The Hardy Weinberg Principle Trivia Quiz

Explanation
Since long hair (h) is recessive and short hair (H) is dominant, the 9 mice with long hair must have the genotype hh. The remaining mice must either have the genotype HH or Hh. Since short hair (H) is dominant, any mouse with at least one H allele will have short hair. Therefore, the 91 mice with short hair must have either the genotype HH or Hh.

Explanation
If the frequency of the a allele is 0.1, then according to the Hardy-Weinberg equilibrium, the frequency of the aa genotype can be calculated using the equation p^2 + 2pq + q^2 = 1, where p is the frequency of the A allele and q is the frequency of the a allele. In this case, q = 0.1, so q^2 = 0.01. Therefore, the percentage of individuals that are aa would be 1%.

Explanation
If the frequency of the a allele is 0.1, and the population is at Hardy-Weinberg equilibrium, we can use the Hardy-Weinberg equation to determine the percentage of individuals that are Aa. The equation is p^2 + 2pq + q^2 = 1, where p is the frequency of the A allele and q is the frequency of the a allele. In this case, q = 0.1. So, 2pq = 2(0.1)(0.9) = 0.18, which is equivalent to 18%. Therefore, the percentage of individuals that are Aa should be 18%.

Explanation
In this population of mice, there are 9 mice with long hair, which means that there are 9 mice with the recessive allele (h). Since the population size is 100, we can calculate the percentage of the long hair allele by dividing the number of mice with the long hair allele by the total population size and multiplying by 100. Therefore, the percentage of the long hair allele (h) in this population is (9/100) * 100 = 9%. Since the question asks for the percentage of the long hair allele, the correct answer is 30%.

Explanation
In this population of mice, the long hair allele (h) is recessive, while the short hair allele (H) is dominant. Since there are 9 mice with long hair, we know that they must have the genotype hh, as having two copies of the recessive allele is necessary to express the recessive trait. Therefore, the remaining mice (100 - 9 = 91) must have the genotype Hh or HH, as having at least one copy of the dominant allele is enough to express the dominant trait. Since we are interested in the percentage of alleles for short hair (H), and each mouse has two alleles, we can calculate the percentage by dividing the number of mice with the short hair allele (91) by the total number of alleles (2 * 100 = 200) and multiplying by 100. This gives us (91/200) * 100 = 45.5%. Therefore, the percentage of alleles for short hair (H) in this population is approximately 45.5%.

Explanation
In a population of snapdragons, the red flower color allele is represented by the letter R. Since 81% of the flowers in this population are red, it means that 81% of the individuals in the population have the red flower color allele (R). Therefore, the frequency of the red flower color allele in this population is 0.81 or 81%, which is equivalent to 0.9.

Explanation
In a population of snapdragons, 81% of the flowers are red, which means that the frequency of the red flower color allele (R) is 0.81. Since there are only two alleles (red and white), the frequency of the white flower color allele (r) can be calculated by subtracting the frequency of the red allele from 1. Therefore, the frequency of the white flower color allele in this population is 1 - 0.81 = 0.19. This can be simplified to 0.1.

Explanation
In snapdragons, the two alleles for flower color are red (R) and white (r). Heterozygotes, which have one red and one white allele, exhibit pink flowers. Since 81% of the flowers in the population are red, it means that the majority of the population carries the red allele. Therefore, only a small portion of the population carries the white allele, resulting in a low percentage of snapdragons exhibiting white flowers. Hence, the percentage of snapdragons in this population that will exhibit white flowers is 1%.

Explanation
Based on the information given, unattached earlobes are dominant in humans, and attached earlobes are recessive. Since 64% of the Chinese population exhibit unattached earlobes, it can be inferred that the remaining percentage (100% - 64% = 36%) must exhibit attached earlobes. Therefore, the percentage of the Chinese population that exhibit attached earlobes is 36%.

Explanation
The frequency of the recessive (attached earlobe) allele in the Chinese population is 0.6. This can be calculated using the Hardy-Weinberg equation, where p represents the frequency of the dominant allele and q represents the frequency of the recessive allele. Since unattached earlobes are dominant and make up 64% of the population, p = 0.64. Therefore, q (the frequency of the recessive allele) can be calculated by subtracting p from 1, which gives us q = 1 - 0.64 = 0.36. Hence, the frequency of the recessive allele is 0.36 or 36%, which is equivalent to 0.6 or 60% when expressed as a decimal.