Solution-Draw the pedigree for this family

1) The following pedigree shows transmission of galactosemia, a rare autosomal recessive disease caused by the inability to metabolize the sugar galactose, resulting in muscle, nerve, and kidney malfunction.

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i) Write genotypes for as many individuals in the pedigree as possible. Write each genotype below the corresponding symbol in the pedigree.

ii) The individuals marked 1 and 2 are planning to have a child. They know about their family history of galactosemia, so they visit a genetic counselor to find out the probability that the child will have galactosemia. What does she tell them?

iii) The couple has a son and he does not have galactosemia. They decide to have a second child and again visit the genetic counselor. What does she tell them is the probability that the second child will have galactosemia?

iv) The couple’s second child is born, and he has galactosemia. They decide to have a third child and again visit the genetic counselor. What does she tell them is the probability that the third child will have galactosemia?

2) A man with brown teeth mates with a woman with normal white teeth. They have four daughters, all with brown teeth, and two sons, both with white teeth. The two sons both mate with women with white teeth, and each couple has one daughter and one son. All of these children have white teeth. One of the four brown-toothed daughters (A) mates with a man with white teeth (B), and they have two brown-toothed daughters, one white-toothed daughter, one brown-toothed son, and one white-toothed son.

i) Draw the pedigree for this family, showing the inheritance of brown teeth.

ii) Explain this pedigree by specifying the most likely mode of inheritance of brown teeth.

iii) Based on your answer to part (ii), what is the chance that the next child of the A-B couple will have brown teeth?

3) In fruit flies, vermilion eye color is a recessive trait determined by an X-linked gene. What genotype and phenotype ratios are expected from the following crosses? You can assume that the parents labeled ‘wild type’ in each cross are homozygous.

i) vermilion male x wild type female

ii) vermilion female x wildtype male

iii) daughter from mating (i) x wild type male

iv) daughter from mating (i) x vermilion male

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