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

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)