Please provide step by step solution. ATP contains two high energy phosphates an

Question

Please provide step by step solution.

ATP contains two high energy phosphates and one whose energy is not high. Many biochemical reactions require more free energy than hydrolysis of a single high energy phosphate can provide. In these cases the ATP hydrolysis removes both high energy phosphates as pyrophosphate, which is then hydrolyzed to make two ordinary inorganic phosphates, Pi. This harnesses double the single phosphate free energy. Pyrophosphate hydrolysis is according to the following reaction, which is catalyzed by the enzyme pyrophosphatase:

=O3P–O–PO3= + H2O + 2Pi. (Pi represents whatever forms of phosphate anion are present in the solution, e.g. H2PO4-, HPO4=, PO4-3 depending on the pH)

kcat = 257 per sec Km = 1.45 x 10-4 M

a) If the enzyme concentration were 7 nM, what would be the maximum hydrolysis rate? (You may use seconds rather than minutes as your time unit if you wish.)

b) If the pyrophosphate concentration were 1.07 x 10-6 M, what would be the reaction rate?

c) Aspartic acid 115 is part of the active site. When it is mutated to glutamic acid, kcat falls to 21 sec-1, and Km falls to 0.6 M. If the mutant enzyme is present at the same concentration as in (a) above, what percent of the wild type maximum rate would the mutant enzyme achieve?

d) If the substrate concentration is the same as in (b) above, what will be the reaction velocity of the mutant enzyme and what percentage of the activity at this substrate concentration of the wild type enzyme will this be?

Explanation / Answer

(a): Given enzyme concentration, [E]t = 7 nM = 7.0 x 10-9 M

Let the maximum hydrolysis rate be Vmax

Turnover number, Kcat = 257 sec-1

Also Kcat = Vmax / [E]t

=> Vmax = [E]t x Kcat = 7.0 nM x  257 sec-1 = 1799 nMsec-1 = 1.80 x 10-6 Msec-1 (answer)

(b): Given the subtrate(pyrophosphate) concentration, [S] = 1.07 x 10-6 M

Km = 1.45 x10-4 M

Now applying Michaelis–Menten equation

V = Vmax x [S] / (Km + [S]) = 1.80 x 10-6 Msec-1 x 1.07 x 10-6 M / (1.45 x10-4 M + 1.07 x 10-6 M)

=> V = 1.32 x 10-8 Msec-1 (answer)

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