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BiochemistryBiochemistry
(Intermediary Metabolism)(Intermediary Metabolism)
Biochemistry: The study of the molecular basis for life.Biochemistry: The study of the molecular basis for life.
Intermediary Metabolism: The pathways of biosynthesis and degradation of the biomolecules.
Intermediary Metabolism: The pathways of biosynthesis and degradation of the biomolecules.
Major Classes of Biomolecules: Proteins, Carbohydrates, Lipids, Nucleic Acids, Vitamins
Major Classes of Biomolecules: Proteins, Carbohydrates, Lipids, Nucleic Acids, Vitamins
StructureStructure
PathwaysPathways
RegulationRegulation
Energy Energy
Water:Water:
Unusual properties result from internal cohesivenessUnusual properties result from internal cohesiveness
Cohesiveness is a direct result of hydrogen bondingCohesiveness is a direct result of hydrogen bonding
OO
HHHHOO
HHHH
::::
++++
__
OOHH
HH............
++
__++ __
HHOO++ HH
++
H Bond: 20 kJ/molH Bond: 20 kJ/mol
Covalent: 420 kJ/molCovalent: 420 kJ/mol
Strong electrostatic attractions of water with Na+ and Cl- ions result in the formation of hydration shells
Strong electrostatic attractions of water with Na+ and Cl- ions result in the formation of hydration shells
Na+Na+Cl-Cl-
Substances that dissolve in water in this manner are hydrophilic
Substances that dissolve in water in this manner are hydrophilic
Hydrophobic molecule surrounded by a “cage” of water molecules
Hydrophobic molecule surrounded by a “cage” of water molecules
clathrateclathrate
Hydrophobic effectHydrophobic effect
• Hydrophobic Effect: Critical in determining the structure of proteins and assembly of biological membranes
• Hydrophobic Effect: Critical in determining the structure of proteins and assembly of biological membranes
• Some compounds contain both hydrophobic and hydrophilic regions and are termed amphipathic
• Some compounds contain both hydrophobic and hydrophilic regions and are termed amphipathic
Ionization of water:Ionization of water: H2O H+ + OH-H2O H+ + OH-
Keq = [H+] [OH-]Keq = [H+] [OH-][H2O][H2O]
• 1L of water has a mass=1000g• 1L of water has a mass=1000g
• MW of water=18.015 g/mol• MW of water=18.015 g/mol
• Therefore, [water]=1000g/L = 55.5M• Therefore, [water]=1000g/L = 55.5M18g/mol18g/mol
55.5Keq=[H+][OH-];55.5Keq=[H+][OH-]; Keq=1.8 x 10-16Keq=1.8 x 10-16
55.5 (1.8 x 10-16) = [H+][OH-] = 1.0 x 10-1455.5 (1.8 x 10-16) = [H+][OH-] = 1.0 x 10-14
ION PRODUCT OF WATER:ION PRODUCT OF WATER:
EQUAL AMOUNTS OF H+ AND OH- IN WATER, SO THERE IS 1 X 10-7M H+
AND 1 X 10-7M OH-
EQUAL AMOUNTS OF H+ AND OH- IN WATER, SO THERE IS 1 X 10-7M H+
AND 1 X 10-7M OH-
• Neutral: Concentration of H+ and OH- is equal
• Neutral: Concentration of H+ and OH- is equal
• Acidic: [H+]>[OH-]• Acidic: [H+]>[OH-]
• Basic: [H+]<[OH-]• Basic: [H+]<[OH-]
• Ion Product of Water forms the basis for the pH scale
• Ion Product of Water forms the basis for the pH scale
• pH = -log[H+]• pH = -log[H+]
• [H+]=1 x 10-7M: pH = -log[1x 10-7] = 7• [H+]=1 x 10-7M: pH = -log[1x 10-7] = 7
