MCAT Biology Review

Chapter 12: Genetics and Evolution

Practice Questions

1.    What is the gene order of linked genes M, N, O, and P, given the following recombination frequencies?

MN: 6%

NO: 18%

MO: 12%

NP: 1%

MP: 5%

OP: 17%

1.    MOPN

2.    NPMO

3.    ONPM

4.    PNMO

2.    Suppose that in humans, the allele for black hair (B) is dominant to the allele for brown hair (b), and the allele for curly hair (C) is dominant to the allele for straight hair (c). When a person of unknown genotype is crossed against a person with straight, brown hair, the phenotypic ratio is as follows:

o   25% curly black hair

o   25% straight black hair

o   25% curly brown hair

o   25% straight brown hair

What is the genotype of the unknown parent?

1.    BbCC

2.    bbCc

3.    Bbcc

4.    BbCc

3.    If a male with hemophilia (XhY) is crossed with a female carrier of both color blindness and hemophilia (XcXh), what is the probability that a female child will be phenotypically normal?

1.    0%

2.    25%

3.    50%

4.    100%

4.    If a test cross on a species of plants reveals the appearance of a recessive phenotype in the offspring, what must be true of the phenotypically dominant parent?

1.    It must be genotypically heterozygous.

2.    It must be genotypically homozygous.

3.    It could be either genotypically heterozygous or homozygous.

4.    It must have the same genotype as the test cross control parent.

5.    Which of the following definitions is FALSE?

1.    Penetrance—the percentage of individuals in the population carrying the allele who actually express the phenotype associated with it

2.    Expressivity—the percentage of individuals in the population carrying the allele who do not express the phenotype associated with it

3.    Incomplete dominance—occurs when the phenotype of the heterozygote is an intermediate of the phenotypes of the homozygotes

4.    Codominance—occurs when multiple alleles exist for a given gene and more than one of them is dominant

6.    In a species of plant, a homozygous red flower (RR) is crossed with a homozygous yellow flower (rr). If the F1 generation is self-crossed and the F2 generation has a phenotypic ratio of red:orange:yellow of 1:2:1, which characteristic accounts for these results?

1.    Codominance

2.    Incomplete dominance

3.    Penetrance

4.    Expressivity

7.    Which of the following statements is INCORRECT regarding inheritance of traits according to the modern synthesis model?

1.    A mutation due to excessive amounts of ultraviolet light occurs in an unfertilized egg; this will affect the child who is born from that egg.

2.    The muscular strength gained by a weight lifter during his lifetime is inherited by his children.

3.    A green-feathered bird that survived all of the predators in the forest will pass on the green feather genes to its offspring.

4.    A flower with a tasty nectar eaten by a butterfly is more likely to pass on its genes through the pollen spread by the butterfly then a flower with less desirable nectar.

8.    Which of the following statements is FALSE based on Darwin’s theory of evolution?

1.    Natural selection is the driving force for evolution.

2.    Favorable genetic variations become more and more common in individuals throughout their lives.

3.    Natural selection can drive organisms living in groups to ultimately become distinct species.

4.    Fitness is measured by reproductive success.

9.    Which of the following is NOT a necessary condition for Hardy–Weinberg equilibrium?

1.    Large population size

2.    No mutations

3.    Monogamous mating partners

4.    No migration into or out of the population

10.As the climate became colder during the Ice Age, a particular species of mammal evolved a thicker layer of fur. What kind of selection occurred in this population?

1.    Stabilizing selection

2.    Directional selection

3.    Disruptive selection

4.    Speciation selection

11.At what point are two populations descended from the same ancestral stock considered to be separate species?

1.    When they can no longer produce viable, fertile offspring.

2.    When they look significantly different from each other.

3.    When they can interbreed successfully and produce offspring.

4.    When their habitats are separated by a significantly large distance so that they cannot meet.

12.In a nonevolving population, there are two alleles, R and r, which code for the same trait. The frequency of R is 30 percent. What are the frequencies of all the possible genotypes?

