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Myoglobin and Hemoglobin Quaternary structure and allosteric properties of proteins Fe
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The importance of quaternary structure in proteins is best illustrated by the binding of O 2 to hemoglobin Philosophy: By comparing hemoglobin with myoglobin, (a single subunit protein), subunits can be seen to regulate the strength of binding and give rise to a property of proteins called “cooperativity” Cooperativity is the basis of another property of proteins called “allosterism” or “allostery”
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Binding of O 2 to Myoglobin Mb + O 2 MbO 2 = MbO 2 MbO 2 + Mb [ MbO 2 ] [Mb][O 2 ] K = Filled Sites Total Sites = the fraction of filled sites = Y O2 Define (Y O2 )
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Derive an expression for as a function of O We are asking: How do the sites filled vary with oxygen pressure? [ MbO 2 ] [Mb][O 2 ] K = MbO 2 = K[Mb][O 2 ] = MbO 2 Mb + MbO 2 Mb + O 2 MbO 2 Sites Filled Total Sites 71% filled 10 out of 14
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Substitute the value for MbO 2 into the equation = K[Mb][O 2 ] [Mb] + Mb 1 = K[O 2 ] 1 + K[O 2 ] K K K 1 K = or K d = O 2 pressure that half fills sites = P 50 When 1/K = [O 2 ], = 0.5 (also called K d ) Therefore: (P 50 distinguishes the specific O 2 -binding protein)
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Redefining O 2 = pO 2 Charles Law: Quantity of gas absorbed by a liquid is proportional to the partial pressure of the gas above the liquid. = pO2pO2 pO2pO2 P 50 + Note: Mb is left out of the final equation. Note: Only connection to Mb is 1/K or P 50
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P50P50 100 50 pO2pO2 BINDING PARAMETERS O2O2 ( x 100) When O 2 binds to myoglobin, the binding increase is hyperbolic with pressure increase Two points on the binding curve are apparent: (1) the point of half saturation (2) the point of full saturation (never attained) Full saturation Half saturation
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P50P50 100 50 pO2pO2 BINDING PARAMETERS O2O2 ( x 100) Weaker Binding Strong Binding The binding is considered weaker when it takes a greater oxygen pressure (larger P 50 ) to reach half-saturation The binding is considered weaker when it takes a greater oxygen pressure (larger P 50 ) to reach half-saturation P 50
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= pO2pO2 pO2pO2 P 50 + Assume P 50 = 10 pO2pO2 10.09 20.17 5 0.33 100.50 200.67 1000.91
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= pO2pO2 pO2pO2 P 50 + n n n = number of binding sites on the protein What about Hemoglobin? For Hb, n = 4
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100 50 pO2pO2 Hemoglobin P50P50 Region of Cooperativity Sigmoidal Curve
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Why does Hemoglobin show Cooperative Binding? Oxygen Transport Inspired air Aveolar air Torrs 158 100 Arterial Blood90 Capillary40 Interstitial30 Cytosol 10 Binding Release Cooperativity is designed to RELEASE O 2 at low Pressures
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EXAM 1 BICH 410 (all sections) 100 pts Tuesday, Oct 4, 11:10 – 12:00, Rm 108 Biochemistry 8 1/2 x 11 scantron (Blue) Calculator Chapters 1-5: Basics of energy and acid-base, amino acids, peptides, proteins, myoglobin, hemoglobin, allostery
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Protein-Ligand Interactions Revisited P + L = PL PL = P + L K a = [PL] [P][L] K d = [P][L] [PL] K d = 1/K a = [L] 0.5 or O = L + L1/K a O = L + LKdKd Association Disassociation
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P + nL = PL n Hb + 4O 2 = Hb(O 2 ) 4 K a = [PL n ] [P][L] n K a = [Hb(O 2 ) 4 ] [Hb] [O 2 ] 4 O = LnLn + L n KdKd O = pO24pO24 +1/K a pO24pO24 Any LigandHemoglobin n=4 O = LnLn + L n 1/K a O = pO24pO24 + K d pO24pO24 Where K d = 1/K a = [L] n 0.5 = P n 50 Where K d = 1/K a = [O 2 ] 4 0.5 = P 4 50 Multi-Ligand Interactions
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ALLOSTERIC BINDING Rearranging the Equation nlog pO 2 - nlog P 50 = n = Hill Coefficient pO2pO2 1 – = P50P50 n n log 1 – O = pO2npO2n + P 50 n pO 2 n
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log 1 - nlog pO 2 - nlog P 50 = y = m x + b n H = 1.0No Cooperativity n H = > 1.0 Positive Cooperativity n H = < 1.0 Negative Cooperativity n H = n Full Cooperativity
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How is n H Determined log 1 - log pO 2 Myoglobin n H = 1 0 -3 +3 01.0 Cooperative Transition n H = 3 Hemoglobin Low Affinity n H = 1 Hemoglobin High Affinity n H = 1 Hemoglobin R-State T-State
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How is n H Determined log 1 - log pO 2 Myoglobin n H = 1 0 -3 +3 01.0 n H = 3 Hemoglobin Low Affinity n H = 1 Hemoglobin High Affinity The higher the O 2 pressure, the stronger the binding, (binding mode) The lower the O 2 pressure, the weaker the binding, (release mode)
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