Question

You are investigating two genes for interaction. You have two lines of the organism that are homozygous at these two loci. Line A has the genotype AAbb and line B has the genotype aaBB. Both lines have an identical phenotype. You cross these lines and collect offspring from the F1 generation which are all double heterozygotes and wild type in appearance. Then you cross these to produce an F2 generation. You have 40 offspring in your F2 generation. The offspring have two phenotypes. There are 24 with the wild type and 16 with the phenotype of the F1 parents. a. Which type of epistasis does this data fit the best?

i. What ratio is expected for that type of epistasis? ii. What ratio is observed?

iii. Calculate the Χ2

iv. Determine the probability that this model fits the data

b. Is there another possibility? If so which type is it?

i.     What ratio is expected for that type of epistasis?

ii.     What ratio is observed

iii.     Calculate the Χ²

iv.     Determine the probability that this model fits the data

Answer 1

Answer: (a) The above given inheritance best fit the data via double recessive epistasis which is also known as complementary gene interaction. Let A and B are the two genes which results in the desired trait (let say wild type flower color which is purple in this case) as a part of interaction of these two dominant genes. In the absence of dominant gene A or B or both, the flowers are white which is the phenotype of the parents. It is assumed that gene A produces an enzyme that catalyzes the formation of necessary raw material for the synthesis of required pigment and gene B produces an enzyme which transforms the raw material into the pigment. The cross of the two homozygous lines of AAbb and aaBB both of which have an identical phenotype (let say white flower color) is shown below.

Parents; AAbb (white flowers) aaBB (white flowers)

Gametes: Ab Ab aB aB

F1 generation: AaBb (wild type purple)

when the F1 is selfed, the resulting genotypes obtained in F2 generation are as follows: AABB (wild purple), AABb (wild purple), AaBb (wild purple), AaBb (wild purple), AABb (wild purple), AAbb (white), AaBb (wld purple), Aabb (white), AaBB (wild purple), AaBb (wild purple), aaBB (white), aaBb (white), AaBb (wild purple), Aabb (white), aaBb (white), aabb (white)

(i) therefore the expected ratio of this type of epistasis is 9 (wild purple) :7 (white)

(ii) But here the number of offspring with wild type phenotype is 24 and 16 with phenotype of F1 parents. So the observed ratio is 3:2.

(iii)

Phenotypes	observed number(o)	Expected number(e)	d=(o-e)	(o-e)2	(o-e)2/e
wild purple	24	22.5	1.5	2.25	0.10
white	16	17.5	-1.5	2.25	0.13
Total	40

Here the degrees of freedom, df=2-1=1 and the chi-square value is given by,

(iv) For df=1 and chi square value=0.23, the Probability or the P value is between 0.50 and 0.70. This means that, when the hypothesis being tested, in 50 to 70 out of 100 trials we could expect chi-square values of such magnitude or greater due to chance. As, the obtained P value is greater than 0.05, the deviation of expected from observed is not considered statistically significant and the data indicate that the hypothesis stands valid and should not be rejected. So there is no possibility to enforce a new hypothesis.