Suppose that Vmax for 1 mg/ml of a pure enzyme is 30 mM/min, i.e., Vmax is 30 (umols/ml)•min. The enzyme is a homodimer with 2 identical active sites, one active site per monomer. The dimer Mr is 60 kilodaltons , i.e., 60,000 g/mol. A. Calculate the turnover number. B. Calculate the average time required for one catalytic turnover (one "round" of catalysis in one

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Can you please help explain how to approach these two problems of calculating the turnover number and average time?
Suppose that Vmax for 1 mg/ml of a pure enzyme is 30 mM/min, i.e., Vmax is 30 (µmols/ml)•min!.
The enzyme is a homodimer with 2 identical active sites, one active site per monomer. The dimer Mr
is 60 kilodaltons , i.e., 60,000 g/mol.
A. Calculate the turnover number.
B. Calculate the average time required for one catalytic turnover (one "round" of catalysis in one
active site) in sec.
Transcribed Image Text:Suppose that Vmax for 1 mg/ml of a pure enzyme is 30 mM/min, i.e., Vmax is 30 (µmols/ml)•min!. The enzyme is a homodimer with 2 identical active sites, one active site per monomer. The dimer Mr is 60 kilodaltons , i.e., 60,000 g/mol. A. Calculate the turnover number. B. Calculate the average time required for one catalytic turnover (one "round" of catalysis in one active site) in sec.
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