Chapter 4: Biochemistry

Question 1

What are the 2 relevant forms of energy in chemsitry?

Answer 1

heat energy (movement of molecules) and potential energy (energy stored in chemical bonds)

Question 2

1st law of thermodynamics

Answer 2

the law of conservation of energy; states that the energy in the universe is constant

(implies that when energy of a system decreases, energy of sorroudings increases and vice vers)

Question 3

2nd law of thermondynamics

Answer 3

disorder (entropy) of the universe tends to increase, can also be said as spontaneous reactions tend to increase the disorder of the universe

Question 4

Gibbs Free Energy

Answer 4

1. ΔG = ΔH – TΔS (H = enthalpy, which is heat or thermodynamic potential)
   a) ΔG > 0 → needs heat, nonspontaneous
   b) ΔG
2. ΔH = ΔE – PΔV (E = bond energy)

ΔH = enthalpy

ΔG increases when ΔH increases

ΔG increases when ΔS decreases

Question 5

ΔG > 0

Answer 5

A positive ΔG is endergonic

Question 6

ΔG

Answer 6

A negative ΔG is exergonic

Question 7

ΔH

Answer 7

exothermic and release heat

Question 8

ΔH > 0

Answer 8

endothermic and require an imput of heat

Question 9

Standard Gibbs free energy

Answer 9

1. ΔGº → all at 1 M concentration
2. ΔGº’ → 1 M concentration and @ pH 7
3. ΔGº’ = -RtlnK’eq (R = gas constant, Keq = ratio of products to reactants at equilibrium)
   a) Keq = [C]eq[D]eq/[A]eq[B]eq
4. Remember: spontaneity says nothing about the reaction rate!

Question 10

Equillibrium

Answer 10

defined as the point where the rate of reaction in one direction equals the rate of reaction in the other

Question 11

Q

Answer 11

ratio of products to reactants in any given set up

Question 12

_{<span>K</span>eq}

Answer 12

ratio at equillibrium

Question 13

How can ΔG be negative if ΔGº’ is positive (which indicates that the reaction is unfavorable at standard conditions)?

Answer 13

The reaction may be favorable (ΔG

Question 14

Does K_eq indicate the rate at which a reaction will proceed?

Answer 14

K_eq indicates only the relative concentration once equillibrium is reached, not the reaction rate (how fast equillibrium is reached)

Question 15

When K_eq is large, which has lower free energy: products or reactants?

Answer 15

A large K_eq means that more products are present at equillibrium. Remember that equillibrium tends toward the lowest energy state. Hence, when K_eq is large, products have lower free energy than reactants.

Question 16

When Q is large, which has lower free energy: products or reactants?

Answer 16

The size of Q says nothing about the properties of the reactants and products. Q is calculated from whatever the initial concentrations happen to be. It is K_eq that says something about the nature of reactants and products since it describes their concentrations after equillibrium has been reached.

Question 17

Which direction, forward or backward, will be favored in a reaction if ΔG = 0?

Answer 17

If ΔG = 0 then neither the forward nor the reverse reaction is favored. Q = K_eq and when this is true we are by definition at equillibrium. Understand and memorize the following: When ΔG = 0, you are at equillibrium: forward reaction equals back reaction and the net concentrations of reactants and products do not change.

Question 18

Spontaneous means that a reaction may proceed without additional energy input BUT it says nothing about what?

Answer 18

It says nothing about the rate of a reaction

Question 19

Thermodynamics will tells you where

Answer 19

a system starts and finishes but nothing about the path traveled to get there. The difference in free energy in a reaction is only a function of the nature of the reactants and products. Therefore ΔG does not depend on the pathway a reaction takes or the rate of reaction, it is only a measurement of the difference in free energy between reactants and products.

Question 20

How does the ΔG for a reaction burning sugar in a furnace compare to the ΔG when sugar is broken down in a human?

Answer 20

The ΔG is the same in both cases. ΔG does not depend on the pathway only on the different energies of the reactants and products.

Question 21

All reaction proceed through a transient intermediate that is unstable and takes a great deal of energy to produce. The energy required to produce the transient intermediate state is called the …….. ?

