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25 June, 23:40

A population with 65 A1A1 individuals, 55 A1A2 individuals, and 10 A2A2 individuals is in Hardy-Weinberg Equilibrium. Hint: For this problem assume if the expected value is within 3 of the observed it's in HWE.

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  1. 26 June, 01:15
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    Yes, the population is in Hardy Weinberg equilibrium.

    Explanation:

    The hardy Weinberg equilibrium must have expected and observed frequency nearly equal.

    The individual has two alleles.

    Expected frequency is as follows:

    65 A1A1 individuals, A1 individuals = 130.

    55 A1A2 individuals, A1A2 individuals = 55 A1 and 55A2.

    10 A2A2 individuals, A2 individuals = 20.

    Total alleles = A1 + A2 = 130 + 55 + 55 + 20 = 260.

    Progeny = 130.

    Fr (A1) = 185/260 = 0.71.

    Fr (A2) = 75/260 = 0.29.

    Observed frequency:

    A1 can be calaculated by = 0.71*0.71 * 130 = 65.53.

    A2 can be calculted by = 0.29*0.29*130 = 11

    A1 and A2 = 2*0.71*0.29*130 = 53.53.

    Total = 130.

    Since, the observed and expected frequency are nearly equal to 130.

    Thus, the population is in Hardy Weinberg equilibrium.
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