To carry out a Schildt plot, the receptor response to increasing concentrations of
agonist for a series of fixed concentrations of antagonist is studied and the dose ratio
for each concentration of antagonist calculated. The maximum response in all cases
should be the same. Equation 17.14 then predicts that a plot of the log(r– 1) against
log[B] should be a straight line of slope unity with an intercept on the abscissa equal
to logKb. If the slope is not unity then either the antagonist is not acting competitively
or more complex interactions are occurring between the antagonist and the receptor,
possibly allosteric in nature. It is important to note from the Schild equation that the
value ofKb, unlike that of IC 50 , is independent of the precise agonist used to generate
the data and is purely a characteristic of the antagonist for the specific receptor.
A limitation of a Schildt plot is that any error in measuring EC 50 (i.e. the response in
the absence of antagonist) will automatically influence the value of all the derived
dose ratios. The intercept of a Schildt plot on they-axis also gives the pA 2 value for
the antagonist (p for negative logarithm; A for antagonist; and 2 for the dose ratio
when the concentration of antagonist equals pA 2 ). The pA 2 will be equal to the value
oflogKbsince at an antagonist concentration that gives a dose ratio of 2, the Schildt
equation reduces to log[B]¼logKb.pA 2 is a measure of thepotencyof the antagonist.
Software programs are commercially available for the analysis of ligand-binding data.
Examples include:Prism(www.graphpad.com/curvefit);Sigmaplot(www.systat.com/
products/sigmaplot);Origin8andOriginPro 8 (www.originlab.com) andCalcusyn
(www.biosoft.com).Example 2 SCHILDT PLOT – CALCULATION OF AKbAND pA 2 VALUE
Question Use the data in Fig. 17.2 to construct a Schildt plot. The left-hand plot is the receptor
response to agonist binding in the absence of antagonist. The next three responses
are in the presence of 10^7 M, 10^6 M and 10^5 M antagonist. Read off from the
graph the concentration of agonist required to produce 50% maximum response in
the absence and presence of the antagonist and calculate the dose ratio (r) at each of
the three antagonist concentrations. Then plot a graph of log(r1) against log
[antagonist] and hence calculate both pA 2 andKb.
[Antagonist] (M) 10 ^710 ^610 ^5
EC 50 (M) 10 ^710 ^610 ^5
r 10 100 1000
r 1 9 99 999
log (r– 1) 0.954 1.9956 2.999
log [B] 7 6 5Answer You will see that the Schildt plot is linear (r¼0.9999) and that it has a slope of 1.02
confirming the competitive nature of the antagonist. The extrapolation of the line to
they-axis gives a value of8.12. This is equal tologKb, henceKb¼7.58 10 ^9
M and pA 2 ¼8.12.679 17.2 Quantitative aspects of receptor–ligand binding