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.-Zs 0.7 JM and for AmpCis Menten plot of the assays is given below M. The Michaelis- ADC-7 Vmax 0.7995 Km 0.09972 0.6 0.4 vmax-O 4008 Km 0.19821 0.7 0.2 AmpC 0.0 0.0 0.2 0.4 0.6 [CENTA], in mM a. Calculate the keat for each B-lactamase. Be sure to show your work. (3 pts) ADC-7 koat5 1945 art AmpC Keat .1462 b. Whi ch enzyme do you think is more truly efficient at turning over the substrate CENTA? Explain your reasoning. (2 pts) c. Determine the catalyticeffiency for the ADC-7 P-actamase. Be sure to show your work. (3 pts)

Explanation / Answer

(a)GIVEN, for ADC-7 [Et]=0.7um, vmax=0.7995 um/s, km=0.09972

for AMP-C [Et]=0.35um, vmax=0.4008 um/s, km=0.19821

we know that, vmax = kcat * [Et]

where vmax is maximum rate of reaction, & [Et] is total enzyme concentration

so kcat for ADC-7 = vmax/[Et] = 0.7995 / 0.7 = 1.14214 per second

so kcat for AmpC = vmax/[Et] = 0.4008 / 0.35 = 1.14514 per second

(b) as Ampc have high kcat value for substrate centa. and kcat can be defined as number of substrate molecules converted to product per unit time so Ampc is more efficient at turning over the substrate centa.

(c) catalytic efficiency for ADC-7 b-lactamase = kcat / km

= 1.14214 / 0.09972

=11.45346

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