# The allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally...

## Question:

The allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally pigmented woman whose father is an albino marries an albino man whose parents are normal. They have three children, two normal and one albino. Give the genotypes for each person listed. Prove the answer.

Paternal grandfather, Paternal grandmother, maternal grandfather, maternal grandmother, father, mother, child 1, child 2, child 3

a. Show the cross of the paternal grandparents.

b. What are the possible genotypes of the maternal grandmother?

c. Show the cross of mother and father.

## Pedigree analysis

Pedigree analysis is done to study the inheritance of a trait through generations. It will allow to depict a relationship between the parents, offsprings and siblings along with their physical trait due to the type of alleles present in them.

We will solve this problem step wise and determine the genotype of all the individuals mentioned in it:

1. Women is normally pigmented but have a father who is albino. So, maternal grandfather will be of genotype aa as albino is a recessive condition which will be evident only when homozygous recessive alleles are present in genotype. Women is normal but will be carrier for albino as one recessive allele will come from the father so genotype of women will be Aa. Maternal grandmother will have genotype A_ as she can be either homozygous dominant or heterozygous dominant for this trait.
2. Man is albino so genotype will be aa. His parents are both carrier as he is albino because he inherited the recessive allele from both the parents and they are normal in phenotype. So, genotype of both paternal grandfather and paternal grandmother will be Aa.
3. Kids produced from the cross of man (aa) and women (Aa) will be as follows {eq}Aa*aa= 1Aa : 1aa {/eq}. Therefore, the genotype of normal children will be Aa and albino child will have genotype aa.

a. The cross of the paternal grandparents will be :

{eq}Aa (grandmother) * Aa (grandfather) = 1AA: 2Aa: 1aa(Man) {/eq}

b. The possible genotypes of the maternal grandmother is A_ as both Aa and AA is possible in this condition.

c. The cross of mother and father will be

{eq}Aa*aa= 1Aa : 1aa {/eq}.