Memo．Translated on May 13, 2025．
Several methods exist for calculating the inhibition constant (\(K_{\mathrm i}\)). Besides the widely known Cheng–Prusoff equation, modified Cheng–Prusoff equation using [L]50 value instead of [L]T, and the exact Munson–Rodbard correction, which allows direct calculation using the total concentration of an unlabeled ligand—namely, the 50% inhibitory concentration \({\rm IC}_{50}\)—without requiring approximations. Additionally, the 50% inhibitory concentration of the free ligand, denoted \([I]_{50}\), can be estimated either via nonlinear regression or linear regression.
We investigates the deviation between the estimated \(K_{\rm i}\) values obtained by these methods and the true \(K_{\rm i}\) values using simulated data.
To simulate realistic assay conditions, typical experimental values were used for the binding between the protein kinase C δ-C1B domain (as receptor) and the radiolabeled ligand [³H]phorbol 12,13-dibutyrate ([³H]PDBu):
- Total receptor concentration: \([R]_{\rm T}\) = 3.45 nM
- Total radioligand concentration: \([L]_{\rm T}\) = 17 nM
- Radioligand dissociation constant: \(K_{\rm d}\) = 0.53 nM
For various true values of \(K_{\rm i}\), the ranges of total unlabeled ligand concentration \(\log_{10} [I]_{\rm T}\) used in the simulations were:
- \(K_{\rm i}\) = 0.05 nM: from −9.021 to −7.457
- \(K_{\rm i}\) = 0.1 nM: from −9.172 to −7.459
- \(K_{\rm i}\) = 0.5 nM: from −8.908 to −6.987
- \(K_{\rm i}\) = 1 nM: from −8.454 to −6.496
- \(K_{\rm i}\) = 10 nM: from −7.495 to −5.500
- \(K_{\rm i}\) = 100 nM: from −6.500 to −4.500
(Five data points spaced approximately 0.5 log units apart, centered around the expected 50% inhibition point.)
There was no significant difference between the results of nonlinear regression (fitting to a standard sigmoid function) and linear regression (after logit transformation) in estimating \({\rm IC}_{50}\):
| True Ki | [I]50 (nM) | True IC50 (nM) | IC50 ( from nonlinear regression) | IC50 (from linear regression) |
| 0.05 nM | 1.51 | 3.24 | 3.18 | 3.17 |
| 0.1 nM | 3.02 | 4.75 | 4.71 | 4.68 |
| 0.5 nM | 15.1 | 16.8 | 16.8 | 16.8 |
| 1 nM | 30.2 | 31.9 | 31.9 | 31.9 |
| 10 nM | 302 | 304 | 303 | 303 |
| 100 nM | 3019 | 3020 | 3015 | 3014 |
| True Ki | Ki (nM) calculated from True IC50 | |||
| Cheng–Prusoff | Cheng–Prusoffa | Munson–Rodbard (exact) correction | IC50 – [R]T/2 correctionb | |
| 0.05 nM | 0.0980 | 0.108 | 0.500 | 0.502 |
| 0.1 nM | 0.143 | 0.159 | 0.100 | 0.100 |
| 0.5 nM | 0.508 | 0.561 | 0.499 | 0.499 |
| 1 nM | 0.964 | 1.07 | 1.00 | 1.00 |
| 10 nM | 9.19 | 10.2 | 10.0 | 10.0 |
| 100 nM | 91.3 | 101 | 100 | 100 |
| a Using [L]50 instead of [L]T 。 b \(K_{\mathrm i} = (\mathrm{IC}_{50} – [R]_\mathrm{T}/2 )/(2 [L]_{50}/[L]_{0} – 1 + [L]_{50}/K_\mathrm{d}) \). | ||||
Conclusions
The original Cheng–Prusoff equation shows unexpectedly large deviations even when the \(K_{\mathrm i}\) value is high.
Modified Cheng–Prusoff equation using \([L]_{50}\) agrees with the true \(K_{\rm i}\) value to two significant digits when \(K_{\rm i}\) ≥ 10 nM.
Nowadays, when spreadsheet software is available, there is no reason not to use the exact Munson–Rodbard correction. At least in cases where \(K_{\mathrm i} < 10\) nM, where the difference between \([I]_{50}\) and \(\mathrm{IC}_{50}\) widens, it is necessary to use the Munson–Rodbard equation.
