Galactosemia is an autosomal recessive disorder caused by the inability to metabolize galactose, a component of the lactose found in mammalian milk. galactosemia can be partially managed by eliminating dietary intake of lactose and galactose. amanda is healthy, as are her parents, but her brother alonzo has galactosemia. brice has a similar family history. he and his parents are healthy, but his sister brianna has galactosemia. amanda and brice are planning a family and seek genetic counseling. based on the information provided, complete the following activities and answer the questions. a) draw a pedigree that includes amanda, brice, their siblings, and parents. identify the genotype of each person, using g and g to represent the dominant and recessive alleles, respectively. b) what is the probability that amanda is a carrier of the allele for galactosemia? what is the probability that brice is a carrier? explain your reasoning for each answer. c) what is the probability that the first child of amanda and brice will have galactosemia? show your work. d) if the first child has galactosemia, what is the probability that two of the next three children will not have galactosemia? show your work and explain the reasoning for your answer.

a)Since, galactosemia is an autosomal recessive disorder, the person with this disorder will have gg genotype (both recessive alleles).

Amanda’s brother Alonzo has gg genotype (one allele from father and one from mother), while her parent are carriers with genotypes Gg.

Briana also has gg genotype and Brice’s parents Gg both.

Amanda’s parents cross: Gg x Gg

Offspring: GG Gg Gg gg  

b)  If both parent are heterozygous that means that there is 25% (1/4 is dominant homozygous) that Amanda is healthy and 50% (2/4 are heterozygous) that she is a carrier of the allele for galactosemia. But if we eliminate gg (because Amanda doesn’t have galactosemia) the ratios are 1/3 and 2/3 respectively. The same thing is with Brice.  

c)The probability that the first child of Amanda and Brice will have galactosemia is 2/3 (probability that Amanda is a carrier - hetrozygous) * 2/3 (probability that Brice is a carrier - heterozygous) * 1/4 (that child having galactosemia) = 1/9

d)If the first child has galactosemia, this means that both of the parents are carriers (Gg).

The chance that one child will not have galactosemia is ¾. The probability that two of the next three children will not have galactosemia is ¾ * ¾ * ¼ .

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