1.    49% RR, 42% Rr, 9% rr

2.    30% RR, 21% Rr, 49% rr

3.    0.09% RR, 0.42% Rr, 0.49% rr

4.    9% RR, 42% Rr, 49% rr

13.In a particular Hardy–Weinberg population, there are only two eye colors: brown and blue. Of the population, 36% have blue eyes, the recessive trait. What percentage of the population is heterozygous?

1.    24%

2.    48%

3.    60%

4.    64%

14.Which of the following was NOT a belief of Darwin’s?

1.    Evolution of species occurs gradually and evenly over time.

2.    There is a struggle for survival among organisms.

3.    Genetic recombination and mutation are the driving forces of evolution.

4.    Those individuals with fitter variations will survive and reproduce.

15.A child is born with a number of unusual phenotypic features and genetic testing is performed. The child is determined to have partial trisomy 21, with three copies of some segments of DNA from chromosome 21, and partial monosomy 4, with only one copy of some segments of DNA from chromosome 4. Which of the following mutations could have occurred in one of the parental gametes during development to explain both findings?

1.    Deletion

2.    Insertion

3.    Translocation

4.    Inversion

PRACTICE QUESTIONS

Answers and Explanations

1.    B

This is a gene-mapping problem. Because there is a correlation between the frequency of recombination and the distance between genes on a chromosome, if we are given the frequencies, we can determine gene order. Remember that one map unit equals 1 percent recombination frequency. The easiest way to begin is to determine the two genes that are farthest apart; in this case, N and O recombine with a frequency of 18%, so they are 18 map units apart on the chromosome:

N and P recombine with 1% frequency, and P and O recombine with 17% frequency, so P must be between N and O:

Finally, M and P recombine with 5% frequency, and M and O recombine with 12% frequency, so M must be between P and O:

2.    D

In this dihybrid problem, a doubly recessive individual is crossed with an individual of unknown genotype; this is known as a test cross. The straight- and brown-haired individual has the genotype bbcc and can thus only produce gametes carrying bc. Looking at the F1 offspring, there is a 1:1:1:1 phenotypic ratio. The fact that both the dominant and recessive traits are present in the offspring means that the unknown parental genotype must contain both dominant and recessive alleles for each trait. The unknown parental genotype must therefore be BbCc. If you want to double-check the answer, you can work out the Punnett square for the cross BbCc × bbcc:

3.    C

The female in this example is a carrier of two sex-linked traits; based on her genotype, the affected alleles are found on different X chromosomes. Drawing out a Punnett square, we see that 25% of the offspring will be female hemophiliacs (XhXh) and 25% will be female carriers of both alleles (XcXh). This question is asking what percentage of females will have a normal phenotype, which would be half of the females (those who are carriers for both traits).

4.    AThe control parent in a test cross is always recessive. Therefore, if the test parent is phenotypically dominant, yet can provide a recessive allele (as evidenced by the presence of recessive children), then the parent must have both a dominant and recessive allele. Therefore, this test parent must by heterozygous.

5.    BThe definition given here for expressivity is a much better match for defining penetrance (or, really, one minus the penetrance). Expressivity refers to the variable manifestations of a given genotype as different phenotypes; the degree to which various phenotypes are expressed. All of the other definitions given are accurate.

6.    BSome progeny in the second generation are apparently blends of the parental phenotypes. The orange color is the result of the combined effects of the red and yellow alleles. An allele is incompletely dominant if the phenotype of the heterozygotes is an intermediate of the phenotypes of the homozygotes.

