TOPIC 1-6 OBJECTIVES

Question 1

1. Define the concepts of a biochemical pathway and cellular metabolism

Answer 1

Biochemical pathway – In a pathway, the product of one reaction serves as the substrate of the subsequent reaction – these are classified as either catabolic or anabolic. Metabolism is best understood by examining its component pathways. Each pathway is composed of multienzyme sequences, and each enzyme, in turn, may exhibit important catalytic or regulatory features
Cellular metabolism – collectively network of pathways is called metabolism

Question 2

2. Explain the difference between catabolism and anabolism

Answer 2

Catabolic – where complex molecules are broken down into simple molecules
Anabolic – where simple molecules are used to build larger more complex molecules

Question 3

3. Explain the concepts of the transition state and activation energy, indicating the effect that enzyme catalysis has on these parameters

Answer 3

Have substrate on left and to go from substrate to product – go through a transition state – activation energy
Activation energy (AG‡) is the amount of energy required to bring all of the molecules in 1 mole of substrate to the transition state
This is critical to life, without it high energy compounds would be too unstable to exist
- To increase rate of reaction, we either need to:
1. Add more energy to the system
2. Decrease AG‡ - we use a catalyst to do this

Question 4

4. Define the terms Vmax and Km as related to the Michaelis-Menten model of enzyme kinetics and explain what these terms tell us about the relationship between an enzyme and its substrate

Answer 4

Vmax – maximum velocity, The rate of an enzyme-catalyzed reaction increases with substrate concentration until a maximal velocity (Vmax) is reached. The leveling off of the reaction rate at high substrate concentrations reflects the saturation with substrate of all available binding sites on the enzyme molecules present.
Km – is the substrate concentration at which the reaction rate is half maximum (Vmax)

Question 5

5. Derive Vmax and Km values from the Lineweaver-Burk plot

Answer 5

To determine Km and Vmax, it is easier to transform the Michaelis-Menten equation
We take the reciprocal (1/v) of both sides:

Question 6

6. Define the five main ways in which enzyme activity is regulated

Answer 6

1. Enzyme production
   - Transcription or translation is switched on or off
   - e.g. production of specific cytochrome P450 enzymes in the liver
2. Compartmentalization
   - This confines specific metabolic pathways to different parts of the cell
   - e.g. enzymes for b-oxidation of fatty acids are only found in the mitochondria
3. Post-translational modification - e.g. phosphorylation and glycosylation
   - e.g. phosphorylation of glycogen synthase helps control blood sugar levels
4. Environment
   - Some enzymes are only active in certain environments
   - e.g. chymotrypsin is produced in the pancreas but only activated by acidic conditions in the stomach
5. Activation and inhibition
   - Specific molecules can bind to enzymes and switch them on or off
   - This is a common target for drug development
   - This is known as allosterism

Question 7

7. Explain the concept of substrate regulation of enzyme activity

Answer 7

At low [S], when one substrate molecule binds to the enzyme, it makes it easier for the next to bind
- This is referred to as homotropic interaction
This tells us that there is more than one active site per enzyme molecule
There is interaction (co-operation) between these active sites

Question 8

8. Distinguish between positive, negative, homotropic and heterotropic allosteric interactions

Answer 8

homotropic interaction - At low [S], when one substrate molecule binds to the enzyme, it makes it easier for the next to bind

heterotropic interaction - regulation of an enzyme by the binding of an effector molecule at the protein’s allosteric site

Positive allosteric effectors (activators) enhance enzyme activity

Negative allosteric effectors (inhibitors) decrease enzyme activity

Question 9

9. Compare and contrast reversible and irreversible enzyme inhibitors

Answer 9

irreversible (binds to enzyme covalently) or reversible (noncovalently)

Question 10

10. Interpret a Lineweaver-Burk plot to demonstrate the effect of competitive, non-competitive, and uncompetitive inhibitors on enzyme kinetics

Answer 10

Competitive
inhibitor binds at same site where substrate would usually bind – competitive and can be overcome by increasing [S], as [S] required for Vmax increases so does [S] required for 1/2Vmax so Km increases
1. effect on Vmax – unchanged
2. effect on Km – increased, more substrate is needed
3. effect on Lineweaver Burk plot

