To solve this problem, we need to determine the conditions under which a population will not be in Hardy–Weinberg equilibrium. The Hardy–Weinberg principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. These conditions include:$\newline \bullet$Random mating: Individuals pair by chance, not according to their genotypes or phenotypes.$\newline \bullet$No mutations: The alleles in the population do not change.$\newline \bullet$No migration: There is no movement of individuals into or out of the population.$\newline \bullet$Large population size: The population is sufficiently large to prevent random changes in allele frequencies.$\newline \bullet$No natural selection: All individuals have an equal chance of survival and reproduction.$\newline$Let's analyze each option:$\newline \bullet$Option 1: Individuals mate selectively.$\newline$ - Selective mating violates the condition of random mating, which is necessary for Hardy–Weinberg equilibrium.$\newline$ - Therefore, this option would cause the population to not be in Hardy–Weinberg equilibrium.$\newline \bullet$Option 2: There are no mutations.$\newline$ - This condition supports Hardy–Weinberg equilibrium, as mutations can alter allele frequencies.$\newline \bullet$Option 3: There is no migration.$\newline$ - This condition supports Hardy–Weinberg equilibrium, as migration can introduce new alleles into the population.$\newline \bullet$Option 4: The population is large.$\newline$ - A large population size supports Hardy–Weinberg equilibrium by minimizing genetic drift.$\newline$Therefore, the correct answer is Option 1: Individuals mate selectively. This condition disrupts Hardy–Weinberg equilibrium.