Question

3. Acid Phosphatase was assayed at [S] =2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% 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 at the end of 3.5 minutes assuming product formation is in 1st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5 minutes? (3 points) c) Determine maximal velocity (Vmax) with the enzyme concentration used. (4 points)

Answer 1

For any reate Co enzyme - substrate reeaction . K2 [€]. It Km km Michaelis - Menten constant [6] concentreation of substrale [E]. concentreation of enzyme S at initial K₂ » Rate constant of the . reaction Here .kmz 5810 4 M [s] = 2x105 m ' 5X10" 4 :: It 1+25=25*25 Ko CE . . Rate =v=- [7.17 Km Rate will be Ist onder with respect to substrate concentration so. k= ť engan where a t Initial conc. of relactant

(a-n cone. of recaetant at timet where l amount is dissociated from ca? At the end of 1 min K = 1 / 2 en 1 o 25) = 1 luogo min =0.0253 a) At the end of 315 min . K = is en aan or 0.0253= 1/35 en an or 0:088 55 = lu a can com a mesma = 1.0925 on and on = 0.915 on (a-x)= a + (0-915-1) on. - - -0.085

on. Th=0.085 :.4085X100)= 8:57. of pNpp will be converted to product b) : Primary concent reation of substreate [5]= 2x10-5 M . Its . 8.5% is convented to product. sooo at 3.5 min. so the rest iid. (100~8.5)). - 9159. of substreate is still remained. - ... Product concenieation at 3.5 min react ant concentration X 8.5% = 2X105 x 85 M. 100 = (1.780095) M - () Umax = kq [€].

NEW At initial [E70= [5] . .. Nmax = 10.0253* 2X105) M.ml 51067107 Minin