Multiple Choice
Identify the
choice that best completes the statement or answers the question.
|
|
|
1.
|
Pea plants were particularly well suited for use in Mendel's breeding
experiments for all of the following reasons except that
a. | peas show easily observed variations in a number of characters, such as pea shape and
flower color. | b. | it is possible to completely control matings between different pea
plants. | c. | it is possible to obtain large numbers of progeny from any given
cross. | d. | peas have an unusually long generation time. | e. | many of the
observable characters that vary in pea plants are controlled by single
genes. |
|
|
|
2.
|
A plant with purple flowers is allowed to self-pollinate. Generation after
generation, it produces purple flowers. This is an example of
a. | hybridization. | b. | incomplete dominance. | c. | true-breeding. | d. | the law of segregation. | e. | polygenetics. |
|
|
|
3.
|
Which of the following statements about Mendel's breeding experiments is
correct?
a. | None of the parental (P) plants were true-breeding. | b. | All of the F2
progeny showed a phenotype that was intermediate between the two parental (P)
phenotypes. | c. | Half of the F1 progeny had the same phenotype as one of the parental (P) plants, and
the other half had the same phenotype as the other parent. | d. | All of the F1
progeny resembled one of the parental (P) plants, but only some of the F2 progeny
did. | e. | none of the above |
|
|
|
4.
|
What is the difference between a monohybrid cross and a dihybrid cross?
a. | A monohybrid cross involves a single parent, whereas a dihybrid cross involves two
parents. | b. | A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two
progeny. | c. | A monohybrid cross involves organisms that are heterozygous for a single character,
whereas a dihybrid cross involves organisms that are heterozygous for two
characters. | d. | A monohybrid cross is performed only once, whereas a dihybrid cross is performed
twice. | e. | A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1
ratio. |
|
|
|
5.
|
A cross between homozygous purple-flowered and homozygous white-flowered pea
plants results in offspring with purple flowers. This demonstrates
a. | the blending model of genetics. | b. | true-breeding. | c. | dominance. | d. | a dihybrid cross. | e. | the mistakes made by
Mendel. |
|
|
|
6.
|
The F1 offspring of Mendel's classic pea cross always looked
like one of the two parental varieties because
a. | one allele was completely dominant over another. | b. | each allele affected
phenotypic expression. | c. | the traits blended together during
fertilization. | d. | no genes interacted to produce the parental phenotype. | e. | different genes
interacted to produce the parental phenotype. |
|
|
|
7.
|
What was the most significant conclusion that Gregor Mendel drew from his
experiments with pea plants?
a. | There is considerable genetic variation in garden peas. | b. | Traits are inherited
in discrete units, and are not the results of "blending." | c. | Recessive genes
occur more frequently in the F1 than do dominant ones. | d. | Genes are composed
of DNA. | e. | An organism that is homozygous for many recessive traits is at a
disadvantage. |
|
|
|
8.
|
Which of the following is (are) true for alleles?
a. | They can be identical or different for any given gene in a somatic
cell. | b. | They can be dominant or recessive. | c. | They can represent alternative forms of a
gene. | d. | Only A and B are correct. | e. | A, B, and C are
correct. |
|
|
|
9.
|
What is genetic cross between an individual showing a dominant phenotype (but of
unknown genotype) and a homozygous recessive individual called?
a. | a self-cross | b. | a testcross | c. | a hybrid
cross | d. | an F1 cross | e. | a dihybrid
cross |
|
|
|
10.
|
How many unique gametes could be produced through independent assortment by an
individual with the genotype AaBbCCDdEE?
|
|
|
11.
|
Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular
trait. This suggests
a. | that the parents were true-breeding for contrasting traits. | b. | incomplete
dominance. | c. | that a blending of traits has occurred. | d. | that the parents
were both heterozygous. | e. | that each offspring has the same
alleles. |
|
|
|
12.
|
Two characters that appear in a 9:3:3:1 ratio in the F2 generation should have
which of the following properties?
a. | Each of the characters is controlled by a single gene. | b. | The genes
controlling the characters obey the law of independent assortment. | c. | Each of the genes
controlling the characters has two alleles. | d. | Only A and C are correct. | e. | A, B, and C are
correct. |
|
|
|
13.
|
A 9:3:3:1 phenotypic ratio is characteristic of which of the following?
a. | a monohybrid cross | b. | a dihybrid cross | c. | a trihybrid
cross | d. | linked genes | e. | both A and D |
|
|
|
14.
|
A sexually reproducing animal has two unlinked genes, one for head shape
(H). and one for tail length (T). Its genotype is HhTt. Which of the following
genotypes is possible in a gamete from this organism?
|
|
|
15.
|
It was important that Mendel examined not just the F1 generation in
his breeding experiments, but the F2 generation as well, because
a. | he obtained very few F1 progeny, making statistical analysis
difficult. | b. | parental traits that were not observed in the F1 reappeared in the
F2, suggesting that the traits did not truly disappear in the
F1. | c. | analysis of the F1 progeny would have allowed him to discover the law of
segregation, but not the law of independent assortment. | d. | the dominant
phenotypes were visible in the F2 generation, but not in the
F1. | e. | all of the above |
|
|
|
16.
|
When crossing a homozygous recessive with a heterozygote, what is the chance of
getting an offspring with the homozygous recessive phenotype?
a. | 0% | b. | 25% | c. | 50% | d. | 75% | e. | 100% |
|
|
|
Use the diagram and description below to answer the following question.
