Question

Q2) Using the data in the below table, which were obtained for enzyme activity without presence of inhibitors. [15 points) IS (MM) 1.3 2.6 6.5 13.0 26.0 VomM/s) 2.50 4.00 6.30 7.60 9.00 A) Draw Lineweaver-Burk plot and determine the K. and Vmax for this uninhibited enzyme. (You may use hand, excel or other program to draw the graph and find the equation. Then copy and paste the graph here, you need to show axis labels) [6 points) B) From the graph or the slope and line equation, find the values for both Vmax and K. [5 points) Vmax Km: C) Draw two extra lines on the chart to represent the enzyme activity in present of competitive (label it as Com.) and uncompetitive (label it as UnCom.) inhibitors.[4 points)

Answer 1

. Cla Book1 - Microsoft Excel Chart Tools Home Insert Page Layout Formulas Data Review View Design Layout Format X Calibri (Bodyl 10 Α Α΄ Wrap Text General 27 Paste * Cut En Copy Format Painter Clipboard BI U a Merge & Center $ % 4.0 .00 .00 0 Insert Delete Format Conditional Format Cell Formatting as Table Styles Styles Σ AutoSum Fill Sort & Find & 2 Clear - Filter - Select Editing Font Alignment Number Cells Chart 1 fix A B E F G H J - K L M N o P Q R S T U 1 Vo S 2 2.5 3 4 C D 1/Vo 1/S 1.3 0.4 0.769231 2.6 0.25 0.384615 6.5 0.15873 0.153846 13 0.131579 0.076923 26 0.111111 0.038462 1/Vo vs 1/S 4 0.45 6.3 7.6 5 y = 0.392x + 0.098 6 9 Chart Area 7 0.35 8 9 10 0.25 11 1/Vo 12 13 0.15 14 1/Vo vs 1/S - Linear (1/Vo vs 1/S) 15 16 0.05 17 18 19 -0.15 -0.05 0.05 0.15 0.25 0.35 0.45 0.55 0.65 0.75 0.85 20 1/S 21 22 -0.15 23 24 25 Sheet1 Sheet2 Sheet3 Ready BO 100% + Type here to search © EL PI 2:45 PM 7/27/2020 E

B) From Lineweaver-Burk plot 1 SA (km )* Ś S Vasan Vasar y intercept = 1 Vimax Gradient km Vasaz From grapb, y = 0.3924 +0.098 Yintercept = 0.098 Vimax V imax 10.204 0.098 Gradient tim = 0.392 Vomar koma = 0.392 X V -0.392X10.204 2 =4

1/Vo vs 1/5 0.45 0.35 1/VO Non-comp Comp 0.25 - 1/Vo vs 1/5 0.2 0.15 0.1 0.05 -1/km 0 -0.25 -0.15 -0.05 0.05 0.15 1/5 0.25 0.35 0.45