Concepts of Genetics,11th Edition By Klug, Cummings, Spencer & Palladino -Test Bank

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Concepts of Genetics,11th Edition By Klug, Cummings, Spencer & Palladino -Test Bank

 

Sample  Questions

 

Chapter 5   Chromosome Mapping in Eukaryotes

 

1) The term normally applied when two genes fail to assort independently is ________.

  1. A) discontinuous inheritance
  2. B) Mendelian inheritance
  3. C) linkage
  4. D) tetrad analysis
  5. E) dominance and/or recessiveness

Answer:  C

Section:  5.2

 

2) Assume that a cross is made between AaBb and aabb plants and that all the offspring are either AaBb or aabb. These results are consistent with ________.

  1. A) complete linkage
  2. B) alternation of generations
  3. C) codominance
  4. D) incomplete dominance
  5. E) hemizygosity

Answer:  A

Section:  5.2

 

3) Assume that a cross is made between AaBb and aabb plants and that the offspring fall into approximately equal numbers of the following groups: AaBb, Aabb, aaBb, aabb. These results are consistent with ________.

  1. A) independent assortment
  2. B) alternation of generations
  3. C) complete linkage
  4. D) incomplete dominance
  5. E) hemizygosity

Answer:  A

Section:  5.2

 

4) Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with ________.

  1. A) sex-linked inheritance with 30% crossing over
  2. B) linkage with 50% crossing over
  3. C) linkage with approximately 33 map units between the two gene loci
  4. D) independent assortment
  5. E) 100% recombination

Answer:  C

Section:  5.2

 

5) Assume that regarding a particular gene, one scored 30 second-division ascospore arrangements and 70 first-division arrangements in Neurospora. What would be the map distance between the gene and the centromere?

  1. A) 15
  2. B) 30
  3. C) 60
  4. D) 70
  5. E) Insufficient information is provided to answer this question.

Answer:  A

Section:  5.10

 

6) The phenomenon in which one crossover increases the likelihood of crossovers in nearby regions is called ________.

  1. A) chiasma
  2. B) negative interference
  3. C) reciprocal genetic exchange
  4. D) positive interference
  5. E) mitotic recombination

Answer:  D

Section:  5.4

 

7) Which of the following methods is involved in determining the linkage group and genetic map in humans?

  1. A) syntenic testing and lod score determination
  2. B) twin spots and tetrad analysis
  3. C) tetrad analysis and bromodeoxyuridine
  4. D) zygotene and pachytene DNA synthesis
  5. E) chiasmatype and classical analyses

Answer:  A

Section:  5.6

 

8) The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 phenotypically wild-type females were mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

Answer:  mahogany = 375; ebony = 375; wild type = 125; mahogany-ebony = 125

Section:  5.2

 

9) Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?

Answer:  48

Section:  5.2

 

10) Draw a diagram of the chromosomal events that will ultimately result in the segregation of alleles (A and a) during meiosis II rather than meiosis I.

Answer:

Section:  5.2

 

11) Given that loci A and B in Drosophila are sex-linked and 20 map units apart, what phenotypic frequencies would you expect in male and female offspring resulting from the following crosses? (Assume A and B are dominant to a and b, respectively.)

 

(a) AaBb (cis) female × ab/Y male

(b) AaBb (trans) female × ab/Y male

(c) aabb female × AB/Y male

Answer:

(a)   AB = 40; ab = 40; Ab = 10; aB = 10 (sexes have the same phenotypes)

(b)   Ab = 40; aB = 40; AB = 10; ab = 10 (sexes have the same phenotypes)

(c)   all males = ab; all females = AB

Section:  5.2

 

12) Phenotypically wild F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed with homozygous light-straw males:

 

Phenotype                 Number

light-straw                           22

wild                                     18

light                                   990

straw                                  970

Total             2000

 

Compute the map distance between the light and straw loci.

Answer:  2 map units

Section:  5.5

 

 

13) Assume that the genes for tan body and bare wings are 15 map units apart on chromosome #2 in Drosophila. Assume also that a tan-bodied, bare-winged female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

Answer:  wild type = 425; tan-bare = 425; tan = 75; bare = 75

Section:  5.5

14) Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe on the second chromosome with a strain homozygous for the second chromosome recessive mutations smooth abdomen and straw body. The F1 Lobe females were then backcrossed with homozygous smooth abdomen, straw body males, and the following phenotypes were observed:

 

smooth abdomen, straw body                   820

Lobe                                                          780

smooth abdomen, Lobe                              42

straw body                                                  58

smooth abdomen                                       148

Lobe, straw body                                      152

 

(a) Give the gene order and map units between these three loci.

(b) What is the coefficient of coincidence?

Answer:

(a) Lobe is in the middle.

 

smooth abdomen——5——Lobe———————15————————straw body

 

(b) zero

Section:  5.5

 

15) In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 15.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the phenotypes and frequencies expected.

Answer:  scute = 425; ruby = 425; wild type = 75; scute-ruby = 75

Section:  5.5

 

16) Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with which arrangement of genes?

Answer:  In the AaBb parent, the dominant alleles are on one homolog, and the recessive alleles are on the other.

Section:  5.2

 

17) In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and the following progeny (1000 total) were observed:

 

Phenotypes                             Number Observed

spineless                                        321

wild                                                 38

claret, spineless                             130

claret                                               18

claret, hairless                               309

hairless, claret, spineless                 32

hairless                                          140

hairless, spineless                            12

 

(a) Which gene is in the middle?

(b) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the phenotypically wild F1 heterozygote?