- In practice, the Ki value obtained using the approximation \([I]_{50} \simeq \mathrm{IC}_{50} – [R]_{\mathrm T}/2 \) is sufficiently accurate.
Contents
Binding Assay Calculations
The binding assay methodology is based on Sharkey & Blumberg (1985), with modifications. Non-specific binding at equilibrium is neglected. Key equations include:
- Specific binding: $$[RL]_0 =[\mbox{pellet(total)}] – k \times [\mbox{supernatant(total)}]\times \frac{437}{50} $$
- Partition Coefficient k: $$\mathbf{\mathit{k}} = \frac{[\mbox{pellet(nonspecific)}]}{[\mbox{supernatant(nonspecific)}] \times \frac{437}{50}}$$
- Total Ligand Concentration: $$\mathbf{[\mathit{L}]_{\mathrm T}} = [\mbox{pellet(total)}] + [\mbox{supernatant(total)}] \times \frac{437}{50}$$
- Corrected Total Ligand Concentration: $$\mathbf{[\mathit{L}]^{*}_{\mathrm T}} =\frac{[L]_{\mathrm T}}{1 + k}$$
- Free Ligand Concentration: $$\mathbf{[\mathit{L}]_0} = [\mbox{supernatant(total)}] = \frac{[L]_{\mathrm T} − [RL]_0}{1 + k}$$
- Ligand Concentration at 50% Binding: $$\mathbf{[\mathit{L}]_{50}} = \frac{[L]_T − \frac{[RL]_0}{2}}{1 + k}$$
- Total Receptor Concentration: $$\mathbf{[\mathit{R}]_\mathrm{T}} = \frac{K_\mathrm{d} [RL]_0}{[L]_0 + [RL]_0}$$
- Conversion Factor N from DPM to nM:: $$N = \frac{1}{60} \times \frac{1}{3.7 \times 10^{10}} \times \frac{1}{\mbox{specific activity (Ci/mmol)}} \times \frac{1}{10^3} \times \frac{1}{250 \times 10^{-6}} \times 10^9$$
Non-Specific Binding Considerations
In membrane preparations, non-specific binding can occur due to low-affinity interactions with membrane proteins, phospholipid partitioning, or filter adsorption. These are typically unsaturable and modeled as the product of the non-specific binding coefficient \(k\) and the free radioligand concentration \([L]\) (Hulme and Trevethick, 2010).
In [3H]PDBu binding assays, polyethylene glycol (PEG) is added to precipitate proteins, with γ-globulin included as a co-precipitant, leading to non-specific adsorption (typically \(k = 3–5\%\)). Calculations show that at \(k = 0\), \([RL]_0 = 3.321\ \mbox{nM}\) and \([L]_0 = 13.7\ \mbox{nM}\); at \(k = 0.04\), \([RL]_0 = 3.316\ \mbox{nM}\) and \([L]_0 = 13.2\ \mbox{nM}\). High ligand concentrations minimize the impact on bound \([RL]\), allowing correction of free \([L]\) using \(k\).
For unlabeled ligands, if the partition coefficient is assumed equal to that of [3H]PDBu, adjusting the added concentration by dividing by \(1 + k\) provides a good estimate of \([I_\mathrm{T}]\). For example, with a true \(K_\mathrm{i} = 10\ \mbox{nM}\) and \(k = 0.04\), the apparent \(K_\mathrm{i}\) calculated using the Munson–Rodbard equation is \(10.47\ \mbox{nM}\) without correction and \(10.06\ \mbox{nM}\) with correction.
