It's seems to me that the definition of relative and absolute fitness is not that clear.

Absolute fitness[edit]

The absolute fitness (W) of a genotype is defined as the proportional change in the abundance of that genotype over one generation attributable to selection. For example, if ${\displaystyle n(t)}$ is the abundance of a genotype in generation t in an infinitely large population (so that there is no genetic drift), and neglecting the change in genotype abundances due to mutations, then[4] ${\displaystyle n(t+1)=Wn(t)}$. An absolute fitness larger than 1 indicates growth in that genotype's abundance; an absolute fitness smaller than 1 indicates decline.

Relative fitness[edit]

Whereas absolute fitness determines changes in genotype abundance, relative fitness (w) determines changes in genotype frequency. If N(t) is the total population size in generation t, and the relevant genotype's frequency is ${\displaystyle p(t)=n(t)/N(t)}$, then ${\displaystyle p(t+1)={\frac {w}{\overline {w}}}p(t)}$, where ${\displaystyle {\overline {w}}}$ is the mean relative fitness in the population (again setting aside changes in frequency due to drift and mutation). Relative fitnesses only indicate the change in prevalence of different genotypes relative to each other, and so only their values relative to each other are important; relative fitnesses can be any nonnegative number, including 0. It is often convenient to choose one genotype as a reference and set its relative fitness to 1. Relative fitness is used in the standard Wright-Fisher and Moran models of population genetics.

Absolute fitnesses can be used to calculate relative fitness, since ${\displaystyle p(t+1)=n(t+1)/N(t+1)=(W/{\overline {W}})p(t)}$ (we have used the fact that ${\displaystyle N(t+1)={\overline {W}}N(t)}$, where ${\displaystyle {\overline {W}}}$ is the mean absolute fitness in the population). This implies that ${\displaystyle w/{\overline {w}}=W/{\overline {W}}}$, or in other words, relative fitness is proportional to ${\displaystyle W/{\overline {W}}}$. It is not possible to calculate absolute fitnesses from relative fitnesses alone, since relative fitnesses contain no information about changes in overall population abundance

  1. That's good when we are talking about genotypes or phenotypes. But when we are using models, e.g. fitness landscapes, how can we translate that? For example, if I have a linear fitness landscape (made with a linear function), then the "fitness" value that I find with the model are "expected fitness" or $\widehat y$. Thus in this case, it's the "fitness of all possible phenotypes, based on the population fitness and phenotypic information". Are we calling this relative or absolute?
  2. When fitness, or one of its components, is calculate from a model (say a mark-recapture model), which type of fitness is computed? In Marc Kéry and Michael Schaub (2012) Bayesian Population Analysis using WinBUGS: A Hierarchical Perspective, they are showing models to compute survival probabilities based on a recapture history. Would that be relative or not?
  3. What are the distinction of the 2 definitions, and are they too restrictive when we want to extend them to models that calculates the fitness of a all the individuals in a population?

See also, Endler, J.A. (1986) Natural Selection in the Wild. On page 168 of the book:

  • Mean absolute fitness: $\overline {W} = [∑ƒ(X)W(X) ]/[∑ƒ(X)]$
  • ƒ(X) frequency of genotype or phenotype X
  • Absolute fitness: $W(X)$
  • Relative fitness: $w(X)= W(X) /\overline {W}$
  • Therefor, $\overline {w}(X)=1$

Relative fitness can also be measured with reference to a particular phenotype (or genotype), in which cas $\overline {w}$ is not necessarily 1; this is the most common method used for polymorphic traits. If the population is sampled twice (or more) within a generation so that individuals in the second sample represent a subset of those sampled in the first sample (as in a capture-recapture or cohort study), then absolute fitnesses can be calculated. Examples are the probability of surviving between samples, or the probability of mating. On the other hand, if samples are made without replacement, or if samples are made of juveniles and adults at a single time, then only relative fitness can be calculaqted; information on total numbers and mean fitness is lost (see discussions in O'Donald 1971, Horns and Harrison 1970, and Manley 1974).)

