Question

In humans, normal pigmentation is dominant to albinism; and wet earwax is dominant to dry earwax....

In humans, normal pigmentation is dominant to albinism; and wet earwax is dominant to dry earwax. A man that is heterozygous for both traits marries a woman that is also heterozygous for both traits. What would be the phenotypic ratio of the offspring?

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Answer #1

Notations used :

A : Normal pigmentation

a : Albinism

B : Wet Earwax

b : Dry Earwax

Since A is dominant over a , and B is dominant over b, so -

Genotype of Heterozygous Man = AaBb

Genotype of Heterozygous Woman = AaBb

Gametes from Man = AB , Ab , aB and ab

Gametes from Woman = AB , Ab , aB and ab

This punnett square will depict that their crossover will yield :

АВ АВА, В APBB ABB, Ao BB ААВЬ АРЬ ААВЬ A. BB Ao Bь по 80 AaBb Aabb aaBb Ь Аовь АльЬ аа Вь aabb а) ab

The Phenotype of Offsprings will be :

Normal Pigmentation + Wet Earwax = 9 ( AABB, AABb, AaBB, AaBb, AABb, AABb, AaBB, AaBb, AaBb)

Normal Pigmentation + Dry Earwax = 3 (AAbb, Aabb, Aabb)

Albinism + Wet Earwax = 3 (aaBB, aaBb, aaBb)

Albinism + Dry Earwax = 1 (aabb)

So the phenotypic ratio will be 9 : 3 : 3 : 1 respectively, for this dihybrid cross.

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