19-452 Metabolic Processes

Dr. R.A.B. Keates

Answers to Mid-Term Examination, Wednesday October 31, 1990

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Answer seven questions (5 marks each) for a maximum of 35 marks.

1. Use the tables of free energies of hydrolysis to determine the standard free energies of the reactions:

MgATP^2- + H₂O MgADP^- + Pi^2- + H⁺ ; -30.3 kJ mol^-1 (equivalent to UTP)
UDP^3- + glucose + H⁺ UDPglucose^2- + H₂O ; +30.5 kJ mol^-1
Glc-1-P^2- + H₂O glucose + Pi^2- ; -20.9 kJ mol^-1
2 Pi^2- + H⁺PPi^3- + H₂O ; +33.4 kJ mol^-1


---

UTP + Glc-1-P UDPGlc + PPi
= - 30.3 + 30.5 - 20.9 + 33.4 kJ/mol = +12.7 kJ/mol

UDPglucose^2- + H₂O UDP^3- + glucose + H⁺ ; -30.5 kJ mol^-1
glycogen_n-1 + glucose glycogen_n + H₂O ; +16.8 kJ mol^-1

---

Glycogen_n + UDPGlc Glycogen_n+1 + UDP
= - 30.5 + 16.8 kJ/mol = -13.7 kJ/mol

for the first reaction is positive. What factors drive this reaction from left to right?

The production of inorganic pyrophosphate PPi in the first reaction provides a hidden coupling to the action of pyrophosphatase. Pyrophosphatase is present at high activity levels in cells, so PPi is rapidly broken down, and the intracellular concentration is exceedingly low, possible as little as 10^-11 M. This makes a huge negative contribution to the concentration term RT ln [products]/[reactants] so that also ends up negative. RT ln 10^-11 = -62 kJ /mol.

In addition, the two reactions presented here occur consecutively in glycogen synthesis. The sum of the two steps has net negative so can be favourable overall even if one step is unfavourable. In this example, the unfavorable step preceeds the favorable, so the pair of reactions can proceed with a very low concentration of the intermediate UDP-glucose.

2. What enzymes are required for the formation of linoleyl-SCoA (18:2 ^9,12) after the normal fatty acid synthase releases its product?

Fatty acid synthase releases palmitate. The reactions needed to make linoleyl-CoA are

Activation to CoA derivative by fatty acyl CoA synthetase (synthetase implies ATP hydrolysis
palmitate + HSCoA + ATPpalmityl-CoA + AMP + PPi

The reactions of the elongase complex
ketoacyl synthase
palmityl-CoA + malonyl-CoA3-ketostearyl-CoA + CO₂ + HSCoA
ketoacyl reductase
3-ketostearyl-CoA + NADPH + H⁺ 3-hydroxystearyl-CoA + NADP⁺
dehydratase
3-hydroxystearyl-CoA ³octadecenoyl-CoA + H₂
enoyl CoA reductase
octadecenoyl-CoA + NADPH + H⁺ stearyl-CoA + NADP⁺

Desaturation mono-oxygenases
mid-chain specific desaturase
stearyl-CoA + O₂ + NADH + H⁺ oleyl-CoA (18:1 ⁹) + NAD⁺ + 2 H₂O
far-chain desaturase (found in higher plants only)
oleyl-CoA + O₂ + NADH + H⁺ linoleyl-CoA (18:2 ^9,12) + NAD⁺ + 2 H₂O

Note that NADH is used as a reducing agent (acting on O₂) even though desaturation is an oxidation reaction.

What steps beyond normal ß-oxidation are required for the breakdown of linoleyl-SCoA?

Enoyl-CoA isomerase can convert ^{3 cis} into ^{2 trans} needed for the ß-oxidation hydratase.
12:2 ^{3, 6 cis} dienoyl-CoA 12:2 ^{2 trans, 6 cis} dienoyl-CoA

At the the start of ß-oxidation cycle 5, 8:2 ^{2 trans, 4 cis} dienoyl CoA will form.
Dienoyl-CoA reductase reduces the diene to ^{3 cis} enoyl CoA.
8:2 ^{2 trans, 4 cis} dienoyl-CoA + NADPH + H⁺ 8:1 ^{3 cis} enoyl-CoA + NADP⁺

Another round with enoyl-CoA isomerase gives

3. Biochemical pathways operate under steady state conditions rather than equilibrium. Explain the distinction.

Equilibrium refers to a state in a single reversible reaction in which the rate of forward reaction is exactly equal to the rate of the reverse reaction. A reaction at equilibrium gives no net conversion of substrate to product.