Answer 21

activation energy , this is the barrier that prevents many reactions from proceeding even though the ΔG for the reaction may be negative. It is the activation energy that determines the kinetics of the reaction.

Question 22

Chemical Kinetics

Answer 22

the study of reaction rates

Question 23

How would the rate of a spontaneous reaction be affected if the activation energy were lowered?

Answer 23

The rate would be increased

Question 24

Transition State

Answer 24

The transition state of a chemical reaction is a particular configuration along the reaction coordinate. It is defined as the state corresponding to the highest potential energy along this reaction coordinate. It exists for a very short time either moving forward to form productor breaking back down into reactants

Question 25

Catalyst

Answer 25

A catalyst lowers the E_a of a reaction without changing the ΔG. The catalyst lowers the E_a by stabilizing the transition state, making existence less thermodynamically unfavorable.

It is also important to note that a catalyst is not consumed in the reaction, it is regenerated with each reaction cycle

Question 26

Enzymes are _________

Answer 26

catalysts, they increase the rate of a reaction by lowering the reactions activation energy but they do not affect ΔG between reactants and products. As catalysts, enzymes have a kinetic role, not a thermodynamic one.

Question 27

Will enzymes alter the concentrations of reagents at equillibrium?

Answer 27

No, it will only affect the rate at which reactants and products reach equillibrium.

Question 28

Hydrolase

Answer 28

hydrolyzes chemical bonds

Question 29

Isomerase

Answer 29

rearranges bonds within a molecule to form an isomer

Question 30

Ligase

Answer 30

forms a chemical bond

Question 31

Lyase

Answer 31

breaks chemical bonds by means other than onxidation or hydrolysis

Question 32

Kinase

Answer 32

transfers a phosphate group to a molecule from a high energy carrier such as ATP

Question 33

Oxidoreductase

Answer 33

runs redox reactions

Question 34

Polymerase

Answer 34

polymerization (adding of nucleotides to leading strand of DNA by DNA polymerase 3)

Question 35

Phosphatase

Answer 35

removes a phosphate group from a molecule

Question 36

Phosphorylase

Answer 36

transfers a phosphate group to a molecule from inorganic phosphate

Question 37

Protease

Answer 37

hydrolyzes peptide bonds

Question 38

Enzymes increase the rate of what type of reactions?

Answer 38

reactions that have a negative ΔG. These reactions would occur on their own without an ezyme but far more slowly than with one

Question 39

Thermodynamically unfavorable reaction in the cell can be driven forward by what?

Answer 39

reaction coupling, in reaction coupling, one very favorable reaction is used to drive an unfavorbale one. This is possible because free energy changes are additive.

Question 40

What is the favorable reaction that the cell can use to drive unfavorable reactions?

Answer 40

ATP Hydrolysis

Question 41

kinetics and activation energy summary

Answer 41

A. Kinetics: the study of reaction rates

B. Activation energy (Ea): energy required to produce a transient intermediate (denoted ‡)

1. ↑ Ea = slower rxn rate
2. The energy for bob to apply for a job is higher than Bob with or without a job

C. Catalysts (enzymes) will lower Ea but will not affect ΔG!!

D. Because enzymes lower Ea, they increase the rate at which favorable rxns occur!

1. Ea is lower with catalyst than without, but notice ΔG does not change!
2. ΔG represents net change of energy from reactants to products

E. ATP as energy source: useful for reactions with + ΔG

1. Reaction coupling: a favorable rxn is used to drive an unfavorable rxn by combining 2 together
2. ATP hydrolysis drives unfavorable reactions
3. Note that reaction coupling will alter total free energy, making an unfavorable rxn proceed!

Question 42

Kinetics Diagram

Answer 42

Understand this diagram

Question 43

Enzymes

Answer 43

Enzymes lower the activation E of a rxn making it easier for the substrate to reach the transition state.

