Biology Reference
In-Depth Information
How are the binding polynomials
X
4
obtained? They are simply
the sum of the concentrations of all the binding species present in
solution. For example, if dimeric hemoglobin contained two identical,
nondistinguishable oxygen-binding sites, then in solution only three
possible oxygenation states would exist; with zero, one, or two O
2
molecules bound. Thus, its binding polynomial could be written as:
X
2
and
X
¼½abþ½ab
O
2
þ½abð
O
2
Þ
2
:
(7-25)
2
Using the law of mass action to express the concentration of the
oxygenated species in terms of the concentrations of the unoxygenated
dimeric hemoglobin and the unbound O
2
concentration, we obtain:
2
X
2
¼½abþ
K
21
½ab½
O
2
þ
K
22
½ab½
O
2
:
(7-26)
In this case, the definition of K
21
is the average (i.e., macroscopic)
Adair-binding constant for the first O
2
being bound to either of the O
2
binding sites. As we shall see below, this is not the binding constant to
either the
a
subunit or the
b
subunit, but rather the sum of the two.
An interesting property of this application of binding polynomials is that
the units of hemoglobin and O
2
concentration are arbitrary. This is a
consequence of the natural logarithms contained in Eqs. (7-21) and
(7-23). We take the concentration of the reference state to be the
unoxygenated dimeric hemoglobin concentration. Because the units of
hemoglobin concentration are arbitrary, we simply express these units as
a fraction of the unoxygenated hemoglobin concentration. In these units,
Eq. (7-25) is transformed into Eq. (7-27), since in these units [
ab
¼
]
1.
X
¼
1
þ½ab
O
2
þ½abð
O
2
Þ
2
2
(7-27)
2
X
¼
1
þ
K
21
½
O
2
þ
K
22
½
O
2
:
2
Notice this is exactly the quantity
X
2
presented in Eq. (7-19).
E
XERCISE
7-6
Use binding polynomials to derive Eq. (7-6).
E
XERCISE
7-7
Assuming, as we did for the dimeric hemoglobin in Eq. (7-25), that the
binding polynomial is defined as the sum of the concentrations of all the
binding species present in solution, show that for the tetrameric
hemoglobin,
2
3
4
X
¼
1
þ
K
41
½
O
2
þ
K
42
½
O
2
þ
K
43
½
O
2
þ
K
44
½
O
2
:
4
