Question

12. Imagine a cross between AaBb x aabb produces progeny in the following proportions: | AaBb 31 aabb | 33. Aabb | 12 aaBb] 13 Based on this information, are genes A and B linked, or do they assort independently (Chi-square test)? IF linked, what is the genetic distance between A and B? Are gene A and B in coupling (cis) or repulsion (trans) conformation in the AaBb parent? How can you tell?

Answer 1

Question 12:

Null hypothesis: The genes are not linked. It means independent assortment is occured.

Please look in the image for chi square test:

Аавь - x a abb — — Punnet square gametes l a b . Expected х39 22:25 - АВ A abb AЬ Aa bb 22-25 ав аавь 22.5 ab Labb 2.5

| Now chi-square table! o €) Genotype Observed | Expected ! A abb aabb Aalbb a Bb 31 T33 i 12 , 13 22-25 22-25 A 22-25 i , 22.25', 3.44 1.5.1917 4.72 3.845 Chi square value is = 17.195 Degree of freedom = n-1=4-1=3 Chisquare value from table at 10=0.05 and af : 3 is 7.815

Now the chi square from experiment (17.195) is more than chi square from table (7.815). That's why null hypothesis is rejected. It means linkage is found between two genes.

Gene A and B are in cis form because parental types offsprings are more than recombinant type offsprings.

The genetic distance between A and B is calculated by recombination great between two genes.

Recombination frequency= ( recombinant/total) ×100

= {(12+13) /(12+13+31+33) }×100= (25/89) ×100=28.08

That's why gene distance between A and B is 28.08 centimorgan.

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