In fruit flies, the phenotype for eye color is determined by a certain locus. E indicates the dominant allele and e indicates the recessive allele. The cross between a male wild-type fruit fly and a female white-eyed fruit fly produced the following offspring.

Wild Type Male Wild Type Female White-eyed Male White-eyed Female Brown-eyed Female
F1 Generation 0 45 55 0 1

The wild-type and white-eyed individuals from the F1 generation were then crossed to produce the following offspring.

Wild Type Male Wild Type Female White-eyed Male White-eyed Female Brown-eyed Female
F2 Generation 23 31 22 24 0

Required:
a. Determine the genotypes of the original parents (P generation) and explain your reasoning.
b. Use a Chi-squared test on the F2 generation data to analyze your prediction of the parental genotypes.

Answer:

The genotypes of the original parents are

The Chi-square analysis leads us to accept the prediction of the parental genotypes.

Explanation:

Due to technical problems, you will find the complete explanation in the attached files.

(a) The genotypes for the original parents (P generation) are

Male: X⁺Y  

Female: X⁻X⁻

(b) The prediction of the parental genotype is accepted by the Chi-squared test analysis.  

Drosophila Melanogaster is a fruit fly with abnormally big, brightly colored eyes.

The range in color from red to sepia to white reveals a lot about the genetic makeup of the fly.

(a)  E = + ⇒ Dominant allele coding for wild-type eyes

     e = - ⇒ Recessive allele coding for white eyes

Now, Crossing the : male wild-type with a female white-eyed

(Parental)   =  X⁺Y x X⁻X⁻

(Phenotype) =   wild   white

(Gametes)  =    X⁺ Y X⁻ X⁻

(Punnett square)  =  X⁺ Y

X⁻   =  X⁺X⁻  X⁻Y

X⁻   =  X⁺X⁻   X⁻Y

In (F1)generation = 2/4 = 1/2 = 50% of the progeny are expected to be wild-type females.

And 2/4 = 1/2 = 50% of the progeny are expected to be white-eyed males.

Thus, the percentages are almost 50:50, as the expected ones.

So, the genotypes of the original parents are

Male:      X⁺Y

Female:  X⁻X⁻

(b) Using Chi-squared test :

Observed fruit fly type:

Wild Type Male - 25

Wild Type Female - 31

White-eyed Male - 22

White-eyed Female - 24

Expected fruit fly type:

Wild Type Male - 25

Wild Type Female - 25

White-eyed Male - 25

White-eyed Female - 25

Now, chi square [tex]= \sum\dfrac{(O- E)^2 }{E}[/tex]

By putting numbers in the formula, we are getting

Wild Type Male - 0.16

Wild Type Female - 1.44

White-eyed Male - 0.36

White-eyed Female - 0.04

Now, X² = 0.16 + 1.44 + 0.36 + 0.04 = 2

Critical value = 7.82

X² (2) < critical value (7.82)

Thus, this is the evidence to accept the given hypothesis.

Learn more about fruit flies here:

https://brainly.com/question/7274152

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