Chapter 14 BIO Flashcards


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1

What do we mean when we use the terms monohybrid cross and dihybrid cross?
A) A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents.
B) A dihybrid cross involves organisms that are heterozygous for two characters that are being studied, and a monohybrid cross involves organisms that are heterozygous for only one character being studied.
C) A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations.
D) A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1ratio.

B

2

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 generation than do dominant ones.
D) Genes are composed of DNA.

B

3

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?
A) 4
B) 8
C) 16
D) 64

B

4

The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason?
A) recurrent mutations forming new alleles
B) crossing over during prophase I
C) different possible assortment of chromosomes into gametes
D) the tendency for dominant alleles to segregate together

C

5

Mendel continued some of his experiments into the F2 or F3 generation to ____.
A) obtain a larger number of offspring on which to base statistics
B) observe whether or not a recessive trait would reappear
C) observe whether or not the dominant trait would reappear
D) distinguish which alleles were segregating

B

6

Which of the following statements about independent assortment and segregation is correct?
A) The law of independent assortment requires describing two or more genes relative to one another.
B) The law of segregation requires describing two or more genes relative to one another.
C) The law of independent assortment is accounted for by observations of prophase I
D) The law of segregation is accounted for by anaphase of mitosis

A

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?
A) Hh
B) HhTt
C) T
D) HT

D

8

Mendel accounted for the observation that traits that 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) members of the F1 generation had only one allele for each trait, but members of the F2 had two alleles for each trait

C

9

The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following?
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.

D

10

Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division?
A) prophase I of meiosis
B) anaphase II of meiosis
C) metaphase II of meiosis
D) anaphase I of meiosis

D

11

Mendel's second law of independent assortment has its basis in which of the following events of meiosis I?
A) synapsis of homologous chromosomes
B) crossing over
C) alignment of tetrads at the equator
D) separation of cells at telophase

C

12
card image

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 boxes marked 1-4 correspond to plants with dark leaves?

A) 1 only

B) 2 and 3

C) 4 only

D) 1, 2, and 3

D

13
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Which of the boxes marked 1-4 correspond to plants with a heterozygous genotype?

A) 1

B) 1, 2, and 3

C) 2 and 3

D) 2, 3, and 4

C

14
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Which of the boxes marked 1-4 correspond to plants that will be true-breeding?

A) 1 and 4 only

B) 2 and 3 only

C) 1, 2, 3, and 4

D) 1 only

A

15

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?

A) 2

B) 4

C) 8

D) 16

B

16

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

A) No genes interacted to produce the parental phenotype.

B) Each allele affected phenotypic expression.

C) The traits blended together during fertilization.

D) One allele was dominant.

D

17

Mendel crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F2 generation. The results were as follows: 6022 yellow and 2001 green (8023 total). The allele for green seeds has what relationship to the allele for yellow seeds?

A) dominant

B) incomplete dominant

C) recessive

D) codominant

C

18

Albinism is an autosomal (not sex-linked) recessive trait. A man and woman are both of normal pigmentation, but both have one parent who is albino (without melanin pigmentation). What is the probability that their first child will be an albino?

A) 0

B) 1/2

C) 1/4

D) 1

C

19

Albinism is an autosomal (not sex-linked) recessive trait. A man and woman are both of normal pigmentation and have one child out of three who is albino (without melanin pigmentation). What are the genotypes of the albino's parents?

A) One parent must be homozygous for the recessive allele; the other parent can be homozygous dominant, homozygous recessive, or heterozygous.

B) One parent must be heterozygous; the other parent can be homozygous dominant, homozygous recessive, or heterozygous.

C) Both parents must be heterozygous.

D) One parent must be homozygous dominant; the other parent must be heterozygous.

C

20

A black guinea pig crossed with an albino guinea pig produced twelve black offspring. When the albino was crossed with a second black animal, six blacks and six albinos were obtained. What is the best explanation for this genetic situation?

A) Albino is recessive; black is dominant.

B) Albino is dominant; black is incompletely dominant.

C) Albino and black are codominant.

D) Albino is recessive; black is codominant.

A

21

Gray seed color in peas is dominant to white. Assume that Mendel conducted a series of experiments where plants with gray seeds were crossed among themselves, and the following progeny were produced: 302 gray and 98 white.

(a) What is the most probable genotype of each parent?

(b) Based on your answer in(a) above, what genotypic and phenotypic ratios are expected in these progeny? (Assume the following symbols: G = gray and g = white.)