(c) What are the map distances between the three genes? A correct formula with the values “plugged in” for each distance will be sufficient.

(d) What is the coefficient of coincidence? A correct formula with the values “plugged in” will be sufficient.

Answer:

(a) hairless

 

(b) cl  h  + /cl  h +  and  +  +  sp/ +  +  sp

 

(c) cl————30————h——10——sp

 

(d) 0.03/0.03 = 1

Section:  5.3, 5.5

 

18) Three loci, mitochondrial malate dehydrogenase that forms a and b (MDHa, MDHb), glucouronidase that forms 1 and 2 (GUS1, GUS2), and a histone gene that forms + and (H+, H-), are located on chromosome #7 in humans. Assume that the MDH locus is at position 35, GUS at position 45, and H at position 75. A female whose mother was homozygous for MDHa, GUS2, and H+ and whose father was homozygous for MDHb, GUS1, and H- produces a sample of 1000 egg cells. Give the genotypes and expected numbers of the various types of cells she would produce. Assume no chromosomal interference.

Answer:

MDHa       GUS2  H+     =     315      MDHa       GUS2  H        =    135

MDHb       GUS1  H       =     315      MDHb       GUS1  H+      =    135

MDHa       GUS1  H       =       35      MDHa       GUS1  H+      =      15

MDHb       GUS2  H+     =       35      MDHb       GUS2  H        =      15

Section:  5.3

 

 

19) To which scientific activities do the terms synkaryon and heterokaryon refer?

Answer:  cell (heterokaryon) and nuclear (synkaryon) fusion and gene mapping in eukaryotes

Section:  5.6

20) (a) In a three-point mapping experiment, which three general classes of offspring are expected (assuming crossovers occur)?

(b) How many genotypic classes are expected?

Answer:

(a) noncrossovers, single crossovers, double crossovers

(b) 8

Section:  5.3

 

21) What is the expected evolutionary significance of genetic recombination?

Answer:  production of genetic variation

Section:  5.1

 

22) Assume that two genes are 80 map units apart on chromosome #2 of Drosophila and that a cross is made between a doubly heterozygous female and a homozygous recessive male. What percent recombination would be expected in the offspring of this type of cross?

Answer:  50% (maximum)

Section:  5.4

 

23) Provide a brief definition for positive interference.

Answer:  A crossover in one region decreases the likelihood of crossovers in nearby regions.

Section:  5.4

 

24) Two lines of work indicated that crossing over actually involves breakage and reunion of chromatid material. Which organisms were involved, and who did the work?

Answer:  Creighton and McClintock (corn) and Stern (Drosophila)

Section:  5.8

 

25) What advantage does BrdU (bromodeoxyuridine) have in the study of chromosome structure and recombination?

Answer:  Chromatids stained with BrdU in both DNA strands are distinguishable from those with BrdU in only one strand of the double helix.

Section:  5.9

 

26) Sister chromatid exchanges increase in frequency in the presence of X-rays, certain viruses, ultraviolet light, and certain chemical mutagens. In which autosomal recessive disorder is there an increase in sister chromatid exchanges?

Answer:  Bloom syndrome

Section:  5.9

 

27) Under what circumstance might two loci be on the same chromosome but behave as if they were independently assorting in crosses?

Answer:  If the genes are far apart, they may show independent assortment.

Section:  5.2

28) In the early 1900s, two scientists noted that there were many more genes than chromosome pairs, thus setting the stage for the suggestion that some gene loci might be linked during meiotic processes. Who were these two scientists?

Answer:  Walter Sutton and Theodor Boveri

Section:  Introduction

29) What is the relationship between the degree of crossing over and the distance between two genes?

Answer:  It is direct; as the distance increases, the frequency of recombination increases.

Section:  5.2

 

30) At what stage of the meiotic cell cycle and during what chromosomal configuration does crossing over occur?

Answer:  at the four-strand stage of meiosis, after synapsis of homologous chromosomes, and before the end of prophase I

Section:  5.2

 

31) What is meant by the term second-division segregation?

Answer:  a condition that gives evidence of a crossover between a gene in question and the centromere: aa++aa++ or aa++++aa, for example

Section:  5.10

 

32) Describe a convenient method for determining gene order from three-point cross results.

Answer:  Compare the double-crossover class with the parental class and ask which gene has switched places. The gene that switched places is in the middle.

Section:  5.3

 

33) Assume that a cross is made between an albino (a) strain of Neurospora and one that is normally pigmented (A). The frequency with which second-division segregation occurs indicates that there are 25 map units between the a locus and the centromere. Given the four asci below, shade in the ascospore patterns with the appropriate frequency that would provide such a map distance (25 map units).

 

Answer:

 

 

Section:  5.10

 

 

34) Describe the general relationship between gene location and crossover frequency between those genes.

Answer:  In general, the more distant the two genes are, the higher is the frequency of crossing over between them.

Section:  5.2

35) Why is it possible to “map” the centromere in Neurospora and not in most other organisms?

Answer:  Ascospores are held in order as meiotic products. This order is dependent on centromere migration, and a comparison of ascospore patterns allows one to determine whether crossovers have occurred. Quantifying such crossovers allows one to estimate map distance between the gene in question and the centromere.

Section:  5.10

 

36) Which yields more genetic variation, meiotic or mitotic crossing over?

Answer:  Meiotic crossing over leads to significant genetic variation.

Section:  Introduction, 5.1

 

37) When considering linked genes, which category of ascospore arrangement occurs most frequently: parental ditype, nonparental ditype, or tetratype?