Non-competitive
causes decrease in Vmax
inhibitor and substrate bind at different sites on the enzyme, inhibition cannot be overcome by increase [S], enzyme-substrate complex cannot convert substrate to product, so enzyme concentration is effectively reduced
1. effect on Vmax – decrease, inhibitor cannot be overcome by increasing the substrate concentration
2. effect on Km – unchanged, no interference with binding
3. effect on Lineweaver Burk plot

Uncompetitive
uncompetitive inhibitors can only bind to ES complex and not to free enzyme
inhibition cannot be overcome by high concentrations of substrate
enzyme-substrate-inhibitor complex cannot convert substrate to product, so enzyme concentration is effectively reduced
binding of inhibitor to ES complex means that takes longer for substance to leave active site
1. effect on Vmax – decreased
2. effect on Km – decreased
3. effect on Lineweaver Burk plot

Question 11

1. Define the five central themes of metabolism

Answer 11

1) Fuel molecules are degraded (catabolism) and large molecules made (anabolism) step-by-step in a series of linked reactions called metabolic pathways
2) The energy currency of all life is ATP
3) The oxidation of pre-existing carbon molecules drives the formation of ATP
4) There are only a limited number of types of reactions in metabolism
5) Metabolic pathways are tightly regulated

Question 12

2. Relate bond formation and breakage to energy capture and release

Answer 12

1) Breaking bonds = releases energy
2) Making bonds = consumes energy

Question 13

3. Understand the central importance of ATP in metabolism

Answer 13

1) Adenosine triphosphate (ATP), energy-carrying molecule found in the cells of all living things. ATP captures chemical energy obtained from the breakdown of food molecules and releases it to fuel other cellular processes.
2) When ATP is hydrolysed, energy is released
3) Adenosine triphosphate is used to transport chemical energy in many important processes, including:
- Aerobic respiration (glycolysis and the citric acid cycle)
- Fermentation
- Cellular division
- Photophosphorylation
- Motility (e.g., shortening of myosin and actin filament cross-bridges as well as cytoskeleton construction)
- exocytosis and endocytosis
- Photosynthesis
- Protein synthesis

Question 14

4. Relate oxidation and reduction to electron transfer reactions

Answer 14

• An oxidation–reduction or redox reaction is a reaction that involves the transfer of electrons between chemical species (the atoms, ions, or molecules involved in the reaction).
• During a redox reaction, some species undergo oxidation, or the loss of electrons, while others undergo reduction, or the gain of electrons.

Question 15

5. Understand the role of NAD+ and FAD in metabolism

Answer 15

• Nicotinamide adenine dinucleotide (NAD+) and flavin adenine dinucleotide (FAD+) are two cofactors that are involved in cellular respiration. They are responsible for accepting “high energy” electrons and carrying them ultimately to the electron transport chain where they are used to synthesize ATP molecules.
• NAD accepts just one hydrogen a single hydrogen and an electron pair is transferred, and the second hydrogen is freed into the medium reduced to NADH
• FAD can accommodate two hydrogen reduced to FADH2

Question 16

6. Calculate concentrations by mass and by moles and use appropriate prefixes to convert concentrations in biological systems

Answer 16

• Concentration by mass : concentration = amount in mass/ volume
• Concentration by moles or molarity = moles/ volume
• Common prefixes
  o 10^-3 M = 1 mM = 1 mmole/litre (1 millimolar)
  o 10^-6 M = 1 µM = 1 µmole/litre (1 micromolar)
  o 10^-9 M = 1 nM = 1 nmole/litre (1 nanomolar)

Question 17

7. Define the three broad classes of hormone

Answer 17

1) Peptide hormones are hormones that are made of small chains of amino acids, water soluble – e.g. prolactin, insulin, glucagon
2) Steroid hormones are steroids that act as hormones, lipid soluble– e.g. progesterone, testosterone, estradiol
3) Amino acid derivatives are hormones derived from amino acids (tyrosine and tryptophan), are water soluble – e.g. adrenaline, thyroxine, triiodothyronine

