introduction of biochemistry
TRANSCRIPT
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Introduction of Biochemistry and importance of H2O
By: Asheesh Kumar Pandey
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Pancreatic cell section
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Universal features of living cells
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Phylogeny of the three domains of life
Introduce by Carl Woese in 1977 on the basis of 16SrRNA
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Classification of Organisms
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Several functional groups in a single biomolecule
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Complementary fit between a macromolecule and a small
molecule
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Stereoisomers have different effects in humans
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Many drugs are racemic mixtures
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Energy Transformations in Living Organisms
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The major carrier of chemical energy in all cells is adenosine
triphosphate (ATP)
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Very polar
Oxygen is highly electronegativeH-bond donor and acceptorHigh b.p., m.p., heat of vaporization,
surface tension
Properties of water
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Water dissolves polar compounds
solvation shellor
hydration shell
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Water
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Hydrogen bonding in
ice
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Hydrogen Bonding of Water
Crystal lattice of ice
One H2O molecule canassociate with 4
other H20 molecules
•Ice: 4 H-bonds per water molecule
•Water: 2.3 H-bonds per water molecule
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Biologically important hydrogen bonds
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Relative Bond Strengths
Bond type kJ/moleH3C-CH3 88H-H 104Ionic 40 to 200H-bond 2 - 20Hydrophobic interaction 3 -10van der Waals 0.4 - 4
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non-covalent interactions
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Directionality of the hydrogen bond
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Water as a solvent
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Non-polar substances are insoluble in water
Many lipids are amphipathic
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How detergents work?
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Amphipathic compounds in aqueous solution
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Dispersion of lipids in water
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Release of ordered water is energetically favorable
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Ionization of Water
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Ionization of water
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pH Scale Devised by Sorenson (1902) [H+] can range from 1M and 1 X 10-14M using a log scale simplifies notation pH = -log [H+]Neutral pH = 7.0
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Conjugate acid-base pairs consist of a proton donor and a proton acceptor
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Titration curve of acetic acid
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[OH-] [H+] Keq = [H2O]
Kw = [OH-] [H+] = 10-14 M2
Pure H2O : [H+] = [OH-] = 10-7 MpH = - log [H+] = -log (10-7) = 7If [H+] < 10-7 M then pH < 7 (acidic)If [H+] < 10-7 M then pH < 7 (basic)
Blood: [H+] = 4 x 10-8 M Blood pH = 7.4
H2O OH- + H+
Equilibriumconstant = = 1.8 x 10-16 M
Ion productof water
=
H2O
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Ionization of WaterH20 + H20 H3O+ + OH-
Keq= [H+] [OH-] [H2O]
H20 H+ + OH-Keq=1.8 X 10-16M
[H2O] = 55.5 M[H2O] Keq = [H+] [OH-]
(1.8 X 10-16M)(55.5 M ) = [H+] [OH-]1.0 X 10-14 M2 = [H+] [OH-] = Kw
If [H+]=[OH-] then [H+] = 1.0 X 10-7
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Acid/conjugate base pairs HA + H2O A- + H3O+ HA A- + H+HA = acid ( donates H+)(Bronstad Acid)A- = Conjugate base (accepts H+)(Bronstad Base)
Ka = [H+][A-] [HA]
Ka & pKa value describe tendency to loose H+
large Ka = stronger acidsmall Ka = weaker acid
pKa = - log Ka
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pKa values determined by titration
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Phosphate has three ionizable H+ and three
pKas
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Buffers
Buffers are aqueous systems that resist changes in pH when small amounts of a strong acid or base are added.
A buffered system consist of a weak acid and its conjugate base.
The most effective buffering occurs at the region of minimum slope on a titration curve
(i.e. around the pKa).Buffers are effective at pHs that are within
+/-1 pH unit of the pKa
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Henderson-Hasselbach Equation1) Ka = [H+][A-] [HA]
2) [H+] = Ka [HA] [A-]3) -log[H+] = -log Ka -log [HA] [A-]
4) -log[H+] = -log Ka +log [A-] [HA]
5) pH = pKa +log [A-] [HA]
HA = weak acid
A- = Conjugate base
* H-H equation describes the relationship between pH, pKa and buffer concentration
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Relationship between pH and pKa
pH = pKa when:
The molar concentration of acid and conjugate base are equal
[H2PO4-] = [HPO42-]
pH = pKa = 6.8
Henderson – Hasselbalch equation
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Physiological pHThe pH in the human body needs to remain ~7. Enzyme catalysis, protein-protein interactions, receptor binding, and other biological processes are very sensitive to pH. pH balance of the blood is maintained using a CO2 - bicarbonate buffer.
CO2 + H2O H2CO3 H+ + HCO3- pKa = 6.1(acid) (hydrated (bicarbonate CO2) base)
There is more than 10-fold more base (HCO3-) than acid (CO2) so pH < pK (pH= 7.4)
CO2 is exhaled by the lungs H+ + HCO3- CO2 + H2O
Breathing rate controls CO2
CO2 balance is controlled by the lungs, HCO3- by the kidneys
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Case where 10% acetate ion 90% acetic acid
pH = pKa + log10 [0.1 ] ¯¯¯¯¯¯¯¯¯¯ [0.9]
pH = 4.76 + (-0.95)pH = 3.81
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pH = pKa + log10 [0.5 ] ¯¯¯¯¯¯¯¯¯¯ [0.5]
pH = 4.76 + 0pH = 4.76 = pKa
Case where 50% acetate ion 50% acetic acid
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pH = pKa + log10 [0.9 ] ¯¯¯¯¯¯¯¯¯¯ [0.1]
pH = 4.76 + 0.95pH = 5.71
Case where 90% acetate ion 10% acetic acid
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pH = pKa + log10 [0.99 ] ¯¯¯¯¯¯¯¯¯¯ [0.01]
pH = 4.76 + 2.00pH = 6.76
pH = pKa + log10 [0.01 ] ¯¯¯¯¯¯¯¯¯ [0.99]
pH = 4.76 - 2.00pH = 2.76
Cases when buffering fails
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Weak Acids and Bases Equilibria
•Strong acids / bases – disassociate completely•Weak acids / bases – disassociate only partially•Enzyme activity sensitive to pH• weak acid/bases play important role in protein structure/function
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LEHNINGER PRINCIPLES OF BIOCHEMISTRY
Sixth Edition
David L. Nelson and Michael M. Cox
© 2013 W. H. Freeman and Company
CHAPTER 1The Foundations of Biochemistry
Reference: