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Example Questions
Example Question #1 : Michaelis Menten Equation
Which of the following is true about the Michaelis constant for any given enzyme?
It increases as the enzyme’s affinity for the substrate decreases
None of the other answers are true
It increases as the enzyme’s specificity for the substrate decreases
It is independent of the type of substrate
It is equal to
It increases as the enzyme’s affinity for the substrate decreases
The Michaelis constant, , is not equal to , but is rather the substrate concentration when the reaction rate is . 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, , the specificity constant. Although and specificity are in an inversely proportional relationship, does not necessarily increase as specificity decreases; rather, , 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.
Example Question #1 : Michaelis Menten Equation
What is the ratio of when ?
Cannot be determined
This question is answered using the Michaelis-Menten equation:
Rearrange the equation to find the ratio of interest.
Plug in for and simplify.
Example Question #2 : Michaelis Menten Equation
What does Michaelis-Menten model assume?
enzyme, substrate, and enzyme-substrate complex are in equilibrium
Enzyme-substrate complex decomposes to product only
The rate limiting step may not be present
No intermediate is formed
There are no enzyme catalyzed reactions
enzyme, substrate, and enzyme-substrate complex are in equilibrium
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.
Example Question #1 : Michaelis Menten Equation
Given an enzyme with of 0.5mM. at what substrate concentration will the velocity of the enzyme reach of the ?
To solve this, we need the solve for in the Michaelis-Menten equation:
We know the following information:
Plug in these numbers and solve for substrate concentration.
Example Question #1 : Michaelis Menten Equation
Suppose that an enzyme mixture contains an enzyme with a michaelis constant of . If the substrate concentration in this mixture is , what is the fractional saturation of this enzyme mixture?
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 and we also know that this enzyme has a Michaelis constant of .
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 . We can also relate these terms by the following equations.
Dividing these equation by each other, we obtain:
Furthermore, we can relate these terms by considering the Michaelis-Menten equation:
Or, written another way:
And combining everything we have done so far, we have:
And plugging in values, we can solve:
Example Question #3 : Michaelis Menten Equation
Which of the following is false about the Michaelis-Menten equation?
Velocity is proportional to enzyme concentration.
The maximum rate of reaction is reached as the substrate concentration increases indefinitely.
Velocity is proportional to substrate concentration.
Velocity is proportional to the turnover number.
Velocity is inversely proportional to enzyme concentration.
Velocity is inversely proportional to enzyme concentration.
The Michaelis-Menten equation can be expressed as:
The velocity is therefore proportional to the enzyme concentration , not inversely so. 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, , has been reached.
Example Question #1 : Michaelis Menten Equation
An enzyme with a value of has a reaction rate of at a substrate concentration of . What is the maximum reaction rate that this enzyme can achieve when it is saturated with substrate?
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.
Then, we can rearrange the equation above in order to isolate the term.
Now, we can plug in the values given to us in the question stem in order to solve for our answer.
Example Question #1 : Michaelis Menten Equation
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
I and III
I only
II only
I and II
II only
Reaction rate of an enzymatic reaction can be calculated using the Michaelis-Menten equation.
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.
Example Question #1 : Michaelis Menten Equation
What is the maximum reaction rate if adding of substrate produces a reaction rate of ?
Cannot be determined from the given information
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 is half the maximum reaction rate; therefore, the maximum reaction rate is .
Example Question #11 : Michaelis Menten Equation
Which of the following is a correct statement with regards to an enzyme-catalyzed reaction that obeys Michaelis-Menten kinetics?
The reaction is first-order with respect to substrate at low substrate concentrations, and zero-order with respect to substrate at higher substrate concentrations
The reaction is always zero-order with respect to substrate regardless of substrate concentration
The reaction is first-order with respect to substrate at high substrate concentrations, and zero-order with respect to substrate at lower substrate concentrations
The reaction is always first-order with respect to substrate regardless of substrate concentration
None of these
The reaction is first-order with respect to substrate at low substrate concentrations, and zero-order with respect to substrate at higher substrate concentrations
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.
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 . As a result of this, we can essentially say that . Thus, at low concentrations of , the equation can change as follows.
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, . 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 Substrate Concentrations:
Now, let's see what happens when substrate concentration is high. When this happens, we can say that . Just as we made an estimation before, we can do so once again by stating that under such conditions. Next, we can take a look at how this changes the original equation.
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.