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Chapter 11 Test Bank
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Each pea-plant gamete has how many alleles for the height gene?
a. 1
b. 2
c. 3
d. 4
____
2. The different forms of a gene are called
a. traits.
b. pollinations.
c. alleles.
d. hybrids.
____
3. Gregor Mendel removed the male parts from the flowers of some plants in order to
a. prevent hybrids from forming.
b. prevent cross-pollination.
c. stimulate self-pollination.
d. control crosses between plants.
____
4. If a pea plant has a recessive allele for green peas, it will produce
a. green peas if it also has a dominant allele for yellow peas.
b. both green peas and yellow peas if it also has a dominant allele for yellow peas.
c. green peas if it does not also have a dominant allele for yellow peas.
d. yellow peas if it does not also have a dominant allele for green peas.
____
5. When Gregor Mendel crossed a tall plant with a short plant, the F1 plants inherited
a. one allele from each parent.
b. two alleles from each parent.
c. three alleles from each parent.
d. four alleles from each parent.
____
6. If a pea plant’s alleles for height are tt, what is true of its parents?
a. Both parents were tall.
b. Both parents were short.
c. Both parents contributed a recessive allele.
d. Both parents contributed a dominant allele.
____
7. When Gregor Mendel crossed true-breeding tall plants with true-breeding short plants, why was it impossible
to observe segregation?
a. Alleles for height do not segregate in the F2.
b. Alleles segregate only in the F2 generation.
c. Alleles segregate best when two tall plants are crossed.
d. Alleles in the F1 must be Tt to have height variety in the F2.
____
8. A tall plant (TT) is crossed with a short plant (tt). If the tall F1 pea plants are allowed to self-pollinate,
a. the offspring will be of medium height.
b. all of the offspring will be tall.
c. all of the offspring will be short.
d. the offspring can be tall or short.
____
9. In the P generation, a tall plant was crossed with a short plant. Short plants reappeared in the F2 generation
because
a. the allele for shortness becomes more common in the F2 generation.
b. the allele for shortness becomes dominant in the F2 generation.
c. the alleles for both heights segregated when the F1 plants made gametes.
d. the alleles for tallness begin to disappear in the F2 generation.
____ 10. When you flip a coin, what is the probability that it will come up tails?
a. 1
b. 1/2
c. 1/4
d. 1/8
____ 11. The principles of probability can be used to
a. predict the traits of the offspring of genetic crosses.
b. determine the actual outcomes of genetic crosses.
c. determine which species should be used in genetic crosses.
d. decide which organisms are best to use in genetic crosses.
____ 12. A heterozygous tall pea plant is crossed with a short plant. The probability that an F1 plant will be tall is
a. 25%.
b. 50%.
c. 75%.
d. 100%.
____ 13. Organisms that have two identical alleles for a particular trait are said to be
a. hybrid.
b. homozygous.
c. heterozygous.
d. dominant.
Tt
T
t
T
TT
Tt
T
TT
Tt
TT
T
=
Tall
t
=
Short
Figure 11–1
____ 14. In the Punnett square shown in Figure 11–1, which of the following is true about the offspring resulting from
the cross?
a. About half are expected to be short.
b. All are expected to be short.
c. About three fourths are expected to be tall.
d. All are expected to be tall.
____ 15. What principle states that during gamete formation genes for different traits separate without influencing each
other’s inheritance?
a. principle of dominance
b. principle of independent assortment
c. principle of probabilities
d. principle of segregation
RrYy
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
RrYy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Seed Shape
R = Round
r = Wrinkled
Seed Color
Y = Yellow
y = Green
Figure 11–2
____ 16. The Punnett square in Figure 11–2 shows that the gene for pea shape and the gene for pea color
a. assort independently.
b. are linked.
c. have the same alleles.
d. are always homozygous.
____ 17. How many different allele combinations would be found in the gametes produced by a pea plant whose
genotype was RrYY?
a. 2
b. 4
c. 8
d. 16
RrYy
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
RRYy
Ry
RRYy
RRyy
RrYy
Rryy
RY
RRYY
RrYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
Figure 11–3
____ 18. Use Figure 11–3 to answer the following question. If a pea plant that is heterozygous for round, yellow peas
(RrYy) is crossed with a pea plant that is homozygous for round peas but heterozygous for yellow peas
(RRYy), how many different phenotypes are their offspring expected to show?
a. 2
b. 4
c. 8
d. 16
____ 19. Gregor Mendel’s principles of genetics apply to
a. plants only.
b. animals only.
c. pea plants only.
d. all organisms.
