1. In lab last year, biochemistry students studied wheat germ acid phosphatase,

Question

1. In lab last year, biochemistry students studied wheat germ acid phosphatase, which has the E.C. number 3.1.3.2. This enzyme has recently been purified from Arabidopsis thaliana and extensively characterized. Here are some of the important findings when studied at pH = 4.50 and T = 37.0° C: KM = 0.703 mM; turnover number (kcat) = 161.9 sec-1; molecular weight = 54,200 Daltons. Answer the following questions using these values.

a. Given the units of the turnover number, is this an apparent first or second order reaction under these conditions? (1 pt)

b. What is the catalytic efficiency of this enzyme (2 pts)?

c. Given your answer in “b,” is this enzyme near the diffusion-controlled limit? (1 pt)

d. What is the value of G for the enzyme-catalyzed reaction at 37.0° C? (Please use a transmission coefficient value of 0.75.) (4 pts)

e. The uncatalyzed rate constant (kuncat) for the hydrolysis of phosphate monoesters has been reported to be 2.0 × 10-20 sec-1! If this is accurate, what is the rate enhancement provided by the acid phosphatase? (2 pts) f. Given this value of rate enhancement, what is the value of Gcat? (2 pts) g. What is the value of Guncat? (2 pt)

h. In lab, students measured a wheat germ acid phosphatase maximum velocity (Vmax) in the range of 2.25 moles of substrate hydrolyzed per hour in a 5.0 mL reaction mixture. What would this value be with the units of mM/sec? (2 pts)

i. Using the Vmax value from “h” with the units of mM/sec, what would be the value of kcat? Assume the protein added as enzyme was at the concentration of 0.0133 mg/mL in the reaction mixture (which it was), but was only 0.20% acid phosphatase (which is a crude, but literature-supported estimate). (Hint: Assume wheat germ acid phosphatase has the same molecular weight as the Arabidopsis acid phosphatase.) How does this value compare to the value of kcat of the Arabidopsis enzyme? (4 pts)

Explanation / Answer

(a): Since the unit of turnover number is sec-1, this is a first order reaction under thase conditions.

(b): The formulae for catalytic efficiency or kinetic efficiency is

catalytic efficiency = Kcat / Km = 161.9 sec-1 / 0.703 mM = 230.3 mM-1sec-1 or 2.303x105 M-1.sec-1 (answer)

(c): The upper limit for diffusion controlled reaction is around 108 - 109 M-1sec-1. Since the calculated amount (2.303x105 M-1.sec-1 ) is much smaller than 108 - 109 M-1sec-1,  this enzyme is not near the diffusion-controlled limit (answer)

(d):

(e): Rate enhancement provided by the acid phosphatase

= Kcat / Kuncat = 161.9 sec-1 / 2.0x10-20 sec-1 = 8.095x1021 (answer)

(h): Given moles of subtrate = 2.25 umol = 2.25x10-3 mmol

Volume = 5.0 mL = 5.0x10-3 L

Hence concentration change of the subtrate = moles of subtrate / volume = 2.25x10-3 mmol / 5.0x10-3 L

= 0.45 mM

total time, t = 1 hr = 1hr x (60 min/1hr)x(60 sec/1 min) = 3600 sec

Hence Vmax = 0.45 mM / 3600 sec = 1.25x10-4 mM /sec (answer)

(i): concentration of acid phosphatase enzyme, [E]t =( 0.0133 mg/mL) x (0.2 / 100) = 2.66x10-5 mg / mL

= 2.66x10-2 mg / L

= 2.66x10-2 mg / 54200 g/mol / L

= 4.91x10-7 mmol/L = 4.91x10-7 mM

Hence

Kcat = Vmax / [E]t = 1.25x10-4 mM /sec / 4.91x10-7 mM [ From (h)]

=> Kcat = 255 sec-1 (answer)

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