Okay, I had to think about this question because you supplied ΔGo’, and not ΔG. So you have to solve for ΔG to solve this problem, and you have to use the concentration of ATP, ADP, and Pi in muscle cells, because the concentrations very by cell type. The concentrations of ATP, ADP, and Pi are 5.0, 0.5, and 1.0mM, respectively. ΔGo’ for the formation of ATP is 7.3 cal/mol or 30.5 kJ/mol, the reverse of its hydrolysis. Converting from mM to M, and plugging the values into the equation, ΔG=ΔGo’+RTlnQ, we get
ΔG=30.5 kJ/mol+(2.58 kJ/mol)(9.21)=54.3kJ/mol
Since, 100kcal/mi*(4.2kJ/1kcal)=420kJ/mi
Determine moles of ATP
420kJ/54.3kJ/mol=7.73 moles of ATP
7.73 moles of ATP *( 507 g of ATP/1 mole of ATP)=3.92 X10^3 g of ATP= 3.92 kg of ATP.
I am not sure about the final number, because different references have different values for ATP, ADP, and Pi for muscle cells.
Since 40% of the ATP can not be used because it is being used to phosphorylate ADP, then the number of moles of ATP needed from the oxidation of ATP is equal to the number of ATP required plus an extra 40%.
7.73 moles of ATP + 40%= 10.8 moles of ATP
The production of 10.8 moles of ATP will require 586kJ
Since The complete oxidation of glucose is equal to 686 kcal/mol or 2,881kJ/mole (check these values),
586kJ/2,881kJ/mole=moles of glucose needed, which is 0.203 moles of Glucose.
0.203 moles of Glucose *(180 g of glucose/1 mole of glucose)= 36.6 g of glucose.
Not sure about the answers, because of the values, but I hope this helps.
1. Walking consumes approximately 100 kcal/mi. In the hydrolysis of ATP (ATP → ADP + Pi), the reaction that drives the muscle contraction, ΔGo’ is -7.3 kcal/mol (-30.5 kJ/mol). Calculate how many grams of ATP must be produced to walk a mile. ATP synthesis is coupled to the oxidation of glucose (ΔGo’ = -686 kcal/mol). How many grams of glucose are actually metabolized to produce this amount of ATP? (Assume that only glucose oxidation is used to generate ATP and that 40% of the energy generated from this process is used to phosphorylate ADP. The gram molecular weight of glucose is 180 g and that of ATP is 507 g.
2 answers
I totally screwed up on the second part.
7.8 moles of ATP is correct, and you need to take into account how much ATP is lost by the oxidation of glucose, which is 40%.
Depending on what textbooks you use, 1 molecule of glucose will yield 30, 34, or 36 ATP. But lets use 1 molecule yields 36 ATP as our conversion factor, and let's subtract 40% from the 36, which gives us 22 free ATP per molecule of glucose.
We need 7.73 moles of ATP, so
7.73 moles of ATP*(6.02 x 10^23 atoms of ATP/1 mole of ATP)= atoms of ATP
Since 1 atom of glucose=22 ATP
atoms of ATP *( 1glucose/22 ATP)=number of glucose molecules
And since 1 mole glucose=180 g of glucose and 1 mole of glucose =6.02 x 10^23 atoms,
number of glucose molecules*(1 mole of glucose/6.02 x 10^23 atoms)*(180 g of glucose/1 mole glucose)= g of glucose needed for oxidation.
I really do apologize for the mixup on the second part. But remember, check the values that I gave you especially for reference values, since they are not all the same in every textbook.
7.8 moles of ATP is correct, and you need to take into account how much ATP is lost by the oxidation of glucose, which is 40%.
Depending on what textbooks you use, 1 molecule of glucose will yield 30, 34, or 36 ATP. But lets use 1 molecule yields 36 ATP as our conversion factor, and let's subtract 40% from the 36, which gives us 22 free ATP per molecule of glucose.
We need 7.73 moles of ATP, so
7.73 moles of ATP*(6.02 x 10^23 atoms of ATP/1 mole of ATP)= atoms of ATP
Since 1 atom of glucose=22 ATP
atoms of ATP *( 1glucose/22 ATP)=number of glucose molecules
And since 1 mole glucose=180 g of glucose and 1 mole of glucose =6.02 x 10^23 atoms,
number of glucose molecules*(1 mole of glucose/6.02 x 10^23 atoms)*(180 g of glucose/1 mole glucose)= g of glucose needed for oxidation.
I really do apologize for the mixup on the second part. But remember, check the values that I gave you especially for reference values, since they are not all the same in every textbook.