The ATP yield for a molecule depends on where it enters glycolysis or the citric acid cycle. The yield can be compared to the net yield of glucose, 30 ATP.
Determine the net ATP yield of each starting molecule when fully oxidized to CO2. Assume that glycolysis, the citric acid cycle, and oxidative phosphorylation are fully active.
net ATP yield from fructose 1,6‑bisphosphate: ATP molecules
net ATP yield from galactose: ATP molecules
net ATP yield from dihydroxyacetone phosphate: ATP molecules
net ATP yield from pyruvate: ATP molecules
net ATP yield from lactate: ATP molecules
net ATP yield from phosphoenolpyruvate: ATP molecules

a) The net ATP yield from fructose 1,6‑bisphosphate is 32 ATP

b) The net ATP yield from galactose is 30 ATP

c) The net ATP yield from dihydroxyacetone phosphate is 16 ATP

d) The net ATP yield from pyruvate is 12.5 ATP

e) The net ATP yield from lactate is 14 ATP

f) The net ATP yield from phosphoenolpyruvate is 13.5 ATP

Explanation:

a) In this case, 2 ATP molecules are not used for the phosphorylation of glucose for the production of fructose bisphosphate 1.6, thus, the net ATP yield is 32 ATP.

b) Galactose is converted to glucose-1-phosphate through reactions that lead to phosphorylation. Glucose-1-phosphate is transformed into glucose-6-phosphate by the enzyme phosphoglucomutase. Later it is used in glycolysis, thus, the net ATP yield is 30 ATP.

c) The fructose 1,6 bisphosphate is decomposed in glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. Dihydroxyacetone phosphate is isomerized to glyceraldehyde -3-phosphate. ATP performance is (2.5+2.5+11) = 16 ATP

d) Pyruvate is decarboxylated oxidatively to produce acetyl-CoA. ATP yields are:

Glycolysis = 0

Oxidative decarboxylation = 2.5 ATP

TCA cycle = 10 ATP

The net yield is 2.5 + 10 = 12.5 ATP

e) Lactate is oxidized to pyruvate by the enzyme lactate dehydrogenase with a reduction of NAD. Pyruvate is brought into the mitochondria, where it is decarboxylated oxidatively to produce acetyl-CoA. If 1.5 ATP is assumed for cytosolic NADH, therefore 1.5 is produced. So, the net yield is: 12.5 + 1.5 = 14

f) Phosphoenolpyruvate is converted to pyruvate by ATP production through phosphorylation. ATP yield is as follows:

Glycolysis = 1

Oxidative decarboxylation = 2.5 ATP

TCA cycle = 10 ATP

The net yield is 1 + 2.5 + 10 = 13.5 ATP

1. loamy sand is about 80% sand with 10% clay and defined mostly by sand particles with some clay to provide structure and fertility. the sand particles are easily visible and the soil holds structure but breaks apart easily.

2. scott sand can improve his soil quality by incorporating more silt and organic matter into the soil. peat moss and organic matter can also be incorporated.

the answers are as following:

1. c. double helix, nucleosome, coil, supercoil, chromosome

2. b. g2 phase, preparation for mitosis

3. d. interphase consists of the g1, s, and g2 phases

4. c. metaphase

5. a. chromosomes are duplicated before cell division so that each new daughter cell has a complete set