Q. 3.69 - A First Course in Probability

Question

3.69. A certain organism possesses a pair of each of $5$ different genes (which we will designate by the first $5$ letters of the English alphabet). Each gene appears in $2$ forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene, in the sense that if an organism possesses the gene pair $xX$, then it will outwardly have the appearance of the $X$ gene. For instance, if stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas one having gene pair xx will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus, 2 organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype.) In a mating between 2 organisms, each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of the organism’s mate. In a mating between organisms having genotypes , bB, cC, dD, eE and aa, bB, cc, Dd, ee what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble

(a) the first parent? 

(b) the second parent? 

(c) either parent? 

(d) neither parent? 

Step-by-Step Solution

Answer

The probability that the progeny will $\left(i\right)$ phenotypically 

a)  $\frac{9}{128}$

b)$\frac{9}{128}$

c) $\frac{9}{64}$ 

d) $\frac{55}{64}$

The probability that the progeny will $\left(ii\right)$ genotypically

a) $\frac{1}{32}$

b) $\frac{1}{32}$

c)  $\frac{1}{16}$

d) $\frac{15}{16}$

Step 1: Given Data

First Parent : $aA bB cC dD eE$

Second Parent : $aa bB cc dD ee$

Events:

$A_i\text{ - }$the progeny received gene $A$ from  the $i\text{-th }$parent, $i=1,2$

$B_i\text{ - }$the progeny received gene $B$ from the $i\text{-th}$ parent, $i=1,2$

The mutual independent events are, $A_1,A_2,B_1,B_2,\dots ,E_2$

Evaluate:

$\left(i\right)$ a) $P("Phenotype as the first parent")$

     b) $P("Phenotype as the second parent")$

     c) $P("Phenotype as either parent")$

     d) $P("Phenotype as neither parent")$

$\left(ii\right)$ a) $P("Genotype as the first parent")$

      b) $P("Genotype as the second parent")$

      c) $P("Genotype as either parent")$

      d) $P("Genotype as neither parent")$

Step 2: Mutually Exclusive

failed to  have same genotype or phenotype they're  mutually exclusive,

$$\begin{gathered} P("Phenotype as either parent") = \\ P("Phenotype as the first parent") + P("Phenotype as the Second parent") \end{gathered}$$

The probability value is,

$P("Phenotype as neither parent") =1- P("Phenotype as either parent")$

To each  pair of alleles, estimate the likelihood from each gender within the offspring:

$P("AA")=P\left(A_1{A}_2\right)$

              $\begin{array}{r}=P\left(A_1\right)P\left(A_2\right)\end{array}$

               $=\frac{1}{2}\times 0$

                $=0$

$\begin{array}{r}P\left("a{a}^{\prime \prime }\right)=P\left(A_1^c{A}_2^c\right)\\ =P\left(A_1^c\right)P\left(A_2^c\right)\end{array}$

           $=\frac{1}{2}\times 1$

           $=\frac{1}{2}$

$P("BB")=\frac{1}{4}P("Bb")=\frac{1}{2}P("bb")=\frac{1}{4}$

$$\begin{gathered} P("CC")=0P("Cc")=\frac{1}{2}P("cc")=\frac{1}{2} \\ P("DD")=\frac{1}{4}P("Dd")=\frac{1}{2}P("dd")=\frac{1}{4} \\ P("EE")=0P("Ee")=\frac{{\displaystyle 1}}{{\displaystyle 2}}P("ee")=\frac{{\displaystyle 1}}{{\displaystyle 2}} \end{gathered}$$

Step 3: Phenotype (part a and b)

Calculation of 

$\left(i\right) \left(a\right)$ $P("Phenotye as the first parent")$

$=P\left[\left("Aa"\cup "A{A}^{\prime \prime }\right)("Bb"\cup "BB")("Cc"\cup "CC")("Dd"\cup "DD")\left("E{e}^{"\mathrm{\prime }}\cup "EE"\right)\right]$

$=P("Aa"\cup "AA")P("Bb"\cup "BB")P("Cc"\cup "CC")P("Dd"\cup "DD")P("Ee"\cup "EE")$

$\begin{array}{r}=\left[P\left("A{a}^{\prime \prime }\right)+P\left("A{A}^{\prime \prime }\right)\right]\left[P\right("Bb")+P("BB"\left)\right]\left[P("Cc")+P\left("C{C}^{\prime \prime }\right)\right]\\ \left[P\left("D{d}^{\prime \prime }\right)+P\left("D{D}^{\prime \prime }\right)\right]\left[P\left("E{e}^{\prime \prime }\right)+P("EE")\right]\end{array}$

$=\left(\frac{1}{2}+0\right)\left(\frac{1}{2}+\frac{1}{4}\right)\left(\frac{1}{2}+0\right)\left(\frac{1}{2}+\frac{1}{4}\right)\left(\frac{1}{2}+0\right)$

$=\frac{9}{128}$

$\left(i\right) b) P("Phenotype as the second parent")$

$=P\left[\left("a{a}^{\prime \prime }\right)("Bb"\cup "BB")("cc")("Dd"\cup "DD")\left("ee"\right)\right]$

$=P("aa")P("Bb"\cup "BB")P("cc")P("Dd"\cup "DD")P("ee")$

$\begin{array}{r}=\left[P{("aa")}^{}\right]\left[P\right("Bb")+P("BB"\left)\right]\left[P("\mathrm{cc}")\right]\\ \left[P\left("D{d}^{\prime \prime }\right)+P\left("D{D}^{\prime \prime }\right)\right]\left[P("ee")\right]\end{array}$

$$\begin{gathered} =\left(\frac{1}{2}\right)\left(\frac{1}{2}+\frac{1}{4}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}+\frac{1}{4}\right)\left(\frac{1}{2}\right) \\ =\left(\frac{{\displaystyle 9}}{{\displaystyle 128}}\right) \end{gathered}$$

$$\begin{gathered} \left(i\right) c) P("Phenotype as either parent") = P("Phenotype as the first parent") \\ + P("Phenotype as the second parent") \end{gathered}$$

$=\frac{9}{128}+\frac{9}{128}$

$=\frac{9}{64}$

$P("phenotype\text{ as neither parent" })=1-P("phenotype\text{ as either parent") }$

                                                               $$\begin{gathered} =1-\frac{9}{64} \\ =\frac{55}{64} \end{gathered}$$

Step 4: Genotype (part c and d)

$II) a) P("geenotype\text{ as the first parent") }$

$=P("A{a}^{\prime \prime })("Bb")("Cc")("Dd")\left("E{e}^{\prime \prime })\right.$

$=P("Aa")P("Bb")P("Cc")P("Dd")P("Ee")$

$$\begin{gathered} =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} \\ =\frac{1}{32} \end{gathered}$$

$b) P("genotype\text{ as the second parent") }$

$=P\left[\right("aa"\left)\right("Bb"\left)\right("cc"\left)\right("Dd"\left)\right("ee"\left)\right]$

$=P("aa")P("Bb")P("cc")P("Dd")P("ee")$

$$\begin{gathered} =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} \\ =\frac{1}{32} \end{gathered}$$

$c) P(\text{ "genotype as either parent" ) =}$$P("genotype as the first parent") + ("genotype as the second parent")$

$$\begin{gathered} =\frac{1}{32}+\frac{1}{32} \\ =\frac{1}{16} \end{gathered}$$

$d) P("genotype\text{ as neither parent") }=1-P("genotype\text{ as either parent") }$

  $$\begin{gathered} =1-\frac{1}{16} \\ =\frac{15}{16} \end{gathered}$$