Question 18

8. Distinguish between the ways in which hormones can exert their effects

Answer 18

1) Those that act at the level of the cell surface – outside
a. Peptide hormones and the catecholamines
b. Growth factors
c. Water-soluble and unable to cross the membrane
d. They exert their effects on the target via intracellular secondary messengers
2) Those that enter the cell to exert their effects – inside
a. Steroid hormones and the thyroid hormones
b. Lipid soluble and can readily penetrate the membrane
c. They exert their effects from within the target cell

Question 19

9. Explain the mechanism of action of G protein-coupled receptor using cAMP as a secondary messenger

Answer 19

• For those agents acting at the cell surface, it has been shown that an essential target in their mode of action is the generation of an identifiable intracellular second messenger
  o This acts to notify the cell that the first messenger (hormone/ growth factor) has bound
  o The first second messenger/ most important to be identified was cyclic AMP
• Cyclic AMP-mediated response:
  o Hormone outside cell binds to receptor – causing change in shape of receptor which activates G protein (GNP) which interacts with GTP and this eventually activates a membrane bound enzyme called adenylate cyclase – which is able to take ATP and convert it into a circular molecule cyclic AMP – this is then able to allosterically activate protein kinase A within in cell – can then phosphorylate enzymes (switch them on or off) because it is a protein

Question 20

10. Distinguish between the terms hypo- and hyperglycemia

Answer 20

Hyperglycemia
Excessive blood glucose in circulating plasma
Generally classified as BG > 10 mM
Has widespread effects on the body: CNS; heart; immune system; skin; vision
Can be caused by diabetes; eating disorders; some drugs; some diseases; and physiological stress

Hypoglycemia
Lower than normal level of circulating blood glucose
Generally classified as BG < 3.6 mM
Has wide-ranging effects: adrenergic system; CNS; neuroglycopenia
In adults, it can be caused by: diabetes’ immunological disorders; problems with the adrenal and pituitary glands; tumours

Question 21

11. Explain why blood glucose levels need to be regulated

Answer 21

• Glucose is the primary source of energy for all cells of the body
• Some cells can only metabolise glucose, so it is required in the body at all times
• Blood glucose levels in the body are tightly regulated within a range of 4 to 6 mM

Question 22

12. Describe the basic structure and site of synthesis of Insulin

Answer 22

• A dimer consisting of two peptide chains (A and B), held together by disulphide bonds
  o 51 amino acids, MW 5808 Da
  o Synthesised in the pancreas

Question 23

13. Explain how insulin release is controlled

Answer 23

Stimulation of insulin secretion - Insulin is stored in granules in the cytosol
- Secretion of insulin and glucagon are tightly regulated to maintain glucose levels
- B-cells transport glucose via GLUT2 and phosphyorylate it via glucokinase
- As levels of phosphorylated glucose increase, it signals release of insulin and decreases release of glucagon
Inhibition of insulin secretion - Insulin secretion is inhibited by lack of dietry fuel or during stress (e.g. infection)
- This is mediated via adrenaline
- Regulated via the sympathetic nervous system
- Allows the body to override glucose-dependent insulin production during emergencies
Insulin: mechanism of action - When insulin binds the alpha-subunits, tyosine kinase is activated and phosphorylates cellular proteins

Question 24

14. Describe the major effect of insulin on metabolism

Answer 24

• Increases glucose uptake into muscle cells and adipocytes - GLUT4 transporters are translocated form the intracellular vesicles to the cell membrane allowing more glucose to be transported from blood to muscle cells
• Increases glycolysis – because will use more glucose
• Decreases hepatic gluconeogenesis – because wont be producing glucose at a time we are trying to decrease it
• Increases glycogen synthesis, decreases glycogenolysis – so glucose isnt being released
• Increases triglyceride synthesis (in adipocytes) – so cells use glucose instead
• Decreases lipolysis
• Increases protein synthesis – so its not available as a feul source, same as trigs
• Decreases protein degradation

Question 25

15. Describe the basic structure and site of synthesis of glucagon

Answer 25

- A single polypeptide chain, unlike insulin - 29 amino acids, MW 3485 Da - Similar to insulin, it is synthesised as a large precursor - This is serially cleaved to produce active glucagon

