Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

You are studying a population of squirrels in a city park in McMinniville Tn.,.

ID: 216518 • Letter: Y

Question

You are studying a population of squirrels in a city park in McMinniville Tn.,. You are interested in this population because it contains some albinos (caused by a recessive allele for fur color). Recent studies in rural areas have shown that albino squirrels suffer predator mortality at a higher rate than normally colored squirrels. You want to know if natural selection acts on albinism in a city park in the same manner that it does in the wild. You go to the park and randomly shot 120 squirrels and sample their DNA. From these samples you determine that: 68 were homozygous dominant (B,B) 28 were heterozygous (B,b) 24 were homozygous recessive (b,b) Use the Chi Square Goodness of Fit Test to test the Null hypothesis that this population is not undergoing natural selection for fur color. If you conclude that it is experiencing natural selection, state whether albinism is being selected for or selected against.

Explanation / Answer

The null hypothesis is: Ho: the population is undergoing natural selection for fur color.

If this is the case, then the expected value of ‘bb’ genotype should be zero. Using this value you will not be able to find out the ‘chi-square analysis’. So, let us consider the null hypothesis for independent assortment.

So, the null hypothesis should be: The population is assorting independently.

The alternate hypothesis will be: The population is not assorting independently.

Observed (O)

Expected (E)

(O-E)^2

?2= (O-E)^2/E

BB

68

30

1444

48.13333333

Bb

28

60

1024

17.06666667

bb

24

30

36

1.2

120

?2=

66.4

Degree of freedom = (3-1)(4-1) = 6

Tabulated value of chi-square = 12.59 (at 0.05 level of significance).

Note that null hypothesis is accepted only if the calculated value is less than the tabulated value.

So, in this case null hypothesis is rejected. This population is not assorting independently. Note that the expected value for albinism (bb) was 30; and the value comes out to be 24. This simply shows that albinism is selected slightly against. But BB is totally selected for.

On the contrary, Bb genotype is selected against (according to the observed and expected values).

This is an example of diversifying selection.

Observed (O)

Expected (E)

(O-E)^2

?2= (O-E)^2/E

BB

68

30

1444

48.13333333

Bb

28

60

1024

17.06666667

bb

24

30

36

1.2

120

?2=

66.4

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote