Question

In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occupied. To model this effect in a simple way, imagine that a hemoglobin molecule has just two sites, either or both of which can be occupied. This system has four possible states (with only oxygen present). Take the energy of the unoccupied state to be zero, the energies of the two singly occupied states to be -0.55 eV, and the energy of the doubly occupied state to be -1.3 eV (so the change in energy upon binding the second oxygen is -0.75 eV). As in the previous problem, calculate and plot the fraction of occupied sites as a function of the effective partial pressure of oxygen. Compare to the graph from the previous problem (for independent sites). Can you think of why this behavior is preferable for the function of hemoglobin? And find average number <N> of oxygen molecules bound.

Answer 1

Page No - 1 Answer diven that - In a real hemoglobin molecule, the tendance of oxygen to bind to a heme site increases as the other three heme site became occupied. The mathematical expression for Gibbs factor is 1. Gibbs factor = ex It (nu) Here kis Bolt a man Constant - Tis Temperature eis energy of the state , u = chemical potentied of system - In the given system of red hemoglobin molecule, oxygen can occupy four different states... The energy of un occupied state is zero - Eo-Oeu The energy of two singly occupied states is t = -0.550V The energy of one doubly occupied state is t2 =-1.3ev with occupation numbers 0,1, 1 and 2 and Corresponding energies o, ti, t, and E2 the grand partition is follows junction Zaltze ferred to - (€2-24) - 1.3ev

PageNo=2 - The average no of oxygen molecules in the system is N = { N(S) = 1.8 (1=1) + 2.P (= 2) КТ 2e + KT 7 The average fraction molecules that are occupied is just half the average number of oxygen present. Hence, the occupancy of system is follows: ñ= -2u) + e - € (,-) KT te (€2-24) KT Hee field te fle-20) The expression for chemical potentid : - U=_kTln Nzint ? il mio use the ided gas equation - PIENET substitute of fort in equation us ki ln [ zite) and solve force For M=.=KT ln (AT A

Solve the equation for ulet Page No - 3 e luft - Pro - Kizint To Simplefy this expression the first Gibbs factor can be written as follows e fer essere rete substitute Pro for exp (a) Pro e KT e KT. To KIzint PVO Azint Here, the Vañable Po= KTZ: ut eft The quantum volume of oxygen is as follows e - Jaimer Substitute 6.626 X10 34 ells forh= 32x1.66x10 ting form= 1.381 +502 16 for K and 310k for T. JTMKT By substituting values, we get Vo = 5.3 x 103 min - The numered evaluation of Po is the same as in the previous Problem, except for the value of t, which is much less negative solving the equation Po= 2.03 bar is obtained

PageNo=4 The Seconds Gibbs factor Can be written as follows: At the room temperature 310k, the exponential of is equal ta79 e = 1190 Interms of P, po and f, the occupancy for oxygen, interms of the partial pressure of oxygen, is as follows i++)t elemente (%)/170() - Titafetest comments As in the independant site model the occupancy in the model is neady loot near the lungs where P = 0.20 bar The figure below shows the graph between the fractions of the occupied sites and the partid pressure of onggen Here pissmall Fraction of occupied STÖ Ő ở Dos of ot's om onts P(bar)