Question 26

16. Explain how glucagon release is controlled

Answer 26

Inhibition of glucagon secretion - Glucagon is inhibited by elevated blood glucose Glucagon: mechanism of action - Glucagon acts through a G protein-coupled receptor - Binding of glucagon causes an increase in cAMP, which activates protein kinase A - This activates a cascade of other enzymes that affect carbohydrate and lipid metabolism

Question 27

17. Describe the major effect of glucagon on metabolism

Answer 27

- Decreases glycolysis – want glucose in blood - Increases hepatic gluconeogenesis – want increase glucose in liver - Decreases glycogen synthesis, increases glycogenolysis (in liver not muscle) – don’t want to be packaging it away - Decreases triacylglycerol synthesis – want to use this as a energy fuel instead of glucose, same for protein below - Increases lipolysis (in adipose tissue) - Decreases protein synthesis - Increases protein degradation

Question 28

18. Relate the levels of Insulin and glucagon to their respective roles in controlling blood glucose concentration

Answer 28

Insulin ↓ Glycogenolysis ↓ Gluconeogenesis ↓ Ketogenesis ↓ Lipolysis Polypeptide hormone, produced by beta cells of the pancreas Primary function is to trigger absorption of glucose form the blood into the liver, skeletal muscle and fat tissue Its many other effects mean that it is the main hormone regulating the use of feuls by the body Glucagon ↑ Glycogenolysis ↑ Gluconeogenesis ↑ Ketogenesis ↑ Lipolysis Polypeptide hormone, produced by alpha cells of the pancreas Primary function is to trigger the release of glucose into the blood from the liver via gluconeogenesis and glycogenolysis Essentially, its main role is to oppose the actions of insulin

Question 29

1. Understand the role of the pathway of glycolysis in metabolism

Answer 29

- Glycolysis is the metabolic pathway that converts glucose into pyruvate. The free energy released during the biochemical reactions in glycolysis is used to generate a net gain of two molecules of ATP.

Question 30

3. Identify the reversible and irreversible steps of the pathway

Answer 30

1) Step 1: Phosphorylation of glucose = irreversible under physiological conditions 2) Step 2: Isomerisation of glucose-6-P = completely reversible 3) Step 3: Phosplorlation of fructose-6P = irreversible under physiological conditions 4) Step 4: Cleavage of fructose-1,6-biphosphate = reversible under physiological conditions 5) Step 5: interconversion of triose phosphate = reversible under physiological conditionsn 6) Step 6 = Oxidation of glyceraldehyde-3-P = reversible 7) Step 7: Transfer of phosphate to ADP = reversible 8) Step 8: Intramolecular rearrangement of 3PG = reversible 9) Step 9: Dehydration of 2PG = reversible 10) Step 10: Transfer of P from PEP to ADP = irreversible

Question 31

4. Explain the physiological significance between the different Km and tissue localization of glucokinase and hexokinase

Answer 31

Hexokinase: - Is active in most cells - Will phosphorylate glucose, as well as other sugars (e.g. mannose, fructose) - Has a K m ~ 0.1 mM (low) high affinity, Vmax gets achieved at very low substrate concentration Glucokinase: - Active in liver and islet cells of the pancreas - Is specific for glucose - Has a K m ~ 10 mM (high) lower affinity for glucose, glucose activity needs to be high before glucokinase activity is saturated - By having a much higher Km, glucokinase has the capacity to respond to the large increase of glucose in the blood after a meal rich in CHO - Very important for the liver, as the liver is the main organ that processes CHO after a meal - But, the low Km of hexokinase is important in most other tissues between meals, as it allows for maximal activity when glucose concentration is lower - Hexo cannot increase rate of phosphorylation – low km, but gluco can because larger km

Question 32

5. Locate the site of glycolysis in a eukaryotic and bacterial cell

Answer 32

- Eukaryotes and prokaryotes = cytoplasm

Question 33

6. Calculate the overall yield of ATP in glycolysis from metabolism of glucose

Answer 33

net ATP production of 2 ATP per glucose

Question 34

7. Define the general features of all membranes

Answer 34

- All cells are surrounded by membranes - A cell can’t be alive if it doesn’t have a membrane of some sort - The membrane is composed of a phospholipid bilayer - It is semipermeable - Large, non-polar molecules (like glucose) cannot simply diffuse into the cell - They need a specific transport system - Fluid mosaic model

