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Myoglobin and Hemoglobin Quaternary structure and allosteric properties of proteins Fe.

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Presentation on theme: "Myoglobin and Hemoglobin Quaternary structure and allosteric properties of proteins Fe."— Presentation transcript:

1 Myoglobin and Hemoglobin Quaternary structure and allosteric properties of proteins Fe

2 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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4 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 )

5 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

6 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)

7 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

8 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

9 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

10  = pO2pO2 pO2pO2 P 50 + Assume P 50 = 10 pO2pO2  10.09 20.17 5 0.33 100.50 200.67 1000.91

11  = pO2pO2 pO2pO2 P 50 + n n n = number of binding sites on the protein What about Hemoglobin? For Hb, n = 4

12  100 50 pO2pO2 Hemoglobin P50P50 Region of Cooperativity Sigmoidal Curve

13 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

14 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

15 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

16 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

17 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

18 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

19 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

20 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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