Answer:  parental ditype

Section:  5.10

 

38) If two genes assort independently, which categories of ascospore arrangements occur in approximately equal frequencies: parental ditype, nonparental ditype, or tetratype?

Answer:  parental ditype and nonparental ditype

Section:  5.10

 

39) Mendel predicted that some genes would be carried in the same chromosome.

Answer:  FALSE

Section:  5.11

 

40) The cross GE/ge × ge/ge produces the following progeny: GE/ge 404, ge/ge 396, gE/ge 97, Ge/ge 103. From these data, one can conclude that the G and E loci assort independently.

Answer:  FALSE

Section:  5.2

 

41) The cross GE/ge × ge/ge produces the following progeny: GE/ge 404, ge/ge 396, gE/ge 97, Ge/ge 103. From these data, one can conclude that the recombinant progeny are gE/ge and Ge/ge.

Answer:  TRUE

Section:  5.2

 

42) The cross GE/ge × ge/ge produces the following progeny: GE/ge 404, ge/ge 396, gE/ge 97, Ge/ge 103. From these data, one can conclude that there are 20 map units between the G and E loci.

Answer:  TRUE

Section:  5.2

43) Linkage (viewed from results of typical crosses) always occurs when two loci are on the same chromosome.

Answer:  FALSE

Section:  5.2

 

44) Positive interference occurs when a crossover in one region of a chromosome interferes with crossovers in nearby regions.

Answer:  TRUE

Section:  5.4

45) If the frequency of parental ditypes is greater than the frequency of nonparental ditypes, then the genes in question are linked.

Answer:  TRUE

Section:  5.10

 

46) In Drosophila, the frequency of crossing over in males is about equal to the frequency of crossing over in females.

Answer:  FALSE

Section:  5.5

 

47) If two gene loci are on nonhomologous chromosomes, genes at these loci are expected to assort independently.

Answer:  TRUE

Section:  5.1

Chapter 7   Sex Determination and Sex Chromosomes

 

1) In Drosophila, sex is determined by a balance between the number of haploid sets of autosomes and the number of ________.

  1. A) telomeres
  2. B) centromeres
  3. C) X chromosomes
  4. D) Y chromosomes
  5. E) nucleolar organizers

Answer:  C

Section:  7.5

 

2) In humans, the genetic basis for determining the sex “male” is accomplished by the presence of ________.

  1. A) a portion of the Y chromosome
  2. B) one X chromosome
  3. C) a balance between the number of X chromosomes and the number of haploid

sets of autosomes

  1. D) high levels of estrogen
  2. E) multiple alleles scattered throughout the autosomes

Answer:  A

Section:  7.3

 

3) Klinefelter syndrome in humans, which leads to underdeveloped testes and sterility, is caused by which chromosomal condition?

  1. A) 47, XXY
  2. B) 47, 21+
  3. C) 45, X
  4. D) 47, XYY
  5. E) triploidy

Answer:  A

Section:  7.3

 

4) The Protenor mode of sex determination is the ________.

  1. A) scheme based on F plasmids inserted into the FMR-1 gene
  2. B) XX/XO scheme
  3. C) XO/YY scheme
  4. D) hermaphroditic scheme
  5. E) scheme based on single translocations in the X chromosome

Answer:  B

Section:  7.1, 7.2

 

 

5) The Lygaeus mode of sex determination is the ________.

  1. A) XY/XX scheme
  2. B) XX/XO scheme
  3. C) XO/YY scheme
  4. D) hermaphroditic scheme
  5. E) scheme based on single translocations in the X chromosome

Answer:  A

Section:  7.1, 7.2

6) The sex of birds, some insects, and other organisms is determined by a ZW chromosomal arrangement in which the males have like sex chromosomes (ZZ) and females are ZW (similar to XY in humans). Assume that a recessive lethal allele on the Z chromosome causes death of an embryo in birds. What sex ratio would result in the offspring if a cross were made between a male heterozygous for the lethal allele and a normal female?

  1. A) 4:1 male to female
  2. B) 2:1 male to female
  3. C) 3:1 male to female
  4. D) 1:2 male to female
  5. E) 1:1 male to female

Answer:  B

Section:  7.2

 

7) A recessive gene for red-green color blindness is located on the X chromosome in humans. Assume that a woman with normal vision (her father is color blind) marries a color-blind male. What is the likelihood that this couple’s first son will be color blind?

  1. A) 0%
  2. B) 25%
  3. C) 50%
  4. D) 75%
  5. E) 100%

Answer:  C

Section:  7.3

 

8) One form of hemophilia is caused by a sex-linked recessive gene. Assume that a man with hemophilia marries a phenotypically normal woman whose father had hemophilia. What is the probability that they will have a daughter with hemophilia? (Note: In this problem, you must include the probability of having a daughter in your computation of the final probability.)

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

Answer:  C

Section:  7.3

 

 

9) One form of hemophilia is caused by a sex-linked recessive gene. Assume that a man with hemophilia marries a phenotypically normal woman whose father had hemophilia. What is the probability that their first son will have hemophilia?

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

Answer:  D

Section:  7.3

10) Assume that a man who carries an X-linked gene has children. Assuming normal meiosis and random combination of gametes, the man would pass this gene to ________.

  1. A) half of his daughters
  2. B) all of his daughters
  3. C) all of his sons
  4. D) half of his sons
  5. E) all of his children

Answer:  B

Section:  7.3

 

11) Assume that you are told that a particular organism, Drosophila, has the XO chromosome complement. You are also told that the autosomal complement is a normal 2n. You know that in humans the XO complement is female determining. Would you be correct in assuming that the Drosophila sex for XO is also female? Choose the answer that includes the correct explanation.

