simplifying the expression:
[I] = 6.67x10-6 M
Therefore, the concentration of the competitive inhibitor required to yield 75% inhibition at a substrate concentration of 1.5x10-3 M is 6.67x10-6 M.
To determine the concentration of a competitive inhibitor required to yield 75% inhibition, you can use the equation for competitive inhibition:
\(V = \dfrac{V_{max} \times [S]}{K_m \times (1 + \dfrac{[I]}{K_i}) + [S]}\)
Where:
- V is the initial reaction velocity
- Vmax is the maximum reaction velocity in the absence of inhibitor
- [S] is the substrate concentration
- Km is the Michaelis constant
- [I] is the inhibitor concentration
- Ki is the inhibition constant
We need to solve for the concentration of the inhibitor ([I]) for 75% inhibition. Since inhibition is equal to 1 - fraction of activity remaining, we can write:
\(75\% = 1 - \dfrac{V}{V_{max}}\)
Substituting this value into the competitive inhibition equation, we get:
\(0.75 = 1 - \dfrac{V_{max} \times [S]}{V_{max} \times ([S] \times (1 + \dfrac{[I]}{K_i}) + K_m)}\)
Simplifying the equation gives us:
\(-0.25 = \dfrac{[I]}{K_i} \times [S] + K_m\)
Now, we can substitute the given values:
Km = 2.9x10^(-4)
Ki = 2x10^(-5)
[S] = 1.5x10^(-3)
Plugging these values into the equation, we solve for [I]:
\(-0.25 = \dfrac{[I]}{2x10^{-5}} \times 1.5x10^{-3} + 2.9x10^{-4}\)
Rearranging the equation, we get:
\[0.25 - 2.9x10^{-4} = \dfrac{[I]}{2x10^{-5}} \times 1.5x10^{-3}\]
Solving for [I], we get:
\[[I] = \dfrac{(0.25 - 2.9x10^{-4}) \times 2x10^{-5}}{1.5x10^{-3}}\]
Evaluating this expression, the concentration of the competitive inhibitor required to yield 75% inhibition at a substrate concentration of 1.5x10^-3 M is approximately equal to 3.43x10^-5 M.