On page 42 of the same book:

Adaptedness and Adaptation. Adaptedness is the degree to which an organism is able to live and reproduce in a given set of environments: the state of being adapted (Dobzhansky 1968a,b). Adaptation is the process of becoming adapted or more adapted (ibid.). Unfortunately, adaptation.is also used in the sense of an adaptive trait (Lewontin 1978), confounding the end product with the process (see also Dunbar 1982). An adaptive trait is "an aspect of the developmental pattern of the organism surviving and reproducing" (Dobzhansky 1956, 1968a). There are problems in defining precisely what adaptednedd is so that it can be measured (Dobzhansky 1956, 1968a,b; Stern 1970; Lewontin 1978; Dunbar 1982). One solution is to define it in the sense of absolute (rather than relative) fitness (Table 2.1). In this case it can be measured by the average absolute lifetime contribution to the breeding population by a phenotype or a class of phenotypes. It thus becomes intimately related to the actual (R) or intrinsic $(r_{m})$ rate of increase, or "Malthusian parameter," and these have actually been used as measures of fitness for populations and species, though there are some problems (Fisher 1930; Dobzhansky 1968a,b; Dunbar 1982). Adaptedness has also been defined as the mean absolute fitness (Sober 1984). [...]

Bonus: I'm trying to recreate the equation from Lande and Arnold 1983. But it's not working. I don't know what is not correct (the line where you see (mean.rel.fit = mean(w.relative.fit)), this result should give 1, but it's not in my case. I don't know why):

# W: Absolute fitness 
# w: relative fitness 

nb.data = 5000
y = rnorm(nb.data,10,2)
x = 1:nb.data

# This should look normal
     breaks = 30)

# To have a mean of 0 for phenotypes scale the phenotypes 
z.raw = y
z = scale(z.raw, 
          center = TRUE,
          scale = TRUE)

     probability = TRUE, # In stead of frequency
     breaks = "FD",      # For more breaks than the default
     col = "grey", border = "black")
lines(density(z),   # Add the kernel density estimate (-.5 fix for the bins)
      col = "firebrick2", lwd = 3)

# Normally distributed trait, Phenotypic distribution (normal)
p.z =function(z) {
  sigma = sd(z)
  1/sqrt(2*pi) * exp((-1/2)*((z-z.mean)/sigma)^2)

# This is not working! It's suppose to be normal 
     breaks = 30)

# Linearly distributed absolute fitness function 
W.z =function(z) {
  slope = 1 
  intercept = 0 

# There is a linear relationship between trait z and Absolute fitness W 

de = density(z,
             n = 600000)
# If n = 600000 in the density function, then it should ~=1 

df.z = as.data.frame(table(round(z,1)))
values = as.numeric(as.character(df.z$Var1))
frq = as.numeric(as.character(df.z$Freq))
frqcy = function(len){frq[1:len]}

f = function(values)(W.z(values)*frqcy(length(values)))
(mean.absolute.fitness = integrate(f,
                                   subdivisions = 2000,
                                   rel.tol = .Machine$double.eps^.05,
                                   lower = min(z), 
                                   upper = max(z))

# Relative fitness 
w.z <- function(z) {
(w.relative.fit = w.z(z))

# This should be 1 
(mean.rel.fit = mean(w.relative.fit))

f2 = function(z)(z*w.z(z)*p.z(z))

(mean.absolute.fitness = integrate(f2,
                                   lower = min(z), 
                                   upper = max(z))
  • $\begingroup$ Before answering, I suggest reading this SE post. $\endgroup$
    – user22020
    Aug 16, 2017 at 14:20
  • $\begingroup$ I'm not totally convinced in this post. The definition on Wikipedia changed. It is no longer "the ratio between the number of individuals with that genotype after selection to those before selection". Also, it seems there is a confusion with the fitness components and the measure of fitness when looking at relative vs absolute. In my mind, the absolute fitness can be larger than 1 for sure. $\endgroup$ Aug 16, 2017 at 14:41
  • $\begingroup$ Relative fitness is "relative" so that it's supposed to be between 0 and 1. That's a ratio. $\endgroup$ Aug 16, 2017 at 14:45
  • $\begingroup$ The reason why relative fitness is between 0 and 1 is because it's in relation to the highest performing individual (genotype). You take the frequency of the best performer and divide all others by that value, where the highest performer has a relative fitness of 1 (because it's being divided by itself), and all others are less than or equal to 1. This process can also be considered as "normalizing" the dataset. $\endgroup$
    – user22020
    Aug 16, 2017 at 14:47
  • $\begingroup$ Exactly, so the relative fitness is based on the absolute fitness in which the most "fit" is used to bound the relative fitness between 0 and 1. It's not necessarily based on the most frequent (the most frequent genotype or phenotype is not necessarily the most fit...) $\endgroup$ Aug 16, 2017 at 14:53


You must log in to answer this question.

Browse other questions tagged .