Exam 1

Question 1

Change in Enthalpy

Answer 1

🔺H=🔺G+T🔺S

🔺H must be <t>
</t>

Question 2

Standard Gibbs Free Energy

Answer 2

Chemical : 🔺G⁰ @ T=25^oC and pH=0

Biological : 🔺G^0’ @ T=25^oC and pH=7

Question 3

Using Gibbs Free Energy

For reversible reaction: A+B yields Y+Z

Answer 3

🔺G=🔺G^0’+RT ln[Y][Z]/[A][B]

@ Equilibrium 🔺G=0 and ratio=Keq

Therefore

🔺G^0’= - RT ln Keq

Question 4

Coupled Reactions

Answer 4

Use free energy released by an exergonic reaction to drive second energonic reaction.

Question 5

Reduction of Pyruvate

Answer 5

Endergonic

Pyruvate + 2H⁺+2e^- —> Lactate

🔺G^0’= 36 kJmol^-1

Question 6

Reduction of NADH/H⁺

Answer 6

Exergonic

NADH + H⁺—>NAD⁺+2e^- +2H⁺

🔺G^0’ = -62 kJmol^-1

Question 7

Coupled Reaction

Answer 7

Exergonic

Pyruvate + NADH + H⁺—> NAD⁺+ Lactate

🔺G^0’ = -26 kJmol^-1

Question 8

Activation Energy

E_a

Answer 8

Amount of energy required to bring all molecules in a mole of reactant at a given temperature to a reactive state.

Increasing T decreases E_a by increasing kinetic energy of all molecules.

Adding enzyme reduces E_a and increasses reaction rates by 10³ to 10⁷ times.

Question 9

Covalent Bonds

1) Van der Waals
2) Hydrogen
3) Ionic
4) Hydrophobic Bonds

Answer 9

1) Interaction between dipolar molecules
2) Sharing of electrons between dipolar molecules
3) Anion/Cation bonding
4) Non-polar water-averse regions or molecules.

Question 10

Weak Bonds and Temperature

Affects and Consequences

Answer 10

Increasing Temperature

• weakens Hydrogen, Van der Waals, and Ionic bonds
• strengthens Hydrophobic bonds

Consequence: denaturing of 3D structures (DNA, proteins, membranes)

Can lead to clumping/strengthening of hydrophobic elements.

Question 11

Water

Answer 11

Unique - Covalent and Polar

Hydrogen bonds inividually weak (1-5 kcal to break 1 mole)

Overall effects are strong - high melting/boiling points and high surface tension/cohesion.

Question 12

Colligative Properties

Answer 12

Depend on concentration of solute(not size or type)

Increasing solute concentration of water increases boiling point, vapor/osmotic pressure, and reduces freezing point.

Question 13

Rate of diffusion

Answer 13

Rate (dQ_s/dt) depends on solute size, electrical charge and solubility.

dQ_s/dt = P(C₁-C₂)

P=permeability constant (cm/s)

Question 14

Partition coefficient (K)

Answer 14

K=[solute]_lipid/[solute]_water

Question 15

pH

Answer 15

pH = -log[H⁺]

In pure water at 25^oC, pH = 7

Question 16

Strong Acids and Bases

vs

Weak Acids and Bases

Answer 16

Strong - readily release ions

Weak - are only partially ionized under biological conditions

Question 17

Dissociation of Acids -

1) General Formula
2) pK
3) Consequences

Answer 17

1) HA <=> H⁺ + A^-
2) pK = -log₁₀K_eq = pH at which [A^-] = [HA]
3) pK (pH at equillibrium) is low, ex: <3 for HCl and Sulfuric Acid

Question 18

Henderson-Hasselbach equation

Answer 18

pH = pK + log[A^-] / [HA]

Question 19

pH importance

Answer 19

Low pH means [H⁺] is high, amino and carboxyl protonated (valence electron is positive).

High pH means [H⁺] is low, amino and carboxyl groups ionized(valence electron is negative).

Question 20

Buffers

1) What are they?
2) Adding Acid
3) Limits
4) Ideal conditions?

