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?
___50 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. 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  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).
vi
9. Why does the graph level off?
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 + 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.
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:
vi
mM/sec
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 G 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 100M 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 G would change by circling your
choice.
vi
G
Decrease the G6P concentration to 5M.
increase
decrease
no change
increase
decrease
no change
Add 1 g/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 M. 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_____
__ Vmax Km_
__EA____
___Keq___
__vi, Vmax_
____G__
__G, EA__
This variable(s) is a ratio and thus has no units.
G
This value(s) can be determined from a graph of vi vs.
substrate concentration.
EA
The addition of an enzyme decreases this value(s).
Keq
For a spontaneous reaction, the value of this variable(s) is
between zero and one.
Km
This variable(s) could have units of g/sec.
vi
In a nonspontaneous reaction, the value of this variable(s)
is positive.
This variable(s) could have units of kcal/mole.
Vmax
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
Sucrose synthase from grass plants
Sucrose synthase from pea plants
Sucrose synthase from corn plants
Sucrose synthase from soybean plants
Km= 0.01
Km= 0.05
Km= 0.07
Km= 0.03
Km= 0.10
Vmax= 105
Vmax= 103
Vmax= 106
Vmax= 102
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).