Exercises Enzymology

Based upon Néstor Torres & Guido Santos (2017): A simple simulator to teach enzyme kinetics dynamics. Application in a problem-solving exercise, Higher Education Pedagogies, 2:114-27, DOI: 10.1080/23752696.2017.1307693. Licensed under a CC BY 4.0 license.

In the simulator there is an embedded excel file. You can use this directly in the book (in a new browser tab) or download it using the button in the footer of the spreadsheet. For this exercise you will only be using the “Experiments” tab. Please do not alter anything else – if you think you’ve accidentally changed something and the Experiments tab doesn’t work, simply download the Excel file again and start again.

In the top of the Experiments tab, you will see the following parameter entry fields:

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	0	0	0	1
Condition 2	5	1	1	0	0	0	1

For this exercise, you will be changing values in the red and blue boxes, which correspond to different parameters in the Michaelis-Menten equation:

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

Underneath the parameter entry field, you will see four graphs that will change in response to the parameters entered:

• Substrate and product concentrations time course

• Initial rates at given substrate concentrations

• Rate time course

• Double inverse plot

You will be analysing these graphs and obtaining values from them. You can obtain (x, y) values by entering the x-value in the cell beneath each plot.

Please also note that the Excel spreadsheet accepts decimals entered as “0,5” and NOT as “0.5”.

Fast hands and careful experiment design are important in the lab!

As Menten and Michaelis demonstrated, measuring substrate concentrations early as possible after initiation of the reaction is important in obtaining accurate values of V₀. This is because they use a linear fit to the initial values of substrate vs. time to obtain the V₀ . Remember that V₀ is used to obtain K_m and V_max values, which are used to understand how the enzyme works.

Remember that in a typical kinetics experiment, you can change S and Vmax and measure P (or S, or both P and S), though within some limits. We’ll explore those limits in this first section.

Start with these values:

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	0	0	0	1
Condition 2	5	1	1	0	0	0	1

In the first graph, you see substrate being consumed from its original concentration at S=5 (solid line) while product accumulates from 0 (dashed line). (The red and blue colored lines overlap so they appear as one line right now.)

The second graph is a plot of V₀ as a function of substrate concentration S(0).

The third graph depicts the reaction velocity as a function of time as the reaction proceeds.

You can ignore the fourth graph!

Exercise 1

As you can see from the third graph, the velocity begins to drop as soon as the reaction is initiated.

1. What is V₀ (V at time = 0 minutes) in terms of V_max? (You can get this from the graph, but you could also calculate this using the Michaelis-Menten equation.)

2. What is the initial velocity at t=0?

3. After the reaction is initiated, how long does the velocity remain within roughly 10% of the initial value?

4. When does the velocity reach 50% of its initial value?

5. Why does the velocity decrease during the course of the reaction?

6. Remember the derivation of the Michaelis-Menten equation. What important assumption was made about the concentration of the ES complex during the course of the reaction, which greatly simplified the derivation?

Now change the S(0) of Condition 2 to 1, so the values look like this (do not change anything else). Now, S(0) = Km.

	S(0)	Vmax	Km
Condition 1	5	1	1
Condition 2	1	1	1

Exercise 2

Consider the graphs: now the conditions diverge so you can distinguish them (condition 1 is blue, condition 2 is red).

1. Consider the second graph. For Condition 2, what is V₀ in terms of V_max? (You can also get this from the MM equation)

2. Consider the third graph. For Condition 2, what is the velocity at t=0?

3. Consider the third graph. For Condition 2, how long does the velocity remain within roughly 10% of the initial value?

4. Consider the third graph. For Condition 2, when does the reaction velocity reach 50% of its initial value?

Now change the S(0) of Condition 2 to 0.5:

	S(0)	Vmax	Km
Condition 1	5	1	1
Condition 2	0.5	1	1

Exercise 3

1. How long does the velocity of Condition 2 remain within roughly 10% of the initial value?

2. When does the reaction velocity of Condition 2 reach 50% of its initial value?

Exercise 4

Now look at the first graph (substrate concentration vs. time): these are the data that you would see in an actual experiment. Remember that in the lab, you will only be able to measure either the product or the substrate concentrations.