• Enzymes increase the rate of a rxn (how fast equilib is reached), but do not alter the position of the equilibrium;
• Do not affect the overall free E or enthalpy of a rxn;
• Are not altered or consumed in the course of a rxn; thus, small amt required.
• Enzymes are pH and T sensitive; ranges of optimal activity.
• may consist of a single polypeptide chain, or several subunits in quaternary structure (globular)

Question 44

Enzyme Specificity

Answer 44

Enzymes are designed to work only on a specific substrate or group of closely related substrates

Question 45

Enzyme prefix

Answer 45

End in -ase

Question 46

Active site

Answer 46

• the reason for the importance of folding in enzyme function is the proper formation of the active site, the region in an ezyme 3D structure that is directly involved in catalysis.
• location on the enzyme where the substrate is held via weak chemical bonds

Question 47

Active Site Model (also referred to as the lock and key theory)

Answer 47

One of two theories about enzymes (the other is ‘Induced Fit’); says that the spatial structure of an enzyme’s active site (lock) is exactly complimentary to the spatial structure of the substrate (key); this theory has been largely discredited.

Question 48

Induced Fit Model

Answer 48

This is the more widely accepted theory; it describes the active site as having flexibility of shape, when the appropriate substrate comes in contact with the active site, the conformation of the active stie changes to fit the substrate.
- release of substrate from active site is exothermic process


Question 49

Regardless of the model, enzymes accelerate the rate of a give reaction by helping to what?

Answer 49

stabilize the transition state

Question 50

Which configurations are important in animals?

Answer 50

L amino acids but D sugars

Question 51

Proteases

Answer 51

Many proteases (protein cleaving enzymes) have an active site wirh a serine residue whose OH group can act as a nucleophile attacking the carbonyl carbon of an amino acid residue in a polypeptide chain. These enzymes usually have a recognition pocket near the active site. This is a pocket in the enzymes structure which attracts certain residues on substrate polypeptides. The enzyme always cuts polypeptides at the same site, just to one side of the recognition residue.

Question 52

Given the importance of active sites, it is clear that small alterations in its structure can drastically alter enymatic activity. What are some examples of this?

Answer 52

Temperature and pH both play critical roles in enzymatic function. As temp increases, the structure starts to get destabilizied. If temp rises sufficiently, the protein denatures and loses orderly structure. pH also impacts protein stability, several AA’s posses R group whose charge that depends on pH. So if pH deviates sufficently, the protein can denature.

Question 53

Cofactors

Answer 53

A nonprotein molecule or metal ion that is required for the proper functioning of an enzyme. Cofactors can be permanently bound to the active site or may bind loosely with the substrate during catalysis.

Question 54

Coenzymes

Answer 54

Organic molecules serving as cofactors (required for the proper functioning of an enzyme).
- Vitamins and ATP function as coenzymes in metabolic reactions.

Question 55

Regulation of Enzyme Activity

Answer 55

1) Covalent Modificiation: a phosporyl group (normally from ATP) can be covalently added to enzymes to either activate or inactivate them.

2) Proteolytic Cleavage: protease can activate inactive forms of enzymes (called zymogens)

3) Association with other polypeptides: some enzymes are regulated by another portion of themselves; either inhibitive or required to function

1. Constitutive activity – continuous activity (when regulatory subunit is removed)

4) Allosteric Rgeulation: enzymes can have multiple active sites and are regulated by the activity of these other sites

1. These allosteric sites are located far away from the active site
2. Binding of the allosteric regulator to these allosteric sites is generally noncovalent and reversible
3. When bound, these alter the conformation of the enzyme to increase or decrease catalysis

Question 56

Feedback inhibition (also called negative feedback)

Answer 56

enzymes are usually part of a pathway, with multiple reactions and enzymes needed to go from reactants to final products

1. In a pathway where the following enzymes catalyze each reaction, too much of the end product “D” will shut off Enzyme E1
2. This is negative feedback, or feedback inhibition
3. There is also feedback stimulation (but this is less common)

Question 57

Feedforward stimulation

Answer 57

stimulation of an enzyme by its substrate, where A might stimulate E3. This makes sense because when a lot of A is around, we want the pathway for utilization of A to be active.

