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Consider albino in Hong Local government where population of the affected individual is 1 in to 100,000, Using Hardy Weinberg formula to calculate 1. Heterozygous 2. Homozygous dominant 3. Recessive allele frequency.

Biology

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Here's how to solve this genetics problem using the Hardy-Weinberg principle:

The Hardy-Weinberg equations are:
Allele frequencies: p+q=1p + q = 1
Genotype frequencies: p2+2pq+q2=1p^2 + 2pq + q^2 = 1

Where:

  • pp = frequency of the dominant allele
  • qq = frequency of the recessive allele
  • p2p^2 = frequency of the homozygous dominant genotype
  • q2q^2 = frequency of the homozygous recessive genotype (affected individuals)
  • 2pq2pq = frequency of the heterozygous genotype (carriers)

Given: The frequency of affected individuals (albinos) is 1 in 100,000. Since albinism is a recessive trait, affected individuals have the homozygous recessive genotype.

  1. Determine the frequency of the homozygous recessive genotype (q2q^2).

    q2=1100,000=0.00001q^2 = \frac{1}{100,000} = 0.00001
  2. Calculate the recessive allele frequency (qq).

    q=q2=0.00001q = \sqrt{q^2} = \sqrt{0.00001} q≈0.003162q \approx 0.003162
  3. Calculate the dominant allele frequency (pp).

    Using the allele frequency equation p+q=1p + q = 1:

    p=1−q=1−0.003162p = 1 - q = 1 - 0.003162 p≈0.996838p \approx 0.996838

    Now, let's calculate the requested values:

    3. Recessive allele frequency
    This is the value of qq.

    q≈0.00316q \approx \boxed{\text{0.00316}}

    2. Homozygous dominant allele frequency
    This refers to the frequency of the homozygous dominant genotype (p2p^2).

    p2=(0.996838)2p^2 = (0.996838)^2 p2≈0.9937p^2 \approx \boxed{\text{0.9937}}

    1. Heterozygous allele frequency
    This refers to the frequency of the heterozygous genotype (2pq2pq).

    2pq=2×p×q2pq = 2 \times p \times q 2pq=2×0.996838×0.0031622pq = 2 \times 0.996838 \times 0.003162 2pq≈0.006302pq \approx \boxed{\text{0.00630}}
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