PROBLEM SET 9. The Citric Acid Cycle

 

Given:

1.Acetyl-CoA is oxidized to carbon dioxide and water by the citric acid cycle.
2. Citric acid is 2-hydroxypropane-1,2,3-tricarboxylic acid.

Predict three citric acid cycles. Which agrees with the data given?

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ANSWERS TO PROBLEM SET 9. The Citric Acid Cycle

 

Given:
1.Acetyl-CoA is oxidized to carbon dioxide and water by the citric acid cycle.
2. Citric acid is 2-hydroxypropane-1,2,3-tricarboxylic acid.
Predict three citric acid cycles. Which agrees with the data given?

ANSWERS to PROBLEM SET 9

For a detailed discussion, see Chapter 9 of CBM.

You are going to grade your answers. There are 100 possible points. >90= A; 75-90 = B, 55-74 = C; 30-54 = D; <30 = F.  If you scored less than 55, you need to study Part 1 of CBM more carefully.

1. Did you write a balanced half reaction that shows 4 oxidation steps are required to oxidize acetyl-CoA to carbon dioxide? (5 points) (section 9.1, pp. 211-212 of CBM).
2. Did you write a net reaction that show the water formed from oxidation of acetyl-CoA comes from reduction of molecular oxygen. (5 points) (p. 212 of CBM).
3. Were you able to write the structure of citric acid from the name you were given? (5 points)(Fig  9.1, p. 211)
4.  Did you recognize that citric acid cannot be oxidized because it is a tertiary alcohol? (5 points).
5. Did you recognize that citric acid can undergo either elimination of water or a retrograde enolate condensation? (5 points).
6. Did you work backwards to succinic acid? (3 points) ( section 9.2, pp. 212-217)
7. Working backwards did you conclude that oxaloacetic acid, malic acid, fumaric acid (or maleic acid) and succinic acid are part of the citric acid cycle? (5 points) (Fig 9.11, p. 217)
8. Did you recognize that conversion of succinic acid to oxaloacetic acid accounts for two of the four oxidations required for oxidation of acetyl-CoA to carbon dioxide and water? (2 points)(Fig 9.11, p. 217).
9. Did you recognize that condensation of acetyl-CoA and oxaloacetic acid gives citroyl-CoA? (5 points)(Fig  9.3, p. 213).
10. Did you recognize that citroyl-CoA can be converted to citric acid by either hydrolysis or substrate level phosphorylation of ADP? (3 points)(Figs 9.4 and 9.5, p 214)
11. Did you conclude that elimination of water from citric acid produces aconitic acid? (5 points)( Fig 9.12, p. 217)
12. Did you conclude that hydration of aconitic acid could produce either citric acid or isocitric acid? (5 points) (p.218).
13. Did you recognize that isocitric acid can undergo either oxidation and decarboxylation to give α-ketoglutaric acid or a retrograde enolate condensation to give succinic acid and glyoxylic acid? (5 points) (p. 218).
14. Did you recognize that oxidation of isocitric acid adds a third oxidation step to the citric acid cycle? (3 points) (Fig 9.14, p 219).
15. Did you recognize that α-ketoglutaric acid  can undergo oxidative decarboxylation  to give succinyl-CoA and carbon dioxide? (5 points)(Fig  9.16,  p. 219)
16. Did you recognize that succinyl-CoA can be converted  to succinic acid by either hydrolysis or substrate phosphorylation? (2 points)(Figs 9.17 and 9.18, p. 220)
17. Did you write a possible citric acid cycle at this point? (10 points)(Fig 9.19, p. 221)
18. Does this cycle agree with the balanced reaction  you wrote at the start? (3 points)(Fig 9.21, p.221)
19. Did you recognize that a retrograde enolate condensation of isocitric acid produces an alternative   citric acid cycle. (5 points)(section 9.6,  p. 223)
20. Did you recognize that a second molecule of acetyl-CoA could react with glyoxylic acid to form malic acid? (10 points)(Fig  9.27, p. 225)
21. Did you recognize that condensation of acetyl-CoA with glyoxylic acid completes another alternative  citric acid cycle? (3 points) (Fig 9.28, p. 225)
22. Did you conclude that neither of the alternative  cycles agree with the information you were given or with the balanced reaction you wrote at the start?  (3 points)

We will explore these reactions in more detail in Problem Sets 10 and 11.

