Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result is 72 black eyed lizards, and 28 yellow-eyed lizards. Calculate the chi-squared value for the null hypothesis that the black eyed parents were heterozygous for eye color. Explain the chi-square test results.

Sagot :

marianaegarciaperedo

Assuming a single diallelic gene coding for the eye color trait, the chi-square test results suggests that there is not enough evidence to reject the null hypothesis. Parents must be heter0zyg0us. 0.48 < 3.841

Available data

  • 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.

Hypothesis: 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.

                             Black eyes                    Yellow eyes    

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 not enough evidence to reject the null hypothesis. Parents must be heter0zyg0us.

You will learn more about chi square problems at

https://brainly.com/question/20566562

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.