A) (a) GG × gg; (b) genotypic = 3:1, phenotypic = 1:2:1

B) (a) Gg × Gg; (b) genotypic = 1:2:1, phenotypic = 3:1

C) (a) GG × Gg; (b) genotypic = 1:2:1, phenotypic = 2:1

D) (a) gg × Gg; (b) genotypic = 1:2, phenotypic = 3:1

B

22

When Mendel crossed yellow-seeded and green-seeded pea plants, all the offspring were yellow seeded. When he took these F1 yellow-seeded plants and crossed them to green-seeded plants, what genotypic ratio was expected?

A) 1:2:1

B) 3:1

C) 1:1

D) 1:1:1:1

C

23

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?

A) 1/16

B) 3/8

C) 1/2

D) 9/16

C

24

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?

A) 1

B) 1/2

C) 1/4

D) 0

D

25

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

A) 1/4

B) 1/8

C) 1/16

D) 1/64

D

26

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 with genotype AABBCc?

A) 1/4

B) 3/4

C) 3/8

D) 1

B

27

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

A) the probability that two or more independent events will both occur

B) the probability that either one of two independent events will occur

C) the probability of producing two or more heterozygous offspring

D) the likelihood that a trait is due to two or more meiotic events

B

28

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

A) Calculate the probability of black offspring from the cross AaBb × AaBb, where B is the symbol for black.

B) Calculate the probability of children with both cystic fibrosis and polydactyly when parents are each heterozygous for both genes

C) Calculate the probability of each of four children having cystic fibrosis if the parents are both heterozygous.

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

D

29

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?

A) 9/16

B) 1/16

C) 3/16

D) 1/4

C

30

A man has extra digits (six fingers on each hand and six toes on each foot). His wife and their daughter have a normal number of digits. Having extra digits is a dominant trait. The couple's second child has extra digits. What is the probability that their next (third) child will have extra digits?

A) 1/2

B) 1/16

C) 1/8

D) 3/4

A

31

Phenylketonuria is an inherited disease caused by a recessive autosomal allele. If a woman and her husband are both carriers, what is the probability that their first child will be a phenotypically normal girl?

A) 1/4

B) 1/16

C) 3/16

D) 3/8

D

32

Assuming independent assortment for all gene pairs, what is the probability that the following parents, AABbCc × AaBbCc, will produce an AaBbCc offspring?

A) 1/2

B) 1/16

C) 1/8

D) 3/4

C

33

Suppose two AaBbCc individuals are mated. Assuming that the genes are not linked, what fraction of the offspring are expected to be homozygous recessive for the three traits?

A) 1/4

B) 1/8

C) 1/16

D) 1/64

D

34

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

B

35

Which of the following describes the ability of a single allele to have multiple phenotypic effects?

A) incomplete dominance

B) multiple alleles

C) pleiotropy

D) epistasis

C

36

Which of the following is an example of polygenic inheritance?

A) pink flowers in snapdragons

B) the ABO blood group in humans

C) white and purple flower color in peas

D) skin pigmentation in humans

D

37

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?

A) the knowledge that multiple alleles are involved

B) the allele for blue hydrangea being completely dominant

C) the alleles being codominant

D) environmental factors such as soil pH

D

38

Which of the following provides an example of epistasis?

A) Recessive genotypes for each of two genes (aabb) results in an albino corn snake.

B) In rabbits and many other mammals, one genotype (ee) prevents any fur color from developing.

C) In Drosophila (fruit flies), white eyes can be due to an X-linked gene or to a combination of other genes.

D) In cacti, there are several genes for the type of spines.

B

39

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?

A) red and long

B) white and long

C) purple and long

D) purple and oval

C

40

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 flower color trait in radishes is an example of which of the following?

A) a multiple allelic system

B) sex linkage

C) codominance

D) incomplete dominance

D

41

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 (S1S3) and its mate has alleles 2 and 4 (S2S4). If each allele confers a unit of color darkness such that S1 has one unit, S2 has two units, and so on, then what proportion of their offspring would be expected to have five units of color?

A) 1/4

B) 1/8

C) 1/2

D) 0

C

42

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 ____.

A) incomplete dominance

B) epistasis

C) pleiotropy

D) codominance

B

43

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. 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) It is impossible to determine the phenotypes of the progeny.

A

44

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?

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) 9 sharp-spined:3 dull-spined:4 spineless

D

45

Feather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. A blue budgie is crossed with a white budgie. Which of the following results is NOT possible?

