Mastering Bio Ch. 14

Front 1

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?

16
32
64
8
4

Back 1

8

Front 2

The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason?

different possible assortment of chromosomes into gametes
segregation of maternal and paternal alleles
recurrent mutations forming new alleles
the tendency for dominant alleles to segregate together
crossing over during prophase I

Back 2

Different possible assortment of chromosomes into gametes

Front 3

Why did Mendel continue some of his experiments to the F2 or F3 generation?

to be able to describe the frequency of recombination
to observe whether or not a recessive trait would reappear
to obtain a larger number of offspring on which to base statistics
to distinguish which alleles were segregating
to observe whether or not the dominant trait would reappear

Back 3

To observe whether or not a recessive trait would appear

Front 4

Which of the following differentiates between independent assortment and segregation?

The law of segregation requires having two or more generations to describe.
The law of independent assortment requires describing two or more genes relative to one another.
The law of segregation is accounted for by anaphase of mitosis.
The law of segregation requires describing two or more genes relative to one another.
The law of independent assortment is accounted for by observations of prophase I.

Back 4

The law of independent assortment requires describing 2 or more genes relative to one another.

Front 5

Mendel's second law of independent assortment has its basis in which of the following events of meiosis I?

alignment of tetrads at the equator
separation of cells at telophase
crossing over
separation of homologs at anaphase
synapsis of homologous chromosomes

Back 5

Alignment of tetrads at the equator

Front 6

Mendel accounted for the observation that traits which had disappeared in the F1 generation reappeared in the F2 generation by proposing that

new mutations were frequently generated in the F2 progeny, "reinventing" traits that had been lost in the F1.
members of the F1 generation had only one allele for each trait, but members of the F2 had two alleles for each trait.
traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1.
the traits were lost in the F1 due to dominance of the parental traits.
the mechanism controlling the appearance of traits was different between the F1 and the F2 plants.

Back 6

Traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1

Front 7

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?

HT
T
tt
HhTt
Hh

Back 7

HT

Front 8

Why did the F1 offspring of Mendel's classic pea cross always look like one of the two parental varieties?

One allele was dominant.
Phenotype was not dependent on genotype.
Each allele affected phenotypic expression.
The traits blended together during fertilization.
No genes interacted to produce the parental phenotype.

Back 8

One allele was dominant

Front 9

Use the figure and the following description to answer the question below.

In a particular plant, leaf color is controlled by gene locus D. Plants with at least one 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 F1 offspring is allowed to self-pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.



Which of the plants will be true-breeding?

1 only
1 and 2 only
1 and 4 only
1, 2, 3, and 4
2 and 3 only

Back 9

1 and 4 only

Front 10

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 crosses BbTt × BBtt will be expected to have black fur and long tails?

3/16
3/8
1/16
1/2
9/16

Back 10

1/2

Front 11

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?

1
1/6
1/4
0
1/2

Back 11

0

Front 12

In the cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC?

1/16
1/8
1/64
1/4
1/32

Back 12

1/4

Front 13

Which of the following calculations require that you utilize the addition rule?

Calculate the probability of each of four children having cystic fibrosis if the parents are both heterozygous.
Calculate the probability of purple flower color in a plot of 50 plants seeded from a self-fertilizing heterozygous parent plant.
Calculate the probability of a child having either sickle-cell anemia or cystic fibrosis if parents are each heterozygous for both.
Calculate the probability of black offspring from the cross AaBb × AaBb, when B is the symbol for black.
Calculate the probability of children with both cystic fibrosis and polydactyly when parents are each heterozygous for both genes.

Back 13

Calculate the probability of a child having either sickle-cell anemia or cystic fibrosis if parents are each heterozygous for both.

Front 14

Which of the following is the best statement of the use of the addition rule of probability?

the probability that two or more independent events will both occur in the offspring of one set of parents
the probability that either one of two independent events will occur
the probability of producing two or more heterozygous offspring
the likelihood that a trait is due to two or more meiotic events
the probability that two or more independent events will both occur

Back 14

The probability that either one or 2 independent events will occur

Front 15

Given the parents AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent?

1/4
3/4
3/8
1/8
1

Back 15

3/4

Front 16

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. The genes for flower color and location assort independently.


If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers?

750
565
65
190
250

Back 16

190

Front 17

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. The genes for flower color and location assort independently.


Among the F2 offspring, what is the probability of plants with white axial flowers?

1/4
1/16
9/16
1/8
3/16

Back 17

3/16

Front 18

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.


These results indicate which of the following?

Epistasis is involved.
Black is dominant to brown and to yellow.
Yellow is dominant to black.
Brown is dominant to black.
There is incomplete dominance.

Back 18

Epistasis is involved

Front 19

Use the following information to answer the question below.

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.


How many genes must be responsible for these coat colors in Labrador retrievers?

5
3
2
1
4

Back 19

2

Front 20

Use the following information to answer the question below.

Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.


In one type cross of black × black, the results were as follows:
9/16 black
4/16 yellow
3/16 brown
The genotype eebb must result in which of the following?

white
a lethal result
yellow
brown
black

Back 20

Yellow

Front 21

Use the following information to answer the question below.

Drosophila (fruit flies) usually have long wings (+) but mutations in two different genes can result in bent wings (bt) or vestigial wings (vg).


If a homozygous bent wing fly is mated with a homozygous vestigial wing fly, which of the following offspring would you expect?

all homozygous + flies
3/4 bent to 1/4 vestigial ratio
1/2 bent and 1/2 vestigial flies
all +bt +vg heterozygotes
1/2 bent and vestigial to 1/2 normal

Back 21

All bt+vg heterozygotes

Front 22

Use the following information to answer the question below.

Drosophila (fruit flies) usually have long wings (+) but mutations in two different genes can result in bent wings (bt) or vestigial wings (vg).


If flies that are heterozygous for both the bent wing gene and the vestigial wing gene are mated, what is the probability of offspring with bent wings only?

1/4
3/8
3/16
1/8
9/16

Back 22

3/16

Front 23

Use the following information to answer the question below.

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait.


If true-breeding red long radishes are crossed with true-breeding white oval radishes, the F1 will be expected to be which of the following?

Back 23

Red and long

Front 24

Use the following information to answer the question below.

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait.


In the F2 generation of the above cross, which of the following phenotypic ratios would be expected?

1:1:1:1:1:1
6:3:3:2:1:1
9:3:3:1
1:1:1:1
9:4:3

Back 24

6:3:3:2:1:1

Front 25

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?

red × roan
white × roan
red × white
roan × roan
The answer cannot be determined from the information provided.

Back 25

RoanXRoan

Front 26

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 which of the following?

environmental factors such as soil pH
the fact that a mutation has occurred
the alleles being codominant
the knowledge that multiple alleles are involved
the allele for blue hydrangea being completely dominant

Back 26

Environmental factors such as soil PH

Front 27

Use the following information to answer the question below.

Tallness (T) in snapdragons is dominant to dwarfness (t), while red (R) flower color is incompletely dominant to white (r). The heterozygous condition results in pink (Rr) flower color.


If snapdragons are heterozygous for height as well as for flower color, a mating between them will result in what ratio?

6:3:3:2:1:1
9:3:3:1
9:4:3
27:9:9:9:3:3:3:1
1:2:1

Back 27

6:3:3:2:1:1

Front 28

Use the following information to answer the question below.

Skin color in a certain species of fish is inherited via a single gene with four different alleles.


How many different types of gametes would be possible in this system?

16
8
2
1
4

Back 28

4

Front 29

Use the following information to answer the question below.

Skin color in a certain species of fish is inherited via a single gene with four different alleles.


One fish of this type has alleles 1 and 3 (S 1 S 3) and its mate has alleles 2 and 4 (S 2 S 4). If each allele confers a unit of color darkness such that S 1 has one unit, S 2 has two units, and so on, then what proportion of their offspring would be expected to have five units of color?

1/4
1/5
1/8
0
1/2

Back 29

1/2

Front 30

Refer to the following information to answer the question below.

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.


The relationship between genes S and N is an example of

pleiotropy.
codominance.
complete dominance.
incomplete dominance.
epistasis.

Back 30

Epistasis

Front 31

Refer to the following information to answer the question below.

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.


If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios?

