A population will not exist in Hardly-Weinberg equilbrium if

Options

A. The population is large
B. Individuals mate selectively
C. There are no mutations
D. There is no migration

Answer

### Step-by-Step Solution:

1. **Understanding Hardy-Weinberg Equilibrium**:
- The Hardy-Weinberg equilibrium describes a state in which the allele and genotype frequencies in a population remain constant from generation to generation, provided that certain conditions are met.

2. **Conditions for Hardy-Weinberg Equilibrium**:
- A large population size (to prevent genetic drift).
- No natural selection (all individuals have equal chances of survival and reproduction).
- Random mating (individuals pair by chance, not by phenotype or genotype).
- No mutations (no new alleles are introduced into the gene pool).
- No migration (no individuals enter or leave the population, preventing gene flow).

3. **Analyzing the Question**:
- The question states that a population will not exist in Hardy-Weinberg equilibrium if certain conditions are met. We need to identify which condition disrupts the equilibrium.

4. **Evaluating Each Condition**:
- **Population is large**: This condition supports Hardy-Weinberg equilibrium, so it does not disrupt it.
- **Individuals mate selectively**: This condition disrupts Hardy-Weinberg equilibrium because selective mating leads to non-random mating, affecting allele frequencies.
- **No mutation**: This condition supports Hardy-Weinberg equilibrium, so it does not disrupt it.
- **No migration**: This condition supports Hardy-Weinberg equilibrium, so it does not disrupt it.

5. **Conclusion**:
- The population will not exist in Hardy-Weinberg equilibrium if individuals mate selectively. Therefore, the correct answer is that selective mating disrupts the equilibrium.