Assuming a single diallelic gene coding for the eye color trait, the chi-square test results suggests that there is <u>not enough evidence</u> to reject the null hypothesis. Parents must be heter0zyg0us. 0.48 < 3.841
<h3 /><h3>Available data</h3>
- Eyes are black or yellow
- Cross: between two black eyed lizards
- F1) 72 black eyed lizards, and 28 yellow-eyed lizards.
Assuming the eye color is coded by a single diallelic gene and that black is dominant over yellow, we will name the alleles B for black and b for yellow.
<u>Hypothesis:</u> Two heter0zyg0us lizards are crossed. Bb x Bb
The expected phenotypic ratio among the progeny is 3:1,
- 3/4 expressing the dominant trait, and
- 1/4 expressing the recessive trait.
So, according to these proportions, we can get the expected numbers
4 ------------------- 100 individuals
3 ------------------- X = 75 black-eyed individuals
1 -------------------X = 25 yellow-eyed individuals
Now, we will compare the observed values and the expected values.
<u> Black eyes Yellow eyes </u>
Observed 72 28
Expected 75 25
(Obs-Exp)²/Exp 0.12 0.36
X² = Σ(Obs-Exp)²/Exp = 0.12 + 0.36 = 0.48
So the chi-square value is 0.48.
Now we will compare this value with the critical table value. To do it, we need to know the degrees of freedom and the significance level.
Degrees of freedom = n - 1 = 2 - 1 = 1
Significance level = 0.05
Critical value = 3.841
X² < Critical value
0.48 < 3.841
The critical value is greater than X², so there is <u>not enough evidence</u> to reject the null hypothesis. Parents must be heter0zyg0us.
You will learn more about chi square problems at
brainly.com/question/20566562
