hi guys, please help me with my homework An enzyme ATPase has a molecular weight

Question

hi guys, please help me with my homework

An enzyme ATPase has a molecular weight of 5 times 10^4 daltons, a K_M value of 10^-4 M, and a k_2 value of k_1 = 10^4 molecules ATP/min molecule enzyme at 37 degree C. The reaction catalyzed is the following: which can also be represented as where S is ATP. The enzyme at this temperature is unstable. The enzyme inactivation kinetics are first order: E = E_0 e^-k_c where E_0 is the initial enzyme concentration and k_d = 0.1 min^-1. In an experiment with a partially pure enzyme preparation, 10 mu g of total crude protein (containing enzyme) is added to a 1 ml reaction mixture containing 0.02 M ATP and incubated at 37 degree C. After 12 hours the reaction ends (i.e., t rightarrow infinity) and the inorganic phosphate (P_i) concentration is found to be 0.002 M, which was initially zero. What fraction of the crude protein preparation was the enzyme?

Explanation / Answer

The reaction rate is represented by

d(P) / dt = k2[E]

[E] = [d(P) / dt] x k2

Initial ATP concentration = 0.02 M

Concentration of inorganic phosphate after 12 hours = 0.002 M

k2 = 10-4 M

Substituting these values in the equation,

[E] = [d(P) / dt] x k2

= [(0.02 - 0.002) / 12] x 10-4

= (0.018/12) x 10-4

= 0.0015 x 10-4

= 0.15 x 10-6

= 0.15 micrograms

Enzyme concentration is 0.15 micrograms

Crude protein is 10 micrograms

Therefore, the fraction of enzyme in crude protein is,

= (0.15/10)

= 0.015 (1.5%)

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