2) (4 point question) Consider a hairy pea plant that was grown from yellow seed

Question

2) (4 point question) Consider a hairy pea plant that was grown from yellow seed (an F1 hybrid). This hybrid had one parent (P1) that was an inbred variety with no hairs and was grown from a green seed. P2 was also an inbred variety, but grown from a yellow seed and had hairy leaves. After a backcross of the F1 hybrid to the P1 parent the following progeny are observed. 29 82 Yellow seed/Smooth Yellow seed /Hairy Green seed /Smooth Green seed /Hairy 30 a) (1 point) What is the null hypothesis for the ratio of progenies based on the backcross described above b) (3 point) Using a Chi square statistical test, do you reject or fail to reject the null hypothesis. Show your calculations

Explanation / Answer

Answer :

A) null hypothesis : there is no significant difference between the observed and expected value after the baccross between the f1and p1 .so the phenotype frequency remain as 9:3:3:1

B) to the check the frequency we do a chi- square test:

Green Smooth(GS) 9

Yellow smooth (YS) 29

Green hairy (GH) 30

Yellow hairy (YH) 82

So the expected frequency are

GS=1/16*150=9.37

YS=3/16*150=28.12

GH=3/16*150=28.12

YH=9/16*150=84.375

Now find out the Chi square value=(O-E)^2/E

So GS it is (9-9.35)^2/9.37=0.015

Similarly for YS= (29-28.12)^2/28.12=0.0275

GH=(30-28.12)^2/28.12=0.126

And for YH = 0.067

So Chi square =0.015+0.0275+0.126+0.067

=0.2355

This value is less than the Chi square table value

So we reject the null hypothesis. that means there is no

Significant difference between the observed and expected value of frequency.

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