Let’s say that the time required to take each measurement of product concentration requires 30 seconds for sampling and analysis, and that 4 samples are required for an accurate measurement. (This means it requires at least 2 minutes to collect all the S(0) vs time values.)

1. Judging from the linearity of the substrate vs. time curves, is it possible to obtain an accurate measurement of V₀ when S(0)= 0.5 and Km = 1 (Condition 2)? (Hint: what values of S(0) would be measured at 30, 60, 90, and 120 seconds? Do these form a straight line?)

2. Judging from the linearity of the substrate vs. time curves, is it possible to obtain an accurate measurement of V₀ when S(0)= 5 and Km = 1 (Condition 1)?

3. Let’s say you used the concentrations at 30, 60, 90, and 120 seconds, fit them to a linear fit, and claimed the slope represented V₀. Would the badly-measured “V₀” you reported be higher or lower than the actual V₀?

4. Now changing only the S(0) values of Condition 2, can you find three S(0) values, each separated by a factor of 2 (e.g. 0.5, 1, 2) for which all maintain a velocity above 75% of the value at t=0 by the time the fourth data point is measured? (Hint: change S(0) and check the velocity values at t = 2 minutes.)

Experiments with a higher Vmax value:

	S(0)	Vmax	Km
Condition 1	5	1	1
Condition 2	5	5	1

(i.e. change Vmax of Condition 2 to 5, set all S(0) to 5)

Exercise 5

1. Compare the blue and red curves in the first and third graphs. Given our slow measurement times, will it be possible to obtain an accurate measurement of V₀ for Condition 2?

2. Imagine that you observed the Condition 2 data in the first graph. Knowing that an accurate measurement is difficult for this condition given our slow measurement times, how would you change the experimental conditions to obtain an accurate measurement of V₀ ?

Effects of Vmax and Km on enzyme kinetics.

Reset the values so your table looks like this (condition 1 and 2 are identical).

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	0	0	0	1
Condition 2	5	1	1	0	0	0	1

Exercise 6

1. Examine the substrate vs. time graph (first graph). A useful measure of reaction speed is the time required to consume half of the available substrate.
   How long does it take for half of the substrate to be consumed in this condition? (Hint: the dashed line depicts the substrate concentrations, while the solid line depicts the product concentrations. As the reaction is simply S 🡪 P, what’s a quick way of seeing when S=0.5*S(0)?)

Change Km of Condition 2 to 5:

	S(0)	Vmax	Km
Condition 1	5	1	1
Condition 2	5	1	5

2. How long does it take for half of the substrate to be consumed in Conditions 1 and 2?

3. Why is the reaction in Condition 2 slower than Condition 1? Which three specific rates (lowercase k) in the Michaelis-Menten scheme might have changed to give this result, and what molecular process would these changes indicate?

4. In Condition 2, what is V₀ in terms of V_max ?

Change Km of Condition 2 to 0.2:

	S(0)	Vmax	Km
Condition 1	5	1	1
Condition 2	5	1	0.2

Exercise 7

1. How long does it take for half of the substrate to be consumed in Condition 2?

2. Why is the reaction in Condition 2 now faster than Condition 1? Which three specific rates (lowercase k) in the Michaelis-Menten scheme might have changed to give this result, and what molecular process would these changes indicate?

3. Why is the change in reaction speed so much smaller when Km is changed from 1 to 0.2 than when it was changed from 1 to 5, while S(0) is maintained in all conditions at 5?

4. Now consider the third graph. How long does velocity remain above 90% of V₀ in Condition 2?

Now, reset Km of Condition 2 to = 1, and change Vmax of Condition 2 to 10:

	S(0)	Vmax	Km
Condition 1	5	1	1
Condition 2	5	10	1

Exercise 8

1. Consider the first graph. How long does it take for half of the substrate to be consumed in Condition 2? How much faster is the reaction in Condition 2 relative to Condition 1?

2. In the simple Michaelis-Menten scheme, what are the two parameters that determine V_max?

3. How could a cell speed up an enzyme reaction by 10-fold?

4. Imagine that Conditions 1 and 2 are comparing two enzymes from two different species that catalyze the same reaction. You measure the concentrations E_T used in the two conditions, and to your surprise, the E_T values of the two conditions are exactly equal. What parameter is different? Why might this be the case?

