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A population would not be in Hardy Weinberg equilibrium for a particular gene if it had:

a. assortative mating
b. immigration
c. A small population
d. All of the above

1 Answer

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Final answer:

Assortative mating, immigration, and a small population size would all violate Hardy-Weinberg equilibrium.

Step-by-step explanation:

Hardy-Weinberg equilibrium is a principle that states that allele frequencies in a population will remain constant in the absence of factors such as natural selection, mutation, genetic drift, and migration (gene flow). Therefore, a population would not be in Hardy-Weinberg equilibrium if any of these factors are present.

Assortative mating is a form of non-random mating where individuals with similar phenotypes are more likely to mate with each other. This would result in a deviation from the expected genotype frequencies and violate Hardy-Weinberg equilibrium.

Immigration, which is gene flow from other populations, can introduce new alleles into a population and change the allele frequencies, thereby violating Hardy-Weinberg equilibrium.

A small population size can lead to genetic drift, which is a random change in allele frequencies due to sampling error. This can cause the population to deviate from the expected Hardy-Weinberg proportions.

Therefore, the correct answer is d. All of the above would violate the conditions of Hardy-Weinberg equilibrium.

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