Bioenergetics & Enzymes Flashcards

Question

What is FAD?

Answer 1

FAD (Flavin adenine dinucleotide) is another important electron acceptor.

Answer 2

This reaction is important for the breakdown of fatty acids.

Answer 3

Biochemical isomerisations involve the intra molecular shift of a hydrogen to change the location of a double bond. The metabolically most important isomerisation is the aldose-ketose interconversion. (NOTION. 2.10)

Answer 4

Rearrangements produce altered carbon skeletons. (NOTION 2.11)

Answer 5

… both degradative and biosynthetic metabolism

Answer 6

The most common group used for C-C bond formation is the electrophile carbonyl group of aldehydes, ketones, esters or CO2. The second compound (nucleophile) are stabilised carbanions. (NOTION 2.12)

Answer 7

(NOTION 2.13)

Answer 8

The aldolase reaction (glycolysis) uses the reverse mechanism.

Answer 9

1. See NOTION 2.14 2. Decarboxylation is an important reaction in the metabolism i.e isocitrate synthases or fatty acid synthase

Answer 10

The Capacity for Work - A dynamic state related to change - It’s presence emerges when a change occurs

Answer 11

Joules = Nm = Ws = (Kg m^2)/ s^2

Answer 12

Calorie (cal) 1 cal = Energy required to heat 1g of water by 1*C

Answer 13

1 cal = 4.184J

Answer 14

Wh or kWh is often used

Answer 15

Kinetic energy = The harnessing of potential energy. (Biosynthesis results from harnessing energy).

Answer 16

Potential energy = bound in a specific form

Answer 17

- Chemical (combustions, batteries, fuel cells) - Mechanical (moving/ turning masses) - Heat - Light - Electric - Nuclear

Answer 18

Mechanical work: - Muscle contraction - Cell division Chemical work: - Synthesis of molecules Transport work: - Diffusion - Active Transport Electrical work: - Transport of charged particles - Action potentials in nerves & muscles

Answer 19

… always include the release of waste heat

Answer 20

Energy cannot be created or destroyed, but is transformed from one form to another without being depleted. Sometimes it appears as if energy is destroyed, it isn’t: - Just the ‘quality’ changes - The energy is ‘diluted’ - Energy can loose its ability to ‘do’ work

Answer 21

All energy transformations ultimately increase the entropy (disorder/ randomness) of the universe. However, building and maintenance of the structure of living organisms involves a decrease in entropy.

Answer 22

NOTION 1.1

Answer 23

NOTION 1.2

Answer 24

(Gibbs) Free Energy = Energy that can be used or ‘harvested’ by organism

Answer 25

Total Energy = Useful Energy + Non-Useful Energy H (Enthalpy) = G (Free energy) + T x S (Temp x Entropy)

Answer 26

In a spontaneous process, a system: - Gives up energy And/or - Becomes more random Taking the previous equation a spontaneous process involves: - A decrease in H And/or - An increase in S This means that any process leading to a decrease in G is spontaneous.

Answer 27

The sign of /\G indicates on which side of the equilibrium the reactant concentrations lie at any particular time. The size of /\G is an indication of how far from the equilibrium the reaction is. NOTION 1.3

Answer 28

Reactions with a positive /\G can occur if linked to a reaction that releases a lot of free energy. NOTION 1.4

Answer 29

No, living organism are open systems and therefore they can never reach an equilibrium. For example: - Ingestion of high enthalpy low entropy food - Conversion to low enthalpy high entropy waste products

Answer 30

They form a steady-state system. In dynamic steady state systems production rate of a product is in balance with its consumption.

Answer 31

Mode of action: - Lock & key mechanism - Enzyme-substrate complex - Enzymes allow - Lower temperatures - Neutral pH

Answer 32

Yes, even exergonic reactions often require a certain activation energy to reach a high energy transition state. NOTION 1.5

Answer 33

The higher the difference between the free energy of the transition stage and the free energy of the reactants, the slower the reaction.

Answer 34

Enzymes reduce the free energy of the transition stage & allow: - Faster reactions - Lower temperatures - Neutral pH

Answer 35

No, the lowered ‘barrier’ increases reaction speed for both forwards & reverse reaction, therefore the equilibrium constant does NOT change.

Answer 36

Catabolic metabolism is the breakdown or oxidation of nutrients - it releases energy. It produces intermediates that are useful for the metabolism.

Answer 37

Anabolic metabolism produces or synthesises new components or substances i.e enzymes, fat, hormones. This consumes energy. It requires different precursors.