Effects of enzyme inhibition on kinetics, and comparing competitive inhibition versus non-competitive inhibition.

Reset the value of Vmax in Condition 2 to 1.

Next, change the values for “Kic” (for “K inhibitor competitive”) for Conditions 1 and 2 to 1.
Change the value of I in Condition 2 to 1.
Your values should look like this:

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	0	1	0	1
Condition 2	5	1	1	1	1	0	1

Exercise 9

1. Consider the first graph. How much longer does it take Condition 2 to convert half of the substrate than Condition 1?

2. Explain the molecular mechanism why Condition 2 is slower than Condition 1.

3. Consider the second graph (V₀ versus S(0)). What is the apparent value of K_M of Condition 2 relative to Condition 1?

4. Keeping all other parameters constant, can you find an S(0) value for Condition 2 such that the initial velocities V₀ are equal? (Hint: this follows from your answer to (c). Also, looking at the Michaelis-Menten equation with competitive inhibition terms included will help here!)

5. Now look at the first graph again. With your new value of S(0) that sets equal initial velocities V₀ (hint: set S(0) of Condition 2 = 10), how long does it take Condition 1 to make 2.5 mM of product? How long does it take Condition 2 to make 2.5 mM of product?

Now change the values of S(0) back to 5, and change the Kic value of Condition 2 to 0.1:

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	0	1	0	1
Condition 2	5	1	1	1	0.1	0	1

Exercise 10

1. How long is required for Condition 2 to convert half the substrate to product? How does this compare to the situation at the beginning of this section, when you set I=1?

2. At the beginning of this section, you used the same inhibitor concentration (I=1) and the reaction was much faster. What is different in this new situation?

3. Look at the graph on the second side. What is the new ratio of apparent Km values for Condition 1 and Condition 2? What does this mean?

Now remove competitive inhibition and replace it with non-competitive inhibition by changing the values to the following: (set Kic to 0, set Kinc for both conditions to 1, set I for Condition 2 to 1)

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	0	0	1	1
Condition 2	5	1	1	1	0	1	1

Exercise 11

1. Consider the first graph. How much longer does it take Condition 2 to convert half of the substrate than Condition 1?

2. Consider the second graph. How much substrate is required for the initial velocity to equal 0.5*Vmax?

3. Given your answer to (b), how can a system overwhelm non-competitive inhibition to restore the speed of an enzymatic reaction to close to values observed in the absence of an inhibitor? What is happening on the molecular level?

Now let’s directly compare competitive with non-competitive inhibition. Set I =1 in both Condition 1 and Condition 2. Now, set Kic in Condition 1 = 1 and = 0 in Condition 2. Set Kinc = 0 in Condition 1 and = 1 in Condition 2. Your numbers should look like this:

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	1	1	0	1
Condition 2	5	1	1	1	0	1	1

Exercise 12

1. How long is required before half the substrate has been converted in these two cases? (You’ve already answered this question above, but your graphs should confirm your earlier results.)

2. Using the third graph: What is the initial velocity at t=0 for these two reactions?

3. Using the third graph: What is the velocity of the reactions when half of the substrate has been consumed?

4. While a fast reaction rate is very impressive and important for cells, another important consideration in metabolic systems is robustness – in other words, maintaining a relatively constant flux (i.e., rate of product conversion) despite changes in conditions like changes in substrate concentrations.
   Looking at the two plots in the third graph, which mode of inhibition leads to a more steady flux despite the same relative reduction in substrate concentrations?

5. Put a number on the robustness: Calculate the relative percentage decrease in velocity from V₀ by the time half of the substrate has been consumed for the two conditions.

6. What’s the percentage decrease in velocity when S has been consumed by 80%? (i.e. P = 4 mM)

Can non-competitive inhibition be as robust as competitive inhibition, without sacrificing the reaction rate? Let’s find out. Set Vmax of Condition 2 to 1.5 while keeping all other values the same:

	S(0)	Vmax	Km	I	Kic	Kinc	n
Condition 1	5	1	1	1	1	0	1
Condition 2	5	1.5	1	1	0	1	1

7. When do the conditions reach half their initial substrate concentrations?

8. What is the percentage drop in velocity by S = 2.5 mM of the two conditions?