Question 35

8. Explain the factors that affect the permeability of a membrane to a given molecule

Answer 35

Key factors which dictate how a molecule is transported are its polarity, size and charge. - In general, membranes are: 1. Permeable to small, non-polar molecules (CO 2 and O 2) up to a MW of approx 100 (eg. EtOH, glycerol) 2. Poorly Permeable to non-polar molecules with MW > 100 (e.g. glucose) 3. Impermeable to polar molecules or ions (pyruvate, Cl-, Na +) – anything with a pos or neg charge

Question 36

9. Distinguish between simple and facilitated diffusion

Answer 36

1. Simple diffusion: - A passive process - The unassisted net movement of a solute from a region of high concentration to a region of low concentration - No protein required to assist as molecule moves solely on the basis of the concentration gradient - Because membranes have a non-polar interior, simple diffusion only works for gases, and small, non-polar molecules - Diffusion of CO 2 and O 2 into and out of erythrocytes - CO2 and O 2 small, non-polar molecules - Diffuse at a high rate across membranes - ONLY from high-low concentrations 2. Facilitated diffusion (aka passive transport) - A passive process - Where a protein acts to facilitate the movement of a molecule across a membrane - Still only moves the molecule down the concentration gradient - But, facilitated diffusion gets around the hydrophobicity of the membrane interior using the protein - The protein can either transport or channel the molecule through the hydrophobic region of the membrane.

Question 37

10. Explain glucose uptake via GLUT transporters

Answer 37

They transport from high concentration to low concentration (facilitated diffusion) changes conformation

Question 38

11. Relate your knowledge of membranes, transport, and enzyme-catalyzed reactions to explain why intermediates of glycolysis are phosphorylated

Answer 38

1. They can’t traverse the cell membrane - There are no transporters for G-6-P on the membrane - Addition of PO 4 3- increases polarity of glucose - ‘Traps’ glucose in the cell 2. Necessary to form substrate-enzyme complex - Only recognised by enzyme after phosphorylated 3. Necessary to phosphorylate ADP, yielding a net gain of ATP

Question 39

12. Explain the biochemical role of fermentation

Answer 39

An enzyme-catalyzed, energy-generating process in which organic compounds act as both donors and acceptors of electrons in the absence of oxygen (Fuel molecules are broken down anaerobically)

Question 40

13. Distinguish between the pathway of fermentation in yeast and muscle cells

Answer 40

Ethanol fermentation Yeast cells and some bacteria (e.g. Zymomonas) - Pyruvate is converted to ethanol: - Results in the oxidation of NADH to NAD+ - NAD+ can then return to glycolysis and be reused - Ethanol fermentation allows cells to recycle NAD +/NADH - Note: First reaction is irreversible Fermentation to ethanol is essential in: Baking: - CO2 released in baking causes dough to rise - Part of the smell of fresh bread is the ethanol Brewing - Both wine and beer are produced from yeast fermenting grape sugar/grain carbohydrate - Ethanol is a poison to yeast at higher concentration (above about 16% v/v) - CO2 released from fermentation is trapped in sparkling wines Lactate fermentation In skeletal muscle cells (and lactic acid bacteria) - When O2 low, pyruvate converted to lactate: - Results in the oxidation of NADH to NAD + - Lactate fermentation provides a crucial means for hypoxic tissue (low oxygen) that cannot undergo respiration to recycle NAD +/NADH - Note that the reaction is reversible, - When [NADH] is high, favours forward reaction In Skeletal muscle cells - Lactate is acidic (i.e. lactic acid) - Accumulation in muscles lowers pH - Ultimately causes muscle contraction to cease Lactic acid bacteria - Metabolise milk sugar (lactose) - Do not possess an electron transport chain - Can’t respire, must ferment - Produce lactate which accumulates - Lowers pH which causes ppt of milk proteins - Thickens milk and gives it a sour taste (yoghurt)

