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: ...