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:

2 ADP =(Mg2+)= ATP + AMP

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 3% 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 3% of the ATP is hydrolyzed to ADP?

Answer 1

Given that the reaction lies is: 2 ADP <=====> ATP + AMP Initial Concentration of ATP = 265 μM Total concentartion of adenine nucleotide = 364μM

Initially at time zero the [ADP] + [AMP] = 364 - 265 = 99μM.

K = 0.82 = [ATP][AMP] / [ADP] ² Eq. 1

when 3% of ATP is hydrolyzed .

[ ATP] = 0.97 × 265 μM = 257.05 μM.

2ADP <========> ATP + AMP

Initial [ADP] 265 [AMP]

Change + 6% = 15.9 -7.95 - 3%= 7.95

Equilibrium [ADP] +15.9 257.05 [AMP] - 7.95

Let's say initial concentration of AMP is x μM,

so [ADP] = 99.0- x μM

After hydrolysis of ATP, [AMP] = x - 7.95 μM,

[ ADP] = 99 - x + 15.9 = 114.9 - x  μM.

Substitute value in equation 1

257.05 × (x - 7.95) / (114.9 - x)² = 0.82

Two value of x comes = 512.66 and 30.613  μM.

Accept value of x = 30.613  μM.

After hydrolysis [AMP] = 30.613 - 7.95 = 22.663 μM.