Mixed Inhibition - Biochemistry

Card 1 of 4

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 43 left

Question

A mixed inhibitor was found to bind unequally to an enzyme and its enzyme-substrate complex. It has a of for the enzyme, and a of for the enzyme-substrate complex.

If the inhibitor decreased $\small $V_{max}$$ by a factor of 3, what concentration of enzyme was used?

Tap to reveal answer

Answer

Recall that the change in $\small $V_{max}$$ is only effected by the binding affinity to the enzyme substrate complex. We can represent the effect on the $\small $V_{max}$$ by this inhibitor by the following equation.

$V_${max}^{apparent}$= $$\frac{V_{max}$$}{\alpha}$

Since the apparent $\small $V_{max}$$ decreased by a factor of 3, we know that $\small \alpha$ is 3.

Solve for in the following equation:

$\small \alpha=1+\frac{[I]}{K_I}$$

Above, $\small \small \small $K_{I}$$ is of the enzyme substrate complex, which is .

$\small 3=1+\frac{[I]}{2mM}$$

Solve.

← Didn't Know|Knew It →

Question

A mixed inhibitor was found to bind unequally to an enzyme and its enzyme-substrate complex. It has a of for the enzyme, and a of for the enzyme-substrate complex.

If the inhibitor decreased $\small $V_{max}$$ by a factor of 3, what concentration of enzyme was used?

Tap to reveal answer

Answer

Recall that the change in $\small $V_{max}$$ is only effected by the binding affinity to the enzyme substrate complex. We can represent the effect on the $\small $V_{max}$$ by this inhibitor by the following equation.

$V_${max}^{apparent}$= $$\frac{V_{max}$$}{\alpha}$

Since the apparent $\small $V_{max}$$ decreased by a factor of 3, we know that $\small \alpha$ is 3.

Solve for in the following equation:

$\small \alpha=1+\frac{[I]}{K_I}$$

Above, $\small \small \small $K_{I}$$ is of the enzyme substrate complex, which is .

$\small 3=1+\frac{[I]}{2mM}$$

Solve.

← Didn't Know|Knew It →

Question

A mixed inhibitor was found to bind unequally to an enzyme and its enzyme-substrate complex. It has a of for the enzyme, and a of for the enzyme-substrate complex.

If the inhibitor decreased $\small $V_{max}$$ by a factor of 3, what concentration of enzyme was used?

Tap to reveal answer

Answer

Recall that the change in $\small $V_{max}$$ is only effected by the binding affinity to the enzyme substrate complex. We can represent the effect on the $\small $V_{max}$$ by this inhibitor by the following equation.

$V_${max}^{apparent}$= $$\frac{V_{max}$$}{\alpha}$

Since the apparent $\small $V_{max}$$ decreased by a factor of 3, we know that $\small \alpha$ is 3.

Solve for in the following equation:

$\small \alpha=1+\frac{[I]}{K_I}$$

Above, $\small \small \small $K_{I}$$ is of the enzyme substrate complex, which is .

$\small 3=1+\frac{[I]}{2mM}$$

Solve.

← Didn't Know|Knew It →

Question

A mixed inhibitor was found to bind unequally to an enzyme and its enzyme-substrate complex. It has a of for the enzyme, and a of for the enzyme-substrate complex.

If the inhibitor decreased $\small $V_{max}$$ by a factor of 3, what concentration of enzyme was used?

Tap to reveal answer

Answer

Recall that the change in $\small $V_{max}$$ is only effected by the binding affinity to the enzyme substrate complex. We can represent the effect on the $\small $V_{max}$$ by this inhibitor by the following equation.

$V_${max}^{apparent}$= $$\frac{V_{max}$$}{\alpha}$

Since the apparent $\small $V_{max}$$ decreased by a factor of 3, we know that $\small \alpha$ is 3.

Solve for in the following equation:

$\small \alpha=1+\frac{[I]}{K_I}$$

Above, $\small \small \small $K_{I}$$ is of the enzyme substrate complex, which is .

$\small 3=1+\frac{[I]}{2mM}$$

Solve.

← Didn't Know|Knew It →

0

Knew It