Answer/Explanation:
Scenario 1
What are the allele frequencies (Y and y)?
490 people have the genotype YY, 420 people have the genotype Yy, and 90 people have the genotype yy.
The allele frequency of Y is 490 (from the YY group) + <em>half </em>of 420 (from the Yy group) = 700. There are 1000 people total, so the allele frequency of Y is 700/1000 = 0.7 (70%)
The allele frequency of y is 90 (from the yy group) + <em>half </em>of 420 (from the Yy group) = 300. There are 1000 people total, so the allele frequency of y 300/1000 = 0.3 (30%)
This checks out, as the equation for the allele frequencies for 2 alleles of a gene is p + q = 1.0. In this case, Y + y = 1 (as 0.7 + 0.3 = 1)
What are the expected genotype frequencies (YY, Yy, and yy)?
The expected genotypes based on genotype frequencies p and q are calculated from the following equation:
p² + 2pq + q² = 1 (or 100%).
Therefore, the expected frequency of YY (p²) is 0.7² = 0.49 (or 49%). For a population of 1000, that means we expect 490 YY.
The expected frequency of Yy (2pq) is 2 x 0.3 x 0.7 = 0.42 (or 42%) For a population of 1000, that means we expect 420 Yy.
The expected frequency of yy (q²) is 0.3² = 0.09 (or 42%) For a population of 1000, that means we expect 90 yy.
Is the population in equilibrium? Yes or no?
This matches up with our observations exactly (and the total of the values 0.49 + 0.42 + 0.09 = 1), meaning that yes, the population is in Hardy Weinberg equilibrium.
Scenario 2
What are the allele frequencies (Y and y)?
60 people have the genotype YY, 40 people have the genotype Yy, and 0 people have the genotype yy.
The allele frequency of Y is 60 (from the YY group) + <em>half </em>of 40 (from the Yy group) = 80. There are 100 people total (60 + 40 + 0), so the allele frequency of Y is 80/100 = 0.8 (80%)
The allele frequency of y is 0 (from the yy group) + <em>half </em>of 40 (from the Yy group) = 20. There are 100 people total (60 + 40 + 0), so the allele frequency of Y is 20/100 = 0.2 (20%)
What are the expected genotype frequencies (YY, Yy, and yy)?
The expected genotypes based on genotype frequencies p and q are calculated from the following equation:
p² + 2pq + q² = 1 (or 100%).
Therefore, the expected frequency of YY (p²) is 0.8² = 0.64 (or 64%). For a population of 100, that means we expect 64 YY.
The expected frequency of Yy (2pq) is 2 x 0.8 x 0.2 = 0.32 (or 32%) For a population of 100, that means we expect 32 Yy.
The expected frequency of yy (q²) is 0.2² = 0.04 (or 4%) For a population of 100, that means we expect 4 yy.
Is the population in equilibrium? Yes or no?
No, the population deviates from Hardy Weinberg equilibrium, because we are missing any individuals with a yy genotype. This means the observed frequencies are not true of the expected frequency
If the population is not in equilibrium, which of the assumptions of Hardy-Weinberg do you think may have been violated?
The assumption that has likely been violated is that the population is large. This population is small, and is a small sample of the original population, meaning genetic variation is reduced.
