1/Vmax
Summary
1/Vmax is the reciprocal of an enzyme's maximum velocity. It is the y-intercept of a Lineweaver-Burk plot (1/v₀ versus 1/[S]). A higher intercept means a lower Vmax, so competitive inhibition leaves the intercept unchanged and other inhibition types raise it.
Detail
Meaning: 1/Vmax is the reciprocal of the maximum reaction velocity. Vmax is reached when the enzyme is saturated with substrate, and it depends on the total enzyme concentration and kcat. On a Lineweaver-Burk (double-reciprocal) plot, which graphs 1/v₀ against 1/[S], the line crosses the y-axis at 1/Vmax.
Equation: 1/v₀ = (Km/Vmax)(1/[S]) + 1/Vmax, where v₀ is initial velocity (mol/(L·s)), [S] is substrate concentration (mol/L), Km is the Michaelis constant (mol/L), and Vmax is the maximum velocity (mol/(L·s)). The y-intercept is 1/Vmax (L·s/mol), the slope is Km/Vmax, and the x-intercept is −1/Km.
On the MCAT: A passage shows Lineweaver-Burk lines with and without an inhibitor, and you compare intercepts. Lines that cross at the same y-intercept mean Vmax is unchanged, which is competitive inhibition. In that case the x-intercept moves toward zero because Km rises. A higher y-intercept means a lower Vmax. Pure noncompetitive inhibition keeps the same x-intercept. Uncompetitive inhibition gives parallel lines. Mixed inhibition changes both intercepts.
Common confusion: Students read a higher intercept as a higher Vmax. The axis shows 1/Vmax, so a higher point means a slower maximum rate. Take the reciprocal of the intercept to get Vmax. Also remember that the x-intercept is −1/Km, not +1/Km.
Sources
- Lehninger Principles of Biochemistry
- Biochemistry (Berg, Tymoczko, Stryer)
Written for the MCAT and checked before it was published. For study reference only. Spotted an error? Let us know.