Answer:
Probability that both children will be phenotypically identical with regards to skin color is 10/16.
Explanation:
The couple has a heterozygous genotype for the recessive allele. Representing the recessive allele with 'a' and the dominant allele with 'A', we can write the heterozygous genotype for the couple as Aa.
The gene pool of the children for this couple can be represented as:
Alleles A a
A AA Aa
a Aa aa
There can result 4 combinations of alleles AA,Aa,Aa and aa. The recessive allele is responsible for albinism so the child with the allele set 'aa' will have albinism while the others, AA,Aa and Aa will show no signs of albinism since the recessive allele is responsible for it.
The probability that both children will be phenotypically identical can be calculated by first considering that the children show albinism and then consider the probability that they have a normal phenotype.
Probability that a child will have albinism (aa) = 1/4
Probability that a child will not have albinism (AA, Aa and Aa) = 3/4
Probability that both children have albinism = 1/4 x 1/4 = 1/16
Probability that both children will have no albinism = 3/4 x 3/4 = 9/16
Probability that both children will be phenotypically identical = 1/16 + 9/16
= 10/16
Probability that both children will be phenotypically identical with regards to skin color is 10/16.
