# Molecular Bases for Conformational Changes

### Molecular Bases for Conformational Changes

Based on crystallographic structures, two main conformational states appear to exists for Hb, the deoxy (or T - taut) state, and the oxy (or R -relaxed) state. The major shift in conformation occurs at the alpha-beta interface, where contacts with helices C and G and the FG corner are shifted on oxygenation.

Figure: Conformation Changes on Oxygen Binding to Deoxy-Hemoglobin (files aligned with DeepView, display with Pymol)

The deoxy or T form is stabilized by 8 salt bridges which are broken in the transition to the oxy or R state.

Figure: Salt Bridges in Deoxy Hb

6 of the salt bridges are between different subunits (as expected from the above analysis), with 4 of those involving the C- or N- terminus.

In addition, crucial H-bonds between Tyr 140 (alpha chain) or 145 (on the beta chain) and the carbonyl O of Val 93 (alpha chain) or 98 (beta chain) are broken. Crystal structures of oxy and deoxy Hb show that the major conformational shift occurs at the interface between the a and b subunits. When the heme Fe binds oxygen it is pulled into the plan of the heme ring, a shift of about 0.2 nm. This small shift leads to larger conformational changes since the subunits are packed so tightly that compensatory changes in their arrangement must occur. The proximal His (coordinated to the Fe) is pulled toward the heme, which causes the F helix to shift, causing a change in the FG corner (the sequence separating the F and G helices) at the alpha-beta interface as well as the C and G helices at the interface, which all slide past each other to the oxy-or R conformation.

### The Molecular Bases for Allosteric Effects

Decreasing pH shifts the oxygen binding curves to the right (to decreased oxygen affinity). Increased [proton] will cause protonation of basic side chains. In the pH range for the Bohr effect, the mostly likely side chain to get protonated is His (pKa around 6), which then becomes charged. The mostly likely candidate for protonation is His 146 (on the b chain - HC3) which can then form a salt bridge with Asp 94 of the b (FG1) chain. This salt bridge stabilizes the positive charge on the His and raises its pKa compared to the oxyHb state. Carbon dioxide binds covalently to the N terminus to form a negatively charge carbamate which forms a salt bridge with Arg 141 on the alpha chain. BPG, a strongly negatively charged ligand, binds in a pocket lined with Lys 82, His 2, and His 143 (all on the beta chain). It fits into a cavity present between the bsubunits of the Hb tetramer in the T state.Notice all these allosteric effectors lead to the formation of more salt bridges which stabilize the T or deoxy state. The central cavity where BPG binds between the bsubunits becomes much smaller on oxygen binding and the shift to the oxy or R state. Hence BPG is extruded from the cavity.

The binding of H+ and CO2 helps shift the equilibrium to deoxyHb which faciliates dumping of oxygen to the tissue. It is in respiring tissues that CO2 and H+ levels are high. CO2 is produced from the oxidation of glucose through glycolysis and the Krebs cycle. In addition, high levels of CO2 increase H+ levels through the following equilibrium:

$H_2O + CO_2 \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$

In addition, H+ increases due to production of weak acids such as pyruvic acid in glycolysis.

Hb, by binding CO2 and H+, in addition to O2, serves an additional function: it removes excess CO2 and H+ from the tissues where they build up. When deoxyHb with bound H+ and CO2 reaches the lungs, they leave as O2 builds and deoxyHb is converted to oxyHb.

### Mathematic Bases for the Sigmoidal Binding Curve for Oxygen

Previously we have shown that the binding of oxygen to Mb, which can be described by the equilibrium, M + L ↔ ML, can be described mathematically by

$Y = \dfrac{L}{K_d +L}. \tag{1}$

This is the equation of a hyperbola. Remember, that this hyperbolic plot can be transformed in a variety of ways, as summarized in the graphs below for Mb.

How does the sigmoidal binding curve for Hb arise? At least three models (Hill, MWC, and KNF) can be developed that give rise to sigmoidal binding curves. Remember, sigmoidal curves imply cooperative binding of oxygen to Hb: as oxygen binds, the next oxygen seems to bind with higher affinity (lower Kd).

Hill Model: In this model, we base our mathematical analysis on the fact that the stoichiometry of binding is not one to one, but rather 4 to 1: Perhaps a more useful equation to express the equilibrium would be M + 4L ↔ ML4. For this equilibrium, we can derive an equation analogous to the equation 1 above. This equation is:

$Y = \dfrac{L^4}{K_d + L^4}. \tag{2}$

For any given L and Kd, a corresponding Y can be calculated. Using this equation, the plot of Y vs L is not hyperbolic but sigmoidal (see next link below). Hence we're getting closer to modeling that actual data. However, there is one problem. The sigmoidal curves don't give a great fit to the actual oxygen binding curve for Hb. Maybe a better fit can be achieved by altering the exponents in equation 2. A more general equation for binding might be M + nL <=> MLn, which gives the following equation:

$Y = \dfrac{L^n}{K_d + L^n}. \tag{3}$

If n is set to 2.8, the theoretical curve of Y vs L gives the best but still not perfect fit to the experimental data. It must seem arbitrary to change the exponent which seems to reflect the stoichiometry of binding. Consider another meaning of the equilibrium described above:

$M + 4L \rightleftharpoons ML_4.$

One interpretation of this is that all 4 oxygens bind at once to Hb. Or, alternatively, the first one binds with some low affinity, which through associated conformational changes changes the remaining 3 sites to very high affinity sites which immediately bind oxygen if the oxygen concentration is high enough. This model implies what is described as infinitely cooperative binding of oxygen.

