In humans, aniridia, a type of blindness, is due to a dominant allele a. migraine headaches are due to another dominant allele m. if a man who suffers from both conditions (aamm) marries a woman who suffers from both (aamm), what are the chances of an offspring expressing both traits?

9/16

Explanation:

This is a typical dihybrid cross involving two genes; one coding for anridia (form of blindness) and the other for Migraine headache, both in humans. The allele for possessing anridia (A) is dominant over the allele for not possessing it (a) in the first gene, while the allele for having Migraine headache (M) is dominant over the allele for not having it (m) in the second gene.

A cross between a man that is heterozygous for both genes (AaMm) and a woman also heterozygous for the two genes (AaMm), this means that both parents are dominant for the two traits i.e. they both possess anridia and migraine headache.

Both parent cells will undergo meiosis to produce four possible combination of gametes in which the alleles are independent of one another in the gametes, according to Mendel's law of independent assortment. The gametes are: AM, Am, aM, am

These gametes are crossed using a punnet square (see attached image) to produce 16 possible offsprings in a phenotypic ratio of 9:3:3:1.

9- Offsprings with both anridia and migraine

3- Offsprings with only anridia but without migraine

3- Offsprings with only migraine but without anridia

1- Offsprings without both anridia and migraine

According to the question, the chance of an offspring expressing both traits i.e. having both anridia and migraine headache is 9/16

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