Question 58

Enzyme Kinetics

Answer 58

the study of the rate of formation of products from substrates in the presence of enzymes

Question 59

Reaction rate (V for velocity)

Answer 59

the total amount of product formed per unit time is moles per second (mol/s)

1. Depends on the concentration of the substrate [S], and the enzyme
2. if [S] is low, doubling it will double V (linear relationship) until all active sites of enzymes are occupied most of the time
3. if the active sites of enzymes are occupied ALL of the time, adding more substrate does not increase reaction rate, and rate is denoted as Vmax
4. Michaelis constant (Km): the substrate concentration at which V is ½ of Vmax
5. A low Km means high enzyme affinity

Question 60

If a small amount of enzyme in a solution is acting at V_max, the substrate concentration is doubled, what is the new reaction rate?

Answer 60

If the enzyme is acted at Vmax, it is saturated with substrate so addding more substrate will not increase the reaction rate, the rate is still V_max

Question 61

Cooperativity

Answer 61

Some enzymes have multiple binding sites, and the binding of 1 site allosterically increases the affinity of other subunits (think about hemoglobin)

1. Tense: the conformation of the above enzyme prior to binding, with low substrate affinity
2. Relaxed: the conformation of the above enzyme with increased affinity, after partial binding
   a) Region 1: at low [S], the enzyme complex has low affinity for substrate (in tense state), adding more substrate doesn’t really increase the reaction rate until…
   b) Region 2: the range of substrate concentrations where adding substrate greatly increases reaction rate because the enzyme is in a relaxed state
   c) Region 3: represents enzyme saturation
   - This graph represents a sigmoidal curve when talking about the 3 different regions

Question 62

Cooperativity is just a special type of ?

Answer 62

allosteric regulation

Question 63

Competitive Inhibtion

Answer 63

this occurs when a molecule that resembles substrate bonds to the active site, thereby blocking the substrate; this occurs with weak bonds; the higher the concentration of the inhibitor, the slower the reaction increasing Km (less affininty) ; overcome by increase in [S].

a) Molecules that compete with the substrate for binding to the active site (Vmax not affected; just increase concentration of substrate to increase competition)
b) Km is affected; takes higher concentration to get to ½V_max Km will increase
c) is generally LINEAR

Question 64

Noncompetivive Inhibition

Answer 64

form of enzyme inhibition where the inhibitor binds to an enzyme at a location other than the active site (allosteric site/ regulatory site); while at this site, the enzyme shape changes, the inhibitor is unable to bind to its substrate, and no product forms.
- Km remains unchanged (no change in [S] while help) but Vmax decreases.

• graph will plateau out

Question 65

Competitive Inhibition vs Noncompetitive Inhibition (graphs)

Answer 65

competive graph will be linear while noncompetitive will have a flat line

Question 66

Uncompetitive Inhibtion

Answer 66

takes place when an enzyme inhibitor binds only to the complex formed between the enzyme and thesubstrate (the E-S complex). So it cant bind before substrate has been found

• No competition. Substrate bind to active site. Decrease in Vmax and Km

Graph: no inhibitor and uncompetitive inhibitor criss cross with each other.

Question 67

Mixed type Inhibition

Answer 67

occurs when an inhibitor can bind to either the unoccupied enzyme or the ES complex. If the enzyme has greater affinity for the inhibitor in its free form, the enzyme will have lower affinity for the substrate (K_m increases) similar to competitive inhibition.If the enzyme substrate complex has greater affinity for the inhibitor, the enzyme will have an apparently greater affinity (K_m decreases) for the substrate similar to uncompetitiv inhibition. In each of these situations, the inhibitor binds to an allosteric site and additional substrate cannot overcome inhibition (V_max decreases).

So overall in mixed type inhibition, K_m varies and V_max decreases

Question 68

Photosynthesis

Answer 68

the process by which plants store energy from the sun in the bond energy of carbohydrates

Question 69

Photoautotrophs

Answer 69

organisms that use the sun to produce their own food (plants)

Question 70

Chemoheterotrophs

Answer 70

organisms that use energy from other living organisms to make their own food

Question 71

Oxidation

Answer 71

Oxidation (3 meanings, opposite of reduction)

a) Attach oxygen
b) Remove hydrogen
c) Remove electrons

Question 72

Reduction

Answer 72

Reduction (3 meanings, opposite of oxidation)

a) Remove oxygen
b) Attach hydrogen
c) Attach electrons
d) *When molecules are reduced, they are “compressed”, have more stored energy, they want to be oxidized and release energy

