contestada

A student makes a monohybrid eross with Drosophila(fruit flies). She crosses two heterozygotes for
the white eye. Wwx Ww. She expects to see a 3:1 phenotypic ratio of Red eyes (WW and Ww) to white
eyes (ww), her null hypothesis. She rears the next generation through to adult flies and counts the
following numbers: white eyes:210. Red eyes:680. (According to the chi-square test, did she get the predicted outcome?. State whether or not you reject the null hypothosis

Respuesta :

marianaegarciaperedo

Assuming a significance level of 0.05, X² < P₀.₀₅, so there is no enought evidence to reject the null hypothesis. We fail to reject it. She got the predicted ratios.

Available data

  • cross between two heter0zyg0us flies for eye color, Ww
  • expected ratio 3:1
  • Observed numbers 210 white-eyed and 680 red-eyed flies.

The total number of individuals is 210 + 680 = 890

Let us first take the expected numbers from the expected ratio,

4 -------------890

3 ------------X = 667.5

1 -------------X = 222.5

                        Red-eyed individuals      White-eyed individuals

Observed                     680                                   210

Expected                      667.5                                222.5

(Obs-Exp)²/Exp             0.234                               0.702

Now we need to get the chi-square value

X² = Σ(Obs-Exp)²/Exp

X² = 0.936

Degrees of freedom = n - 1 = 2 - 1 = 1

Significance level = 0.05

Critical value = 3.841

X² < P₀.₀₅

0.936 < 3.841

p value is greater than 0.05, so there is not enough evidence to reject the null hypothesis. The genotypes might be in equilibrium, and there might be independent assortment.

You will learn more about chi square at

https://brainly.com/question/20566562

https://brainly.com/question/25113136

Otras preguntas