You are studying the genes Rtl1 and Mtl2 that are unlinked autosomal genes known

Question

You are studying the genes Rtl1 and Mtl2 that are unlinked autosomal genes known to affect tail length in mice. Recessive knockout mutations of each gene individually cause the tail to be long. However, the double knockout (rrmm) has no tail. Let's say you cross two heterozygous mice (RrMm x RrMm) and select one of the wild type offspring (normal tail length) and then cross offspring with a heterozygous mouse (wild type times RrMm). What is the probability that an offspring of this second cross will have no tail?

Explanation / Answer

When you cross two heterozygous mice, i.e RrMm * RrMm, 9 among the 16 offspirings will have atleast one allele of both the genes R and M as dominant. therefore, these 9 will have normal tail length (wildype).

therefore, probabilty of getting a wildtype offspring= 9/16

The 9 wildtype mice will have the genotype as 1 RRMM, 2RrMM, 2RRMm and 4RrMm

The next cross is between one of these 9 wildtype offspring with a heterozygous mouse, and we need to find the probabilty of getting an offspring with no tail in this cross (i.e rrmm). In order to get an offspring with no tail, the gametes produced by both the parents should of the type 'rm'. So among the 9 wildtype offspirings of the first cross, only RrMm type can be chosen for the next cross.

the probability of chosing the parent for 2nd cross = (9/16) *( 4/9).

in the second cross, i.e between a wildtype with a heterozygous mice, i.e RrMm * RrMm, only one among the 16 ofsprings will tailless and will have the genotype of 'rrmm' .

hence the probabilty of getting a tailless offspring = (9/16) * (4/9) *(1/16) = 1/64

NB: * Represents crossing

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