You are a researcher with an interest in understanding the mechanism of serine protease enzymes. To test the effect of the role
of Asp102 in the efficiency of catalysis, you make targeted substitutions of that residue and compare the rate of hydrolysis of p-nitrophenyl acetate with the rate on the unmutated enzyme. Answer the following questions: a. (4) You observe a bi-phasic plot, i.e. a quick burst of product (p-nitrophenol) release (seen as a steep curve), followed by a slower (less steep) curve. What step of the reaction corresponds to each phase of the chymotrypsin reaction? And which corresponds to the Michaelis-Menten Steady State? Explain. b. (4) If you mutate Asp102 to create an enzyme with Asn102, describe (or draw) changes you would expect to see in p-nitrophenol release – and, of course, explain your answer. c. (4-8) For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the slower phase only) of 0.216 (in a cuvette with a pathlength of 1 cm) over the course of two minutes. Use the Beer-Lambert Law to calculate the concentration of product released over this time; the extinction coefficient of p-nitrophenol at 405 nm is 18,000 M-1 cm-1. For extra credit: If your assay is done with 1 x 10-9 M enzyme, what is the turnover number for this enzyme?
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Please find the complete question in the attached file.
Explanation:
It would only be radioactive if the DNA molecule that employed the poly-T rand as templates. Its other molecule of the daughter would not have been radioactive as it did not need dATP for its replication. While each strand of the second molecule includes t, simultaneous reproduction dATP from both daughter molecules is needed so that each of those is radioactive.
