A true-breeding black mouse is crossed with an albino mouse, producing F1 offspr

Question

A true-breeding black mouse is crossed with an albino mouse, producing F1 offspring that are all agouti (grizzled fur). The F1s are intercrossed, producing F2s with the following phenotypes: 48 agouti, 28 albino, and 18 black. Actual research has revealed that fur color in this population of mice is controlled by two genes.
a. What is the observed phenotypic ratio for the F2 generation?

b. The researchers believe that dominant epistasis is occurring in this cross. Use the chi test to test their hypothesis. Are they correct?

c. Do you think another type of epistasis is occurring? Test your hypothesis with another chi square. Are you correct?

Explanation / Answer

a)The observed phenotype ratio for the F2 generation is 1.5:2:1. (1.5 albino, 2 agouti and 1 black). the ratio is almost equivalent to the ratio of epistasis. This is a an example of epistasis where the genotype at one locus (homozygous recessive albino gene aa) masks the phenotype that results from the genotype at a different locus (black gene BB, Bb or bb). For the two loci black (B) and albino (A), there are nine different possible genotypes and three different phenotypes.  

Black mouse - genotype - BB AA, Bb AA, Bb Aa and Bb Aa

Albino mouse - genotype - BB aa, Bb aa and bb cc &

Agouti mouse - genotype - bb AA and bbAa

b & c) The ratio is an example of 'recessive epistasis' where the recessive epistatic gene supresses the expression of non allelic gene only when the former gene is in homozygous recessive condition.

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