PKU Carrier Frequency Notes (Hardy-Weinberg)
Autosomal recessive inheritance of PKU
- PKU (phenylketonuria) is an autosomal recessive genetic disorder.
- Affected individuals have genotype aa (two recessive alleles).
- Carriers are heterozygous Aa (one PKU allele, one normal allele). The dominant normal allele does not cause PKU when paired with a recessive allele.
Hardy-Weinberg basics for allele frequencies
- Let q be the frequency of the recessive PKU allele in the population.
- Let p be the frequency of the normal allele, where p + q = 1.
- Under Hardy-Weinberg equilibrium, genotype frequencies are:
- p^2 for homozygous normal (AA)
- 2pq for heterozygous carriers (Aa)
- q^2 for homozygous recessive affected (aa)
- The sum p^2 + 2pq + q^2 = 1.
Given allele frequencies and calculation used in the example
- In the example, the PKU allele frequency is given as q = 0.01 (1%).
- Therefore, p = 1 - q = 0.99.)
- The carrier frequency is 2pq = 2\cdot 0.99 \cdot 0.01 = 0.0198\,.
- This equals approximately 1.98\%, which is about 2\% of the population.
- The affected frequency is q^2 = (0.01)^2 = 0.0001 = 0.01\%.
Step-by-step calculation
- Start with the allele frequencies: p = 1 - q = 1 - 0.01 = 0.99, \, q = 0.01.
- Compute carrier frequency: 2pq = 2(0.99)(0.01) = 0.0198 \approx 1.98\% \approx 2\%.
- Compute affected frequency: q^2 = (0.01)^2 = 0.0001 = 0.01\%.
- Summarize: about 2% carriers, about 0.01% affected, under Hardy-Weinberg assumptions.
Numerical examples and interpretation
- If you apply this to a population of N individuals:
- Expected number of carriers ≈ 0.0198 \times N.
- Example: In 100,000 people, carriers ≈ 1,980; affected ≈ 10.
- If two random individuals who are both carriers have a child, the probabilities are:
- AA (unaffected, non-carrier): \tfrac{1}{4}
- Aa (carrier): \tfrac{1}{2}
- aa (affected): \tfrac{1}{4}
- Therefore, if two carriers mate, there is a 25% chance the child is affected.
Genotype frequencies under Hardy-Weinberg for this scenario
- With p = 0.99 and q = 0.01:
- Homozygous normal: p^2 = (0.99)^2 = 0.9801 \,(98.01\%)
- Heterozygous carriers: 2pq = 0.0198 \,(1.98\%)
- Homozygous recessive affected: q^2 = 0.0001 \,(0.01\%)
Assumptions and caveats
- Assumes Hardy-Weinberg equilibrium: large population, random mating, no selection, no migration, no mutation, no genetic drift.
- Real populations may deviate due to:
- Population substructure and ethnic variation in allele frequencies.
- Non-random mating patterns.
- Selection pressures (though carriers often have no PKU phenotype).
- Migration and demographic changes.
- The 1% allele frequency and the resulting 2% carrier frequency are approximate for illustrative calculation.
Connections to foundational principles and real-world relevance
- Demonstrates the use of Hardy-Weinberg to estimate carrier and affected frequencies from allele frequencies.
- Connects to population genetics, genetic screening, and public health planning.
- Practical implications include:
- Screening programs to identify carriers in populations with higher frequencies.
- Genetic counseling for prospective parents who are carriers.
- Prenatal and newborn screening for PKU to enable early dietary management.
Ethical, philosophical, and practical implications
- Screening access and equity: ensuring diverse populations have access to testing.
- Privacy and potential discrimination based on genetic information.
- Reproductive decision-making influenced by carrier status and available options.
- The balance between public health benefits and individual autonomy.
Quick recap and takeaways
- PKU is autosomal recessive; carriers are Aa, affected are aa.
- Under Hardy-Weinberg: p^2 + 2pq + q^2 = 1 with p = 1 - q.
- For q = 0.01: carrier frequency ≈ 2pq = 2(0.99)(0.01) = 0.0198 \approx 1.98\% \approx 2\%.
- Affected frequency ≈ q^2 = (0.01)^2 = 0.0001 \approx 0.01\%.
- Real-world values vary by population; use as approximate guidance for population-level expectations.