____ 20. Why did Thomas Hunt Morgan use fruit flies in his studies?
a. Fruit flies produce a large number of offspring.
b. Fruit flies take a long time to produce offspring.
c. Fruit flies share certain characteristics with pea plants.
d. Fruit flies have a long life span.
____ 21. A male and female bison that are both heterozygous for normal skin pigmentation (Aa) produce an albino
offspring (aa). Which of Mendel’s principles explain(s) why the offspring is albino?
a. dominance only
b. independent assortment only
c. dominance and segregation
d. segregation only
____ 22. Roan cattle show codominance for the color of their hair. There are alleles for red hair and white hair. What
would you expect a heterozygous roan bull to look like if the trait showed incomplete dominance instead?
a. It would be red.
b. It would be white.
c. It would be spotted.
d. It would be pink.
____ 23. A breed of chicken shows codominance for feather color. One allele codes for black feathers, another codes
for white feathers. The feathers of heterozygous chickens of this breed will be
a. black.
b. white.
c. gray.
d. speckled.
____ 24. Situations in which one allele for a gene is not completely dominant over another allele for that gene are
called
a. multiple alleles.
b. incomplete dominance.
c. polygenic inheritance.
d. multiple genes.
____ 25. In rabbits, there are four different versions of the gene for coat color. What pattern of inheritance is this?
a. incomplete dominance.
b. polygenic inheritance.
c. codominance.
d. multiple alleles.
____ 26. Variation in human skin color is an example of
a. incomplete dominance.
b. codominance.
c. polygenic traits.
d. multiple alleles.
____ 27. What determines the color of western white butterflies?
a. genes alone.
b. the environment alone
c. temperature and genes
d. exposure to sunlight and genes
____ 28. Which of the following supports the claim that the environment can affect genetic traits?
a. Oak trees get taller as they grow.
b. Hydrangea flower color varies with soil pH.
c. Dandelion plants are self pollinating.
d. Pinion trees bear cones every other year.
____ 29. The arctic fox is blue-gray in the summer and white in the winter. What most likely influence(s) this change?
a. genes and the environment
b. dominant alleles
c. the environment alone
d. codominant alleles
____ 30. The number of chromosomes in a gamete is represented by the symbol
a. Z.
b. X.
c. N.
d. Y.
____ 31. If an organism’s diploid number is 12, its haploid number is
a. 12.
b. 6.
c. 24.
d. 3.
____ 32. Gametes have
a. homologous chromosomes.
b. twice the number of chromosomes found in body cells.
c. two sets of chromosomes.
d. one allele for each gene.
____ 33. Gametes are produced by the process of
a. mitosis.
b. meiosis.
c. crossing-over.
d. replication.
Figure 11–4
____ 34. What is shown in Figure 11–4?
a. independent assortment
b. anaphase I of meiosis
c. crossing-over
d. replication
____ 35. Chromosomes form tetrads during
a. prophase I of meiosis.
b. metaphase I of meiosis.
c. interphase.
d. anaphase II of meiosis.
____ 36. A tetrad consists of
a. a homologous pair of chromosomes, each made of two chromatids.
b. the four copies of a chromosome that are normally present in cells.
c. two sister chromatids that have each been replicated during interphase.
d. a parental chromosome that was replicated to form a pair, then replicated again.
____ 37. Unlike mitosis, meiosis results in the formation of
a. diploid cells.
b. haploid cells.
c. 2N daughter cells.
d. body cells.
____ 38. What is formed at the end of meiosis?
a. two genetically identical cells
b. four genetically different cells
c. four genetically identical cells
d. two genetically different cells
____ 39. One way that meiosis I is different from mitosis is that
a. replication occurs during interphase before mitosis, but not before meiosis I.
b. meiosis I produces 2 haploid daughter cells, but mitosis produces 2 diploid daughter cells.
c. homologous chromosomes are segregated during mitosis, but remain together during
meiosis I.
d. sister chromatids are pulled apart during meiosis I, but not during mitosis.