3 sharp-spined:1 spineless
1 sharp-spined:2 dull-spined:1 spineless
1 sharp-spined:1 dull-spined
9 sharp-spined:3 dull-spined:4 spineless
1 sharp-spined:1 dull-spined:1 spineless

Back 31

9 Sharp-spined: 3 Dull-spined: 4 Spineless

Front 32

Use the following information to answer the question below.

A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.


Which of the following is a possible phenotype for the father?

B positive
A negative
AB negative
O negative
impossible to determine

Back 32

B positive

Front 33

Use the following information to answer the question below.

A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.


If both children are of blood type M, which of the following is possible?

Both children are heterozygous for this gene.
Each parent must be type M.
The MN blood group is recessive to the ABO blood group.
Neither parent can have the N allele.
Each parent is either M or MN.

Back 33

Each parent is either M or MN

Front 34

Refer to the following information to answer the question below.

Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.


One species of a small birdlike animal has an extremely variable tail length, an example of polygenic inheritance. Geneticists have come to realize that there are eight separate genes for tail length per haploid genome, with each gene having two alleles. One allele for each gene (a 1, b 1, and so on) increases the length by 1 cm, whereas the other allele (a 2, b 2, and so on) increases it by 0.5 cm. One bird was analyzed and found to have the following genotype:
a 1 a 1 b 2 b 2 c 1 c 2 d 1 d 2 e 2 e 2 f 1 f 2 g 1 g 1 h 1 h 2
What is the length of its tail?

6 cm
8 cm
12 cm
36 cm
24 cm

Back 34

12 CM

Front 35

Refer to the following information to answer the question below.

Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.


One species of green plant, with frondlike leaves, a spine-coated stem, and purple cup-shaped flowers, is found to be self-pollinating. Which of the following is true of this species?

The species must be haploid.
Its reproduction is asexual.
All members of the species have the same genotype.
All of its dominant traits are most frequent.
Some of the seeds would have true-breeding traits.

Back 35

Some of the seeds would have true breeding traits

Front 36

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.



What is the genotype of individual II-5?

ww
WW or ww
ww or Ww
Ww
WW

Back 36

ww

Front 37

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.



What is the probability that individual III-1 is Ww?

3/4
1/4
2/3
1
2/4

Back 37

1

Front 38

Use the following pedigree (the figure) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.



In each generation of this family after generation I, the age at diagnosis is significantly lower than would be found in nonfamilial (sporadic) cases of this cancer (~ 63 years). What is the most likely reason?

Affected members of this family are born with colon cancer, and it can be detected whenever they are first tested.
This is pure chance; it would not be expected if you were to look at a different family.
Hereditary (or familial) cases of this cancer typically occur at earlier ages than do nonfamilial forms.
This cancer requires mutations in more than this one gene.
Members of this family know to be checked for colon cancer early in life.

Back 38

Hereditary (or familial) cases of this cancer typically occur at earlier ages than do nonfamilial forms.

Front 39

Use the following pedigree (the figure) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.



From this pedigree, how does this trait seem to be inherited?

as an autosomal dominant
from mothers
as an autosomal recessive
as an incomplete dominant
as a result of epistasis

Back 39

As an autosomal dominant

Front 40

Use the following pedigree (the figure) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.



The affected woman in generation IV is thinking about her future and asks her oncologist (cancer specialist) whether she can know whether any or all of her children will have a high risk of the same cancer. The doctor would be expected to advise which of the following?
I. genetic counseling
II. prenatal diagnosis when/if she becomes pregnant
III. testing to see whether she has the allele
IV. testing to see whether her future spouse or partner has the allele

I only
III and IV only
I, II, and IV only
II only
I, II, and III only

Back 40

1, 2, and 4 only

Front 41

Marfan syndrome in humans is caused by an abnormality of the connective tissue protein fibrillin. Patients are usually very tall and thin, with long spindly fingers, curvature of the spine, sometimes weakened arterial walls, and sometimes ocular problems, such as lens dislocation. Which of the following would you conclude about Marfan syndrome from this information?

It has a late age of onset (> 60).
It is pleiotropic.
It is dominant.
It is epistatic.
It is recessive.

Back 41

It is pleiotropic