Question 41

14. Be able to calculate both enzyme activity and specific activity and distinguish between these two terms

Answer 41

Enzyme activity = a) how much substrate is consumed (or how much product is formed) by an enzyme, b) over a known period of time c) from a given sample or reaction volume Specific activity = a) enzyme activity/ mass of protein present in a sample

Question 42

1. Understand the role of the TCA cycle in metabolism

Answer 42

Required to fully metabolize pyruvate, serve as a source of precursors for biosynthesis and route to breakdown macromolecules

Question 43

3. Identify the reversible and irreversible steps of the pathway of TCA cycle

Answer 43

1) Step 1: Acetyl CoA – Citrate (6C) = irreversible 2) Step 2: Citrate – isocitrate = reversible 3) Step 3: Isocitre – a-ketoglutarate = irreversible 4) Step 4: a-ketoglutarate – succinyl CoA (4C) = irreversible 5) Step 5: Succinyl CoA (4C) – succinate (4C) = reversible 6) Step 6: Succinate (4C) – fumarate (4C) = reversible 7) Step 7: Fumarate (4C) – malate (4C) = reversible 8) Step 8: Malate (4C) - oxaloacetate (4C) = reversible

Question 44

4. Locate the site of the TCA cycle in a eukaryotic and bacterial cell

Answer 44

In bacteria and archaea, the TCA cycle enzymes are in the cell cytoplasm (as for glycolysis) But, in eukaryotes, the TCA cycle enzymes are in the mitochondria So, the product of glycolysis (i.e. pyruvate) must make its way into the mitochondrion before being able to enter the TCA cycle

Question 45

5. Distinguish between the permeability of the inner and outer mitochondrial membrane

Answer 45

The outer membrane of the mitochondrion is freely permeable to small molecules and ions But the inner membrane is not Need a specific transport system to allow access to mitochondrion

Question 46

6. Distinguish between primary and secondary active transport

Answer 46

Types of active transport: Active transport systems primary or secondary - Primary: energy used to drive solute movement against concentration gradient - Secondary: a gradient already established by primary active transport is used to cotransport a second solute

Question 47

7. Explain how pyruvate enters mitochondria

Answer 47

Pyruvate is transported into the mitochondria by the pyruvate transport system (aka MCP or monocarboxylate carrier protein)

Question 48

1. Locate the site of the electron transport chain in a eukaryotic and bacterial cell

Answer 48

Eukaryotes = mitochondria Bacterial = plasma membrane – no mito

Question 49

2. Identify the four complexes of the mitochondrial ETC and describe which complexes receive electrons from NADH and FADH2 and which transfer protons

Answer 49

Complex I: Step 1 - The first reaction is the oxidation of NADH + H + by NADH oxidoreductase - A tightly-bound prosthetic group, flavin mononucleotide (FMN) becomes reduced Complex I: Step 2 - NADH oxidoreductase also contains non-heme Fe - This is probably involved in the transfer of electrons to coenzyme Q (aka ubiquinone) Complex II: Step 3 - Reduced FADH 2 is generated by FAD-linked dehydrogenases - e.g. succinate dehydrogenase from TCA cycle; fatty acyl CoA dehydrogenase from -oxidation - FADH 2 directly reduces coenzyme Q Complex III: Steps 4-6 - Coenzyme Q donates electrons to complex III - Complex III consists of a series of cytochromes - Electron transport proteins that contain a heme prosthetic group - Iron alternates between ferric (Fe3+) and ferrous (Fe2+) states Complex IV: Final Step - Cytochromes (a + a3) are the terminal members of the chain - They exist as a complex called cytochrome oxidase - This complex also contains copper, which undergoes cupric (Cu2+)/cuprous (Cu +) redox reaction - This reaction is important in the final transfer of electrons to O2 - Of all the members of the ETC, only cytochrome (a + a3) can react directly with O2

Question 50

3. Describe how the ETC functions to generate a proton gradient

Answer 50

- Flow of electrons down the ETC drives H + ions across the mitochondrial membrane into the intermembrane space - This creates an electrochemical (proton) gradient - This gradient is a potential source of energy - Cells harness it to generate ATP - Proton transfer occurs at complex I, III and IV - Called the chemiosmotic theory

Question 51

4. Explain how the proton gradient is harnessed to derive ATP

Answer 51

The proton gradient produced by proton pumping during the electron transport chain is used to synthesize ATP. Protons flow down their concentration gradient into the matrix through the membrane protein ATP synthase, causing it to spin (like a water wheel) and catalyze conversion of ADP to ATP.