  1. A) Yes, because sex determination in humans and insects is essentially the same.
  2. B) No, sex determination in Drosophila is dependent on the presence or absence of the Y chromosome.
  3. C) No, the chromosomal basis for sex determination in Drosophila based on the balance between the number of X chromosomes and haploid sets of autosomes.
  4. D) Yes, the presence of an X chromosome determines a female in both organisms.
  5. E) No, it takes two X chromosomes to produce a female in humans and a Y chromosome to produce a male in Drosophila.

Answer:  C

Section:  7.6

 

12) For an individual with the XXY chromosomal composition, the expected number of Barr bodies in interphase cells is ________.

  1. A) variable
  2. B) one
  3. C) two
  4. D) three
  5. E) zero

Answer:  B

Section:  7.5

 

 

13) Glucose-6-phosphate dehydrogenase (G6PD) deficiency is inherited as an X-linked recessive gene in humans. A woman whose father suffered from G6PD marries a normal man.

 

(a) What proportion of their sons is expected to be G6PD?

 

(b) If the husband were not normal, but were G6PD deficient, would you change your answer in part (a)?

Answer:  (a) 1/2          (b) no

Section:  7.3

14) In Drosophila, an individual female fly was observed to be of the XXY chromosome complement (normal autosomal complement) and to have white eyes as contrasted with the normal red eye color of wild type. The female’s father had red eyes, and the mother had white eyes. Knowing that white eyes are X-linked and recessive, present an explanation for the genetic and chromosomal constitution of the XXY, white-eyed individual. It is important that you state in which parent and at what stage the chromosomal event occurred that caused the genetic and cytogenetic abnormality.

Answer:  Nondisjunction could have occurred either at meiosis I or meiosis II in the mother, thus giving the XwXwY complement in the offspring.

Section:  7.6

 

15) In Drosophila, an individual female fly was observed to be of the XXY chromosome complement (normal autosomal complement) and to have white eyes as contrasted with the normal red eye color of wild type. The female’s mother and father had red eyes. The mother, however, was heterozygous for the gene for white eyes. Knowing that white eyes are X-linked and recessive, present an explanation for the genetic and chromosomal constitution of the XXY, white-eyed individual. It is important that you state in which parent and at what stage the chromosomal event occurred that caused the genetic and cytogenetic abnormality.

Answer:  Nondisjunction would have occurred at meiosis II in the mother, thus giving the XwXwY complement in the offspring.

Section:  7.6

 

16) A color-blind, chromatin-positive male child (one Barr body) has a maternal grandfather who was color blind. The boy’s mother and father are phenotypically normal. Construct and support (using appropriately labeled diagrams) a rationale whereby the chromosomal and genetic attributes of the chromatin-positive male are fully explained.

Answer:  The female (mother) must be heterozygous and undergo nondisjunction at meiosis II to produce the XrgXrgY boy.

Section:  7.5

 

17) A color-blind woman with Turner syndrome (XO) has a father who is color blind.

Given that the gene for the color-blind condition is recessive and X-linked, provide a likely explanation for the origin of the color-blind and cytogenetic conditions in the woman.

Answer:  The woman inherited an Xrg chromosome from the father. Nondisjunction in the female (either at meiosis I or II) produced an egg with no X chromosome, which, when fertilized by the Xrg-bearing sperm, produced the Turner syndrome condition.

Section:  7.3

18) Give the sex-chromosome constitution (X and Y chromosomes) and possible genotypes of offspring resulting from a cross between a white-eyed female (Xw XwY) and a wild-type male (normal chromosome complement) in Drosophila melanogaster. Include all zygotic combinations whether viable or inviable.

Answer:

X+XwXw   =        inviable (dies at third instar stage)

XwXwY      =        white-eyed female

X+Y            =        wild-type male

YY               =        inviable (dies at egg stage)

X+XwY      =        wild-type female

XwYY         =        white-eyed male

X+Xw         =        wild-type female

XwY            =        white-eyed male

Section:  7.6

 

19) A cross is made between a female calico cat and a male cat having the gene for black fur on his X chromosome. What fraction of the offspring would one expect to be calico?

Answer:  1/4

Section:  7.5

 

20) Give the sex of the following organisms assuming that the autosomes are present in the normal number.

 

Sex Chromosome                                     Organism

Complement               Humans                Drosophila

XX                              _____                    _____

XY                              _____                    _____

XO                              _____                    _____

XXX                           _____                    _____

XXY                           _____                    _____

 

Answer:

Sex Chromosome                                     Organism

Complement               Humans                Drosophila

XX                              female                   female

XY                              male                       male

XO                              female                   male

XXX                           female                   female

XXY                           male                       female

Section:  7.3, 7.6

 

21) Dosage compensation leads to a variety of interesting coat color patterns in certain mammals. For instance, a female cat that is heterozygous for two coat color alleles, say black and orange, will usually have the “calico” or mosaic phenotype. Describe the chromosomal basis for the mosaicism (calico) in the female. Explain why chromosomally normal male cats do not show the mosaic phenotype, but XXY male cats can be calico.

Answer:  Because of dosage compensation, one of the X chromosomes randomly “turns off” early in development. Once such a chromosome is inactivated, it remains so in daughter cells. Recessive alleles on the remaining active X chromosome are expressed because their normal allele (on the inactive X chromosome) is not capable of expression. Because males typically have only one X chromosome, X chromosome inactivation does not occur; however, in XXY males that are heterozygous for certain coat color genes, such inactivation and mosaicism is possible.

