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 $-0.55eV$, and the energy of the doubly occupied state to be $-1.3eV$(so the change in energy upon binding the second oxygen is $-0.75eV$). 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?

Short answer

Occupancy rises gradually for smaller values of $P$. This shows that with cooperative binding, hemoglobin reacts with oxygen much fast in cells when $P$ is small.

Step-by-step solution

Step 1. Formula Gibb's Factor

Gibb's Factor is given by formula:

$Gibb\text{'}sFactor=e^{\left[\frac{-(\epsilon -n\mu )}{kT}\right]}$

where, $k$is Boltzmann's constant, $T$is temperature, $\mu$is chemical potential, and $\epsilon$is energy of state .

As, in hemoglobin molecule, oxygen can have four different states given by:

localid="1647144603783" $$\begin{gathered} {\epsilon }_o=0eV \\ {\epsilon }_1=-0.55eV(two\sin glyoccupied) \\ {\epsilon }_2=-1.3eV \end{gathered}$$

The average number of oxygen molecules is given by,

$$\begin{gathered} \overline{N}=\sum _{S}N\left(S\right) \\ =\frac{2e^{{\displaystyle \frac{-({\epsilon }_1-\mu )}{kT}}}}{Z}+\frac{2e^{{\displaystyle \frac{-({\epsilon }_2-2\mu )}{kT}}}}{Z} \end{gathered}$$

Step 2. Formula occupancy

Occupancy of system is given by:

$$\begin{gathered} \overline{n}=\frac{\overline{N}}{Z} \\ =\frac{1}{Z}\left(e^{\frac{-({\epsilon }_1-\mu )}{kT}+\frac{-({\epsilon }_2-2\mu )}{kT}}\right) \end{gathered}$$

Chemical Potential formula is :

$\mu =-kT\times \ln \left(\frac{V{Z}_{in}}{N{\nu }_Q}\right)$

Substitute $\frac{V}{N}=\frac{kT}{P}$in above formula,

$\mu =-kT\times \ln \left(\frac{kT{Z}_{in}}{P{\nu }_Q}\right)$

$e^{\frac{\mu }{kT}}=\frac{P{\nu }_Q}{kT{Z}_{in}}$

Gibb's Factor can be written as:

$$\begin{gathered} e^{\left[\frac{-({\epsilon }_1-\mu )}{kT}\right]}=e^{\frac{-{\epsilon }_1}{kT}}\times \frac{P{\nu }_Q}{kT{Z}_{in}} \\ =\frac{P{\nu }_Q}{kT{Z}_{in}{e}^{\frac{{\epsilon }_1}{kT}}} \\ =\frac{P}{P_o} \end{gathered}$$

Step 3. Calculation

Oxygen molecule quantum volume is:

${\nu }_Q={\left(\frac{h}{\sqrt{2\mathrm{\pi mkT}}}\right)}^3$

Substitute the values of the variables in above formula,

$$\begin{gathered} {\nu }_Q={\left(\frac{6.63\times {10}^{-34}J·s}{\sqrt{2\pi (32\times 1.66\times {10}^{-27}kg)(1.38\times {10}^{-23}J/K)(310K)}}\right)}^3 \\ =5.3\times {10}^{-33}{m}^3 \end{gathered}$$

In equation, $\overline{n}=\frac{1}{Z}\left(e^{\frac{-({\epsilon }_1-\mu )}{kT}+\frac{-({\epsilon }_2-2\mu )}{kT}}\right)$

Substitute $e^{\frac{\epsilon }{KT}}=1790$ and $P_o=2.03bar$at room temperature,

$\overline{n}=\frac{\left({\displaystyle \frac{P}{2.03}}+1790{\left({\displaystyle \frac{P}{2.03}}\right)}^2\right)}{\left(1+\frac{P}{2.03}+1790{\left(\frac{P}{2.03}\right)}^2\right)}$

Occupancy in this model is nearly $100\%$near lungs given $P=0.20bar$

Step 4. Table and Graph

The table is created between Pressure $P$and fraction of occupied sites:

$\overline{P}$	$\overline{n}$
0.01	0.0461
0.02	0.1554
0.03	0.2899
0.04	0.4204
0.05	0.5332
0.06	0.6257
0.07	0.7001
0.08	0.7598
0.09	0.8080

The graph between fraction of occupied sites and oxygen partial pressure is created as,

We see that the occupancy rises gradually for smaller values of $P$. This shows that with cooperative binding, hemoglobin reacts with oxygen much fast in cells when $P$is small.