Question

Suppose that in Example 2.3.5, the observed individual has the recessive trait. Determine the posterior probability that the parents have the genotypes of the event \({{\bf{B}}_{\bf{4}}}\).

Short answer

The posterior probability that the parents have the genotypes of the event \({B_4}\)is \(\frac{1}{4}\)

Step-by-step solution

Given information

There are two phenotypes, one is a dominant trait, and another one is a recessive trait. Here said that the observed individual has the recessive trait.

State the events

Referring to example 2.3.5,

Eis the event that the observed individual has a dominant trait. Now, consider an event F;the observed individual has a recessive trait. So,\(F = 1 - E\).

Therefore, the probability of the observed individual has a recessive trait given that the individual is from\({B_1}\)genotype is-

\(\begin{aligned}{}\Pr \left( {F\left| {{B_1}} \right.} \right) = 1 - \Pr \left( {E\left| {{B_1}} \right.} \right)\\ = 1 - 1\\ = 0\end{aligned}\)

Similarly,

For i=2,\(\Pr \left( {F\left| {{B_2}} \right.} \right) = 0\)

For i=3,\(\Pr \left( {F\left| {{B_3}} \right.} \right) = 0\)

For i42,\(\Pr \left( {F\left| {{B_4}} \right.} \right) = \frac{1}{4}\)

For i=5,\(\Pr \left( {F\left| {{B_5}} \right.} \right) = \frac{1}{2}\)

For i=6,\(\Pr \left( {F\left| {{B_6}} \right.} \right) = 1\)

Calculate the total probability of a recessive trait

The probability of the event Foccurring is,

\(\begin{aligned}{}{\bf{Pr}}\left( {\bf{F}} \right){\bf{ = }}\sum\limits_{{\bf{i = 1}}}^{\bf{6}} {{\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{\bf{F}}\left| {{{\bf{B}}_{\bf{i}}}} \right.} \right)} \\ = \left( {\frac{1}{4} \times \frac{1}{4}} \right) + \left( {\frac{1}{4} \times \frac{1}{2}} \right) + \left( {\frac{1}{{16}} \times 1} \right)\\ = \frac{1}{{16}} + \frac{1}{8} + \frac{1}{{16}}\\ = \frac{1}{4}\end{aligned}\)

Calculate the required posterior probability

The posterior probability that the parents have the\({B_4}\)category genotype, given that the observed individual has a recessive trait is,

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{B}}_{\bf{4}}}\left| {\bf{F}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{B}}_{\bf{4}}}} \right){\bf{Pr}}\left( {{\bf{F}}\left| {{{\bf{B}}_{\bf{4}}}} \right.} \right)}}{{{\bf{Pr}}\left( {\bf{F}} \right)}}\\ = \frac{{\frac{1}{4} \times \frac{1}{4}}}{{\frac{1}{4}}}\\ = \frac{1}{4}\end{aligned}\)

Thus, the posterior probability is \(\frac{1}{4}\).