Section:  7.5

 

22) Below is a pedigree of a fairly common human hereditary trait; the boxes represent males and the circles represent females. Filled in symbols indicate the abnormal phenotype. Given that one gene pair is involved,

 

 

(a) Is the inheritance pattern X-linked or autosomal?

 

(b) Give the genotype of each individual in the pedigree. If more than one genotypic possibility exists, present all possible alternatives.

Answer:

(a) autosomal recessive

 

(b)

 

Section:  7.3

 

23) List three abnormalities involving numbers of X chromosomes.

Answer:  Klinefelter syndrome, Turner syndrome, XXXX syndrome

Section:  7.3

 

 

24) Individuals have been identified who have two different karyotypes, such as 45,X/46,XY or 45,X/46,XX. Such individuals are called ________.

Answer:  mosaics

Section:  7.3

25) What particular karyotype was at one time considered to be related to criminal predisposition?

Answer:  XYY

Section:  7.3

 

26) A small part of the human Y chromosome contains the gene that is responsible for determining maleness. What is the name of this gene?

Answer:  SRY (sex-determining region Y)

Section:  7.3

 

27) Under what condition might a human female have the XY sex chromosome complement?

Answer:  This female would have one complete X chromosome and a Y chromosome that lacks SRY.

Section:  7.3

 

28) How many Barr bodies would one expect to see in cells of Turner syndrome females and Klinefelter syndrome males?

Answer:  zero and one, respectively

Section:  7.5

 

29) What can cause phenotypic mosaicism for X-linked genes in female mammals?

Answer:  dosage compensation involving the X chromosome

Section:  7.5

 

30) Assuming a normal number of autosomes, what would be the sex of the following: XXY mouse, XXY Drosophila?

Answer:  male and female, respectively

Section:  7.3, 7.6

 

31) Data produced by C. Bridges in the early part of this century indicate that sex in Drosophila is determined by ________.

Answer:  a balance between the number of X chromosomes and the number of haploid sets of autosomes

Section:  7.6

 

32) In Drosophila, the sex of a fly with the karyotype XO:2A is ________.

Answer:  male

Section:  7.6

 

33) Although triple-X human females typically have normal offspring, what kinds of gametes, with respect to the X chromosomes, would you expect from such XXX females? Draw meiotic stages that show the gametes that are expected to be produced.

Answer:

 

Section:  7.3

 

34) Dosage compensation in mammals typically involves the random inactivation of one of the two X chromosomes relatively early in development. Such X-chromosome inactivation often leads to phenotypic mosaicism. Assume that black fur in cats is due to the X-linked recessive gene b, whereas its dominant allele B produces yellow fur. A Bb heterozygote is a mosaic called “tortoise shell” or “calico.” Using appropriate gene symbols, diagram a mating between a black male and a calico female. Give the phenotypes and genotypes of all the offspring.

Answer:

bY × Bb =

 

Bb (calico female), bb (black female), BY (yellow male), bY (black male)

Section:  7.5

 

35) Klinefelter and Turner syndromes have how many chromosomes, respectively?

Answer:  47, 45

Section:  7.3

 

36) Studies done in the 1960s suggested that individuals with the XYY condition were prone to criminal behavior. What conclusions presently seem appropriate concerning this chromosomal condition?

Answer:  There is a high, but not constant, correlation between the extra Y chromosome and the predisposition of males to have behavioral problems.

Section:  7.3

 

37) Describe an experiment in which transgenic mice were used to identify the male-determining region of the Y chromosome.

Answer:  When DNA containing only the mouse SRY is injected into normal XX mouse eggs, most of the offspring develop into males.

Section:  7.3

 

 

38) Describe three distinct genetic regions of the human Y chromosome.

Answer:  PARs = pseudoautosomal regions, NRY = nonrecombining region of the Y, SRY = sex-determining region

Section:  7.3

39) What is the composition of a Barr body?

Answer:  X chromosome—with associated proteins

Section:  7.5

 

40) Normally in humans, all the sons of a male showing a sex-linked phenotype will inherit the trait.

Answer:  FALSE

Section:  7.3

 

41) Normally in humans, all the sons of a female homozygous for a sex-linked recessive gene will inherit that trait.

Answer:  TRUE

Section:  7.3

 

42) Sex-influenced genes are those that cause males to be males and females to be females.

Answer:  FALSE

Section:  7.3

 

43) Sex-limited genes are those that cause males to be males and females to be females.

Answer:  FALSE

Section:  7.3

 

44) In Drosophila melanogaster, sex is determined by the ratio of the number of X chromosomes to the number of haploid sets of autosomes.

Answer:  TRUE

Section:  7.6

 

45) An individual with Klinefelter syndrome generally has one Barr body.

Answer:  TRUE

Section:  7.3

 

46) An individual with Turner Syndrome has no Barr bodies.

Answer:  TRUE

Section:  7.5

 

47) A typical XX human female has one Barr body.

Answer:  TRUE

Section:  7.5

 

48) In humans, the male is the homogametic sex.

Answer:  FALSE

Section:  7.2

 

49) In Drosophila, the female is the heterogametic sex.

Answer:  FALSE

Section:  7.2, 7.6

 

50) Dosage compensation is accomplished in humans by inactivation of the Y chromosome.

Answer:  FALSE

Section:  7.5

Chapter 11   DNA Replication and Recombination

 

1) Which of the following terms accurately describes the replication of DNA in vivo?

