Question

Oculocutaneous albinism results from a mutation in one of several genes which codes for proteins important in the manufacturing of melanin. It is inherited via traditional Mendelian inheritance patterns in which normal melanin production is dominant and abnormal melanin production (or low/no melanin production) is recessive. Using the letters ‘A’ and ‘a’ to represent the 2 forms of this gene, complete the following. Show your work! You may use Punnett squares or probabilities.

1. What is the probability of a child having albinism when one parent has albinism and the other does not? Be sure to state both the expected genotype and phenotype ratios. (4pts)

2. What is the probability of a child having albinism if both parents are heterozygotes? Be sure to state both the genotype and phenotype ratios. (3pts)

Answer 1

It is given in the question that 'A' is the dominant allele for normal melanin production and 'a' is the recessive allele for the abnormal melanin production.

As it is following simple Mendelian inheritance,

AA/ Aa - will be individuals with normal melanin production

aa - will be individuals with abnormal melanin production

a) parents - phenotype-- albinism x normal melanin individual

genotype-- aa x Aa (because the child is having albinism, so must contribute 'a' allele)

gametes- a A, a

cross

	a
A	Aa - normal
a	aa - albinism

genotypic ratio - 1 Aa : 1 aa

phenotypic ratio - 1 normal melanin individual: 1 albinism

b) parents-- genotype-- Aa x Aa

(* given in the question)

gametes-- A, a A, a

cross

	A	a
A	AA- normal	Aa- normal
a	Aa - normal	aa- albinism

genotypic ratio - 1 AA : 2 Aa : 1 aa

phenotypic ratio - 3 normal individuals : 1 albnism

probability of child with albinism in the cross = 1/4 x 100 = 25%