In a particular plant, leaf color is controlled by gene D. Plants with the
dominant allele D have dark green leaves, and plants with the homozygous recessive dd
genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a
light-leaved one, and the F 1 offspring is allowed to self-pollinate. The predicted outcome
of this cross is diagrammed in the Punnett square shown below, where 1, 2, 3, and 4 represent the
genotypes corresponding to each box within the square. 
|
|
|
17.
|
Which of the boxes marked 1-4 correspond to plants with dark leaves?
a. | 1 only | b. | 1 and 2 | c. | 2 and
3 | d. | 4 only | e. | 1, 2, and 3 |
|
|
|
18.
|
Which of the boxes correspond to plants with a heterozygous genotype?
a. | 1 | b. | 1 and 2 | c. | 1, 2, and
3 | d. | 2 and 3 | e. | 2, 3, and 4 |
|
|
|
19.
|
Which of the plants will be true-breeding?
a. | 1 and 4 | b. | 2 and 3 | c. | 1-4 | d. | 1 only | e. | none |
|
|
|
20.
|
P = purple, pp = white. The offspring of a cross between two
heterozygous purple-flowering plants ( Pp  Pp) results
in
a. | all purple-flowered plants. | b. | purple-flowered plants and white-flowered
plants. | c. | two types of white-flowered plants: PP and Pp. | d. | all white-flowered
plants. | e. | all pink-flowered plants. |
|
|
|
21.
|
Mendel accounted for the observation that traits which had disappeared in the F1
generation reappeared in the F2 generation by proposing that
a. | new mutations were frequently generated in the F2 progeny, "reinventing"
traits that had been lost in the F1. | b. | the mechanism controlling the appearance of
traits was different between the F1 and the F2 plants. | c. | traits can be dominant or recessive, and the
recessive traits were obscured by the dominant ones in the F1. | d. | the traits were lost
in the F1 due to blending of the parental traits. | e. | members of the F1 generation had only one
allele for each character, but members of the F2 had two alleles for each
character. |
|
|
|
22.
|
What are Punnett squares used for?
a. | predicting the result of genetic crosses between organisms of known
genotypes | b. | determining the DNA sequence of a given gene | c. | identifying the gene
locus where allelic variations are possible | d. | testing for the presence of the recessive
allele | e. | more than one of the above |
|
|
|
23.
|
Which of the following is false, regarding the law of segregation?
a. | It states that each of two alleles for a given trait segregate into different
gametes. | b. | It can be explained by the segregation of homologous chromosomes during
meiosis. | c. | It can account for the 3:1 ratio seen in the F2 generation of
Mendel's crosses. | d. | It can be used to predict the likelihood of
transmission of certain genetic diseases within families. | e. | It is a method that
can be used to determine the number of chromosomes in a plant. |
|
|
|
24.
|
The fact that all seven of the pea plant traits studied by Mendel obeyed the
principle of independent assortment means that
a. | none of the traits obeyed the law of segregation. | b. | the diploid number
of chromosomes in the pea plants was 7. | c. | all of the genes controlling the traits were
located on the same chromosome. | d. | all of the genes controlling the traits behaved
as if they were on different chromosomes. | e. | the formation of gametes in plants occurs by
mitosis only. |
|
|
|
25.
|
Black fur in mice ( B) is dominant to brown fur ( b) Short tails
( T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt  BBtt will have black fur and long tails? a. | 1/16 | b. | 3/16 | c. | 3/8 | d. | 1/2 | e. | 9/16 |
|
|
|
26.
|
In certain plants, tall is dominant to short. If a heterozygous plant is crossed
with a homozygous tall plant, what is the probability that the offspring will be short?
|
|
|
27.
|
A couple has three children, all of whom have brown eyes and blond hair. Both
parents are homozygous for brown eyes (BB) but one is a blond (rr) and the other is a
redhead (Rr). What is the probability that their next child will be a brown-eyed
redhead?
|
|
|
28.
|
Two true-breeding stocks of pea plants are crossed. One parent has red, axial
flowers and the other has white, terminal flowers; all F1 individuals have red, axial
flowers. If 1,000 F2 offspring resulted from the cross, approximately how many of them
would you expect to have red, terminal flowers? (Assume independent assortment).
