Chapter 5.1, Problem 47CQ

a.

Summary Introduction

To propose:

The hypothesis to explain the inheritance of kitey, black, and red feather color in pegion.

Explanation of Solution

To investigate the relation between kitey, black and red feather color, crosses were made which are given in following table:

Cross	Offspring
$\text{Kitey}\times \text{kitey}$	16 kitey, 5 black, 3 red
$\text{Kitey}\times \text{black}$	6 kitey, 7 black
$\text{Red}\times \text{kitey}$	18 red . 9 kitey, 6 black

Feather color in pigeons are determined by four alleles: $K^1,K^2,K^B,\text{and}{K}^R$

$\begin{array}{l}{K}^1:\text{Responsible}\text{for}\text{kitey}\text{color}\text{in}\text{pigeon}\\ K^2:\text{Responsible}\text{for}\text{kitey}\text{color}\text{in}\text{pigeon (homozygous}\text{condition}\text{produces}\text{red}\text{color)}\\ K^B:\text{Responsible}\text{for}\text{black}\text{color}\text{in}\text{pigeon}\\ K^R:\text{Responsible}\text{for}\text{red}\text{color}\text{in}\text{pigeon}\end{array}$

Cross between two pegions carrying kitey phenotype:

$\begin{array}{l}{K}^1{K}^R\times K^2{K}^B\\ \downarrow \end{array}$

	$K^1$	$K^R$
$K^2$	$K^1{K}^2$ (Kitey)	$K^2{K}^R$ (Red)
$K^B$	$K^1{K}^B$ (Kitey)	$K^B{K}^R$ (Black)

Total number of offsprings observed in the cross $K^1{K}^R\times K^2{K}^B\text{are:}$

$\begin{array}{c}\text{=16}\text{kitey}\text{progeny}+\text{5 black}\text{progeny + 3}\text{red}\text{progeny}\\ =\text{24}\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{kitey}\text{progenies}\text{produced is}\\ =\left(\frac{16}{24}\right)\times 4\\ =2.67\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{black}\text{progenies}\text{produced}\text{is}\\ =\left(\frac{5}{24}\right)\times 4\\ =0.83\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{red}\text{progenies}\text{produced}\text{is}\\ =\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$24$}\right.\right)\times 4\\ =0.5\end{array}$

Cross between two pegions carrying black kitey phenotype:

$\begin{array}{l}{K}^1{K}^B\times K^B{K}^B\\ \downarrow \end{array}$

	$K^1$	$K^B$
$K^B$	$K^1{K}^B$ (Kitey)	$K^B{K}^B$ (Black)
$K^B$	$K^1{K}^B$ (Kitey)	$K^B{K}^B$ (Black)

Total number of offspring observed in the cross $K^1{K}^B\times K^B{K}^B\text{are:}$

$\begin{array}{c}\text{= 6}\text{kitey}\text{progeny}\text{+}7\text{black}\text{progeny}\\ \text{=13}\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{kitey}\text{progenies}\text{produced}\text{is}\\ =\left(\frac{6}{13}\right)\times 4\\ =1.85\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\mathrm{black}\text{progenies}\text{produced}\text{is}\\ =\left(\frac{7}{13}\right)\times 4\\ =2.15\end{array}$

Cross between two pegions carrying red and kitey phenotype;

$\begin{array}{l}{K}^2{K}^R\times K^2{K}^B\\ \downarrow \end{array}$

	$K^2$	$K^R$
$K^2$	$K^2{K}^2$ (Red)	$K^2{K}^R$ (Red)
$K^B$	$K^2{K}^B$ (Kitey)	$K^B{K}^R$ (Black)

Total number of offspring observed in the cross $K^2{K}^R\times K^2{K}^B\text{are:}$

$\begin{array}{c}\text{=18}\text{red}\text{progeny}\text{+}9\text{kitey}\text{progeny}+\text{6}\text{black}\text{progeny}\\ \text{=33}\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{red}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{18}{33}\right)\times 4\\ =2.18\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{kitey}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{9}{33}\right)\times 4\\ =1.09\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{black}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{6}{33}\right)\times 4\\ =0.73\end{array}$

On the basis of the results obtained from the crosses , it can be hypothesized that “$K^2$ allele is dominant over other alleles, and can be termed as epistatic as it masks the expression of other alleles”. Dominance of the alleles follow the given pattern during expression:

$\begin{array}{l}{K}^1>K^B>K^R\\ K^R>K^2\\ K^2>K^B\\ K^2{K}^2>K^R{K}^R\end{array}$

b.

Summary Introduction

To test:

The hypothesis for each of the preceding crosses using a chi square test.