  1. A) conservative
  2. B) dispersive
  3. C) semidiscontinuous
  4. D) nonlinear
  5. E) nonreciprocal

Answer:  C

Section:  11.1

 

2) Which term(s) accurately reflect(s) the nature of replication of the chromosome in E. coli?

  1. A) bidirectional and fixed point of initiation
  2. B) unidirectional and reciprocal
  3. C) unidirectional and fixed point of initiation
  4. D) multirepliconic and telomeric
  5. E) bidirectional and multirepliconic

Answer:  A

Section:  11.2, 11.3, 11.4

 

3) DNA polymerase III adds nucleotides ________.

  1. A) to the 3′ end of the RNA primer
  2. B) to the 5′ end of the RNA primer
  3. C) in the place of the primer RNA after it is removed
  4. D) to both ends of the RNA primer
  5. E) to internal sites in the DNA template

Answer:  A

Section:  11.2

 

4) DNA polymerase I is thought to add nucleotides ________.

  1. A) to the 5′ end of the primer
  2. B) to the 3′ end of the primer
  3. C) in the place of the primer RNA after it is removed
  4. D) on single-stranded templates without need for an RNA primer
  5. E) in a 5′ to 5′ direction

Answer:  C

Section:  11.2, 11.3

 

5) Structures located at the ends of eukaryotic chromosomes are called ________.

  1. A) centromeres
  2. B) telomerases
  3. C) recessive mutations
  4. D) telomeres
  5. E) permissive mutations

Answer:  D

Section:  11.7

6) Which cluster of terms accurately reflects the nature of DNA replication in prokaryotes?

  1. A) fixed point of initiation, bidirectional, conservative
  2. B) fixed point of initiation, unidirectional, conservative
  3. C) random point of initiation, bidirectional, semiconservative
  4. D) fixed point of initiation, bidirectional, semiconservative
  5. E) random point of initiation, unidirectional, semiconservative

Answer:  D

Section:  11.2, 11.3, 11.4

 

7) The discontinuous aspect of replication of DNA in vivo is caused by ________.

  1. A) polymerase slippage
  2. B) trinucleotide repeats
  3. C) the 5′ to 3′ polarity restriction
  4. D) topoisomerases cutting the DNA in a random fashion
  5. E) sister-chromatid exchanges

Answer:  C

Section:  11.3, 11.4, 11.5

 

 

8) Assume that a culture of E. coli was grown for approximately 50 generations in 15N (provided in the medium in the ammonium ion), which is a heavy isotope of nitrogen (14N). You extract the DNA from the culture, and it has a density of 1.723 gm/cm3 (water = 1.00 gm/cm3). From the literature, you determine that DNA containing only the common form of nitrogen, 14N, has a density of 1.700 gm/cm3. Bacteria from the 15N culture were washed in buffer and transferred to 14N medium for one generation immediately after which the DNA was extracted and its density determined.

 

(a) What would be the expected density of the extracted DNA?

 

(b) After you heat the extracted DNA until it completely denatures (95oC for 15 minutes), what would you expect the density of the DNA in the denatured extract to be? For the purposes of this question, assume that DNA has the same density regardless of whether it is single- or double-stranded.

 

(c) Assuming that the molar percentage of adenine in the extracted DNA was 20%, what would be the expected molar percentages of the other nitrogenous bases in this DNA?

 

(d) Assume that a fraction of the extracted DNA was digested to completion with the enzyme snake venom diesterase. This enzyme cleaves between the phosphate and the 3′ carbon. Present a “simplified” diagram that would illustrate the structure of the predominant resulting molecule.

Answer:

(a) approximately 1.712

(b) 1.723 and 1.700

(c) thymine = 20%, guanine = 30%, cytosine = 30%

(d)

 

 

Section:  11.1

 

 

9) Refer to the following diagram of a generalized tetranucleotide to answer questions (a) through (e).

 

 

(a) Is this a DNA or an RNA molecule? ________

 

(b) Place an “X” (in one of the circles in the diagram) at the 3′ end of this tetranucleotide.

 

(c) Given that the DNA strand, which served as a template for the synthesis of this tetranucleotide, was composed of the bases 5′-ACAG-3′, fill in the parentheses (in the diagram) with the expected bases.

 

(d) Suppose that one of the precursors for this tetranucleotide was a 32P-labeled guanine nucleoside triphosphate (the innermost phosphate containing the radioactive phosphorus).Circle the radioactive phosphorus atom as it exists in the tetranucleotide.

 

(e) Given that spleen diesterase (breaks between the phosphate and the 5′ carbon) digests the pictured tetranucleotide, which base(s) among the breakdown products would be expected to be attached to the 32P?

Answer:

(a) DNA

(b) place in bottom circle

(c) 3′-TGTC-5′

(d) phosphate on the 5′ side of the guanine

(e) thymine closer to the 5′ end

Section:  11.2, 11.3

 

10) Assume that you are microscopically examining mitotic metaphase cells of an organism with a 2n chromosome number of 4 (one pair metacentric and one pair telocentric). Assume also that the cell passed through one S phase labeling (innermost phosphate of dTTP radioactive) just prior to the period of observation. Assuming that the circle below represents a cell, draw its chromosomes and the autoradiographic pattern you would expect to see.

Answer:

 

Section:  11.1

 

11) The Meselson and Stahl experiment provided conclusive evidence for the semiconservative replication of DNA in E. coli. What pattern of bands would occur in a CsCl gradient for conservative replication?

Answer:  After one generation in the 14N, there would be two bands, one heavy and one light (no intermediate). After the second generation in the 14N, there would also be two bands, one heavy and one light (no intermediate).

