Enzyme Catalysis and Regulation

Bio102 Problems

Enzyme Catalysis and Regulation

1. After weeks in the laboratory, you have discovered that the chemical 3-AT activates the enzyme Histidine Synthase. Which statement might be true about how this enzyme activation works?

A. 3-AT might bind to the active site of Histidine Synthase.

B. 3-AT might lower the Km of Histidine Synthase.

C. 3-AT might bind to the Histidine Synthase’s substrate.

D. 3-AT might lower the Vmax of Histidine Synthase.

E. 3-AT might raise the Km of Histidine Synthase.

2. Which one statement is true when the concentration of the substrate is increased in an enzyme-catalyzed reaction?

A. The vi increases.

B. The Km increases.

C. The Km decreases.

D. The Vmax increases.

E. The Vmax decreases.

3. The ability of an enzyme to bind to its substrate is best measured by its

A. DG.

B. vi.

C. Vmax.

D. Km.

E. Keq.

4. The ability of an enzyme to increase the rate of a reaction is explained by the

A. induced fit model.

B. endosymbiosis model.

C. lock-and-key model.

D. fluid mosaic model.

E. surface area-to-volume model.

5. Which one of the following could be the units of a Km measurement?

A. g/mole

B. mmoles

C. mg

D. mmoles/minute

E. mM

6. Aconitase is an enzyme that catalyzes the conversion of citrate à isocitrate. Without the enzyme, this reaction has a DG value of +0.8 kcal/mole. We isolate the enzyme from either plant or animal cells and compare how well the two enzymes work in the lab.

The animal enzyme has a Vmax 100mmoles/sec of and a Km of 20mM.

The plant enzyme has a Vmax 80mmoles/sec of and a Km of 2.5mM.

6A. The equilibrium constant (Keq) for this reaction …

is > 1.0.

is = 1.0.

is between 0 and 1.0.

is < 0.

can’t be determined from the given information.

6B. What is the initial velocity of the reaction at 20mM citrate when catalyzed by the animal enzyme? ___5mmoles/sec ______

6C. Why is it essential to measure the initial velocity of the reaction in the lab and not the velocity after a minute or two?

We will be approaching equilibrium and the back-reaction will be occuring as well.

6D. Does the animal enzyme or the plant enzyme bind citrate more tightly? How do you know?

The plant enzyme binds more tightly because it has the lower Km value.

6E. In the presence of the plant enzyme, what is the DG value of this reaction? __ +0.8 kcal/mole

7. An inhibitor that binds an enzyme’s active site would be expected to

A. raise the Km.

B. lower the Km.

C. raise the Vmax.

D. be a noncompetitive inhibitor.

E. be an allosteric activator.

8. We are interested in studying how an enzyme catalyzes the reaction:

2-phosphoglycerate à 3-phosphoglycerate.

8A To determine the vi, we need to draw a graph of conc. of 3-phosphoglycerate on the y-axis vs. time on the x-axis.

8 Sketch this graph below and indicate how you will find vi.

the slope of the straight

part of the line is the

vi

8 To determine the Vmax, we need to draw a graph of ______vi______on the y-axis vs. _conc. of 2-phosphoglycerate_ on the x-axis.

9 On the right is a Michaelis-Menten plot for

the enzyme Maltase (which catalyzes the reaction

Maltose + H20 à 2 Glucose).

9. Why does the graph level off? vi

All of the active sites are saturated, or are occupied

all the time.

9. Suggest appropriate units for vi.

mmoles per minute Concentration of Maltose (mM)

9. This graph can be used to find the Km and Vmax of Maltase. What units would Km have?

mM

9D. Suppose that we added an allosteric activator of Maltase which affects only the binding of Maltose to the active site. Draw a new line on the graph that would correspond to how Maltase acts in the presence of the activator.

The problem describes a change in the active site so the substrate fits better. This means a decrease in the Km.

