Michaelis-Menten Equation - Biochemistry

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Question

For a given enzyme catalyzed reaction, the Michaelis constant is 0.6mM and the substrate concentration is 1.0mM. What is the fractional saturation of the enzyme under these conditions?

Answer

The fractional saturation of an enzyme is defined as the amount of enzyme that is bound to substrate divided by the total amount of enzyme. To calculate the fractional saturation, we'll need to use the Michaelis-Menton equation:

$V_{o}=\frac{V_{max}\left [ S\right ]}{K_{M}+\left [ S\right ]}$

In addition, we'll need to define the rate and maximum rate in terms of enzyme concentrations:

$V_{o}=k_{cat}\left [ ES\right ]$

$V_{max}=k_{cat}\left [ E\right ]_{T}$

From the above equations, we can calculate the fractional saturation of the enzyme:

$Fractional:Saturation=\frac{\left [ ES\right ]}{\left [ E\right ]_{T}}=\frac{V_{o}}{V_{max}}=\frac{\left [ S\right ]}{K_{M}+\left [ S\right ]}$

$\frac{\left [ S\right ]}{K_{M}+\left [ S\right ]}=\frac{1.0: mM}{0.6: mM+1.0: mM}=0.625=62.5%$

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Question

What is the ratio of $\frac{V_o}{V_{max}}$ when ?

Answer

This question is answered using the Michaelis-Menten equation:

$V_o = \frac{V_{max}[S]}{ [S] + K_M}$

Rearrange the equation to find the ratio of interest.

$\frac{V_o}{V_{max}} = \frac{[S]}{ [S] + K_M}$

Plug in for and simplify.

$\frac{V_o}{V_{max}}=\frac{5}{6}$

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Question

Which of the following is true about the Michaelis constant for any given enzyme?

Answer

The Michaelis constant, , is not equal to $\frac{V_{max}}{2}$ , but is rather the substrate concentration when the reaction rate is $\frac{V_{max}}{2}$ . is an inverse measure of a substrate’s affinity for the enzyme. So as the affinity decreases, increases. Enzyme specificity is measured by a different constant, $\frac{K_{cat}}{K_M}$ , the specificity constant. Although and specificity are in an inversely proportional relationship, does not necessarily increase as specificity decreases; rather, $K_{cat}$ , also known as the catalytic constant, could decrease proportionally for a given enzyme. The Michaelis constant, being a measure of affinity, is going to differ for different types of substrates, depending on their shape and other features that influence their ability to bind to an enzyme.

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Question

Given an enzyme with of 0.5mM. at what substrate concentration will the velocity of the enzyme reach $\frac{1}{4}$ of the $V_{max}$ ?

$V_{max}=200\frac{mmol}{s}$

Answer

To solve this, we need the solve for in the Michaelis-Menten equation:

$V_o= \frac{V_{max} [S]}{K_m+[S]}$

We know the following information:

$V_{max}= 200 \frac{mmol}{s}$

$V_o=\frac{1}{4}V_{max}= 50 \frac{mmol}{s}$

Plug in these numbers and solve for substrate concentration.

$50= \frac{200[S]}{.5+[S]}$

$\frac{1}{4}= \frac{[S]}{.5+[S]}$

$\frac{1}{6}mM=[S]\approx 0.17mM$

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Question

What does Michaelis-Menten model assume?

Answer

In this model, an intermediate is formed when the substrate binds to an enzyme. The intermediate decomposes to an enzyme and a product, not just the product alone. The model requires enzyme-catalyzed reactions that include a rate limiting step of the enzyme-substrate complex breaking down into the enzyme and product.

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Question

Suppose that an enzyme mixture contains an enzyme with a michaelis constant of $5.010^{-6}M$ . If the substrate concentration in this mixture is $4.510^{-6}M$ , what is the fractional saturation of this enzyme mixture?

Answer

This question is asking us to determine the fractional saturation of a solution containing enzyme and substrate. Let's start by considering what we have, and what we are trying to solve for.

We know that the substrate concentration is $4.510^{-6}M$ and we also know that this enzyme has a Michaelis constant of $5.010^{-6}M$ .

