G To answer this question, you may reference the Metabolic Map. Select the steps of glycolysis in which ATP is produced. phosphoenolpyruvate → pyruvate phosphoenolpyruvate→pyruvate glucose → glucose 6‑phosphate glucose→glucose 6‑phosphate glyceraldehyde 3‑phosphate → 1,3‑bisphosphoglycerate glyceraldehyde 3‑phosphate→1,3‑bisphosphoglycerate 1,3‑bisphosphoglycerate → 3‑phosphoglycerate 1,3‑bisphosphoglycerate→3‑phosphogl ycerate fructose 6‑phosphate → fructose 1,6‑bisphosphateSuppose 13 glucose molecules enter glycolysis. Calculate the number of inorganic phosphate molecules required as well as the number of pyruvate molecules produced. Pi = Pyruvate =

a) Total of ATP produced in glycolysis is equal to 2

ATP produced:

i) 1,3‑bisphosphoglycerate → 3‑phosphoglycerate (2 ATP are produced)

ii) phosphoenolpyruvate→pyruvate (2 ATP are produced)

b) 26 pyruvate molecules.

Explanation:

a) Glycolysis consists of two steps in which ATP is consumed and two steps in which ATP is released.

ATP produced = 4

ATP consumed = 2

The total of ATP produced in glycolysis is equal to 2 (4-2)

ATP produced:

i) 1,3‑bisphosphoglycerate → 3‑phosphoglycerate (2 ATP are produced)

ii) phosphoenolpyruvate → pyruvate (2 ATP are produced)

b) In the sixth step of glycolysis, inorganic phosphate molecules are consumed:

glyceraldehyde-3‑phosphate → 1,3‑bisphosphoglycerate

2 pi are added in glycolysis. As each glucose molecule undergoes glycolysis, the pi required by the 13 glocosa molecules is 26 (13 * 2).  Glycolysis of each glucose produces 3 pyruvates, so 13 glucose molecules will produce 26 pyruvate molecules.

i believe it would be the lungs.

answer: the blood-brain barrier is a layer of specialized membranes and cell processes designed to keep undesirable substances from passing from the bloodstream into the precious tissues of the brain. while the blood-brain barrier does keep most bad things out of the central nervous system, it is not impermeable.

explanation: