Question

HELP! or help guide!

Answer 1

Scenario #2

Jan Mayen Island is a tiny island in the Arctic Ocean, approximately 400 miles northeast of Iceland. It had an original population of 1,000, but due to a series of volcanic eruptions and earthquakes in the year 2010, much of the island’s population was killed, leaving only 200. It is now 80 years later and the population on Jan Mayen Island has reached 500. In the current population, 20 people are homozygous recessive for hemophilia C, which is autosomal (hh genotype).

q2 –
q –
p –
p2 –
2pq –
Scenario #2 Conclusion Questions

Use this information to help you answer the following questions.
The original population’s actual genotype frequencies were:

HH – 0.58; Hh – 0.26; hh – 0.16

1. Is the current population in Hardy-Weinberg equilibrium? Yes or No

2. Is the current population evolving? Yes or No

3. If the current population is evolving, what type of effect is responsible for genetic drift?

Answer 2

@jherrera57

Answer 3

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Answer 4

Don't know why your teachers keep giving these problems. You can't really do them unless you assume equilibrium. But assuming equilibrium, here goes:
20 out of 500 people are hh
that means that q2 is 20 divided by 500 is 0.04.
and q is the square root of that or 0.2.
q2 is the frequency of the recessive genotype
q is the frequency of the recessive allele
you have to keep genotype frequencies vs. allele frequencies straight.
all allele frequencies must add up to 100% or 1 because all the alleles make up all of that particular gene in the gene pool.
so, if you only have two alleles H and h, q stands for the h and p stands for the H, therefore
if q = 0.2 then p has to = 0.8 because they have to add up to 1.
So p2 = 0.64, that is 0.8 squared. And, that is the frequeny of HH genotype
All of the genotype frequencies also have to add up to 1.
So 2pq will be 1 - 0.04 - 0.64
That is 1 - 0.68 = 0.32 (or you can just multiply out 2pq

Answer 5

Then you compare your genotype frequencies to the original. If they are not the same then evolution is occurring. So, do HH at 0.64 and hh at 0.04 and Hh at 0.32 match the original or not?

Answer 6

OK, so the frequency of the recessive is going down and the frequency of the dominant is going up, so h is being selected against.
But again, you can't really do these problems unless you assume equilibrium and the population is NOT in equilibrium.