• pH>7 Alkaline = [OH-]>[H+]• pH>7 Alkaline = [OH-]>[H+]
• pH<7 Acidic = [H+]>[OH-]• pH<7 Acidic = [H+]>[OH-]
• pH scale is logrithmic• pH scale is logrithmic
Water:Water:
• Hydrogen bonding• Hydrogen bonding
• Ionization• Ionization
• Reactant• Reactant
Acids and BasesAcids and Bases
• Lowry/Bronsted: Acid is a substance capable of donating a proton; Base is a substance which accepts protons
• Lowry/Bronsted: Acid is a substance capable of donating a proton; Base is a substance which accepts protons
• A proton donor and its corresponding proton acceptor = conjugate acid-base pair
• A proton donor and its corresponding proton acceptor = conjugate acid-base pair
CH3COOHCH3COOH CH3COO-CH3COO-H+ +H+ +
Keq = [H+][CH3COO-] / [CH3COOH]Keq = [H+][CH3COO-] / [CH3COOH]
Dissociation constants expressed as pKa:Dissociation constants expressed as pKa:
log 1/Ka = -log Kalog 1/Ka = -log Ka
pHpH
[NaOH][NaOH]
0. 50. 5
22
44
66
88
Titration of Acetic AcidTitration of Acetic Acid
CH3COOCH3COOHH
[NaOH][NaOH]
0. 50. 5
22
44
66
88CH3COOCH3COOHH
pHpH
[NaOH][NaOH]
0. 50. 5
22
44
66
88CH3COOCH3COO
__
pHpH
[NaOH][NaOH]
0. 50. 5
22
44
66
88[CH3COO[CH3COO
__
[CH3COOH][CH3COOH]
==]]
pHpHMidpoint
pKa = 4.76
[NaOH][NaOH]
0. 50. 5
22
44
66
88CH3COO-CH3COO-
pHpH
Ka = [H+] [A-] / [HA]Ka = [H+] [A-] / [HA]
Solve for [H+]: [H+] = Ka [HA] / [A-]Solve for [H+]: [H+] = Ka [HA] / [A-]
Take -log of both sides:
-log[H+] = -log Ka - log [HA] / [A-]
Take -log of both sides:
-log[H+] = -log Ka - log [HA] / [A-]
pH = pKa - log [HA] / [A-]pH = pKa - log [HA] / [A-]
pH = pKa + log [A-] /[HA]pH = pKa + log [A-] /[HA]
Henderson-Hasselbach EquationHenderson-Hasselbach Equation
Review the following problems:Review the following problems:
The examples on pages 45 & 47The examples on pages 45 & 47
Problems 1,2,3,4 at the end of the chapterProblems 1,2,3,4 at the end of the chapter
A- = X/V X=equivalents of OH- added; V=volume of the solutionA- = X/V X=equivalents of OH- added; V=volume of the solution
Co = equivalents of conjugate acidCo = equivalents of conjugate acid
[HA]=Co - X/V[HA]=Co - X/V
Incorporating into Henderson-Hasselbach:Incorporating into Henderson-Hasselbach:
pH = pKa + log[x / Co-x]pH = pKa + log[x / Co-x]
[NaOH][NaOH]0. 50. 5
22
44
66
88
pHpHpK1pK1
pK2pK2
Titration Curve for a Polyprotic AcidTitration Curve for a Polyprotic Acid
One pK for each ionization stepOne pK for each ionization step
The pK’s of two closely associated acid-base groups can influence one another:The pK’s of two closely associated acid-base groups can influence one another:
H-O-C-C-O-HH-O-C-C-O-H
OOOO
oxalic acidoxalic acidH-O-C-CH
2-CH
2-C-O-HH-O-C-CH
2-CH
2-C-O-H
OOOO
succinic acidsuccinic acid
pK1 = 1.27pK
1 = 1.27
pK2 = 4.27pK
2 = 4.27
differ by 3 pH unitsdiffer by 3 pH units
pK1 = 4.21pK
1 = 4.21
pK2 = 5.64pK
2 = 5.64
differ by 1.4 pH unitsdiffer by 1.4 pH units
Bicarbonate: Major Blood BufferBicarbonate: Major Blood BufferCO2CO2
CO2CO2
H2OH2O
H2CO3H2CO3
H+H+
HCO3-HCO3- bicarbonatebicarbonateMetabolic reactionsMetabolic reactions
