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9 November, 14:03

In a large, randomly mating population with no forces acting to change gene frequencies, the frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000, or 0.09. What percentage of the population carries this trait but displays the dominant phenotype, short eyelashes

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  1. 9 November, 16:14
    0
    16.38%

    Step-by-step explanation:

    Given

    Frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000

    Let p = Probability of homozygous recessive individuals having character extra-long eyelashes = 0.09

    p + q = 1 where q = Probability of homozygous recessive individuals not having character extra-long eyelashes

    0.09 + p = 1

    p = 1 - 0.09

    p = 0.91

    The frequency of the carrier individual is calculated by npq

    Where n = mating partners = 2

    Frequency = 2 * 0.09 * 0.91

    Frequency = 0.1638 or 16.38%
  2. 9 November, 17:37
    0
    The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals which is

    0.42 or 42 %

    Step-by-step explanation:

    To solve the question, we note that

    p²+2pq+q²=1 and p + q = 1

    where

    p = frequency of the dominant allele

    q = frequency of the recessive allele

    p² = frequency of homozygous dominant allele

    q² = frequency of homozygous recessive allele

    2·p·q = frequency of hetrozygous individuals

    The frequency of the homozygous recessive individuals q² is 0.09

    Therefore the frequency of q = √ (0.09) = 0.3, therefore p = 1 - 0.3 = 0.7

    The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals or 2*p*q = Ae = 2*0.3*0.7 = 0.42

    → 0.42 * 100 = 42 %
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