10. You and your lab partner are studying the enzyme sucrase which catalyzes the sucrose + H2O à glucose + fructose. When taking your lab measurements, why it is essential to find the reaction rate immediately after mixing the substrate and enzyme?

We will be approaching equilibrium and the back-reaction will be occuring as well.

11. We are able to break open some spinach cells and measure the activity of Acyltransferase. After running a series of experiments, we generate the Michaelis-Menten curve shown below:

Concentration of Substrate

11A. Label the y-axis of this graph. Be sure to include some appropriate units.

11B. To generate this curve, why is it essential to measure the enzymatic rate immediately after the reaction begins?

We want to measure the rate of the forward reaction. As time progresses, the reverse reaction (which is also catalyzed) begins to occur as we approach equilibrium and thus the amount of product is dependent upon both the forward and reverse reactions.

11C. If we break open some yeast cells, we can measure the activity of Acyltransferase from this organism too. We discover that yeast Acyltransferase is more active, so it doesn’t become saturated with substrate as quickly. However, the yeast Acyltransferase binds to its substrates just as tightly as spinach Acyltransferase. Draw a new line on the graph to represent the yeast Acyltransferase and explain below how you knew where to draw the line.

The Vmax of the enzyme is higher as it is more active but the Km is the same (since it binds the substrate equally well).

12. The enzyme Phosphoglucoisomerase (abbreviated PGI) catalyzes a reaction changing G6P to F6P. (It doesn’t matter for the purpose of this problem exactly what these molecules are.) The DG for this reaction is +0.4kcal/mol. This enzyme is activated by a small molecule called 4TT. We purchase pure PGI enzyme, dissolve it in a pH8.0 buffer, add 100mM G6P and measure a vi.

12A. Suggest appropriate units for vi. any concentration unit divided by any time unit

such as mM/sec

12B. What can you conclude about the Keq for conversion of G6P to F6P? Keq<1

12C. For each of the following changes, describe how vi and DG would change by circling your choice.

vi / D G
Decrease the G6P concentration to 5mM. / increase decrease no change / increase decrease no change
Add 1 mg/mL of 4TT. / increase decrease no change / increase decrease no change
Decrease the pH of the solution to 2. / increase decrease no change / increase decrease no change

12D. The activator 4TT will bind to the enzyme’s ____allosteric______site.

12E. The addition of 4TT changes the enzyme’s Km value. What kind of units will the Km have? Will the Km increase or decrease? How do you know?

Km will be in mM. It will decrease. 4TT is an activator and thus increases the enzyme’s activity. Enzymes with lower Km values bind to their substrates better and are thus activated.

12F. Outline an experiment to estimate the Vmax of PGI. What units would Vmax have?

As described above, we have found one vi value for this enzyme. Now we’ll repeat the experiment using a variety of different substrate (G6P) concentrations. We’ll plot the resulting vi values vs. [G6P]. The vi value of the plateau will be the Vmax. It’s units will be the same as the vi units (shown in the answer to 14A)

13. In our discussions of energy and enzymes, we looked at several important variables that are shown in the right-hand column below. For each of the following statements, which variable (or variables) is being described? On each line, list all the variables that are accurately described by that statement.

__Keq_____ / This variable(s) is a ratio and thus has no units. / DG
__ Vmax Km_ / This value(s) can be determined from a graph of vi vs. substrate concentration. / EA
__EA____ / The addition of an enzyme decreases this value(s). / Keq
___Keq___ / For a spontaneous reaction, the value of this variable(s) is between zero and one. / Km
__vi, Vmax_ / This variable(s) could have units of mg/sec. / vi
____DG__ / In a nonspontaneous reaction, the value of this variable(s) is positive. / Vmax
__DG, EA__ / This variable(s) could have units of kcal/mole. / None of these

14. The enzyme sucrose synthase is present in many different cell types and connects the molecules glucose and fructose together to make sucrose. However, sucrose synthase acts a little differently depending on the cells from which it is isolated. Shown below are the Km and Vmax values for sucrose synthase isolated from five different plants.