Steady state refers to a series of consecutive reactions in which the net rate of production of intermediate X is exactly equal to the net rate of consumption of X by the next step (rather than by reversal of the production reaction). In a steady state, the net conversion or flux rate is the difference between forward and reverse rates reaction of reaction at any step. It is a general principle of steady state systems that the flux rate is the same for all steps of an unbranched pathway.

If initially, the rate of production of intermediate X were to exceed the rate of consumption, the concentration of X would gradually rise, speeding up the consumption reaction until both match. If the rate of consumption were to exceed the rate of production, the concentration of X would gradually fall, causing the rate of consumption to drop back. Thus there is an inherent tendency to assume a stable flux rate for all steps of the pathway.

In a hypothetical pathway

Steady state reaction have net conversion of A to D, but true equilibrium implies no net conversion. Reactions may be far from equilibrium so that forward rate = flux rate, reverse rate = 0, or close to equilibrium, implying that the reverse rate is almost equal to the forward rate. The net difference between forward and reverse rates must be equal to the flux rate.

If an analysis is performed to determine concentrations [A], [B], [C], [D] what clues would you look for to detect a potential regulatory step?

The only way to reverse the direction of the non-equilibrium steps is to provide an alternative reaction with different cofactors so that the new reaction is favoured in the opposite direction. An enzymes can't shift the position of reaction equilibrium, nor can inhibitors or activators selectively change the catalysis in one direction but not the other. So we have to reverse the action of phosphofructokinase (which consumes ATP) by using fructose bisphosphatase, which does not regenerate ATP.

In the test tube, glyceraldehyde-3-phosphate dehydrogenase (glycolysis/gluconeogenesis pathway) is inhibited by one of its products, NADH. Why is this regulation considered irrelevant in vivo?

4. The reactions of ß-oxidation are clearly directed toward breakdown of long chains into acetyl-SCoA. What strategies are used by fatty acid biosynthesis to reverse the net direction of the cycle?

a) The use of NADPH provides a more powerful reducing agent. For the enoyl-ACP reductase, NADPH replaces FAD, which has a much less negative reduction potential (more negative implies a better reducing agent). NADPH is also present in large excess over NADP⁺, and this contributes to a more negative value of for reduction of the substrate ([NADPH] will be on the reactant side). NAD⁺ is present in excess over NADH, enhancing its reaction as an oxidizing agent, as needed in ß-oxidation.

b) The condensation step must reverse the energetically favourable thiolase reaction of ß-oxidation. By using malonyl-CoA as the condensing agent, the condensation is coupled to decarboxylation of the malonyl group. This favours the reaction in two ways: carbon dioxide is a lower energy structure than carboxylate due to resonance stabilization, so the change makes a negative contribution (typically about -30 to -35 kJ/mol) to H and thus to also.

Reactions occur in an atmosphere which is typically 300 ppm CO₂, so if CO₂ is a reaction product, it is present orders of magnitude less than standard state, and this makes a negative contribution to .

RT ln 0.00003 = -25.8 kJ/mol.

The energy of one ATP was invested in making malonyl CoA; this investment is returned by making the condensation more favourable.

Note that we are making a reaction more favourable not by selective activation of the forward direction or selective inhibition of the reverse reaction, both of which are impossible, but by changing the chemical nature of the reaction by including the decarboxylation.

What is the benefit of using Acyl Carrier Protein instead of coenzyme A? Why does the use of acyl carrier protein require three steps to replace the single thiolase reaction of ß-oxidation?

This makes the fatty acid synthase catalytically more efficient, but has no influence on the direction. Comment: the full fatty acid biosynthesis process was included in 19-452 in 1990. This is now handled in Chem*3560.

A round of ß-oxidation makes 5 ATP, after NADH and UQH₂ are utilized for oxidative phosphorylation. How many ATP equivalents are needed to elongate a growing fatty acid chain by two carbons in one complete cycle of fatty acid synthase? (To compare the ATP equivalence of NADPH and NADH, compare the interconversion of malate and pyruvate via malate dehydrogenase and malic enzyme.)

NADPH is not utilized by oxidative phosphorylation so it's energy equivalence in ATP units must be estimated by other means.

