Question

One common application for the chi-square test is a genetic cross. In this case, the statistical null hypothesis is that the observed results from the cross are the same as those expected, for example, the 3:1 ratio or 1:2:1 ratio for a Mendelian trait.

Dr. William Cresko, a researcher at the University of Oregon, conducted several crosses between marine stickleback fish and freshwater stickleback fish. All marine stickleback fish have spines that protrude from the pelvis, which presumably serve as protection from larger predatory fish. Many freshwater stickleback populations lack pelvic spines. Dr. Cresko wanted to find out whether the presence or absence of pelvic spines behaves like a Mendelian trait, meaning that it is likely to be controlled mainly by a single gene.

In one cross, marine stickleback with spines were crossed with stickleback from Bear Paw Lake, which don’t have pelvic spines. All the progeny fish from this cross, the so-called F₁generation, had pelvic spines. Dr. Cresko then took the F₁ offspring and conducted several crosses between them to produce the F₂ generation.

F₂ Generation: Cross of F1 Generation Individuals

Cross Number	Total # of F2 Fish	F2 Fish with Spines	F2 Fish without Spines
1	98	71	27
2	79	62	17
3	62	49	13
4	34	28	6
5	29	24	5
6	23	17	6
7	21	17	4
8	19	18	1
9	15	11	4
10	12	10	2
11	12	10	2
12	4	3	1
Total	408	320	88

**If the presence of pelvic spines is controlled by a single gene and the presence of pelvic spines is the dominant trait as suggested by the F₁ results, you would expect a ratio of 3:1 for fish with pelvic spines to fish without pelvic spines in the F₂ generation. For a total of 408 fish, the expected results would be 306:102. The results from Dr. Cresko’s crosses are 320:88.

The null hypothesis is that there is no real difference between the expected results and the observed results, and that the difference that we see occurred purely by chance. The statistical alternative hypothesis is that there is a real difference between observed and expected results.

To test this statistic using chi-square, you must first fill in the following table based on information provided above.

Stickleback Spine Chi-Square Value Calculations

Phenotype	Observed (o)	Expected (e)
Spines present
Spines absent

SHOW WORK

What are the hypotheses (null and alternative) that you will test with chi-square?

What is the chi-square -value?

What is the critical value?

Do you reject or fail to reject the null hypothesis?

What is your conclusion?

Answer 1

hypotheses (null and alternative) :

H0​: The variables are independent

Ha​: The variables are dependent

Test stat :

Phenotype	Observed	Expected	O-E	(O-E)^2	(O-E)^2/E
Spines present	320	306	14	196	0.640523
Spines Absent	88	102	-14	196	1.921569
Total	408	408		X^2 =	2.562092

X^2 = 2.56209

Critical value = 3.8415 at df = 1 ((r-1)*(c-1)) and significance level = 0.05

X^2 = 2.56209 < Critical value = 3.8415

So we do not reject the null hypothesis

Hence we conclude that there is no significant difference between observed and expected frequencies, that means The variables are independent