4. The graph depicts the reaction of CENTA (substrate) turnover by two different

Question

4. The graph depicts the reaction of CENTA (substrate) turnover by two different B-lactamase enzymes. The final concentration for ADC-7 is 0.7 uM and for AmpC is 0.35 ?M. The Michaelis- Menten plot of the assays is given below ADC-7 Vmax 0.7995 Km 0.09972 0.6 0.4 Vmax-0.4008 Km 0.19821 5 0.2 AmpC 0.0 0.0 0.2 [CENTA], in mM 0.4 0.6 a. Calculate the kcat for each ß-lactamase. Be sure to show your work. (3 pts) ADC-7 Kcat AmpC kcat b. Which enzyme do you think is more truly efficient at turning over the substrate CENTA? Explain your reasoning. (2 pts) c. Determine the catalytic efficiency for the ADC-7 B-lactamase. Be sure to show your work. (3 pts)

Explanation / Answer

Ans. #a. ADC-7 Kcat = Vmax / [E]

                                    = 0.7995 uM s-1 / 0.7 uM

                                    = 1.1421 s-1

# AmpC Kcat = Vmax / [E] = 0.4008 uM s-1 / 0.35 uM = 1.1451 s-1

#b. When determining the relative efficiency of the two enzymes towards same substrate, the term “catalytic efficiency” is preferred. Greater is the v catalytic efficiency of an enzyme, the more efficient is the enzyme towards catalysis of the substrate.

Catalytic efficiency = Kcat / Km

# ADC-7 Catalytic efficiency = 1.1421 s-1 / 0.09972 mM = 11.4531 mM-1 s-1

# AmpC Catalytic efficiency = 1.1451 s-1 / 0.19821 mM = 5.7772 mM-1 s-1

Conclusion: Since catalytic efficiency of ADC-7 is highest among the given list, it is the most efficient enzyme towards catalysis of CENTA.

#c. Catalytic efficiency is rate of catalysis of a substrate per unit time to unit concertation of enzyme under specified reaction conditions.

            Catalytic efficiency = Kcat / Km

# ADC-7 Catalytic efficiency = 1.1421 s-1 / 0.09972 mM = 11.4531 mM-1 s-1

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