1. Cystic Fibrosis is due to abnormal mucus-secreting tissue. Abnormal pancreatic secretions and respiratory problems will occur. This is an autosomal recessive disease. The mother and father are carriers for the disease. A. What is their genotype? B. What is their phenotype? C. What are the possible genotypes of offspring? D. What is the probability their children will have the disease? Show your work.
To answer these questions, we must first understand the genetics of cystic fibrosis (CF). CF is an autosomal recessive disease, which means that both copies of the gene responsible for CF must be mutated for an individual to have the disease.
A. To determine the genotype of the mother and father, we need to know that carriers of CF have one normal allele (denoted as "A") and one mutated allele (denoted as "a"). Since CF is autosomal recessive, carriers do not show any signs or symptoms of the disease.
Given that both the mother and father are carriers, their genotypes can be represented as follows:
Mother: Aa
Father: Aa
B. The phenotype refers to the observable traits or characteristics of an individual. In this case, the mother and father, being carriers, do not show any signs or symptoms of CF. Therefore, their phenotype can be described as "normal" or "healthy."
C. Now, let's consider the possible genotypes of their offspring. Since the mother and father are both carriers and each contributes one allele to their offspring, there are four possible combinations of alleles:
Offspring genotype possibilities:
1) AA (no CF mutation from either parent)
2) Aa (carriers like the parents)
3) Aa (carriers like the parents)
4) aa (inherited CF mutation from both parents)
D. To determine the probability of their children having CF, we need to consider the possible genotypes of the offspring from the previous answer.
The probability that a child inherits two normal alleles (AA) is 25% or 1/4.
The probability that a child inherits one normal allele and one mutated allele (Aa or Aa) is 50% or 2/4.
The probability that a child inherits two mutated alleles (aa), resulting in CF, is 25% or 1/4.
Therefore, the probability that their children will have CF is 1/4 or 25%.
This calculation is based on the assumption that the mother and father each only carry one CF mutation and that the alleles segregate independently during gamete formation and fertilization.