____ 40. Which of the following assort independently?
a. chromosomes
b. linked genes
c. multiple alleles
d. codominant alleles
____ 41. Linked genes
a. are never separated.
b. assort independently.
c. are on the same chromosome.
d. are always recessive.
Gene Map of Chromosome 2 of the Fruit Fly
Figure 11–5
____ 42. Which trait is most likely linked to having a curved wing in the fruit fly in Figure 11–5?
a. dumpy wing
b. vestigial wing
c. arc (bent wings)
d. speck wing
____ 43. Gene maps are based on
a. the frequencies of crossing-over.
b. independent assortment.
c. genetic diversity.
d. the number of genes in a cell.
____ 44. If two genes are on the same chromosome and rarely assort independently,
a.
b.
c.
d.
crossing-over never occurs between the genes.
crossing-over always occurs between the genes.
the genes are probably located far apart from each other.
the genes are probably located close to each other.
____ 45. The farther apart two genes are located on a chromosome, the
a. less likely they are to be inherited together.
b. more likely they are to be linked.
c. less likely they are to assort independently.
d. less likely they are to be separated by crossing over.
Modified True/False
Indicate whether the statement is true or false. If false, change the identified word or phrase to make the statement true.
____
1. A trait is a specific characteristic that can vary from one individual to another. _________________________
____
2. True-breeding plants that produced axial flowers were crossed with true-breeding plants that produced
terminal flowers. The resulting offspring all produced terminal flowers because the allele for terminal flowers
is recessive. _________________________
____
3. During the formation of gametes in a hybrid tall plant, the tall allele and the short allele stay together.
_________________________
____
4. If the alleles for a trait did not segregate during gamete formation, offspring would always show the trait of at
least one of the parents. _________________________
____
5. The principles of probability can explain the numerical results of Mendel’s experiments.
_________________________
____
6. The probability that a gamete produced by a pea plant heterozygous for stem height (Tt) will contain the
recessive allele is 100%. _________________________
____
7. If Mendel had found that an F2 cross of plants that were heterozygous for two traits had made offspring with
two phenotypes, this finding would have supported the theory of independent assortment.
_________________________
____
8. A trait in an unidentified plant is controlled by one gene that has two alleles. One allele is dominant over the
other. According to Mendel’s principles, one fourth of the offspring made from a cross between two
heterozygous plants will show the recessive trait. _________________________
____
9. If two speckled chickens are mated, according to the principle of codominance, 25% of the offspring are
expected to be speckled. _________________________
____ 10. For the trait of blood type in humans, there is an allele for Type A, an allele for Type B, and an allele for Type
O. Each person inherits one of these alleles from each of their parents, and their blood type is determined by
what combination of these alleles they receive. Blood type is inherited as a polygenic trait.
____ 11. If an organism has 16 chromosomes in each of its egg cells, the organism’s diploid number is 32.
_________________________
____ 12. If an organism is heterozygous for a particular gene, the two different alleles will be separated during
anaphase II of meiosis, assuming that no crossing-over has occurred. _________________________
____ 13. Mitosis results in two cells, whereas meiosis results in one cell. _________________________
____ 14. If an organism has four linkage groups, it has eight chromosomes. _________________________
____ 15. Genes in the same linkage group are usually inherited separately. _________________________
Completion
Complete each statement.
1. The plants that Gregor Mendel crossed to produce the F1 generation made up the ____________________
generation.
2. Due to the process of segregation, alleles separate during the production of ____________________ .
3. An organism has 38 chromosomes in a body cell. After mitosis each cell has 38 chromosomes. After meiosis
each gamete has ____________________ chromosomes.
4. What is the probability of flipping a coin and getting heads 5 times in a row?
Tt
T
t
T
TT
Tt
T
TT
Tt
TT
T
=
Tall
t
=
Short
Figure 11–1
5. In the Punnett square shown in Figure 11–1, the genotypes of the offspring are ____________________.
6. The principle of independent assortment states that ____________________ for different traits can segregate
independently during the formation of gametes.
7. If pea plants that are homozygous for round, yellow seeds (RRYY) were crossed with pea plants that are
heterozygous for round, yellow seeds (RrYy), the expected phenotype(s) of the offspring would be
_________________________.
8.
