[vande060]

my prof asks me

the following data were obtained in a study of an enzyme known to follow michaelis menton kinetics

Vo (mmol/min)
217
325
433
488
647



substrate added(mm/l)
.8
2
4
6
1000


find the approximate km for this enzyme regarding this substrate
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the formulas for michaelis menton are variable, and i dont know what to use. the general formula is given here http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics

i did some searching in my book.
and found that michaelis menton kinetics has a practical rule km = [s*] when v0 = 1/2vmax, but i dont know how to use this

if it follows this wouldnt km just be [s*] as v0 is 1/2 vmax? i suppose for that i would need to know what vmax is. 
 
the astrixs are next to the s because the text interprets it without the astrix to be crossed out text


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edit: i think i got it, for an approximation it is probably safe to assume that v0=647 is close to vmax because at a huge concentration of 1000 substrates added, the v0 changed very little from when it was just 6 substrates added. 1/2 of 647 is approx 325, so km is equal to the concentration of substrate at v0 =325, which is 2.

does it sound right

[SheVa]

you can use lineweaver-burk method for analyzing michaelis-menten equation.
1/v = Km/Vmax (1/[ S]) + 1/Vmax (reciprocal of michaelis-menten eqn).

then just make a regretion from your data with that lineweaver-burk.
y = 1/v

m = Km/Vmax
x = 1/[ S]
c = 1/vmax

see..you can get that Km from there..its not well enough but good approximation.