Cheng–Prusoff equation
Parameters Used
Total Binding Group:
$$[R]_0 + [L]_0 \rightleftharpoons [RL]_]$$ $$K_{\rm d} = \frac{[R]_0[L]_0}{[RL]_0}$$ $$[R]_{\rm T} = [R]_0 + [RL]_0$$ $$K_{\rm d} = \frac{([R]_{\rm T} – [RL]_0)[L]_0}{[RL]_0}$$ $$[RL]_0 = \frac{[R]_{\rm T}[L]_0}{K_{\rm d} + [L]_0}$$
Competitive Inhibition Group:
\([I]\): Concentration of free, non-labeled ligand
$$[RI] \rightleftharpoons [I] + [R] + [L] \rightleftharpoons [RL]$$ $$K_{\rm i} = \frac{[R][I]}{[RI]}$$ $$[R]_{\rm T} = [RI] + [R] + [RL]$$ $$[R]_{\rm T}= \frac{[R][I]}{K_{\rm i}} + [R] + [RL]$$ $$[R]_{\rm T}= \left(1+\frac{[I]}{K_{\rm i}}\right)[R] + [RL]$$ $$[R]_{\rm T} = \left(1+\frac{[I]}{K_{\rm i}}\right)\frac{K_{\rm d}[RL]}{[L]}+ [RL]$$ $$[RL] = \frac{[R]_{\rm T}[L]}{\left(1+\frac{[I]}{K_{\rm i}}\right)K_{\rm d} + [L]}$$
At 50% Inhibition of Binding
Let us consider the situation at 50% inhibition of binding: $$\frac{1}{2}[RL]_0 = [RL]_{50}$$ $$\frac{1}{2}\frac{[R]_{\rm T}[L]_0}{K_{\rm d} + [L]_0} = \frac{[R]_{\rm T}[L]_{50}}{\left(1+\frac{[I]_{50}}{K_{\rm i}}\right)K_{\rm d} + [L]_{50}}$$ Multiply both sides and rearrange: $$2\frac{K_{\rm d} }{[L]_0} + 2 = \left(1+\frac{[I]_{50}}{K_{\rm i}}\right)\frac{K_{\rm d} }{[L]_{50}}+ 1$$ $$2\frac{[L]_{50}}{[L]_0} + \frac{[L]_{50}}{K_{\rm d}} = 1+\frac{[I]_{50}}{K_{\rm i}}$$ Solving for \(K_{\rm i}\): $$K_{\rm i} = \frac{[I]_{50}}{2\frac{[L]_{50}}{[L]_0} – 1 + \frac{[L]_{50}}{K_{\rm d}}} \;\;\;\mbox{(Goldstein and Barrett, 1987)}$$ $$= \frac{[I]_{50}}{1 + \frac{[L]_{50}}{K_{\rm d}} + 2\frac{[L]_{50}-[L]_0}{[L]_0}}$$
Conversion to Cheng–Prusoff Equation
If the ligand concentration \([L]\) is sufficiently large compared to receptor concentration \([R]\), then \([L]_0 \simeq [L]_{50} \simeq [L]^{*}_{\rm T}\), so $$K_{\rm i} = \frac{[I]_{50}}{1 + \frac{[L]^{*}_{\rm T}}{K_{\rm d}}}\;\; .$$
If the inhibitor concentration \([I]\) is sufficiently large compared to the receptor concentration \([R]\), \([I]_{50}\) can be approximated as the total ligand concentration at 50% inhibition, i.e., \(\rm{IC}_{50}\):
$$K_{\rm i} = \frac{\textrm{IC}_{50}}{1+\frac{[L]^{*}_{\rm T}}{K_{\rm d}}} \;\;\;\mbox{(Cheng–Prusoff equation)}\;\; .$$
Estimating Ki Using Nonlinear Regression
The ideal equation above corresponds to the Hill equation in the case where there is no cooperativity in ligand binding (i.e., \(n=1\)).
Let \(\theta = [RL]/[RL]_0\), then:
$$\theta = \frac{[R]_{\rm T}[L]}{\left(1+\frac{[I]}{K_{\rm i}}\right)K_{\rm d} + [L]} \times \frac{K_{\rm d} + [L]_0}{[R]_{\rm T}[L]_0}$$ $$\theta = \frac{K_{\rm d} + [L]_0}{\frac{K_{\rm d}}{K_{\rm i}}[I] + K_{\rm d} + [L]} \times \frac{[L]}{[L]_0}\;\; .$$
Then, the horizontal axis in Figure 1 is changed to \(\log [I]_{\rm T}\), the logarithmic total concentration of unlabeled ligand.
data.csv
logIT,theta -9.495,0.999 -8.995,0.997 -8.495,0.990 -7.995,0.968 -7.496,0.906 -6.996,0.759 -6.498,0.512 -5.999,0.259 -5.500,0.102 -5.000,0.035 -4.500,0.011 -4.000,0.004
Nonlinear Regression Using R
1) \(\theta = f([I])\)
The value of \([L]\) in $$\theta = f([I]) = \frac{K_{\rm d} + [L]_0}{\frac{K_{\rm d}}{K_{\rm i}}[I] + K_{\rm d} + [L]} \times \frac{[L]}{[L]_0}$$ changes as \([I]\) changes. To address this, we substitute the variable \([L]\) with the constant \([L]_{50}\). Also, although \(\theta\) should ideally be a function of the free unlabeled ligand concentration \([I]\), since \([I]\) is unknown, we approximate it using the added total concentration \([I]_{\rm T}\).