Answer 38

1. Glycolysis 2. Krebs Cycle 3. Electron Transport Chain & Oxidative Phosphorylation NOTION 3.1

Answer 39

NAD+ NADP+ FAD

Answer 40

ATP consists of 2 main components: - Phosphates - Adenosine NOTION 3.2

Answer 41

Energy is stored in the phospho-anhydride bond

Answer 42

ATP is produced from ADP by substrate level phosphorylation or oxidative phosphorylation.

Answer 43

/\G = around - 31 kJ/mol NOTION 3.3

Answer 44

1. Electron density in the anhydride bonds 2. Negative charges in close proximity 3. Resonance stabilisation of inorganic phosphate NOTION 3.4

Answer 45

NOTION 3.5

Answer 46

NOTION 3.6

Answer 47

The actual energy released (from ATP hydrolysis) under physiological conditions has to consider the concentrations. /\G = /\G* + RT ln ( [ADP][Pi]/ [ATP][H2O]) T = absolute temperature (k) R = Gas constant = 8.31451 J/K/mol

Answer 48

The metabolites at the top of the following table can lead to the phosphorylation of those below: NOTION 3.7

Answer 49

A combustion is an uncontrolled reaction that produces heat. In the metabolism the energy is released in a chain of controlled reactions to make it usable for other processes.

Answer 50

C6H12O6 + 6O2 -> 6CO2 + 6H2O

Answer 51

NOTION 3.8

Answer 52

The redox potentials can be calculated using the Nernst equation. Redox potentials are described with the identifier: E The unit for redox potential is: V.

Answer 53

/\G = -n F /\E n = electrons transferred F = faraday constant = 96, 494 J/V /\E = redox potential Note: the negative sign means that positive redox potentials are linked to a negative ∆G, therefore half-reactions with a positive potential are spontaneous.

Answer 54

NOTION 3.9

Answer 55

NOTION 3.10/ 3.11

Answer 56

Electrons are transferred from the high chemical potentials to lower chemical potentials. Each transfer is linked to a release of free energy: NOTION 3.12

Answer 57

The ETC consists of many redox complexes that are located in the inner membrane of the mitochondria and transfer electrons from NADH + H+ or FADH2 to the oxygen. The transport of electron pairs (2 e-) through these redox complexes is linked to the production of ATP.

Answer 58

An enzyme is a biological catalyst which: - Is made up of a globular protein - Catalyses very high reaction rates - Shows great reaction specificity - Work in mild temperature/ pH conditions - Can be regulated

Answer 59

Some enzymes are termed ribozymes - catalytic RNA molecules with no protein component

Answer 60

Cofactor = non-protein component needed for activity Fe2+ or Fe3+ = Catalase, Peroxidase K+ = Pyruvate Kinase Mg2+ = Glucose-6-phosphate

Answer 61

Coenzyme = Complex organic molecule, usually produced from a vitamin FAD = Riboflavin NAD+ = Niacin Coenzyme A = Pantothenate

Answer 62

Prosthetic group = Cofactor covalently bound to the enzyme or very tightly associated with the enzyme

Answer 63

Apoenzyme = The protein component of an enzyme that contains a cofactor Holoenzyme = “Whole enzyme” - The Apoenzyme plus the cofactor(s)

Answer 64

Nearly all enzymes end in “-ase” with the name of their substrate or activity.

Answer 65

1. Oxidoreductases = Transfer e- 2. Transferases = Group Transfers 3. Hydrolases = Hydrolysis 4. Lyases = Form, or add groups to double bonds 5. Isomerases = Transfer groups within molecules (form isomers) 6. Ligases = Formation of C-C, C-S, C-O and C-N bonds

Answer 66

They do: - Increase rate of spontaneous reactions - Lower the activation energy of biochemical reactions - Accelerate movement towards reaction equilibrium They do not: - Move reaction equilibria - Make a non spontaneous reaction spontaneous

Answer 67

Gibes Free-Energy (G)

Answer 68

Spontaneous reactions aren’t instantaneous because of the energy barrier (i.e the energy required to position chemical groups correctly, bond rearrangements, e- rearrangements, etc). NOTION 4.1

Answer 69

- Addition of an enzyme lowers the activation energy. - An enzyme will facilitate the change of S to P through intermediates. - Enzymes allow the reaction to proceed via a different route. - Enzymes form non-covalent bonds with substrate molecules, called the “binding energy” allowing them to take the reaction through a different path of reaction intermediates. NOTION 4.2