|
|
|
29.
|
In a cross AaBbCc  AaBbCc, what is the
probability of producing the genotype AABBCC? a. | 1/4 | b. | 1/8 | c. | 1/16 | d. | 1/32 | e. | 1/64 |
|
|
|
30.
|
Given the parents AABBCc  AabbCc, assume
simple dominance and independent assortment. What proportion of the progeny will be expected to
phenotypically resemble the first parent?
|
|
|
31.
|
A 1:2:1 phenotypic ratio in the F2 generation of a monohybrid cross
is a sign of
a. | complete dominance. | b. | multiple alleles. | c. | incomplete
dominance. | d. | polygenic inheritance. | e. | pleiotropy. |
|
|
|
Refer to the result below to answer the following questions.
A
tall plant is crossed with a short plant, and the progeny are all intermediate in size between the
two parental plants.
|
|
|
32.
|
This could be an example of
a. | incomplete dominance. | b. | polygenic inheritance. | c. | complete
dominance. | d. | A and B | e. | B and C |
|
|
|
33.
|
If the intermediate F1 progeny were allowed to self-pollinate, and
the F2 progeny were also intermediate in size, but following a normal distribution, this
would suggest
a. | incomplete dominance. | b. | polygenic inheritance. | c. | complete
dominance. | d. | a strong environmental influence. | e. | codominance. |
|
|
|
34.
|
If the intermediate F1 progeny were allowed to self-pollinate, and
25% of the F2 progeny were tall, 50% were intermediate in size, and 25% were short, this
would suggest
a. | incomplete dominance. | b. | polygenic inheritance. | c. | complete
dominance. | d. | pleiotropy. | e. | multifactorial
inheritance. |
|
|
|
35.
|
In snapdragons, heterozygotes have pink flowers, whereas homozygotes have red or
white flowers. When plants with red flowers are crossed with plants with white flowers, what
proportion of the offspring will have pink flowers?
a. | 0% | b. | 25% | c. | 50% | d. | 75% | e. | 100% |
|
|
|
36.
|
Tallness (T) is dominant to dwarfness (t), while red (R)
flower color is dominant to white (r). The heterozygous condition results in pink
(Rr) flower color. A dwarf, red snapdragon is crossed with a plant homozygous for tallness and
white flowers. What are the genotype and phenotype of the F1 individuals?
a. | ttRr-dwarf and pink | b. | ttrr-dwarf and white | c. | TtRr-tall and
red | d. | TtRr-tall and pink | e. | TTRR-tall and
red |
|
|
|
37.
|
Skin color in a fish is inherited via a single gene with four different alleles.
How many different types of gametes would be possible in this system?
|
|
|
38.
|
In cattle, roan coat color (mixed red and white hairs) occurs in the
heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which
of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white?
a. | red  white | b. | roan 
roan | c. | white  roan | d. | red 
roan | e. | The answer cannot be determined from the information
provided. |
|
|
|
39.
|
The relationship between genes S and N is an example of
a. | incomplete dominance. | b. | epistasis. | c. | complete
dominance. | d. | pleiotropy. | e. | codominance. |
|
|
|
40.
|
A cross between a true-breeding sharp-spined cactus and a spineless cactus would
produce
a. | all sharp-spined progeny. | b. | 50% sharp-spined, 50% dull-spined
progeny. | c. | 25% sharp-spined, 50% dull-spined, 25% spineless progeny | d. | all spineless
progeny | e. | It is impossible to determine the phenotypes of the
progeny. |
|
|
|
41.
|
If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the
F2 would segregate in which of the following ratios?
a. | 3 sharp-spined : 1 spineless | b. | 1 sharp-spined : 2 dull-spined : 1
spineless | c. | 1 sharp spined : 1 dull-spined : 1 spineless | d. | 1 sharp-spined : 1
dull-spined | e. | 9 sharp-spined : 3 dull-spined : 4 spineless |
|
|
|
Use the information below to answer the following questions.
Feather color in budgies is determined by two different genes Y and B. YYBB, YyBB, or
YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white.
|
|
|
42.
|
A blue budgie is crossed with a white budgie. Which of the following results is
not possible?
a. | green offspring | b. | yellow offspring | c. | blue
offspring | d. | A and B | e. | A, B, and C |
|
|
|
43.
|
Two blue budgies were crossed. Over the years, they produced 22 offspring, 5 of
which were white. What are the most likely genotypes for the two blue budgies?
a. | yyBB and yyBB | b. | yyBB and yyBb | c. | yyBb and
yyBb | d. | yyBB and yybb | e. | yyBb and
yybb |
|
|
|
44.
|
Three babies were mixed up in a hospital. After consideration of the data below,
which of the following represent the correct baby and parent combinations?
