1) When 10ug of an enzyme (MW 50,000) is added to a reaction mixture containing

Question

1) When 10ug of an enzyme (MW 50,000) is added to a reaction mixture containing its substrate at a concentration one hundred times the Km, it catalyzes the conversion of 75umol of substrate into product in 3 min. What is the enzyme's turnover number?

2) You want to load 15g of protein in 10L into one of the 12% polyacrylamide gel well. The protein needs to be in 1X buffer and in a total volume of 0.100 ml. You are given a 4.5 mg/ml protein solution, a 20X sample buffer, and distilled water. How much of each would you mix together to make required volume?

3) The Km of an enzyme of an enzyme-catalyzed reaction is 6.5uM. What substrate concentration will be required to obtain 45% of Vmax for this enzyme?

Explanation / Answer

Ans. 1. It’s assumed that, at [S] >> Km, velocity of enzyme catalysis is equal to its Vmax.

So, Vmax = rate of catalysis per unit time

            = 75 umol / 3 min = 25 umol min-1

Now, using,

Turnover number (Kcat) = Vmax / [ET]         ; where, [ET] = concentration of enzyme.

                                                = 25 umol min-1 / 10 ug

                                                = 2.5 umol ug-1 min-1

Ans. 2. Given stock solution

            Protein = 4.5 mg/ mL = 4500 ug/ mL

            Buffer = 20 X

            Total reaction volume = 0.100 mL = 100 uL

Required protein = 15 ug

            Volume of protein stock solution containing 15 ug protein = (1/ 4500 ug per mL) x 15 ug

                                    = (1/4500 ug ml-1) x 15 ug

                                    = 0.00333 mL = 3.33 uL

Buffer stock solution = 20 X

Amount of buffer required to form 1 mL 1 X buffer is calculated as follow-

1 mL 1X buffer = (1/20X) = 0.05 mL = 50 uL

Amount of distilled water = 100 uL (total reaction volume) – (3.3 uL of protein + 50 uL of buffer)

                                    = 100 uL- 53.3 uL = 56.7 uL

Ans. 3. Given, Km = 6.5 uM

                        Vmax = 1

                        V0 = 0.45                      [V0 = 45% Vmax]

                        [S] = ?

Using MM equation

            V0 = Vmax [S] / (Km + [S])

Or, 0.45 = (1 x [S] ) / (6.5 uM + [S])

Or, 0.45 = [S] / (6.5 uM + [S])

Or, 0.45 x (6.5 uM + [S]) = [S]

Or, 2.925 uM = [S] – 0.45 [S] = 0.55 [S]

Or, [S] = 2.925 uM/ 0.55 = 5.318 uM

Hence, [S] = 5.318 uM

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