Chemistry 301 Introduction to Biochemistry

Study Guide :: UNIT 6

Metabolism I

Overview

There are virtually no biochemical reactions in a cell that are not connected to other reactions. This statement makes sense if you realize that reaction A will consume certain biomolecules (reactants), and produce other biomolecules (products). Unless there is a process to produce the reactants for reaction A, the reaction will not occur. Unless there is another process to consume the products of reaction A, these products will accumulate and overload the system. The complex sequences of chemical processes that produce products and consume reactants, are called “metabolic pathways.” This unit provides an overview of metabolism and then presents a number of metabolic pathways needed for cellular function.

Unit 6 is divided into six lessons:

Learning Objectives

After completing this unit, you should be able to:

1. Explain the meaning of the term “metabolic pathway.”
2. List the principle characteristics of metabolic pathways, and explain the purpose of each of the two major types of pathways.
3. Identify the types of chemical reactions that are important in biochemistry.
4. Describe the three stages of glucose metabolism.
5. Explain the major mechanisms by which the three stages of glucose metabolism are regulated.
6. Describe fermentation.
7. Compare glycolysis and gluconeogenesis.
8. Describe the citric acid cycle.
9. Outline the purpose of the glyoxylate cycle.
10. Give an general overview of acetyl‑CoA and cholesterol metabolism.
11. Discuss fatty acid oxidation and synthesis.
12. Demonstrate an understanding of ketogenesis.

Glossary

Lesson 1: Overview of Metabolic Concepts

Overview/Objectives

After completing this lesson, you should be able to:

1. Explain the meaning of the term “metabolic pathway.”
2. List the principle characteristics of metabolic pathways, and explain the purpose of each of the two major types of pathways.
3. Identify the types of chemical reactions that are important in biochemistry.
4. Define “irreversible metabolism” and “steady state” as they apply to metabolism.
5. Explain the source of the “energy” in high‑energy compounds.
6. Relate the metabolic pathways in eukaryotic cells to the specific cellular locations of these metabolic events.

Readings and Activities

1. In Chapter 6 Metabolism I, Read the introduction, and “Definitions” and “Perspective” (pages 141–143 of the textbook).
2. You can also watch three video lectures on Metabolism and Glycolysis:
   • #21 Biochemistry Glycolysis Lecture for Kevin Ahern’s BB 450/550
   • #22 Biochemistry Glycolysis II Lecture for Kevin Ahern’s BB 450/550

   • #23 Biochemistry Glycolysis III Lecture for Kevin Ahern’s BB 450/550

     (Links to these are also provided on page 142 of the textbook.)

Commentary

Metabolic Pathways

A major metabolic pathway in vivo is the degradation of glucose (a sugar) to pyruvate. Under normal conditions, the body regulates the concentration of glucose in the blood, keeping it constant at ~5 mM. Glucose flows into a cell and is converted to pyruvate in the cytoplasm. There are ten sequential reactions between glucose and pyruvate (see the figure on page 144). This sequence or chain of reactions is called a metabolic pathway. The concentrations of the ten “products” do not vary, because this metabolic pathway is controlled by:

• how fast glucose flows into the cell
• how quickly pyruvate can be used up
• whether any of the ten “products” are diverted to other uses
• the concentration of the catalysts that control the ten reactions
• feedback loops in which products of the ten reactions regulate preceding reactions.

Metabolic pathways are degradative (catabolism or catabolic pathways) or synthetic (anabolism or anabolic pathways). Usually, in catabolic pathways, larger molecules are broken down to smaller ones with a release of energy which the cell can harvest as ATP. In anabolic pathways, energy is consumed to produce larger molecules from small precursors.

One or more of the intermediate products in a metabolic pathway can be diverted to enter another pathway. It is a little like rush hour. Cars will flow down a street (pathway) only if a more convenient, faster street (pathway) is not available. From day to day, the traffic down a street is pretty much the same, and so is the speed. The same can be said of metabolic pathways.

We examine various metabolic pathways in subsequent units. However, a main interest in biochemistry is to understand how an organism acquires and uses the building blocks and free energy it needs to carry out its various functions; that is, to understand metabolism. So, right from the beginning, it is useful to have a framework for the biomolecules presented in the next units.

The organic reactions that occur in vivo fall into three categories, outlined below.

• Oxidation and reduction reactions: On page 142 of the textbook, read the paragraph that begins with “Catabolic processes are the primary sources of energy…”.

  Both organic and metallic redox reactions occur. In every case, a pair of electrons, rather than a single electron, is transferred.

  

• Elimination, isomerization, and rearrangement reactions: The most commonly eliminated species are H₂O, ammonia, primary amines (RNH₂), and alcohols (ROH). Isomerization usually involves the shift of a hydrogen atom within the same molecule. Rearrangement reactions usually involve breaking and reforming a C$\ce{-}$C bond to produce a molecule with a different geometry.

• Group transfer reactions: Acyl, phosphoryl, and glycosyl groups are the ones most commonly transferred from one biomolecule to another (or to water).

One of the most important features of carbohydrate metabolism is that it is not an equilibrium process. The process is, however, maintained in a steady state. Many students (and working biochemists) find this a difficult distinction, so let us say it another way:

• When a reaction is at equilibrium, levels of reactants and products do not change, and the free energy change is zero.

• When a reaction is in a steady state, the levels of reactants and products do not change, but the free energy change may be far from zero.

  Δ G o′ =−RTln K eq

  Δ G o′ =−RTln [ Product ] [ Reactant ]

Consider a bucket of water with a hole in the bottom. If the bucket is filled from a tap at the same rate as water runs out through the hole, then the level of water in the bucket will remain constant. It is not at equilibrium because, if the tap is shut off, all the water will drain away; but, as long as the influx is the same as the outflow, then the contents of the bucket (equivalent to reactants and products of reactions of a metabolic pathway) will stay at the same level—that is, in a steady state.

There are two types of reactions in a metabolic pathway in a steady state. Many reactions function close to their actual equilibrium values, and are therefore readily reversible with small free energy changes. A few reactions function far from equilibrium, and are therefore essentially irreversible, since very large free energy changes are involved. It is these irreversible reactions that usually function as the control points (allosteric enzymes) in a metabolic pathway (i.e., they are the taps that regulate the flow of intermediates into the bucket and out of the bucket).

What is a high‑energy compound? This is determined by:

• the magnitude of ΔG^o′ of hydrolysis of the compound (i.e., the amount of heat—energy—produced when the compound is split by water).
• the redox potential of the reaction.

Standard Free Energies

The three hydrolysis reactions below show that phosphocreatine is the highest energy compound, and glycerol 3‑phosphate is the lowest. That is, the more negative ΔG^o′, the greater the energy release, and therefore the greater the energy in the original bond.

The energy released by these compounds comes from three sources:

• resonance: More resonance forms can be written for the products than for the reactants. (Remember benzene and resonance stabilization.)
• electrostatic repulsion: Electrostatic repulsion in the products is lower than in the reactants.
• entropy: Entropy (ΔS) is greater for the products than for the reactants.

