3. a. Enzyme X is a muscle enzyme that catalyzes the following reaction: A -> B

Question

3. a. Enzyme X is a muscle enzyme that catalyzes the following reaction: A -> B The turnover number (catalytic rate constant kcat) for the reaction is 50 sec-1 . How long does it take for one molecule of enzyme X to produce one molecule of B?

b. You grind up a muscle tissue sample and assay its rate with a concentration of A which is 4Km. A reaction rate of 8 µmole/min/ml muscle extract is obtained. What is the concentration of enzyme X in the extract? (Hint: start with the Michaelis-Menten equation and use the turnover number from part a above).

c. Enzyme Y is an isoform of X that catalyzes the same reaction (A B), but with different kinetics. Enzyme Y has a molecular weight of 40,000. Under zero order conditions, a 1 mg/ml solution of pure enzyme Y gives a velocity of 800 µmole B/min/ml. How many molecules of B are produced per molecule of Enzyme Y per min?

Any help is really appreciated! This one is tough.

Thank you

Explanation / Answer

3.(a): Given the turnover number, Kcat = 50 sec-1.

i.e the enzyme X produces 50 molecules of B in 1 second.

Hence time required by enzyme X to produce 1 molecule of B = 1/50 sec-1 = 0.02 sec (answer)

(b): Given subtrate, A concentration, [A] = 4Km

Reaction rate, V = 8 µmole/min/ml

Applying Michaelis-Menten equation

V = Vmax x [A] / (Km + [A])

=> 8 µmole/min/ml =  Vmax x 4Km / (Km + 4Km)

=>  8 µmole/min/ml = 4VmaxKm / 5Km

=> Vmax = 40 / 4 = 10 µmole/min/ml

Also given turnover number, Kcat = 50 sec-1 = (50 / sec) x (60 sec / 1 min) = 3000 min-1

=> Kcat = 3000 min-1 = Vmax / [E]T

=> 3000 min-1 = (10 µmole/min/ml) / [E]T

=> [E]T = (10 µmole/min/ml) / 3000 min-1 = 3.33x10-3 µmole / ml (answer)

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