Question

show work please!

with inhibitor (umoles/min/mg) [S] (mm) v- no inhibitor (umoles/min/mg) V - with inhibitor (umoles 3.0 2.29 x 103 1.83 x 103 5.0 or ble 3.20 x 103 Boristeder s obre o 2.56 x 103 7.0 3.86 x 103 USD 3.09 x 103 0 9. 4 .36 x 100W 3.49 x 103 0 11. 4 .75 x 108 3.80 x 103 Draw Lineweaver-Burk plots for the enzyme data shown above. When present, I = 0.200 m a) What are Km and Vmax for (i) the uninhibited enzyme and (ii) the inhibited reaction b) What is the inhibition type? c) What is K; for the inhibitor?

Answer 1

Okay, this is an enzyme kinetics exercise. You already have the enzyme's activity values, the speed in which the enzyme can work with and without an inhibition. Now, the lineweaver-burk plot is the inverse curve, actually you have to invert every value to obtaine the values for our lineweaver-burk plot, for example, the first value (2.29x103) would be obtained like this:

2.29x1023 - = 4.36x10л – 4

And so on. Have in mind you have to make the same calculations for the substrate cocnentration values (your independent variable). The values then are:

1/S	1/V no in	1/V with in
0.33333333	0.00043668	0.00054645
0.2	0.0003125	0.00039063
0.14285714	0.00025907	0.00032362
0.11111111	0.00022936	0.00028653
0.09090909	0.00021053	0.00026316

Now we plot such values with the S inverted in the X axis:

Lineweaver-burk plot 0.0006 0.0005 0.0004 0.0003 0.0002 0.0001 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 IN 1/V noin 1/V with in

The values are not enough to reach and cross both Y and X axis, but this is the basis to calculate such intersections. You have to indicate your program (Excel is enough) to draw a trendline (we'll answer question b with this) and give you the line equation for each line (will answer question a with this).

Lineweaver-burk plot 0.0006 0.0005 y=0.0012x + 0.0002 0.0004 y = 0.0009x+0.0001 0.0003 0.0002 s/t 0.0001 -0.3 -09 - 0.1 0.2 - 0.3 0.4 -0.1-0.0001 0 -0.0002 1N 1/V no in ...... Lineal (1/V no in) — 1/V with in ....... Lineal (1/V with in)

a) What are Km and Vmax for the uninhibited enzyme and the inhibited reaction?

Km is the value that intersects the X axis in the plot we made, while Vmax is the value that intersects Y axis. We will calculate them with the equation provided in the plot above

- Uninhibited Vmax: Take the equation y=0.0009x + 0.0001, Vmax is the y value where the x value is 0, so:

y = 0.0009(0) + 0.0001 = 0.0001

- Uninhibited Km: Take the equation y=0.0009x + 0.0001, Km is the x value where the y value is 0, so:

x = (y - 0.0001)/0.0009 = (0-0.0001)/0.0009 = -0.1111

- Inhibited reaction Vmax: Take the other equation y=0.0012x + 0.0002, Vmax is the y value where the x value is 0, so:

y=0.0012(0) + 0.0002 = 0.0002

- Inhibited reaction Km: Take the other equation y=0.0012x + 0.0002, Km is the x value where the y value is 0, so:

x = (y - 0.0002)/0.0012 = (0-0.0002)/0.0012 = -0.1666

b) What is the inhibition type?

This can be seen in the plot, it is indicated by the point in which both lines intersect. If they intersect in the Y axis then it is a Competitive inhibition, when it occurs in the X axis the it's Noncompetitive, and if they never intersect it is uncompetitive.

In this case, even when the intersection in X axis is not exact, the inhibition is Noncompetitive

c) What is Ki for the inhibitor?

Ki is the inhibitor concentration needed to produce half the maximum inhibition. But to calculate such value we need to measure the effect of such inhibitor on different concentrations, we have only values for a single inhibitor concentration (0.2 mM)