The greater the gain in resonance stabilization, the greater the relief from electrostatic repulsion; and the greater the gain in entropy (disorder) produced by a chemical change, the greater the energy released.

Oxidation‑Reduction Reactions

Redox reactions are those in which pairs of electrons are exchanged. Just as given compounds (see above) can be hydrolyzed, producing varying amounts of energy, so other compounds have different tendencies to accept or donate electrons.

NADH ↔ NAD⁺ + H⁺ + 2e⁻   ε^o′ = + 0.315 V *

where ε^o′ represents the reduction potential of all components in standard state.

H₂O ↔ ½ O₂ + 2H⁺ + 2e⁻ + 2H⁺ + 2e⁻   ε^o′ = −0.815 V

NADH has a strong tendency to release a pair of electrons (high positive ε^o′). Water has a very weak tendency to give up a pair of electrons and is a stable compound. If the second reaction is written in reverse:

½ O₂ + 2H⁺ + 2e⁻ ↔ H₂O   ε^o′ = +0.815 V *

Oxygen has a strong tendency to accept a pair of electrons (high positive ε^o′).

Where does carbohydrate metabolism occur in vivo? Everywhere. Some cells (e.g., muscle) efficiently metabolise enormous quantities of glucose; some cells metabolise lower amounts of glucose more or less inefficiently (e.g., gut epithelial cells). But all cells are capable of carbohydrate metabolism.

Glucose is found throughout the body (inside and outside the cells). Glycolysis occurs in the cytosol of cells. The tricarboxylic acid cycle enzymes are in the matrix of mitochondria. The electron transport chain enzymes are on the inner membranes of the mitochondria.

Study Questions

1. For metabolism in multicellular organisms to proceed efficiently, it is important that the final products be gases, water, or both. Why?
2. Look at the two starred reactions in the lesson just covered (the redox reactions involving NADH and O₂). Compare the tendency of NADH to donate electrons and the tendency of oxygen to accept them. If NADH and oxygen are mixed, will the electrons stay with NADH or go to oxygen? Explain.
3. What structural feature do the “high‑energy” compounds ATP, FADH₂, and NADH share with acetyl‑CoA?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 2: Glycolysis, Fermentation, and Gluconeogenesis

Overview/Objectives

Carbohydrate metabolism (the breakdown of sugars) is considered the central metabolic pathway in vivo. The overall reaction for glucose can be written as:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Considerable energy is released during carbohydrate metabolism. Part of the released energy escapes as heat (which is why we are at 37°C while the temperature of the room around us is 19–20°C). However, ~67% of the released energy is used to synthesize “high‑energy” phosphate compounds, such as ATP and creatine phosphate. ATP and creatine phosphate, in turn, are hydrolyzed to provide the energy for all biological needs, such as motion, ion transport, and biosynthesis.

Carbohydrate metabolism can be divided into two stages:

1. glycolysis (glucose → pyruvate + ATP)
2. tricarboxylic acid cycle (TCA or Citric Acid Cycle) (pyruvate → NADH/FADH₂ + CO₂)

A third stage uses the energy stored in the above two processes and converts it into ATP:

3. electron transport chain (O₂ + NADH/FADH₂ → ATP + H₂O)

Note that each stage produces high‑energy compounds (ATP, NADH, FADH₂, or some combination), but only one stage (the electron transport chain) uses oxygen to do so. We covered the electron transport chain and oxidative phosphorylation in Unit 2.

As shown above, glycolysis is the first important step in carbohydrate metabolism. This lesson will cover the reactions and principles of glycolysis.

After completing this lesson, you should be able to:

1. Explain the difference between standard free energies (ΔG^o′) and physiological free energies (ΔG ).
2. Demonstrate an understanding of the reactions in glycolysis.
3. Define “substrate‑level phosphorylation.”
4. Explain the chemical strategy by which glycolysis yields a “profit” of two ATP molecules per glucose molecule oxidized.
5. Use standard free energies (ΔG^o′) and physiological free energies (ΔG ) to determine control points in a multi‑enzyme process.
6. Name and describe two processes for anaerobic replenishment of NAD⁺.
7. Name the enzymes, reactants, and products for the three irreversible reactions of glycolysis.
8. Write the overall equation for glycolysis.
9. Compare the pathways of glycolysis and gluconeogenesis and discuss the regulation of these pathways.
10. Explain the basis of the Cori Cycle.

Readings and Activities

1. Read “Glycolysis,” “Intermediates,” “Reactions,” “Enzymes/Control,” “Pyruvate Metabolism,” “Gluconeogenesis,” and “Cori Cycle” (pages 143–153 of the textbook).

2. You can also watch three video lectures on Metabolism and Glycolysis, and two on Gluconeogenesis.

   Gluconeogenesis:

   • #24 Biochemistry Gluconeogenesis Lecture for Kevin Ahern’s BB 450/550

   • #25 Biochemistry Glycogen Metabolism I Lecture for Kevin Ahern’s BB 450/550

     (Links to these are also provided on page 142 of the textbook.)

3. Metabolism and Glycolysis:
   • #21 Biochemistry Glycolysis Lecture for Kevin Ahern’s BB 450/550
   • #22 Biochemistry Glycolysis II Lecture for Kevin Ahern’s BB 450/550

   • #23 Biochemistry Glycolysis III Lecture for Kevin Ahern’s BB 450/550

     (Links to these are also provided in Lesson 1, and on page 145 of the textbook.)

Glycolysis

Glycolysis is the name given to the ten enzyme‑controlled reactions by which glucose is converted to pyruvate. The overall reaction is:

Glucose + 2NAD⁺ + 2ADP + P_i ↔ 2NADH + 2 pyruvate + 2ATP + 2H₂O + 4H⁺

Note that, overall, two ATP molecules are produced by glycolysis. In the first stage, two ATP molecules are hydrolyzed, but in the second stage, four ATP molecules are generated, giving the overall equation in which two ATP are produced. This division is an artificial one, designed to help the student to understand the process. In the cytosol, the reactions in the two stages are continuous.

The controlled cleavage of glucose (six carbons) into two pyruvate molecules (three carbons each) provides a net gain of two high‑energy ATP molecules. Since no oxygen is involved, glycolysis is an anaerobic process.

Glycolysis is a universal metabolic pathway; that is, it is used by both aerobes and anaerobes. However, the fate of the end product, pyruvate, varies. If oxygen is available, pyruvate enters the tricarboxylic acid cycle (see Lesson 4, below). In the absence of oxygen, pyruvate is converted to lactic acid by aerobes and to ethanol by anaerobes. Both fates of pyruvate are vital for the continuation of glycolysis, because they regenerate the cystolic NAD⁺ needed as a reactant for glycolysis.

The table below lists both the standard free energy changes (ΔG^o′) and the physiological free energy changes (ΔG) for the glycolysis reactions in erythrocytes.

Adapted from: Glycolysis, https://en.wikipedia.org/wiki/Glycolysis

A “standard free energy change” is the amount of energy released when a reaction is considered in isolation and allowed to come to equilibrium (say, in a test tube). The standard free energy change is determined as follows:

Δ G o ′ =−RTln [ Product ] [ Reactant ]

R is a thermodynamic constant, and T is the temperature; as you know, the symbol [ ] indicates molar concentration of the bracketed species.

