The images attached show a parental cross that is homozygous wild female x white male. F1 were intercrossed to produce the F2 generation as indicated below:Wild Female: 416 Wild Male: 192 White Female: 0 White Male: 192A chi-square analysis was done and an image was also attached.Please discus what conclusions can be made based on the data/findings. What inheritance pattern is this? Phenotypes FractionExpected2/4WildFemaleWild MaleWhiteFemaleWhiteMaleTotal1/41/41Observed Expected(0)(E)4164001920192800200200800Deviation(D)16-8-80D²2566464D²/E0.640.320.32X² = 1.28 P:F₁:F₂:X+X+XXXwY1 X+XW 1 X+Y2 X+X 1 X+Y :1 XWYFigure 1. Monohybrid cross for the inheritance of the white mutation. Homozygous wild-type (red-eyed) females were crossed with white-eyed males. The resulting F₁ generation were then crossed togenerate the F₂ generation, resulting in a ratio of 2 red-eyed females (X*X* and X*X³): 1 red-eyed male(X+Y): 1 white-eyed male (XY).

Name: The images attached show a parental cross that is homozygous wild fema
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Description: The images attached show a parental cross that is homozygous wild female x white male. F1 were intercrossed to produce the F2 generation as indicated below:Wild Female: 416 Wild Male: 192 White Female: 0 White Male: 192A chi-square analysis was done

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Human Anatomy & Physiology (11th Edition)

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Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

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The images attached show a parental cross that is homozygous wild female x white male. F1 were intercrossed to produce the F2 generation as indicated below:

Wild Female: 416 Wild Male: 192 White Female: 0 White Male: 192

A chi-square analysis was done and an image was also attached.

Please discus what conclusions can be made based on the data/findings. What inheritance pattern is this?

Phenotypes Fraction
Expected
2/4
Wild
Female
Wild Male
White
Female
White
Male
Total
1/4
1/4
1
Observed Expected
(0)
(E)
416
400
192
0
192
800
200
200
800
Deviation
(D)
16
-8
-8
0
D²
256
64
64
D²/E
0.64
0.32
0.32
X² = 1.28

P:
F₁:
F₂:
X+X+
X
X
XwY
1 X+XW 1 X+Y
2 X+X 1 X+Y :1 XWY
Figure 1. Monohybrid cross for the inheritance of the white mutation. Homozygous wild-type (red-
eyed) females were crossed with white-eyed males. The resulting F₁ generation were then crossed to
generate the F₂ generation, resulting in a ratio of 2 red-eyed females (X*X* and X*X³): 1 red-eyed male
(X+Y): 1 white-eyed male (XY).

Expert Solution

Step 1: Introduce about the chi-square test

Chi-square (X²) test is performed to compare the observed data of an experiment with the expected data. This is helpful in genetical analysis specially in case of inheritance patterns of genes.

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