Print Options

Card layout: ?

← Back to notecard set|Easy Notecards home page

Instructions for Side by Side Printing
  1. Print the notecards
  2. Fold each page in half along the solid vertical line
  3. Cut out the notecards by cutting along each horizontal dotted line
  4. Optional: Glue, tape or staple the ends of each notecard together
  1. Verify Front of pages is selected for Viewing and print the front of the notecards
  2. Select Back of pages for Viewing and print the back of the notecards
    NOTE: Since the back of the pages are printed in reverse order (last page is printed first), keep the pages in the same order as they were after Step 1. Also, be sure to feed the pages in the same direction as you did in Step 1.
  3. Cut out the notecards by cutting along each horizontal and vertical dotted line
To print: Ctrl+PPrint as a list

51 notecards = 13 pages (4 cards per page)

Viewing:

Chapter 14 Notecards

front 1

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 monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny.

C) A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid cross involves only one.

D) A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations.

E) A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.

back 1

Answer: C

front 2

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

  1. A) No genes interacted to produce the parental phenotype.
  2. B) Each allele affected phenotypic expression.
  3. C) The traits blended together during fertilization.
  4. D) One phenotype was completely dominant over another.
  5. E) Different genes interacted to produce the parental phenotype.

back 2

Answer: D

front 3

3) What was the most significant conclusion that Gregor Mendel drew from his experiments with pea plants?

  1. A) There is considerable genetic variation in garden peas.
  2. B) Traits are inherited in discrete units, and are not the results of "blending."
  3. C) Recessive genes occur more frequently in the F1generation than do dominant ones.
  4. D) Genes are composed of DNA.
  5. E) An organism that is homozygous for many recessive traits is at a disadvantage.

back 3

Answer: B

front 4

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

A) 4

B) 8

C) 16

D) 32

E) 64

back 4

Answer: B

front 5

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

A) segregation of maternal and paternal alleles

B) recurrent mutations forming new alleles

C) crossing over during prophase I

D) different possible alignments of chromosomes

E) the tendency for dominant alleles to segregate together

back 5

Answer: D

front 6

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

  1. A) to obtain a larger number of offspring on which to base statistics
  2. B) to observe whether or not a recessive trait would reappear
  3. C) to observe whether or not the dominant trait would reappear
  4. D) to distinguish which alleles were segregating
  5. E) to be able to describe the frequency of recombination

back 6

Answer: B

front 7

7) Which of the following differentiates between independent assortment and segregation?

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 segregation requires having two or more generations to describe.

D) The law of independent assortment is accounted for by observations of prophase I.

E) The law of segregation is accounted for by anaphase of mitosis.

back 7

Answer: A

front 8

8) Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular trait. What does this suggest?

  1. A) that the parents were true-breeding for contrasting traits
  2. B) that the trait shows incomplete dominance
  3. C) that a blending of traits has occurred
  4. D) that the parents were both heterozygous for a single trait
  5. E) that each offspring has the same alleles for each of two traits

back 8

Answer: D

front 9

9) 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) tt

B) Hh

C) HhTt

D) T

E) HT

back 9

Answer: E

front 10

10) When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype?

A) 0%

B) 25%

C) 50%

D) 75%

E) 100%

back 10

Answer: C

front 11

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

  1. A) new mutations were frequently generated in the F2progeny, "reinventing" traits that had been lost in the F1.
  2. B) the mechanism controlling the appearance of traits was different between the F1and the F2
  3. C) traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1.
  4. D) the traits were lost in the F1due to dominance of the parental traits.
  5. E) members of the F1generation had only one allele for each trait, but members of the F2had two alleles for each trait.

back 11

Answer: C

front 12

12) 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.

E) The formation of gametes in plants occurs by mitosis only.

back 12

Answer: D

front 13

13) Mendel was able to draw his ideas of segregation and independent assortment because of the influence of which of the following?

A) His reading and discussion of Darwin's Origin of Species.

B) The understanding of particulate inheritance he learned from renowned scientists of his time.

C) His discussions of heredity with his colleagues at major universities.