The standard free energy values (ΔG^o′) indicate how much product relative to reactant there will be at equilibrium. That is:

negative (ΔG^o′) $\ce{#}$\ more product
positive (ΔG^o′) $\ce{#}$\ more reactant

A “physiological free energy change” (ΔG) is the amount of energy released when a reaction is considered inside the cell (at steady state). The actual concentrations of product and reactant inside the cell are measured, and then the following formula is used to determine the physiological free energy change:

ΔG=Δ G o ′ +RTln [ Product ] [ Reactant ]

If the physiological free energy is at or near zero, the reaction is at equilibrium. If the physiological free energy is large and negative, the reaction is spontaneous in a forward direction.

Of the ten reactions that make up glycolysis, seven are approximately at equilibrium; that is, the in vivo “steady‑state” concentrations are very close to their thermodynamic equilibrium concentrations.

The other three reactions are functioning far from equilibrium, with large, negative physiological free energy values, and so are the spontaneous or control points for glycolysis (allosteric enzymes). Glucose moves through glycolysis to pyruvate, and never the other way around, because of the overall physiological free energy change for the whole process.

The glycolytic pathway is shown on page 144 of the textbook. (You are not required to know the substrates, products, and enzymes of the entire pathway, but you should know some important points about this pathway, which overall involves the breakdown of glucose into pyruvate and the production of ATP.) As mentioned above, 2 molecules of ATP are needed to drive this pathway and they are used in two early steps, for the formation of glucose‑6‑phosphate from glucose and then for the formation of fructose‑1,6‑bisphosphate. ATP is generated later on in the pathway, through the formation of 3‑phosphoglycerate and then pyruvate. The conversion of phosphoenolpyruvate (PEP) to pyruvate and ATP is an example of substrate‑level phosphorylation, which we covered in Unit 2.

Gluconeogenesis is essentially the pathway opposite of glycolysis, and results in the formation of glucose from pyruvate. Glycolysis is a catabolic pathway and gluconeogenesis is anabolic. See the figure of glycolysis and gluconeogenesis on page 152 of the textbook. You will see that these pathways run the reverse of each other for the most part; that, while many of the same enzymes are used, the reactions run in the opposite direction.

Glycolysis and gluconeogenesis are regulated in a reciprocal fashion. This is important to prevent a futile cycle. The major enzymes that are regulated in glycolysis are hexokinase, phosphofructokinase (PFK), and pyruvate kinase. As shown in the figure on page 147 of the textbook, ATP inhibits PFK and pyruvate kinase. However, these enzymes are activated by AMP. This ensures that, when ATP is abundant, glycolysis is not active and the ATP can be used to synthesize glucose; but, if ATP is low, the cell will switch to glycolysis and the formation of ATP. Pyruvate kinase is allosterically activated by fructose‑1,6‑bisphosphate (F1,6BP) and PFK is activated by F2,6BP.

Gluconeogenesis uses 7 of the same enzymes that are used in glycolysis, which operate at around zero ΔG. This makes sense, because the direction of the reaction can therefore be controlled by changing product and substrate concentrations. However, gluconeogenesis needs four reactions to get around the three reactions that are regulated in glycolysis because of the high ΔG values. These are listed in the figure on page 152 of the textbook.

Pyruvate, produced in glycolysis, has many fates in the cell, as shown in the figure on page 150 of the textbook. Pyruvate can be used for:

• gluconeogenesis
• citric acid cycle and electron transport
• fermentation
• alanine synthesis
• synthesis of oxaloacetate.

Aerobic vs. Anaerobic Production of ATP (Fermentation)

Anaerobic production of ATP is simply glycolysis in which pyruvate is converted to the waste product, lactic acid; one glucose molecule metabolized yields two ATP molecules.

Aerobic production of ATP comprises glycolysis, the citric acid cycle, and the electron transport chain; 38 ATP molecules are produced from one glucose molecule.

Despite the greater efficiency of aerobic ATP production, an organism under stress (you running the 100 metre dash?) will revert to anaerobic ATP production. Because of the high concentrations of glycolytic enzymes, anaerobic production of ATP is faster than aerobic production of ATP. When the stress is relieved, the organism can convert lactic acid back to pyruvate, and hence through the citric acid cycle and the electron transport chain. (Breathing hard after exercise provides extra O₂ to convert all the lactate to CO₂ and H₂O.) The lactate is reoxidized back to pyruvate by the liver in the Cori Cycle, which is shown on page 153 of the textbook.

Study Questions

1. Add up the ten physiological free energy values for the ten glycolysis reactions in erythrocytes (see the table in the commentary above). What is the overall ΔG for glycolysis? Why can the reaction never come to equilibrium in vivo?
2. Glycolysis is inhibited by iodoacetic acid through inactivation of the enzyme glyceraldehyde‑3‑phosphate dehydrogenase (GAPDH). As a result, there is an accumulation of fructose‑1,6‑biphosphate. Why is this product more prevalent than glyceraldehyde‑3‑phosphate and dihydroxyacetone phosphate, the products that form immediately preceding GAPDH?
3. The “Pasteur effect” is the dramatic decrease in glucose consumption when oxygen is introduced to an anaerobic fermentation broth. Why do the yeast cells use less glucose after oxygen is introduced? How much less glucose do they use after oxygen is introduced?
4. Why do you get hot when you exercise?
5. What are the three enzymes that are regulated in glycolysis? How does AMP affect glycolysis?
6. List the possible uses of pyruvate.
7. What does the liver do with the lactate that is produced during heavy exercise?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 3: Citric Acid and Glyoxylate Cycles

Overview/Objectives

After completing this lesson, you should be able to:

1. Explain the importance of the regeneration of oxaloacetate to TCA cycle function.
2. Explain how the energy released when glucose is oxidized through the TCA cycle is chemically “stored.”
3. Explain why NADH and FADH₂ are considered high‑energy compounds.
4. List the chemical transformations that occur with one turn of the TCA cycle.
5. Describe the metabolic importance of acetyl CoA.
6. Identify the enzymes involved in the regulation of the TCA cycle.
7. Identify the mechanisms by which the TCA cycle is regulated.
8. Explain how the functioning of the electron transport chain keeps the TCA cycle going.

Readings and Activities

1. Read “Citric Acid Cycle” and “Glyoxylate Pathway” (pages 153–157 of the textbook).
2. You can also watch the video lectures on The Citric Acid and Glyoxylate Cycles.
   • #28 Biochemistry Citric Acid Cycle Lecture for Kevin Ahern’s BB 451/551

   • #29 Biochemistry Citric Acid Cycle II Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on page 153 of the textbook.)

Commentary

The Citric Acid Cycle

The Citric Acid Cycle (CAC), or Tricarboxylic Acid Cycle (TCA Cycle), is the next step in the aerobic respiration of glucose.

