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

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

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Answer

AA: 360 (0.36), Aa: 480 (0.48), aa: 160 (0.16)

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.

Step 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*} AA: &p^2 N \\ Aa: &2pq N \\ 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*} AA: &(0.6)^2 \times 1000 = 0.36 \times 1000 = 360 \\ Aa: &2 \times 0.6 \times 0.4 \times 1000 = 0.48 \times 1000 = 480 \\ aa: &(0.4)^2 \times 1000 = 0.16 \times 1000 = 160 \end{align*}

Frequencies: AA = 3601000=0.36\frac{360}{1000} = 0.36, Aa = 0.480.48, aa = 0.160.16.
AA: 360 (0.36), Aa: 480 (0.48), aa: 160 (0.16)

Step 2: New allele frequencies after 100 Aa die before reproducing.
Survivors:

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

Step 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*} AA: &(p')^2 \\ Aa: &2p'q' \\ 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*} AA: &\left( \frac{11}{18} \right)^2 = \frac{121}{324} \approx 0.373 \\ Aa: &2 \times \frac{11}{18} \times \frac{7}{18} = \frac{154}{324} = \frac{77}{162} \approx 0.475 \\ 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}

Step 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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Hardy-Weinberg principle: In a large randomly-mating population experiencing no evolutionary forces (no selection, mutation, migration, or genetic drift), allele frequencies (p for A, q = 1 - p for a) and genotype frequencies (p^2 for AA, 2pq for Aa,…

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 (p for A, q = 1 - p for a) and genotype frequencies (p^2 for AA, 2pq for Aa, q^2 for aa) remain constant across generations. Step 1: Calculate initial genotype numbers and frequencies (N = 1000, p = 0.6, q = 0.4). Formula: AA: &p^2 N \\ Aa: &2pq N \\ aa: &q^2 N Substitute: AA: &(0.6)^2 × 1000 = 0.36 × 1000 = 360 \\ Aa: &2 × 0.6 × 0.4 × 1000 = 0.48 × 1000 = 480 \\ aa: &(0.4)^2 × 1000 = 0.16 × 1000 = 160 Frequencies: AA = (360)/(1000) = 0.36, Aa = 0.48, aa = 0.16. AA: 360 (0.36), Aa: 480 (0.48), aa: 160 (0.16) Step 2: New allele frequencies after 100 Aa die before reproducing. Survivors: AA: &360 \\ Aa: &480 - 100 = 380 \\ aa: &160 Total survivors N' = 360 + 380 + 160 = 900. Total alleles = 2 × 900 = 1800. A alleles: 2 × 360 + 380 = 720 + 380 = 1100. p' = (1100)/(1800) = (11)/(18) ≈ 0.6111, q' = (700)/(1800) = (7)/(18) ≈ 0.3889 p' = (11)/(18) ≈ 0.611, q' = (7)/(18) ≈ 0.389 Step 3: Expected genotype frequencies after random mating (p' = (11)/(18), q' = (7)/(18)). Formula: AA: &(p')^2 \\ Aa: &2p'q' \\ aa: &(q')^2 Substitute and simplify: AA: &( (11)/(18) )^2 = (121)/(324) ≈ 0.373 \\ Aa: &2 × (11)/(18) × (7)/(18) = (154)/(324) = (77)/(162) ≈ 0.475 \\ aa: &( (7)/(18) )^2 = (49)/(324) ≈ 0.151 AA: (121)/(324) ≈ 0.373, Aa: (77)/(162) ≈ 0.475, aa: (49)/(324) ≈ 0.151 Step 4: Is the population in Hardy-Weinberg equilibrium after selection (before mating)? Genotype frequencies after selection: AA: (360)/(900) = 0.400, Aa: (380)/(900) ≈ 0.422, aa: (160)/(900) ≈ 0.178. Expected under HWE (using p' = (11)/(18)): AA ≈ 0.373, Aa ≈ 0.475, aa ≈ 0.151. Observed ≠ 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)