Question 73

Redox pair

Answer 73

when one atom gets reduced, another one must get oxidized

Question 74

Oxidizing agents allow

Answer 74

for the oxidation of other chemicals by getting reduced themselves

Question 75

Catabolism

Answer 75

breakdown of complex molecules to smaller molecules to release energy (exothermic).
e.g. Glucose to CO2, H2O and ATP

Question 76

Anabolism

Answer 76

Consumes energy (endothermic) to create a complex molecule from simple ones (e.g. photosynthesis)

Question 77

Oxidative Catabolism

Answer 77

The way we extract energy from glucose is oxidative catabolism. We break down glucose by oxidizing it. The oxidative catabolism of glucose involves 4 steps

1) Glycolysis
2) Pyruvate Dehydrogenase Complex
3) Krebs Cycle
4) Electron Transport Chain / cxidative phosphorylation

Question 78

Celluar respiration is theoretically just a

Answer 78

big coupled reaction, we make the unfavorable synthesis of ATP happen by coupling it to a very favorable oxidation of glucose.

Question 79

Glycolysis

Answer 79

GLYCOLYSIS: splits 6C glucose into 2 3 carbon pyruvates + 2ATP + 2NADH

so overall 2 pyruvate, 2 ATP and 2 NADH are formed

a) All cells do glycolysis
b) Location: cytoplasm
c) Oxygen independent!!
(1) F6P → F 1,6-bP (done by phospofructokinase) is practically irreversible
(2) This is a key regulatory point of glycolysis
(3) Known as a “committed step”; F1,6-bP can only go down glycolysis, where F6P can be used elsewhere
(4) allosterically regulated by ATP
(5) 5th rxn: NADH is produced when an aldehyde is oxidized to COOH
d) NET RXN: Glucose + 2ADP + 2Pi + 2NAD+ → 2PYR + 2ATP + 2NADH + 2H2O + 2H+
(1) Glycolysis produces 2 NADH

Question 80

More important facts about glycolysis

Answer 80

• Hexokinase catalyzes the first step of glycolysis, the phosphorylation of glucose to G6P. G6P ffedback inhibits hekokinase
• PFK catalyzes the 3rd step (F16P to F16BP) Important because this reaction is thermodynamically very favorable so it is practically irreversible. Is a commited step. Key regulatory point in glycolysis.
• NADH is produced in only 1 step
• ATP is converted to ADP every time a phosphat is added to a substrate

Question 81

Fermentation

Answer 81

a) Anaerobic conditions only
b) Goal: regenerate NAD+ so it can be used back in glycolysis to make ATP
c) Location: cytoplasm
d) Eukaryotes: lactic acid fermentation

start: pyruvate
product: lactate, NAD+

e) Prokaryotes: alcoholic fermentation (ethanol)

start: pyruvate
product: ethanol, NAD+

• pyruvate is reduced to lactate, NADH is oxidized to NAD +

pyruvate is reduced to ethanol, NADH is oxidized to NAD +

f) Excessive lactate is brought back to the liver where it is converted back to pyruvate

Question 82

Pyruvate Dehydrogenase Complex

Answer 82

(PDC): Pyruvic acid into Acetyl CoA makes 2 NADH per glucose, one per pyruvate

a) Pyruvate gets oxidatively decarboxylated (oxidized to release CO2 and make NADH)
b) Location: mitochondrial matrix
c) Goals: 3C → 2C, NADH. The PDC changes pyruvate into an activated acetyl unti and activated means its attached to the carrier coenzyme A.
(1) Once pyruvate is decarboxylated, the acetyl group is “activated” by attachment to coA, which is basically a long handle with sulfur attached. When loaded with acetyl unit, it is called acetyl CoA
(2) The S-C bond in acetly CoA is high energy, making it easy to transfer the acetyl group to the Krebs cycle.
(3) Prosthetic group: non protein molecule covalently bonded to an active site of an enzyme. Example: Vitamins can serve as prosthetic groups.
(4) TPP (thiamine pyrophosphate) is a prosthetic group of the PDC and Krebs. The α ketogluturate dehydrogenase complex which catalyzes the 3rd step in the Krebs Cycle is very similar to the PDC
(5) Cofactors: various substances necessary to the function of the enzyme, but never interact with the enzyme

