Case Study – A Sickeningly Sweet Baby Boy
Part I Questions 1. What additional information would you want to know to understand Emma and Jacob’s panic?
To understand the cause of the panic that was brought to Jacob and Emma, you would need to know more about the state of the baby. This article just describes that the boy was having difficulty feeding, and after seven days he stopped feeding. This isn’t a situation that brings upon panic right away, but for Emma and Jacob it did. The panic was partly because they had already had a child that died from unknown reasons in the first nine days of his life, and didn’t want to lose another. The state of the current baby, such as if it was premature, or if it was very sick looking and…show more content…
Part II Questions
Pedigree charts are useful tools used by genetic counselors to look for the incidence of disease within multiple generation families. Each generation is shown on a separate row. 1. Label the pedigree chart below to explain the relationships and the disease incidence within this family. Be sure to include Emma, Jacob, Samuel, Matthew, Emma’s father, Emma’s mother, Emma’s aunts, Jacob’s mother, Jacob’s father, and Jacob’s aunt.
Please see Pedigree sheet: Case Study: A Sickeningly Sweet Baby Boy 2. Indicate on your pedigree chart the individuals who are carriers by shading half of each circle or square.
Please see Pedigree sheet: Case Study: A Sickeningly Sweet Baby Boy 3. Define the terms genotype, phenotype, homozygous and heterozygous.
Genotype: The make-up of alleles for a particular trait in an organism. For example, if black was a recessive trait in chickens, coded for by b, and there was a black chicken, its genotype would be bb.
Phenotype: The physical appearance of a trait in an organism. The phenotype of the above chicken is that it is black. A phenotype could be that a person has blue eyes, or curly hair.
Homozygous: Refers to expressing the same alleles for a particular trait. If black was dominant for the above chicken, and the alleles were coded for by B, if it was homozygous, the chicken’s genotype would be BB.