Question

vo " No effect or a slight increase B. A decrease in the an C. An increase in the rate D. An increase in the rate of E. A complete loss of enzyme crease in the affinity for substrate coup e in the rate of mentide hond cleavage are ne rate of peptide-bond cleavage due to an incre of enzyme activity due to the inability to bind substrate rmation below. of test tubes, es the enzyme in each In questions 16-19, calcula You wish to characterize a new M enzyme and various substrate calculate the indicated parameters using the information bel e a new enzyme that you have isolated. You set up a series of test to 15 substrate concentrations. You measure the activity of the enzyme i me activity (V) vs. ISL The plot is shown below VouM/min) 4 6 10 12 14 8 [S] (mm) 16. What is meg for your enzyme with this substrate? A. 10 m/min D. 0.05 mm/min Β. 30 μM/min E. 50 mm/min C. 0.8 m/min 17. What is the of your enzyme for this substrate? A. 1.5 M D. 1.0 mm B. 10 M E. 1.5 mm C. 0.5 mm 18. What is the cat (turnover number) for your enzyme? A. 5 X 10°/s D. 5 X 10 /min B. 1.0/min E. 1 X 10°/s C5 x 10°/min 19. Why does your plot eventually plateau? The enzyme becomes inhibited at high substrate concentrations B. The enzyme affinity for substrate changes. C. The enzyme is being degraded by proteases D. The active site is saturated with substrate E. The substrate is being degraded.

Answer 1

The Michaelis-Menten Equation is shown below: d[P] Vmax [S] v= dt KM + [S] According to the above equation, on plotting a graph between Ve and concentration of substrate, the maximum rate is Vmax. According to the given graph, the Vmax is 50uM/min. The conversion of 50uM/min to mM/min is shown below: lumol/min = 0.001 mmol/min 50umol/min = 50 <0.001 mmol/min = 0.05 mmol/min Therefore, the correct option is (D).