Question 52

5. Define respiratory control and explain its biological significance

Answer 52

- Regulation of the rates of ETC and oxidative phosphorylation by ADP levels is known as respiratory control - This is of obvious physiological importance - Electrons do not flow from fuel molecules to O2 unless ATP synthesis is needed - This means that fuel molecules are not catabolised unnecessarily

Question 53

6. Describe other processes which are powered by proton gradients in cells

Answer 53

- Electron transport chains couple redox reactions to generation of proton gradients. - Proton gradients can in turn be harnessed by ATP synthase to generate ATP. - Proton gradients are essential to life - Important for many processes other than Ox Phos o Transport processes o Chemotaxis o Photosynthesis - Some organisms (e.g. lactic acid bacteria) do not have ETCs o They use ATP to generate a proton gradient (i.e. They turn the ‘ATPase around’ and pump protons out)

Question 54

7. Distinguish between the roles of the malate-aspartate and glycerol-3-phosphate shuttles and explain the similarities and differences in their operation

Answer 54

In the malate–aspartate shuttle, 2.5 molecules of ATP are produced for each molecule of cytosolic NADH, rather than 1.5 ATP in the glycerol– phosphate shuttle, a point that affects the overall yield of ATP in these tissues. Re-oxidation of cytoplasmic NADH 1. Glycerol 3-Phosphate shuttle: Glycerol-3-P Shuttle - Mitochondrial G3P DH is actually part of the electron transport chain - 1 NADH is converted to 1 FADH 2. - This shuttle is present in skeletal muscle cells and nerve cells 2. Malate-Aspartate shuttle Malate-Aspartate Shuttle - 1 cytosolic NADH is converted to 1 mitochondrial NADH - There is a cytosolic malate DH and the normal mitochondrial Malate DH (TCA cycle) - This shuttle is most active in Liver, kidney and heart cells.

Question 55

8. Explain how the oxidation of 1 mole NADH and 1 mole FADH2 in the electron transport chain yields 2.5 and 1.5 moles ATP respectively in oxidative phosphorylation

Answer 55

- Remember that as the electrons from each NADH and FADH2 move down the ETC, protons are transferred into the intermembrane space - These protons are used to drive ATP synthesis by ATP synthase - The number of H+ translocated is different for NADH and FADH2 - NADH results in 10H+ transported out - FADH2 results in 6H+ transported out - ATP synthesis requires 3H+ in for synthesis - But, we need to account for protons required for ATP, ADP and Pi movement

Question 56

1. Identify the sites of gluconeogenesis and explain its importance in metabolism.

Answer 56

- The major sites of gluconeogenesis are the liver, the kidney and the intestine - Maintain glucose level during fasting state

Question 57

2. Describe the four key precursors for gluconeogenesis in humans.

Answer 57

- Lactate - liver, kidney - Glycerol – intestine - Alanine – liver - Glutamine - kidney, intestine

Question 58

3. Identify the three irreversible steps of glycolysis and the four enzymes required to bypass these steps and allow gluconeogenesis to occur.

Answer 58

The three irreversible steps are: 1) Glucose to glucose-6-P by hexokinase/glucokinase 2) Fructose-6-P to fructose-1,6-bisp by phosphofructokinase-1 3) Phospheonolpyruvate to pyruvate by pyruvate kinase • To bypass hexokinase/glucokinase: Glucose-6-phosphatase (ER) • To bypass phosphofructokinase-1: fructose-1,6-bisphosphatase (cytoplasm) • To bypass pyruvate kinase: Pyruvate carboxylase (mitochondria); phosphoenolpyruvate carboxykinase (cytoplasm)

