1. A population which meets the hardy- weinberg requirements:
a. Evolves
b. is small and usually isolated
c. has allele frequency in equilibrium
d. changes genotypic distribution from generation to generation
e. always has 75% A and 25% a allele frequencies
2. In a hypothetical population, the frequency of the dominant A allele is 80% and the frequency of the recessive allele a is 20%. that percentage of the population would you expect to be heterozygous(Aa) in genotype?? can you please show work or explain how you did it.
A population that meets Hardy-Weinberg requirements is in genetic equilibrium, and neither its allele frequencies nor its genotype frequencies changes from generation to generation. If allele and genotypes frequencies are not changing from generation to generation, can the population be evolving? What is the requirement for H-W concerning population size?
In H-W equilibrium, the 3 genotype frequencies are
p^2 + 2pq + q^2 = 1
where p is the frequency of the dominant allele (A, which is 0.8) and q is the frequency of the recessive alleled (a, which is 0.2).
(0.8)^2 + 2(0.8)(0.2) + (0.2)^2 = 1
0.64 + 0.32 + 0.04 = 1
F(AA) = 0.64
F(Aa) = 0.32
F(aa) = 0.04
Now just multiply the above 3 frequencies by the total number of individuals in the population to determine how many AA, Aa, and aa individuals there should be.
1. The correct answer is c. a population that meets the Hardy-Weinberg requirements has allele frequency in equilibrium. This means that the frequency of alleles remains constant from generation to generation when certain conditions are met, including a large population size, random mating, no mutation, no migration, and no natural selection.
2. To determine the percentage of the population that would be heterozygous (Aa) in genotype, we can use the Hardy-Weinberg equation: p^2 + 2pq + q^2 = 1.
In this equation:
- p^2 represents the frequency of homozygous dominant genotype (AA).
- 2pq represents the frequency of heterozygous genotype (Aa).
- q^2 represents the frequency of homozygous recessive genotype (aa).
Given that the frequency of the dominant A allele is 80% (or 0.8) and the frequency of the recessive allele a is 20% (or 0.2), we can substitute these values into the equation:
p^2 + 2(0.8)(0.2) + q^2 = 1
p^2 + 0.32 + q^2 = 1
Since p^2 + q^2 represents all the possible genotypes in the population, we can subtract this from 1 to find the percentage of heterozygous individuals:
2pq = 1 - p^2 - q^2
2pq = 1 - (p^2 + q^2)
In this case, we do not know the value of p or q, so we will need to calculate them first using the allele frequencies.
p + q = 1
0.8 + 0.2 = 1
p = 0.8
q = 0.2
Now we can substitute these values back into the equation to find the percentage of heterozygous individuals:
2pq = 1 - (p^2 + q^2)
2(0.8)(0.2) = 1 - (0.8^2 + 0.2^2)
0.32 = 1 - (0.64 + 0.04)
0.32 = 1 - 0.68
0.32 = 0.32
Therefore, the percentage of the population expected to be heterozygous (Aa) in genotype is 32%.
1. The correct answer is c. A population that meets the Hardy-Weinberg requirements has allele frequency in equilibrium. This means that the frequencies of different alleles in the population remain constant from generation to generation, and there is no overall change in the genetic makeup of the population.
To determine whether a population meets the Hardy-Weinberg requirements, you need to check if certain conditions are met. These conditions include a large population size, random mating, no migration, no mutation, and no natural selection. If all these conditions are met, the population is said to be in Hardy-Weinberg equilibrium, and allele frequencies remain constant.
2. To calculate the percentage of the population that would be heterozygous (Aa) based on the allele frequencies provided, you can use the Hardy-Weinberg equation: p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele (A) and q is the frequency of the recessive allele (a). The term 2pq represents the proportion of heterozygous individuals in the population.
Given that the frequency of the dominant allele (A) is 80% (0.80) and the frequency of the recessive allele (a) is 20% (0.20), you can substitute these values into the equation:
2pq = 2 * 0.8 * 0.2 = 0.32
Therefore, 32% of the population would be expected to be heterozygous (Aa) in genotype.
It's important to note that the Hardy-Weinberg equation assumes that the population is in equilibrium and that the conditions for it to be valid are met. If the population does not meet these conditions, the observed genotype frequencies may deviate from those predicted by the equation.