How many ATP are needed to transfer the reducing electrons of NADH to NADP⁺?

pyruvate carboxylase
pyruvate + ATP + CO₂ oxaloacetate + ADP + Pi

malate dehydrogenase (in reverse)
oxaloacetate + NADH malate + NAD⁺

malic enzyme (or malate dehydrogenase (decarboxylating) by its full name)
malate + NADP+ pyruvate + CO₂ + NADPH

Hence NADPH is energetically equivalent to NADH + 1 ATP. This difference is entirely due to the different concentration ratios [NAD(P)H] / [NAD(P)⁺] since the standard reduction potentials are almost identical.

In 1990 we were counting 3 ATP per NADH and 2 per FADH₂ or UQH₂. Hence 2 NADPH as needed per fatty acid synthase cycle would be tallied as 8 ATP equivalents. One additional ATP is needed for citrate lyase, and one more for acetyl CoA carboxylase, for a total of 10 ATP consumed per fatty acid synthase cycle.

These days there's substantial evidence that each NADH generates 2.5 ATP and each FADH ₂ or UQH₂ generates 1.5 ATP in oxidative phosphorylation, but the subject is still open to debate.

5. Summarize the regulatory steps that direct the carbohydrate from a donut to fatty acid biosynthesis in a well fed, underexercised person. Assume that starch is converted to glucose in the digestive tract and distributed to the body in the blood.

A well fed, underexercised person has higher carbohydrate intake than energy demand. Regulation of key glycolytic enzymes in muscle is strongly influenced by AMP levels, so glycolysis activity is low. Once adequate glycogen is stored, glucose-6-P is unused and builds up. The peripheral tissue hexokinase is subject to product inhibition by glucose-6-P, so glucose levels rise to their maximum, limiting addtional uptake from blood.

Meanwhile glucose incorporation in the liver is dominated by glucokinase, which is not product inhibited, to the excess glucose is metabolized almost exclusively by the liver. In addition, liver phosphofructokinase is positively regulated by fructose-2,6-bisphosphate, which is produced when fructose-6-phosphate levels rise. Thus glycolysis maintains activity in the liver, even though energy demand is low, and pyruvate is produced.

At what point are metabolites redirected from the usual breakdown pathways into biosynthetic pathways?

Pyruvate enters the liver mitochondria, and may be directed either to pyruvate dehydrogenase to make acetyl CoA or to pyruvate carboxylase to maintain oxaloacetate levels. In both cases, citrate is formed, and some is converted to isocitrate by aconitase.

Mitochondrial isocitrate dehydrogenase is inhibited by high ATP, but because the aconitase equilibrium is moderately unfavourable, citrate accumulates. High citrate levels (>10 mM) allow efflux from mitochondria to cytoplasm, which is the site of fatty acid biosynthesis. In the cytoplasm, citrate breaks down releasing acetyl CoA which is the starting point for fatty acid (and cholesterol) biosynthesis. Presence of citrate in the cytoplasm activates acetyl-CoA carboxylase, which is a key commitment and regulatory point for the fatty acid biosynthesis pathway.

In many cases, end products exert a negative effect on the initial steps of a pathway, and long chain acyl CoA is a negative regulator of acetyl CoA carboxylase. Unfortunately, the liver uses up the acyl CoA for triacylglycerol synthesis, so the negative regulation fails to keep the reaction in check, and fatty acid biosynthesis can proceed limited only by substrate availability

6. How is the phospholipid phosphatidyl choline formed i) if choline is being recycled, and ii) if free choline and ethanolamine are unavailable.

This is material which is now covered in Chem*3560.

acyl transferase:
Acyl CoA + glyceryl-3-P monoacylglyceryl-3-P + HSCoA
Acyl-CoA + monoacylglyceryl-3-P diacylglyceryl-3-P + HSCoA

The recyling pathway:
diacylglycerol phosphatase or phosphatidate phosphatase
diacylglyceryl-3-P + H₂O diacylglycerol + Pi

phosphocholine transerase
CDP-choline + diacylglycerol phosphatidyl choline + CMP

choline kinase
Choline + ATP phosphocholine + ADP

Phosphocholine cytidylyl transferase
CTP + phosphocholine CDP-choline + PPi

The de novo synthesis pathway in animals
phosphatidylethanolamine + serinephosphatidylserine + ethanolamine (can be recycled)
phosphatidylserine decarboxylase
phosphatidylserine phosphatidylethanolamine + CO₂
S-adenosylmethionione methyltransferase (in 3 separate steps)

phosphatidylethanolamine + 3 S-adenosylmethionine

phosphatidylcholine + 3 S-adenosylhomocysteine.

