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6 Cards in this Set

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  • Back
Proteins are thermodynamically unstable and the delta G for protein hydrolysis is negative, yet proteins can be quite stable. Explain this apparent paradox.
The activation energy for protein hydrolysis is high, thus this hydrolysis reaction cannot take place in the absence of an enzyme.
Enzymatic velocities have been measured for the reaction of chymotrypsin with its substrate at six different concentrations. Use the data below to make a reasonable estimate of Vmax and Km value for this substrate
mM [S]
0.00125 0.01 0.04 0.10 2.0 10
14 35 56 66 69 70
Vmax is 70mM/min, and Km is 0.01 mM
Based on the Michalis-Menton equation (shown below), explain why Vmax is the V(naught) when [S] is MUCH greater than Km, i.e., not just 3-4 times higher than Km as most of the curves show.
V(0) = Vmax * [S]/([S]+Km)
The equation can be rearranged as:
V0/Vmax = [S]/[S] + Km
When V0 reaches Vmax, V0 will be equal to Vmax
Thus [S] = [S] + Km
This is unlikely until [S] is MUCH greater than Km, rendering Km
Note: This is the reason researchers plot the data using the Lineweaver-
Burk double reciprocal plot to determine Vmax and Km. The traditional
enzyme kinetics curves do not yield very precise Vmax and Km values.
For a one-substrate enzyme catalyzed reaction, the Lineweaver-Burk double
reciprocal plots were determined for three different enzyme concentrations.
Which of the following three families of curves would you expect to be observed.
Explain your answer.
It is the middle curve because the rate of an enzymatic reaction is proportional to the amount of active enzyme as long as there is a sufficient supply of substrate. Thus, higher amounts of enzyme lead to higher Vmax without affecting the affinity between enzyme and substrate
This question is related to the “turnover number” of an enzyme (also called Kcat),
which is the number of molecules of substrate being converted to product per unit
of time. This is a parameter reflecting how efficiency of an enzyme.
Carbonic anhydrase catalyzes the following reaction:
CO2 + H2O  HCO3
- + H+
A carbonic anhydrase isolated from liver mitochondria has 300 amino acids, and a
solution containing 0.033 ug/uL (ug is 10-6 g, and ul is 10-6 L) of this enzyme can
catalyze the formation of 0.6 M HCO3
- per second when the enzyme is fully
saturated with the substrates, CO2 and H2O. What is the turnover number, Kcat,
of this enzyme?
The turnover number of an enzyme is the number of product molecules
produced by one enzyme molecule per unit time. Thus, when the substrate
concentration is unlimited, all enzyme molecules will be occupied, and the
enzymatic rate reaches Vmax.
Hence: Vmax = Kcat [E] or Kcat = Vmax/[E]
In this case, the mol wt of carbonic anhydrase is 300 amino acids X 110 D =
33,000 D
The concentration of the enzyme is 0.033 ug/uL, or 0.033 g/L, or 10-6 mole of
enzyme/L, or 10-6 M.
Kcat = Vmax/[E] = 0.6 M s-1/10-6 M = 6 X 105 s-1
[Note: This is a very fast enzymatic rate, one of the largest Kcat known, i.e.
taking less than 2 uSec for each reaction to take place.]
The following questions are related to an enzyme subject to allosteric regulation.
a) An allosteric enzyme usually has two forms, active and inactive. If a
competitive inhibitor (NOT an allosteric inhibitor) is added to a solution
containing this enzyme, the ratio of active form of this enzyme to inactive
form actually increases. Explain how this can happen.
b) What are the two most important parameters in enzyme kinetics? Which of
these two is likely affected by competitive inhibition?
a.) An allosteric enzyme has two forms: an active form is maintained when
the active site is occupied, and an inactive form is maintained when the
regulatory site is occupied by an allosteric inhibitor. A non-allosteric
competitive inhibitor competes against the substrate for the same active
site, thus it actually helps maintain the “active” form.
b.) Vmax and Km are the two most important parameter in enzyme kinetics.
Km is likely affected by competitive inhibition while Vmax in NOT likely
to be affected (Vmax is the V0 when [S] is very high, thus S can out
compete the inhibitor).