Couple #
Blood Groups
| I
A and A | II
A and B | III
B and O | Baby #
Blood
Groups
| 1
B | 2
O | 3
AB | | | | |
a. | I-3, II-1, III-2 | b. | I-1, II-3, III-2 | c. | I-2, II-3,
III-1 | d. | I-2, II-1, III-3 | e. | I-3, II-2,
III-1 |
|
|
|
Use the following information to answer the questions below.
A
woman who has blood type A, has a daughter who is type O positive and a son who is type B negative.
Rh positive is a simple dominant trait over Rh negative.
|
|
|
45.
|
Which of the following is a possible genotype for the son?
a. | IBIB | b. | IBIA | c. | ii | d. | IBi | e. | IAIA |
|
|
|
46.
|
Which of the following is a possible genotype for the mother?
a. | IAIA | b. | IBIB | c. | ii | d. | IAi | e. | IAIB |
|
|
|
47.
|
Which of the following is a possible phenotype for the father?
a. | A | b. | O | c. | B | d. | AB | e. | impossible to
determine |
|
|
|
48.
|
Which of the following is the probable genotype for the mother?
a. | IAIARR | b. | IAIARr | c. | IAirr | d. | IAiRr | e. | IAiRR |
|
|
|
49.
|
Which of the following is a possible phenotype of the father?
a. | A negative | b. | O negative | c. | B
positive | d. | A positive | e. | O positive |
|
|
|
50.
|
What is the chromosomal system for determining sex in mammals?
|
|
|
51.
|
What is the chromosomal system for sex determination in grasshoppers and certain
other insects?
|
|
|
52.
|
What is the chromosomal system for sex determination in birds?
|
|
|
53.
|
What is the chromosomal system of sex determination in most species of ants and
bees?
|
|
|
Use the terms listed below to answer the following questions. Each term may
be used once, more than once, or not at all.
| A.
incomplete dominance | | B. multiple
alleles | | C. pleiotropy | | D. epistasis | |
|
|
|
54.
|
the ability of a single gene to have multiple phenotypic effects
|
|
|
55.
|
the ABO blood group system
|
|
|
56.
|
the phenotype of the heterozygote differs from the phenotypes of both
homozygotes
|
|
|
57.
|
cystic fibrosis affects the lungs, the pancreas, the digestive system, and other
organs, resulting in symptoms ranging from breathing difficulties to recurrent infections
|
|
|
58.
|
Which of the following is an example of polygenic inheritance?
a. | pink flowers in snapdragons | b. | the ABO blood groups in
humans | c. | Huntington's disease in humans | d. | white and purple flower color in
peas | e. | skin pigmentation in humans |
|
|
|
59.
|
Hydrangea plants of the same genotype are planted in a large flower garden. Some
of the plants produce blue flowers and others pink flowers. This can be best explained by
a. | environmental factors such as soil pH. | b. | the allele for blue hydrangea being completely
dominant. | c. | the alleles being codominant. | d. | the fact that a mutation has
occurred. | e. | acknowledging that multiple alleles are involved. |
|
|
|
Use the information below to answer the following questions.
A
woman and her spouse both show the normal phenotype for pigmentation, but both had one parent who was
an albino. Albinism is an autosomal recessive trait.
|
|
|
60.
|
What is the probability that their first child will be an albino?
|
|
|
61.
|
If their first two children have normal pigmentation, what is the probability
that their third child will be an albino?
|
|
|
62.
|
Huntington's disease is caused by a dominant allele. If one of your parents
has the disease, what is the probability that you, too, will have the disease?
|
|
|
63.
|
A woman has six sons. The chance that her next child will be a daughter
is
a. | 1. | b. | 0. | c. | 1/2. | d. | 1/6. | e. | 5/6. |
|
|
|
The pedigree chart below is for a family, some of whose members exhibit the
recessive trait, wooly hair. Affected individuals are indicated by an open square or circle. Use the
chart to answer the following questions.
|
|
|
64.
|
What is the genotype of individual B-5?
a. | WW | b. | Ww | c. | ww | d. | WW or ww | e. | ww or
Ww |
|
|
|
65.
|
What is the likelihood that the progeny of D-3 and D-4 will have wooly
hair?
a. | 0% | b. | 25% | c. | 50% | d. | 75% | e. | 100% |
|
|
|
66.
|
What is the probability that individual C-1 is Ww?
|
|
|
67.
|
People with sickle-cell trait
a. | are heterozygous for the sickle-cell allele. | b. | are usually
healthy. | c. | have increased resistance to malaria. | d. | produce normal and abnormal
hemoglobin. | e. | all of the above |
|
|
|
68.
|
When a disease is said to have a multifactorial basis, it means that
a. | many factors, both genetic and environmental, contribute to the
disease. | b. | it is caused by a gene with a large number of alleles. | c. | it affects a large
number of people. | d. | it has many different
symptoms. | e. | it tends to skip a generation. |
|
|
|
69.
|
Which of the following terms is least related to the others?
a. | pedigree | b. | karyotype | c. | amniocentesis | d. | chorionic villus sampling | e. | epistasis |
|
|
|
Use the answers below to answer the following questions. Each answer may be
used once, more than once, or not at all.
| A.
