CRAN status


The main content of pedprobr is an implementation of the Elston-Stewart algorithm for pedigree likelihoods given marker genotypes. It is part of the pedsuite, a collection of packages for pedigree analysis in R.

The pedprobr package does much of the hard work in several other pedsuite packages:

The workhorse of pedprobr is the likelihood() function, supporting a variety of situations:


To get the current official version of pedprobr, install from CRAN as follows:


Alternatively, you can obtain the latest development version from GitHub:

# install.packages("devtools") # install devtools if needed

Getting started

#> Loading required package: pedtools

To set up a simple example, we first use pedtools utilities to create a pedigree where two brothers are genotyped with a single SNP marker. The marker has alleles a and b, with frequencies 0.2 and 0.8 respectively, and both brothers are heterozygous a/b.

# Pedigree with SNP marker
x = nuclearPed(nch = 2) |> 
  addMarker(geno = c(NA, NA, "a/b", "a/b"), afreq = c(a = 0.2, b = 0.8))

# Plot with genotypes
plot(x, marker = 1)

The pedigree likelihood, i.e., the probability of the genotypes given the pedigree, is obtained as follows:

likelihood(x, marker = 1)
#> [1] 0.1856

Genotype probability distributions

Besides likelihood(), other important functions in pedprobr are:

In both cases, the distributions are computed conditionally on any known genotypes at the markers in question.

To illustrate oneMarkerDistribution() we continue our example from above, and consider the following question: What is the joint genotype distribution of the parents, conditional on the genotypes of the children?

The answer is found as follows:

oneMarkerDistribution(x, ids = 1:2, partialmarker = 1, verbose = F)
#>            a/a        a/b       b/b
#> a/a 0.00000000 0.01724138 0.1379310
#> a/b 0.01724138 0.13793103 0.2758621
#> b/b 0.13793103 0.27586207 0.0000000

The output confirms the intuitive result that the parents cannot both be homozygous for the same allele. The most likely combination is that one parent is heterozygous a/b, while the other is homozygous b/b.