The de novo synthesis pathway in bacteria
phosphatidate cytidylyl transferase
diacylglyceryl-3-P + CTP CDP-diacylglycerol + PPi
phosphoserine synthase
CDP-diacylglycerol + serine phosphatidyl-serine + CMP
then as for the de novo pathway in animals.

7. If [pyruvate] is 50 µM, pCO₂ is 0.05 atm, [ATP] is 1.0 mM, [ADP] is 0.20 mM and [Pi] is 1.0 mM in the mitochondrion, what is the maximum concentration of oxalacetate that may accumulate as a result of the pyruvate carboxylase reaction? What is the consequence to the direction of malate dehydrogenase in the mitochondrion?

The carboxylation of pyruvate to oxaloacetate

		pyruvate-	+ CO2		H+ +	oxaloacetate2-
		- 474.5 kJ/mol	- 394.4 kJ/mol		- 39.9 kJ/mol	- 797.2 kJ/mol	f values

For direct carboxylation: = +31.8 kJ/mol. For carboxylation with ATP hydrolysis: = -30.3 + 31.8 = +1.5 kJ/mol The maximum [oxaloacetate] is obtained when the reaction reaches equilibrium, so is set to zero:

Comment: the value for pCO₂ is high compared with the atmosphere, but could be found in the mitochondrion, when CO₂ is actively being produced by TCA oxidation.

The presence of such high levels of oxaloacetate would force malate dehydrogenase, which has a large positve standard free energy change in the oxidation direction, into reverse so that malate would accumulate and be exported from the mitochondrion to the cytoplasm.

8. Estimate the overall standard free energy for conversion of 3 moles of ribulose-5-P into two moles of fructose-6-P and one mole of glyceraldehyde-3-P by the pentose phosphate pathway.

3 ribulose-5-P				2 fructose-6-P + glyceraldehyde-3-P
3 x -1597.6				2 x -1758.3 - 1285.6		moles x f

Overall = -9.4 kJ/mol.

What is the minimum [ribulose-5-P] required for this conversion if [Fru-6-P] = 15 µM and [glyceraldehyde-3-P] = 20 µM?

The minimum [ribulose-5-P] is the reactant concentration that just brings the reaction to equilibrium, so is set to zero:

9. Given values for the reduction potential of oxalacetate to malate and appropriate values from the table of_f, estimate a value for the reduction potential E' for the half reaction:

pyruvate^- + CO₂ + 2 e^- + H⁺ malate^2-

Strategy:
i) use reduction potentials to estimate for oxaloacetate reduction using NADPH.
ii) use _f to estimate for carboxylation of pyuvate (without ATP)
iii) Sum the two to get the free energy change for carboxylating pyruvate to malate with NADPH.
iv) Convert back to E^o', then add back the reduction potential of NADP⁺ to get the half reaction potential for the equation above.

step i:
E^o' = -0.166 - (- 0.324) = 0.158 V
= -nFE^o' = -2 x 96.485 x 0.158 = -30.48 kJ/mol

step ii:

		pyruvate-	+ CO2		H+ +	oxaloacetate2-
		- 474.5 kJ/mol	- 394.4 kJ/mol		- 39.9 kJ/mol	- 797.2 kJ/mol	f values

For direct carboxylation without ATP: = +31.8 kJ/mol.

step iii:
for the equation above driven by NADPH reduction, overall = -30.48 + 31.8 = +1.32 kJ/mol

step iv:
E^o' = -/nF

E^o' = - 1.38 / 192.97 = -0.007 V

E^o'_{half rxn} = -0.007 - 0.324 V = -0.331 V.

Comment: we did not deal with reduction potentials this year.

Given your estimate of E,' is malate (with the help of malic enzyme for catalysis) capable of driving the reduction of NADP⁺ to a large value of [NADPH]/[NADP⁺]?

The standard reduction potential of carboxylation of pyruvate and its reduction to malate is more negative than the reduction potential for NADP⁺, so malate will tend to be oxidized and NADP⁺ will tend to be reduced. However, the difference is small, so the reaction is only marginally favourable. What does allow the reaction to proceed to generate excess NADPH is that CO₂ is a product in the favoured direction. The overall direction will be governed by the relative level of CO₂ in the environment, which tends to be low, allowing a large excess of NADPH to be formed before equilibrium is reached.

