Question

LAST ONE! HELP OR HELP QUIDE!

Answer 1

Sickle cell anemia is most common in sub-Saharan regions of Africa, where malaria is prevalent, than it is in regions where malaria is not common. This is because being heterozygous for sickle cell anemia in malaria-prone regions carries a fitness.

A new malaria vaccine was introduced and given to all the people of Population A, which is located in one of the regions where malaria is most prevalent. Since the government wished to test the effectiveness of the vaccine, Population A was isolated; therefore, there was no migration in or out of the population.

Every citizen of Population A was vaccinated when the experiment began, and every new baby was vaccinated immediately after birth. Allele frequencies were calculated every 20 years. After 100 years, 360 out of 1,000 people in Population A are homozygous recessive for sickle cell anemia (ss genotype).
q2 –
q –
p –
p2 –
2pq –

Answer 2

@jherrera57

Answer 3

What is the question????

Answer 4

Again these problems are really bad and can't be done. If a population is undergoing change (evolution) then the Hardy Weinberg equilibrium does not apply and you can't use the equation unless you know the frequency of one of the homozygotes AND the heterozygotes. You cannot assume that the homozygous recessives necessarily equal q2.

But, if you do go ahead and assume that they do = q2 (and all these twos are exponents not multipliers).

So, assuming that the homozygous recessives actual do = q2 then q2 is the proportion of that genotype in the population or 360 divided by 1,000 which is equal to 0.36.


IF q2 is 0.36, then the square root of that is 0.6 and that is q.

There are two alleles at this genetic locus in this population, so if one of them is 60%, then the other must be 40%, so p is 0.4. And, that is the frequency of the dominant allele.

If p is 0.4, then p2 is 0.16 and that is the frequency of the homozygous dominant.

The H-W equation is p2 + 2pq + q2 = 1

That is adding the frequencies of all the possible genotypes makes up all the population.

So 1 - 0.36 - 0.16 = 0.48 and that is the frequency of the heterozygotes.

OR, you can multiply 2pq and come up with the same answer.

You can't really say whether or not the population is changing unless you compare it at two different generations.