What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5x10-3 if Km=2.9x10-4 and Ki=2x10-5?
To determine the concentration of the competitive inhibitor required to yield 75% inhibition, we can use the following equation:
\[I_eff = \frac{I}{(1+(S/Km)(1+(I/Ki)))}\]
Where:
- I_eff is the effective concentration of the inhibitor
- I is the concentration of the inhibitor
- S is the substrate concentration
- Km is the Michaelis constant
- Ki is the inhibition constant
In this case, we want to find the inhibitor concentration (I) required to yield 75% inhibition. We are given the substrate concentration (S = 1.5x10^(-3)), Km (2.9x10^(-4)), and Ki (2x10^(-5)).
To solve for I, we can rearrange the equation:
\[I_eff = \frac{I}{(1+(S/Km)(1+(I/Ki)))}\]
\[I_eff((1+(S/Km)(1+(I/Ki)))) = I\]
\[I_eff + I_eff(S/Km) + I_eff(S/Km)*(I/Ki) = I\]
\[I_eff + I_eff(S/Km) + I_eff(S/Km)*(I/Ki) - I = 0\]
Now we can substitute the given values:
I_eff = 0.75 (75% inhibition)
S = 1.5x10^(-3)
Km = 2.9x10^(-4)
Ki = 2x10^(-5)
\[0.75 + 0.75(S/Km) + 0.75(S/Km)*(I/Ki) - I = 0\]
\[0.75 + 0.75(1.5x10^(-3))/(2.9x10^(-4)) + 0.75(1.5x10^(-3))/(2.9x10^(-4))*(I/2x10^(-5)) - I = 0\]
Now we can solve this equation using numerical methods or software to find the value of I.
Using this approach, we can find the concentration of the competitive inhibitor required to yield 75% inhibition at a substrate concentration of 1.5x10^(-3) with Km = 2.9x10^(-4) and Ki = 2x10^(-5).
To find the concentration of competitive inhibitor required to yield 75% inhibition, we can use the following equation:
I = (Ki / (Ki + [S])) × 100
Where:
I is the percentage inhibition
Ki is the dissociation constant of the inhibitor
[S] is the substrate concentration
Given information:
I = 75% (percentage inhibition)
Ki = 2x10^-5 (dissociation constant of the inhibitor)
[S] = 1.5x10^-3 (substrate concentration)
Km = 2.9x10^-4 (Michaelis constant)
Since this is a competitive inhibition, we can consider:
Km = Ki + [S]
Let's substitute the values and calculate the concentration of the competitive inhibitor:
2.9x10^-4 = 2x10^-5 + 1.5x10^-3
2.9x10^-4 - 2x10^-5 = 1.5x10^-3
2.7x10^-4 = 1.5x10^-3
[(Ki + [S]) - Ki] = [S]
1.5x10^-3 - 2x10^-5 = [S]
1.48x10^-3 = [S]
Now, let's substitute the values into the equation for percentage inhibition:
I = (Ki / (Ki + [S])) × 100
I = (2x10^-5 / (2x10^-5 + 1.48x10^-3)) × 100
I = (2x10^-5 / 1.48x10^-3 + 2x10^-5) × 100
I = (2x10^-5 / 1.48x10^-3 + 2x10^-5) × 100
I ≈ 0.0135 × 100
I ≈ 1.35%
To achieve 75% inhibition, a concentration higher than the given Ki (2x10^-5) is required. Alternatively, if the inhibitor concentration is fixed at the given Ki, it would result in approximately 1.35% inhibition.