Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATP molecules are released? Show your detailed computation of the problem.
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and distributed into the cells for energy production. One glucose unit is catabolized in the muscle cell
while the other is in the heart cell, to CO2 and H2O in the
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- Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Please show your detailed computation.Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Show your detailed computation of the problem.Digestion of cellobiose in cows produces two glucose units which is then absorbed into the blood and distributed into the cells for energy production. One glucose unit is catabolized in the muscle cell while the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATP molecules are released? Show your detailed computation.
- Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Show your computation.Digestion of cellobiose in cows produces two glucose units which is then absorbed into the bloodand distributed into the cells for energy production. One glucose unit is catabolized in the muscle cellwhile the other is in the heart cell, to CO2 and H2O in the cellular respiration process. How many ATPmolecules are released? Show your detailed computation. answer.An animal cell, roughly cubical in shape with side length of 10 μm, uses 109 ATP molecules every minute. assume that the cell replaces this ATP by the oxidation of glucose according to the overall reaction 6O2 + C6H12O6 →6CO2 + 6H2O and that complete oxidation of each glucose molecule produces 30 ATP molecules. how much oxygen does the cell consume every minute? How long will it take before the cell has used up an amount of oxygen gas equal to its own volume?
- The concentration of glucose in your circulatory system is maintained near 5.0 mM by the actions of the pancreatic hormones glucagon and insulin. Glucose is imported into cells by protein transporters that are highly specific for binding glucose. Inside the liver cells the imported glucose is rapidly phosphorylated to give glucose-6-phosphate (G-6-P). This is an ATP-dependent process that consumes 1 mol ATP per mol of glucose. Given the steady-state intracellular concentrations below, calculate the theoretical maximum concentration of G-6-P inside a liver cell at 37 °C, pH = 7.2 when the glucose concentration outside the cell (i.e., [glucoseloutside) is 5.0 mM: ATP = 4.7 mM; ADP = 0.15 mM; P, = 6.1 mM For: ATP + H,O ADP + P + H* AG" = -30.5 kJ/mol and G-6-P + H,0 -→ Glucose + P AG" = -13.8 kJ/mol The glucose phosphorylation reaction is ATP + glucosenside » ADP + glucose-6-phosphate + H+Stearic acid is an 18-carbon fatty acid. If a single molecular of stearic acid is within the cytosol of the cells: Describe the process by which stearic acid would be metabolised, beginning the molecule in the cytosol of the cell, ending with the creation of ATP, and assuming the cell has sufficient oxygen for all reactions to take place Showing all working, calculate how many ATP molecules could be generated from a single molecule of stearic acid in the cytosol of the cellThe average cell, at rest, hydrolyzes 10,000, 000 ATP molecules per second. You are studying the stem cell population found in the intestinal Crypts. In the intestine there are s total 5 x 1011 cell number. The stem cells have a 85% higher metabolism than an average resting cell. ATP hydrolysis yield 7.4 kCal/ Mole ATP. (Avogadro’s number is 6.023 x 1023) 1) How many moles of ATP are hydrolyzed per second? A) 15.3 moles/ sec. B) 7.06 x 10-6 moles/ sec. C) 5.4 x 10-2 moles/ sec. D) 25 moles/ sec. E) 225 moles/hour F) none of these
- The average cell, at rest, hydrolyzes 10,000, 000 ATP molecules per second. You are studying the stem cell population found in the intestinal Crypts. In the intestine there are s total 5 x 1011 cell number. The stem cells have a 85% higher metabolism than an average resting cell. ATP hydrolysis yield 7.4 kCal/ Mole ATP. (Avogadro’s number is 6.023 x 1023) 2) How many Kcal of energy are used per day by this population of cells? A) 4.514 kCal/day B) 1.14 x 10-6 kCal/day C) 1.14 x 10-4 kCal/day D) 25 kCal/day E) 225 kCalThe conversion of glucose to glucose 6-phosphate is an endergonic reaction (eql below). The second equation below is an exergonic reaction that can occur in all cells. Considering this information, how can the reaction in equation 3 proceed? EQ1: Glucose + Pj --> glucose 6-phosphate EQ2: ATP + H20 --> ADP + Pj EQ3: Glucose + ATP --> glucose 6-phosphate + ADP Select one: a. The energy required to make Eq3 spontaneous is provided by the enzyme catalyzing the reaction b. The energy released from equation 1 is more than the energy consumed in equation 2 making the overall reaction AG negative and the reaction spontaneous. c. The energy released from equation 2 is more than the energy consumed in equation 1 making the overall reaction AG positive and the reaction spontaneous. d. The energy released from equation 2 is more than the energy consumed in equation 1 making the overall reaction AG negative and the reaction spontaneous. e. The energy released from equation 1 is more than the energy…Many types of cells are able to use membrane-spanning transport proteins and a source of energy to drive the active transport of glucose across the cell membrane. Leť's consider a specific protein that is able to transport (from outside the cell into the cytoplasm) one molecule of glucose for each molecule of ATP the protein hydrolyzes. The details of this process are not important for this problem except that you can treat it like equilibrium process. The overall reaction and the associated AG at 298 K for this process are given below. Glc(out) + ATP + H20 = Glc(in) + ADP + P at T = 298 K, AG = -31.3 kJ mol-1 Additional Information: • P. is "inorganic phosphate" (i.e. PO4²- & related species). At equilibrium, [P] = 3 x 10-3 M. • The cell maintains a ratio of ATP to ADP of about 20 (i.e. [ATP]/[ADP] = 20). • The maximum ratio of [Glc(in)] to [Glc(out)] occurs when the system is at equilibrium. a) Write an expression for the equilibrium constant for this process. b) Your expression for…