Modeling the regulation of enzymes
Rana Yuksel
CPR 1 questions + answer
Exercise 1 - Product inhibition at high Keq
Open the excel file and the sheet “Product inhibition_a”. It plots rate as a function of [S] at three
different concentrations of P.
a) The sheet gives a calculated value of Vb . How is that value calculated from the parameters in
yellow above it? Why should we not just substitute any value for Vb?
Vf ∗Kp
Haldane relation! The formula for the Haldane relation: =Keq. We know all the
Vb∗Ks
values of the parameters except for Vb, so it is easy to calculate Vb by rearranging the
formula and putting all the values in. We cannot substitute any value for Vb, because it
depends on the [S] and [P] which has to do with the rate at which the enzyme works.
b) You see hardly any effect of [P] in this graph at this value of Keq. Explain why. You can check
your answer by changing stuff in the sheet, but do not change Keq.
Since the value of Keq is high, you will hardly see any effect of [P]. Important to know is that
in enzymes, the effect of [P] is ALWAYS stronger when the value of Keq is low. In this case,
Keq is 5300 which is extremely high so therefore you will hardly see any effect of [P].
Go to sheet “Product inhibition_b”, and make sure that the values in the yellow boxes are the same as
in the first sheet. This sheet plots fraction of rate as a function of [P] relative to the condition [P] = 0 at
three different concentrations of S.
c) Here you do see an effect of [P], especially at low concentrations of S. Why is the effect
particularly large at low concentrations of S?
When [P] increases, the net forward rate decreases and therefore the fraction activity will
decrease as well ([P] inhibits the net forward rate) low concentrations means that [S] has a
lower ability to compete with [P[ for the enzyme.
Go back to sheet “Product inhibition_a”. Decrease the KP to a value 0.5.
d) Now you do see an effect of P, even at very high concentrations of S. Is the apparent Michaelis
constant for S the same in the three graphs?
No, because the rate will decrease when [P] is high so when Kp decreases, susbtrate affinity of
the enzyme is greater.
e) Referring to eq. (4) and the calculated value of Vb in the sheet, do you think that the rate of the
backward reaction catalysed by the enzyme will be high under these conditions?
No, because the backward reaction catalysed by the enzyme will be slower/lower under these
conditions, due to the fact that Vb will be low when Kp decreases, meaning that the numerator
(under the line) will be higher and denominator (above the line) lower.
f) Can you explain your answer about the apparent KS, for example by referring to eq. (4) and using
your previous answer to make an approximation of v?
The forward reaction will remain the same, but the backward reaction will be faster/higher due to
the fact that the [P] will bind slower when Ks is high, in other words, increasing [P] = decreasing
substrate affinity (thereby the function of the enzyme).
Exercise 2 - Using product inhibition for regulation
In this exercise we want to find out what the function of product inhibition in metabolism could be. In
principle, product inhibition is disadvantageous for the function of an enzyme. It decreases its rate of
catalysis, which seems to be a waste of protein and energy that went into synthesizing this protein.
However, in a pathway it can have a very important function. Go to sheet Product inhibition_a, and
restore the default settings, if necessary.
Rana Yuksel
CPR 1 questions + answer
Exercise 1 - Product inhibition at high Keq
Open the excel file and the sheet “Product inhibition_a”. It plots rate as a function of [S] at three
different concentrations of P.
a) The sheet gives a calculated value of Vb . How is that value calculated from the parameters in
yellow above it? Why should we not just substitute any value for Vb?
Vf ∗Kp
Haldane relation! The formula for the Haldane relation: =Keq. We know all the
Vb∗Ks
values of the parameters except for Vb, so it is easy to calculate Vb by rearranging the
formula and putting all the values in. We cannot substitute any value for Vb, because it
depends on the [S] and [P] which has to do with the rate at which the enzyme works.
b) You see hardly any effect of [P] in this graph at this value of Keq. Explain why. You can check
your answer by changing stuff in the sheet, but do not change Keq.
Since the value of Keq is high, you will hardly see any effect of [P]. Important to know is that
in enzymes, the effect of [P] is ALWAYS stronger when the value of Keq is low. In this case,
Keq is 5300 which is extremely high so therefore you will hardly see any effect of [P].
Go to sheet “Product inhibition_b”, and make sure that the values in the yellow boxes are the same as
in the first sheet. This sheet plots fraction of rate as a function of [P] relative to the condition [P] = 0 at
three different concentrations of S.
c) Here you do see an effect of [P], especially at low concentrations of S. Why is the effect
particularly large at low concentrations of S?
When [P] increases, the net forward rate decreases and therefore the fraction activity will
decrease as well ([P] inhibits the net forward rate) low concentrations means that [S] has a
lower ability to compete with [P[ for the enzyme.
Go back to sheet “Product inhibition_a”. Decrease the KP to a value 0.5.
d) Now you do see an effect of P, even at very high concentrations of S. Is the apparent Michaelis
constant for S the same in the three graphs?
No, because the rate will decrease when [P] is high so when Kp decreases, susbtrate affinity of
the enzyme is greater.
e) Referring to eq. (4) and the calculated value of Vb in the sheet, do you think that the rate of the
backward reaction catalysed by the enzyme will be high under these conditions?
No, because the backward reaction catalysed by the enzyme will be slower/lower under these
conditions, due to the fact that Vb will be low when Kp decreases, meaning that the numerator
(under the line) will be higher and denominator (above the line) lower.
f) Can you explain your answer about the apparent KS, for example by referring to eq. (4) and using
your previous answer to make an approximation of v?
The forward reaction will remain the same, but the backward reaction will be faster/higher due to
the fact that the [P] will bind slower when Ks is high, in other words, increasing [P] = decreasing
substrate affinity (thereby the function of the enzyme).
Exercise 2 - Using product inhibition for regulation
In this exercise we want to find out what the function of product inhibition in metabolism could be. In
principle, product inhibition is disadvantageous for the function of an enzyme. It decreases its rate of
catalysis, which seems to be a waste of protein and energy that went into synthesizing this protein.
However, in a pathway it can have a very important function. Go to sheet Product inhibition_a, and
restore the default settings, if necessary.