____________________’s principles can be used to study heredity in dogs, cats and sheep.
9. The reddish-brown pigment that gives color to a fruit-fly’s eye is controlled by three genes, so a fruit fly’s eye
color is a __________________.
10. Western white butterflies that hatch in springtime have more pigment in their wings than those that hatch in
summer. The darker wings help the butterflies stay warmer by absorbing more ____________________ than
the lighter-colored wings.
11. The characteristics of an organism are determined by two factors: ____________________.
12. In four o’clock plants, flower color is controlled by two alleles that show _________________________.
13. An organism’s gametes have ____________________ the number of chromosomes found in the organism’s
body cells.
14. Crossing-over occurs during the stage of meiosis called ____________________.
15. The relative locations of each known gene can be shown on a ____________________ map.
Short Answer
1. What attributes of the garden pea plant made it an excellent organism for Gregor Mendel’s genetic studies?
2. Explain how allowing the F1 generation plants to self-pollinate in producing the F2 generation allowed
Mendel to reach his conclusions about inheritance.
3. Explain how a trait might seem to “disappear” for a generation, and then “reappear” in the following
generation.
RrYy
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
RrYy
rY
RrYY
RrYy
rrYY
rrYy
Seed Shape
R = Round
r = Wrinkled
Seed Color
Y = Yellow
y = Green
ry
RrYy
Rryy
rrYy
rryy
Figure 11–2
4. What is the phenotype ratio of the offspring of the plants in the Punnett square in Figure 11–2?
5. A tall pea plant with yellow seeds is heterozygous for height and seed color (TtYy). This plant is crossed with
a pea plant heterozygous for height but homozygous recessive for seed color (Ttyy). If 80 offspring are
produced, how many are expected to be tall and have yellow seeds?
6. Mendel chose to use pea plants in his studies and Morgan chose to use fruit flies in his work. What
characteristics of these species make them good organisms to use for the study of genetics?
7. What is the probability that a cross between parents who are both homozygous recessive for trait will have
offspring that are homozygous recessive for that trait?
8. Arctic foxes have blue-gray fur in summer and change to white fur in winter. What would be one way to test
whether this change is influenced by cooling seasonal temperatures?
9. How many sets of chromosomes are in a diploid cell?
10. Define homologous chromosomes.
11. What happens to the number of chromosomes per cell during meiosis?
12. Contrast the cells produced by mitosis with those produced by meiosis.
13. What are the only kinds of cells that undergo meiosis?
14. Why did Gregor Mendel not observe gene linkage during his experiments with pea plants?
Gene Map of Chromosome 2 of the Fruit Fly
Figure 11–5
15. Figure 11–5, the gene map of a fruit fly’s chromosome 2, shows the relative locations of the star eye, dumpy
wing, and black body genes to be 1.3, 13.0, and 48.5, respectively. Between which two genes does crossingover occur most frequently?
Science Skills
Heterozygous male guinea pigs with black, rough hair (BbRr) are crossed with heterozygous female guinea
pigs with black, rough hair (BbRr). The incomplete Punnett square in Figure 11–6 shows the expected results
from the cross.
BbRr
BR
Br
bR
br
BR
BBRR
BBRr
BbRR
BbRr
Br
BBRr
BBrr
BbRr
Bbrr
BbRr
bR
BbRR
BbRr
?
bbRr
br
BbRr
Bbrr
bbRr
bbrr
Figure 11–6
Hair Color
B = Black
b = White
Hair Texture
R = Rough
r = Smooth
1. Interpret Tables Identify the genotype of the offspring that would be represented by the question mark in
Figure 11–6.
2. Interpret Tables Identify the phenotype of the offspring represented by the question mark in Figure 11–6.
3. Analyze Data In Figure 11–6, what are the different phenotypes of the offspring?
4. Analyze Data In Figure 11–6, what are the genotypes of the offspring that have black, rough hair?
5. Calculate What fraction of the offspring in Figure 11–6 would be expected to have white, smooth hair?
Figure 11–7
6. Infer What do the letters R and I represent in Figure 11–7?
7. Interpret Visuals In Figure 11–7, what is the genotype of the pink-flowered snapdragons?
8. Infer Explain whether the alleles in Figure 11–7 show dominance, incomplete dominance, or codominance.
9. Infer According to Figure 11–7, if red-flowered snapdragons and ivory-flowered snapdragons are crossed,
what percentage of their offspring would be expected to be pink-flowered?
