Mixed Inhibition - Biochemistry

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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?

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.

Compare your answer with the correct one above

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?

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.

Compare your answer with the correct one above

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?

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.

Compare your answer with the correct one above

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?

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.

Compare your answer with the correct one above

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