IGA 8/e Chapter 2

Chapter Two 41independent variables in the data. Generally, the degrees of freedom can becalculated as the number of phenotypes in the problem minus one. In thisexample, there are two phenotypes (purple and blue) and therefore, one degreeof freedom. Using the table, you will find that the data have a p value that isbetween 0.025 and 0.01. The p value is the probability that the deviation of theobserved from that expected is due to chance alone. The relative standardcommonly used in biological research for rejecting a hypothesis is p < 0.05. Inthis case, the data do not support the hypothesis.Unpacking the Problem71. A man’s grandfather has galactosemia, a rare autosomal recessive diseasecaused by the inability to process galactose, leading to muscle, nerve, andkidney malfunction. The man married a woman whose sister had galactosemia.The woman is now pregnant with their first child.a. Draw the pedigree as described.b. What is the probability that this child will have galactosemia?c. If the first child does have galactosemia, what is the probability that asecond child will have it?Solution to the ProblemAnswer:a. Galactosemia pedigree?b. Both parents must be heterozygous for this child to have a 1 / 4 chance ofinheriting the disease. Since the mother’s sister is affected withgalactosemia, their parents must have both been heterozygous. Since themother does not have the trait, there is a 2 / 3 chance that she is a carrier(heterozygous). One of the father’s parents must be a carrier since hisgrandfather had the recessive trait. Thus, the father had a 1 / 2 chance ofinheriting the allele from that parent. Since these are all independent events,the child’s risk is:1/ 4 2 / 3 1 / 2 = 1 / 12c. If the child has galactosemia, both parents must be carriers and thus thoseprobabilities become 100 percent. Now all future children have a 1 / 4 chanceof inheriting the disease.