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

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

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

Do you fail to reject or reject the null hypothesis? __________

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

Given:

tall(TT) is dominant to short (tt)

purple(WW) flowers are dominant to white(ww)

When  One is homozygous for white flowers and the other is heterozygous for purple flowers. Both are heterozygous for being tall plants are crossed whith each other we get following table.

Ttww crossed with TtWw

Allel	TW	Tw	tW	tw
Tw	TTWw	TTww	TtWw	Ttww
Tw	TTWw	TTww	TtWw	Ttww
tw	TtWw	TtWw	ttWw	ttww
tw	TtWw	TtWw	ttWw	ttww

Phenotype	Probability
tall purple	7/16=0.4375
short purple	2/16=0.125
tall white	5/16=0.3125
short white	2/16=0.125

Null Hypothesis states that the two genes are independently assorted.

While

Alternative hypothesis states the two genes are dependent.

Phenotype	Probability	Observed(O)	Expected(E)=n*Pi	O-E	(O-E)^2	(O-E)^2/E
tall purple	7/16=0.4375	150	109.375	40.625	1650.391	15.08929
short purple	2/16=0.125	18	31.25	-13.25	175.5625	5.618
tall white	5/16=0.3125	62	78.125	-16.125	260.0156	3.3282
short white	2/16=0.125	20	31.25	-11.25	126.5625	4.05
	n=	250			Sum=	28.08549

Here degrees of freedom =4-1=3

Pvalue is =3.485005e-06

### By using R command

> 1-pchisq(28.0855, df=3)
[1] 3.485005e-06

Since Pvalue is less that 0.05 we reject the null hypothesis at 5% level of significance.

As a result of this statistical analysis, it is possible to conclude that pea plant height and pea plant blossom color are linked traits.