What Do You Know About Population Genetics? Trivia Quiz

1. Hardy-Weinberg equilibrium assumes that in the population.

No selection is occurring.

No mutation is occurring

No migration is occurring

Mating is random

The mating population is very large

All of the above

The Hardy-Weinberg equilibrium assumes that in the population, there is no selection occurring, meaning that there is no advantage or disadvantage for certain traits. It also assumes that no mutation is occurring, meaning that there are no new genetic variations being introduced. Additionally, it assumes that no migration is occurring, meaning that there is no movement of individuals in or out of the population. It also assumes that mating is random, meaning that individuals choose their mates without any preference for specific traits. Lastly, it assumes that the mating population is very large, which helps ensure that genetic drift does not significantly change the allele frequencies.

Explanation

The Hardy-Weinberg equilibrium assumes that in the population, there is no selection occurring, meaning that there is no advantage or disadvantage for certain traits. It also assumes that no mutation is occurring, meaning that there are no new genetic variations being introduced. Additionally, it assumes that no migration is occurring, meaning that there is no movement of individuals in or out of the population. It also assumes that mating is random, meaning that individuals choose their mates without any preference for specific traits. Lastly, it assumes that the mating population is very large, which helps ensure that genetic drift does not significantly change the allele frequencies.

2. If the frequency of the a allele is 0.1, then, if the population is at Hardy-Weinberg equilibrium, the percentage of individuals that are AA should be ______%

In a population at Hardy-Weinberg equilibrium, the frequency of the homozygous genotype AA can be calculated by squaring the frequency of the A allele. In this case, the frequency of the A allele is 0.1, so squaring it gives 0.01. To convert this to a percentage, we multiply by 100, resulting in 1%. Therefore, the correct answer is 1%, not 81,81%.

Explanation

In a population at Hardy-Weinberg equilibrium, the frequency of the homozygous genotype AA can be calculated by squaring the frequency of the A allele. In this case, the frequency of the A allele is 0.1, so squaring it gives 0.01. To convert this to a percentage, we multiply by 100, resulting in 1%. Therefore, the correct answer is 1%, not 81,81%.

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3. If the frequency of the a allele is 0.1, then, if the population is at Hardy-Weinberg equilibrium, the percentage of individuals that are Aa should be ______%

If the frequency of the "a" allele is 0.1, then in a population at Hardy-Weinberg equilibrium, the percentage of individuals that are heterozygous (Aa) can be calculated using the equation 2pq, where p is the frequency of the dominant allele (A) and q is the frequency of the recessive allele (a). In this case, p = 0.9 (1 - 0.1) and q = 0.1. Plugging these values into the equation, we get 2 * 0.9 * 0.1 = 0.18 or 18%. So, the percentage of individuals that are Aa is 18%.

Explanation

If the frequency of the "a" allele is 0.1, then in a population at Hardy-Weinberg equilibrium, the percentage of individuals that are heterozygous (Aa) can be calculated using the equation 2pq, where p is the frequency of the dominant allele (A) and q is the frequency of the recessive allele (a). In this case, p = 0.9 (1 - 0.1) and q = 0.1. Plugging these values into the equation, we get 2 * 0.9 * 0.1 = 0.18 or 18%. So, the percentage of individuals that are Aa is 18%.

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4. If the frequency of the a allele is 0.1, then, if the population is at Hardy-Weinberg equilibrium, the percentage of individuals that are aa should be ______%

If the frequency of the a allele is 0.1, then in a population at Hardy-Weinberg equilibrium, the percentage of individuals that are aa can be calculated using the equation p^2, where p is the frequency of the a allele. In this case, p = 0.1. Therefore, the percentage of individuals that are aa would be 0.1^2 = 0.01, which is equivalent to 1%.

Explanation

If the frequency of the a allele is 0.1, then in a population at Hardy-Weinberg equilibrium, the percentage of individuals that are aa can be calculated using the equation p^2, where p is the frequency of the a allele. In this case, p = 0.1. Therefore, the percentage of individuals that are aa would be 0.1^2 = 0.01, which is equivalent to 1%.

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5. If, in a population at Hardy-Weinberg equilibrium, 4% of the individuals are aa, then the percentage of AA individual should be ______%

In a population at Hardy-Weinberg equilibrium, the percentage of AA individuals can be determined by subtracting the percentage of aa individuals from 100%. Since 4% of the individuals are aa, the remaining percentage of individuals must be made up of AA and Aa genotypes. Therefore, the percentage of AA individuals is 100% - 4% = 96%.

Explanation

In a population at Hardy-Weinberg equilibrium, the percentage of AA individuals can be determined by subtracting the percentage of aa individuals from 100%. Since 4% of the individuals are aa, the remaining percentage of individuals must be made up of AA and Aa genotypes. Therefore, the percentage of AA individuals is 100% - 4% = 96%.

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