Question 59

4. Explain which step is the primary regulatory point for gluconeogenesis and how this is controlled.

Answer 59

- Fructose-1,6-bisphosphatase - Acetyl CoA

Question 60

5. Explain why the ATP yields of glycolysis and gluconeogenesis differ

Answer 60

- 4 ATP required + 2 GTP required - The conversion of glucose to pyruvate is an energy-releasing reaction, therefore heat is lost - If we wish to reverse the overall process, then the total energy will need to be put into the system (6 ATP)

Question 61

6. Describe the role of the Cori cycle

Answer 61

- A process in the liver that regenerates glucose from lactate released by muscles

Question 62

7. Explain why acetyl CoA is not a gluconeogenic precursor in animals

Answer 62

- In animals, acetyl CoA enters the TCA cycle and yields 2 CÓ with no net production of oxaloacetate or any other TCA intermediate. Therefore, no gluconeogenesis can come from acetyl CoA

Question 63

8. Understand the importance of regulation of biochemical pathways

Answer 63

- Although there are many thousands of biochemical conversions possible in an organism, there are usually only specific circumstances when they need to happen. - That is: the reactions need to be regulated in a way that reflects the needs of the cell - Regulation of biochemical pathways allows for metabolic coordination and avoids wasting energy

Question 64

9. Identify the regulatory points in glycolysis and the TCA cycle

Answer 64

Glycolysis - Step 1: Phosphorylation of Glucose. Enzymes are Hexokinase and Glucokinase - Step 3: Phosphorylation of Fructose-6-P. Enzyme is phosphofructokinase. - Step 10: Transfer of P form PEP to ADP. Enzyme is Pyruvate Kinase. TCA Cycle - Step 1: Acetyl CoA + Oxaloacetate --> Citrate. Enzyme is Citrate Synthase - Step 3: Isocitrate --> a-ketoglutarate. Enzyme is Isocitrate dehydrogenase - Step 4: a-ketoglutarate --> succinyl CoA. Enzyme is a-ketoglutarate dehydrogenase  These enzymes are allosterically regulated and catalyse the irreversible steps of the TCA cycle, which are the main point of regulation.

Question 65

10. Describe how glucokinase and hexokinase regulate entry of glucose intro glycolysis

Answer 65

- Hexokinase is allosterically regulated by one of its products (glucose-6-phosphate), whereas glucokinase is hormonally controlled by insulin Hexokinase: - Is active in most cells - Will phosphorylate glucose, as well as other sugars (e.g. mannose, fructose) - Has a K m ~ 0.1 mM (low) high affinity, Vmax gets achieved at very low substrate concentration Glucokinase: - Active in liver and islet cells of the pancreas - Is specific for glucose - Has a K m ~ 10 mM (high) lower affinity for glucose, glucose activity needs to be high before glucokinase activity is saturated

Question 66

11. Explain how pyruvate kinase regulates glycolytic activity

Answer 66

- An allosteric enzyme - High ATP allosterically inhibits the enzyme’s affinity for PEP - Fructose-1,6-bisphosphate is an activator of pyruvate kinase - Also inhibited by long chain fatty acids and acetyl CoA

Question 67

12. Describe how phosphofructokinase and Fructose-1,6 bisphosphatase are regulated

Answer 67

- By alloseteric and hormonal control - Allosteric enzyme - Concentration is regulated by glucagon - Fructose-2,6-bisphosphate is a compound produce in liver cells - is a stimulatory modulator

Question 68

13. Explain the regulation of pyruvate dehydrogenase

Answer 68

- High ATP allosterically inhibits the enzyme's affinity for PEP F-1,6-bis is an activator of pyruvate kinase Also inhibited by long fatty acids, acetyl CoA and Alanine

Question 69

14. Identify stimulators and/or inhibitors of citrate synthase, isocitrate dehydrogenase and α-ketoglutarate dehydrogenase

Answer 69

TCA cycle: Control of the pathway - Citrate synthase o Inhibited by NADH, ATP, succinyl CoA and Citrate o Stimulated by ADP - Isocitrate dehydrogenase o Inhibited by ATP and NADH o Stimulated by ADP - α ketoglutarate dehydrogenase o Stimulated by Ca2+ o Inhibited by NADH and Succinyl CoA