Note: the word cut off on number 3a is “enzyme.” 2) Data for an enzyme mediated

Question

Note: the word cut off on number 3a is “enzyme.” 2) Data for an enzyme mediated reaction were measured for several different substrate concentrations. ISL (umol/ (umol/L-min) 10 20 50 100 200 39 65 102 120 135 a) Estimate Vmax and Km from a Michaelis-Menten graph of v versus [S]. b) Now, use a Lineweaver-Burk plot to analyze the same data. Determine the same values from this plot. 3a) If the total enzyme concentration above was 1 nmol/L, how many molecules of substrate can each e molecule turnover each minute? b) Calculate the keat/Km value for this enzyme reaction.

Explanation / Answer

(2) km is the substrate concentration at which velocity is half of its maximum.

so, vmax =135 micromol/l-min; km = 20 micromol/l

(3) (a) ETOTAL = 1 nmol/l = 10-3 micromol/l

kcat or turnover no. is the no. of substrate molecules converted to product in a given unit of time on a single enzyme molecule when the enzyme is saturated with substrate.

vmax = kcat ETOTAL

135 = kcat * 10-3 so kcat = 1.35 * 105 per min

(3) (b) so catalytic efficiency kcat/km = (1.35* 105 per min) / (20 micromol/l) = 6750

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

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