Map Lehninger Principles of Biochemistry Nelson 1 Cox SEvE MHE/Freeman preented

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Map Lehninger Principles of Biochemistry Nelson 1 Cox SEvE MHE/Freeman preented by Sapting Leaning For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B For substrate A, she determined that Km-30 AM and kca.-20 min-1. Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows Michaelis-Menten kinetics. In one experiment where [A+ 4 mM, she found b) In another experiment with [EJ = 0.2she a that the initial velocity, Vo was 624 nM min1 found that Vo 3 M min What was the [A] she used in this experiment? What was the [El she used in this experiment? E,l= nM The compound Z was found to be a very strong competitive inhibitor of the enzyme, with an of 10. In an experiment with the same [Etl as in part (a), but Menten curve at low [S], the second order rate a different [Al, an amount of Z is added that reducesconstant Keal/ Km A "perfect enzyme, limited the rate Vo to 312 nM min-1 What was the [A] in this experiment? A measure of the catalytic efficiency of an enzyme is the initial slope of the Michaelis- by diffusion, would have a value of 108 or 0Ms 1 Calculate Kea/Km for this enzyme. Number Number M's Tools O Previous Give Up & View Solution e Check Answer Next Exit

Explanation / Answer

Ans. #a. Using the equation-          kcat = Vo / [Et]                     

            Where, Vo = Vmax attainable in given reaction mixture

            Or, 20 min-1 = (624 nM min-1) / [Et]

            Or, [Et] = (624 nM min-1) / (20 min-1) = 31.2 nM

Hence, [Et] = 31.2 nM

#b. Step 1: Note from #a: [A] = 4 mM = 4 x 103 uM                ; Km = 3.0 uM

Note that substrate concertation [A] >> Km, so the reaction velocity attained in #a is the maximum reaction velocity, Vmax for the enzyme.

That is,            Vmax = 624 mM min-1      at [Et] = 31.2 nM

Step 2: Correlating #a and #b

Given, [Et] = 0.2 uM = 200 nM        ; Vo = 3 uM min-1 = 3000 uM min-1

Vmax is proportional to enzyme concertation.

So, (Vmax” at [Et] = 200 nM) must be equal to (Vmax at [Et] = 31.2 nM) per unit enzyme concentrations in both cases-

            Or, (Vmax” / 200 nM) for #b = (Vmax / 31.2 nM) for #a

            Or, Vmax” = (624 nM min-1 / 31.2 nM) x 200 nM = 4000 nM min-1

So, Vmax” for #b = 4000 nM min-1 = 4.0 uM min-1

Step 3: Calculate [A] using MM equation-

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

            3.0 uM min-1 = 4.0 uM min-1 [A] / (3.0 uM + [A])

            Or, (3.0 uM + [A]) = 4.0 uM min-1 [A] / (3.0 uM min-1) = 1.33 [A]

            Or, 3.0 uM = 1.33 [A] – [A] = 0.33 [A]

            Or, [A] = 3.00 uM / 0.33 = 9.091 uM

Hence, [A] = 9.091 uM

#c. For a completive inhibitor,       Km,app = a (Km)

            Or, Km,app = 10 x (3.0 uM) = 30.0 uM

Note that a completive inhibitor does not affect the Vmax. So, Vmax remains constant at 624 nM min-1 = 0.624 uM min-1 as in #a at [Et] = 31.2 nM

# Using MM equation with following values-

Vmax = 0.624 uM min-1

Vo = 312 nM min-1 = 0.312 uM min-1

Km,app = 30.0 uM

Now,

0.312 uM min-1 = 0.624 uM min-1 [A] / (30.0 uM + [A])

            Or, (30.0 uM + [A]) = 0.624 uM min-1 [A] / (0.312 uM min-1) = 2.0 [A]

            Or, 30.0 uM = 2.0 [A] – [A] = [A]

            Hence, [A] = 30.0 uM

Therefore, required [A] = 30.0 uM

#d. Kcat/ Km = 20 min-1 / 3.0 uM

                        = 20 min-1 / (3.0 x 10-6 M)                                       ; [1 uM = 10-6 M]

                        = 6.667 x 106 M-1 min-1                                            ; [1 min = 60 s]

                        = 1.11 x 105 M-1 s-1

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