Question

Two pea plants are crossed. One is homozygous for white flowers and the other is heterozygous for purple flowers. Both are heterozygous for being tall plants. In pea plants, tall is dominant to short, and purple flowers are dominant to white.

Fill out the table below for the probability of each possible phenotype. Report probability as a decimal rounded to four places (e.g. 0.1250, not 1/8 or 12.5%).

Phenotype	Probability
tall purple
short purple
tall white
short white

In a population of 150 pea plants, there are 50 tall-purple plants, 18 short-purple plants, 62 tall-white plants, and 20 short-white plants. In order to test if the two traits are experiencing independent assortment researchers would perform a chi squared test.

1.) The (null/alternative)  hypothesis states that the two genes are independently assorted while the (null/alternative)  hypothesis states the two genes are dependent.

2.) What is your calculated Chi Squared statistic?

• When performing a contingency table, do not round your expected values!!
• Report your calculated X² rounded to four decimal places

3.) What is the corresponding P value?

• Use the formula =1-(CHISQ.DIST(X²,df,TRUE)) to convert calculated X^{2 _{into a P value}}
• Report your answer rounded to 4 decimal places

4.) Do you fail to reject or reject the null hypothesis?

5.) As a result of this statistical analysis, it is possible to conclude that pea plant height and pea plant blossom color (are or are not)  linked traits.

Answer 1

Probability of each possible phenotype:

Phenotype	Probability
Tall purple	50 / 150 = 0.3333
Short purple	18 / 150 = 0.1200
Tall white	62 / 150 = 0.4133
Short white	20 / 150 = 0.1333

1) The null hypothesis states that the two genes are independently assorted while the alternative hypothesis states the two genes are dependent.

Observed Frequencies
	Tall	Short	Total
Purple	50	18	68
White	62	20	82
Total	112	38	150
Expected Frequencies
	Tall	Short	Total
Purple	112 * 68 / 150 = 50.7733	38 * 68 / 150 = 17.2267	68
White	112 * 82 / 150 = 61.2267	38 * 82 / 150 = 20.7733	82
Total	112	38	150
	(fo-fe)²/fe
Purple	(50 - 50.7733)²/50.7733 = 0.0118	(18 - 17.2267)²/17.2267 = 0.0347
White	(62 - 61.2267)²/61.2267 = 0.0098	(20 - 20.7733)²/20.7733 = 0.0288

2) Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 0.0851

df = (r-1)(c-1) = 1

3) p-value = CHISQ.DIST.RT(0.0851, 1) = 0.7706

4) p-value > α, fail to reject the null hypothesis.

5) As a result of this statistical analysis, it is possible to conclude that pea plant height and pea plant blossom color are not linked traits.