Question 1

At least some of the requirements for a Hardy-Weinberg equilibrium would be in effect on the island.For instance there is probably

Accepted Answer

The answer of At least some of the requirements for...

Question 2

Which of the following would be expected to produce the smallest evolutionary change in a given period of time in a population of birds?

Accepted Answer

Mutations occur randomly and typically at a low rate, leading to small evolutionary changes over time compared to other mechanisms like natural selection or gene flow.

Question 3

The recessive phenotype of a trait occurs in 20% of a population.There are no selection pressures affecting this trait.What would be the expected frequency of the dominant allele after six generations of continued non-selection?

Accepted Answer

The recessive phenotype frequency (q^2) is 0.2, so q = sqrt(0.2) = 0.447. The frequency of the dominant allele (p) is 1 - q = 1 - 0.447 = 0.553. Since there are no selection pressures, allele frequencies remain constant over generations.

Question 4

Which one of the following could not be involved in gene flow?

Accepted Answer

Gene flow is the movement of genes from one population to another. Disassortive mating within a population does not involve the movement of genes from one population to another and therefore could not be involved in gene flow. The other options all involve some form of dispersal or movement of genetic material between populations.

Question 5

Assortive and disassortive mating are similar in that both

Accepted Answer

Both assortive and disassortive mating affect the genotype frequencies by increasing or decreasing the frequency of homozygotes or heterozygotes, respectively, but they do not change allele frequencies.

Question 6

In some instances environmental change causes a situation where one phenotype is favored for a period of time,and then a different phenotype is favored.This oscillating selection causes

Accepted Answer

Oscillating selection favors different phenotypes at different times, which can maintain genetic variation in the population. This is because different genotypes can have different advantages in different environmental conditions, leading to the persistence of multiple genotypes in the population.

Question 7

The text discusses sickle-cell anemia,which is a classic example of

Accepted Answer

The answer of The text discusses sickle-cell anemia,which is a...

Question 8

In directional selection,over time

Accepted Answer

In directional selection, the environment selects for individuals that exhibit a particular trait, resulting in a shift in the population towards that trait. This means that the population is strongly selected for in one direction, such as larger beak size. Thus, option C is the correct answer. Options A, B, D, and E are not correct because they do not accurately describe what happens in directional selection.

Question 9

The disease,sickle-cell anemia is common in malaria-infested areas because individuals that are heterozygous for the gene (AS)have enhanced resistance to malaria over normal individuals (AA).Individuals with severe sickle-cell anemia (SS)usually die before reproduction.Assume that a population under Hardy-Weinberg equilibrium has allele frequencies of A = 0.8 and S = 0.2.Assume that this population is then subject to the following selection forces:
AA: 15% die of malaria
AS: 5% die of malaria
SS: 100% die of severe sickle-cell anemia
What will be the frequency of the S allele after selection occurs?

Accepted Answer

The frequency of the S allele after selection can be calculated by considering the survival rates of each genotype. The AA genotype has a survival rate of 85% (1 - 0.15), AS has a survival rate of 95% (1 - 0.05), and SS has a survival rate of 0% (1 - 1.00). Using these survival rates, the new allele frequencies can be calculated, resulting in an S allele frequency of approximately 0.18.

Question 10

You are studying a population of geese in which there are two color phases,brown and gray.Color in this species is controlled by a single gene,with brown dominant to gray.A random sample of 250 geese shows that 210 are brown.What percentage of the brown geese are heterozygous? (Assume that the population is in Hardy-Weinberg equilibrium. )

Accepted Answer

The answer of You are studying a population of geese...

Question 11

In negative frequency-dependent selection,such as in the study done on the water boatman insect,the incidence of predation leads to an

Accepted Answer

Negative frequency-dependent selection favors rare genotypes, leading to an increase in their frequency. In the study on water boatman insects, those with rare ventral coloration had a survival advantage because they were more difficult for predators to detect. As a result, the frequency of the rare genotype increased over time.

Question 12

The recessive phenotype of a trait occurs in 20% of a population.There are no selection pressures affecting this trait.What would be the expected frequency of the heterozygote genotype after six generations of continued non-selection?

Accepted Answer

The recessive phenotype frequency (q^2) is 0.2, so q = √0.2 ≈ 0.447. The frequency of the dominant allele (p) is 1 - q ≈ 0.553. The heterozygote frequency (2pq) is 2 * 0.553 * 0.447 ≈ 0.494.

