Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate = 2.5x1

Question

Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate = 2.5x10-3 M. 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: (12 points)

a) What percent of pNPP will be converted to pNP at the end of 4 minutes assuming product formation is in 1st order of reaction?

b) What will be a product concentration at the end of 4 minutes?

c) Determine maximal velocity (Vmax) with the enzyme concentration used.

Explanation / Answer

             for fisr order reaction V= Vmax[S]/KM

                       -dS/dt= K'[S]     K' = Vmax/KM

when integrated gives lnS= lnSO- K't,SO= initial substrate concentration and S= Substrate concentration at time , t

at 1 minute   1-S/SO= 2.2/100

S/SO= 1-2.2/100 =0.978

S= 0.978*SO= 0.978*2*10-5=0.00001956M

from lnS= lnSO- K't

ln(0.00001956)= ln(2*10-5)-K'*1

-10.84= -10.82-k'*1

K'= -10.84+10.82= 0.02/min

at 4 min

lnS= ln(2*10-5)- 0.02*4

S= 1.846*10-5

Product formed = amount of S reacted= 2*10-5- 1.846*10-5 =1.54*10-6M

K' = Vmax/KM

0.02= Vmax/2.5*10-3

Vmax= 0.02*2.5*10-3 M/mmin=0.00005 M/min

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