Chapter 4, Problem 12EQ

As mentioned in Experimental Question E11, red eyes is the wild-type phenotype. Several different genes (with each gene existing in two or more alleles) are known to affect eye color. One allele causes purple eyes, and a different allele causes sepia eyes. Both of these alleles are recessive to red eye color. When flies with purple eyes were crossed to flies with sepia eyes, all of the ${\text{F}}_{\text{1}}$ offspring had red eyes. When the ${\text{F}}_{\text{1}}$ offspring were allowed to mate with each other, the following data were obtained:  

146 purple eyes  

151 sepia eyes  

50 purplish sepia eyes  

444 red eyes

Explain this pattern of inheritance. Conduct a chi square analysis to see if the experimental data fit your hypothesis.

Summary Introduction

To analyze:

The inheritance pattern when the number of F₁ offspring with purple, sepia, purplish sepia and red eyes was 146, 151, 50, 444 respectively. Conducting a chi-square analysis to test whether. the data fits in the proposed hypothesis.

Introduction:

Chi-square (often written ${\text{χ}}^{\text{2}}$) is a test that is used to determine if the actual data observed from the cross is obeying the earlier predicted Mendelian hypothesis or not. The general formula of the chi-square is:

$\begin{array}{l}{\text{χ}}^{\text{2}}\text{= Σ}\frac{{\left(\text{O}-\text{E}\right)}^{\text{2}}}{\text{E}}\\ \text{O = observed data in each category}\\ \text{E = expected data in each category based on the experimenter’s hypothesis}\\ \text{Σ means to sum the data values for each category}\text{.}\end{array}$

Explanation of Solution

Red-eye is the wild type phenotype. Purple color and sepia eye color also exist, but are recessive to the red eye color. Flies with purple eyes were crossed with sepia eyed individuals. This cross yielded $\text{444 red eyes : 146 purple eyes : 151 sepia eyes : 50 purplish sepia eye}$.

This approximately equals to $9:3:3:1$ ratio. Eye color in Drosophila is governed by two genes and none of those genes are X-linked. pr⁺ is the allele of red eye color and pr is the allele of purple eye color. sep⁺ is the allele for red eye color and sep is the allele for sepia eye color.

Cross 1: Individual with purple eye color prprsep⁺sep⁺XIndividual with sepia eye color pr⁺pr⁺sepsep

prsep+	prsep+	prsep+	prsep+
pr+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep
pr+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep
pr+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep
pr+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep	pr+prsep+sep

All the individuals in the F₁ generation will be pr⁺prsep⁺sep.Both genes are present in dominant heterozygous condition thus, the color of the eye will be red.

The F₁ offspring are mated with each other in the following cross:

Cross 2: pr+prsep+ sep (red eye) Xpr+ prsep+ sep (red eye)

pr+sep+	pr+sep	prsep+	prsep
pr+sep+	pr+pr+sep+sep+ red eye	pr+pr+sep+sep red eye	pr+prsep+sep+ red eye	pr+prsep+sep red eye
pr+sep	pr+pr+sepsep+ red eye	pr+pr+sepsep sepia eye	prpr+sepsep+ red eye	prpr+sepsep sepia eye
prsep+	pr+prsep+sep+ red eye	pr+prsep+sep red eye	prprsep+sep+ purple eye	prprsep+sep purple eye
prsep	pr+prsep+sep red eye	pr+prsepsep sepia eye	prprsepsep+ purple eye	prprsepsep purplish sepia

This type ofpattern of inheritance is known as the complement system. The protein product of both pr⁺ and sep⁺genesis needed for red coloring, but if one of them is absent, the phenotype changes. The individual having homozygous alleles for purple will have purple eyes and the individual having the homozygous allele for sepia will have sepia eyes. The individual having both the genes in homozygous recessive condition will have purplish sepia eyes.

There isa total of four phenotypes in the F₂ generation. A total number of individual in F₂ generation is $146+151+50+444=791$. A chi-square test needs to be done to determine whether, the data obtained from the cross is consistent with the hypothesis or not.