7.    BTo find the correct answer, we have to read each choice and eliminate the ones that fit with the modern-day theories of inheritance, which state that the genes that make an organism most fit for its environment will be passed to offspring. This can be seen in choice (C) and (D), both of which demonstrate that an organism with improved fitness will pass those genes to offspring. Choice (A) mentions a gamete being exposed to mutagens; a zygote created from this gamete would contain any mutations in the egg and would be affected by them. Thus, choice (B)must be the correct answer: indeed, acquired characteristics not encoded in the genome should not be passed to offspring according to the modern synthesis model.

8.    BDarwin’s theory of natural selection argues that chance variations between organisms can help certain organisms survive to reproductive age and produce many offspring, transmitting their variations to the next generation. Thus, natural selection would drive the process of evolution forward, enabling the survival of characteristics that impart an advantage in the environment, eliminating choice (A). In Darwin’s theory, fitness is measured in terms of reproductive success, as choice (D) states. Through natural selection, organisms may be separated in groups depending on environmental pressures, and these groups can eventually separate to the point of becoming distinct species, eliminating choice (C)Choice (B) is the correct answer because the theory of natural selection applies to a population of organisms, not to a particular individual. As such, favorable genetic variations become more and more common from generation to generation, not during the lifetime of an individual.

9.    CHardy–Weinberg equilibrium exists in certain ideal conditions that, when satisfied, allow one to calculate the gene frequencies within a population. The Hardy–Weinberg equation can be applied only under these five conditions: (1) the population is very large; (2) there are no mutations that affect the gene pool; (3) mating between individuals in the population is random; (4) there is no migration of individuals into or out of the population; (5) the genes in the population are all equally successful at reproduction. Thus, from the given choices, only choice (C) is false: monogamy is not a necessary condition for the Hardy–Weinberg equations to be applied.

10.BThe situation described in the question stem is an example of directional selection. In directional selection, the phenotypic norm of a particular species shifts toward an extreme to adapt to a selective pressure, such as an increasingly colder environment. Only those individuals with a thicker layer of fur were able to survive during the Ice Age, thus shifting the phenotypic norm.

11.AA species is defined as the largest group of organisms that can interbreed to produce viable, fertile offspring. Therefore, two populations are considered separate species when they can no longer do so.

12.DLet’s use the information provided by the question stem to set up our equations. We are told that the frequency of R equals 30%, and as such, p = 0.30. The frequency of the recessive gene r = 100% – 30% = 70%; thus, q = 0.70. The frequency of the genotypes, according to the Hardy–Weinberg equilibrium, are given by p2 = RR, 2pq = Rr, and q2 = rr. Therefore, the frequency of the genotypes are (0.3)2 = 0.09 = 9% RR, 2 × 0.3 × 0.7 = 0.42 = 42% Rr, and (0.7)2 = 0.49 = 49% rr.

13.BUsing the information given in the question stem, we can determine that the percentage of the population with blue eyes (genotype bb) = 36% = 0.36 = q2; therefore, q = 0.6. Because this is a Hardy–Weinberg population, we can assume that p + q = 1, so p = 1 – 0.6 = 0.4. The frequency of heterozygous brown eyes is therefore 2pq = 2 × 0.4 × 0.6 = 0.48 = 48%.

14.CDarwin’s main argument was that natural selection is the driving force for evolution. Darwin did not consider the role of genetic mutation and recombination, which were unknown at the time. These aspects were added to Darwin’s theory as part of the modern synthesis model, which did not occur until significantly after Darwin’s time.

15.CThis scenario—a deletion of some DNA and a duplication of other DNA—would be consistent with a translocation between chromosomes 4 and 21 during development of an egg or sperm. If part of chromosome 21 was swapped with part of chromosome 4, then a gamete resulting from meiosis in this cell would result in a daughter cell with two copies of some of the DNA from 21 and no copies of some of the DNA from 4. Therefore, after fertilization, there would be partial trisomy 21 and partial monosomy 4. While a deletion or insertion could explain one of the findings, it cannot explain both, eliminating choices (A) and (B). An inversion should not lead to partial trisomy or partial monosomy because the DNA is simply reversed, eliminating choice (D).