Answer 70

1. Entropy reduction 2. Desolvation 3. Induced fit

Answer 71

Entropy reduction – Molecules in free solution will only react by “bumping” into one another – Enzymes “force” the substrate(s) to be correctly orientated by binding them in the formation they need to be in for the reaction to proceed

Answer 72

Desolvation – Weak bonds between the substrate and enzyme essentially replace most or all of the H-bonds between substrate and aqueous solution

Answer 73

Induced fit – Conformational changes occur in the protein structure when the substrate binds

Answer 74

The active site needs to be compatible with the transition state (but not the substrate itself). This is is because if the active site was compatible with the substrate itself, the enzyme substrate complex would be highly stable, and more energy would then be required to reach the transition state.

Answer 75

1. Enzyme kinetics 2. Mutagenesis (Manipulating DNA sequences to swap one amino acid with another amino acid, to then determine enzyme function) 3. 3D Structure (By X-Ray Crystallography etc)

Answer 76

If we changed the substrate concentration [S] we would change the initial rate of a reaction. More substrate = higher initial rate of reaction. As the reaction proceeds, the substrate is used up and the rate of reaction changes. NOTION 4.3

Answer 77

- At low [S] you get an almost linear increase in Vo as [S] increases - At higher [S] the Vo changes very little in response to an increase in [S] - When [S] becomes so large that V0 changes are vanishingly small you get maximum reaction velocity, Vmax NOTION 4.4

Answer 78

It considers: - Maximum velocity of the enzyme Vmax - Stability of the enzyme-substrate complex - Km

Answer 79

They proposed a model to account for the hyperbolic curve seen when you plot Vo against [S]

Answer 80

Critical to this model is the formation of an enzyme substrate complex (ES). NOTION 5.1

Answer 81

- Model states that the first part of the reaction (to produce ES) occurs reversibly - Second part of the equation (to produce E and P) occurs more slowly than the first part

Answer 82

Several assumptions: – If 2nd part is slower it must limit the rate of the overall reaction, so the overall rate of reaction must be proportional to the amount of ES – In other words, more ES would give a higher overall reaction rate and less would give a slower overall reaction rate

Answer 83

Vo usually equates to the steady state of a reaction, so study of these initial rates of reaction is termed “steady state kinetics”.

Answer 84

NOTION 5.2 [S] = Substrate concentration Vo = Initial reaction velocity Vmax = Maximum reaction velocity Km = Michaelis constant

Answer 85

Michaelis constant is calculated from the hyperbolic reaction curve at half of the Vmax. Km is also calculated another way (to be discussed later). NOTION 5.3

Answer 86

NOTION 5.4

Answer 87

NOTION 5.5

Answer 88

NOTION 5.6

Answer 89

“Km is equivalent to the substrate concentration at which the initial reaction rate is half of the maximum reaction rate”

Answer 90

- If we have some purified enzyme we can set up an experiment in which we know [S] - We can measure V0 and therefore be able to plot this on a graph to measure the Vmax and Km - One problem with this – it can be hard to determine exactly what the Vmax is because the graph essentially continues to infinity - The “better way” of experimentally defining the Vmax and the Km is to draw a Lineweaver-Burk plot using the same data as was used to draw the M-M hyperbolic curve.

Answer 91

Lineweaver-Burk (double-reciprocal) plot can allow a straight line graph to be produced: NOTION 5.7

Answer 92

NOTION 5.8

Answer 93

Km = (k_-1)/(k_1) This can also be termed the dissociation constant, Kd of the ES complex.

Answer 94

So, Km is the ratio of rate constant for breakdown of ES to E + S compared to the rate constant for formation of ES from E + S

Answer 95

Therefore, – larger Km values indicate a less stable ES complex – smaller Km values indicate a more stable ES complex Km gives you a clue to the affinity of the enzyme with its substrate.

Answer 96

Vmax tells you how fast a reaction is proceeding when the enzyme is saturated with substrate. It can also be termed a measure of the enzymes catalytic rate.

Answer 97

NOTION 5.9

Answer 98

They can sometimes change in response to cell conditions (e.g the same enzyme may function differently in different cells).

Answer 99

These two enzymes are isozymes (they are different proteins but they catalyse the same reaction). They both catalyse: Glucose + ATP -> Glucose-6-phosphate + ADP

Answer 100

NOTION 5.10

Answer 101

When [blood glucose] goes UP after a meal the glucokinase activity increases but hexokinase activity does not respond as it is already working at its Vmax – this property allows glucokinase to respond proportionally depending on the [blood glucose].