10. Infer Could the red snapdragons shown in the P generation of Figure 11–7 ever have white offspring?
Explain your answer.
Figure 11–8
11. Interpret Visuals In Figure 11–8, what is the structure labeled X in stage A?
12. Interpret Visuals In Figure 11–8, during which stage might new allele combinations form? Identify the
stage.
13. Infer If the stages shown in Figure 11–8 were taking place in a female animal, how many eggs would
generally result from stage G? Explain your answer.
14. Interpret Visuals List the stages in Figure 11–8 in which the cells are 2N and those in which the cells are N.
15. Infer In Figure 11–8, in which stage does each cell have a single copy of each gene? Identify the stage.
Essay
1. You wish to determine whether a tall pea plant is homozygous or heterozygous for tallness. What cross
should you perform to arrive at your answer? Explain your choice of cross.
2. A pea plant with yellow seeds was crossed with a plant with green seeds. The F1 generation produced plants
with yellow seeds. Explain why green seeds reappeared in the F2 generation.
3. Why are the results of genetic crosses shown in Punnett squares interpreted as probabilities, not certainties?
Give some specific reasons.
4. A cross between two organisms heterozygous for two different genes (AaBb) likely results in a 9 : 3 : 3 : 1
phenotype ratio among the offspring. Is the offspring’s genotype ratio the same? Explain your answer.
5. A plant nursery owner wants to guarantee that the seeds she sells produce only pink-flowered four o’clock
plants. How should she obtain the seeds?
6. Suppose you have western white butterflies from a population that hatched in spring, and those butterflies
then had offspring that hatched in the summer. What would you expect the offspring born in the summer to
look like? Explain your answer.
7. Which has more genetic information, a body cell or a gamete? Explain your answer.
8. The stages of meiosis are classified into two phases: meiosis I and meiosis II. Compare and contrast these two
phases.
9. Explain why the daughter cells produced by meiosis are genetically different from each other, whereas the
daughter cells produced by mitosis are not.
10. Define linkage, and explain how linkage is used to make gene maps.
Chapter 11 Test Bank
Answer Section
MULTIPLE CHOICE
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knowledge
knowledge
evaluation
synthesis
comprehension
application
analysis
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analysis
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analysis
synthesis
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comprehension
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synthesis
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comprehension
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MODIFIED TRUE/FALSE
1. ANS: T
2. ANS: F, dominant
PTS: 1
3. ANS: F, separate
BLM: application
PTS: 1
4. ANS: T
5. ANS: T
6. ANS: F
BLM: knowledge
PTS: 1
BLM: comprehension
PTS: 1
PTS: 1
BLM: evaluation
BLM: comprehension
PTS: 1
BLM: analysis
PTS: 1
BLM: analysis
50%
PTS: 1
BLM: application
7. ANS: F, contradicted
PTS: 1
8. ANS: T
9. ANS: F, 50%
BLM: evaluation
PTS: 1
BLM: analysis
10. ANS: F, trait controlled by multiple alleles
PTS: 1
BLM: application
11. ANS: T
12. ANS: F, anaphase I
PTS: 1
13. ANS: F, four cells
BLM: analysis
PTS: 1
14. ANS: F, four
BLM: knowledge
PTS: 1
15. ANS: F, together
BLM: analysis
PTS: 1
COMPLETION
BLM: application
1. ANS: P
PTS: 1
BLM: application
2. ANS: gametes, sex cells
PTS: 1
3. ANS: 19
BLM: comprehension
PTS: 1
BLM: analysis
4. ANS: 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32
PTS: 1
5. ANS: TT and Tt
BLM: analysis
PTS: 1
BLM: application
6. ANS: genes, chromosomes
PTS: 1
BLM: knowledge
7. ANS: round yellow seeds only
PTS: 1
8. ANS: Mendel
BLM: analysis
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BLM: application
9. ANS: polygenic trait
PTS: 1
10. ANS: light energy
BLM: comprehension
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BLM: evaluation
11. ANS: genes and environmental conditions
PTS: 1
BLM: knowledge
12. ANS: incomplete dominance
PTS: 1
13. ANS: half
BLM: analysis
PTS: 1
14. ANS: prophase I
BLM: comprehension
PTS: 1
15. ANS: gene
BLM: analysis
PTS: 1
SHORT ANSWER
1. ANS:
BLM: knowledge
Garden pea plants produce many offspring, they have traits that come in only two forms, and crosses between
the plants can be controlled easily.
