12. (20 pts.) A sensitive assay for ATP is based on its participation in the lig
ID: 636577 • Letter: 1
Question
12. (20 pts.) A sensitive assay for ATP is based on its participation in the light-producing reaction of the firefly. The reaction catalyzed by the enzyme luciferase is as follows HO Luciferase ATP2 Mg PP+AMPCO2+ Light Oxvluciferin When the reactants are mixed, the solution gives off light in the form of chemiluminescence. The light intensity decays slowly with time due to product inhibition of the reaction. Otherwise, the light would have steady intensity because the rate at which reactants are consumed is negligible. That is, ATP and luciferin maintain their original concentrations throughout the few minutes the reaction is monitored. A sensitive assay for ATP is based on its participation in the light- producing reaction of the firefly. Some typical results are shown below. Initial intensity s+X 5 min Time Let the initial concentration of ATP in the reaction be denoted [S]. Suppose that additional ATP is added, increasing the concentration in the reaction to [S1+ The kinetic description of the reaction predicts that the increase in light intensity after the addition will be given by, K[S] where K is constant. Suppose that [S]- 250. uM and after 5 minutes Is -58.7 arbitrary intensity units. Then a standard addition of [X]- 200. uM is made, and s+x74.5 (a) Find a value for K. (b) When the intensity of this mixture had decayed to 63.5 units, an unknown aliquot of ATP was added to the reaction and the intensity increased to 74.6 units. Determine the concentration of ATP in the unknown aliquotExplanation / Answer
By putting the values in the formula given in the question:
58.7/74.5= (1/250+200){K(250)/K+250}+250/K+250
0.787= 0.0022(250K/K+250)+250/K+250
0.78= (0.55K/K+250)+250/K+250
0.787= 0.55K+K/K+250
0.787K+19.67 = 1.55K
0.77K = 19.67
K = 25.54
2. In this question: [S] = ?
IS+X= 74.6
IS = 63.5
K is constant = 25.54 from above solution
By putting values in above equation
63.5/74.6 = {1/[S]+200(25.54/25.54+[S]}+[S]/25.54+[S]
0.851 = {25.54/S2+25.54S+5108+200S}+S/25.54+S
0.851-(S/25.54+S) ={25.54/S2+25.54S+5108+200S}
(21.73+0.851S-S)/S = 25.54/ S2+225.54S+5108
(21.73-0.149S) S2+225.54S+5108= 25.54S
From this equation we can find out the value S
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