D) His experiments with the breeding of plants such as peas and fuchsia.

E) His reading of the scientific literature current in the field.

back 13

Answer: E

front 14

14) 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 I of meiosis

D) anaphase I of meiosis

E) anaphase of mitosis

back 14

Answer: D

front 15

15) 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 homologs at anaphase

E) separation of cells at telophase

back 15

Answer: C

front 16

16) 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) 3/16

C) 3/8

D) 1/2

E) 9/16

back 16

Answer: D

front 17

17) 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) 1/6

E) 0

back 17

Answer: E

front 18

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

A) 1/4

B) 1/8

C) 1/16

D) 1/32

E) 1/64

back 18

Answer: E

front 19

19) 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?

A) 1/4

B) 1/8

C) 3/4

D) 3/8

E) 1

back 19

Answer: C

front 20

20) 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 two or more independent events will both occur in the offspring of one set of parents

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

D) the probability of producing two or more heterozygous offspring

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

back 20

Answer: C

front 21

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

A) Calculate the probability of black offspring from the cross AaBb × AaBb, when 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.

E) Calculate the probability of purple flower color in a plot of 50 plants seeded from a self-fertilizing heterozygous parent plant.

back 21

Answer: D

front 22

22) 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.

back 22

Answer: B

front 23

23) Which of the following describes the ability of a single gene to have multiple phenotypic effects?

A) incomplete dominance

B) multiple alleles

C) pleiotropy

D) epistasis

back 23

Answer: C

front 24

24) Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this?

A) incomplete dominance

B) multiple alleles

C) pleiotropy

D) epistasis

E) codominance

back 24

Answer: C

front 25

25) Which of the following is an example of polygenic inheritance?

A) pink flowers in snapdragons

B) the ABO blood group in humans

C) Huntington's disease in humans

D) white and purple flower color in peas

E) skin pigmentation in humans

back 25

Answer: E

front 26

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?

A) the knowledge that multiple alleles are involved

B) the allele for blue hydrangea being completely dominant

C) the alleles being codominant

D) the fact that a mutation has occurred

E) environmental factors such as soil pH

back 26

Answer: E

front 27

27) 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) The allele b17 produces a dominant phenotype, although b1 through b16 do not.

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

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

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

back 27

Answer: C

front 28

28) Most genes have many more than two alleles. However, which of the following is also true?

A) At least one allele for a gene always produces a dominant phenotype.

B) Most of the alleles will never be found in a live-born organism.

C) All of the alleles but one will produce harmful effects if homozygous.

D) There may still be only two phenotypes for the trait.

E) More than two alleles in a genotype is considered lethal.

back 28

Answer: D

front 29

29) How could you best predict the maximum number of alleles for a single gene whose polypeptide product is known?

A) Search the population for all phenotypic variants of this polypeptide.

B) Count the number of amino acids in the polypeptide.

C) Mate all known genotypes and collect all possible offspring different from the parents.

D) Measure the rate of new mutations in the species and estimate the number since it first evolved.

E) Count the number of DNA nucleotides that are in the code for the polypeptides.

back 29

Answer: E

front 30

30) An ideal procedure for fetal testing in humans would have which of the following features?

A) the procedure that can be performed at the earliest time in the pregnancy

B) lowest risk procedure that would provide the most reliable information

C) the procedure that can test for the greatest number of traits at once

D) a procedure that provides a three-dimensional image of the fetus

E) a procedure that could test for the carrier status of the fetus

back 30

Answer: A

front 31

31) A scientist discovers a DNA-based test for one allele of a particular gene. This and only this allele, if homozygous, produces an effect that results in death at or about the time of birth. Of the following, which is the best use of this discovery?

A) Screen all newborns of an at-risk population.

B) Design a test for identifying heterozygous carriers of the allele.

C) Introduce a normal allele into deficient newborns.

D) Follow the segregation of the allele during meiosis.

E) Test school-age children for the disorder.

back 31

Answer: B

front 32

32) 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) CVS

B) ultrasound imaging

C) amniocentesis

D) blood transfusion

E) X-ray

back 32

Answer: C

front 33

33) 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) Darwin's observations of competition

E) the malarial parasite changing the allele

back 33

Answer: C

front 34

34) Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize a particular amino acid. The amino acid is not otherwise 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) Transfuse the patients with blood from unaffected donors.

