Michaelis–Menten describes how fast an enzyme works as you add more substrate. The shape of that relationship is the same for nearly every enzyme, and the two numbers that define it, Vmax and Km, come from the two ends of the curve. Most of the marks in this topic are lost by people who can plug into the equation but cannot say what the numbers mean.
Michaelis–Menten is the equation behind every enzyme assay. It runs forwards when you know Vmax and Km, and backwards when all you have is two measured points — which is how real kinetics data gets turned into Km. Both directions are here.
Enter two measurements at different substrate concentrations. The tool takes a Lineweaver–Burk line through them: the y intercept is 1/Vmax and the x intercept is −1/Km.
The equation, and what each part is doing
v = Vmax[S] / (Km + [S])
Vmax is the rate the enzyme would reach if every active site were permanently saturated, so it depends on how much enzyme you have. Km is the substrate concentration at which the rate is exactly half of Vmax.
That second sentence is not a derivation, it is the definition. [S] = Km always gives v = Vmax/2. If you remember one thing, make it this, because it lets you reason about a curve without the equation at all.
Which part of the curve you are on
Substrate concentration relative to Km is what determines the behaviour, and it is worth getting a feel for the three regimes.
- [S] well below Km — the curve looks like a straight line through the origin. Rate rises almost in step with substrate, and the enzyme is nowhere near saturated. This is the region where the classic linear plot is least reliable.
- [S] = Km — half of Vmax. The single most useful point on the curve.
- [S] well above Km — the enzyme is saturated, the curve flattens, and adding more substrate barely changes anything. This is why real assays are usually run well above Km, so small substrate errors stop mattering.
Going backwards: solving for Vmax and Km
Real kinetics data arrives as substrate concentrations and measured rates, and you have to extract the two constants from it. The standard method is the Lineweaver–Burk plot, a double reciprocal:
1/v = (Km/Vmax)(1/[S]) + 1/Vmax
Take 1/v against 1/[S] and the Michaelis–Menten hyperbola becomes a straight line, which is the entire point of the transform. The two intercepts give the constants directly:
- y intercept = 1/Vmax, so Vmax = 1 divided by the y intercept.
- x intercept = −1/Km, so Km is −1 divided by the x intercept.
- slope = Km/Vmax, so Km is the slope times Vmax.
This tool will solve from two points directly, which is the minimum the method needs.
Why everyone criticises Lineweaver–Burk
It is still the standard taught method and you still need to recognise the plot, but it has a bad reputation it has earned. Taking a reciprocal of a small number amplifies its error enormously, so the points furthest from the axis, which are the least precise measurements, get the most visual weight. The transform also turns a gentle curve into a noticeably bent line, and that bend is where the interesting information about inhibitors lives. Hanes–Woolf and Eadie–Hofstee plots weight the points better, and non-linear fitting is what a careful lab actually does.
Two points also give you a straight line with no way to judge whether the data was curved in the first place. If you have more than two points, plot them rather than trusting a two-point fit.
Worked example
Suppose Vmax = 100 and Km = 2. At [S] = 2 the rate is 100×2/(2+2) = 50, exactly half of Vmax, as it must be.
At [S] = 100 the rate is 100×100/102 = 98.04. That is 98% of Vmax while using only 98 of the 100 substrate units available, which is exactly why saturated assays are run at high substrate: the rate becomes insensitive to the substrate concentration you measured.
Now solve it backwards from data instead. At [S] = 1 the measured rate is 33.33, and at [S] = 4 it is 66.67. Those two points are consistent with a true Vmax of 100 and Km of 2, and the solver recovers both. If your own data does not give positive values for Vmax and Km, either a measurement is wrong or the two points are not compatible with any Michaelis–Menten enzyme.
What Km does and does not tell you
A low Km means the enzyme reaches half-maximum rate at a low substrate concentration, which normally means tight binding and high apparent affinity. A high Km means you need a lot of substrate to get going.
What it is not is a binding constant, and equating Km with Kd is one of the most persistent errors in the topic. Km is an empirical shape parameter. It equals Kd only under restrictive conditions: a single substrate, rapid equilibrium, and no cooperativity. Under those conditions the simplified scheme gives Km = (k−1 + kcat)/k1, so it includes the catalytic step as well as binding. For anything approaching physiological conditions, treat Km as a description of the curve rather than a molecular measurement.
Assumptions this model makes
- One substrate, one binding site, one product. Multi-substrate enzymes need a different treatment.
- The enzyme–substrate complex concentration is steady, not a transient.
- No cooperativity. Allosteric enzymes do not follow this curve, and sigmoid kinetics are the clue.
- Free substrate concentration is approximately equal to total, so the substrate is not being depleted as the reaction runs.
- Product does not inhibit, or inhibition is being ignored.
The last one bites in practice. When product accumulation slows an enzyme, the Lineweaver–Burk pattern changes: a shared y intercept with a steeper slope for competitive inhibition, a shared x intercept for non-competitive, and lines crossing left of the y axis for uncompetitive. The model itself is unchanged; it is the inhibition that moves the lines.
Frequently Asked Questions
What does Km mean?
Km is the substrate concentration at which the rate is exactly half of Vmax. That is the definition, not a consequence, so [S] = Km always gives v = Vmax/2 for any enzyme.
How do I find Vmax from a Lineweaver-Burk plot?
Read the y intercept, which equals 1/Vmax, then take the reciprocal. The x intercept equals minus 1/Km, so Km is minus one divided by that intercept.
Is Km the binding constant?
No, and this is a persistent error. Km equals Kd only for a single substrate with rapid equilibrium and no cooperativity. Otherwise it includes the catalytic step as well as binding, so it is a description of the curve rather than a molecular measurement.
Why is Lineweaver-Burk criticised?
Taking a reciprocal of a small number amplifies its error, so the least precise measurements get the most visual weight. It also makes a gentle curve visibly bent, and that bend is where inhibitor information lives.
What does a high Km mean about an enzyme?
It takes a lot of substrate to reach half-maximum rate, so the enzyme has low apparent affinity for that substrate. A low Km means the opposite.
Why are assays usually run well above Km?
When the enzyme is saturated the rate stops depending much on the exact substrate concentration, so small errors in your measurement stop mattering. Below Km the rate rises steeply and substrate errors show up directly in the result.
How do inhibitors change the Lineweaver-Burk plot?
Competitive inhibition leaves the y intercept unchanged and steepens the slope. Non-competitive leaves the x intercept unchanged. Uncompetitive gives parallel lines that cross left of the y axis.