Section:  11.1

12) Given that the nature of DNA replication in eukaryotes is not as well understood as in prokaryotes, (a) present a description of DNA (chromosome) replication as presently viewed in eukaryotes and (b) state the differences known to exist between prokaryotic and eukaryotic DNA replication.

Answer:

(a) Eukaryotic DNA is replicated in a manner very similar to that in E. coli: bidirectional, continuous on one strand and discontinuous on the other, and similar requirements for synthesis (four deoxyribonucleoside triphosphates, divalent cation, template, and primer).

 

(b) Okazaki fragments are about one-tenth the size of those in bacteria. Different portions of the chromosome (euchromatin, heterochromatin) replicate at different times. There are multiple replication origins in eukaryotic chromosomes.

Section:  11.6

13) Each of the following terms refers to the replication of chromosomes. Describe the role (relationship) of each in (to) chromosome replication.

 

(a) Okazaki fragment

(b) Lagging strand (c)Bidirectional

Answer:

(a) Okazaki fragment is a short single-stranded stretches of DNA on the lagging strand. See figures in the Klug/Cummings text.

 

(b) Lagging strand is the side of the replication fork where synthesis is discontinuous. See figures in the Klug/Cummings text.

 

(c) Bidirectional indicates that from the point of initiation, replication occurs in both directions along the DNA. See figures in the Klug/Cummings text.

Section:  11.3, 11.4

 

14) Assume that you were growing cells in culture and had determined the cell-cycle time to be 24 hours. You introduce 3H thymidine and prepare autoradiographs of metaphase chromosomes after 48 hours. Of the chromosomes that are labeled, you expect two classes: one class that had completed one S phase in the label, and a second class that had completed a cellular division and an additional S phase in the label. Draw the DNA (double-stranded) labeling pattern for each chromosome that you would expect to find in these two types of metaphase chromosomes. (Use a broken line {- – -} for labeled single strands of DNA and a solid line for unlabeled single strands of DNA.)

 

(a) metaphase chromosome having replicated once in label

(b) metaphase chromosome having gone through two S phases in label

Answer:

(a)

 

(b)

 

 

Section:  11.1

 

15) Below is a diagram of DNA replication as currently believed to occur in E. coli. From specific points, arrows lead to numbers. Answer the questions relating to the locations specified by the numbers.

 

 

(1) Which end (5′ or 3′) of the molecule is here?

(2) Which enzyme is probably functioning here to deal with supercoils in the DNA?

(3) Which enzyme is probably functioning here to unwind the DNA?

(4) Which nucleic acid is probably depicted here?

(5) What are these short DNA fragments usually called?

(6) Which enzyme probably functions here to couple these two newly synthesized fragments of DNA?

(7) Is this strand the leading or lagging strand?

(8) Which end (5′ or 3′) of the molecule is here?

Answer:

(1)        5′

(2)        gyrase

(3)        helicase

(4)        RNA

(5)        Okazaki fragments

(6)        ligase

(7)        lagging

(8)        5′

Section:  11.2, 11.3, 11.4

 

 

16) Assume that you grew a culture of E. coli for many generations in medium containing 15N (from the ammonium ion), a heavy isotope of nitrogen. You extract DNA from a portion of the culture and determine its density to be 1.723 gm/cm3 (call this sample A). You then wash the remaining E. coli cells and grow them for one generation in 14N, extract the DNA from a portion of the culture, and determine its density to be 1.715 gm/cm3 (call this sample B). You let the culture grow for one more generation in 14N, and extract the DNA (call this sample C). Each sample of DNA (A, B, and C) is then heated to completely denature the double-stranded structures, cooled quickly (to keep the strands separate), and subjected to ultracentrifugation. Present the centrifugation profiles for heat-denatured DNA (samples A, B, and C) that you would expect. Use the graph below. (Note: Although not the case, assume that single-stranded DNA has the same density as double-stranded DNA.)

 

 

 

Answer:

 

 

Section:  11.1

 

17) Assume that you grew a culture of E. coli for many generations in medium containing 15N (from the ammonium ion), a heavy isotope of nitrogen. You extract DNA from a portion of the culture and determine its density to be 1.723 gm/cm3 (call this sample A). You then wash the remaining E. coli cells and grow them for one generation in 14N, and extract the DNA from a portion of the culture (call this sample B). You let the culture grow for one more generation in 14N, and extract the DNA (call this sample C). Each sample of DNA (A, B, and C) is then subjected to ultracentrifugation. Present the centrifugation profiles that you would expect under (a) semiconservative replication and (b) conservative replication. (Note: Assume that unlabeled [14N] DNA has a density of 1.700 gm/cm3.)

Answer:

 

Section:  11.1

 

18) List four enzymes known to be involved in the replication of DNA in bacteria.

Answer:  Appropriate answers would include any four of the following: DNA polymerase I, III, ligase, RNA primase, helicase, gyrase

Section:  11.2, 11.3, 11.4

 

19) Which structural circumstance in DNA sets up the requirement for its semidiscontinuous nature of replication?

Answer:  5′ > 3′ polarity restrictions of DNA synthesis and the antiparallel orientation of the DNA strands in DNA

Section:  11.2, 11.3, 11.4

 

20) As unwinding of the helix occurs during DNA replication, tension is created ahead of the replication fork. Describe the nature of this tension and state the manner in which this tension is resolved.

Answer:  supercoiling; DNA gyrase

Section:  11.2, 11.3, 11.4

 

21) The complex of proteins that is involved in the replication of DNA is called a(n) ________.