> simulation <- read.csv("data.csv")
> simulation$IT <- 10^(simulation$logIT + 9)
> result1 = nls(theta~(0.53 + 13.679)/(0.53/Ki * IT + 0.53 + 15.339) * 15.339/13.679,data=simulation,start = c(Ki = 10))
> summary(result1)
Output:
Formula: theta ~ (0.53 + 13.679)/(0.53/Ki * IT + 0.53 + 15.339) * 15.339/13.679 Parameters: Estimate Std. Error t value Pr(>|t|) Ki 11.0184 0.1833 60.12 3.34e-15 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.006362 on 11 degrees of freedom Number of iterations to convergence: 4 Achieved convergence tolerance: 1.598e-06
Visualizing the Fit
> plot(theta~IT, data=simulation, log="x") > concpre = seq(-10, -3.5, length=100) > concpre <- 10^(concpre + 9) > lines(concpre, predict(result1, newdata=list(IT=concpre)))
Due to the fixed value of \([L]\), the fit of the curve is not very good. However, the estimated \(K_{\rm i} = 11.0184\ \mbox{nM}\) closely matches the theoretical value (11 nM), indicating that the estimation is quite accurate.
Estimating \({\rm IC}_{50}\)by Regression
Nonlinear Regression
1) Fitting to the Complementary Error Function (erfc)
$${\rm erfc}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{\infty}e^{-t^{2}}\,dt$$
> simulation <- read.csv("data.csv")
> erfc <- function(x) 2 * pnorm(x * sqrt(2), lower=FALSE)
> result2 = nls(theta~1/2*erfc((logIT - logIC50)/a), data = simulation, start = c(logIC50=-6.5, a=1))
> summary(result2)
Formula: $$\theta = \frac{1}{2}\cdot\mbox{erfc}\left(\frac{\log[I]_{\rm T} - \log{\rm IC}_{50}}{a}\right)$$
(Note: The complementary error function is multiplied by 1/2 to constrain the range between 0 and 1)
Output:
Formula: theta ~ 1/2 * erfc((logIT - logIC50)/a)
Parameters:
Estimate Std. Error t value Pr(>|t|)
logIC50 -6.476263 0.006851 -945.27 < 2e-16 ***
a 1.082295 0.013710 78.94 2.6e-15 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.005889 on 10 degrees of freedom
Number of iterations to convergence: 5
Achieved convergence tolerance: 6.534e-07
Visualizing Fit:
> concpre = seq(-10,-3.5, length=100) > plot(theta~logIT,data=simulation,xlim=c(-10,-3.5),ylim=c(0,1)) > lines(concpre, predict(result2, newdata=list(logIT=concpre)))
The fit is not very good.
2) Standard Sigmoid Function \(1/(1 + \exp(-x))\):
$$f(x) = \frac{1}{1 + e^{-x}}$$
> sigmoid = function(x) 1/(1+exp(-x)) > result3 = nls(theta~1-sigmoid((logIT - logIC50)/a), data = simulation, start = c(logIC50=-6.5, a=1)) > summary(result3)
Formula: $$\theta = 1 - \frac{1}{1 + e^{-\frac{\log[I]_{\rm T} - \log{\rm IC}_{50}}{a}}}$$
Output:
Formula: theta ~ 1 - sigmoid((logIT - logIC50)/a)
Parameters:
Estimate Std. Error t value Pr(>|t|)
logIC50 -6.476238 0.001243 -5209.7 <2e-16 ***
a 0.450384 0.001096 411.1 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.001071 on 10 degrees of freedom
Number of iterations to convergence: 5
Achieved convergence tolerance: 4.447e-07
The sigmoid function seems to fit perfectly. This is because, as explained in the linear regression section below, \(\theta\) is related to the logit function, and the standard sigmoid function is the inverse of the logit function.
