Genetics Review: Pedigree Analysis and Mendelian Genetics

Q16. The following pedigree tracks a recessive condition (r).

Background

Topic: Pedigree Analysis & Mendelian Genetics

This question tests your ability to interpret a pedigree chart to determine genotypes and predict the probability of inheriting a recessive genetic condition. Pedigrees are visual representations of family relationships and the inheritance of traits across generations.

Key Terms and Concepts:

• Pedigree: A diagram that shows the occurrence and appearance of phenotypes of a particular gene or organism and its ancestors.

• Recessive allele (r): An allele that only expresses its phenotype when two copies are present (homozygous recessive).

• Carrier: An individual who has one dominant and one recessive allele (heterozygous) and does not show the trait but can pass it on.

• Genotype: The genetic makeup of an individual (e.g., RR, Rr, rr).

• Phenotype: The observable trait (e.g., affected or unaffected).

Step-by-Step Guidance

1. Examine the pedigree chart and identify which individuals are affected by the recessive condition (usually shaded shapes).

2. Recall that affected individuals must have the genotype $rr$ (homozygous recessive).

3. Unaffected individuals could be either $RR$ (homozygous dominant) or $Rr$ (heterozygous carriers). If they have affected children, they must be carriers ($Rr$).

4. For part (a), use the information from the pedigree to deduce the genotypes of individuals I-1 and I-2 based on their offspring's phenotypes.

5. For part (b), set up a Punnett square to determine the probability that a child of III-8 (whose genotype you deduced) and a carrier will inherit the condition.

Try solving on your own before revealing the answer!

Final Answer:

For part (a): I-1 and I-2 are both likely carriers ($Rr$) because they have affected children.

For part (b): The probability that their child will have the condition is 25% ($rr$) based on the Punnett square between III-8 (Rr) and a carrier (Rr).

This is because each parent can pass either the dominant or recessive allele, and only the combination $rr$ results in the condition.