Explanation of Solution

To investigate the relation between kitey, black and red feather color, crosses were made which are given in the following table:

Cross	Offspring
$\text{Kitey}\times \text{kitey}$	16 kitey, 5 black, 3 red
$\text{Kitey}\times \text{black}$	6 kitey, 7 black
$\text{Red}\times \text{kitey}$	18 red . 9 kitey, 6 black

Feather color in pigeons are determined by four alleles: $K^1,K^2,K^B,\text{and}{K}^R$

$\begin{array}{l}{K}^1:\text{Responsible}\text{for}\text{kitey}\text{color}\text{in}\text{pigeon}\\ K^2:\text{Responsible}\text{for}\text{kitey}\text{color}\text{in}\text{pigeon,}(\text{homozygous}\text{condition}\text{produces}\text{red}\text{color)}\\ K^B:\text{Responsible}\text{for}\text{black}\text{color}\text{in}\text{pigeon}\\ K^R:\text{Responsible}\text{for}\text{red}\text{color}\text{in}\text{pigeon}\end{array}$

Cross between two pegions carrying kitey phenotype;

$\begin{array}{l}{K}^1{K}^R\times K^2{K}^B\\ \downarrow \end{array}$

	$K^1$	$K^R$
$K^2$	$K^1{K}^2$ Kitey	$K^2{K}^R$ Red
$K^B$	$K^1{K}^B$ Kitey	$K^B{K}^R$ Black

Total number of offspring expected in the cross $K^1{K}^R\times K^2{K}^B$

$\text{2}\text{Kitey}\text{:}\text{1}\text{Red}\text{:}\text{1}\text{Black}$

Total number of offspring produced in the cross $K^1{K}^R\times K^2{K}^B\text{are:}$

$\begin{array}{c}\text{=16}\text{kitey}\text{progeny}\text{+}\text{5}\text{black}\text{progeny}\text{+}\text{3}\text{red}\text{progeny}\\ \text{=24}\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{kitey}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{16}{24}\right)\times 4\\ =2.67\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{black}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{5}{24}\right)\times 4\\ =0.83\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{red}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{3}{24}\right)\times 4\\ =0.5\end{array}$

To compare the observed number of progenies with the expected number of progenies in cross $K^1{K}^R\times K^2{K}^B$, a chi-square test is used:

$\begin{array}{c}{x}^2={\displaystyle \sum \frac{{(\text{Observed}-\text{Expected})}^2}{\text{Expected}}}\\ x^2=\left[\frac{{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{kitey}\text{phenotype}\text{with}\text{kitey}\text{phenotype}\end{array}\right)}^2}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{kitey}\text{phenotype}}\right]\\ +\left[\frac{{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{red}\text{phenotype}\text{with}\text{red}\text{phenotype}\end{array}\right)}^2}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{red}\text{phenotype}}\right]\\ +\left[\frac{{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{black}\text{phenotype}\text{with}\text{black}\text{phenotype}\end{array}\right)}^2}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{black}\text{phenotype}}\right]\\ =\left[\frac{{\left(2.67-2\right)}^2}{2}\right]+\left[\frac{{\left(0.83-1\right)}^2}{1}\right]+\left[\frac{{\left(0.5-1\right)}^2}{1}\right]\\ =\left[\frac{{\left(0.67\right)}^2}{2}\right]+{\left(-0.17\right)}^2+{\left(0.25\right)}^2\\ =0.2245+0.0289+0.25\\ =0.5034\end{array}$

The difference between expected number of progenies and observed number of progenies in $K^1{K}^R\times K^2{K}^B$ cross is $0.5034$, which shows the difference occurred only due to chance.

Cross between two pegions carrying black kitey phenotype:

$\begin{array}{l}{K}^1{K}^B\times K^B{K}^B\\ \downarrow \end{array}$

	$K^1$	$K^B$
$K^B$	$K^1{K}^B$ (Kitey)	$K^B{K}^B$ (Black)
$K^B$	$K^1{K}^B$ (Kitey)	$K^B{K}^B$ (Black)

Total number of offspring expected in the cross $K^1{K}^B\times K^B{K}^B\text{are:}$

$\text{2}\text{Kitey}\text{:}2\text{Black}$

Total number of offspring produced in the cross $K^1{K}^B\times K^B{K}^B\text{are:}$

$\begin{array}{c}\text{=6}\text{kitey}\text{progeny}\text{+}7\text{black}\text{progeny}\\ \text{=13}\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{kitey}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{6}{13}\right)\times 4\\ =1.85\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\mathrm{black}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{7}{13}\right)\times 4\\ =2.15\end{array}$

To compare observed number of progenies with expected number of progenies in cross$K^1{K}^B\times K^B{K}^B$, a chi-square test is used:

$\begin{array}{c}{x}^2={\displaystyle \sum \frac{{(\text{Observed}-\text{Expected})}^2}{\text{Expected}}}\\ x^2=\left[\frac{\begin{array}{l}{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{kitey}\text{phenotype}\text{with}\text{kitey}\text{phenotype}\end{array}\right)}^2\\ \end{array}}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{kitey}\text{phenotype}}\right]\\ +\left[\frac{\begin{array}{l}{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{black}\text{phenotype}\text{with}\text{black}\text{phenotype}\end{array}\right)}^2\\ \end{array}}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{black}\text{phenotype}}\right]\\ =\left[\frac{{\left(1.85-2\right)}^2}{2}\right]+\left[\frac{{\left(2.15-2\right)}^2}{2}\right]\\ =\left[\frac{{\left(-0.15\right)}^2}{2}\right]+\left(\frac{{\left(0.15\right)}^2}{2}\right)\\ =\frac{0.225}{2}+\frac{0.225}{2}\\ =0.113+0.113\\ =0.226\end{array}$

The difference between expected number of progenies and observed number of progenies in $K^1{K}^B\times K^B{K}^B$ cross is $0.226$, which shows the difference occurred only due to chance.

Cross between two pegions carrying red and kitey phenotype;

$\begin{array}{l}{K}^2{K}^R\times K^2{K}^B\\ \downarrow \end{array}$

	$K^2$	$K^R$
$K^2$	$K^2{K}^2$ (Red)	$K^2{K}^R$ (Red)
$K^B$	$K^2{K}^B$ (Kitey)	$K^B{K}^R$ (Black)

Total number of offspring expected in the cross $K^2{K}^R\times K^2{K}^B$

$\text{2}\text{Red}\text{:}\text{1}\text{Kitey}\text{:}\text{1}\text{Black}$

Total number of offspring observed in the cross $K^2{K}^R\times K^2{K}^B\text{are:}$

$\begin{array}{c}\text{=18}\text{red}\text{progeny}\text{+}9\text{kitey}\text{progeny}\text{+}\text{6}\text{black}\text{progeny}\\ \text{=33}\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{red}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{18}{33}\right)\times 4\\ =2.18\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{kitey}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{9}{33}\right)\times 4\\ =1.09\end{array}$

$\begin{array}{c}\text{In}\text{a}\text{ratio}\text{out}\text{of}\text{4,}\text{the}\text{number}\text{of}\text{black}\text{progenies}\text{produced}\text{are}\\ =\left(\frac{6}{33}\right)\times 4\\ =0.73\end{array}$

To compare the observed number of progenies with the expected number of progenies in cross$K^2{K}^R\times K^2{K}^B$, a chi-square test is used;

$\begin{array}{c}{x}^2={\displaystyle \sum \frac{{(\text{Observed}-\text{Expected})}^2}{\text{Expected}}}\\ x^2=\left[\frac{\begin{array}{l}{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{kitey}\text{phenotype}\text{with}\text{kitey}\text{phenotype}\end{array}\right)}^2\\ \end{array}}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{kitey}\text{phenotype}}\right]\\ +\left[\frac{\begin{array}{l}{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{red}\text{phenotype}\text{with}\text{red}\text{phenotype}\end{array}\right)}^2\\ \end{array}}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{red}\text{phenotype}}\right]\\ +\left[\frac{\begin{array}{l}{\left(\begin{array}{l}\text{Observed}\text{number}\text{of}\text{progeny}-\text{Expected}\text{number}\text{of}\text{progeny}\\ \text{with}\text{black}\text{phenotype}\text{with}\text{black}\text{phenotype}\end{array}\right)}^2\\ \end{array}}{\text{Expected}\text{number}\text{of}\text{progenywith}\text{black}\text{phenotype}}\right]\\ =\left[\frac{{\left(1.09-2\right)}^2}{1}\right]+\left[\frac{{\left(2.18-2\right)}^2}{2}\right]+\left[\frac{{\left(0.73-1\right)}^2}{1}\right]\\ ={\left(0.09\right)}^2+\left[\frac{{\left(0.18\right)}^2}{2}\right]+{\left(-0.27\right)}^2\\ =0.081+0.162+0.729\\ =0.972\end{array}$

The difference between expected number of progenies and observed number of progenies in $K^2{K}^R\times K^2{K}^B$ cross is $0.972$, which shows the difference occurred only due to chance.

Conclusion

Considering the crosses and their results, it can be hypothesized that $K^2$ allele is dominant over other alleles. The comparison between the observed number of progenies with the expected number using a chi-square test: $K^1{K}^R\times K^2{K}^B$ cross is $0.5034$, $K^1{K}^B\times K^B{K}^B$ cross is $0.226$, and $K^2{K}^R\times K^2{K}^B$ cross is $0.972$.