IGA 8/e Chapter 2

20 Chapter Twopedigree, however, the genotypes can be stated only as G/– for both Johnand Martha.The probability that John is carrying the allele for galactosemia is 2 / 3 ,rather than the 1 / 2 that you might guess. To understand this, recall thatJohn’s parents must be heterozygous in order to have a child with therecessive disorder while still being normal themselves (the assumption ofnormalcy is based on the information given in the problem). John’sparents were both G/g. A Punnett square for their mating would be:FatherGgMotherGgG/GG/gG/gg/gThe cross is:P G/g G/gF 1 g/g John’s brotherG/– John (either G/G or G/g)The expected ratio of the F 1 is 1 G/G : 2 G/g : 1 g/g. Because John doesnot have galactosemia (an assumption based on the information given inthe problem), he can be either G/G or G/g, which occurs at a ratio of 1:2.Therefore, his probability of carrying the g allele is 2 / 3 .The probability that Martha is carrying the g allele is based on thefollowing chain of logic. Her great-grandmother had galactosemia, whichmeans that she had to pass the allele to Martha’s grandparent. Because theproblem states nothing with regard to the grandparent’s phenotype, it mustbe assumed that the grandparent was normal, or G/g. The probability thatthe grandparent passed it to Martha’s parent is 1 / 2 . Next, the probabilitythat Martha’s parent passed the allele to Martha is also 1 / 2 , assuming thatthe parent actually has it. Therefore, the probability that Martha’s parenthas the allele and passed it to Martha is 1 / 2 1 / 2 , or 1 / 4 .In summary:John p(G/G) = 1 / 3p(G/g) = 2 / 3Martha p(G/G) = 3 / 4p(G/g) = 1 / 4This information does not fit easily into a Punnett square.