As glycolysis is the conversion of:

glucose → pyruvate

The citric acid cycle generates high energy compounds from pyruvate:

pyruvate → NADH/FADH₂

The end product of glycolysis, pyruvate, proceeds from the cytosol into the mitochondrial matrix via a specific symport (i.e., exchange route). Inside the mitochondrion, pyruvate is converted to acetyl‑CoA plus one carbon dioxide molecule.

Acetyl‑CoA is central in linking glycolysis with the citric acid cycle. Fatty acids and amino acids, as well as carbohydrates, can be metabolized to acetyl‑CoA. The bulk of the ATP molecules which result from “metabolism” come from the processing of acetyl‑CoA through the citric acid cycle, and then into the electron transport chain.

The citric acid cycle is also known as the tricarboxylic acid (TCA) cycle. As its name implies, it is a cyclic process. Oxaloacetate (a four carbon molecule) combines with acetyl‑CoA and, eight reactions later, oxaloacetate is regenerated while two molecules of CO₂ are released. The eight reactions are shown on page 154 of the textbook.

The output of the citric acid cycle is one high‑energy phosphate compound: GTP (similar to ATP) and four high‑energy reduced molecules: 3 NADH + FADH₂.

Thus, the pyruvate bonding energy (electrons) is used principally to produce compounds that contain “high‑energy electrons.”

Note that NAD⁺ contains an aromatic (benzene‑like) ring structure. This ring structure is said to be “resonance stabilized.” To reduce the ring (i.e., add to it a pair of electrons) requires energy to overcome the resonance stabilization forces. Thus NADH has a strong tendency to donate its new electron pair to regain ring resonance. FAD, similarly, is resonance stabilized, while its reduced form, FADH₂, is not. It is this reducing power of NADH and FADH₂ that will drive the electron transport chain (see the next lesson) in the production of ATP.

It is more difficult to estimate the physiological free energy changes for reactions of the TCA cycle than for glycolysis. However, estimated values indicate three enzymes most likely to be regulatory). The citric acid cycle is controlled by the action of its own products, intermediates, and substrates of the three enzymes.

The electron transport chain, which we discussed in Unit 2, Energy, is the next step after the CAC and this is where ATP is generated (34 ATP) through oxidative phosphorylation. The NADH and FADH₂ generated in the CAC are needed for electron transport. There must be redox balance in vivo. This fact means that the reduced compounds, NADH and FADH₂, must donate their electron pairs to recycle as NAD⁺ and FAD to keep the citric acid cycle going. Without the electron transport chain, the citric acid cycle cannot function.

Electron Transport Chain (NADH/FADH₂ → ATP)

The Glyoxylate Cycle

The glyoxylate cycle is closely related to the CAC because many of the steps are the same. The difference between this cycle and the CAC is that this pathway generates 2 molecules of oxaloacetate, one of which can be used to generate glucose in gluconeogenesis. This pathway occurs only in plants, protists, fungi, and bacteria because animals lack two of the enzymes, isocitrate lyase and malate synthase. Isocitrate lyase catalyzes the conversion of isocitrate into succinate and glyoxylate so that none of the carbons end up as CO₂ which occurs in the CAC. Animals, therefore cannot produce glucose from acetyl‑CoA in net amounts; however, the benefit of the CAC is that it generates more of the high energy compounds NADH (3) and FADH₂ (1) as well as GTP (1). The glyoxylate cycle only generates 1 NADH and 1 FADH₂.

The differences between the CAC and the glyoxylate cycle are logical if you consider the metabolic requirements of animals versus other organisms. Fatty acids from lipids are an important energy source for animals. As we will see in subsequent lessons, fatty acids are degraded through beta oxidation into acetate. This acetate binds to the active thiol group of coenzyme A and enters the CAC where it is oxidized to CO₂, allowing cells to obtain energy as the CAC proceeds to the electron transport chain. Animals require large amounts of energy to function. In contrast, organisms such as plants, fungi, and bacteria contain cell walls which require constant synthesis of complex structural carbohydrates such as cellulose, glucans, and chitin. The glyoxylate cycle allows the synthesis of glucose from lipids when acetate is formed in beta oxidation. This allows for the production of large amounts of polysaccharides, a priority in these organisms.

Study Questions

1. Can one get a net synthesis of oxaloacetate if one adds acetyl‑CoA to a system that contains only the enzymes and intermediates of the citric acid cycle? (Consider one complete cycle.)
2. What is the purpose of the production of NADH and FADH₂ in the citric acid cycle (CAC or TCA)?
3. In general, how is the CAC connected to many other metabolic pathways? What compound acts as the link to these pathways?
4. What two enzymes are present in the glyoxylate cycle that animals lack?
5. Compare the overall outcomes of the CAC and the glyoxylate cycles.
6. Why is the glyoxylate cycle important for plants, fungi, protists, and bacteria? Why would the CAC be more important for animals?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 4: Acetyl‑CoA and Cholesterol Metabolism

Overview/Objectives

After completing this lesson, you should be able to:

1. Demonstrate an understanding of the role of acetyl‑CoA in metabolism.
2. Describe the high energy thiolester bond in acetyl‑CoA.
3. Explain how cholesterol is transported, and describe how it is taken up by cells.
4. List the functions of cholesterol in the body.
5. Provide a general overview of cholesterol synthesis.
6. Demonstrate an understanding of the regulation of cholesterol synthesis.
7. Discuss the importance of bile acids in digestion.

Readings and Activities

1. Read “Acetyl‑CoA Metabolism,” “Cholesterol Metabolism,” and “Bile Acid Metabolism” (pages 157–161 of the textbook).
2. You can also watch the videos on Steroid and Lipid Metabolism, and Lipid Movement in the Body (2 links, on page 158).
   • #35 Biochemistry Phosphoglycerides, Cholesterol, Steroids Lecture for Kevin Ahern’s BB 451/551

   • #36 Biochemistry Cholesterol in the Body Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on page 158 of the textbook.)

Commentary

Acetyl coenzyme A or acetyl‑CoA is an important molecule in metabolism, as it is used in many reactions. These include fatty acid oxidation and reduction, pyruvate oxidation, the citric acid cycle (CAC), amino acid anabolism and catabolism, ketone body metabolism (which we will cover in the next lesson), steroid and bile acid synthesis, and prostaglandin synthesis. The main function of this molecule is to move C atoms in the acetyl group to the CAC for oxidation and energy production.

The acetyl group of acetyl‑CoA is linked by a high energy thioester bond. Hydrolysis of this bond is exergonic (−31.5 kJ). The structure of acetyl‑CoA is shown below:

Adapted from: Bryan Derksen (original) and DMacks (colour change). Manually changed some colours of File:Acetyl‑CoA‑2D.svg, Public domain, https://commons.wikimedia.org/w/index.php?curid=47102777

The acetyl group is indicated in blue in the diagram. This acetyl group is linked by a high energy thioester bond to the sulhydryl of the β‑mercaptoethylamine group. Coenzyme A (CoASH or CoA) consists of the β‑mercaptoethylamine group linked to the vitamin pantothenic acid by an amide linkage.