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Problem Set 8 with answers

ANSWERS TO PROBLEM SET 8.
Oxidation of Fatty Acids

The first seven Problem Sets dealt with the fundamental principles you need to predict the products of any biochemical reaction. In Problem Sets 8 to 25, you will use these principles to predict the intermediates in complicated metabolic pathways.
I am having trouble getting the necessary structures into answer for problem 8. Sorry!

Given: Fatty acids with an even number of carbon atoms are oxidized to acetyl-CoA.

1. Predict the intermediates in this pathway.
HINT: “In solving a problem of this sort, the grand thing is to be able to reason backwards. That is a very useful accomplishment, and a very easy one, but people do not practice it much.” Sherlock Holmes (From A Study in Scarlet by A. Conan Doyle)
ANS:
a. Did you write a balanced equation? (CBM p.192)
b. What does the balanced equation tell you? (CBM p. 192)
c. Did you work backwards as suggested by the world’s greatest detective?
d. What is the justification for working backwards? (rule 6.2, p. 130)
e. Did you use electron pushing to predict the intermediates?
It is very important that you proceed stepwise using the rules of electron pushing established in Chapter 1. If you just write down an intermediate from previous knowledge, you are using memory recall. The purpose of these problem sets is to give you practice in working out reactions logically rather than relying on memory.
f. Did you arrive at diagram 8.12 shown on p. 202 of CBM?
g. Did you recognize the fundamental reaction types occurring in each case?

ANS: (see Fig 8.12)

2. Write two different mechanisms for the initial reaction. From a metabolic point of view, which is the most likely?

ANS:
a. Did you write the mechanisms shown in Figs. 8.9 and 8.10 of CBM?
b. Did you choose the second mechanism because hydrolysis of PPi pulls the reaction to completion? (see p. 201 of CBM)

3. Look in the databases( http://www.chem.qmul.ac.uk/iubmb/ and (http://www.brenda-enzymes.info/) to see if you can find the enzyme required for the reactions you have written.
ANS:
a. Did you find the enzymes shown in Fig. 8.14?
b. Did you confirm your choice for the first reaction?

4. How many ATPs are produced by each turn of the oxidation spiral?
ANS:
There are two oxidation steps. When you looked up the enzymes, did you find that one uses FAD and the other uses NAD? You know that oxidation of FADH2 by the electron transport system produces 2 ATPs and that oxidation of NADH produces 3 ATPs. Did you conclude that each turn of the cycle produces 5 ATPs?

5. How much energy is produced by oxidation of stearic acid to acetyl-CoA?
ANS:
a. Stearic acid has 18 carbons. Did you conclude that it takes (18/2) – 1 = 8 turns of the cycle to convert stearic acid to acetyl-CoA?
b. One ATP is used to initiate beta oxidation. Did you conclude that oxidation of stearic acid to acetyl-CoA produces (5 x 8) –1 =39 ATPs?
c. Each ATP is stores 7,300 cal/mol at pH 7 (Table 1, p. 135 of CBM). Did you conclude that oxidation of stearic acid to acetyl-CoA produces 39 x 7,300 = 285,000 cal/mol

6. Fatty acyl dehydrogenase produces the trans isomer. What can you conclude about how the enzyme works?
ANS:
a. Did you conclude that either two HR’s or two HS’s must be removed by the enzyme?
b. Did you conclude that these hydrogens can be either syn periplanar or anti periplanar so we cannot predict whether the elimination will be syn or anti?

7. Naturally occurring fatty acids are always cis. What problem does that present for oxidation of oleic acid (9-cis-octadecenoic acid)?
ANS:
Enoyl hydratase (EC 4.2.1.17) requires a trans double bond. The cis double bond produceby three oxidation steps (Fig. 8.20 of CBM) must isomerize to trans.

8. Write two different mechanisms for isomerization of a double bond in a fatty acid.
Which of these mechanisms works in the oxidation of oleic acid? (HINT: Look for the appropriate enzyme(s).
a. Redox:

b. Removal and replacement of an α-hydrogen:

The enzyme, dodecenoyl-CoA Δ3-isomerase (EC 5.3.3.8) catalyzes mechanism “b”.

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Answers to Problem Set 7

PROBLEM SET 7
Oxygen and Transition Complexes

This problem set continues with principles electron transport. It focuses on the structure of oxygen and transition metal complexes.

1. Electrons are passed one at a time through the electron transport system of oxygen. Using your MO diagram for oxygen (Problem Set 6), predict the types of molecules that will form when electrons are added one at a time.