Question 83

Krebs Cycle

Answer 83

a) Acetyl CoA gets oxidized, lots of NADH and FADH2 are produced
b) Location: mitochondrial matrix
(1) OVERVIEW OF KREBS
(2) STAGE 1 KREBS
(a) 4 carbon OAA (oxaloacetate) combines with the 2 carbon acetyl group of Acetyl-CoA to form 6 carbon citric acid
(3) Stage 2
(a) Citrate (isocitrate) is further oxidized to release CO2 and produce NADH from NAD+; the product is alpha ketoglutarate (5C)
(b) Alpha ketoglutarate is oxidatively decarboxylated to produce succinyl CoA, releasing another CO2 and NADH
(4) Stage 3: OAA is regenerated; don’t worry about the individual steps

Krebs Cycle: 6 NADH, 2 FADH₂, and 2 GTP per glucose

Question 84

Stage 1 of Krebs Cycle

Answer 84

4-C OAA (oxaloacetate) combines with the acetyl group of Acetyl-CoA to form 6-C citric acid

Question 85

Stage 2 and Stage 3 of Krebs Cycle

Answer 85

Stage 2:

(a) Citrate (isocitrate) is further oxidized to release CO2 and produce NADH from NAD+; the product is alpha ketoglutarate (5C)
(b) Alpha ketoglutarate is oxidatively decarboxylated to produce succinyl CoA, releasing another CO2 and NADH

Stage 3: OAA is regenerated; don’t worry about the individual steps

Question 86

The structure of the mitochondrion

Answer 86

The mitochondrion contains 2 membranes, an outer membrane and an inner membrane each composed of a lipid bilayer. The outer membrane is smooth and contains large pores formed by porin protein. The inner mebrane is impermeable even to very small items like H+ andis densely folded into structures termed cristae. The cristae extend into the matrix, which is the innrmost space of the mitochondrion. The space between the 2 membranes, the intermembrane space is continuous with the cytoplasm due to the large pores in the outer membrane. The enzymes of the Krebs Cycyle and PDC are located in the matrix and those of the ETC and ATP synthase involved in oxidative phosporylation are bound to the inner mitochondrial membrane.

Question 87

What are the 2 goals of electron transport / oxidative phosphorylation?

Answer 87

1) reoxide all the electron carriers reduced in glycolysis, PDC, and the Krebs Cycle and
2) store energy in the form of ATP in the process

Question 88

Oxidative Phosphorylation

Answer 88

the oxidation of the high energy electron carriers NADH and FADH₂ coupled to the phosphorylation of ADP to produce ATP. The energy releases through oxidation of NADH and FADH₂ by the ETC is used to pump protons out of the mitochondrial matrix. This proton gradient is the source of energy used to drive the phosphorylation of ADP to ATP.

Question 89

Electron Transport Chain

Answer 89

a) Location: along inner mitochondrial membrane (eukaryotes) or cell membrane (prokaryotes)
b) NADH and FADH2 are oxidized (e– removed), and the energy released is used to pump protons out of the mitochondrial matrix (I think the e– passing through drive this pump)
c) Proton gradient drives ADP → ATP
d) ATP synthase allows H+ back into mitochondrial matrix, and it captures their energy to make ATP

Question 90

Electron Transport Chain and Oxidative Phosphorylation Summary

Answer 90

• eukaryotes use inner mitochondrial membrane and bacteria uses cell membrane
• Oxidative phosphorylation is the oxidation of NADH and FADH₂ coupled with the phosphorylation of ADP to make ATP

Electron Transport Chain is a group of 5 electron carriers. Each member of the chain reduces the next member down the line

• 3 are large protein complexes called cytochromes. They contain heme prosthetic groups or iron-sulfur electron-transfer systems
• 2 are small protein complexes

First large carrier is called _NADH dehydrogenase (_also known as coenzyme Q reductase). ( NADH → NAD+) Passes electrons to one of the small carriers called ubiquinone or aka coenzyme Q. Ubiquinone passes electrons to second large compex known as cytochrome c reductase. → small carrier cytochrome c → large carrier cytochrome c oxidase → oxygen to get reduced to water

Each of the 3 large complexes pumps protons across the inner mitochondrial membrane every time electrons flow past. Since inner mitochondrial membrane is highly impermeable to protons, the ETC creates a large proton with pH being much higher inside the matrix than rest of the cell

• The passage of proteins from intermembrane space through the ATP synthase channel causes it so synthesize ATP from ADP + Pi

* pumping of protons to form pH gradient has positive ΔG

Question 91

protons are pumped from ________ to ________ ?

