NOTE: for actual data analysis if values are ±5 no change has occurred. 1. An en
ID: 507146 • Letter: N
Question
NOTE: for actual data analysis if values are ±5 no change has occurred. 1. An enzyme was found that catalyzes the reaction between students, S and academic success, P (for pass); the enzyme was called studyase. An enzyme assay was run on studyase. 100g of studyase (molecular weight 20,000D) was placed in a set of test tubes each of which contained 1mL of the substrate student at varying concentrations and the amount of product, academic success (P) was determined. The following results were determined: S, (mM) Velocity, V (mole/mL/sec) 2.50 30.0 5.00 50.0 10.0 70.0 20.0 85.0 30.0 100. Note: 1mM = 1millimole/L = 1mol/mL a. Plot a Lineweaver-Burk plot of 1/V versus 1/S b. Determine Vmax and Km for the enzyme studyase. c. Calculate Kcat and the catalytic constant (kcat/km) for the enzyme. (find [E]T as total moles/mL) d. Another enzyme called partyase was analyzed and when 0.050 moles of this enzyme was assayed it was found to have a Vmax of about 124 mol/mL/sec and a Km of 10 mM. Calculate Kcat and the catalytic constant for this enzyme. i. Compare the 2 enzymes on the basis of affinity for the binding site and kinetic efficiency.
Explanation / Answer
1/V= (KM/Vmax)*1/S +1/Vmax
so a plot of 1/V vs 1/S gives a straight line whose intercept is 1/Vmax and slope is KM/Vmax. The plot along with data points is shown below
from the p;lot 1/Vmax= 0.008, Vmax= 1/0.008=125 umol/L.sec
KM/Vmax= 0.062, KM= 0.062*125 mM=7.8 mM
Kcat= Vmax/ET=
ET= concentration of enzyme= 100*10-6/ 20000/1 = 5*10-9 gmole/L
Vmax= 125*10-9 moles/L.sec
Kcat= 125*10-9/(5*10-9)= 25/sec
for the second enzyme, KCat= 124*10-9/0.05*10-9 = 2480/sec
Since the turn over number for second enzyme is more, it is more active
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