Question

Consider the following data for an enzyme catalyzed hydrolysis reaction in the presence and absence of inhibitor I:

[Substrate] M                                    v_o (µmol/min)                    v_oI (µmol/min)

6x10^-6 20.8 4.2

1x10^-5                                                    29                                                           5.8

2x10^-5                                                    45                                                           9

6x10^-5                                                    67.6                                                        13.6

1.8x10^-4                                                87                                                           16.2  

Use the above data, do the following:

a. Generate Lineweaver-Burk plots of the data.

b. Explain the significance of the x-interecpt, y-intercept, and the slope.

c.   Identify the type of inhibition.

Answer 1

Ans. 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’.

# Determination of V_max and K_m 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/ V_max           ,

Slope, m = K_m/ V_max   

# Trendline (linear regression) equation for “No- inhibitor” from LB plot y = 2.27x + 0.0108 in absence of inhibitor

Now, from Intercept, c = 1/ V_max

            Or, 0.0108= 1/ V_max

            Or, V_max = 1/ 0.0108 = 92.593

Hence, V_max = 92.953 umol min^-1

Now,

Km = m x Vmax = 2.27 x 92.593 = 210.186

Thus, Km = 210.186 uM = 2.102 x 10^-4 M

# Trendline (linear regression) equation for “No- inhibitor” from LB plot y = 11.084x + 0.0563 in presence of inhibitor.

Now, from Intercept, c = 1/ V_max

            Or, V_max = 1/ 0.0563 = 17.762

Hence, V_max = 17.762 umol min^-1

Now,

Km = m x Vmax = 11.084 x 17.762 = 196.874

Thus, Km = 196.874 uM = 1.97 x 10^-4 M

# Type of inhibitor: Note that Km remains constant whereas the Vmax is lowered in presence of inhibitor.

It is the characteristic of a completive inhibitor.

Thus, the inhibitor is a competitive inhibitor.