As shown in the text, acetyl‑CoA is needed for both cholesterol and ketone body synthesis. Two acetyl‑CoA molecules combine to form acetoacetyl CoA. Two more carbons are added from another acetyl‑CoA to form hydroxyl‑methyl‑glutaryl‑CoA or HMG‑CoA.

Cholesterol has many functions in the body, in particular, as a component of cell membranes. It is also a precursor to steroid hormones, bile acids, and vitamin D. The cholesterol synthesis pathway is known as the isoprenoid pathway. The formation of cholesterol starts with the synthesis of HMG‑CoA from acetyl‑CoA, and then the formation of mevalonate by the enzyme HMG‑CoA reductase. (We do not expect you to know all the steps of this pathway, but is it important to know the overall reactants and products and other general features of this pathway.)

The isoprenoid pathway requires a large input of energy. This explains why the pathway is transcriptionally regulated (i.e., the HMG‑CoA reductase gene is turned on and transcription increases when cholesterol levels fall and production of cholesterol is required, as we learned in Units 4 and 5). The enzyme itself is also regulated by two methods:

• One of them is feedback inhibition, meaning that when cholesterol is present the enzyme is inhibited due to a feedback loop that signals cholesterol production is not needed.
• The other is covalent modification of the enzyme. An AMP‑activated protein kinase phosphorylates HMG‑CoA reductase when cell energy is low and concentrations of AMP rise, which inhibits HMG‑CoA reductase and halts cholesterol production.

Cholesterol (shown below) is an unusual, ring‑structured, biochemical alcohol. As have other lipids, it has acquired a bad name in health circles. However, it has important biological functions, and it can be synthesized by mammals as well as absorbed from the diet. Furthermore, it can be eliminated as a bile salt.

Cholesterol is transported to tissues mainly in low density lipoprotein (LDL) complexes in the blood stream. LDL not only transports the cholesterol ester to peripheral tissues, but it also helps to regulate synthesis of this lipid in vivo. Cholesterol is also found in chylomicrons (dietary, non‑esterified cholesterol) and in HDL complexes. HDL lipoprotein complexes scavenge cholesterol from dying cells and from membrane turnover. After esterification in the HDL complexes, cholesterol is cycled to LDL complexes for transport or storage.

Cholesterol in the LDL lipoprotein complexes is absorbed into cells via endocytosis. The LDL receptors on cell surfaces are key to cholesterol entry into cells. When a cell is rich in cholesterol, LDL receptor synthesis is suppressed; when cholesterol is low in the cell, LDL receptor synthesis is enhanced.

Regulation of cholesterol levels in vivo is also accomplished by the rate of cholesterol synthesis (in the liver, as described above), and by the rate of esterification of cholesterol. This latter point is important, since cholesterol in a membrane is non‑esterified. Cholesterol is esterified only for storage or transport.

Study Questions

1. Why is acetyl‑CoA such an important molecule? Name three pathways that require this molecule.
2. What type of bond is the high energy bond in acetyl‑CoA?
3. List three functions of cholesterol in the body.
4. What is the name of the pathway involved in the production of cholesterol? Why is this pathway regulated? How do levels of AMP affect the regulation of this pathway?
5. Why do you think HMG‑CoA reductase is an ideal target for cholesterol‑reducing medications?
6. Why are bile acids important in metabolism? What molecule is converted to bile acids?
7. Why do those on a strict vegetarian diet rarely suffer from diet–induced hypercholesterolemia?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 5: Ketone Bodies

Overview/Objectives

After completing this lesson, you should be able to:

1. Name the ketone bodies.
2. Explain the normal physiological role of ketone bodies.
3. Describe the circumstances under which ketone bodies are synthesized, and outline the chemical reactions involved in their synthesis.

Readings and Activities

1. Read “Ketone Body Metabolism” on page 161 in the textbook.
2. You can also watch the video lecture on Ketone Bodies (1 link, page 161).

   • #37 Biochemistry Fat/Fatty Acid Metabolism I Lecture for Kevin Ahern’s BB 451/551

     (A link to this is also provided on page 161 of the textbook.)

Commentary

Ketone bodies are acetoacetate, acetone, and β‑hydroxybutyrate. In a process called ketogenesis, ketone bodies are formed in the liver from excess acetyl‑CoA.

Acetyl‑CoA from fatty acid breakdown enters the citric acid cycle. However, oxaloacetate must be present to combine with acetyl‑CoA or citrate will not form. As in the citric acid cycle illustrated in Lesson 3, oxaloacetate was necessarily regenerated after citrate was metabolized to CO₂ and NADH/FADH₂. Since perpetual motion machines do not exist in vivo any more than they do anywhere else, oxaloacetate is not eternally regenerated. Approximately every seven turns of the citric acid cycle, oxaloacetate fails to be regenerated because one or the other of the intermediates has been trapped for another purpose. Intermediates of the citric acid cycle can be drawn off for the biosynthesis of amino acids, cholesterol, glucose, and the porphyrins. This “drawing off” can lead to a deficit in these intermediates.

When there is insufficient oxaloacetate to react with acetyl‑CoA, the excess acetyl‑CoA is converted to acetoacetate or β‑hydroxybutyrate. These ketone bodies are not necessarily waste products. In fact, they are an important source of fuel for muscle tissue. β‑hydroxybutyrate can be oxidized back to acetoacetate with conversion to acetyl‑CoA. Acetoacetate and β‑hydroxybutyrate can cross the blood‑brain barrier and provide energy for the brain when glucose is limiting. This is shown in the figure on page 161 of the text.

Because (even‑numbered) lipids are metabolized solely to acetyl‑CoA, lipids are considered an imperfect fuel. Carbohydrates, on the other hand, are considered to be a perfect fuel, because they produce acetyl‑CoA through pyruvate. When oxaloacetate is in short supply, pyruvate can be diverted into the synthesis of additional oxaloacetate; acetyl‑CoA cannot.

An important role for one of the ketone bodies, acetoacetate, is to inhibit further lipolysis (breakdown of triacylglycerols to free fatty acids).

Study Questions

1. Name the ketone bodies. What is the role of these molecules in metabolism? When are ketone bodies a source of energy?
2. What is the connection between the citric acid cycle and oxaloacetate production and ketone body formation?
3. How much energy can you obtain from the oxidation of palmitate, CH₃ (CH₂)₁₄COOH, in the liver under conditions of ketosis? How does this compare to the amount you can obtain when on a balanced diet providing glucogenic fuel molecules?
4. Excess acetyl‑CoA, produced by the citric acid cycle or by fatty acid breakdown, is channelled into ketone bodies. Would it be more sensible to provide a feedback mechanism to slow down the production of acetyl‑CoA itself?
5. It was stated that one oxaloacetate molecule must be regenerated approximately every seven turns of the citric acid cycle. Since glycerol (from triacylglycerols) can produce oxaloacetate, and there are only seven to eight acetyl‑CoAs produced per fatty acid, why are ketone bodies produced at all?
6. How are ketone bodies funnelled back into central metabolism if citric acid cycle intermediates are synthesized?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 6: Fatty Acid Oxidation and Synthesis

Overview/Objectives

Like carbohydrates, lipids are important biochemical fuel molecules. Contrast the metabolism of a six carbon lipid (below) with the overall equation for the metabolism of glucose (Lesson 2).

