Question

In: Chemistry

50. The table represents the rates of reaction at specific substrate concentrations for an enzyme that...

50. The table represents the rates of reaction at specific substrate concentrations for an enzyme that displays classical Michaelis-Menten kinetics. Two sets of inhibitor data are also included. Determine the Km and Vmax for the uninhibited enzyme.

[s](mM) Without inhibitor Vo (µM/s) With inhibitor A With Inhibitor B

1.3 2.50 1.17 0.62

2.6 4.00 2.10 1.42

6.5 6.30 4.00 2.65

13.0 7.60 5.70 3.12

26.0 9.00 7.20 3.58

Please Please Please....show detailed steps so I can repeat this with other problems. Need to be able to do without making a graph, thank you

Expert Solution

Ans. #Part 1. Instructions for plotting LB Plot:

(The process here in for MS Word 2016) -

1. Enter [S] and V values in excel sheet in separate rows. The unit of [S] has been changed from M to uM accordingly.

2. Generate 1/ [S] and 1/V values in excel sheet.

3. Select 1/[S] and 1/V columns and click on “Insert” tab right to ‘Home’ tab.

4. Select ‘scatter plot’ -displayed as few dots in the graph

5. Select trendline option

6. Add linear trendline and check the option for ‘trendline equation’.

#Part 2. Determination of Vmax and Km using LB Plot”

Lineweaver-Burk plot gives an equation in from of y = mx + c

where, y = 1/ Vo, x = 1/ [S],

Intercept, c = 1/ Vmax ,

Slope, m = Km/ Vmax

#a. Enzyme kinetics in absence of inhibitor [y = 0.3921x + 0.0987]:

Vmax = 1 / c = 1 / 0.0987 = 10.1317

Hence, Vmax = 10.1317 uM/s

Now,

Km= m x Vmax = 0.3921 x 10.1317 = 3.9726

Hence, Km= 3.9726 mM

Note: there is no need of unit conversion. Do calculations without using units. Put the respective units as mentioned in question after the final cacluation.

#b. Enzyme kinetics in presence of inhibitor A [y = 0.9802x + 0.1001]:

The Km and Vmax are calculated as in #a.

Vmax = 9.990 uM/s ; Km = 9.7927 mM

#c. Enzyme kinetics in presence of inhibitor B [y = 1.8246x + 0.1397]:

The Km and Vmax are calculated as in #a.

Vmax = 7.1582 uM/s ; Km = 13.0608 mM

# Part 3: Type of Inhibitor:

Inhibitor A: Vmax remains almost the same whereas Km is increased. It is the characteristic of competitive inhibitor A. Therefore, Inhibitor A is a competitive inhibitor.

Inhibitor B: Vmax is lowered but Km is increased in presence of inhibitor B. It is the characteristic of non-competitive inhibitor. Therefore, Inhibitor B is a non-competitive inhibitor.

Note: By seeing the values you say simply say if inhibition takes place or not. However, to determine the type of inhibitor, you must plot the graph (LB plot or others).

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