During periods of intense activity, when glycolysis is used in the generation of

Question

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 M, and the total adenine nucleotide concentration (the concentration of ATP, ADP, and AMP) is 3.90 × 102 M. The equilibrium constant K is 0.82 What is the concentration of AMP after 3% of the ATP is hydrolyzed to ADP? Please solve

Explanation / Answer

Given that the reaction lies is:

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

Initial Concentration of ATP = 265 M

Total concentartion of adenine nucleotide = 3.90 X 102 M = 390M

Initially at T zero

the concentration of ATP = 265M

therefore concentration of AMP =390-265=125M because at time zero the concen, of ADP = 0M

After 3 % hydrolysis of ATP, the ADP produced will be 3% of ATP conc.

=3% of 265 =7.95M

And the amount of ATP after hydrolysis =265-3% of 265 = 265-7.95 =257.05.

According to the question ATP hydrolysed = 3%

therefore ADP produced will be 3% of concn. of ATP

therefore [ADP] = 3% of 265 M

= 7.95

And that instant ATP = 265-3% of 265

=265-7.95 = 257.05

Equilibrium constant = [Product]n/[Reactant]

for foward reaction K = 0.82

but hydrolysis of ATP is reverse reaction

therefore K = 1/0.82 =1.22

therefore 1.22 = [7.95]2/[257.05][AMP]

[AMP]= 7.95X7.95/257.05X1.22

=0.201M

therefore accumulation concentration of AMP = 0.201M

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.