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

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

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

back 34

Answer: C

front 35

35) 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 approximately 13. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption?

A) All cases must occur in relatives; therefore, there must be only one mutant allele.

B) Successive generations of a family will continue to have more and more cases over time.

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

D) Each patient will have had at least one affected family member in a previous generation.

E) The disease is autosomal dominant.

back 35

Answer: C

front 36

36) 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. If a young child is the first in her family to be diagnosed, which of the following is the best explanation?

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

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

C) The condition skipped a generation in the family.

D) The child has a different allele of the gene than the parents.

back 36

Answer: B

front 37

Use Figure 14.1 and the following description to answer the questions 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 Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

37) 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

back 37

Answer: E

front 38

Use Figure 14.1 and the following description to answer the questions 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 Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

38) 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

back 38

Answer: D

front 39

Use Figure 14.1 and the following description to answer the questions 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 Figure 14.1, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

39) Which of the plants will be true-breeding?

A) 1 and 4 only

B) 2 and 3 only

C) 1, 2, 3, and 4

D) 1 only

E) 1 and 2 only

back 39

Answer: A

front 40

The following questions refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

40) What is the genotype of individual II-5?

A) WW

B) Ww

C) ww

D) WW or ww

E) ww or Ww

back 40

Answer: C

front 41

The following questions refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

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

A) 0%

B) 25%

C) 50%

D) 75%

E) 100%

back 41

Answer: C

front 42

The following questions refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

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

A) 3/4

B) 1/4

C) 2/4

D) 2/3

E) 1

back 42

Answer: E

front 43

Use the following pedigree (Figure 14.3) 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.

43) What is the genotype of the deceased individual in generation II?

A) homozygous for a gene for colon cancer

B) homozygous for both cancer alleles from his mother

C) heterozygous for a gene for colon cancer

D) affected by the same colon cancer environmental factor as his mother

E) carrier of all of the several known genes for colon cancer

back 43

Answer: C

front 44

Use the following pedigree (Figure 14.3) 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.

44) 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?

A) Members of this family know to be checked for colon cancer early in life.

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

C) This is pure chance; it would not be expected if you were to look at a different family.

D) This cancer requires mutations in more than this one gene.

E) Affected members of this family are born with colon cancer, and it can be detected whenever they are first tested.

back 44

Answer: B

front 45

Use the following pedigree (Figure 14.3) 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.

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

A) from mothers

B) as an autosomal recessive

C) as a result of epistasis

D) as an autosomal dominant

E) as an incomplete dominant

back 45

Answer: D

front 46

Use the following pedigree (Figure 14.3) 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.

46) 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?

  1. genetic counseling
  2. prenatal diagnosis when/if she becomes pregnant
  3. testing to see whether she has the allele
  4. testing to see whether her future spouse or partner has the allele

A) I only

B) II only

C) I and II only

D) I, II, and III only

E) III and IV only

back 46

Answer: C

front 47

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.

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

  1. A) 65
  2. B) 190
  3. C) 250
  4. D) 565
  5. E) 750

back 47

Answer: B

front 48

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.

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

  1. A) 9/16
  2. B) 1/16
  3. C) 3/16
  4. D) 1/8
  5. E) 1/4

back 48

Answer: C

front 49

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.

49) These results indicate which of the following?

A) Brown is dominant to black.

B) Black is dominant to brown and to yellow.

C) Yellow is dominant to black.

D) There is incomplete dominance.

E) Epistasis is involved.

back 49

Answer: E

front 50

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.

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

A) 1

B) 2

C) 3

D) 4

E) 5

back 50

Answer: B

front 51

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.

51) In one type cross of black × black, the results were as follows:

9/16 black

4/16 yellow

3/16 brown

The genotype aabb must result in which of the following?

  1. A) black
  2. B) brown
  3. C) yellow
  4. D) a lethal result
  5. E) white

back 51

Answer: C