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11111nata11111 [884]
3 years ago
12

In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occu

pied. 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?

Physics
1 answer:
sasho [114]3 years ago
7 0
Since there are four states, then the grand partition function of the system is

Z = 1+2 e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT}

where α is the chemical potential

Then, the occupancy of the system is

bar \ n=(e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT})/Z

Then using this equation, \alpha =-kT(V Z_{int} /N v_{Q}) and approximating Z_int to be kT/0.00018 eV, the model would look as that attached in the figure. That is the occupancy vs. pressure graph. 

There are more occupancies when the oxygen is high (high pressure) especially in the lungs. Heme sites tend to be occupied by oxygen. 

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Answer:

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                Ey = -Ax –C

c) the electric field on the z axis

              dv / dz = 0

              Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

              0 = = - A y + 2B x

              y = 2B / A x

Axis y

             0 = -Ax –C

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We substitute this value in the previous equation

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6 0
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