Answer 102

When [blood glucose] is LOW, gluconeogensis releases glucose from the liver, but glucokinase cannot catalyse glucose back into glucose-6 phosphate under these conditions allowing glucose to be used by the body.

Answer 103

If we take the previous example of hexokinase, it catalyses the production of glucose-6-phosphate from glucose and ATP. Glucose and ATP are both substrate molecules for the hexokinase enzyme and can be analysed using M-M kinetics. NOTION 5.11

Answer 104

This can occur in several ways: – Random order or Ordered with a ternary complex – No ternary complex formation NOTION 5.12

Answer 105

Lactate dehydrogenase exhibits an ordered sequential mechanism to its catalysis of pyruvate to lactate. The coenzyme (NADH) binds first and the lactate is always released first, which can be drawn like this (this diagram is called Cleland notation): NOTION 5.13

Answer 106

Creatine kinase catalyses the formation of phosphocreatine from creatine and exhibits a random sequential mechanism. In this mechanism it doesn’t matter whether creatine or the cofactor (ATP) bind to the enzyme first, or which product is released first. This can be drawn like this: NOTION 5.14

Answer 107

Reactions were amino groups are shuttled between amino acids and ketoacids are classic examples of reactions that have no ternary complex formation. Aspartate aminotransferase shows a double displacement or ping-pong reaction pathway when it transfers an amino group from aspartate to α-ketoglutarate. This double displacement reaction can be drawn in Cleland notation like this: NOTION 5.15

Answer 108

Allosteric enzymes do not follow M-M kinetics. Allosteric enzymes are made up of many subunits, which contain many active sites. The kinetics of an allosteric enzyme looks like: - One substrate binding to an enzyme subunit can cause changes in other active sites on other subunits - This can lead to the concept of “cooperative binding” of substrate molecules - Haemoglobin is a good example of cooperative binding of a substrate NOTION 5.16

Answer 109

NOTION 5.17/5.18

Answer 110

1. Temperature 2. pH 3. Inhibitors

Answer 111

- Increase in temperature = Increase in molecule collisions - Increase in temperature = Increase in internal energy of molecules - Increase in temperature = eventually denatures enzymes

Answer 112

1. pH changes the charge of amino acids 2. If the active site amino acids charge changes the enzyme will cease to function correctly 3. Extreme pH will denature most enzymes 4. pH will also affect the substrates of the reaction, some of which may require H+ or OH- groups to be involved in the reaction

Answer 113

1. Competitive Inhibitor 2. Uncompetitive Inhibitor 3. Non competitive inhibitor NOTION 6.1

Answer 114

1. Competitive Inhibitors bind to enzymes non-covalently and will usually resemble the substrate molecule, therefore competing with the active site 2. This leads to a decrease in the affinity between the active site and the substrate, so the Km of the substrate-enzyme complex increases 3. Increasing substrate concentration can overcome this inhibition, so the same Vmax can be achieved 4. Therefore competitive inhibitors exhibit increased Km values but the Vmax remains unchanged – this gives a Lineweaver-Burke plot that look like.. NOTION 6.2

Answer 115

1. Controversial drug developed in 1990s 2. AZT = Azidothymidine 3. AZT acts by competitive inhibition of the reverse transcriptase enzyme 4. Reverse transcriptase is used by HIV to produce a dsDNA molecule from it’s ssRNA 5. AZT undergoes triphosphorylation in the body, and thus mimics the ordinary DNA precursor thymidine triphosphate (TTP)

Answer 116

1. Oseltamivir (Tamiflu) is an example of a transition state analogue 2. In the body oseltamivir is hydrolysed in the liver to it’s active form 3. The active form of Oseltamivir is then able to block the activity of neuraminidase enzyme 4. Neuraminidase normally cleaves sialic acid (found on the surface of cells) that allows the release of new virus particles from the cells

Answer 117

1. Antibodies can be generated against potentially any biological molecule. Therefore if you wanted, you could make an antibody that was specific to a transition state molecule This process would allow the production of an antibody with a structure that resembles that of the active site of the original enzyme 2. One difficulty with this method is that transition states are notoriously difficult to isolate as they are the intermediate step between an enzyme-substrate complex and an enzyme-product complex.

