1. Active yet responsive. What is the biochemical advantage of having a
KM approximately equal to the substrate con- centration normally
available to an enzyme?
2. Destroying the Trojan horse. Penicillin is hydrolyzed and thereby
rendered inactive by penicillinase (also known as �-lactamase), an
enzyme present in some resistant bacteria. The mass of this enzyme in
Staphylococcus aureus is 29.6 kd. The amount of penicillin hydrolyzed
in 1 minute in a 10-ml solution containing 10�9 g of purified
penicillinase was measured as a function of the concentration of
penicillin. Assume that the concentration of penicillin does not change
appreciably during the assay.
(a) Plot V0 versus [S] and 1/V0 versus 1/[S] for these data. Does
penicillinase appear to obey Michaelis–Menten kinetics? If so,
what is the value of KM?
(b) What is the value of Vmax?
(c) What is the turnover number of penicillinase under- these
experimental conditions? Assume one active site per enzyme
3. A fresh view. The plot of 1/V0 versus 1/[S] is sometimes called a
Lineweaver–Burk plot. Another way of expressing the kinetic data
is to plot V0 versus V0/[S], which is known as an Eadie–Hofstee plot.
(a) Rearrange the Michaelis–Menten equation to give V0 as a
function of V0/[S].
(b) What is the significance of the slope, the vertical intercept,
and the horizontal intercept in a plot of V0 versus V0/[S]?
4. Competing substrates. Suppose that two substrates, A and B,
compete for an enzyme. Derive an expression relat- ing the ratio of the
rates of utilization of A and B, VA/VB, to the concentrations of these
substrates and their values of kcat and KM. (Hint: Express VA as a
function of kcat/KM for substrate A, and do the same for VB.) Is
specificity determined by KM alone?
5. More Michaelis–Menten. For an enzyme that follows simple
Michaelis–Menten kinetics, what is the value of Vmax if V0 is equal to 1
mol minute-1 when [S] = 1/10 KM?
6. Angry biochemists. Many biochemists go bananas, and justifiably,
when they see a Michaelis–Menten plot like the one shown here. To see
why they go bananas, determine the V0 as a fraction of Vmax when the
substrate concentration is equal to 10 KM and 20 KM. Please control
Data Interpretation Problems
7. KM matters. The amino acid asparagine is required by cancer cells to
proliferate. Treating patients with the enzyme asparaginase is
sometimes used as a chemotherapy treatment. Asparaginase hydrolyzes
asparagine to aspartate and ammo- nia. The adjoining illustration shows
the Michaelis–Menten curves for two asparaginases from different
sources, as well as the concentration of asparagine in the environment
(indi- cated by the arrow). Which enzyme would make a better
8. Varying the enzyme. For a one-substrate, enzyme-cat- alyzed
reaction, double-reciprocal plots were determined for three different
enzyme concentrations. Which of the following three families of curve
would you expect to be obtained? Explain.
9. Too much of a good thing. A simple Michaelis–Menten enzyme, in
the absence of any inhibitor, displayed the fol- lowing kinetic behavior.
The expected value of Vmax is shown on the y-axis.
(a) Draw a double-reciprocal plot that corresponds to the velocityversus-substrate curve. (b) Provide an explanation for the kinetic
10. Rate-limiting step. In the conversion of A into D in the following
biochemical pathway, enzymes EA, EB, and EC have the KM values
indicated under each enzyme. If all of the substrates and products are
present at a concentration of 10-4 M and the enzymes have
approximately the same Vmax, which step will be rate limiting and why?
11.Paradoxical at first glance. Phosphonacetyl-L-aspartate (PALA) is
a potent inhibitor of the allosteric enzyme aspartate transcarbamoylase
(ATCase), which has six active sites, because it mimics the two
physiological substrates. ATCase controls the synthesis of pyrimidine
nucleotides. However, low concentrations of this unreactive bisubstrate
analog increase the reaction velocity. On the addition of PALA, the
reaction rate increases until an average of three molecules of PALA are
bound per molecule of enzyme. This maximal velocity is 17-fold
greater than it is in the absence of PALA. The reaction rate then
decreases to nearly zero on the addition of three more molecules of
PALA per molecule of enzyme. Why do low concentrations of PALA
12. R versus T. An allosteric enzyme that follows the con- certed
mechanism has a T/R ratio of 300 in the absence of substrate. Suppose
that a mutation reversed the ratio. How would this mutation affect the
relation between the rate of the reaction and the substrate
13. Distinguishing between models. The following graph shows the
fraction of an allosteric enzyme in the R state ( fR) and the fraction of
active sites bound to substrate (Y) as a function of substrate
concentration. Which model, the con- certed or sequential, best
explains these results?
14. Reporting live from ATCase 1. The allosteric enzyme aspartate
transcarbamoylase (ATCase) has six active sites, arranged as two
catalytic trimers. ATCase was modified with tetranitromethane to form
a colored nitrotyrosine group (�max � 430 nm) in each of its catalytic
chains. The absorp- tion by this reporter group depends on its
immediate environment. An essential lysine residue at each catalytic
site also was modified to block the binding of substrate. Cat- alytic
trimers from this doubly modified enzyme were then combined with
native trimers to form a hybrid enzyme. The absorption by the
nitrotyrosine group was measured on addition of the substrate analog
succinate. What is the sig- nificance of the alteration in the absorbance
at 430 nm?
15. Reporting live from ATCase 2. A different ATCase hybrid was
constructed to test the effects of allosteric acti- vators and inhibitors.
Normal regulatory subunits were combined with nitrotyrosinecontaining catalytic subunits. The addition of ATP, an allosteric
activator of ATCase, in the absence of substrate increased the
absorbance at 430 nm, the same change elicited by the addition of
succinate (see the graph in problem 14). Conversely, CTP, an allosteric
inhibitor, in the absence of substrate decreased the absorbance at 430
nm. What is the significance of the changes in absorption of the