In practice, the binding ratio \(\theta\) (i.e., [3H]PDBu binding) is usually kept within the range of approximately 0.2–0.8 (20–80%). Thus, using the complementary error function instead of the sigmoid function makes little difference to the results. However, from a modeling perspective, the standard sigmoid function is superior and should be used.
Linear Regression
Next, to visually evaluate the goodness of fit of the fitted curve, a transformation to a linear graph (e.g., for use in Microsoft Excel) is shown.
From the following expression: $$\frac{\theta}{1-\theta} = \frac{\frac{[L]}{[L]_0}(K_{\rm d} + [L]_0)}{\frac{K_{\rm d}}{K_{\rm i}}[I] + K_{\rm d} + [L] - \frac{[L]}{[L]_0}(K_{\rm d} + [L]_0)} = \frac{\frac{[L]}{[L]_0} K_{\rm d} + [L]}{\frac{K_{\rm d}}{K_{\rm i}}[I] + \left(1 - \frac{[L]}{[L]_0}\right) K_{\rm d}} $$
Assuming \([L]_0 \simeq [L]\), it simplifies to: $$\frac{\theta}{1-\theta} = \frac{K_{\rm d} + [L]}{\frac{K_{\rm d}}{K_{\rm i}}[I]}$$
Taking the logarithm (base 10): $$\log \left(\frac{\theta}{1-\theta}\right) = -\log [I] -\log K_{\rm d} + \log K_{\rm i} + \log (K_{\rm d} + [L])$$ This results in a linear relationship with a slope of \(-1\) when plotted against \(\log[I]\) (free ligand concentration) (logit function) (Figure 3).
Although the horizontal axis in the figure above is \(\log[I]\) (free unlabeled ligand), this value is not experimentally accessible. Instead, we use the added total ligand concentration \(\log[I]_{\rm T}\), and obtain:
- \(\log{\rm IC}_{50} = -6.4765\) from the linear plot
- \(\log{\rm IC}_{50} = -6.4762\) from nonlinear regression using the standard sigmoid function.
These values match up to four significant digits. In nM units:
- 333.8 nM (from linear regression),
- 334.0 nM (from nonlinear regression).
Simple Conversion from Binding Ratio to \(K_{\rm i}\)
For weak ligands that do not inhibit binding by more than 50%, \(K_{\rm i}\) can be calculated from a single data point using the following rearranged form:
$$\frac{\theta}{1-\theta} = \frac{K_{\rm d} + [L]}{\frac{K_{\rm d}}{K_{\rm i}}[I]} \rightarrow K_{\rm i} = \left(\frac{\theta}{1-\theta}\right) \frac{K_{\rm d}[I]}{K_{\rm d} + [L]}\;\; .$$
If the unlabeled ligand concentration \([I]\) is sufficiently large compared to the receptor concentration \([R]\) (which is the case for weak ligands), \([I]\) can be approximated by the total ligand concentration \([I]_{\rm T}\).
$$K_{\rm i} = \left(\frac{\theta}{1-\theta}\right) \frac{K_{\rm d}[I]_{\rm T}}{K_{\rm d} + [L]}$$
Correction for Tight-Binding Unlabeled Ligands
Correction of IC₅₀
When the concentration of the inhibitor \([I]\) is not sufficiently high relative to the receptor concentration \([R]\) — for instance, when the binding affinity of the unlabeled ligand is equal to or greater than that of the labeled ligand PDBu — it is not appropriate to approximate \([I]\) with the total ligand concentration \([I]_{\rm T}\).
Therefore,
$$ [I]_{50} = \mathrm{IC}_{50} - [RI]_{50} $$
and \([RI]_{50} \) must be calculated.
Experimentally, \([RL]_0\) and \([L]_0\) can be determined from the measured radioactivity. Using these values and a known \(K_{\mathrm d}\), the following quantities can be calculated:
$$[L]_{50} = [L]_{0} + [RL]_{0}/2\;\; ,$$
$$[R]_{\mathrm{T}} = [RL]_0 (1 + K_{\mathrm d}/[L]_0)\;\; ,$$
$$[RL]_{50} = [RL]_{0}/2\;\; ,$$
First, consider the following two conservation equations.