Answer 20

1) Usually weak acids
2) Liberated protons can bind with buffer and effect is reduced
3) Will only work over limited pH range
4) When pH=pK because half of buffer works against each direction of pH

Question 21

Events of Enzyme Lowering E_a

3 Steps

Answer 21

1) Non-covalent bonding leads to formation of Enzyme-Substrate Complex
2) Non-covalent bonding leads to formation of Enzyme-Product Complex
3) Dissociation yields free Enzyme & Product

Question 22

How Do Enzymes Work?

Answer 22

Substrate binds to active site and induces a change in the substrate of the structure:

Shifit in electron distributions, change in shape, pairs of destabilized reactants can be brought into proper conformity and proximity.

These changes reduce E_a

Question 23

What Influenes rates of Enzymatic Reactions?

Answer 23

Increase in concentration of products creates competition.

Initial concentration of substrate since addition of substrate leads to max velocity at which nothing else happens.

Physiochemical environment such as temperature, pH and hydrostatic pressure.

Question 24

1) Michaelis-Menten Kinetics

Hint : Involves V, S, and K_m

2) Exception

Answer 24

1) V = V_max x [S] / ([S] + K_m)

where V_max is the velocity at which enzyme saturation occurs.

2) K_m = [S] required to reach 1/2 V_max
3) enzymes with multiple binding sites cooperate because binded enzyme experiences change from substrate too, which promotes more binding.

Question 25

1) Factors Determining Vmax
2) k_cat

Answer 25

1) Number and catalytic effectiveness of enzyme molecules
2) Turnover number - Substrate molecules to product molecules per second, per saturated enzyme molecule

Question 26

Biomolecules

Answer 26

Energy Carriers

Proteins

Carbohydrates

Lipids

Nucleic Acids

Question 27

Biomolecule :

Energy Carriers

Answer 27

Adenotriphosphate (ATP) to ADP yields 🔺G^0’ = -30.5 kJ/mol

Acetyl CoA produced within mitochondria can ber used to produced ATP, GTP and Heat

Question 28

Acetyl CoA and Ketone Transport

Answer 28

When blood-glucose levels are low, insulin production inhibited.

When Insulin levels are low, increase in fat metabolism produces Acetyl CoA.

Liver cells convert Acetyl CoA —> ketones (ketogenesis)

In the CNS, neural cells convert ketones to Acetyl CoA

Question 29

Biomolecules :

Proteins

Answer 29

Amino Acids bonded together with peptide bonds.

Can have 4 structural levels, each shape related to function.

Question 30

Biomolecules :

Carbohydrates

Answer 30

Monosaccharide: Glucose, Fructose and Galactose modify macromolecules

Disaccharides: Obtained by diet but need to be broken down to use

Polysaccharides: Glycogen and Startch serve in energy storage

Question 31

Biomolecules :

Lipids

Answer 31

Diverse group that includes :

Fatty Acids (saturated or unsaturated)

Triglycerides (glycerol backbone with three fatty acids)

Phospholipids which are amphipathic

Steroids (lipid soluble so can diffuse through cell membrane)

Question 32

Biomolecules :

Nucleic Acids

Answer 32

Nucleotide polymers made of sugar, phosphate, and a base

Question 33

Dephosphorylation of Phosphocreatine

Answer 33

Creatine Phosphokinase(CPK) drives Phosphocreatine and ADP reaction to yield ATP, creatine and heat

🔺G^0’ = -12.6 kJ/mol

Question 34

Pros and Cons of CP hydrolysis

Answer 34

Pros : Max muscle power, no harmful by-products, no effect on pH, reaction is reversible.

Cons: Short duration (muscle stores rapidly exhausted), no beneficial by-product.

Question 35

Anaerobic Glycolysis

Answer 35

Glucose + ATP

Glucose 6-phosphate + Glycogen

Fructose bisphosphate splits

Glyceraldehyde 3-phosphate AND Dihydroxyacetone phosphate (which forms a second G3p

Both G3ps combine with one NaD⁺ each to yield 2 NADH

(2) 1, 3 Biphosphogylcerates combine with one ADP each to yield 2 ATP
(2) 3-Phosphoglycerates —–>2-Phosphoglycerates yield H₂0
(2) Phosphoenolpyruvates combine with one ADP each to yield 2 ATP ENDING IN 2 PYRUVATES