Answer 91

protons are pumped from matrix into intermembrane space

Question 92

Matrix has a ____ pH and ____ [H+] ?

Answer 92

The matrix has a high pH and low [H+]

Question 93

The intermembrane space has a ____ pH and ____ [H+]?

Answer 93

The intermembrane space has a low pH and high [H+]

Question 94

ATP production is dependent on what?

Answer 94

It is dependent on a proton gradient, the overall process of electron transport and ATP production is said to be coupled by the proton gradient. Together, electron transport and ATP production are known as oxidative phsophorylation.

Question 95

The pumping of protons to form a pH gradient has a ____ ΔG

Answer 95

POSITIVE (+) ΔG . Creation of the proton gradient is dependent upon the very negative ΔG of electron transport.

Question 96

Energetics of Glucose Catabolism

Answer 96

1. Each NADH that is oxidized to NAD+ results in 10 H+ pumped across inner mitochondrial membrane
2. Each FADH₂ → FADH results in 6 H+ pumped across the IM membrane
3. Each ATP made by ATP synthase requires 4 H+ to pass through
4. FADH₂ → FAD = 1.5 ATP
5. NADH → NAD+ = 2.5 ATP
   a) However, because these molecules need to be transported to the mitochondria (which requires energy) by the GLYCEROL PHOSPHATE SHUTTLE, there is a slight discrepancy in energy produced
   (2. 5 ATP per NADH from mitochondrial matrix, 1.5 ATP per NADH from cytoplasm, and 1.5 ATP per FADH₂ )

Question 97

Chart Summary

Answer 97

See attached image

Question 98

Chart Summary # 2

Answer 98

see attached image

Question 99

How many ATP do eukaryotes and prokaryotes makes?

Answer 99

eukaryotes: 30 ATP
prokaryotes: 32 ATP

Question 100

Other Metabolic Pathways of the Cell

Answer 100

1. Glycogenolysis: breakdown of glycogen, responds only to glucagon

2. Gluconeogenesis: conversion of non-carbohydrate precursor molecules into oxalate then glucose; occurs only when there are no dietary or liver stores of glucose (precursors include carbon skeleton of AAs, lactate, pyruvate)
3. β-oxidation: fatty acids are broken down this way in hepatocytes
   a) 2 C are removed at a time, which are then converted into acetyl-CoA and then enter the Krebs
   b) The glycerol backbone gets converted to glucose and enters cellular respiration
4. AA catabolism: AAs are broken down, often converted to urea for excretion; the remaining carbon skeleton (α-keto acid) can be broken down into H2O and CO2 or converted to glucose or acetyl-CoA

Question 101

Glycogenolysis

Answer 101

• term for glycogen breakdown
• glycogen is found in liver and muscle cells
• the synthesis of glycogen (glycogenesis) and glycogenolysis are opposing processes, controlled by hormones that regulate blood sugar levels and energy

Question 102

Gluconeogenesis

Answer 102

when glucose is unavailable so it converts non carbohydrate precursor molecules into oxaloacetate and then glucose

Question 103

β-oxidation

Answer 103

fatty acids made in cytoplasm of hepatocytes and stored in adipocytes (fat cells) as triglycerides. Fatty acids can be broken down in the hepatocyte mitochondria via fatty acid β oxidation in response to metabolic need. Each round removes 2 carbons rom a fatty acid and converta them to acetyl CoA. β oxidation generates 1 NADH and 1 FADH and 1 H20 for each 2 carbon group removed. Acetyl CoA enters Krebs Cycle. This requires 2 ATP.

