Question

In: Statistics and Probability

Two pea plants are crossed. One is homozygous for white flowers and the other is heterozygous...

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 X2 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.

Solutions

Expert Solution

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.


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