University:
Amherst College
Course:
CHEM 330 | Biochemical Principles of Life at the Molecular Level
Academic year:

2011
Views:

310

Pages:

15
Author:

kulitzkih16orp

BioChem 330 - Course Outline September 27, 2011 • Bio-molecular Structure/Function (I) – PROTEINS • Structure – – – – Chemistry of amino acid building blocks Primary, secondary and tertiary structure Protein folding, thermodynamics and kinetics Predictions of protein folding, dynamics • Function – Binding ….a tale of two globins (hemoglobin and immunoglobulin) Protein Function – Binding of Ligands • A+B⇆C • Small Scale: Usually, A is a large macromolecule (protein, enzyme, receptor) and B is a small molecule (ligand, substrate, drug). – P + L ⇆ PL – E + S ⇆ ES – R + D ⇆ RD • Large Scale: It is also possible for both A and B to be large proteins which is what you see when calmodulin binds to a target protein, a transcription factor binds to DNA, or when an enzyme such as trypsin binds to a protein to initiate digestion. Protein Function – Binding of Ligands • In General [C ] 1 Ka   [ A][ B] K d – A + nB ⇆ C Ka = [C]/[A][B]n – Ka  = tighter binding • First consider, n=1 – Dissociation Eq. Constant Kd –C⇆A+B – – – – Kd = [A][B]/[C] Kd inversely related to binding constant, Ka Kd = weaker binding; units of concentration usually [Atot] fixed and small compared to [B] Protein is Saturated with ligand when [Btot] > Kd Protein Function – Binding of Ligands • P + L ⇆ P-L • Po + L ⇆ PB [ Pb ] 1 Ka   [ Po ][ L] K d Eqn1 • Fractional saturation – [PT]= [PB ] + [Po] – [Po] = [PT] - [PB] • Substitute in expression for Po, get , Eqn 2: [ Pb ] 1  ([ PT ]  [ Pb ])[ L] K d Eqn 2 • Simple rearrangement: – f ≡ [Pb]/[PT]= saturation [ Pb ] [ L] f   [ PT ] [ L]  K d Eqn 3 Protein Function – Binding of Ligands • Dissociation: Pb ⇆ Po + L • Dissociation Constant Kd – Plot of f vs [L] is hyperbolic – When [L] = Kd, f = ½ – The Kd is equal to the [L] when the protein is half saturated. – At conc of [L] above the Kd, protein approaches saturation – At conc of [L] below the Kd, the protein in unsaturated. [ Pb ] [ L] f   [ PT ] [ L]  K d f Kd Concentration L Protein Function – Binding of Ligands • Dissociation: Pb ⇆ Po + L • Dissociation Constant Kd – Lower Kd tighter the binding [ Pb ] [ L] f   [ PT ] [ L]  K d • Tight binding Kd < 10-9 M – Example estrogen to ER • Medium binding Kd < 10-6 M – Ca(II) binding to calmodulin • Loose binding Kd <10-3 M – ATP binding to hexokinase Kd Concentration B Protein Function – Binding of Ligands • At equilibrium • Mb-O2 ⇆ Mb + O2 • Kd = [Mb]free [O2] / [Mb-O2] • Kd = [Mb]free pO2 / [Mb-O2] • [Mb]free = [Mb]T - [Mb-O2] • Define Y ≡ fractional saturation of Mb • Y = [Mb-O2] / [Mb]T • Substitute and Rearrange: – Y = pO2 / Kd + pO2 – hyperbola, when Y = ½; Kd = pO2 • Here Y=f and pO2 = [L] Binding Curve for O2 Binding to Mb Protein Function – Binding of Ligands • Pb ⇆ Po + L • Dissociation Constant Kd – Lower Kd tighter the binding • Tight binding Kd < 10-9 M – Example estrogen to Receptor – kcal/mole binding energy? • Medium binding Kd < 10-6 M – Ca(II) binding to calmodulin – kcal/mole binding energy? • Loose binding Kd <10-3 M – ATP binding to hexokinase – kcal/mole binding energy? Calculate binding energies from Kd using ∆Go = - RT ln 1/Kd Protein Function – Binding of Ligands • ∆G = ∆H - T∆S • Po + L ⇆ P1 + L ⇆ P2 • ∆S an entropic “cost” (decrease in entropy = - ∆S ) whenever two molecules combine to make one molecule • comes from the loss of the translational and rotational movement of two molecules when they become one.……. • Each molecule has 3 deg of freedom from trans in x, y, and z directions, and 3 deg of freedom from rot about Ix, Iy, and Iz moments of inertia. If this is true, then the two reactant molecules have 12 deg of freedom and the single product molecule has 6, for a net LOSS of 6 deg of freedom. • In reality, this is not COMPLETELY true because six NEW degrees of freedom are introduced due to new vibration and librations (rocking) in dimer Hemoglobin Structure: (ab)2 – Hemoglobin is a dimer of dimers, a1 b1 dimer a2 b2 Hemoglobin – Elaine Lin PDB code of structure: 1HHO Name each molecule in complex: 4 subunits each with 1 heme prosthetic group (which itself has a Fe2+ atom), 1 oxygen for each subunit (when bounded) Functionally relevant residues and features of the structure you will explore: Differences in conformation between oxygen bound and no oxygen bound (residue involved: histidine) Sickle cell hemoglobin: glutamate 6 in the beta chain mutated to valine Fetal hemoglobin versus maternal hemoglobin Knobby Surface View Charged Surface Hydrophobic/Polar Tyr 140 Chain = A Element = C Atom = C Arg 141 Chain = A Element = C Atom = C Salt Bridges + Cation Pi interactions + Core Residue Distance • 3.85 A between C-alpha’s of Try 140 and Arg 141 • Salt Bridges – – – – – – – Glu 7-Lys 132 Glu 27-Arg31 Asp 74-Lys 7 Glu 116-Lys 16 Glu 26-Arg 30 Glu 121-Lys 17 Asp 21-Lys 65 • Cation Pi: – Arg 40-Phe 41 (only one) Arg 141 Contacts H Bonds Hydrophobic

BioChem 330 Outline: Bio-molecular Structure Function (Part 1)

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