Question

The velocity of an enzymatic reaction (expressed in nmoles x liter-1 x min-1 ) for different concentrations of substrate (S, expressed in moles x liter–1 )) are summarized in the following table.

s	v
8.33 x 10^-6	13.8
1.00x 10^-5	16.0
1.25 x 10^-5	19
1.67 x 10^-5	23.6
2.00 x 10^-5	26.7
2.50 x 10^-5	30.8
3.33 x10^-5	36.3
4.00 x10^5	40.0
5.00 x 10^-5	44.4
6.00 x 10^-5	48.0
8.00 x 10^-5	53.4
1.00 x 10^-4	57.1
2.00 x 10^-4	66.7

1. The Hanes-Woolf plot represents [S]/v (y axis) versus [S] (x axis). Rearrange the Lineweaver-Burk to determine the linear equation for the Hanes-Woolf plot (y = ax + b, with y= [S]/v and x = [S]). What do the intercepts with the [S]/v and the [S] axis, and the slope of the straight line represent? (For example in the Lineweaver-Burk plot, the intercept of the straight line and the y-axis represent 1/Vmax) (6 pts)

2. The Woolf-Augustinsson-Hofstee plot represents v versus v/[S]. Rearrange the HenriMichaelis-Menten equation to determine the linear equation for the Woolf-AugustinssonHofstee plot. What do the intercepts with the v and the v/[S] axis, and the slope of the straight line represent? (6 pts)

3. Compare the numerical values of Km and Vmax calculated from each plot. (3 pts)

Answer 1

Ans. #0. 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 = 1/ V₀,       x = 1/ [S],     

Intercept, c = 1/ V_max           ,

Slope, m = K_m/ V_max   

Trendline (linear regression) equation for “No- inhibitor” from LB plot y = 5E-07x + 0.0125

Now, from Intercept, c = 1/ V_max

            Or, 0.0125 = 1/ V_max

            Or, V_max = 1/ 0.0125 = 80.0

Hence, V_max = 80.0 nmol L^-1 min^-1

Now,

Km = m x Vmax = (5.00 x 10^-7) x 80 = 4.00 x 10^-5

Thus, Km = 4.00 x 10^-5 moles L^-1

Note: There is no need of converting the units of concentration. Just do calculations and put the unit after the value at last as mentioned in question.

#1. Hanes- Woolf Plot:

Hanes- Woolf plot gives an equation in from of Y = m X + c

where, y-axis = [S]/ Vo        ; x-axis = [S]

Intercept on Y-axis, c = Km/ V_max

Slope, m = 1/ Vmax   

From the trendline equation for Hanes- Woolf plot “ y = 0.0125x + 5E-07” -

# Calculate Vmax

            m = 1/ Vmax

            or, Vmax = 1 / m = 1/ 0.0125 = 80.0

            Hence, V_max = 80.0 nmol L^-1 min^-1

# Intercept on Y-axis, c = Km/ V_max

Or, 5.00 x 10^-7 = Km/ 80.0

Or, Km = (5.00 x 10^-7) x 80 = 4.00 x 10^-5

Thus, Km = 4.00 x 10^-5 moles L^-1

#2. Woolf- Augustinsson- Hofstee Plot (or, Eadie- Hofstee Plot)

Rearrangement of MM equation to yield EH plot is-

            Vo = -Km (Vo/ [S]) + V_max

In Eadie- Hofstee Plot-

            Y-intercept = Vmax = C

            X-intercept = Vmax/ Km

            Slope = -Km

The trendline equation “y = -4E-05x + 79.979” is in from of Y = mX + C

So, Vmax = C = 79.979 = 80.0

Therefore, Vmax = V_max = 80.0 nmol L^-1 min^-1

Using slope = -Km,

-(4.00 x 10^-5) = - Km

Or, Km = 4.00 x 10^-5

Therefore, Km = 4.00 x 10^-5 moles L^-1

#3. Comparison: The values in Km and Vmax is the same in all the three plots.