The three situation at A, B, and C on hyperbola curve can be verified from the M–M equation.

1. At point ASubstrate concentration is much less than K_m. If [S] is added, its value changes very little; hence [S] can be ignored from the denominator of M–M equation. V_o = V_max [S]/K_m + [S]
   V_o = V_max [S]/K_m as [S] is negligible.Since V_max and K_m are both constants, V_max/K_m can be replaced by a new constant K, so we have V_o = K[S]V_o therefore depends on [S] concentration
2. At point B[S] is equal to K_m, hence V_o = V_max [S]/K_m + [S]
   V_o = V_max [S]/[S] + [S] as K_m = [S] at B
   V_o = V_max [S]/2[S]
   V_o = V_max /2K_m, therefore, is the substrate concentration, at which V_max is half.
3. At point CThe substrate concentration is far greater than K_m, so by adding K_m to [S], the value of [S] does not change much; hence, K_m may be dropped. Then, V_o = V_max [S]/K_m + [S]
   V_o = V_max [S]/[S] as [S] >> > K_m
   V_o = V_maxWhen E is fully saturated with [S], the reaction velocity V will increase to the maximal velocity, V_max; that is, at this point, any further increase in the substrate concentration will have no effect on reaction velocity.