Livestock Breeding and Genomics - Notebook 6

Problem 1: Parent Offspring Breeding Values

As shown in the course notes, the breeding value \(u_i\) of animal \(i\) can be decomposed into the average of the parent breeding values plus a mendelian sampling term (\(m_i\)). This means

\[u_i = {1\over 2}u_s + {1\over 2}u_d + m_i\]

where animal \(i\) has parents \(s\) and \(d\). The mendelian sampling term \(m_i\) is the deviation of the single breeding value \(u_i\) from the parent average breeding value. Because \(m_i\) is modelled as a deviation, it follows that for a large number (\(N\)) of offspring from parents \(s\) and \(d\), the average over all mendelian sampling terms must be \(0\).

Your Task

Show that the average mendelian sampling term over a large number of offspring is \(0\) using a single locus model for the following cases.

Case 1: Homozygous and Heterozygous Parents

Parent \(s\) with genotype \(G_1G_1\) and parent \(d\) with genotype \(G_1G_2\)

Case 2: Homozygous and Heterozygous Parents

Parent \(s\) with genotype \(G_2G_2\) and parent \(d\) with genotype \(G_1G_2\)

Case 3: Heterozygous Parents

Both parents \(s\) and \(d\) have genotype \(G_1G_2\)

Your Solution

• For each of the following cases compute the parent average of breeding values.
• Compute the difference between the breeding values of every possible offspring and the parent average
• Compute the average over all mendelian sampling terms

Case 1: Homozygous and Heterozygous Parents

Case 2: Homozygous and Heterozygous Parents

Case 3: Heterozygous Parents

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Latest Changes: 2023-11-10 07:36:54 (pvr)