Question 13

Antigens on red blood cells are hereditary traits that allow blood to be typed in different ways.One system is based on a gene with two alleles,M and N.If the frequency of the M allele in a population is 0.4,then according to the Hardy-Weinberg rule,the expected frequency of the heterozygous MN genotype is ____.

Accepted Answer

According to the Hardy-Weinberg rule, the frequency of the MN genotype can be calculated using the equation: 2pq, where p is the frequency of the M allele and q is the frequency of the N allele (p + q = 1). Here, p = 0.4 and q = 0.6 (1 - 0.4). Therefore, 2pq = 2(0.4)(0.6) = 0.48 or 48%. Thus, the expected frequency of the heterozygous MN genotype is 0.48 or 48%. The closest answer choice is D) 0.48.

Question 14

If the frequency of an autosomal recessive trait in humans is 1 out of 4000 births,what would be the expected frequency of heterozygote carriers for the trait if we assume that the gene is in Hardy-Weinberg equilibrium?

Accepted Answer

The frequency of the recessive allele (q) is the square root of 1/4000, which is 0.0158. The frequency of heterozygotes (2pq) is approximately 2 * 0.9842 * 0.0158 = 0.031.

Question 15

Which of the following would be expected to produce the largest evolutionary change in a given period of time in a population of birds?

Accepted Answer

The answer of Which of the following would be expected...

Question 16

A gene is represented in the gene pool of a population of 10,000 individuals by a dominant (W)and recessive allele (w).If the initial frequency of W in the pool is 0.7,how many generations of complete selection against it would be required to eliminate all W alleles from the population? Assume that there is no mutation and the population meets all other Hardy-Weinberg conditions.

Accepted Answer

The answer of A gene is represented in the gene...

Question 17

The genotype frequencies in a population with no evolution occurring are AA = 0.25,Aa = 0.5,aa = 0.25.By removing all AA individuals before they reproduce and assuming the population will remain under Hardy-Weinberg conditions,what will be the frequency of aa individuals in the next generation?

Accepted Answer

First, we need to calculate the frequency of the Aa heterozygotes in the population. Since Aa individuals have a frequency of 0.5, and AA individuals are being removed, we can simply divide 0.5 by the frequency of the remaining alleles (a):

Aa frequency = 0.5 / (2*0.25 + 0.5) = 0.5 / 1 = 0.5

Next, we can use the equation p + q = 1 to solve for the frequency of the aa homozygotes:

q = frequency of aa = 1 - p

Since we know the frequency of Aa is 0.5, we can use that to solve for the frequency of the A allele:

p = frequency of A = 1 - q

Substituting in the values we know:

0.5 = 1 - q

q = 0.5

So the frequency of aa individuals in the next generation would be 0.5 or 50%. Therefore, the correct answer is D (0.67 is incorrect since it would be the frequency of Aa after AA individuals are removed, not the frequency of aa).

Question 18

In disruptive selection,over time

Accepted Answer

Disruptive selection favors extreme values for a trait over intermediate values, leading to a population being selected in two directions.

Question 19

Coloration in the peppered moth (Biston betularia)is determined by a single gene with two alleles showing complete dominance.Dark moths are homozygous dominant or heterozygous for the gene,light moths are homozygous recessive.In a sample of 100 moths,you determine that 64 of the moths are dark.According to the Hardy-Weinberg rule,the expected frequency of the dominant allele is ____.

Accepted Answer

The answer of Coloration in the peppered moth (Biston betularia)is...

Question 20

In a large population of randomly reproducing rabbits,a recessive allele r comprises 80% of the alleles for a gene,while a dominant allele R comprises the remaining 20%.What percentage of the rabbits would you expect to have the recessive phenotype?

Accepted Answer

Since the recessive allele comprises 80% of the alleles, we can use the Hardy-Weinberg equation to calculate the expected proportion of rabbits with the recessive phenotype. The equation is:
p^2 + 2pq + q^2 = 1
where p is the frequency of the dominant allele and q is the frequency of the recessive allele. We know that q = 0.8 and p = 0.2, so:
(0.8)^2 + 2(0.2)(0.8) + (0.2)^2 = 0.04 + 0.32 + 0.04 = 0.4
So, 40% of the rabbits would be expected to have the recessive phenotype, which is equivalent to 64% (100% - 40%) having the dominant phenotype. Therefore, the correct answer is C.

Quiz 20: Genes Within Populations