10. Complete the reaction mechanisms below. Return the sheet with the rest of your answers.

a)
b)
c)
d)
e)

11 a)

Which of the two mechanisms shown on the right is correct? Briefly explain why the other is wrong. Draw the product of this step of the reaction.

The mechanism on the right is correct, because it shows proton loss from the hydroxyl group and hydride transfer from the CH to the pyridine ring of NAD⁺.

The representation on the left breaks and forms the appropriate bonds, but removes a hydride ion into the medium, which is not realistic in an aqueous environment, and shows electrons leaving a positive center at the N-atom of pyridine. No electrons are transfereed to the pyridine along with the proton, so this would not be a reduction of the pyridine.

11 b)

Which carboxylate is most easily removed by the reaction initiated by the arrow shown? Draw the immediate product of this reaction, and identify it. What pathway uses this reaction, and what additional step completes the reaction?

The natural pathway for electron flow results in easy loss of the lower carboxylate. Attempting to remove the upper carboxylate would result in an electron deficient neutral C atom, or a carbanion bonded to the O, both of which ar chemically improbable processes.

This represents the decarboxylation of oxaloacetate, which produces enolpyruvate as the immediate product.

When this reaction occurs in PEP carboxykinase, the enolpyruvate is relatively easily phosphorylated to give PEP. In contrast, phosphorylation of the normal keto- form of pyruvate is energetically difficult because of the unfavourable keto-enol transition.

PEP carboxykinase produces PEP for gluconeogenesis.

Table of standard free energies of hydrolysis, .

MgATP^2- + H₂O MgADP^- + Pi^2- + H⁺ ; -30.3 kJ mol^-1

ATP^4- + H₂O AMP^2- + PPi^3- + H⁺ ; -37.4 kJ mol^-1

ADP^3- + H₂O AMP^2- + Pi^2- + H⁺ ; -36.3 kJ mol^-1

AMP^2- + H₂O adenosine + Pi^2- ; - 9.6 kJ mol^-1

PPi^3- + H₂O 2 Pi^2- + H⁺ ; -33.4 kJ mol^-1

PEP^3- + H₂O pyruvate^- + Pi^2- ; -61.9 kJ mol^-1

3-P-glyceroyl-1-P^4- + H₂O 3-P-glycerate^3- + Pi^2- + H⁺ ; -49.3 kJ mol^-1

Glc-1-P^2- + H₂O glucose + Pi^2- ; -20.9 kJ mol^-1

Fru-6-P^2- + H₂O fructose + Pi^2- ; -14.0 kJ mol^-1

Glc-6-P^2- + H₂O glucose + Pi^2- ; -13.8 kJ mol^-1

glycerol-P^2- + H₂O glycerol + Pi^2- ; - 9.2 kJ mol^-1

glycogen_n + H₂O glucose + glycogen_n-1; -16.8 kJ mol^-1

UDPglucose^2- + H₂O UDP^3- + glucose + H⁺ ; -30.5 kJ mol^-1

acetyl-SCoA + H₂O acetate^- + HSCoA + H⁺ ; -36.6 kJ mol^-1

succinyl-SCoA^- + H₂O succinate^2- + HSCoA + H⁺; -43.5 kJ mol^-1

Table of standard reduction potentials, E'.

acetate^- + 2e^- + H⁺ acetaldehyde -0.58 V

-ketoglutarate^2- + CO₂ + 2e^- + H⁺ isocitrate^3- ; -0.38 V

NADP⁺ + 2e^- + 2H⁺ NADPH + H⁺ ; -0.324 V

NAD⁺ + 2e^- + 2H⁺ NADH + H⁺ ; -0.321 V

1,3-bisP-glycerate^4- + 2e^- glyceraldehyde-3-p^2- + Pi^2- ; -0.29 V

3-ketoacyl-SCoA + 2e^- + 2H⁺ 3-hydroxyacyl-SCoA; -0.238 V

pyruvate^- + 2e^- + 2H⁺ lactate^- ; -0.185 V

oxalacetate^2- + 2e^- + 2H⁺ malate^2- ; -0.166 V

enoyl-SCoA + 2e^- + 2H⁺ acyl-SCoA ; -0.015 V

ubiquinone + 2e^- + 2H⁺ ubiquinol ; +0.010 V

fumarate^2- + 2e^- + 2H⁺ succinate^2- ; +0.031 V

O₂ + 4e + 4H⁺ 2 H₂O ; +0.816 V

F Faraday's constant = 96,487 coulombs mol^-1 or J mol^-1 V^-1

RT = 2.478 kJ mol^-1