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 enzymecatalyzed 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. G.
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. moles
C. g
D. moles/minute
E. 
6. Aconitase is an enzyme that catalyzes the conversion of citrate  isocitrate. Without the enzyme,
this reaction has a G 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 100moles/sec of and a Km of 20M.
The plant enzyme has a Vmax 80moles/sec of and a Km of 2.5M.
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 20M citrate when catalyzed by the animal
enzyme?
___5moles/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 G value of this reaction? __ +0.8 kcal/mole
7. Including a competitive inhibitor in an enzyme-catalyzed reaction will change some of the
variables of that reaction and leave others unchanged.
Will adding a competitive inhibitor affect the Km? Please explain.
Yes. The inhibitor will bind the active site making it harder for the substrate to bind and thus
raising the Km.
Will adding a competitive inhibitor affect the G? Please explain.
No. The enzyme (either inhibited or uninhibitied) won’t affect the G.
Will adding a competitive inhibitor affect the enzyme concentration? Please explain.
No. Inhibitors don’t destroy (or make) proteins.
Will adding a competitive inhibitor affect the primary structure of the protein? Please explain.
No. The inhibitor will bind to the active site but won’t rearrange the sequence of amino acids in
the protein.
Will adding a competitive inhibitor affect the tertiary structure of the protein? Please explain.
Yes. When the inhibitor binds to the active site, it will slightly push on the surface of the enzyme,
gently changing the three-dimensional structure.
8.
We are interested in studying how an enzyme catalyzes the reaction:
2-phosphoglycerate  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  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. This week in lab, we studied the enzyme tyrosinase, which catalyzed the chemical reaction of
pyrocatechol  quinone that we were able to easily follow by monitoring the absorbance at 450nm.
Kojic acid is a small molecule that binds to the enzyme tyrosinase and inhibits the activity of the
enzyme.
10A. Kojic acid is expected to have what effect on the apparent activation energy of this reaction?
increase
decrease
no effect
10B. Kojic acid would be expected to have what effect on the Keq of this reaction?
increase
decrease
no effect
10C. Your enthusiastic lab instructor asks you to measure how 10M kojic acid affects the Km and
Vmax of tyrosinase. Briefly describe how you would set up the necessary experiment(s).
We’ll need multiple measurements of vi while holding the concentration of enzyme constant and
varying the concentration of pyrocatechol. A plot of the resulting vi values vs. [pyrocatechol] will
allow us to find the Vax and Km in the absence of the inhibitor. Then we’ll repeat the entire
experiment with 10M kojic acid.
10D. When setting up your experiment, you can’t find any pyrocatechol in the lab. However you
are able to find a bottle of pure quinone. Your lab partner insists that you can still do your
experiment using the quinone as a substrate. Is your partner right? Please briefly explain why or
why not.
Yes, my partner is correct. The enzyme will catalyze the reverse reaction (quinone 
pyrocatechol) just as well as the forward reaction.
10E. After a series of careful measurements, you determine that kojic acid has no effect on the
Vmax of tyrosinase. What effect (if any) do you think that kojic acid will have on the Km of the
enzyme? Please briefly explain your answer.
Since it’s an inhibitor, it must affect the Km if it doesn’t affect the Vmax. The Km will be increased,
which inicates that the enzyme binds the substrate less well.
11. You and your lab partner are studying the enzyme sucrase which catalyzes the sucrose + H 2O 
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.