The net yield of ATP from one turn of the citric acid cycle is 1 ATP molecule through substrate-level phosphorylation. However, additional ATP can be generated indirectly through the electron transport chain and oxidative phosphorylation using reducing equivalents generated during the citric acid cycle.
To calculate the number of moles in 342g of sucrose, divide the given mass by the molar mass of sucrose. The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. Therefore, 342g of sucrose is equal to 1 mole.
There are 12 moles of C atoms in 1 mole of sucrose because each molecule of sucrose (C12H22O11) contains 12 carbon atoms.
1 M = 1 mole/liter. so 2 M sucrose would contain 2 moles of sucrose in 1 liter of solution.So to find the volume of solution with 1 mole sucrose:V = (1 mole) / ( 2 mole/liter) = 0.5 liter ; 0.5 liter * (1000 mL/liter) = 500 mL
To make a 1 molar solution of sucrose, you would weigh out 342.3 grams of sucrose (molecular weight of sucrose is 342.3 g/mol) and dissolve it in water to make a final volume of 1 liter. This would give you a solution where 1 mole of sucrose is dissolved in 1 liter of water.
The Krebs cycle generates 1 ATP molecule per turn through substrate-level phosphorylation. Due to the cycle occurring twice per glucose molecule, a total of 2 ATP molecules are produced per glucose molecule entering the cycle.
In glycolysis, one glucose molecule produces a net yield of two ATP molecules at the end of the process.
The net yield of ATP from one turn of the citric acid cycle is 1 ATP molecule through substrate-level phosphorylation. However, additional ATP can be generated indirectly through the electron transport chain and oxidative phosphorylation using reducing equivalents generated during the citric acid cycle.
To calculate the number of moles in 342g of sucrose, divide the given mass by the molar mass of sucrose. The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. Therefore, 342g of sucrose is equal to 1 mole.
1 ATP is equal to 59 KJ
Consuming 1g of fat yields about 9 kcal/g and can produce around 38 ATP molecules. Consuming 1g of carbohydrate provides about 4 kcal/g and can produce around 36-38 ATP molecules. So, in this case, consuming 1g of fat would yield slightly more ATP compared to 1g of carbohydrate.
The heat of combustion of sucrose is 3840 kJ per deg.mol and Molecular weight of sucrose is 342 mol -1.so the calorific value of sucrose is 11.228 kJ/deg.gm , provided the temperature is kept constant.
1
64 net...68 are produced overall but 2 ATP's are used in the reaction per molecule of glucose.
The most ATP are produced during the last stage of cellular respiration-- the electron transport chain which involves chemiosmosisThe theoretical yield for eukaryotes is 36 ATP per glucose moleculeOne ATP generated for each proton pump activatedMultiplied by 2=2 two pyruvates from glucose= 10 NADH = 2 glycolysis + 2 oxidation of pyruvate (2) + 6 from Krebs2 FADH22 ATP from glycolysis + 2 ATP from Krebs10 x3=30 ATP from NADH + 2x2=4 ATP from FADH2 + 4 ATP= 38 ATP producedAccurate for bacteria not eukaryotes= NADH in cytoplasm from glycolysis need 1 ATP/NADH molecule36 potential yield
4 ATP molocules are made from 1 NADH and 1 FADH2 MO
There are 12 moles of C atoms in 1 mole of sucrose because each molecule of sucrose (C12H22O11) contains 12 carbon atoms.