Reducing the substrate concentration to 90mM from 180mM would result in very little change in the reaction rate.

            In the antithrombin III assay, thrombin catalyzes the reaction:

            For this substrate, thrombin has K_m=3.24 μM

The recommended substrate concentration is 55 times higher than the K_m (180μM/3.24μM), so the enzyme will be nearly saturated with substrate even if the substrate concentration is cut in half.

If you wish, you can use the Michaelis-Menten equation to come up with a quantitative answer to the question.

      Quantitative answer:
            V_i=Vmax[S]/(Km+[S]).
            At 180 μM substrate V_i is 98% of V_max.
            At 90μM substrate V_i is 96.5% of V_max.
            Consequently, the reduction in substrate concentration will only cause about a 2% decrease in V_i.