Final answer:
Only Population 1 is in Hardy-Weinberg equilibrium because its observed genotype frequencies match the expected frequencies (p² for MM, 2pq for MN, q² for NN) calculated based on allele frequencies of M and N.
Step-by-step explanation:
To determine if a population is in Hardy-Weinberg equilibrium (HWE), we need to calculate the expected genotype frequencies based on allele frequencies and compare them to the observed frequencies. A population in HWE should meet the condition that the observed genotype frequencies match the expected frequencies derived from the allele frequencies.
Here's how the equilibrium is tested:
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- Calculate the allele frequencies (p and q) using the observed genotypic frequencies.
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- Use these allele frequencies to calculate the expected genotypic frequencies: p² for MM, 2pq for MN, and q² for NN.
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- Compare the expected frequencies with the observed frequencies; if they are roughly equal, the population is in HWE.
For each population given, let's calculate the allele frequency for M (p) using the formula p = (2*MM + MN) / (2*(MM + MN + NN)).
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- For Population 1, p = (2*25 + 50) / (2*(25 + 50 + 25)) = 0.5. Therefore, q = 1 - p = 0.5. The expected frequencies are p² = 0.25 for MM, 2pq = 0.5 for MN, and q² = 0.25 for NN. The observed frequencies match the expected frequencies, indicating Population 1 is in HWE.
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- For Population 2, it's impossible to calculate p as there are no MN individuals, which means this population is not in HWE.
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- For Population 3 and Population 4, the allele frequencies are calculated by the same method. The observed frequencies don't match the calculated expected frequencies, suggesting these populations are not in HWE.