Quantitative Genetics

Inbreeding is the probability that an individual inherits two identical alleles from a common ancestor (Falconer and Mackay, 1996). There are several ways to obtain inbreeding coefficients using pedigree or genomic information. One method is via the diagonal elements of the genetic relationship matrix. Let A be the numerator (pedigree) relationship matrix and $a_{ii}$ its ith diagonal element. The inbreeding coefficient of the ith animal is: $$ F_i = a_{ii} - 1. $$ ...

I’ve heard this question a few times, and I thought it would be helpful to address it in a short post. The short answer is: Yes! But why? Here’s why pedigree information still matters: Not all animals are genotyped, and we still need to perform genetic evaluations for non-genotyped animals. Pedigree-based genetic relationships between genotyped and non-genotyped animals, and among non-genotyped animals themselves are essential for this. Even if we genotype all current and future animals, we can’t change the fact that many animals from previous generations cannot be genotyped. Non-genotyped animals contribute to the genetic evaluation of genotyped animals through their phenotypes. They also provide valuable insights into population structure, such as breed composition and genetic lines. Pedigree information is crucial for genotype imputation, helping fill in missing genetic data with greater accuracy.

You might have heard about the QP transformation. This transformation is named after Quaas and Pollak (1981). These two scientists are among the pioneers of Quantitative Genetics and Animal Breeding. They have published numerous articles that are well ahead of their time. When I was in the University of Nebraska-Lincoln, I had the chance to meet Dr. John Pollak. However, I have not met Dr. Richard Louis Quaas. Later, I realised that he died in October 19, 2021, at the age of 77. The QP transformation enables genetic groups to be included in the pedigree, in contrast to previous approaches where genetic groups are included as a distinct fixed effect in the model. ...

In this short blog post, I’m gonna show you a super simple way to calculate genotype frequencies. Let’s say we’re looking at a single spot in the DNA (a biallelic situation) with two possible versions, “A” and “a”. That means we can have three possible genotype combos: “AA”, “Aa”, and “aa”. What we want to do is count how many times each of these genotypes shows up at each SNP. ...

Here, I introduce an R function for calculating minor allele frequencies (MAF). calcmaf <- function(M, col1ID = TRUE) { if(col1ID) { maf = colMeans(M[,-1])/2 } else { maf = colMeans(M)/2 } maf[maf > 0.5] <- 1 - maf[maf > 0.5] return(unname(maf)) } The calcMAF function takes arguments M and col1ID. M is the genotype data frame with genotypes coded as 0:2. col1ID takes TRUE or FALSE. If TRUE (default) the 1st column of M is animal ID. Let’s create an example genotype data frame for 10 genotypes and 20 SNPs, where the first column is animal ID. ...

Heritability ($\text h^2$) has biological and statistical definitions. Statistically, heritability is the proportion of the phenotypic variation explained by the genetic variation in the population. Let’s start with an example! I took example 3.1 from Mrode (2005). This example is about estimating fixed effects (sex) solutions and predicting breeding values of animals for pre-weaning gain (WWG) of beef calves. The data is presented in the following table: Calf Sex Sire Dam WWG (kg) 4 Male 1 Unknown 4.5 5 Female 3 2 2.9 6 Female 1 2 3.9 7 Male 4 5 3.5 8 Male 3 6 5.0 In an Animal Model BLUP, the inverse of the relationship matrix ($\mathbf A^{-1}$) is multiplied by $\lambda = (1 - \text h^2)/\text h^2$ (equal to $(4 - \text h^2)/\text h^2$ in a Sire Model BLUP). Considering $\text h^2$ = 1/3 (λ = 2), the following solutions are obtained from BLUP: ...