Huntington's disease | | B. Tay-Sachs
disease | | C. phenylketonuria | | D. cystic fibrosis | | E. sickle-cell disease | |
|
|
|
70.
|
Substitution of the "wrong" amino acid in the hemoglobin protein
results in this disorder.
|
|
|
71.
|
Individuals with this disorder are unable to metabolize certain lipids,
affecting proper brain development. Affected individuals die in early childhood.
|
|
|
72.
|
This is caused by a dominant single gene defect and generally does not appear
until the individual is 35-45 years of age.
|
|
|
73.
|
Effects of this recessive disorder can be completely overcome by regulating the
diet of the affected individual.
|
|
|
74.
|
This results from a defect in membrane proteins that normally function in
chloride ion transport.
|
|
|
75.
|
Which of the following techniques involves the preparation of a
karyotype?
a. | amniocentesis | b. | chorionic villus sampling | c. | fetoscopy | d. | A and B only | e. | A, B, and
C |
|
|
|
76.
|
Chromosomes and genes share all of the following characteristics except
that
a. | they are both present in pairs in all diploid cells. | b. | they both undergo
segregation during meiosis. | c. | their copy numbers in the cell decrease after
meiosis, and increase during fertilization. | d. | they are both copied during the S phase of the
cell cycle. | e. | they both pair up with their homologues during prophase of
mitosis. |
|
|
|
77.
|
The improvement of microscopy techniques in the late 1800s set the stage for the
emergence of modern genetics because
a. | it revealed new and unanticipated features of Mendel's pea plant
varieties. | b. | it allowed biologists to study meiosis and mitosis, revealing the parallels between
the behaviors of genes and chromosomes. | c. | it allowed scientists to see the DNA present
within chromosomes. | d. | it led to the discovery of
mitochondria. | e. | All of the above are true. |
|
|
|
78.
|
When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to
each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the
white-eyed flies were male. What was the explanation for this result?
a. | The involved gene was on the X chromosome. | b. | The involved gene
was on the Y chromosome. | c. | The involved gene was on an
autosome. | d. | Other male-specific factors influence eye color in flies. | e. | Other
female-specific factors influence eye color in flies. |
|
|
|
79.
|
Which of the following statements is (are) true?
a. | The closer two genes are on a chromosome, the higher the probability that a crossover
will occur between them. | b. | The observed frequency of recombination of two
genes that are far apart from each other has a maximum value of 50%. | c. | Two of the traits
that Mendel studied-seed color and flower color-are linked on the same
chromosome. | d. | Only B and C are correct. | e. | A, B, and C are
correct. |
|
|
|
80.
|
How would one explain a testcross involving F1 dihybrid flies in which more
parental-type offspring than recombinant-type offspring are produced?
a. | The two genes are linked. | b. | The two genes are unlinked. | c. | Recombination did
not occur in the cell during meiosis. | d. | The testcross was improperly
performed. | e. | Both of the characters are controlled by more than one
gene. |
|
|
|
81.
|
New combinations of linked genes are due to which of the following?
a. | nondisjunction | b. | crossing over | c. | independent
assortment | d. | mixing of sperm and egg | e. | both A and C |
|
|
|
82.
|
What does a frequency of recombination of 50% indicate?
a. | The two genes likely are located on different chromosomes. | b. | All of the offspring
have combinations of traits that match one of the two parents. | c. | The genes are
located on sex chromosomes. | d. | Abnormal meiosis has
occurred. | e. | Independent assortment is hindered. |
|
|
|
The following questions refer to the data and figures below.
CROSS
I. Purebred lines of wild-type fruit flies (gray body and normal wings) are mated to flies with black
bodies and vestigial wings.  F1 offspring all have a normal
phenotype. 
CROSS II. F1 flies are crossed
with flies recessive for both traits (a testcross). | Resulting Offspring | Normal | Percentage | | Gray body; normal
wings | 575 | 25.1 | | Black body; vestigial wings | 571 | 24.9 | | Black body;
normal wings | 577 | 25.2 | | Gray body; vestigial wings | 568 | 24.8 | | | |
KEY: A. CROSS I
results give evidence supporting the statement. B. CROSS I results give evidence against the
statement. C. CROSS II results give evidence supporting the statement. D. CROSS II results give
evidence against the statement. E. Neither CROSS I nor CROSS II results support the
statement.
|
|
|
83.
|
Vestigial wings are a recessive trait.
|
|
|
84.
|
The genes for body color and wing shape are linked.
|
|
|
85.
|
An F1 cross should produce flies that will fall into a Mendelian 9:3:3:1
ratio.
|
|
|
86.