Also, let's consider what fractional saturation is. Fractional saturation refers to the proportion of enzyme molecules in a solution that are bound to substrate. This value can range from (all enzymes are not bound to substrate) to (all enzymes are bound to, or saturated with, substrate).

So, we are looking to calculate the number of enzyme-substrate complexes divided by the total amount of enzyme in solution, which we can express as $\frac{\left [ ES\right ]}{\left [ E\right ]_{T}}$ . We can also relate these terms by the following equations.

$V_{0}=k_{cat}\left [ ES\right ]$

$V_{max}=k_{cat}\left [ E\right ]_{T}$

Dividing these equation by each other, we obtain:

$\frac{V_{0}}{V_{max}}=\frac{[ES]}{[E]_{T}}$

Furthermore, we can relate these terms by considering the Michaelis-Menten equation:

$V_{0}=\frac{V_{max}[S]}{K_{M}+[S]}$

Or, written another way:

$\frac{V_{0}}{V_{max}}=\frac{[S]}{K_{M}+[S]}$

And combining everything we have done so far, we have:

$\frac{V_{0}}{V_{max}}=\frac{[ES]}{[E]_{T}}=\frac{[S]}{K_{M}+[S]}$

And plugging in values, we can solve:
$Fractional:Saturation=\frac{[ES]}{[E]_{T}}=\frac{4.510^{-6}:M}{5.010^{-6}:M+4.5*10^{-6}:M}=47.37: %$

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Question

Which of the following is false about the Michaelis-Menten equation?

Answer

The Michaelis-Menten equation can be expressed as:

$V=\frac{k{_{cat}}[E_{0}][S]}{K_{m}+[S]}$

The velocity is therefore proportional to the enzyme concentration $E_{0}$ , not inversely so. $k_{cat}$ is also referred to as the turnover number. As the substrate concentration keeps increasing, then we end up with a steady state in which all the enzyme is bound. At this point, the maximum velocity, $V_{max}$ , has been reached.

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Question

Which of the following will increase the reaction rate of an enzymatic reaction?

I. Adding a competitive inhibitor

II. Increasing the substrate concentration

III. Decreasing the affinity of substrate to the enzyme

Answer

Reaction rate of an enzymatic reaction can be calculated using the Michaelis-Menten equation.

$V = \frac{Vmax [S]}{K_m + [S]}$

where is the reaction rate, is the maximum reaction rate, is the substrate concentration and is the Michaelis constant.

Recall that adding a competitive inhibitor will increase the but will not alter the . From the equation, we can see that increasing will decrease the reaction rate; therefore, adding a competitive inhibitor will decrease reaction rate.

Substrate concentration is found in both the numerator and denominator of the equation; however, the substrate concentration is multiplied in the numerator whereas it is added in the denominator. This means that increasing substrate concentration will increase the numerator more than the denominator; therefore, increasing substrate concentration will increase reaction rate.

Affinity between enzyme and substrate is determined by . This constant is defined as the substrate concentration required to reach half the . Lowering suggests that a lower substrate concentration is needed to reach the same half (lower substrate concentration is needed because the affinity between enzyme and substrate increases and a more efficient reaction is carried out). This implies that lowering increases the affinity between substrate and enzyme; therefore, and affinity are inversely related. Decreasing the affinity will increase the and, subsequently, decrease reaction rate.

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Question

What is the maximum reaction rate if adding $3\textup{ M}$ of substrate produces a reaction rate of $15\ \frac{\textup{mol}}{\textup{s}}$ ?

$(\textup{K}_\textup{m} = 3\textup{ M})$

Answer

The trick to this question is to notice that the and the substrate concentration are the same. Recall that is the substrate concentration required to reach half the maximum reaction rate. Since we are told that the substrate concentration and are the same, we can conclude that that reaction rate of $15\frac{mol}{s}$ is half the maximum reaction rate; therefore, the maximum reaction rate is $30\frac{mol}{s}$ .

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Question

An enzyme with a $\textup{K}_{\textup{M}}$ value of $5\textup{ mM}$ has a reaction rate of $200\ \frac{\textup{mmol}}{\textup{min}}$ at a substrate concentration of $0.5 \textup{ mmol}$ . What is the maximum reaction rate that this enzyme can achieve when it is saturated with substrate?