ANS: See p. 174 section 7.12 of The Chemical Basis of Metabolism for the answer.

2. Iron is a transition metal. It has 18 electrons. Two of the valance electrons are in the 4s orbital. The remainder are in the 3 d orbitals. Show the orbital distribution of the valance electrons of Fe and Fe+2.
Hint: The 4s electrons are lost first.

ANS: See Fig. 7.27

3. Explain how Fe+2 can have coordination numbers of either 4 or 6.
ANS: Fe+2 has an empty 4s orbital and three empty 4p orbitals. The energy levels of these atomic orbitals are close enough so they can hybridize to form four sp3 orbitals that can combine with four ligands, each carrying an electron pair, to form four bonding and four antibonding MOs. Since sp3 orbitals for a tetrahedron, the coordination complex will be tetrahedral.
Alternatively, all the electrons in the 3d orbitals can be paired. This leaves two empty 3d orbitals. The 3dx2-y2 and3 dz2 orbitals have the correct symmetry to combine with the 4s and 4p orbitals to form six hybrids with octahedral geometry. These hybrids can combine with six ligands to give a coordination number of 6.

4. Desulforedoxin type proteins, which occur in complex I, and rubredoxin type proteins, which are found in complex II, form a transition complex with either ferrus or ferric ions and four cysteine residues from the proteins.
a. Show what these complexes look like.
b. What do the roman numerals mean?
c. What are the charges on the ferrus and ferric complexes?
d.Explain how oxidation/reduction can occur without disturbing the coodination complex.

ANS:
4a. See CBM Fig 7.29
4b. The roman numerals indicate the oxidation level.
4c. Each cysteine residue contributes a negative charge (total –4). Ferrus ion (FeII) is +2. Therefore, the ferrus complex has a net charge of –2. Ferric ion (FeIII) has a charge of +3. Therefore, the ferric complex has a net charge of –1.
4d. The coordination complex is formed from four 4s4p hybrid orbitals. Fe+2 has 7 valance electrons; Fe+3 has 6. They are located in 3d orbitals:

Since the valance electrons are not involved in making the coordination complex gain or loss of a valance electron does not disrupt the coordination complex.

5. Some of the iron-sulfur proteins in the electron transport system are of the general type Fe2S2. They use an inorganic sulfide ion and four cysteine residues to form a coordination complex. Show what the complex look like.

ANS: See Fig. 7.30 of CBM

6. Cytochrome c is an iron phorphyrin protein. It uses methionine 80, histidine 18 and the porphyrin ring of protophorphyrin IX to form a transition complex. Disregarding the side chains of the porphyrin ring, show what the complex looks like.

ANS: See Fig. 7.38

7. Copper is a transition metal with 11 valance electrons. Show the electron distribution of Cu, Cu+ and Cu++.

ANS: See Fig. 7.28 of CBM

8. Cytochrome oxidase uses His 181 and 224, Met 221, the peptide bond of Glu 218, CyS 216 and 220 to for a Cu2CyS2 type coordinate complex with either Cu+ or Cu++. Show what the complex looks like.

ANS: See section of 7.18 of BM

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ANSWERS TO PROBLEM SET 6-Energetics

 

All the answers refer to the 2nd Edition of  CBM

1. Assume the energy of A>B and  A = ΔG1, B = ΔG2. Draw an energy diagram that shows the progress of A→Β. Label both axes. Show ΔG and ΔG* when

a. There are no unstable intermediate. (3 points)

ANS: Fig 6.3, p.129

b. When there is an unstable intermediate. (3 points)

ANS: Fig. 6.7, p. 131

2. How does a transition state differ from an unstable intermediate? (3 points)

ANS:

A transition state occurs at an energy maximum. An unstable intermediate occurs at an energy minimum that is higher than the final energy level.  (Fig. 6.7, p.131)

3. Define the Principle of Microscopic Reversibility. (3 points)

ANS: Rule 6.2, p. 132

4. What does the following equation tell us? (By convention, the energy level of the reactants is subtracted from the energy level of the products).