|
There are 25 centimorgans (map units) between the genes for body color and wing
shape.
|
|
|
87.
|
A 0.1% frequency of recombination is observed
a. | only in sex chromosomes. | b. | only on genetic maps of viral
chromosomes. | c. | on unlinked chromosomes. | d. | in any two genes on different
chromosomes. | e. | in genes located very close to one another on the same
chromosome. |
|
|
|
88.
|
The following is a map of four genes on a chromosome:  Between which two genes would you expect the highest frequency of
recombination? a. | A and W | b. | W and E | c. | E and
G | d. | A and E | e. | A and G |
|
|
|
89.
|
The reason that linked genes are inherited together is that
a. | they are located on the same chromosome. | b. | the number of genes
in a cell is greater than the number of chromosomes. | c. | chromosomes are
unbreakable. | d. | alleles are paired. | e. | genes align that way during metaphase
I. |
|
|
|
90.
|
What is the mechanism for the production of genetic recombinants?
a. | X inactivation | b. | methylation of cytosine | c. | crossing over and
independent assortment | d. | nondisjunction | e. | deletions and
duplications during meiosis |
|
|
|
91.
|
There is good evidence for linkage when
a. | two genes occur together in the same gamete. | b. | a gene is associated
with a specific phenotype. | c. | two genes work together to control a specific
characteristic. | d. | genes do not segregate independently during meiosis. | e. | two characteristics
are caused by a single gene. |
|
|
|
Refer to the figure below to answer the following questions.
|
|
|
92.
|
In a series of mapping experiments, the recombination frequencies for four
different linked genes of Drosophila were determined as shown in the figure. What is the order
of these genes on a chromosome map?
a. | rb-cn-vg-b | b. | vg-b-rb-cn | c. | cn-rb-b-vg | d. | b-rb-cn-vg | e. | vg-cn-b-rb |
|
|
|
93.
|
Which of the following two genes are closest on a genetic map of
Drosophila?
a. | b and vg | b. | vg and cn | c. | rb and
cn | d. | cn and b | e. | b and
rb |
|
|
|
X, Y, and Z are three genes in Drosophila . The recombination
frequencies for two of the three genes are shown below.
|
|
|
94.
|
Genes X and Y could be
a. | located on different chromosomes. | b. | located very near to each other on the same
chromosome. | c. | located far from each other on the same chromosome. | d. | both A and
B | e. | both A and C |
|
|
|
95.
|
If the recombination frequency for Y and Z was found to be 50%, this would mean
that
a. | genes X and Y are on the same chromosome. | b. | genes X and Y are on
different chromosomes. | c. | genes Y and Z are on different
chromosomes. | d. | both A and C. | e. | both B and C |
|
|
|
96.
|
Which of the following is true regarding linkage maps? They
a. | always have a total of 100 map units. | b. | can be used to pinpoint the precise physical
position of a gene on a chromosome. | c. | are a genetic map based on recombination
frequencies. | d. | require preparation of karyotypes. | e. | reflect the frequency of crossing over between
X and Y chromosomes. |
|
|
|
97.
|
The frequency of crossing over between any two linked genes is
a. | higher if they are recessive. | b. | different between males and
females. | c. | determined by their relative dominance. | d. | the same as if they
were not linked. | e. | proportional to the distance between them. |
|
|
|
98.
|
Sturtevant provided genetic evidence for the existence of four pairs of
chromosomes in Drosophila by showing that
a. | there are four major functional classes of genes in
Drosophila. | b. | Drosophila genes cluster into four
distinct groups of linked genes. | c. | the overall number of genes in
Drosophila is a multiple of four. | d. | the entire Drosophila genome has
approximately 400 map units. | e. | Drosophila genes have, on average, four
different alleles. |
|
|
|
99.
|
Map units on a linkage map cannot be relied upon to calculate physical distances
on a chromosome because
a. | the frequency of crossing over varies along the length of the
chromosome. | b. | the relationship between recombination frequency and map units is different in every
individual. | c. | physical distances between genes change during the course of the cell
cycle. | d. | the gene order on the chromosomes is slightly different in every
individual. | e. | all of the above |
|
|
|
100.
|
A map of a chromosome that includes the positions of genes relative to visible
chromosomal features, such as stained bands, is called a
a. | linkage map. | b. | physical map. | c. | recombination
map. | d. | cytogenetic map. | e. | banded map. |
|
|
|
101.
|
Males are more often affected by sex-linked traits than females because
a. | males are hemizygous for the X chromosome. | b. | male hormones such
as testosterone often exacerbate the effects of mutations on the X chromosome. | c. | female hormones such
as estrogen often compensate for the effects of mutations on the X. | d. | X chromosomes in
males generally have more mutations than X chromosomes in females. | e. | mutations on the Y
chromosome often exacerbate the effects of X-linked mutations. |
|
|
|
102.
