Question

 

Enzymes Kinetics:
The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of two

different inhibitors (2) and (3) at 5 mM concentration. Assume [Etotal] is the same in each experiment.

[S] (mM)	1 ν(μmol/mL. sec)	2 ν(μmol/mL. sec)	3 ν(μmol/mL. sec)
1	12	4.3	5.5
2	20	8	9
4	29	14	13
5	35	21	16
12	40	26	18

a. From a double-reciprocal plot of the data, Determine Vmax and Km.

b. Determine the type of inhibition that has occurred in 2 and 3.

NO GRAPHS FROM EXCEL PLEASE

Answer 1

SOLUTION:

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’.

#1 Determination of Vmax and Km using LB Plot”

Lineweaver-Burk plot gives an equation in from of Y = m X + c   

where,

y-intercept = 1/ V0,   

x- intercept = 1/ [S], is same in presence and absence of inhibitor. It indicates that Km remains constant for both the cases.   

Intercept, c = 1/ Vmax ,

Slope, m = Km/ Vmax

# No Inhibitor : Trendline (linear regression) equation for “No- inhibitor” from LB plot y = 0.0652x + 0.0178in absence of inhibitor

Now, from Intercept, c = 1/ Vmax

Or, 0.0178 = 1/ Vmax

Or, Vmax = 1/ 0.0178 = 56.18

Hence, Vmax = 56.18 umol mL-1 min-1

Now,

Km = m x Vmax = 0.0652 x 56.18 = 3.66

Thus, Km = 3.66 mM

## Inhibitor 1: Trendline (linear regression) equation for “No- inhibitor” from LB plot y = y = 0.2182x + 0.0143 in presence of inhibitor 1

Now, from Intercept, c = 1/ Vmax

Or, Vmax = 1/ 0.0143 = 69.93

Hence, Vmax = 69.93 umol mL-1 min-1

Now,

Km = m x Vmax = 69.93 x 0.2182 = 15.26

Thus, Km = 15.26 mM - In presence of Inhibitor 1

## Inhibitor 2: Trendline (linear regression) equation for “No- inhibitor” from LB plot y = 0.1414x + 0.0401 in presence of inhibitor 2

Now, from Intercept, c = 1/ Vmax

Or, Vmax = 1/ 0.0401 = 24.94

Hence, Vmax = 24.94 umol mL-1 min-1

Now,

Km = m x Vmax = 24.94 x 0.1414 = 3.52

Thus, Km = 3.52 mM - In presence of Inhibitor 2

# Type of inhibitor:

In presence of Inhibitor 1: Vmax remains same, Km is increased. It’s a competitive inhibitor.

In presence of Inhibitor 2: Vmax lowered, Km remains the same. It’s a non-competitive inhibitor.