[vande060]

the turnover number for an enzyme is know to be 5000/min. from the following set of data, calculate km and the total amount of enzyme present in these experiments

(these two columns of numbers should be right next to each other, to that reading horizontally you can see both mM [s*] and initial velocity for a given experiment, but it is difficult to do, so i will place them one after the other.)

[s*] = concentration s but the computer thinks that is a strikeout so i needed the asterisk

substrate concentration(mM)
1
2
4
6
100
1000

Initial velocity (micromol/min)

167
250
334
376
498
499

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I know the turnover number is kcat and i thought it safe to call 499 vmax

then i used the formula kcat = vmax/[E]t

5000/min = (499micromol/min)/[E]t

[E]t = .0998 micromol

for km i was a little confused,

i want to use the formula v= (kcat/km)([E]t[S*]  and solve for km

but i will get a different number for each of the reactions, and that doesnt seem right.