(Notice that this equation becomes: Y = L/[Kd + L], when n =1 (as in the case with myoglobin, and in any equilibrium expression of the form: $$M + L \rightleftharpoons ML$$. Remember plots of ML vs L or Y vs L gives hyperbolas, with Kd = L at Y = 0.5.)

#### Does Kd = L at Y = 0.5?

The oxygen concentration at which Y = 0.5 is defined as P50. We can substitute this value into equation 3 which gives an operational definition of Kd in terms of P50.

Y = 0.5 = P50n/[Kd + P50n] - multiple both sides by 2
1 = 2P50n/[Kd + P50n]
Kd + P50n = 2P50n
$K_d = P_{50}n \tag{4}$

Note that for equation 3, $$K_d$$ is not the ligand concentration at half-saturation as we saw in the case with hyperbolic binding curves.

Now consider another model:

$M + L \rightleftharpoons ML \rightleftharpoons L + ML_2 + L \rightleftharpoons ML_3 + L \rightleftharpoons ML_4$

where the binding of each oxygen to the unligated or increasing ligated Hb has the same Kd. That is, the affinity of each binding site for oxygen does not increase as more sites are bound to oxygen. In this model, n in equation 3 is 1, and the resulting graph is completely hyperbolic. The fact that the experimental data fits the equilibrium M + 2.8L ↔ ML2.8 implies that the binding is cooperative but not infinitely cooperative. Graphs of Y vs. L showing these three cases (n=1, 2.8, and 4) are shown below:

The general equation 3), Y = Ln/[Kd + Ln] can be rearranged as shown below:

1 - Y = [Kd + Ln]/[Kd + Ln] - Ln/[Kd + Ln] =
$1 - Y = \dfrac{K_d}{K_d + L^n} \tag{5}$

where 1 - Y is the fraction not bound. Solving for Y/[1-Y] by using equations 3 and 5 gives:

$\dfrac{Y}{1-Y} = \dfrac{\dfrac{L^n}{K_d + L^n}}{ \dfrac{K_d}{K_d + L^n}} = L^n/K_d \tag{6}$

Taking the log of both sides gives:

log (Y/1-Y) = log ( Ln/Kd) =

$\log (\dfrac{Y}{1-Y}) = n\log \, L - \log \, K_d \tag{7}$

A plot of log (Y/1-Y) vs. log L is called a Hill plot, where n is the Hill coefficient. This equation is of the form: y = mx + b which is a straight line with slope n and y intercept of $$-\log\, K_d$$. When n = 1, as it would be with Mb or Hb when oxygen binds to each site with the same affinity irrespective of the number of other oxygens bound to other sites, the Hill plot is linear with a slope of 1. Solving for the x intercept (when the y axis variable is 0) in equations 7 gives:

$$0 = n\, \log\, L -\log\, K_d$$ , or $$n\, \log\, L = \log\, K_d,$$ or $$\log\, L = \dfrac{\log \, K_d}{n}. \tag{8}$$

The X intercept is when the dependent variable "y" value is 0. This occurs when Y/(1-Y) = 1, which occurs at half fractional saturation. (Remember log 1 = log 100 = 0). Substituting equation 4 ($$K_d = P_{50}^n$$) into (7) and (8) gives

$\log \left(\dfrac{Y}{1-Y}\right) = n\,\log \,L - n \,\log \, P_{50} \tag{9}$

which is the Hill Equation with $$P_{50}$$ instead of $$K_d$$,

$0 = n\log \, L -n \log\, P_{50},$

or

$n\log\, L = n \log \, P_{50},$

or

$\log L = \log\, P_{50}.\tag{10}$

Even when n does not equal 1, the Hill plot is linear, since it has the form y=mx+b. If n = 2.8 or 4, the plot is linear, but has a slope of 2.8 and 4, respectively. This can be seen in the graph below which shows HIll plots with n = 1, 2.8, and 4.

Figure: Hill Plot for Mb (n =1)

Hill Plots: n = 1, 2.8, and 4

However, the affinity of dioxygen for Hb changes, so that there must be more than one effective $$K_d$$. Hence, the actual Hill plot of Hb, log (Y/1-Y) vs. log L, can not be linear over all ranges of dioxygen. A linear plot, such as for Mb, crosses the x axis at one point, with a value of (log Kd)/n = logKd since n = 1. In contrast for Hb, since the $$K_d$$ seems to change with L concentration, there can not be just 1 value of Kd, as given by the x intercept. The Hill plot of actual Hb binding data is curvilinear, and cross the x axis only once. However, the ends of the curve (at low and high dioxygen) approach straight lines with slopes of 1 (i.e. n=1). If extrapolated through the x axis, these lines would give the Kd for the binding of the first and last dioxygens, which bind noncooperatvely. LogL values near the region of the curve that crosses the x axis approximate a straight line with slope of 2.8. This implies there is maximal cooperativity in the middle of the binding curve. The graphs shows that the $$K_d$$ for the first oxygen binding is much higher than the $$K_d$$ for the last oxygen binding. Hence the Hill Plots supports our ideas than cooperativity is caused by conformational changes in Hb which occur on oxygen binding such that as progressively more oxygen is bound, the affinity for the remaining sites increases.

Figure: Hill Plots For Hb Showing straight lines for n=2.8 and for n's=1 which model the low and high affinity sites.

Figure: Hill Plot for Hb: Black line showing hypothetical actual curve