Question

The following data has been obtained for an enzyme-catalyzed reaction with two different initial enzyme concentrations of 0.025g/L and 0.015g/L. V (g/L.min) [S] (g/L) Eo = 0.025g/L Ep = 0.015g/L 56.8 1.75 1.05 28.4 1.58 0.97 19.1 1.47 0.89 14.3 1.38 0.83 11.4 1.31 0.77 9.4 1.22 0.73 8.2 1.15 0.69 7.1 1.08 0.66 (a) Using the Lineweaver-Burke plot, determine Km and Vm for Eo = 0.025g/L (b) Using the Lineweaver-Burke plot, determine Km and Vm for Ep = 0.015g/L (c) Are there any observed differences in Km and Vm values found in (a) and (b)? Explain why this is so.

Answer 1

S Enzyme 1 56.8 28.4 1 0.9 1/S 0.017606 0.035211 0.052356 0.06993 0.087719 0.106383 y = 2.7915x + 0.5279 R2 = 0.9975 19.1 14.3 V(Eo=0.025) (Eo=0.015 1/v 1/v 1.75 1.05 0.571429 0.952381 1.58 0.97 0.632911 1.030928 1.47 0.89 0.680272 1.123596 1.38 0.83 0.724638 1.204819 1.31 0.77 0.763359 1.298701 1.22 0.73 0.819672 1.369863 1.15 0.69 0.869565 1.449275 1.08 0.66 0.925926 1.515152 11.4 9.4 0.8 0.7 0.6 30.5 0.4 0.3 8.2 0.121951 7.1 0.140845 0.2 Enzyme 2 0.1 1.8 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 1.6 y = 4.6482x+0.8759 R2 = 0.9977 1.4 1/S 1.2 1 Axis Title 0.8 0.6 0.4 0.2

Equation Sol From Micheal- Menton V = Vmons kmts. km + I Vmos Vmot "Line- Burk Plot 15 -Weore plot b/w z usta & Intercept=t km Which gives Slope = Vmet Vma D for 6 = 0·0889/ R-009975: Y = 207916 X +0.5779 ㅗ & Km 207915 Vinot 20.5279 Timex 10 m = 5+2618 > | Vmox = 1.894 mull ☺ For Co= 0.015 g/L y = yebrou 406482 X:+ 0-8759 RP-0-99. Vmax = 1 a110141 mint Oog 759 km= 1014144-6484 15-306 g/L 15 is Km & Vm & this different both cases which because to is differenti in & hence Vm= WC. km also Changes which is concentichon where V = Vmax.

using given data and line weaver burk plot we can find the Km and Vm values

for any doubts write in comments