Assume that flower color in petunia is controlled by two genes. After making a dihybrid cross,...

Assume that flower color in petunia is controlled by two genes. After making a dihybrid cross, the following
progeny were observed: 9 purple-flowered petunia : 7 white-flowerd petunias
a) What form of epistasis is working to control this trait?
b) One of the heterozygous individuals used in the above dihybrid cross was test-crossed. What phenotypic
classes do you expect in the progeny and at what frequencies? Show your work/thought process.

Solutions

Expert Solution

a. The ratio 9:7 is indicative thay this epistasis is Complementary gene action where the loss of function of any one of the gene shows a similar phenotype as would loss of both the genes required to produce colour. In petunias if the colour is controlled by two genes A and B then loss of function of either A and B will produce a similar phenotype. So, only A_B_ individuals will be pigmented and A_bb, aaB_ and aabb will all be colourless.

b. In a test cross when a heterozygote is mated with a recessive parent it will always produce the phenotype in the ratio of 1:1:1:1. However in this case, epistatic gene interaction is present. So if we consider AaBb crossed with aabb we get gametes of Ab, ab, aB, ab.

	Ab	ab	aB	ab
Ab	AAbb	Aabb	AaBb	Aabb
ab	Aabb	aabb	aaBb	aabb
aB	AaBb	aaBb	aaBB	aaBb
ab	Aabb	aabb	aaBb	aabb

Based on the above punett square we can see that only 2 progenies out of 16 having genotype AaBb is coloured while the rest are all colour less since the colour is produced due to complimentary gene action. Hence the test cross ratio would be 2:14 or 1:7.

P.S. Please rate this answer Thumbs Up if you liked it.

gladiator answered 1 year ago

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