In the absence of inhibitor \(I\) : $$ [R]_{\mathrm{T}} = [R]_{0} + [RL]_{0} $$
When the concentration of \(I\) is \([I]_{50}\): $$ [R]_{\mathrm{T}} = [R]_{50} + [RL]_{50} + [RI]_{50} $$
Considering the equilibrium of the labeled ligand \(L\),
$$K_{\mathrm d} = \frac{[R]_0[L]_0}{[RL]_0} = \frac{[R]_{50}[L]_{50}}{[RL]_{50}}\;\; , $$
Substituting
$$ [R]_{0} = [R]_{\mathrm{T}} - [RL]_{0}$$
and
$$ [R]_{50} = [R]_{\mathrm{T}} - [RL]_{50} - [RI]_{50}$$
into this equation and rearranging yields
$$ [RI]_{50} = [R]_{\mathrm{T}} - \frac{[L]_0}{[L]_{50}}\frac{[RL]_{50}}{[RL]_0}R_{\mathrm{T}} + \left( \frac{[L]_0}{[L]_{50}} - 1\right) [RL]_{50} \;\; .$$
Since
$$\frac{[RL]_{50}}{[RL]_0} = 1/2\;\; ,$$
this simplified to
$$ [RI]_{50} = \left( 1 - \frac{1}{2} \frac{[L]_0}{[L]_{50}} \right)[R]_{\mathrm{T}} + \left( \frac{[L]_0}{[L]_{50}} - 1\right) [RL]_{50} \;\; .$$
Thus, the correction formula
$$ K_{\rm i} = \frac{\mathrm{IC}_{50} - \left( 1 - \frac{1}{2} \frac{[L]_0}{[L]_{50}} \right)[R]_{\mathrm{T}} - \left( \frac{[L]_0}{[L]_{50}} - 1\right) [RL]_{50} }{2\frac{[L_{50}]}{[L_0]} - 1 + \frac{[L_{50}]}{K_{\rm d}}} $$
is obtained.
The Ki value obtained from this correction is exact and is identical to the value obtained using the Munson–Rodbard correction described in the following section.
If the concentration of the labeled ligand is sufficiently in excess of the receptor concentration such that the approximation
$$ [L]_0 \approx [L]_{50}$$
can be used, the numerator on the right-hand side becomes simpler, and the equation can be written as
$$ K_{\rm i} = \frac{\mathrm{IC}_{50} - \frac{[R]_{\mathrm{T}}}{2}} {2\frac{[L_{50}]}{[L_0]} - 1 + \frac{[L_{50}]}{K_{\rm d}}} \;\; .$$
In practice, the Ki values obtained using this simpler correction are sufficiently accurate.
Munson–Rodbard Equation
The Munson–Rodbard correction also allows you to calculate \(K_{\rm i}\) using the actual \({\rm IC}_{50}\) (total concentration), without relying on the approximation \([I]_{50} \simeq \textrm{IC}_{50}\) (Munson and Rodbard, 1988; Huang, 2003).
If we define \(y_0 = [RL]_0/[L]_0\), then: $$K_{\rm i} = \frac{\textrm{IC}_{50}}{1+ \frac{[L]_{\rm T} (y_0 + 2)}{2 K_{\rm d} (y_0 + 1)} + y_0} - K_{\rm d} \left(\frac{y_0}{y_0 + 2}\right) \;\;\;\;\mbox{(Eq.26; Munson–Rodbard equation) }$$
When \(y_0\) is very small, this equation reduces to the Cheng–Prusoff equation.
Derivation of the Munson–Rodbard Equation
Let \(K_1 = 1/K_{\rm d}\), \(K_2 = 1/K_{\rm i}\). Then: $$[RL] = K_1 [R] [L]\;\;\;\;\mbox{(Eq.27)}$$ $$[RI] = K_2 [R] [I]\;\;\;\;\mbox{(Eq.28)}$$ From the conservation laws: $$[L]_{\rm T} = [RL] + [L]\;\;\;\;\mbox{(Eq.29)}$$ $$[I]_{\rm T} = [RI] + [I]\;\;\;\;\mbox{(Eq.30)}$$ $$[R]_{\rm T} = [RI] + [R] + [RL] = K_2 [R] [I] + [R] + K_1 [R] [L] = [R] (1+ K_1[L] + K_2 [I])\;\;\;\;\mbox{(Eq.31)}$$
Now consider the conditions without inhibitor and at 50% inhibition.
