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Describe the duties of God preparation of Moses for leadership Genesis 2

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Hardy-Weinberg principle:
In a large randomly-mating population experiencing no evolutionary forces (no selection, mutation, migration, or genetic drift), allele frequencies (pp for A, q=1−pq = 1 - p for a) and genotype frequencies (p2p^2 for AA, 2pq2pq for Aa, q2q^2 for aa) remain constant across generations.

  1. Calculate initial genotype numbers and frequencies (N=1000N = 1000, p=0.6p = 0.6, q=0.4q = 0.4).

    Formula:

    AA: p2NAa: 2pqNaa: q2N\begin{align*} \text{AA: } &p^2 N \\ \text{Aa: } &2pq N \\ \text{aa: } &q^2 N \end{align*}

    Substitute:

    AA: (0.6)2×1000=0.36×1000=360Aa: 2×0.6×0.4×1000=0.48×1000=480aa: (0.4)2×1000=0.16×1000=160\begin{align*} \text{AA: } &(0.6)^2 \times 1000 = 0.36 \times 1000 = 360 \\ \text{Aa: } &2 \times 0.6 \times 0.4 \times 1000 = 0.48 \times 1000 = 480 \\ \text{aa: } &(0.4)^2 \times 1000 = 0.16 \times 1000 = 160 \end{align*}

    Frequencies: AA = 3601000=0.36\frac{360}{1000} = 0.36, Aa = 0.48, aa = 0.16.
    AA: 360 (0.36), Aa: 480 (0.48), aa: 160 (0.16)

  2. New allele frequencies after 100 Aa die before reproducing.

    Survivors:

    AA: 360Aa: 480−100=380aa: 160\begin{align*} \text{AA: } &360 \\ \text{Aa: } &480 - 100 = 380 \\ \text{aa: } &160 \end{align*}

    Total survivors N′=360+380+160=900N' = 360 + 380 + 160 = 900.
    Total alleles = 2×900=18002 \times 900 = 1800.
    A alleles: 2×360+380=720+380=11002 \times 360 + 380 = 720 + 380 = 1100.

    p′=11001800=1118≈0.6111,q′=7001800=718≈0.3889p' = \frac{1100}{1800} = \frac{11}{18} \approx 0.6111, \quad q' = \frac{700}{1800} = \frac{7}{18} \approx 0.3889

    p' = \frac{11}{18} \approx 0.611, \quad q' = \frac{7}{18} \approx 0.389

  3. Expected genotype frequencies after random mating (p′=1118p' = \frac{11}{18}, q′=718q' = \frac{7}{18}).

    Formula:

    AA: (p′)2Aa: 2p′q′aa: (q′)2\begin{align*} \text{AA: } &(p')^2 \\ \text{Aa: } &2p'q' \\ \text{aa: } &(q')^2 \end{align*}

    Substitute and simplify:

    AA: (1118)2=121324≈0.373Aa: 2×1118×718=154324=77162≈0.475aa: (718)2=49324≈0.151\begin{align*} \text{AA: } &\left( \frac{11}{18} \right)^2 = \frac{121}{324} \approx 0.373 \\ \text{Aa: } &2 \times \frac{11}{18} \times \frac{7}{18} = \frac{154}{324} = \frac{77}{162} \approx 0.475 \\ \text{aa: } &\left( \frac{7}{18} \right)^2 = \frac{49}{324} \approx 0.151 \end{align*}

    AA: \frac{121}{324} \approx 0.373, \quad Aa: \frac{77}{162} \approx 0.475, \quad aa: \frac{49}{324} \approx 0.151

  4. Is the population in Hardy-Weinberg equilibrium after selection (before mating)?

    Genotype frequencies after selection:
    AA: 360900=0.400\frac{360}{900} = 0.400, Aa: 380900≈0.422\frac{380}{900} \approx 0.422, aa: 160900≈0.178\frac{160}{900} \approx 0.178.
    Expected under HWE (using p′=1118p' = \frac{11}{18}): AA ≈0.373\approx 0.373, Aa ≈0.475\approx 0.475, aa ≈0.151\approx 0.151.
    Observed ≠\neq expected, so not in H-W equilibrium (selection disrupted genotype frequencies). After random mating, next generation will be in HWE.
    No (selection changed genotype frequencies; random mating restores HWE next generation)

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