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 7% of the ATP in a hypothetical cell is hydrolyzed to ADP. In this cell, the initial concentration of ATP is 267 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 7% of the ATP is hydrolyzed to ADP?

Explanation / Answer

During glycolysis 4ATPs are produce but 2ATPs are used during process so net production of ATP in Glycolysis is as follows

2ADP + 2Pi à 2ATP

When one ATP is hydrolysed completely,

ATP--àADP + Pi-------1

ADP--àAMP + Pi--------2

ATP-àAMP + 2Pi--------combining above two equations------3

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.

2 ADP ----> ATP + AMP

Generally in the body [ATP]>>[ADP]>[AMP]

We are supposed to determine concentration of [AMP] when 7% of [ATP] à   [AMP] + Pi

Initial [ATP ]= 267 M

And total [ATP]+ ]+[ADP]+[AMP] = 3.90 × 102 M

K (equilibrium constant) = 0.82 is the concentration of AMP after ATP is hydrolysed to ADP

7% of total initial [ATP] concentration = (7/100)* 267 = 18.69 M

At equilibrium , [ADP]+[AMP] - [ATP ] = 3.90 × 102 M - 267 M = 123

[ADP]+[AMP] = 123

[ADP] = 123- [AMP]

Equilibrium constant K since two ATPs are hydrolysed in complete reaction of glycolysis

K= [ATP][AMP]/2[ADP]

0.82 = (248.31)[AMP]/2(123-[AMP]+(2 ×18.69))

Solving for we get, [AMP] = 1.0692 M

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