Practice Problem 16 Cationic trypsinogen is an enzyme made i

Practice Problem 16 Cationic trypsinogen is an enzyme made in the pancreas that helps to digest food. It is encoded for by the PRSS1 gene. Individuals with a disorder called hereditary pancreatitis have mutations in the PRSS1 gene, which result in inflammation of the pancreas, possibly leading to permanent tissue damage. The disorder is inherited in an a utosomal dominant fashion. Submit My Answers Give Up Correct Penetrance is the frequency with which individuals of a given genotype manifest at least some degree of a specific mutant phenotype associated with a trait. Penetrance does not refer to the range of expression of the mutant genotype hat is expressivity. Therefore, if a mutation is 86% penetrant, then 86% of individuals that possess the predisposing genotype would be expected to express the associated mutant phenotype to some degree, Part B Imagine that you are part of a research team interested in determining the penetrance of a different PRSS1 mutation, Ala16Val. Your team has identified 120 individuals with the PRSS1 Ala16Val allele and found that 48 of the individuals did not display any evidence of pancreatitis From your data, what is the penetrance of the PRSS1 Ala16Val mutation? Express your answer as a whole number. My Answers Give Up Part C If a man who is heterozygous for a PRSS1 Ala16Val mutation mates with a woman who is homozygous for normal cationic trypsinogen, what is the probability that their offspring will have hereditary pancreatitis, given the penetrance value obtained in Part B? Express your answer as a whole number. Submit My Answers Give Up

Solution

Cationic trypsinogen is the pancreatic enzyme encoded by the gene PRSS1

Hereditary pancreatitis people will have mutations in PRSS1 gene

Inflammation of pancreas leads to permanent tissue damage

Autosomal dominant inheritance

86% penetrance means 86% of individuals will have the predisposing phenotype expressed along with the mutant phenotype expressed to some degree

Part B

48 out of 120 individuals did not display any evidence of pancreatitis

120-48 = 72 individuals have penetrance of mutation

72 is 7200/120= 60% of the total number of individuals

So, the pancreatitis mutation Ala16Val, is 60% penetrant

Part C

Allele “M” represents mutation Ala16Val of pancreatitis

Cationic trypsinogen represents T allele

Male is heterozygous for the mutation, which is TM

Female is homozygous for normal gene, which is TT

TM X TT

TT -25%

TT-25%

MT – 25%

MT – 25%

According to the conventional method, 50% of the offsprings would have mutation inherited in full form.

But, if the pancreatitis mutation is 60% penetrant

Then

The remaining 10% of the offsprings would also have the mutation expressed in some degree apart from the 50% of the offsprings.

Therefore, 6 in 10 offsprings will have the mutation inherited.