CH₃ (CH₂)₄COOH + 8 O₂ → 6 CO₂ + 6H₂O

Gram per gram, lipids provide almost twice as much ATP as do carbohydrates, and in some tissues—heart is a good example—lipids are the preferred metabolic fuel. Lipids are the major stored fuel in human beings.

In this lesson, we will consider the conversion of fatty acids to acetyl‑CoA, the molecule that enters the citric acid cycle in the middle stage of carbohydrate metabolism. We will also consider fatty acid biosynthesis.

Many of the chemical strategies that have evolved to metabolize lipids are similar to those we have discussed for carbohydrates. Furthermore, metabolism is convergent: all fuel molecules are directed either through the citric acid cycle or directly into the electron transport chain. Since we have already discussed the citric acid cycle and the electron transport chain, you know half of lipid metabolism.

After completing this lesson, you should be able to:

1. Explain how lipids are degraded to acetyl‑CoA in vivo.
2. Explain how fatty acid chains are synthesized in vivo.
3. Explain how lipids (which are insoluble in water) are digested, absorbed, and transported.
4. Describe β‑oxidation (degradation) of lipids to acetyl‑CoA.
5. Explain why the oxidation of fatty acids yields more ATP than would be expected from the amount of acetyl‑CoA produced.
6. Compare the metabolism of odd‑chain fatty acids and even‑chain fatty acids, and explain why the metabolic process is slightly different for each of these groups.
7. Compare the chemical reactions of β‑oxidation with those of the TCA cycle.
8. Describe the differences between the catabolic and synthetic pathways for lipids.
9. Explain how fatty acid metabolism is regulated.

Readings and Activities

1. Read “Prostaglandin Synthesis,” “Fatty Acid Oxidation,” “Enzymes of Beta Oxidation,” “Oxidation of Odd Chain Fatty Acids,” “Unsaturated Fatty Acid Oxidation,” “Alpha Oxidation,” “Fatty Acid Synthesis,” “Enzymes of Fatty Acid Synthesis,” “Fatty Acid Elongation,” “Desaturation of Fatty Acids,” “Metabolism of Fat,” “Glycerophospholipid Metabolism,” and “Connections to Other Pathways” (pages 161–171 in the textbook).
2. You can also watch three video lectures on Fat, Fatty Acid, and Prostaglandin Metabolism:.
   • #36 Biochemistry Cholesterol in the Body Lecture for Kevin Ahern’s BB 451/551
   • #37 Biochemistry Fat/Fatty Acid Metabolism I Lecture for Kevin Ahern’s BB 451/551

   • #38 Biochemistry Fat/Fatty Acid Metabolism II Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on pages 162 and 164 of the textbook.)

Commentary

Dietary lipids, in the form of triacylglycerols or phospholipids, are converted enzymatically to their fatty acids in the intestine. The fatty acids are transported as protein‑lipid complexes through the blood. The lipids, with (or sometimes without) the carrier proteins, are absorbed into cells for degradation. Uptake of lipids by a cell is mediated by receptors on the cell surface. Genetic alteration of the receptors leads to some forms of heart disease. More commonly, however, Western diets provide such a load of dietary lipid that these water‑insoluble molecules cannot all be efficiently transported and then absorbed by cells.

The hydrocarbon chain that makes up most of the fatty acid mass is relatively unreactive. Therefore, lipid metabolism begins at the acid end of the chain. Lipid metabolism, called β‑oxidation, is a four‑step process. The products are acetyl‑CoA, the lipid chain shortened by two carbons, and a “new” acid functional group on the shortened chain. One NADH and one FADH₂ are also produced for each round of β‑oxidation. The reactions all occur between carbons 2 and 3 (C1 is the one linked to the CoA). The steps of β‑oxidation are shown below and include:

1. dehydrogenation to create FADH₂ and a fatty acyl group with a double bond in the trans configuration
2. hydration across the double bond to put a hydroxyl group on carbon 3 in the L configuration
3. oxidation of the hydroxyl group to make a ketone
4. thiolytic cleavage to release acetyl‑CoA and a fatty acid two carbons shorter

The four steps of β‑oxidation are then repeated. Eventually, the fatty acid chain is degraded to (n/2) acetyl‑CoA fragments.

The four steps of β‑oxidation of a lipid are shown below:

Coenzyme A (CoA) is attached to both the cleaved acetyl fragment and the shortened fatty acid chain. CoA is similar to the terminal phosphate in ATP: it is half of a high‑energy bond.

Some of the chemical reactions of the citric acid cycle (and the intermediates of the cycle shown below) bear a striking resemblance to those of β‑oxidation (shown above). Note the three features of β‑oxidation (double bond, alcohol, and ketone), and where they appear in the citric acid cycle intermediates.

Most lipids have an even number of carbons. For these lipids, β‑oxidation will yield (n/2) acetyl‑CoA fragments from a C_n fatty acid. Fatty acids with an odd number of carbons are also metabolized by β‑oxidation to acetyl‑CoA fragments. However, the final fragment is propionyl‑CoA (a 3‑carbon fragment). Through a complex set of reactions (shown in the first figure on page 165) propionyl‑CoA is converted to succinyl‑CoA. Succinyl‑CoA is a 4‑carbon carboxylic acid, which is part of the citric acid cycle.

Thus, just as for carbohydrates, the entire mass of a fatty acid can be funnelled through the tricarboxylic acid cycle to produce useful chemical energy.

Synthesis of fatty acids is similar to β‑oxidation. The biosynthesis of a fatty acid is the sequential addition of 2‑carbon units to a growing chain. There are two chemical problems Mother Nature had to overcome for this addition to be accomplished:

• The to‑be‑added carbon fragment must have a reactive functional group at each end. This structure allows the fragment to add to the growing chain with one end and to attach the next incoming fragment at its other end. However, all but one of the carbon atoms in a completed fatty acid are nonreactive alkyl groups.
• Most of the fatty acids in vivo are 16 or 18 carbons in length. How is the growing chain to be kept in one place long enough for eight or nine 2‑carbon fragments to be pieced together?

The first problem (producing a hydrocarbon chain from reactive starting material) is overcome by using a 3‑carbon fragment (malonyl‑ACP) as the building block. The third carbon atom is cleaved when malonyl‑ACP adds to the growing chain, leaving one (nonreactive) CH₂ and a carbonyl group (C$\ce{=}$O) as part of the new fatty acid.

The second problem (anchoring the growing fatty acid) is solved by attaching the growing fatty acid to a large multi‑enzyme complex (fatty acid synthase). The six chemical reactions required to add and modify each new 2‑carbon fragment are performed by components of fatty acid synthase. You have seen this strategy (anchoring a molecule as it is being constructed) before, when studying protein synthesis in Unit 5.

The activating group in fatty acid biosynthesis is the acyl carrier protein (ACP), not coenzyme A as in fatty acid degradation. ACP and CoA are similar because both contain phosphopantetheine as the attachment site for the 2‑carbon fragment. In addition, CoA contains adenosine monophosphate (small and soluble), while ACP contains a 10kD protein (reasonably large and anchored in fatty acid synthase).