|
SRY is
a. | a gene present on the Y chromosome that triggers male
development. | b. | a gene present on the X chromosome that triggers female
development. | c. | an autosomal gene that is required for the expression of genes on the Y
chromosome. | d. | an autosomal gene that is required for the expression of genes on the X
chromosome. | e. | required for development, and males or females lacking the gene do not survive past
early childhood. |
|
|
|
103.
|
In cats, black fur color is caused by an X-linked allele; the other
allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring
would you expect from the cross of a black female and an orange male?
a. | tortoiseshell female; tortoiseshell male | b. | black female; orange
male | c. | orange female; orange male | d. | tortoiseshell female; black
male | e. | orange female; black male |
|
|
|
104.
|
Red-green color blindness is a sex-linked recessive trait in humans. Two people
with normal color vision have a color-blind son. What are the genotypes of the parents?
a. | XcXc and XcY | b. | XcXc and XCY | c. | XCXC and
XcY | d. | XCXC and XCY | e. | XCXc and
XCY |
|
|
|
105.
|
In the following list, which term is least related to the others?
a. | Duchenne muscular dystrophy | b. | autosome | c. | sex-linked
genes | d. | color blindness | e. | hemophilia |
|
|
|
106.
|
Cinnabar eyes is a sex-linked recessive characteristic in fruit flies. If a
female having cinnabar eyes is crossed with a wild-type male, what percentage of the F1 males will
have cinnabar eyes?
a. | 0% | b. | 25% | c. | 50% | d. | 75% | e. | 100% |
|
|
|
107.
|
Most calico cats are female because
a. | a male inherits only one of the two X-linked genes controlling hair
color. | b. | the males die during embryonic development. | c. | the Y chromosome has
a gene blocking orange coloration. | d. | only females can have Barr
bodies. | e. | multiple crossovers on the Y chromosome prevent orange pigment
production. |
|
|
|
108.
|
A recessive allele on the X chromosome is responsible for red-green color
blindness in humans. A woman with normal vision whose father is color-blind marries a color-blind
male. What is the probability that a son of this couple will be color-blind?
|
|
|
109.
|
In birds, sex is determined by a ZW chromosome scheme. Males are ZZ and females
are ZW. A lethal recessive allele that causes death of the embryo is sometimes present on the Z
chromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male that is
heterozygous for the lethal allele and a normal female?
a. | 2:1 male to female | b. | 1:2 male to female | c. | 1:1 male to
female | d. | 4:3 male to female | e. | 3:1 male to
female |
|
|
|
110.
|
A man who carries an X-linked allele will pass it on to
a. | all of his daughters. | b. | half of his daughters. | c. | all of his
sons. | d. | half of his sons. | e. | all of his
children. |
|
|
|
Refer to the information below to answer the following questions.
An achondroplastic male dwarf with normal vision marries a color-blind woman of normal
height. The man's father was six-feet tall, and both the woman's parents were of average
height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked
recessive.
|
|
|
111.
|
How many of their daughters might be expected to be color-blind dwarfs?
a. | all | b. | none | c. | half | d. | one out of four | e. | three out of
four |
|
|
|
112.
|
How many of their sons would be color-blind and of normal height?
a. | all | b. | none | c. | half | d. | one out of four | e. | three out of
four |
|
|
|
113.
|
They have a daughter who is a dwarf with normal color vision. What is the
probability that she is heterozygous for both genes?
a. | 0 | b. | 0.25 | c. | 0.50 | d. | 0.75 | e. | 1.00 |
|
|
|
114.
|
Male calico cats could be the result of
a. | sex-linked inheritance. | b. | nondisjunction, leading to the male calico
having two X chromosomes. | c. | incomplete dominance of multiple
alleles. | d. | recessive alleles retaining their fundamental natures even when
expressed. | e. | a reciprocal translocation. |
|
|
|
115.
|
A Barr body is normally found in the nucleus of which kind of human cell?
a. | unfertilized egg cells only | b. | sperm cells only | c. | somatic cells of a
female only | d. | somatic cells of a male only | e. | both male and female somatic
cells |
|
|
|
116.
|
Which of these syndromes afflicts mostly males?
a. | Turner syndrome | b. | Down syndrome | c. | Duchenne muscular
dystrophy | d. | cri du chat syndrome | e. | chronic myelogenous
leukemia |
|
|
|
117.
|
If a human interphase nucleus of a person contains three Barr bodies, it can be
assumed that the person
a. | has hemophilia. | b. | is a male. | c. | has four X
chromosomes. | d. | has Turner syndrome. | e. | has Down
syndrome. |
|
|
|
118.
|
If nondisjunction occurs in meiosis II during gametogenesis, what will be the
result at the completion of meiosis?
