Question

32. 33. Determine the total yield of ATP from the complete oxidation of palmitolelc acid, diunsaturated fatty acid. Show your work. Determine the total yield of ATP from the complete oxidation of 1 molecule of phosphoenolpyruvate. Show your work. :rnn و 11 دعاء عء

Answer 1

32.- This fatty acid has 16 carbons and only one insaturation. Let us address each step:

Dehydrogenation produces one FADH2 per round, we will have 7 rounds, so that is 7 FADH2

Hydration does not produce energetic molecules

Oxidation produces 1 NADH per round, we will have 7 rounds, so that is 7 NADH

Thyolisis will produce 1 acetyl CoA per round, we have 7 rounds so that is 7 acetyl CoA + 1 because the final cleave of the 4 carbon acid produces 2 acetyl CoA, then we have 8 acetyl CoA molecules at the end.

Acetyl CoA enter Krebs cycle and finally yield 3 NADH, 1 ATP/GTP and 1 FADH each

The NADH and FADH2 will enter ETC producing 3 ATP and 2 ATP respectively, so we have a total of:

NADH: 7 + 24 = 31. These 31 will produce 93 ATP

FADH2: 7 + 1 = 8. These 8 will produce 16 ATP

Now we sum 93 + 16 + 8 = 117 ATP as the final yield

33.- This molecule requires 1 ATP molecule to be converted into pyruvate, then pyruvate is oxidized into acetyl CoA , this reactions yields 1 NADH molecule. Then acetyl CoA can enter krebs cycle in which will yield a total of 3 NADH, 1 ATP/GTP and 1 FADH2.

These reductive molecules will enter ETC, each NADH will produce 3 ATP and each FADH2 only 2 ATP, so that si:

NADH: 4(3) = 12 ATP

FADH2: 1(2) = 2 ATP

We have a total of 14 ATP from ETC, 1 ATP from the transition reaction, 1 ATP from Krebs Cycle and minus 1 ATP from the pyruvate first reaction, that is:

14 + 1 + 1 -1 = 15 ATP

The final yield of one PEP molecule is 15 ATP

36.- Hydrolysis is the cleavage of a molecule through the addition of water, if you undergo hydrolysis many times until you just cannot cleave using water any further you obtain the following molecules:

a) Nucleotides

b) Glucose

c) Fatty acids and glycerol

d) Amino acids

37.- Remember that the ions will have different charges according to pH and according to their pka. If the pH value is lower tha the pKa value of a certain ion then this ion will be protonated, and if the pH is higher than the pKa then the ion will be deprotonated, let us analyze each ion in this peptide:

- Met amino terminus: pKa= 9.21

- His side chain amino group: pKa= 6

- Cys carboxyl terminus: pKa= 1.96

Now, let us state the charge in each of the required pH values:

- pH 7: Met will be protonated (+) while His and Cys will be deprotonated (0 and -, respectively). In this pH the peptide will have a net charge of 0. This is the drawing:

S SH NH 0 +H3N NH NH NE

- pH 4: Met and His will be protonated (+ and +) while Cys will be deprotonated (-). In this pH the peptide will have a net charge of +1. This is the drawing:

S SH NH 0- +H3N NH NH +HNV