Question 36

Glycolysis Yields

1) ATP
2) 🔺G^0’
3) Total Energy
4) Bonus
5) Problem

Answer 36

1) 2 ATP per Glucose (61.0 kJ/mol)
2) -73 kJ/mol
3) 134 kJ/mol
4) 2 moles of water per Glucose
5) 2 moles of NAD⁺ consumbed per Glucose

Question 37

Anaerobic Solution for use of NAD⁺ in Glycolysis

and 🔺G^0’

Answer 37

Pyruvate + NADH + H⁺<==>NAD⁺ + Lactate

🔺G^0’ = -25.1 kJ/mol

Question 38

Energetic Effect of Glycolysis

1) Glucose —> 2 Pyruvate
2) Glucose —> 2 Lactate
3) Glucosyl —2 Pyruvate

Answer 38

1) -134 kJ/mol
2) -184 kJ/mol
3) -152 kJ/mol

Question 39

Pros and Cons of Anaerobic Glycolysis

Answer 39

Pros : Maximum muscle power possible, no Oxygen needed, 2 moles of water per glucose.

Cons : Short duration, low yield, Lactic acid, lowers pH, not reversible.

Question 40

Oxidative phosphorylation

Four steps

Answer 40

1) Glycolysis but then the resulting 2 Pyruvate to Lactate
2) TCA cycle x two
3) Electron Transport System
4) Proton Motive Force

Question 41

Tricarboxylic Acid Cycle

Answer 41

Acetyl CoA –> Citrate –> Isocitrate –>

2-Oxoglutarate (yielding NADH and CO2) –>

Succinyl CoA (yielding NADH and CO2) –>

Succinate (yielding GTP) —>

Fumarate (yielding FADH2) –>

Malate –>

Oxalaoacetate (yielding NADH) –>

Citrate

Question 42

Electron Transport System

Answer 42

Complex 1 collects electrons from NADH

Complex 2 transfers electrons from succinate to FAD (ubiqunone) yielding Fumarate

Complex 3 transfers H+ out as electron passes through then onto cytochrome c

Complex 4 transfers H+ out and also creates H2O from O2

Complex 5 transfers a H+ in and creates an ATP from ADP

Question 43

Proton Motive Force

Answer 43

Electrochemical gradient created by ETS allows H+ to return via Complex V (energy transformed into heat) and ATP created from ADP.

Question 44

Oxidative phosphorylation theoretical yield

Glycolysis: 2 ATP + 2H2O + 2 NADH/H+

Pyruvate -> Acetyl CoA: 2 NADH/H+

TCA: 2 GTP + 2 FADH2 + 6 NADH/H+

10 NADH/H+ -> NAD+: 30 ATP + 10 H2O

2 FADH2 -> FAD: 4ATP + 2 H2O

Total: 36 ATP + 2 GTP & total Gibbs: -2854kJ/mol & 14 moles of H2O per glucose

Question 45

Pros and Cons of Aerobic Glycolysis

Answer 45

Pros: High yield (2854 kJ/mol glucose), long duration, beneficial by-products.

Cons: Max muscle power not possible, requires O2, harmful by-products, lowers pH, overall not reversible.

Question 46

Lipid Profile

Answer 46

Phosphoglycerides - phosphatidylcholine, phosphatidylserine, phosphatydlethanolamine.

Sphingolipids

Glycolipids

Cholesterol

Question 47

Cholesterol

Answer 47

Strengthens interactions between polar heads and disrupts interactions between polar tails.

Reduces permeability, alters membrane fluidity

Question 48

Membrane Heterogenity

Answer 48

Inner and outer layers different - outer PC and glycolipids, inner PE and PS

Lipid raftes - increased concentration of Cholesterol and glycolipids. Increases rigidity and thickness, reduce fluidity.

Question 49

Passive Diffusion

Answer 49

Needs concentration gradient.

Does not need special transporters.

May enter if - can break H-bonds with water or can dissolve in lipid

Question 50

Rate of Diffusion

Answer 50

J= D(C1-C2)/X

useful when interest in the effect of variation with respect to thickness

Question 51

Facilitated Diffusion

Answer 51

Needs Electrochemical gradient AND specialized membrane proteins.

Used Channels, pores and carrier proteins.

Question 52

Calculating Equilibrium - Nernst Equation

Answer 52

E_x = (RT/zF) ln ( {X} outside / {X} inside)