PTS: 1
BLM: synthesis
2. ANS:
Allowing the F1 pea plants to self-pollinate caused the recessive phenotype to reappear in the F2 generation.
Self-pollination of the F1 plants also allowed the 3:1 phenotype ratios to occur, supporting Mendel’s theory.
Self-pollination showed that traits controlled by recessive alleles could reappear in the F2 generation.
PTS: 1
BLM: synthesis
3. ANS:
If all of the individuals in a generation receive one dominant allele and one recessive allele, then they would
all show the dominant trait. If they are bred, then they will pass on the dominant allele to some of their
offspring and the recessive allele to others. Offspring receiving two recessive alleles will show the recessive
trait, so it will reappear.
PTS: 1
BLM: analysis
4. ANS:
The phenotype ratio is 9 round, yellow seeds : 3 round, green seeds : 3 wrinkled, yellow seeds : 1 wrinkled,
green seed.
PTS: 1
BLM: application
5. ANS:
Thirty of the offspring are expected to be tall and have yellow seeds.
PTS: 1
BLM: analysis
6. ANS:
Both pea plants and fruit flies are small organisms and can be easily manipulated. Both can produce large
numbers of offspring in a relatively short period of time.
PTS: 1
7. ANS:
100%
BLM: synthesis
PTS: 1
BLM: analysis
8. ANS:
Keep an arctic fox warm when its native environment would be cool.
PTS: 1
BLM: evaluation
9. ANS:
A diploid cell has two sets of chromosomes.
PTS: 1
BLM: knowledge
10. ANS:
Homologous chromosomes are the two sets of chromosomes found in a body cell—one set inherited from the
male parent and the other inherited from the female parent.
PTS: 1
BLM: comprehension
11. ANS:
The number of chromosomes is cut in half.
PTS: 1
BLM: knowledge
12. ANS:
Mitosis produces diploid body cells, whereas meiosis produces haploid gametes.
PTS: 1
13. ANS:
sex cells, gametes
BLM: analysis
PTS: 1
BLM: knowledge
14. ANS:
The genes that Mendel studied were located on different chromosomes or were located far apart on the same
chromosome.
PTS: 1
BLM: evaluation
15. ANS:
Crossing-over occurs most frequently between the star eye gene and the black body gene.
PTS: 1
BLM: synthesis
SCIENCE SKILLS
1. ANS:
The genotype of the offspring is bbRR.
PTS: 1
BLM: application
2. ANS:
The phenotype of the offspring is white, rough hair.
PTS: 1
BLM: application
3. ANS:
The phenotypes of the offspring are black, rough hair; black, smooth hair; white, rough hair; and white,
smooth hair.
PTS: 1
BLM: application
4. ANS:
Offspring with black, rough hair have the genotypes BBRR, BBRr, BbRR, and BbRr.
PTS: 1
BLM: analysis
5. ANS:
One sixteenth of the offspring would be expected to have white, smooth hair.
PTS: 1
BLM: analysis
6. ANS:
R represents the allele for red flowers. I represents the allele for ivory flowers.
PTS: 1
BLM: knowledge
7. ANS:
The genotype of the pink-flowered snapdragons is RI.
PTS: 1
BLM: application
8. ANS:
The alleles show incomplete dominance, because a cross between red-flowered snapdragons and ivoryflowered snapdragons produces snapdragons with an in-between trait—pink flowers.
PTS: 1
BLM: application
9. ANS:
One hundred percent of the offspring would be expected to be pink-flowered.
PTS: 1
BLM: application
10. ANS:
It is not possible for the red snapdragon to have white offspring. They are homozygous (RR) and therefore do
not carry any white alleles. The red snapdragons in Figure 11–7 will always pass the R allele on to offspring
and therefore the offspring can only be red or pink.
PTS: 1
BLM: evaluation
11. ANS:
The structure is a tetrad.
PTS: 1
BLM: analysis
12. ANS:
New allele combinations might form during stage A, which is prophase I.
