Pharmacodynamics: Dose–Response

Pharmacodynamics (PD) is what the drug does to the body — the quantitative relationship between drug concentration and the size of its effect. Where pharmacokinetics tells us the concentration over time, PD converts that concentration into an effect, and for antimicrobials the “effect” is the rate at which bacteria are killed.

Sigmoid <span class=EmaxE_{max} (Hill) dose–response curves for increasing Hill coefficients." />

The EmaxE_{max} model

Almost every dose–response relationship saturates: doubling the concentration does not double the effect once receptors approach full occupancy. The workhorse EmaxE_{max} model captures this with a rectangular hyperbola:

E=E0+Emax CEC50+CE = E_0 + \frac{E_{max}\, C}{EC_{50} + C}

Here E0E_0 is the baseline effect with no drug, EmaxE_{max} is the maximal drug-induced effect (efficacy), and EC50EC_{50} is the concentration producing half of EmaxE_{max} (potency). As C→∞C \to \infty the effect approaches E0+EmaxE_0 + E_{max}, and at C=EC50C = EC_{50} it sits exactly halfway.

The sigmoid EmaxE_{max} (Hill) model

Many systems respond more steeply than a simple hyperbola because of cooperativity or multi-step binding. Adding a Hill coefficient hh produces the sigmoid EmaxE_{max} model (writing the baseline as zero for clarity):

E=Emax ChEC50h+ChE = \frac{E_{max}\, C^{h}}{EC_{50}^{h} + C^{h}}

The exponent hh controls steepness: h=1h = 1 recovers the plain hyperbola, h>1h > 1 gives a switch-like S-curve, and h<1h < 1 a shallow one. Plotted against log⁡C\log C — the natural axis for dose–response, since concentrations span orders of magnitude — the curve is a symmetric sigmoid centered on log⁡EC50\log EC_{50}, a shape you may recognize from functions and graphs and from the logistic curve.

Solving for a target effect

Because the Hill equation is invertible, we can find the concentration producing any fractional effect. Setting E/Emax=pE/E_{max} = p and solving,

Cp=EC50(p1−p)1/hC_p = EC_{50}\left(\frac{p}{1-p}\right)^{1/h}

so the concentration for 90% of maximal effect is EC90=EC50 (9)1/hEC_{90} = EC_{50}\,(9)^{1/h}. When h=1h=1 this means EC90EC_{90} is nine times EC50EC_{50}; a steep curve with large hh narrows that gap dramatically.

Worked example

Suppose an antibiotic has Emax=4 h−1E_{max} = 4\ \mathrm{h^{-1}} (maximal kill rate), EC50=2 mg/LEC_{50} = 2\ \mathrm{mg/L}, and Hill coefficient h=2h = 2.

At a concentration C=4 mg/LC = 4\ \mathrm{mg/L} the effect is

E=4×4222+42=4×164+16=6420=3.2 h−1E = \frac{4 \times 4^{2}}{2^{2} + 4^{2}} = \frac{4 \times 16}{4 + 16} = \frac{64}{20} = 3.2\ \mathrm{h^{-1}}

i.e. 80% of the maximal kill rate. The concentration for 90% of maximum is

EC90=EC50 (9)1/h=2×91/2=2×3=6 mg/LEC_{90} = EC_{50}\,(9)^{1/h} = 2 \times 9^{1/2} = 2 \times 3 = 6\ \mathrm{mg/L}

Because h=2h=2, tripling the concentration from EC50EC_{50} takes the effect from 50% to 90% — steeper than the ninefold factor a hyperbola would require.

In code

We plot the Hill curve and then recover its parameters by nonlinear least squares from noisy data — the estimation problem at the heart of optimization.

R

Emax <- 4; EC50 <- 2; h <- 2
hill <- function(C, Emax, EC50, h) Emax * C^h / (EC50^h + C^h)

set.seed(1)
C <- c(0.25, 0.5, 1, 2, 4, 8, 16)
y <- hill(C, Emax, EC50, h) + rnorm(length(C), 0, 0.1)

fit <- nls(y ~ Emax * C^h / (EC50^h + C^h),
           start = list(Emax = 3, EC50 = 1, h = 1))
coef(fit)                       # ~ Emax 4, EC50 2, h 2
hill(4, Emax, EC50, h)          # 3.2 : effect at C = 4

Python

import numpy as np
from scipy.optimize import curve_fit

Emax, EC50, h = 4.0, 2.0, 2.0
hill = lambda C, Emax, EC50, h: Emax * C**h / (EC50**h + C**h)

rng = np.random.default_rng(1)
C = np.array([0.25, 0.5, 1, 2, 4, 8, 16])
y = hill(C, Emax, EC50, h) + rng.normal(0, 0.1, C.size)

p, _ = curve_fit(hill, C, y, p0=[3, 1, 1])
print(p)                        # ~ [4, 2, 2]
print(hill(4, *p))              # ~3.2 : effect at C = 4

[4.02057431 2.02736333 1.98503466]
3.1921646202469307

Julia

using LsqFit

Emax, EC50, h = 4.0, 2.0, 2.0
hill(C, p) = @. p[1] * C^p[3] / (p[2]^p[3] + C^p[3])

C = [0.25, 0.5, 1, 2, 4, 8, 16]
y = hill(C, [Emax, EC50, h]) .+ 0.1 .* randn(length(C))

fit = curve_fit(hill, C, y, [3.0, 1.0, 1.0])
fit.param                       # ~ [4, 2, 2]
hill([4.0], [Emax, EC50, h])[1] # 3.2 : effect at C = 4

Why it matters

The Hill curve is the bridge from a measured concentration to a predicted bacterial kill rate, and its two key numbers have direct clinical meaning: EmaxE_{max} says how good a drug can be, EC50EC_{50} says how much you need. Coupling this dose–response with the time-varying concentration from pharmacokinetics is precisely what defines the PK/PD indices used to design antibiotic regimens.