Answer

For this question, we're provided with the michaelis constant for an enzyme, as well as the reaction rate for that enzyme at a particular substrate concentration. We're asked to determine the maximum possible reaction rate that this enzyme can achieve when it is saturated with substrate.

To solve this problem, we'll need to use the michaelis-menten equation, which is expressed as follows.

$V_{o}=\frac{V_{max}[S]}{K_{M}+[S]}$

Then, we can rearrange the equation above in order to isolate the $V_{max}$ term.

$V_{max}=\frac{V_{o}(K_{M}+[S])}{[S]}$

Now, we can plug in the values given to us in the question stem in order to solve for our answer.

$V_{max}=\frac{(200: \frac{mmol}{min})(5: mmol+0.5: mmol)}{0.5: mmol}$

$V_{max}=2200: \frac{mmol}{min}$

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Question

Which of the following is a correct statement with regards to an enzyme-catalyzed reaction that obeys Michaelis-Menten kinetics?

Answer

To answer this question, it is essential to have an understanding of the Michaelis-Menten kinetics of enzymes.

When plotting a graph of initial reaction rate as a function of substrate concentration, the resulting plot shows a hyperbolic relationship. At first, the rate increases linearly with a fairly steep slope. But as substrate concentration rises, the graph begins to level off.

The question, however, is to determine the order of the reaction with respect to substrate (the effect substrate has on reaction rate). One helpful way to determine this is to make use of the Michaelis-Menten equation.

$V_{0}=\frac{V_{max}[S]}{K_{M}+[S]}=\frac{k_{cat}[E]_{T}[S]}{K_{M}+[S]}$

With this equation in mind, we can make predictions on how substrate concentration will affect the reaction order.

Low Substrate Concentrations:

When substrate concentration is very low, we can assume that $K_{M}>>[S]$ . As a result of this, we can essentially say that $K_{M}+[S]\approx K_{M}$ . Thus, at low concentrations of , the equation can change as follows.

$V_{0}=\frac{V_{max}[S]}{K_{M}}=\frac{k_{cat}[E]_{T}[S]}{K_M}=(\frac{k_{cat}}{K_{M}})[E]_{T}[S]$

As we can see above, under conditions of low substrate concentration, the reaction is first-order with respect to enzyme and first-order with respect to substrate. This means that any slight change in the concentration of substrate will proportionately affect the reaction rate (for instance, if the substrate concentration increases by , then the reaction rate will also increase by , assuming that enzyme concentration is held constant).

* As a side note, it's also worth noting that the above expression also includes a new reaction rate constant, one that is a ratio of two other constants, $\frac{k_{cat}}{K_{M}}$ . This term is actually what is used to assess an enzyme's efficiency, since it states the reaction rate under conditions of very low substrate concentration.

High Substr ate Concentrations :

Now, let's see what happens when substrate concentration is high. When this happens, we can say that $[S]>>K_{M}$ . Just as we made an estimation before, we can do so once again by stating that $[S]+K_{M}\approx [S]$ under such conditions. Next, we can take a look at how this changes the original equation.

$V_{0}=\frac{V_{max}[S]}{[S]}=\frac{k_{cat}[E]_{T}[S]}{[S]}=k_{cat}[E]_{T}=V_{max}$

As we can see, under high substrate concentrations (saturating conditions), the reaction is at its maximum value. Furthermore, notice that the term cancels out of the expression. This means that, under saturating conditions, the reaction is zero-order with respect to substrate and first-order with respect to enzyme.** The consequence of this is that a change in substrate concentration is unable to change the reaction rate; only a change in enzyme concentration (or temperature) can affect the rate at saturating substrate concentrations.

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Question

Based on Michaelis-Menten enzyme kinetics, if a system has a substrate concentration that is significantly more then the Km, which of the following is a sound inference?

Answer

With Michaelis-Menten Kinetics, The Vmax is the maximum rate of the enzyme mediated reaction. When a system has a concentration of substrate well above Km (which is the concentration of substrate at which the reaction is proceeding at one-half Vmax), then it is said that the system is saturated. In this state, the reaction is occurring basically at Vmax, since there is plenty of substrate to react with the enzyme, and adding more substrate would not increase the reaction rate. Rather, the concentration of enzyme is the limiting factor of how much reaction product will be produced per unit time.

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