A) green offspring only

B) yellow offspring only

C) blue offspring only

D) green and yellow offspring

D

46

Feather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. Two blue budgies were crossed. Over the years, they produced twenty-two offspring, five 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

C

47

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

A) A negative

B) O negative

C) B positive

D) AB negative

C

48

A 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?

A) Each parent is either M or MN.

B) Each parent must be type M.

C) Both children are heterozygous for this gene.

D) Neither parent can have the N allele

A

49

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?

A) It is recessive.

B) It is dominant.

C) It is pleiotropic.

D) It is epistatic

C

50

In rabbits, the homozygous CC is normal, Cc results in deformed legs, and cc results in very short legs. The genotype BB produces black fur, Bb brown fur, and bb white fur. If a cross is made between brown rabbits with deformed legs and white rabbits with deformed legs, what percentage of the offspring would be expected to have deformed legs and white fur?

A) 25%

B) 33%

C) 100%

D) 50%

A

51

In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A glycoprotein, The IB allele codes for the B glycoprotein, and the i allele doesn't code for any membrane glycoprotein. IA and IB are codominant, and i is recessive to both IA and IB. People with type A blood have the genotypes IAIA or IAi, people with type B blood are IBIB or IBi, people with type AB blood are IAIB, and people with type O blood are ii. If a woman with type AB blood marries a man with type O blood, which of the following blood types could their children possibly have?

A) A and B

B) AB and O

C) A, B, and O

D) A, B, AB, and O

A

52

An obstetrician knows that one of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is detectable biochemically in fetal cells. The obstetrician would most reasonably offer which of the following procedures to her patient?

A) karyotyping of the woman’s somatic cells

B) X-ray

C) amniocentesis or CVS

D) blood transfusion

C

53

In some parts of Africa, the frequency of heterozygosity for the sickle-cell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following?

A) Mendel's law of independent assortment

B) Mendel's law of segregation

C) Darwin's explanation of natural selection

D) the malarial parasite changing the allele

C

54

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Therefore, the most efficient and effective treatment is which of the following?

A) Feed them the substrate that can be metabolized into this amino acid.

B) Regulate the diet of the affected persons to severely limit the uptake of the amino acid.

C) Feed the patients the missing enzymes in a regular cycle, such as twice per week.

D) Feed the patients an excess of the missing product.

B

55

Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility and death, usually from coronary artery disease, at an average age of 13 years. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption?

A) The disease is autosomal dominant.

B) The disorder will increase in frequency in successive generations within a family.

C) The disorder may be due to mutation in a single protein-coding gene.

D) Each patient will have had at least one affected grandparent or parent.

C

56

One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. Which of the following is the best explanation for why a young, affected child is the first in her family to be diagnosed?

A) The mother carries the gene but does not express it.

B) One of the parents has a mild expression of the gene.

C) The condition skipped a generation in the family.

D) The child has one more chromosome than either of the parents.

B

57
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What is the genotype of individual II-5?

A) WW

B) Ww

C) ww

D) ww or Ww

C

58
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What is the likelihood that the progeny of IV-3 and IV-4 will have the trait?

A) 0%

B) 50%

C) 75%

D) 100%

B

59

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

A) 3/4

B) 1/4

C) 2/4

D) 1

D

60
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From this pedigree, this trait seems to be inherited ____.

A) from mothers

B) as an autosomal recessive

C) as a result of epistasis

D) as an autosomal dominant

...

61
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Which of the following statements is a correct explanation for the observation that all offspring exhibit a phenotype for a particular trait that appears to be a blend of the two parental varieties?

A) Neither of the parental genes is dominate over the other.

B) The genes for the trait are dominant in both of the parents.

C) The genes are linked and do not separate during meiosis.

D) The genes for the trait are recessive in both of the parents.

A

62

The pattern of inheritance (monohybrid, dihybrid, sex-linked, and genes linked on the same chromosomes) can be predicted from data if one is given the parent or offspring genotypes or phenotypes. Two organisms, with genotypes BbDD and BBDd, are mated. Assuming independent assortment of the B/b and D/d genes, determine the genotypic ratios in offspring that would occur.

A) 1/2 BBDD 1/2 bbdd

B) 1/4 BBDD 1/4 BbDD 1/4 BBDd 1/4 BbDd

C) 9/16 BBDD 3/16 BbDD 3/16 BBDd 1/16 bbdd

D) 1/4 BBDD 1/2 BbDd 1/4 bbdd

B