The six steps in the biosynthesis of fatty acids are:

1. conversion of acetyl‑CoA to malonyl‑CoA.
2. conversion of malonyl‑CoA to malonyl‑ACP.
3. condensation of malonyl‑ACP with the growing chain (with cleavage of CH₂).
4. reduction of a carbonyl group to an alcohol group.
5. reduction of the alcohol group to a trans‑α, β double bond.
6. reduction of the C$\ce{=}$C bond to a CH₂$\ce{-}$CH₂ single bond.

The reducing agent for steps 4, 5, and 6 is NADPH.

The chemical strategies of fatty acid biosynthesis and degradation are similar, although used in opposite directions.

• Biosynthesis:

  malonyl‑ACP + FA(C_n) → alcohol → trans‑α, β C$\ce{=}$C → FA(C_n+2)

• Degradation:

  FA(C_n+2) → trans‑α, β C$\ce{=}$C → alcohol → ketone → FA(C_n) + acetyl‑CoA

Mammals can synthesize some unsaturated fatty acids by the use of enzymes called “terminal desaturases.” These enzymes reduce a saturated C$\ce{-}$C bond to a C$\ce{=}$C double bond in an already formed fatty acid. However, mammals cannot form linoleic acid (Δ⁹, ¹²). Since linoleic acid is necessary for the biosynthesis of prostaglandins (local hormones), it is an “essential” fatty acid. The designation “essential” means that the substance must come from the diet.

Fatty acid biosynthesis is regulated (as is carbohydrate metabolism) by hormones, substrate availability, covalent modification, allosteric control of the biosynthetic enzymes, or some combination of these strategies. In addition, lipid biosynthesis is controlled over the long term by the availability of the biosynthetic enzymes. Both starvation and the presence of fatty acids in the diet can decrease the production of the lipid biosynthetic enzymes.

Fatty acid biosynthesis occurs in the cytosol, while fatty acid degradation occurs in the mitochondrion.

Study Questions

1. What biological advantage does storing major fuel reserves as fat rather than glycogen confer on nonplant species?
2. Why are fat stores in mammalian adipose tissue sources of intracellular water?
3. What are the benefits of using β‑oxidation to metabolize lipids?
4. Marathon runners are said to “hit the wall” at about 20 miles, approximately the time the runner converts from carbohydrate to fat stores for energy. Why is this process so difficult that it is called a “wall”?
5. A victim of starvation has to be fed small amounts of food and has to be very careful about introducing fat into the diet. Why?
6. You add radiolabelled malonyl‑CoA to a cell extract which is actively synthesizing palmitate. (Each carbon of the malonyl unit is labelled by the isotope ¹⁴C.) You stop the reaction one minute later by altering the pH. Then you analyze the palmitate for radioactivity. Which of the carbons will be labelled in palmitate?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Answers to Unit 6 Study Questions

Lesson 1: Overview of Metabolic Concepts

1. To maintain metabolism in a steady state, there must be no possibility of a build–up of final products. The easiest disposal products are the multicellular organism’s universal solvent, gases, or both.

2. If NADH and oxygen are mixed, electrons will be transferred from NADH to oxygen with the release of considerable energy:

   NADH + ADH + ½ O₂ + H⁺ ↔ NAD⁺ + H₂O ε^o′ = 1.13 V

3. The “high‑energy” compounds share an ADP unit or, in the case of acetyl‑CoA, a closely related derivative.

Lesson 2: Glycolysis and Gluconeogenesis

1. ΔG = −74.0 kJ/mol for the overall glycolytic process. This large negative value indicates that the overall reaction is spontaneous (i.e., the final product, pyruvate, is being removed fast enough, and glucose is being added constantly enough), so that there is no back reaction:

   Pyruvate → Glucose

2. The reaction that converts fructose 1,6‑bisphosphate to glyceraldehyde‑3‑phosphate and dihydroxyacetone phosphate is near equilibrium in vivo (i.e., ΔG near zero), while the preceding reaction producing fructose 1,6‑bisphosphate is not. Therefore it is F1,6BP that will accumulate, not glyceraldehyde‑3‑phosphate and dihydroxyacetone phosphate.
3. The introduction of oxygen allows yeast to convert from anaerobic to aerobic metabolism. Since aerobic metabolism provides more ATP, the amount of glucose that must be used to nourish the yeast is much less. Approximately 6% of the glucose metabolized anaerobically is needed to provide the same amount of energy under aerobic conditions.
4. PEP is a very high‑energy compound. The reaction that converts PEP to pyruvate is so highly energetically favourable (very negative ΔG) that there is almost enough energy in PEP to stimulate production of a second ATP through substrate level phosphorylation, but it is not used. The excess energy is lost as heat.
5. The three enzymes that are regulated in glycolysis are: hexokinase, phosphofructokinase (PFK), and pyruvate kinase. High amounts of AMP activate PFK and pyruvate kinase, which stimulate glycolysis because ATP is needed.
6. The possible uses of pyruvate are:
   • ATP production and NAD⁺ regeneration through TCA cycle and electron transport
   • glucose synthesis by gluconeogenesis
   • ATP production and NAD⁺ generation and lactate or ethanol production by fermentation
   • alanine synthesis
   • oxaloacetate synthesis
7. During heavy exercise, the body needs ATP and generates lactate through anaerobic respiration faster than the blood can deliver oxygen to continue through aerobic respiration. Muscle cells cannot use the lactate produced anaerobically during exercise, so the lactate is dumped into the blood. The liver converts the lactate to pyruvate in the Cori Cycle. The enzyme lactate dehydrogenase catalyzes this conversion. The pyruvate is then used to make glucose by gluconeogenesis in the liver, and can go back into the blood to be taken up by muscles and used for energy.

Lesson 3: Citric Acid and Glyoxylate Cycles

1. No, one cannot get a net synthesis of oxaloacetate in these circumstances. Acetyl‑CoA is a 2‑carbon species and two molecules of CO₂ are given off for each turn of the cycle. Therefore no net synthesis is possible. In fact, this is the point of the citric acid cycle. All the atoms of glucose are discarded and the glucose energy is conserved in one molecule of GTP and in high‑energy electrons.
2. The production of NADH and FADH₂ is one of the most important features of the CAC. These compounds are needed to drive the electron transport chain for energy (ATP) production.
3. Acetyl‑CoA is central in linking glycolysis with the citric acid cycle, but many other metabolic pathways can be fed through the CAC and on to electron transport for energy production. Fatty acids and amino acids, as well as carbohydrates, can be metabolized to acetyl‑CoA. The bulk of the ATP molecules that result from “metabolism” come from the processing of acetyl‑CoA through the citric acid cycle, and then into the electron transport chain.
4. The two enzymes present in the glyoxylate cycle that animals lack are isocitrate lyase and malate synthase.
5. The CAC and the glyxoylate cycle differ in that the CAC produces 1 oxaloacetate per turn of the cycle and the glyoxylate cycle produces 2. This means that there is no net production of oxaloacetate in the CAC; the one that is made balances out the one used for the cycle. The extra one produced by the glyoxylate cycle in plants and bacteria means that, unlike animals, these organisms can use the oxaloacetate to make glucose and other molecules. Plants and bacteria can turn acetyl‑CoA into glucose, while animals cannot. It also means that plants and bacteria can turn acetyl‑CoA from fat into glucose but animals cannot. The advantage of the CAC in animals, though, is the higher production of NADH and FADH₂. One turn of the glyoxylate cycle produces 1 NADH and 1 FADH₂, whereas one turn of the CAC results in 3 NADH, 1 FADH₂, and 1 GTP. This makes sense considering the high energy requirements of animals.
6. The higher energy output of the CAC for animals provides the requirements for the production of higher amounts of ATP through the electron transport chain, which is needed for the mobility of animals. The extra oxaloacetate produced through the glyoxylate cycle in plants, bacteria, and protists is necessary to produce carbohydrate for structure and storage, which are necessary functions for these organisms.

