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) | Control | With inhibitor A | With inhibitor B |
|---|---|---|---|
| 1 | 20 | 11 | 10 |
| 2 | 33 | 20 | 17 |
| 4 | 50 | 33 | 25 |
| 8 | 67 | 50 | 33 |
| 16 | 80 | 67 | 40 |
| 64 | 94 | 89 | 47 |
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?
This is the rate at 4 mM, which is half of the maximum rate and therefore corresponds to Km, not Vmax.
This is the rate at 16 mM, but the rate is still increasing at that concentration, reaching 94 µmol/min at 64 mM.
The rates approach a plateau well below 200. The rate at 64 mM is already 94.
Why the answer is what it is
The control rate levels off as [S] increases and reaches 94 µmol/min at 64 mM. The curve is approaching a maximum of about 100 µmol/min.
Review: Michaelis–Menten Kinetics · Enzyme Inhibition & Kinetic Modulation · Enzymes as Structured Catalysts
Question 2 of 6
What is the Km of the uninhibited enzyme?
At 1 mM the rate is only 20, one fifth of Vmax. Km is the concentration that gives half of Vmax.
The rate at 8 mM is 67, well above half of Vmax. An apparent Km of 8 mM appears later, with inhibitor A.
The rate at 16 mM is 80, which is much more than half of Vmax.
Why the answer is what it is
With Vmax ≈ 100 µmol/min, half of Vmax is 50. The control reaches 50 µmol/min at [S] = 4 mM, so Km ≈ 4 mM.
Review: Michaelis–Menten Kinetics · Enzyme Inhibition & Kinetic Modulation · Enzymes as Structured Catalysts
Question 3 of 6
Which statement best describes the effect of inhibitor A?
That pattern is the one seen with inhibitor B. With A, the rate at 64 mM is 89, close to the control value.
The data show no drop in Km. With A, the rate needs 8 mM to reach 50, twice as much substrate as the control.
The rates with A are lower than the control at every concentration up to 64 mM, so A does have an effect.
Why the answer is what it is
With A, the rate approaches the same maximum (89 at 64 mM compared with 94 for the control), but it takes about 8 mM substrate to reach half of Vmax instead of 4 mM. Vmax is unchanged and the apparent Km is higher, which is the signature of competitive inhibition.
Review: Michaelis–Menten Kinetics · Enzyme Inhibition & Kinetic Modulation · Enzymes as Structured Catalysts
Question 4 of 6
Which statement best describes inhibitor B?
B lowers the rate at every concentration, so it does not activate the enzyme.
With B the maximum rate is clearly lower than the control, so Vmax has changed.
B’s rate at 64 mM is only 47, about half of the control, so a high substrate concentration does not reverse the effect.
Why the answer is what it is
With B, the rate approaches about 50 µmol/min, half of the control Vmax, and it reaches half of that value (25) at 4 mM, the same Km as the control. Lower Vmax with unchanged Km is the pattern for a noncompetitive inhibitor, which reduces the amount of working enzyme without blocking substrate binding.
Review: Michaelis–Menten Kinetics · Enzyme Inhibition & Kinetic Modulation · Enzymes as Structured Catalysts
Question 5 of 6
On a Lineweaver–Burk plot of these data, how would the lines for the control and for inhibitor A compare?
The lines that share an x-intercept are control and B, whose Km values are the same. A has a higher apparent Km, so its x-intercept is different.
Parallel lines occur when both Vmax and Km decrease by the same factor. For A only Km changes.
The two lines have the same y-intercept, so they cross there even though their slopes differ.
Why the answer is what it is
The y-intercept is 1/Vmax and the x-intercept is −1/Km. Because A leaves Vmax unchanged, its line crosses the y-axis at the same point as the control, but with a steeper slope because the apparent Km is larger.
Review: Michaelis–Menten Kinetics · Enzyme Inhibition & Kinetic Modulation · Enzymes as Structured Catalysts · Equations: Reactions
Question 6 of 6
The researcher repeats the experiment with [S] = 1000 mM. Which prediction is most consistent with the results in Table 1?
Inhibitor strength for a competitive inhibitor decreases as [S] increases, since the substrate wins the competition for the active site.
Only the competitive inhibitor can be outcompeted by extra substrate. B lowers Vmax, so it does not reach the control level.
A’s rate at 64 mM (89) is already close to the control, and rising with [S]. A would not stay at 50.
Why the answer is what it is
Raising [S] overcomes competitive inhibition because substrate wins the competition for the active site, so A approaches the same Vmax as the control (about 100). A noncompetitive inhibitor lowers the number of working enzymes, so B stays near its lower Vmax of about 50, no matter how much substrate is added.
Review: Michaelis–Menten Kinetics · Enzyme Inhibition & Kinetic Modulation · Enzymes as Structured Catalysts