Let the ratio without inhibitor be defined as \(y_0 = [RL_0]/[L_0]\). From Eq.29: $$[L]_{\rm T} = [L]_0 (1 + y_0)\;\;\;\;\mbox{(Eq.32)}$$ $$[L]_0 =\frac{[L]_{\rm T}}{1 + y_0}\;\;\;\;\mbox{(Eq.33)}$$ $$[RL]_0 = \frac{ y_0 [L]_{\rm T}}{1 + y_0}\;\;\;\;\mbox{(Eq.34)}$$
At 50% inhibition: $$[RL]_{50} = \frac{y_0 [L]_{\rm T}}{2 (1 + y_0)}\;\;\;\;\mbox{(Eq.35)}$$ From Eq.29: $$[L]_{50} = [L]_{\rm T} - [RL_{50}] = \frac{ [L]_{\rm T} (2 + y_0) }{2 (1 + y_0)}\;\;\;\;\mbox{(Eq.36)}$$
Using Eq.27, the free receptor at 50% inhibition is: $$[R]_{50}= \frac{[RL]_{50}}{K_1 [L]_{50}} = \frac{y_0}{K_1 (y_0 + 2)}\;\;\;\;\mbox{(Eq.37)}$$
From Eq.28 and Eq.30: $$\textrm{IC}_{50} = [I]_{{\rm T} 50} = [RI]_{50} + [I]_{50} = [I]_{50}(1+K_2 [R]_{50})\;\;\;\;\mbox{(Eq.38)}$$ Substituting Eq.37 into Eq.38: $$[I]_{50} = \frac{\textrm{IC}_{50}}{1+\left(\frac{K_2}{K_1}\right)\left(\frac{y_0}{y_0+2}\right)}\;\;\;\;\mbox{(Eq.39)}$$ Solving Eq.31 for \([R]\) and substituting into Eq.27: $$[RL] = \frac{K_1 [R]_{\rm T} [L] }{1+ K_1 [L] + K_2 [I]}\;\;\;\;\mbox{(Eq.40)}$$
Without inhibitor, using Eq.34 and Eq.33: $$[R]_{\rm T} = y_0 \left(\frac{1}{K_1} + \frac{[L]_{\rm T}}{1 + y_0}\right)\;\;\;\;\mbox{(Eq.41)}$$
At 50% inhibition: $$[RL]_{50} = \frac{K_1 [R]_{\rm T} [L]_{50} }{1+ K_1 [L]_{50} + K_2 [I]_{50}}\;\;\;\;\mbox{(Eq.42)}$$
Solving Eq.42 for \(1/K_2\): $$\frac{1}{K_2} = \frac{[I]_{50}}{\frac{K_1[R]_{\rm T}[L]_{50}}{[RL]_{50}} - K_1[L]_{50} - 1}$$
Substitute Eq.39 into this and simplify: $$\frac{1}{K_2} + \frac{1}{K_1}\left(\frac{y_0}{y_0+2}\right) = \frac{\mathrm{IC}_{50}}{\frac{K_1[R]_{\rm T}[L]_{50}}{[RL]_{50}} - K_1[L]_{50} - 1}$$
Now, substitute:
- Eq.36 into \([L]_{50}\)
- Eq.41 into \([R]_{\rm T}\)
- Use \([L]_{50}[RL]_{50} = \frac{(y_0 + 2)}{2y_0}\)
Then the denominator on the right becomes: $$K_1 \times y_0\left(\frac{1}{K_1} + \frac{[L]_\mathrm{T}}{y_0+1}\right) \times \frac{y_0+2}{y_0} - K_1\frac{[L]_\mathrm{T}(y_0+2)}{2(y_0+1)} - 1$$ $$= 2+ y_0 + K_1\frac{[L]_\mathrm{T}(y_0+2)}{y_0+1} - K_1\frac{[L]_\mathrm{T}(y_0+2)}{2(y_0+1)} - 1 $$ $$= 1+ K_1\frac{[L]_\mathrm{T}(y_0+2)}{2(y_0+1)} + y_0 $$
Therefore: $$\frac{1}{K_2} = \frac{\textrm{IC}_{50}}{1+ \frac{[L]_{\rm T} (y_0 + 2)}{2 \left(\frac{1}{K_1}\right) (y_0 + 1)} + y_0} - \left(\frac{1}{K_1}\right) \left(\frac{y_0}{y_0 + 2}\right)\;\;\;\;\mbox{(Eq.43)}$$
Finally, replacing \(1/K_2 = K_{\rm i}\), \(1/K_1 = K_{\rm d}\), we recover Eq.26.
