Question

Adenosine monophosphate (AMP) is a regulatory molecule in metabolic processes such as glycolysis and gluconeogenesis. For example, it stimulates the glycolytic enzyme phosphofructokinase, and therefore ATP production, and it inhibits the gluconeogenic enzyme fructose 1,6-bisphosphatase. Adenylate kinase catalyzes the reversible reaction shown here: During periods of intense activity, when glycolysis is used in the generation of ATP, the reaction lies to the right, decreasing [ADP], generating ATP, and accumulating AMP. However, [ATP] is usually much greater than [ADP], and [ADP] is greater than [AMP]. Determine [AMP] when 5% of the ATP in a hypothetical cell is hydrolyzed to ADP. In this cell, the initial concentration of ATP is 265 μM, and the total adenine nucleotide concentration (the concentration of ATP, ADP, and AMP) is 364 μM. The equilibrium constant K is 0.82 What is the concentration of AMP after 5% of the ATP is hydrolyzed to ADP?

Answer 1

Given the initial concentration of ATP, [ATP]_i = 265 M

[ATP]i + [ADP]i + [AMP]i = 364 M

Suppose the initial concentration of AMP be 'Y' M i.e [AMP]i = Y M

=> 265 M + [ADP]i + Y = 364 M

=> [ADP]i = (99 - Y) M

Given 5% of ATP is converted to ADP.

5% of initial concentration of ATP = 265 M x (5/100) = 13.25 M

The balanced chemical reaction is

--------------------- ATP + AMP <=======> 2 ADP; K = 0.82

Initl.con.(M): 265, ------- Y, ----------------- (99 - Y)

Change(M): -13.25, -13.25 ---------------- +2x13.25

eqm.con.(M): 251.75, (Y - 13.25), -------- (99 - Y + 26.5) = (125.5 - Y)

K = 0.82 = [ADP]² / [ATP]x[AMP]

=> 0.82 = (125.5 - Y)² / [251.75 x (Y - 13.25)]

=> Y² - 457.435Y + 18485.514 = 0

=> Y = 44.8 M

Hence equilibrium concentration of AMP, [AMP] = (Y - 13.25) = (44.8 - 13.25) = 31.5 M (answer)