3. The protease papain (isolated from papaya fruit) is a monomer with 1 active s

Question

3. The protease papain (isolated from papaya fruit) is a monomer with 1 active site per protein molecule. The enzyme was assayed at a series of substrate concentrations and intial velocity was measured. Use the Lineweaver-Burk plot to determine the following. Include units with all answers.

b)              Calculate Km for this substrate.

c)              Calculation the concentration of substrate required for the rate to proceed at 75% of Vmax.

d)              If kcat = 4 × 105 s1 for this enzyme, determine the value of the specificity constant in units of                        M1s1.

Explanation / Answer

#1. Determination of Vmax and Km 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/ Vmax           ,

Slope, m = Km/ Vmax   

Trendline (linear regression) equation for “No- inhibitor” from LB plot is y = 0.50x + 4.0.

Now, from Intercept, c = 1/ Vmax

            Or, 4.0= 1/ Vmax

            Or, Vmax = 1/ 4.0= 0.25

[Given, unit of (1/ Vmax = mM-1min). So, unit of Vmax = 1 / (mM-1min) = mM min-1]

Hence, Vmax = 0.25 mM min-1     

Now, Km = m x Vmax = 0.50 x 0.25 = 0.125

Thus, Km = 0.125 mM

#C. Given, Vo = 75% of Vmax = 0.75 Vmax

Km = 0.125 mM = 0.000125 M                            ; [1 mM = 0.001 M]

MM equation -

            Vo = Vmax [S] / (Km + [S])                                 - equation 1

Putting the values in equation 1, assuming [S] to be unknown-

            0.75 Vmax = Vmax [S] / (0.125 mM + [S])

            Or, 0.75 Vmax x (0.125 mM + [S]) = Vmax [S]

            Or, 0.75 x (0.125 mM + [S]) = [S]

            Or, 0.09375 mM + 0.75[S] = [S]

            Or, 0.09375 mM = [S] – 0.75[S] = 0.25[S]

            Or, [S] = 0.09375 mM/ 0.25

            Or, [S] = 0.375 mM

Thus, required [S] = 0.375 mM

#D. Specificity constant = Kcat / Km

                                                = (4 x 10-5 s-1) / (0.000125 M)

                                                = (4 x 10-5 s-1) / (0.000125 M)

                                                = 0.32 M-1s-1

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