Block

Monohybrid Cross:

Crossing vg+vg+ x vgvg

The Variables:

p{}_{1}:=
 Proportion of flies with normal wings

p{}_{2}:=
 Proportion of flies with vestigial wings

Hypotheses:

H_{0}:
 p_{1}=0.75,\:p_{2}=0.25


H{}_{a}:\neg H_{0}\quad\alpha=0.05


The test statistic is:

\chi^{2}=
 \sum\frac{(O-E)^{2}}{E}


Computation:

\chi^{2}=\frac{(100-150)^{2}}{150}+\frac{(100-50)^{2}}{50}


\chi^{2}=50


2 Categories:

\chi^{2}CDF\:
  from (50,\infty)\:
 with df=1


p=1.53E-12


p<\alpha\therefore H_{0}=
 False

Conclusion:

The probability that the sampled result can arise from the original expected value is far below 5%.

Therefore we can conclude the 3:1 hypothesis is false.

Dihybrid Cross:

Crossing vg+vg+ bl+bl+ x vgvgblbl

Expected Ratio: 9:3:3:1

Respective Percents: 56.25%, 18.75%, 18.75%, 6.25%

Expected Counts:

338 Normal Wings Normal Body

112 Normal Wing Black Body

112 Vestigial Wings Normal Body

38 Vestigial Wings Black Body

The Variables:

p{}_{1}:=
 Proportion of flies with normal wings and a normal body

p{}_{2}:=
 Proportion of flies with normal wings and a black body

p{}_{3}:=
 Proportion of flies with vestigial wings and a normal body

p{}_{3}:=
 Proportion of flies with vestigial wings and a black body

Hypotheses:

H_{0}:
 p_{1}=0.5625,\:p_{2}=0.1875,\:
 p_{3}=0.1875,\:p_{4}=0.0625


H{}_{a}:\neg H_{0}\quad\alpha=0.05


The test statistic is:

\chi^{2}=
 \sum\frac{(O-E)^{2}}{E}


Computation:

\chi^{2}=\frac{(425-338)^{2}}{338}+\frac{(40-112)^{2}}{112}+\frac{(40-112)^{2}}{112}+\frac{(95-38)^{2}}{38}


\chi^{2}=\frac{237150}{1183}\approx200.465


4 Categories:

\chi^{2}CDF\:
  from (200.465,\infty)\:
 with df=3


p=3.47E-43


p<\alpha\therefore H_{0}=
 False

Conclusion:

The probability that the sampled result can arise from the original expected value is far below 5%.

Therefore we can conclude the 9:3:3:1 hypothesis is false.