PTS: 1
BLM: synthesis
13. ANS:
Generally, one egg would result. One of the four haploid cells would form an egg.
PTS: 1
BLM: comprehension
14. ANS:
The cells in stages A, B, and C are 2N. The cells in stages D, E, F, and G are N.
PTS: 1
BLM: synthesis
15. ANS:
Each cell in stage G, telophase II, has a single copy of each gene.
PTS: 1
BLM: analysis
ESSAY
1. ANS:
The tall pea plant should be crossed with a short pea plant. If the tall pea plant is homozygous, all of the
offspring will be tall. If the tall pea plant is heterozygous, it is likely that about half of the offspring will be
tall and half will be short.
PTS: 1
2. ANS:
BLM: analysis
When the heterozygous yellow-seed F1 plants produced gametes, their dominant allele for yellow seeds
segregated from their recessive allele for green seeds. As a result, some of their gametes had the dominant
allele, and others had the recessive allele. When the F1 plants self-pollinated, some male gametes with the
recessive allele fused with female gametes with the recessive allele during fertilization. Some of the offspring
that resulted had two alleles for green seeds and therefore had green seeds.
PTS: 1
BLM: application
3. ANS:
The kinds of offspring produced by genetic crosses are the results of chance. For example, the number of
gametes produced that contain particular alleles is not certain. Likewise, the fusion of two gametes with
particular alleles is not certain. Thus, the results of genetic crosses shown in Punnett squares are just probable
results.
PTS: 1
BLM: evaluation
4. ANS:
The genotype ratio of the offspring is not the same as their phenotype ratio. The same phenotype can be
produced by several different genotypes. For example, offspring that are heterozygous for both traits (AaBb)
will have the same phenotype as offspring that are homozygous for both traits (AABB).
PTS: 1
BLM: synthesis
5. ANS:
The alleles that determine flower color in four o’clock plants show incomplete dominance. She should use
pollen from white-flowered four o’clock plants to pollinate red-flowered four o’clock plants, or vice versa.
She should then collect seeds from the offspring. All of these hybrid seeds will produce only pink-flowered
four o’clock plants.
PTS: 1
BLM: synthesis
6. ANS:
I would expect the offspring to be lighter-colored than their parents. The phenotype for color in these
butterflies is determined by two components: genes and environmental conditions. The offspring would have
their parents’ genetic material, but they would have a different environmental influence. The butterflies born
in the summer would have lower levels of pigment and so would be lighter.
PTS: 1
BLM: evaluation
7. ANS:
A body cell has more genetic information than a gamete. A gamete is haploid, so it has just one allele for each
gene. A body cell is diploid. It has two alleles for each gene. So, a body cell has twice as much genetic
information as a sex cell.
PTS: 1
BLM: evaluation
8. ANS:
Both meiosis I and meiosis II contain a prophase, a metaphase, and an anaphase. However, chromosomes
replicate prior to meiosis I but not prior to meiosis II. Also, during meiosis I, tetrads form and align along the
center of the cell. Then, the homologous chromosomes are separated and two haploid daughter cells form.
During meiosis II, sister chromatids align along the center of the cell and are then separated. Four haploid
daughter cells form.
PTS: 1
9. ANS:
BLM: analysis
During meiosis, the pairs of homologous chromosomes in the parent cell form tetrads and then separate. As a
result, each daughter cell receives only one chromosome from each homologous pair, and the particular
chromosomes that it receives are random. Thus, each daughter cell has a different combination of
chromosomes. Also, crossing-over may occur during meiosis and may result in new combinations of alleles
on the chromosomes in the daughter cells. In contrast, during mitosis, homologous chromosomes usually do
not form tetrads and separate, and therefore crossing-over usually does not occur.
PTS: 1
BLM: synthesis
10. ANS:
Linkage is the condition in which two genes are located on the same chromosome. As a result, the genes’
alleles are usually inherited together. However, if crossing-over occurs between the alleles of the linked
genes, those alleles no longer are inherited together. The frequency of crossing-over between linked genes (or
the frequency in which linked genes are inherited separately) is used to determine the relative locations of
genes on the same chromosome, resulting in a gene map. The greater the frequency of separate inheritance,
the farther apart the genes on the chromosome.
PTS: 1
BLM: analysis