Ex. 12 carbon saturated FA will convert into 5 FADH2, 5 NADH, and 6 acetyl Co A which can then eneter Krebs Cycyle

Question 104

AA catabolism

Answer 104

AAs are broken down, often converted to urea for excretion; the remaining carbon skeleton (α-keto acid) can be broken down into H₂O and CO₂ or converted to glucose or acetyl-CoA

Question 105

Pentose Phosphate Pathway

Answer 105

• diverts glucose-6-phosphate from glycolysis in order to form ribose-5-phosphate which can be used to synthesize nucleotides
• pathway referred to as a shunt composed of oxidative phase (also generates NADPH)
• the first enzyme in the PPP, glucose-6-phosphate dehydrogenase is the primary point of regulation and generates NADPH
• produces NADPH, a reducing agent for biosynthesis, aka anobolic pathway
• can be used to produce nucleic acids and NAD+

Question 106

Regulation of glycolysis and gluconeogenesis

Answer 106

Phosphofructokinase and F-1,6-BPase are also reciprocally controlled by fructose 2,6-bisphosphate in the liver The level of F-2,6-BP is low during starvation and high in the fed state, because of the antagonistic effects of glucagon and insulin on the production and degradation of this signal molecule. Fructose 2,6-bisphosphate strongly stimulates phosphofructokinase and inhibits fructose 1,6-bisphosphatase. Hence, glycolysis is accelerated and gluconeogenesis is diminished in the fed state. During starvation, gluconeogenesis predominates because the level of F-2,6-BP is very low. Glucose formed by the liver under these conditions is essential for the viability of brain and muscle.

Question 107

Regulaton of Krebs Cycle

Answer 107

Following glycolysis, pyruvate can either continue through celluar respiration or be transfomed into other useful metabolic intermediates. If pyruvate continues through celluar respiration and is converted to acetyl CoA by the PDC, it can no longer be utilized by gluconeogenesis to form glucose. This step is tightly regulated in response to the energy demands of the cell (high concentrations of NAD+ indicate an energy defecit and increase acetyl CoA production). When acetyl CoA enters the Krebs Cycle,those enzymes that catalyze exergonic steps are also regulated. Importantly, the activity of isocitrate dehydrogenase changes with the energy needs of the cell (elevated levels of ATP inhibit the enzyme)

Question 108

Regulation of Glycogen Synthesis and Glycogenolysis

Answer 108

Glycogen (stores glucose in liver and muscle) must be synthesized and broken down in response to changes in blood glucose and metabolic demand. Glycogen Synthase (enzyme responsible for glycogen generation from G-1-P) and glycogen phosphorylase (serves to catabolize glycogen) are reciprocally controlled. After eating, elevated levels of insulin activate glycogen synthase and inhibit glycogen phsophorylase. This stumulates glycogen synthesis while inhibiting its breakdown. Glugacon results in the supression of glycogen synthesis and stimulation of glycogenosis from the liver (not muscle)

Question 109

Overview of regulation

Answer 109

1) In a pathway, those enzymes which cataylze irreversible (exergonic) rxns are frequently sites of regulation.
2) Increased concentrations of intermediates in a pathway generally served to decrease the activity of that pathway (citrate decreases activity of PFK in glycolysis)
3) Each pathway responds to the energy state of the cell. Celluar respiration is stmulated by energy defecits or inhibited energy surpluses (high ADP:ATP or high NAD+:NADH ratios)

Question 110

Ketogenesis

Answer 110

• Produced from acetyl CoA in the liver during fatty acid catabolism
• Produced during fasting or when on low carb diets (ketosis/ketoacidosis)
• Acetoacetic acid and hydroxybutyrate can be used for energy.
• Acetone is excreted in the urine and can be smelled on one’s breath.

Question 111

FAS

Answer 111

Anabolism of Fats (non-template synthesis)

• Citrate is transported from the mitochondria to the cytoplasm.
• It is broken down into acetyl CoA and oxaloacetate.
• The acetyl CoA is used by fatty acid synthases to construct fatty acids of variable lengths.
• repeated addition of 2 carbon units
• done by acetyl CoA synthase to generate malonyl CoA

Question 112

Proteases

Answer 112

enzymes that will irreversibly modify substrate (pernament)

Question 113

fatty acid, AA, and glycogen synthesis are __________ driven processses unlike protein or nucleic acid synthesis

Answer 113

non template