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B/B 02 · Enzyme kinetics and inhibition

Two Inhibitors of One Enzyme

Read a rate table to find Vmax and Km and to tell competitive from noncompetitive inhibition.

Passage

Many enzymes follow Michaelis–Menten kinetics. At a fixed enzyme concentration, the initial rate v of product formation depends on the substrate concentration [S] according to v = Vmax[S]/(Km + [S]). Vmax is the rate when every enzyme molecule is saturated with substrate, and Km is the substrate concentration at which the rate is half of Vmax. Because the initial rate is measured before much substrate has been used, the experiments in this passage can be treated as steady-state measurements.

Reversible inhibitors lower the rate of an enzyme-catalyzed reaction without permanently damaging the enzyme. They can be told apart by how they change the apparent Vmax and the apparent Km of the enzyme. Biochemists often plot 1/v against 1/[S], a double-reciprocal or Lineweaver–Burk plot, because the points then fall on a straight line, and the slope and intercepts of the line are easy to compare between conditions.

Many drugs are enzyme inhibitors. Statins, for example, bind the active site of HMG-CoA reductase and compete with its natural substrate, and methotrexate competes with the natural substrate of dihydrofolate reductase. Whether a drug can be overcome by a rise in the substrate level is an important practical question, because substrate concentrations inside a cell are not fixed and can change when a pathway is partly blocked.

A researcher studies a purified enzyme that converts substrate S into product P. She measures the initial rate of the reaction at six substrate concentrations under three conditions: without inhibitor (control), with 10 µM of inhibitor A, and with 10 µM of inhibitor B. The enzyme concentration is the same in every tube. The two inhibitors have different structures. Neither is converted to product, and the activity returns when either one is removed by dialysis. Table 1 shows the initial rates, in µmol/min, rounded to the nearest whole number.

Product formation is followed by measuring the absorbance of the solution during the first minute of each reaction, when less than a tenth of the substrate has been used. Each rate is the average of three tubes.

Table 1. Initial rate (µmol/min) at each substrate concentration.

[S] (mM)ControlWith inhibitor AWith inhibitor B
1201110
2332017
4503325
8675033
16806740
64948947

The researcher plans to use the results to decide which inhibitor might be a better starting point for a drug. A drug that works by competing with the natural substrate would be affected by the concentration of the substrate in the cell, while a drug that reduces the amount of active enzyme would not be. She also wants to understand how each inhibitor would look on a Lineweaver–Burk plot before she repeats the measurements with a wider range of substrate concentrations.

Questions

Question 1 of 6

Based on Table 1, which value is the best estimate of Vmax for the uninhibited enzyme?

Question 2 of 6

What is the Km of the uninhibited enzyme?

Question 3 of 6

Which statement best describes the effect of inhibitor A?

Question 4 of 6

Which statement best describes inhibitor B?

Question 5 of 6

On a Lineweaver–Burk plot of these data, how would the lines for the control and for inhibitor A compare?

Question 6 of 6

The researcher repeats the experiment with [S] = 1000 mM. Which prediction is most consistent with the results in Table 1?