a. | All the gametes will be diploid. | b. | Two gametes will be n + 1, and two will
be n - 1. | c. | One gamete will be n + 1, one will be n - 1, and two will be
n. | d. | There will be three extra gametes. | e. | Two of the four gametes will be haploid, and
two will be diploid. |
|
|
|
119.
|
The figure below represents the stained nucleus from a cheek epithelial cell of
an individual whose genotype would probably be  a. | XX. | b. | XY. | c. | XYY. | d. | XXX. | e. | XXY. |
|
|
|
120.
|
If a pair of homologous chromosomes fails to separate during anaphase of meiosis
I, what will be the chromosome number of the four resulting gametes with respect to the normal
haploid number (n)?
a. | n + 1; n + 1; n - 1; n - 1 | b. | n + 1; n - 1; n; n | c. | n + 1; n - 1; n -
1; n - 1 | d. | n + 1; n + 1; n; n | e. | n - 1; n - 1; n;
n |
|
|
|
121.
|
A cell that has 2n + 1 chromosomes is
a. | trisomic. | b. | monosomic. | c. | aneuploid. | d. | polyploid. | e. | both A and
C |
|
|
|
122.
|
If a chromosome lacks certain genes, what has most likely occurred?
a. | disjunction | b. | an inversion | c. | a
deletion | d. | a translocation | e. | a
nonduplication |
|
|
|
123.
|
One possible result of chromosomal breakage is for a fragment to join a
nonhomologous chromosome. This is called a (an)
a. | deletion. | b. | disjunction. | c. | inversion. | d. | translocation. | e. | duplication. |
|
|
|
124.
|
In the following list, which term is least related to the others?
a. | trisomic | b. | monosomic | c. | aneuploid | d. | triploid | e. | nondisjunction |
|
|
|
125.
|
A nonreciprocal crossover causes which of the following products?
a. | deletion | b. | duplication | c. | nondisjunction | d. | A and B | e. | B and
C |
|
|
|
126.
|
One possible result of chromosomal breakage can be that a fragment reattaches to
the original chromosome in a reverse orientation. This is called
a. | disjunction. | b. | translocation. | c. | deletion. | d. | inversion. | e. | aneuploidy. |
|
|
|
127.
|
A human individual is phenotypically female, but her interphase somatic nuclei
do not show the presence of Barr bodies. Which of the following statements concerning her is probably
true?
a. | She has Klinefelter syndrome. | b. | She has an extra X
chromosome. | c. | She has Turner syndrome. | d. | She has the normal number of sex
chromosomes. | e. | She has two Y chromosomes. |
|
|
|
128.
|
The karyotype shown below is associated with which of the following genetic
disorders?  a. | Turner syndrome | b. | Down syndrome | c. | Klinefelter
syndrome | d. | hemophilia | e. | male-pattern
baldness |
|
|
|
129.
|
In humans, male-pattern baldness is controlled by a gene that occurs in two
allelic forms. Allele Hn determines nonbaldness, and allele Hb determines pattern
baldness. In males, because of the presence of testosterone, allele Hb is dominant over
Hn. If a man and woman both with genotype HnHb have a son, what is the chance that he will
eventually be bald?
|
|
|
130.
|
Of the following human trisomies, the one that generally has the most severe
impact on the health of the individual is
a. | trisomy 21. | b. | Klinefelter syndrome (XXY). | c. | trisomy
X. | d. | XYY. | e. | All of the above have equal
impact. |
|
|
|
131.
|
What do all human males inherit from their mother?
a. | mitochondrial DNA | b. | an X chromosome | c. | the SRY
gene | d. | A and B only | e. | A, B, and C |
|
|
|
132.
|
Which of the following statements is true regarding genomic
imprinting?
a. | It explains cases in which the gender of the parent from whom an allele is inherited
affects the expression of that allele. | b. | It is greatest in females because of the larger
maternal contribution of cytoplasm. | c. | It may explain the transmission of Duchenne
muscular dystrophy. | d. | It involves an irreversible alteration in the
DNA sequence of imprinted genes. | e. | All of the above are
correct. |
|
|
|
133.
|
The pedigree in the figure below shows the transmission of a trait in a
particular family. Based on this pattern of transmission, the trait is most likely  a. | mitochondrial. | b. | autosomal recessive. | c. | sex-linked
dominant. | d. | sex-linked recessive. | e. | autosomal
dominant. |
|
|
|
134.
|
Which of the following statements about mitochondria is false?
a. | Because of the role of the mitochondria in producing cellular energy, mitochondrial
diseases often affect the muscles and nervous system. | b. | Because mitochondria are present in the
cytoplasm, mitochondrial diseases are transmitted maternally. | c. | Like nuclear genes,
mitochondrial genes usually follow Mendelian patterns of inheritance. | d. | Mitochondria contain
circular DNA molecules that code for proteins and RNAs. | e. | Many mitochondrial
genes encode proteins that play roles in the electron transport chain and ATP
synthesis. |
|