Sucrose synthase from oak leaves / Km= 0.01 / Vmax= 105
Sucrose synthase from grass plants / Km= 0.05 / Vmax= 103
Sucrose synthase from pea plants / Km= 0.07 / Vmax= 106
Sucrose synthase from corn plants / Km= 0.03 / Vmax= 102
Sucrose synthase from soybean plants / Km= 0.10 / Vmax= 104

14A. If we give each of these enzymes a very high concentration of their substrates, which one will most rapidly catalyze the formation of sucrose? Please explain how you know.

Sucrose synthase from pea plants has the highest Vmax so it will yield the highest vi value when the substrate concentrations are high.

14B. Which one of these enzymes binds to its substrates most tightly? Please explain how you know.

Sucrose synthase from oak leaves binds the substrates most tightly because it has the lowest Km value. This essentially measures the ability of the enzyme to interact with the substrates in a very dilute solution.

14C. If we give each of these enzymes a very high concentration of their substrates and a very long time to catalyze the reaction, which enzyme will result in the most sucrose being made? Please explain your answer.

All will be the same! If enough time passes, we’ll reach equilibrium which is entirely independent of the enzyme (or even whether any enzyme at all is present or not).

15. Reasonable units for a vi would be

A. mM

B. mM/sec

C. micromoles

D. kcal/mole

E. %

16. The fact that an enzyme binds to only one substrate molecule is best explained by the

A. cell theory.

B. lock-and-key model.

C. second law of thermodynamics.

D. endosymbiosis hypothesis.

E. induced fit model.

17. The product of a metabolic pathway binds to an allosteric site on the first enzyme in the pathway. What effect would you expect this binding to have on the enzyme?

A. Denaturation

B. It will compete with the substrate for binding to the enzyme

C. Lower the Km

D. Raise the Km

E. None

18. When studying an enzyme in the lab, why do we restrict our observations to early times after the reaction starts?

Otherwise, the reverse reaction begins to happen (as we approach equilibirum) and this obscures the rate of the forward reaction

19. The enzyme Acyl Desaturase catalyzes the formation of unsaturated fatty acids, using saturated fatty acids as a substrate. The DG value for this reaction is +0.4 kcal/mole, with an EA of 0.2 kcal/mol. Taking an interest in dietary lipids, we set out to study this enzyme.

19A. Is this reaction endergonic, exergonic or neutral? Is it spontaneous or nonspontaneous?

The reaction is endergonic and nonspontaneous

19B. Suppose we run this reaction in the lab but change the reaction temperature from 25oC to 37oC. How will this affect the DG value? How will it affect the EA value? How will this affect the reaction rate? Please explain your answers.

The temperature will affect neither the DG nor EA values. However, the elevated temperature will increase the reaction rate as more fatty acid molecules will have sufficient energy to overcome the EA barrier.

19C. After running many assays, we plot the graph on the right from which we can determine the Km and the Vmax of this enzyme. Label both axes of the graph.

vi

19D. Why does this graph start to become flat at the right side?

[Substrate]

The enzyme is approaching Vmax, where the active sites are filled all of the time.

19E. VFG is a small molecule that acts as a competitive inhibitor of this enzyme. At what position on the enzyme does VFG bind?

The active site

19F. Does the addition of VFG affect the primary or tertiary structure of the enzyme? Please briefly explain your answer.

It will affect the tertiary structure of the protein by altering the 3D structure, not the order in which the amino acids are connected.

19G. Draw the energy diagram for the uncatalyzed reaction in the chart below. Be sure to label your axes.

0.7

0.6

0.5

G

(kcal/mol)

0.4

0.3

0.2

0.1

Unsaturated Saturated Fatty Acid

19H. Now we add the enzyme Acyl Desaturase to the reaction. Use a dashed line to show how the enzyme changes the energy diagram. Based on your diagram, record the new DG and EA below (with units!)

DG = ____+0.4kcal/mol______EA = ___0.525 kcal/mol______