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

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

V = Vmax[S]/Km+[S]

a) At the end of 4 min

ln[A]=ln[A]o-kt

ln[A]=ln(2x10^-5)-2.5x10^-3x4

[A]=1.98x10^-5M

percent of pNPP converted to pNP = 1.99x10^-5M

b) for a first order reaction,

the product concentration after 4 min =1.99x10^-7M

c)  Km = 50%Vmax[S]

2.5x10^-3x2 = Vmax, Vmax = 5x10^-3

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