Question #43225

1 Answer
Apr 10, 2015

The initial velocity for your enzyme will be "0.417 nmol L"^(-1)"s"^(-1).

Once again, use the MIchaelis-Menten equation to solve for V_0. Keep in mind, however, that you must convert mu"mol L"^(-1)"s"^(-1), the units used for V_"max", to "nmol L"^(-1)"s"^(-1), the units required for V_0.

500cancel(mu"mol")"L"^(-1)"s"^(-1) * (10^(-3)"nmol")/(1cancel(mu"mol")) = 500 * 10^(-3)"nmol L"^(-1)"s"^(-1)

So, the initial velocity will be equal to

V_0 = (V_"max" * [S])/(K_m + [S])

V_0 = (500 * 10^(-3)"nmol L"^(-1)"s"^(-1) * 5.00cancel("mmol L"^(-1)))/((1.00 + 5.00)cancel("mmol L"^(-1))

V_0 = 5/6 * 500 * 10^(-3)"nmol L"^(-1)"s"^(-1) = "0.417 nmol L"^(-1)"s"^(-1)

SIDE NOTE Since V_"max" is given with one sig fig, while all other variables are given with three, I'll leave the answer rounded to three sig figs.

If, instead of 500 you'd have 500. for V_"max", then the answer should indeed have three sig figs; if not, it should only have one.