The enzyme dragonase was assayed at increasing substrate concentrations with and

Question

The enzyme dragonase was assayed at increasing substrate concentrations with and without a reversible inhibitor. The following shows the Lineweaver-Burk plots of the results: i. Calculate Km and Vmax in the absence of inhibitor and in the presence of 50 nM inhibitor. Show your work ii. You know this is a reversible inhibitor and it is a competitive, mixed or uncompetitive inhibitor. What type of inhibitor is this enzyme based on your results? Explain. iii. In each assay there was 1.5 times 10^-6 u mole of enzyme. What is the turnover number for the enzyme in the absence of inhibitor? iv. What is the catalytic efficiency of that reaction in the absence of inhibitor?

Explanation / Answer

1 - Lineweaver-Burk equation is given by

1/Vo = (Km/Vmax) 1/S+ 1/Vmax

Comparing this equation with Y=aX + c

where a is the slope and c is the intercept

I/Vmax = intercept

so V max = 1/intercept

For no-inhibited reaction =

1/2.032 = 0.492 micromol/min

For 50nm inhibitor =

1/2.6649 = 0.375 micromol/min

Km/Vmax is equal to slope

Km = Vmax * slope

For no-inhibitor enzyme

Km = 0.492 * 0.9997= 0.491 micromol

For 50nm inhibitor

0.375 * 0.9474

= 0.355 micromol

2-

From the above calculation we can see that Km and Vmax are decreased in case of 50nM inhibitor.

If we compare the decrease in Km and Vmax with the given types of inhibitor we will find that the inhibition is uncompetitive inhibition.

3- Given that the initial enzyme concentration is 1.5 * 10-4 micromol

Turnover is given by the

Kcat = Vmax/ Enzyme concentration

= 0.492 / 1.5 * 10-4

0.328 * 104

3.28 * 103 min-1

4- Catalytic efficiency is given by

Kcat/ Km

= 3.28 * 103 / 0.491

6.68 * 103 micromol-1 min-1

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.