A large population of laboratory animals has
been bred randomly for a number of generations.
There was an interesting revelation.
The scientists noted that there was neither
an increase nor a decrease in the frequency
of animals displaying the recessive trait ‘aa’;
i.e., it remained at 60% during several generations.
The remaining animals were dominant
homozygotes and heterozygotes for the dominant
phenotype.
What conclusion would you draw, if no
change in the frequency of the recessive trait
(aa) is observed over a period of time involving
several generations?
1. There has been sexual selection favoring
allele a.
2. The two phenotypes are about equally
adaptive under laboratory conditions. (correct answer)
3. There has been a high rate of mutation of
allele A to allele a.
4. The genotype is lethal.
5. The population is undergoing genetic
drift.
044 (part 2 of 3) 5.0 points
What would be the estimated frequency of
allele ‘a’ in the gene pool?
045 (part 3 of 3) 5.0 points
What would be the frequency of the dominant
heterozygotes (Aa) for this trait?
don't understand how to get answers for parts two and three.
The population is in equilibrium, so the frequency of the recessive genotype aa, 0.6, taking the square root to get the frequency of the allele (.775)
frequency of the dominant Aa then would be you need to work it yourself, see this sample:
http://www.k-state.edu/parasitology/biology198/answers1.html
ggbyty
To answer parts 2 and 3 of the question, we need to understand the concept of Hardy-Weinberg equilibrium. The Hardy-Weinberg principle describes a mathematical relationship between the frequencies of alleles in a population and the genotypes they produce. It assumes that certain conditions are met, such as random mating, large population size, no mutations, no migration, and no natural selection.
In the case described, the frequency of the recessive trait 'aa' remains constant at 60% over several generations. This provides a clue that the population is in equilibrium for that trait. According to the Hardy-Weinberg principle, if a population is in equilibrium, the frequencies of the alleles will remain constant from generation to generation.
Part 2: To find the estimated frequency of allele 'a' in the gene pool, we can use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
In this equation, p^2 represents the frequency of homozygous dominant individuals (AA), q^2 represents the frequency of homozygous recessive individuals (aa), and 2pq represents the frequency of heterozygous individuals (Aa).
Given that the frequency of aa individuals (q^2) is 60%, we can substitute 0.6 for q^2 in the equation and solve for q:
0.6 = q^2
q = √0.6 ≈ 0.775
Since q represents the frequency of allele 'a' in the gene pool, the estimated frequency of allele 'a' would be approximately 0.775.
Part 3: To find the frequency of the dominant heterozygotes (Aa), we can substitute the value of q we obtained in Part 2 into the equation:
2pq
From the equation, we know that p represents the frequency of allele 'A' in the gene pool. Since p + q = 1, we can calculate p:
p = 1 - q
p = 1 - 0.775
p ≈ 0.225
Now we can substitute the values of p and q into the equation:
2pq = 2(0.225)(0.775)
2pq ≈ 0.347
Therefore, the frequency of the dominant heterozygotes (Aa) for this trait would be approximately 0.347.