Lesson 4: Acetyl‑CoA and Cholesterol Metabolism

1. Acetyl‑CoA is an important molecule, because it is used in many reactions and links many different metabolic pathways. The function of this molecule is to move the C atoms in the acetyl group to the CAC and then transport electrons for ATP production. Three pathways that require this molecule are: fatty acid oxidation and reduction, pyruvate oxidation, the citric acid cycle (CAC), (also amino acid anabolism and catabolism, ketone body metabolism).
2. The high energy bond in acetyl‑CoA is a thioester bond. Hydrolysis of this bond is highly exergonic (−31.5 kJ).
3. The three functions of cholesterol in the body are: cell membrane structure, precursor of steroid hormones, precursor of vitamin D (also bile acid precursor).
4. The pathway involved in cholesterol synthesis is the isoprenoid pathway. Because it requires a large energy input, this pathway is regulated so that it will not run unless it is needed. When concentrations of AMP are high, the cell doesn’t have a lot of energy to produce cholesterol. An AMP‑activated protein kinase inhibits HMG‑CoA reductase by phosphorylating it, which inhibits HMG‑CoA reductase and halts cholesterol production.
5. HMG‑CoA reductase is an ideal target for cholesterol‑reducing medications because it is a key enzyme in the synthesis of cholesterol and it is regulated in vivo. Statins are used clinically for treatment. They are fungal products that act as a competitive inhibitor of HMG‑CoA reductase.
6. Insoluble products such as dietary fibre, and waste products such as heme degradation products, must be efficiently eliminated by combination with a detergent. A detergent molecule is one that combines with a non‑soluble molecule, giving a complex which can be suspended in solution rather than clumping with like insoluble molecules. The bile salts, cholic acid and deoxycholic acid, are the major biological detergents. Cholesterol is the precursor of both cholic acid and deoxycholic acid.
7. Cholesterol is not found in plants; therefore, strict vegetarians rarely suffer from diet‑induced hypercholesterolemia.

Lesson 5: Ketone Bodies

1. The ketone bodies are acetoacetate, acetone, and D‑β‑hydroxybutyrate. The role of these molecules is to provide energy when glucose levels are low. They can be converted to acetyl‑CoA for entry into the citric acid cycle (CAC) and then ATP synthesis through the electron transport chain. Acetoacetate and β‑hydroxybutyrate can be converted to acetyl‑CoA. They can both cross the blood‑brain barrier and provide energy for the brain when glucose is limiting.
2. The connection between the CAC, oxaloacetate production, and ketone body formation is that acetyl‑CoA from fatty acid breakdown enters the CAC, but oxaloacetate must be present to combine with acetyl‑CoA to form citrate. Oxaloacetate is usually regenerated each turn of the CAC, but in some cases it is used for other metabolism (drawn off) and is not present. If there is insufficient oxaloacetate, the acetyl‑CoA is converted to acetoacetate or β‑hydroxybutyrate. These ketone bodies can cross into the brain to provide energy and can be converted back into acetyl‑CoA as well.
3. Estimates of energy release are most easily made by comparing the number of high‑energy electron carriers (NADH and FADH₂) produced. Palmitate releases eight acetyl‑CoA fragments by seven rounds of β‑oxidation. Each round of β‑oxidation also yields one NADH and one FADH₂ (a total of 14 high‑energy electron pairs). This number is the same for normal and ketogenic conditions. Under normal conditions each acetyl‑CoA, via the citric acid cycle, will reduce one FAD and 3NAD⁺ (a total of 32 high‑energy electron pairs, since there are eight acetyl‑CoA units). This latter process is not possible under ketogenic conditions. Therefore ketogenic conditions provide ~30% of the energy that is available under normal conditions. Of course, the acetyl‑CoA units produced under ketogenic conditions are still available should conditions improve.
4. Such a feedback mechanism would and would not be more sensible. Ketone bodies are produced under normal conditions and are the preferred fuel of, for example, heart tissue. The problem occurs when citric acid intermediates are in short supply and therefore the ketone bodies accumulate. There is some regulation of acetyl‑CoA production, but this molecule is at the heart of metabolism. Without it, you would die, so its synthesis is never shut down completely.
5. It is a matter of mathematics. A triacylglycerol produces one glycerol and three fatty acids. That is, one oxaloacetate for 21–24 acetyl‑CoA units.
6. Acetoacetate and β‑hydroxybutyrate are reconverted to acetyl‑CoA.

Lesson 6: Fatty Acid Oxidation and Synthesis

1. Fats are insoluble, and so are not hydrated. Adipose cells—fat storage depots—contain droplets of triacylglycerol, much like drops of fat floating on the surface of water. Glycogen is a well‑hydrated molecule. To maintain the equivalent amount of energy reserves, a glycogen storer would be considerably heavier than a fat storer. This situation is not a good one if you are mobile and trying to escape predators.
2. The complete oxidation of lipids yields carbon dioxide and water: roughly 1 mL water per gram of fat. This relationship is a great advantage to camels, whose humps are fat depots. Hydrolysis of the fat provides some water to the animal during dry desert work.
3. Conversion of a non‑reactive, insoluble hydrocarbon chain to a useful energy intermediate presents an immediate problem of solubility. The steps of β‑oxidation provide soluble fragments, as well as a high‑energy bond on each fragment, and an intermediate that can channel into an existing metabolic pathway (the citric acid cycle).
4. Fats, which are less soluble than carbohydrates, cannot be marshalled for use as quickly as can carbohydrates. The lipases in adipose tissue are soluble enzymes which must nibble at the edges of the fat droplets. As the fatty acids enter metabolism, their energy content, which is greater than that of carbohydrates, partially compensates for the slow start. This fact means that you can “get over the wall.”
5. A victim of starvation must be careful because the digestive and biosynthetic enzymes are produced on demand: no food = no enzymes. It normally takes three or more days for the enzymes to be synthesized (assuming amino acids are available to do so). The fatty acid degradative and biosynthetic enzymes are more slowly synthesized.
6. ¹⁴C is carbon‑1; that is, carbon‑1 is radiolabelled, because a malonyl unit is added to the growing chain by insertion at the front.

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.