Inheritance of a Rare Kidney Disease

Determining the Inheritance Pattern

• Observation: Several individuals in the pedigree are affected, even though their parents are not.
  • Inference: The trait is likely a recessive trait.
• Equal Numbers Affected: Both males and females show the disease.
  • Inference: This suggests an autosomal recessive trait rather than an X-linked recessive.

Probability of Child Having the Disease

• Parents: Individuals 1 and 2 are considering having a child.
• Goal: Determine the probability that their child will have the disease.

Genotype Analysis

1. Definitions:

   • Alleles:
     • A1 - Normal
     • A2 - Disease
   • Genotypes:
     • A1/A1 - Normal phenotype
     • A2/A2 - Affected by disease
     • A1/A2 - Carrier, normal phenotype

2. Pedigree Observations:

   • Grandmother of Individual 1: A2/A2
   • Father of Individual 2: A2/A2
   • Individuals 1 & 2: Phenotypically normal (A1/_)
   • Individual 2: Must be A1/A2 since their father is A2/A2.
   • Individual 1's Father: Must have A2 allele due to affected grandmother.
   • Individual 1's Mother: Genotype uncertain, presumed A1/A1 given rarity of disease._

Probability Calculations

• Events Required for Child to be A2/A2:

  1. Mother (Individual 2) is A1/A2 (Probability = 1)
  2. Father (Individual 1) is A1/A2 (Probability = 1/2)
  3. Child receives A2 from both parents (Probability = 1/4)

• Combined Probability:

  • Using product rule:
    • Probability that both parents are A1/A2 and child is A2/A2:
    • (1 \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{8})

Scenario Analysis

• Rare Disease Assumption: Individual 1's Mother is A1/A1:
  • Probability child has disease: (\frac{1}{8})
• Alternate Scenario: If Individual 1's Mother is A1/A2:
  • Use Punnett square to determine probability of Individual 1 being a carrier (A1/A2):
    • New probability Individual 1 is A1/A2: (\frac{2}{3})
  • Probability child has disease: (\frac{1}{6})