LBG - FS2024 – Solution 3
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 \(G_1G_2\)
Solution
Case 1: Homozygous and Heterozygous Parents
Parent \(s\) with genotype \(G_1G_1\) and parent \(d\) with genotype \(G_1G_2\). The average breeding value of parents \(s\) and \(d\) is
\[{1\over 2}\left(u_s + u_d\right) = {1\over 2}\left(2q\alpha + (q-p)\alpha \right) = {3 \over 2}q\alpha - {1\over 2}p\alpha \]
Offspring of parents \(s\) and \(d\) will either be homozygous \(G_1G_1\) or heterozygous both with probability of \(1/2\). For a homozygous offspring \(i\), the mendelian sampling term \(m_i\) can be computed as
\[m_i = u_i - {1\over 2}\left[u_s + u_d \right] = 2q\alpha - {1\over 2}\left[2q\alpha + (q-p)\alpha \right] = 2q\alpha - \left[{3 \over 2}q\alpha - {1\over 2}p\alpha \right] = {1\over 2}q\alpha + {1\over 2}p\alpha = {1\over 2}\alpha\]
For a heterozygous offspring \(j\), the mendelian sampling term \(m_j\) is computed as
\[m_j = u_j - {1\over 2}\left[u_s + u_d \right] = (q-p)\alpha - \left[{3 \over 2}q\alpha - {1\over 2}p\alpha \right] = -{1\over 2}q\alpha - {1\over 2}p\alpha = -{1\over 2}\alpha\]
Because homozygous and heterozygous offspring occur with equal probability, the average mendelian sampling term over a large number of offspring of parents \(s\) and \(d\) is
\[{1\over 2} * {1\over 2}\alpha + {1\over 2} * (-{1\over 2}\alpha) = 0\]
Case 2: Homozygous and Heterozygous Parents
Parent \(s\) with genotype \(G_2G_2\) and parent \(d\) with genotype \(G_1G_2\). The average breeding value of parents \(s\) and \(d\) is
\[{1\over 2}\left(u_s + u_d\right) = {1\over 2}\left(-2p\alpha + (q-p)\alpha \right) = {1\over 2}q\alpha - {3\over 2}p\alpha\]
Offspring of parents \(s\) and \(d\) are either homozygous \(G_2G_2\) or heterozygous both with probability \(1/2\). For a homozygous offspring \(i\), the mendelian sampling term \(m_i\) is computed as
\[m_i = u_i - {1\over 2}\left[u_s + u_d \right] = -2p\alpha - \left[{1\over 2}q\alpha - {3\over 2}p\alpha \right] = -{1\over 2}p\alpha - {1\over 2}q\alpha = -{1\over 2}\alpha\]
For a heterozygous offspring \(j\), the term \(m_j\) is computed as
\[m_j = u_j - {1\over 2}\left[u_s + u_d \right] = (q-p)\alpha - \left[{1\over 2}q\alpha - {3\over 2}p\alpha \right] = {1\over 2}q\alpha + {1\over 2}p\alpha = {1\over 2}\alpha\]
Because homozygous and heterozygous offspring occur both with equal probability, the average mendelian sampling term over a large number of offspring is
\[{1\over 2} * (-{1\over 2}\alpha) + {1\over 2} * {1\over 2}\alpha = 0\]
Case 3: Heterozygous Parents
Both parents \(s\) and \(d\) have genotype \(G_1G_2\). Both parents \(s\) and \(d\) are heterozygous and have therefore the same genotype and the same breeding value. Hence the average breeding value of the parents is also the same which is
\[{1\over 2}\left[u_s + u_d \right] = (q-p)\alpha\]
Offspring of parents \(s\) and \(d\) will be homozygous \(G_1G_1\) or \(G_2G_2\) both with probability \(1/4\) or heterozygous with probability \(1/2\). For offspring \(i\) with genotype \(G_1G_1\), the mendelian sampling term \(m_i\) is
\[m_i = u_i - {1\over 2}\left[u_s + u_d \right] = 2q\alpha - \left[(q-p)\alpha \right] = \alpha\]
For offspring \(j\) with genotype \(G_1G_2\), the mendelian sampling term \(m_j\) is
\[m_j = 0\]
For offspring \(k\) withe genotype \(G_2G_2\), the mendelian sampling term \(m_k\) is
\[m_k = u_k - {1\over 2}\left[u_s + u_d \right] = -2p\alpha - \left[(q-p)\alpha \right] = -\alpha\]
In the computation of \(m_i\), \(m_j\) and \(m_k\), the definition of the allele substitution effect was used. The average mendelian sampling term over a large number of offsprings of parents \(s\) and \(d\) corresponds to
\[{1\over 4} * \alpha + {1\over 2} * 0 + {1\over 4} * (-\alpha) = 0\]