Question 152827: Please Help! I am not sure how to put this into an equations
Address the following Hardy-Weinberg Equilibrium problem.
1)A population of iguanas on the Galapagos Islands has 20 individuals with webbed feet and 180 individuals with the dominant trait of non-webbed feet. What are the frequencies of webbed feet and non-webbed feet traits in this population?
2)How is the dominant allele frequency calculated from this population?
3)Calculate the genotype frequencies: p2 -the frequency of the dominant homozygotes; 2pq –the frequency of heterozygotes; and q2 –the frequency of the recessive homozygotes (hint: insert the values calculated for p and q such that p2+2pq+q2=1.0). Explain why p2 and q2 represent the proportion of dominant and recessive homozygotes, respectively, and why 2pq represents the proportion of heterozygotes (hint: consider the law of independent assortment and the laws of probability).
4)Is the population in Hardy-Weinberg Equilibrium? Why or why not?
Answer by mducky2(62)   (Show Source):
You can put this solution on YOUR website! This problem requires some algebra and biology.
Part 1:
First, we have to find the total population:
20 iguanas with webbed feet + 180 iguanas with non-webbed feet = 200 iguanas total
Now we can find the frequencies of each trait, which is the number of iguanas with that trait over the total number of iguanas:
webbed feet: 20/200 = 0.1 or 10%
non-webbed feet: 180/200 = 0.9 or 90%
Part 2:
The dominant allele of non-webbed feet (F) can be calculated when we find the frequency of the recessive allele of webbed feet (f). The recessive allele frequency can be calculated from the genotype frequency using the Hardy-Weinberg equation.
We already know q2 is 0.1, since any iguanas that show the recessive gene must have both recessive alleles. Since the allele frequency is q, we just need to solve for q:
Now we can find the frequency of non-webbed feet by solving for p. Since there are only two types of feet, we know that the sum of both is 1. And we can solve:
Part 3:
Solving for the other genotypes is fairly easy now that we know p and q.
We already know that the frequency of recessive homozygotes, q2, is 0.1 from before.
Plugging in the numbers to solve for the frequency of dominant homozygotes:
Plugging in the numbers to solve for the frequency of heterozygotes:
The reason why they are this way is because of the way the genes mix in the next generation with the Punnett square:
Basically, p2 is FF, or the frequency of the dominant homozygotes. In the same way, q2 is ff, or the frequency of the recessive homozygotes. The Ff can be found on the two other corners as the sum of pq + pq, or 2pq.
Part D:
We don't know what the next generation actually was like, so we can't answer this question. If you knew there was non-random mating, selection, mutation, migration, or genetic drift, then you would know that the population would not be in Hardy-Weinberg equilibrium.
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