14.6 The Immune System

195

Consider a situation in which upper RR antibodies are exposed to upper XX antigens. The mean

exposure per receptor (i.e., antibody) is

lamda equals upper X divided by upper R periodλ = X/R.

(14.5)

The Poisson distribution well describes the spatial distribution of objects scattered

at random over a grid. ⁴² Hence, the probability p Subscript kpk that a receptor receives exactly kk

antigens is

p Subscript k Baseline equals e Superscript negative lamda Baseline lamda Superscript k Baseline divided by k factorial periodpk = e^−λλ^k/k!.

(14.6)

Let us define a threshold upper TT —this is the minimum number of antigens that must be

received and bound in order to trigger a response; we assume it equals the valence of

the antibody (i.e., 2, 2, 4 and 10 for IgG, IgE, IgA, and IgM antibodies, respectively).

The output yy per receptor is then

y left parenthesis upper T right parenthesis equals sigma summation Underscript k equals upper T Overscript normal infinity Endscripts p Subscript ky(T ) =

∞

E

k=T

pk

(14.7)

and the measured output upper YY equals y upper RyR.

Example. Let there be upper R equals 10 000R = 10 000 receptors and let the exposure be upper X equals 10 000X = 10 000

antigens, hencelamda equals 1.0λ = 1.0. Consider first receptors withupper T equals 1T = 1. The sum in Eq. (14.7)

is easy to calculate since it equals 1 minus p Subscript k equals 01 −pk=0. Hence the output upper Y equals 6321Y = 6321. This is

like carrying out upper R equals 10 Superscript 4R = 10⁴ Bernoulli trials with a success probability German p equals 0.6321p = 0.6321.

The outcomes of the trials are binomially distributed, hence the standard deviation

sigma Subscript upper Y Baseline equals StartRoot upper R German p left parenthesis 1 minus German p right parenthesis EndRoot equals 48σY = ^√Rp(1 −p) = 48.

We assume that the minimum detectable increment Delta upper Y/\Y of output equals sigma Subscript upper YσY.

Hence, the minimum detectable increment Delta upper X/\X of input is one that yields an output

of 6,321 + 48 = 6369 = upper Y primeY ^' (the argument is unchanged in essence were we to use a

multiple of sigmaσ). Going back to Eq. (14.7), we need to solve

upper Y prime equals upper R left parenthesis 1 minus e Superscript minus lamda prime Baseline right parenthesisY ^' = R(1 −e^−λ')

(14.8)

for lamda primeλ^', which equals left parenthesis upper X plus Delta upper X right parenthesis divided by upper R(X + /\X)/R, and which in this example equals 1.013; i.e.,

Delta upper X equals 130/\X = 130 objects.

On the other hand, if the antibody requires 2 or more antigens to trigger a response,

then the output will be, using Eq. (14.7) with upper T equals 2T = 2,

y equals 1 minus e Superscript negative lamda Baseline left parenthesis 1 plus lamda right parenthesisy = 1 −e^−λ(1 + λ)

(14.9)

42 It might be argued that the arrival of individual antigens at a surface via diffusion in solution is not

random, because successive arrival attempts are spatially and temporally correlated (cf. the random

sequential addition model). However, hydrodynamic randomizing exactly (but fortuitously) cancels

out this effect (Bafaluy et al. 1993).