Proposed Hypothesis: The pattern of inheritance is complementation type and both genes are required to produce red eye coloration in flies.The cross between pr⁺prsep⁺sep (red eye) and pr⁺prsep⁺sep(red eye) will yield 9:3:3:1 ratio in the F₂ generation.

Based on this hypothesis, calculating the expected values of two phenotypes:

$\begin{array}{l}\frac{\text{9}}{\text{16}}\text{= red color}\\ \frac{\text{3}}{\text{16}}\text{= purple color}\\ \frac{3}{16}\text{= sepia color}\\ \frac{1}{16}=\text{purplish sepia}\end{array}$

Observed phenotype (O)	Expected phenotypes(E)
444 red eyes	$\frac{791\times 9}{16}=445\text{ red eyes}$
146 purple eyes	$\frac{791\times 3}{16}=\text{148 purple eyes}$
151 sepia eyes	$\frac{791\times 3}{16}=\text{148 sepia eyes}$
50 purplish sepia eyes	$\frac{791\times 1}{16}=49\text{ purplish sepia eyes}$

The population data falls into four sets in this question so, the chi-square calculation wouldbe done by the formula:

${\chi }^2=\frac{{\left(O_1-E_1\right)}^2}{E_1}+\frac{{\left(O_2-E_2\right)}^2}{E_2}+\frac{{\left(O_3-E_3\right)}^2}{E_3}+\frac{{\left(O_4-E_4\right)}^2}{E_4}$

The values of Observed phenotypes (O) and Expected phenotypes (E) are

$\begin{array}{l}{\text{O}}_{\text{1}}{\text{= 444 , E}}_{\text{1}}\text{= 445}\\ {\text{O}}_{\text{2}}{\text{=146 , E}}_{\text{2}}\text{=148}\\ {\text{O}}_{\text{3}}{\text{=151 , E}}_{\text{3}}\text{=148}\\ {\text{O}}_{\text{4}}{\text{=50 , E}}_{\text{4}}\text{=49}\end{array}$

Applying the formula for chi-square,

$\begin{array}{l}{\chi }^2=\frac{{\left(444-445\right)}^2}{445}+\frac{{\left(146-148\right)}^2}{148}+\frac{{\left(151-148\right)}^2}{148}+\frac{{\left(50-49\right)}^2}{49}\\ {\chi }^2=\frac{{\left(-1\right)}^2}{445}+\frac{{\left(-2\right)}^2}{148}+\frac{{\left(3\right)}^2}{148}+\frac{{\left(1\right)}^2}{49}\\ {\chi }^2=\frac{1}{445}+\frac{4}{148}+\frac{9}{148}+\frac{1}{49}\end{array}$

$\begin{array}{l}{\chi }^2=\text{ 0}.002+0.027+0.060+0.020\\ {\chi }^2=0.109\end{array}$

The formula for measuring the degrees of freedom is $\left(\text{n -1}\right)\text{ n = number of categories}$. In this question, there are four categories. So, the degrees of freedom will be given by, $4-1=3$.

$\text{Degree of freedom = 3}$ and the chi-square value that is calculated is $0.109$, which is approximately equal to $0.11$. Therefore, according to the chi-square values and probability table, which gives probability (P value) value of $0.99or99\%$. So, the hypothesis that the alleles for sepia and purple eyes are in two distinct genes and both of them are recessive cannot be rejected and hence, the proposed hypothesis may be correct.

Conclusion

Therefore, it can be concluded that the eye color in Drosophila is governed by two genes sep⁺ and pr⁺, when both of them are present, the eye color is red. When two red-eyed individuals withpr+prsep+ sep (red eye) and pr+ prsep+ sep (red eye) genotype are crossed together, $\text{9 red eyed : 3 purple eyed : 3 sepia eyed :1 purplish sepia eyed}$ ratio is observed in the F₂ generation. After conducting chi-square test with $\text{degree of freedom = 3}$ the chi-square values are $0.109$, which corresponds to the probability (P value) value of $0.99or99\%$ and hence the hypothesis cannot be rejected.So, may be the proposed hypothesis is correct.