ΔG = ΔGº + RT ln[P]/[R]

where R = 1.986 cal/mol/ºK  (8.314 joules/mol/ºK, T = ºK,  [P] = molar concentration of the product(s), [R] = molar concentration of the reactants

ANS:

a.  ΔG is negative for a spontaneous reactions (by convention). (Section 6.3, p 132) (2 points)

b. The reaction is as equilibrium when ΔG = 0 (Rule 6.4) (2 points)

c. At equilibrium,  ΔGº = – RT ln Keq  (Equation 2, p. 133) (2 points)

This equation  tells us that if we measure the equilibrium concentrations of reactants and products at a specified temperature and pressure, we can evaluate  ΔGº (R = 1.987 calories/mol/degree K). Indeed, this is the way many of the ΔGº values we find in the literature were measured.

d. When a reaction is at equilibrium, the ratio of concentrations of reactants to those of the products is a constant that is characteristic of the reaction in question at a specified temperature and pressure. (Rule 6.5, p.133) (2 points)

5. What is the difference between ΔG, ΔGº, ΔG’ and ΔGº’?

ANS:

ΔG = the free energy under the actual reaction conditions (2 points)

ΔGº = the free energy under standard conditions (2 points)

ΔG’ = the free energy of the actual reaction concentrations  when the pH = 7.0 (2 points)

ΔGº’ = the free energy under standard conditions except pH = 7.0. (2 points)

6. Define standard thermodynamic conditions. (4 points)

ANS:

T = 273º K (0º C), [reactants] = [products] = 1 M.

7. How many cal/mol are available when [H+] is diluted from 1 M to 10-7 M at 25º?  (10 points)

ANS:

RTln {[10-7]/[1]} = (1.986)(273º + 25º) (-1.612) = -9,539 cal/mol.

8. Explain why ATP has such a high ΔGº’ (- 7,l300 cal/mol) of hydrolysis. (5 points)

ANS:

This is primarily an entropy effect due to dilution of protons. (section 6.10, p.143)

9. Explain why phosphoenol pyruvate has such a high ΔGº’ (- 13,800 cal/mol) of hydrolysis. (5 points)

ANS:

This is primarily an entropy effect involving keto and enol tautomers.(section 6.8, p. 141)

10. Write a mechanism for the formation of glucose 6-phosphate from glucose and ATP. (Phosphates usually react by substitution) (4 points)

ANS: Fig. 6.25, p. 150

11. Define a coupled reaction and give an example. (4 points)

ANS:

Reactions with a common intermediate are said to be coupled. (Rule 6.8, p. 152)

12. What is the difference between a “closed system” and an “open system”?  (5 points)

ANS:

A closed system exchanges heat but not mass with the environment. An open system exchanges both heat and mass with the environment. (section 6.16)

13. Assume the pathway A→ B→C→D occurs in an open system at steady state at 37º C where

 

a. From these data what can you conclude about the pathway under standard conditions?

ANS:

The standard free energy of B→C is positive so this reaction cannot proceed under standard conditions. (4 points)

b. Calculate ΔG’ for each reaction at 37ºC if the steady state concentrations of [A] = 6mM and [D] = 0. 1 mM.  [Remember all concentrations must be M, T is in ºK and R = 1.986 cal/molºK]

ANS:

The velocity of an enzymatic reaction is given by: v= Vmax[S]/{KM + [S]} We are given that [A] = 0.006 M. Therefore, the initial velocity of A→B is v = 90(0.006 M/(5 x 10-3 M + 0.006 M) = 49 arbitrary units. (4 points)

At steady state all reactions run at the same velocity. Therefore, 49 = 80[B]/(0.01M + [B] and [B] =0.016 M (4 points)

also 49 = 100[C]/(10-3 + [C]) and [C] ≈ 0.001M (4 points)

Using these concentrations, we can calculate ΔG’ for each reaction. (T able 3, p..157) ( 3 points for each reaction = 9 points)

The discussion also shows how to determine the concentrations graphically.

14. The Nomenclature Committee of the International Union of Biochemistry has defined ligases as “—enzymes catalyzing the joining of two molecules coupled with the hydrolysis of a pyrophosphate bond in ATP or a similar triphosphate. The bonds formed are often “high-energy bonds”. (p. 153)

What is wrong with this definition? (10 points)

ANS:

Ligases catalyze coupled reactions between two substrates and ATP or a similar triphosphate. The common intermediate is usually a mixed  anhydride derived from the reaction of ATP and one of the starting materials. The mixed anhydride then reacts with the second starting material to form the final product. Hydrolysis of ATP is never involved. Furthermore, the term “high energy bond” is unacceptable in this context since most of the energy in a compound of this sort is due to entropy and is not concentrated in any bond p. 153).

Grade Yourself:  90-100 = A, 75-89 = B, 50-74 = C, 25-49 = D, <25 = F

 

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