Allostery & Structure

The binding of oxygen to hemoglobin is accompanied by a change in the structure of hemoglobin. The structural change has been the subject of extensive analyses (1, 2). All of the structural analyses rely on the assumption of two stable quaternary structural states for hemoglobin - fully oxygenated and completely deoxygenated. The fully oxygenated state is designated as the R state and the fully deoxygenated state is designated T. This assumption was supported by the observation of only two structual forms by xray crystallography. However, in recent years a different structural form for the fully oxygenated molecule (3, 4) has been reported. These structures have been designated as Y and R2 and are very similar to one another. Unlike previously determined oxy structures, which were crystallized under high salt conditions, the recent structures were determined from crystals grown under physiological salt conditions. The discovery of these structure leads to the question: Do these structures lie along the pathway from T to R? An analysis (5) of the Y form led to the conclusion that it does not lie along the path from T to R.

Comparing Conformations

To understand the relationship of one conformation to another we need a suitable measure by which the two conformations can be related. One widely used measure for comparing conformations is the root mean squared deviation. This method relies on obtaining an optimal superposition of the two conformations. The method works in the following fashion:

Step 1.
Compute the centroid for each conformation, C1 and C2.
Step 2.
Translate conformation 2 so that C1 and C2 are coincident.
Step 3.
Apply some sequence of rotations and translations to conformation 2 so that the distance between corresponding points/atoms in the two conformations is minimized.
Step 4.
Compute the root mean square deviation as
[ RMS= SQRT( (Sum(dx^2 + dy^2 + dz^2)/N )]
where, [(dx^2 + dy^2 + dz^2)] is the distance between corresponding points, summed over all N atoms.

While the method of performing an optimal superimposition and calculating a rms difference is useful for assesing similarity, it is not as useful in comparing conformational changes. Consider for example the superposition in figure 1. The optimal superposition results in atoms 3 and 4 showing the largest deviation. However, looking at the individual conformations in the top frame of the figure, it is evident that the difference between the two structures is:

  1. The vectors 1-2 and 3-4 are on the same side in the conformation on the right while they are on opposite sides in the other.
  2. The vectors 5-6 and 3-4 are on the same side in the conformation on the right while they are on opposite sides in the other.

In fact, one could superimpose the central portion, namely atoms 3, 4, 5 & 6 exactly. Thus, the superposition, by its very nature, averages out differences. Consequently, small but important differences could be submerged.

[See Figure Caption]


Figure 1.Illustration of least squares superimposing of two conformations of a molecule with six atoms. The two conformations are shown in the top frame and the result in the lower frame. The two conformations differ in the orientation of the vector 1-2 vs. 3-4 and 3-4 vs. 5-6. For the conformation on the left the relationship is anti while for the conformation on the right the relationship is syn. The central four points 2, 3, 4 and 5 are exactly superimposable.

An alternative way of comparing conformations is the use of difference distance matrices. In this method one first constructs a distance matrix for each conformation. The elements of the distance matrix are the distance between pairs of atoms. For the conformations illustrated in figure 1. the individual distance matrices are as follows: 

Conformation 1                      Conformation 2
_________________________________   _________________________________

       1    2    3    4    5    6          1    2    3    4    5    6 
_________________________________   _________________________________

1   df11 df12 df13 df14 df15 df16   1   ds11 ds12 ds13 ds14 ds15 ds16
_________________________________   _________________________________

2   df21 df22 df23 df24 df25 df26   2   ds21 ds22 ds23 ds24 ds25 ds26
_________________________________   _________________________________

3   df31 df32 df33 df34 df35 df36   3   sf31 sf32 sf33 sf34 sf35 sf36
_________________________________   _________________________________

4   df41 df42 df43 df44 df45 df46   4   ds41 ds42 ds43 ds44 ds45 ds46
_________________________________   _________________________________

5   df51 df52 df53 df54 df55 df56   5   ds51 ds52 ds53 ds54 ds55 ds56
_________________________________   _________________________________

6   df61 df62 df63 df64 df65 df66   6   ds61 ds62 ds63 ds64 ds65 ds66
_________________________________   _________________________________
Since the distance between any two atoms,say 1 and 3 is the same as the distance between 3 and 1, only the upper half of each matrix needs to be computed. Hence, the matrices are reduced to the following.
Conformation 1                      Conformation 2
_________________________________   _________________________________

       1    2    3    4    5    6          1    2    3    4    5    6 
_________________________________   _________________________________

1      0 df12 df13 df14 df15 df16   1      0 ds12 ds13 ds14 ds15 ds16
_________________________________   _________________________________

2           0 df23 df24 df25 df26   2           0 ds23 ds24 ds25 ds26
_________________________________   _________________________________

3                0 df34 df35 df36   3                0 sf34 sf35 sf36
_________________________________   _________________________________

4                     0 df45 df46   4                     0 ds45 ds46
_________________________________   _________________________________

5                          0 df56   5                          0 ds56
_________________________________   _________________________________

6                               0   6                               0
_________________________________   _________________________________
The next step is to construct the difference distance matrix. This is done by subtracting one matrix form the other as shown below.
Difference distance matrix 
____________________________________________________________________

       1           2           3           4           5           6 
____________________________________________________________________

1      0   df12-ds12   df13-ds13   df14-ds14   df15-ds15   df16-ds16
____________________________________________________________________

2                  0   df23-ds23   df24-ds24   df25-ds25   df26-ds26
____________________________________________________________________

3                              0   df34-ds34   df35-ds35   df36-ds36
____________________________________________________________________
  
4                                          0   df45-ds45   df46-ds46
____________________________________________________________________

5                                                      0   df56-ds56
____________________________________________________________________

6                                                                  0
____________________________________________________________________
For the example in figure 1 this matrix reduces as follows.
Difference distance matrix 
____________________________________________________________________

       1           2           3           4           5           6 
____________________________________________________________________

1      0           0           0   df14-ds14   df15-ds15   df16-ds16
____________________________________________________________________

2                  0           0           0           0   df26-ds26
____________________________________________________________________

3                              0           0           0   df36-ds36
____________________________________________________________________
  
4                                          0           0           0
____________________________________________________________________

5                                                      0           0
____________________________________________________________________

6                                                                  0
____________________________________________________________________
It is evident from the difference distance matrix that the atoms 2, 3, 4 and 5 form a subset for which all entries in the difference distance matrix are zero. Similarly, atoms 1,2 and 3 form a subset as do 4,5 and 6. Thus, the difference distance matrix shows that the positions of atoms 1 and 6 are different in the two conformations. If we now superimpose the two conformations using only atoms 2, 3, 4 and 5, the superposition is as shown in figure 2, and the difference between the two conformations is obvious.
[See Figure Caption]

Figure 2.Illustration of least squares superimposing of two conformations using only atoms 2, 3, 4 and 5.

One disadvantage of the distance matrix apporach is that mirror images will have the same distance matrix and will hence appear identical.

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