About this experiment
At low substrate concentration an enzyme works faster the more substrate it gets, but eventually every active site is busy and the rate levels off. In 1913 Leonor Michaelis and Maud Menten described this with two numbers: Vmax, the top speed, and Km, the substrate concentration at half that speed. Change both, add three kinds of inhibitor and see how each one moves the curve and the Lineweaver-Burk plot, then fit noisy assay data the old way and the modern way.
Learning goals
- Derive the meaning of Vmax and Km from the Michaelis-Menten equation
- Read Km and Vmax from a saturation curve and from a Lineweaver-Burk plot
- Distinguish competitive, uncompetitive and noncompetitive inhibition by their effect on the apparent constants
- Explain why the double reciprocal plot distorts measurement errors
- Compare a linearised fit with a nonlinear least squares fit
How to use it
- 1Drag along the curve. At what [S] is the rate half of Vmax? Check it against Km.
- 2Set [S] to 10 Km. What fraction of Vmax is reached? Why does the curve never quite reach Vmax?
- 3Add a competitive inhibitor and increase [I]/Ki. Which constant changes? Where do the lines meet on the Lineweaver-Burk plot?
- 4Repeat for uncompetitive and noncompetitive inhibition. Sketch the three Lineweaver-Burk patterns.
- 5Run several new assays with measurement error 0.5. Compare the Lineweaver-Burk estimates with the nonlinear fit. Which is more reliable?
- 6Look at which points sit furthest right on the Lineweaver-Burk plot. Why do they have such a big influence on the line?
The model
Assumptions
- Steady-state kinetics of a single-substrate enzyme: substrate is in large excess over enzyme and only initial rates are measured.
- Noncompetitive here means pure noncompetitive inhibition: the inhibitor binds the free enzyme and the complex equally well. Mixed inhibition is not shown.
- Each assay measures eight substrate concentrations with a constant random error, the same for every point, from a seeded generator so a run can be repeated.
- The nonlinear fit is ordinary least squares on v against [S], solved by Gauss-Newton from the Lineweaver-Burk estimate.
Try this in class
- Q1An enzyme has Vmax = 12 µM/s and Km = 3 mM. What is the rate at [S] = 1 mM and at [S] = 9 mM?
- Q2A Lineweaver-Burk line has intercept 0.08 s/µM and slope 0.24 s·mM/µM. Find Vmax and Km.
- Q3Why does a competitive inhibitor have no effect at very high substrate concentration?
- Q4Many drugs are enzyme inhibitors. Why might a drug designer prefer a noncompetitive inhibitor?
- Q5Explain why the low [S] points dominate a Lineweaver-Burk fit.
Frequently asked questions
Is Km the same as the binding constant of the substrate?
Only when the catalytic step is much slower than the substrate leaving again (k_cat much smaller than k₋₁). In general Km also includes k_cat, so a low Km means efficient binding and turnover together, not only tight binding.
Why is the Lineweaver-Burk plot still taught if it is unreliable?
It turns the curve into a straight line, which makes the patterns of different inhibitors easy to recognise by eye. For actual numbers, researchers fit the original curve with nonlinear regression.
Where does the curve stop following this equation?
Allosteric enzymes whose subunits cooperate, such as aspartate transcarbamoylase, give S-shaped curves instead. Very high substrate concentrations can also inhibit some enzymes.