Answer 118

One naturally occurring catalytic antibody is that produced in the disease lupus erythematosus – this disease is characterised by the autoantibodies attacking the connective tissue of the joints, skin, kidneys, heart and lungs.

Answer 119

1. Non competitive inhibitors bind to enzymes non covalently and will usually attach to a site other than the active site of the enzyme. 2. The substrate is usually still able to bind the active site, so the Km of the substrate-enzyme complex remains unchanged. 3. Increasing substrate concentration does not change the inhibition so the Vmax will decrease. 4. Therefore non-competitive inhibitors exhibit unchanged Km values but the Vmax decreases – this gives a Lineweaver-Burke plot that look like... NOTION 6.3

Answer 120

This type of inhibitor will usually bind to the enzyme in a covalent, and therefore irreversible way.

Answer 121

CN- binds to Fe3+ of cytochrome c oxidase and disrupts the terminal respiratory system. Blocking the terminal respiratory system will effectively “starve” cells of ATP causing the individual with cyanide poisoning to exit the carbon cycle post haste.

Answer 122

1. Biological reactions occur in sequences or pathways. Key enzymes in these pathways can be termed regulatory enzymes as they can have the greatest effect on the overall pathway. 2. Often find that the first enzyme in a particular pathway holds the regulatory step for that pathway – makes sense as you don’t want to regulate something half way down a pathway.

Answer 123

Two main ways regulatory enzymes modulate reactions: - Allosteric enzymes - Covalently modified enzyme

Answer 124

Some biochemical pathways regulate the enzymes involved through feedback inhibition. A build up of the end product of a pathway, or a key junction in a pathway, can ultimately slow the entire pathway.

Answer 125

1. Bacterial threonine dehydratase can be inhibited by the final product of a five step pathway (L-isoleucine). 2. L-isoleucine binding to a different site other than the active site, changes the conformation of the active site blocking the enzymes action (non competitive inhibition) 3. A reduction in L-isoleucine releases the block on threonine dehydratase, so the system is adjustable depending on the needs of the bacteria

Answer 126

1. Allosteric effectors are usually cell metabolites that bind noncovalently to a site on the enzyme that is not the active site. This changes the enzyme structure. 2. Some effectors are activators & some are inhibitors. 3. Consequently, binding of the effector will increase or decrease the efficiency of substrate binding and processing. 4. Allosteric effectors are examples of non competitive inhibitors.

Answer 127

- Increase in [S] between A and B gives small increase in Vo - Same Increase in [S] between B and C gives much larger increase in Vo - This suggests that low [S] sensitises the enzyme so it responds more efficiently at higher [S] NOTION 6.4

Answer 128

2 models explaining allosteric enzyme kinetics: - Concerted model - Sequential model

Answer 129

Concerted model: - Each subunit can exist in two different conformations - One conformation binds to the substrate well, while the other doesn’t - When there is no substrate present, the enzyme flips between the two conformations - All subunits must be in the same conformation!! NOTION 6.5

Answer 130

This model assumes: - No flipping between different conformation states - Sub unit exists in a conformation that can bind S, activators, inhibitors - It is the binding that causes a conformational change - Substrate binding causes a change in one sub unit - This causes a change in another sub unit, allowing it to bind S more readily NOTION 6.6

Answer 131

One of the most widely studied, and one of the most important covalent modifications is phosphorylation. Approximately 30% of all eukaryotic proteins are phosphorylated: - At a single site - Multiple sites - Multiple Phosphorylations at one site

Answer 132

Enzymes catalyse the phosphorylation of enzymes: - Protein kinases - Add phosphoryl groups to proteins - Protein phosphatases - Remove phosphoryl groups

Answer 133

Multiple phosphorylation sites allow very fine control for enzyme function depending on the requirement of the particular enzyme at a given time. The enzyme never really exists in an “on” or “off” state, but instead has finely tuned activity dependant on the signals it receives. NOTION 6.7

Answer 134

1. Enzymes can exist as an inactive precursor protein, called a proprotein or proenzyme. Proteolytic cleavage can lead to the conversion of the inactive precursor protein, into its active form. 2. Pro proteins can be cleaved to give active enzymes by proteases 3. Digestive enzymes are regulated in this way - if they’re weren’t they would digest the parts of the gut where they are made 4. Digestive enzymes tend to be called zymogens (Proteases) 5. Cleavage of other proteins can also occur (e.g Insulin)

Answer 135

Covalent modification: - Ubiquitination - ADP Ribosylation - Methylation - Phosphorylation - Adenylylation - Acetylation - Myristoylation