Dandliker's Equation
You can calculate \(K_{\rm i}\) from each data point in an inhibition experiment (Dandliker, 1981).
Define: $$\chi = \frac{[RL]}{[L]}$$ (Note: \([L]\) is computed as \([L] = [L]_{\rm T} - [RL]\))
Then: $$K_i = \frac{[I]_{\rm T} K_d \chi}{[R]_{\rm T} - \frac{[L]_{\rm T}\chi}{1 + \chi} - K_d \chi} - K_d \chi\;\;\;\;\mbox{(Eq.44)}$$
Also, define \(f_b = [RL]/[L]_{\rm T}\)/ Since, \(\chi = f_b/(1-f_b)\), then: $$K_{\rm i} = \frac{[I]_{\rm T}K_{\rm d} f_b}{[R]_{\rm T}(1 - f_b) - K_{\rm d} f_b - [L]_{\rm T} f_b (1 - f_b)} - \frac{K_{\rm d} f_b}{ 1 - f_b}\;\;\;\;\mbox{(Eq.45)}$$
Eq.45 is a rearranged form of Equation (11) in Huang, 2003.
Derivation of Dandliker’s Equation
As before, define \(K_1 = 1/K_{\rm d}\), \(K_2 = 1/K_{\rm i}\). $$\chi = \frac{[RL]}{[L]} = K_1 ([R]_{\rm T} - [RL] - [RI])\;\;\;\;\mbox{(Eq.46)}$$ $$\frac{[RI]}{[I]} = K_2 ([R]_{\rm T} - [RL] - [RI])\;\;\;\;\mbox{(Eq.47)}$$ $$[L]_{\rm T} = [RL] + [L]\;\;\;\;\mbox{(Eq.48)}$$ $$[I]_{\rm T} = [RI] + [I]\;\;\;\;\mbox{(Eq.49)}$$ $$[RL] = \frac{[L]_{\rm T}\chi}{1 + \chi}\;\;\;\;\mbox{(Eq.50)}$$ $$[L] = \frac{[L]_{\rm T}}{1 + \chi}\;\;\;\;\mbox{(Eq.51)}$$ From Eq.46, express \([RL]\) and \([L]\) in terms of \(\chi\): $$[RI] = [R]_{\rm T} - [RL] - \frac{\chi}{K_1} = [R]_{\rm T} - \frac{[L]_{\rm T}\chi}{1 + \chi} - \frac{\chi}{K_1}\;\;\;\;\mbox{(Eq.52)}$$
Solving Eq.47 for \([I]\): $$[I] = \frac{[RI]}{K_2} \times \frac{1}{[R]_{\rm T} - [RL] - [RI]}\;\;\;\;\mbox{(Eq.53)}$$ Since \([RI] = [R]_{\rm T} - [RL] - \frac{\chi}{K_1}\), then: $$[I] = \frac{[R]_{\rm T} - [RL] - \frac{\chi}{K_1}}{K_2} \times \frac{1}{\frac{\chi}{K_1}} = \frac{K_1}{K_2\chi}\left([R]_{\rm T} - \frac{[L]_{\rm T}\chi}{1 + \chi} - \frac{\chi}{K_1}\right)\;\;\;\;\mbox{(Eq.54)}$$ Substitute Eqs. 52 and 54 into Eq.49: $$[I]_{\rm T} = \left(1 + \frac{K_1}{K_2\chi}\right) \left([R]_{\rm T} - \frac{[L]_{\rm T}\chi}{1 + \chi} - \frac{\chi}{K_1}\right)\;\;\;\;\mbox{(Eq.55)}$$
Solve Eq.55 for \(1/K_2\), giving: $$K_i = \frac{1}{K_2} = \frac{[I]_{\rm T} K_d \chi}{[R]_{\rm T} - \frac{[L]_{\rm T}\chi}{1 + \chi} - K_d \chi} - K_d \chi\;\;\;\;\mbox{(Eq.56)}$$
References
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