Acid Phosphatase was assayed at [S] = 2x10-5M. The KM for the substrate = 2.5x10

Question

Acid Phosphatase was assayed at [S] = 2x10-5M. The KM for the substrate = 2.5x10-3M. At the end of one minute 2.2% of the substrate.pNPP, had been converted to pNP (a product). Answer the following questions using this information: a.) What percent of pNPP will be converted to pNP at the end of 4minutes assuming product formation is in 1st order of reaction? b.) What will be the product concentration at the end of 4minutes? c.) Determine maximal velocity (Vmax) with the enzyme concentration used.

Explanation / Answer

) for a first order reaction,

ln[S] = ln[S0] - Kmt

with,

[So] = 2 x 10^-5

Km = 2.5 x 10^-3 M

t = 4 min

we get,

ln[S] = ln(2 x 10^-5) - 2.5 x 10^-3 x 4

[S] = 1.98 x 10^-5 M

So the percent of pNPP converted to pNP = (2 x 10^-5 - 1.98 x 10^-5) x 100 = 2 x 10^-5%

b) Product concentration at the end of 4 min = 2 x 10^-7 M

c) Vmax = 2 x 2.5 x 10^-5 = 5 x 10^-5 M

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