The site and ligand conservation equations are:
The four-step algebraic solution (without the math) is:
1. Eliminate [Q] and [R] in Eq (11.26) using Eq (11.27).
2. Solve for the concentrations of occupied sites and eliminate [Ro] and [Qo] using Eq (11.25); solve for the ratios of bound sites to [mmtot]
3. Substitute the result from step 2 into Eq (11.28); with Eq (11.21), arrive at:
[L] K R+ [L] K q + [L] multiplying by KR + [L] and dividing by KR yields:
Which is of the same form as the single-site version Eq (11.22) with an additional term containing the effect of the second (Q) site. In order to permit plotting of the left-hand side of Eq (10.29b) against v, the dependence of [L] on v is needed. This is obtained by cross-multiplying Eq (11.29a) by the product of the three denominators, which yields the quadratic equation:
from which [L] can be calculated as a function of v. Figure 11.13 shows the Scatchard plot for the 2-independent-site model analyzed above. R is the stronger binding of the two sites. The dashed line neglects the presence of the Q sites. The effect of including these is to extend the v/[L] plot to larger values of v. With both sites active, the right hand curve shows the variation of [L] with the binding ratio expressed as a combination of Eqs (11.21) and (11.28):
Because (as in Fig. 11.13) there are 4 sites per macromolecule, this equation shows why the saturation curve ends at v = 4. In addition, to achieve complete saturation, an infinite concentration of ligand in solution is required. This can be seen from Eq (11.26); for the free site concentrations [R] and [Q] to approach zero, [L] must approach «.
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