A couple decides to have children. Both have the same genotype: they are fully heterozygous at...

A couple decides to have children. Both have the same genotype: they are fully heterozygous at the following loci: albino (Aa), MN blood group (L^M L^N), and Rh factor (Rh+/Rh-). The Rh+ allele is dominant to the Rh- allele. Assuming that the genes encoding these traits are located on different autosomes, determine the excepted genotypic and phenotypic of their ratios in their offspring.

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Expert Solution

In albino a person having a compleate loss of pigment melanin in the skin, hair and the eyes. Albinism causes due to inheritance of recessive gene alleles aa. therefore cross between heterozygous cross between Aa * Aa produces 1/4 AA, 1/2 Aa, 1/4aa individual with genotype aa will have albino. therefore phenotypic ratio will be 3 : 1 and genotypic ratio will be 1:2:1.

In case of MN blood group, is the case of codominance where both M and N alleles are equally dominant. therefore cross between L^M L^N * L^M L^N produces 1/4 L^ML^M, 1/2 L^ML^N, 1/4 L^NL^N, Where individual having genotype with L^ML^N will have MN blood group. Hence, both genotypic and phenotypic ratio will be 1:2:1.

in 3rd case Rh factor, an individual having Rh factor on the surface of blood are Rh + and an individual without Rh factor is Rh-. Therefore, Rh + may recieve both Rh+ and Rh- but, Rh- individual only receive Rh- blood group. hence, expected phenotypic ratio of heterozygous cross will be 3:1 and genotypic ratio will be 1:2:1.

gladiator answered 1 year ago

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