Answer:  replisome

Section:  11.2

 

 

22) Given that the origin of replication is fixed in E. coli, what signals the location of the origin?

Answer:  a region called oriC, which consists of about 250 base pairs characterized by repeating sequences of 9 and 13 bases (9mers and 13mers)

Section:  11.3

23) Which protein is responsible for the initial step in unwinding the DNA helix during replication of the bacterial chromosome?

Answer:  DnaA

Section:  11.3

 

24) During DNA replication, what is the function of RNA primase?

Answer:  RNA primase provides a free 3′-OH upon which DNA polymerization depends.

Section:  11.3, 11.4

 

25) Compare the rate of DNA replication in prokaryotes and eukaryotes.

Answer:  Eukaryotic DNA polymerases synthesize DNA at a rate 25 times slower (about 2000 nucleotides per minute) than do prokaryotes.

Section:  11.6

 

26) What is the name of the replication unit in prokaryotes, and how does it differ in eukaryotes?

Answer:  replicon; one replicon in prokaryotes, multiple replicons in eukaryotes

Section:  11.1, 11.3, 11.6

 

27) Describe the DNA base sequence arrangement at the end of the Tetrahymena chromosome and the resolution of DNA replication at the end of a linear DNA strand.

Answer:  Telomeres terminate in a 5′-TTGGGG-3′ sequence, and telomerase is capable of adding repeats to the ends, thus allowing the completion of replication without leaving a gap and shortening the chromosome following each replication.

Section:  11.7

 

28) Describe a somewhat extraordinary finding related to the Tetrahymena telomerase enzyme.

Answer:  The enzyme contains a short piece of RNA that is essential for its catalytic activity.

Section:  11.7

 

29) What term is used to describe genetic exchange at equivalent positions along two chromosomes with substantial DNA sequence homology?

Answer:  general or homologous recombination

Section:  11.8

 

30) Describe the function of the RecA protein.

Answer:  The RecA protein promotes the exchange of reciprocal single-stranded DNA molecules by enhancing hydrogen bond formation during strand displacement.

Section:  11.8

 

 

31) What three possible models were suggested to originally describe the nature of DNA replication?

Answer:  conservative, semiconservative, dispersive

Section:  11.1

32) Given the diagram below, assume that a G1 chromosome (left) underwent one round of replication in 3H-thymidine and the metaphase chromosome (right) had both chromatids labeled. Which of the following replicative models (conservative, dispersive, semiconservative) could be eliminated by this observation?

 

 

Answer:  conservative

Section:  11.3

 

33) Meselson and Stahl determined that DNA replication in E coli is semiconservative. What additive did they initially supply to the medium in order to distinguish “new” from “old” DNA?

Answer:  15N

Section:  11.1

 

34) Briefly describe what is meant by the term autoradiography and identify a classic experiment that used autoradiography to determine the replicative nature of DNA in eukaryotes.

Answer:  Autoradiography is a technique that allows an isotope to be detected within a cell; the Taylor, Woods, and Hughes (1957) experiment used 3H-thymidine.

Section:  11.1

 

35) What primary ingredients, coupled with DNA polymerase I, are needed for the in vitro synthesis of DNA?

Answer:  dNTP, DNA template, primer DNA or RNA, Mg++ (appropriate buffering, temperature, and salt concentrations might be considered “secondary” ingredients)

Section:  11.2

 

36) DNA replication in vivo requires a primer with a free 3′ end. What molecular species provides this 3′ end, and how is it provided?

Answer:  The free 3′ end is provided by an RNA primer; it is provided by the enzymatic activity of RNA primase.

Section:  11.2, 11.3

 

 

37) DNA replication occurs in the 5′ to 3′ direction; that is, new nucleoside triphosphates are added to the 3′ end.

Answer:  TRUE

Section:  11.2, 11.3

 

38) DNA replicates conservatively, which means that one of the two daughter double helices is “old” and the other is “new.”

Answer:  FALSE

Section:  11.2

39) DNA strand replication begins with an RNA primer.

Answer:  TRUE

Section:  11.3, 11.4

 

40) In general, DNA replicates semiconservatively and bidirectionally.

Answer:  TRUE

Section:  11.1

 

41) In ligase-deficient strains of E. coli, DNA and chromosomal replication are unaltered because ligase is not involved in DNA replication.

Answer:  FALSE

Section:  11.5

 

42) During replication, primase adds a DNA primer to RNA.

Answer:  FALSE

Section:  11.2, 11.3, 11.4

 

43) An endonuclease is involved in removing bases sequentially from one end of DNA or the other.

Answer:  FALSE

Section:  11.2, 11.3, 11.4

 

44) In the Meselson and Stahl (1958) experiment, bean plants (Vicia faba) were radioactively labeled so that autoradiographs could be made of chromosomes.

Answer:  FALSE

Section:  11.1

 

45) A nucleosome is a structure associated with the nuclear membrane. It helps maintain a stable relationship between the extracellular matrix and the membrane itself.

Answer:  FALSE

Section:  11.6

 

46) A characteristic of aging cells is that their telomeres become shorter.

Answer:  TRUE

Section:  11.7

 

 

47) Telomerase is an RNA-containing enzyme that adds telomeric DNA sequences onto the ends of linear chromosomes.

Answer:  TRUE

Section:  11.7

 

48) Bacteria are dependent on telomerase to complete synthesis of their chromosome ends.

Answer:  FALSE

Section:  11.7

 

49) Chromatin assembly factors (CAFs) move along with the replication fork and assemble new nucleosomes.

Answer:  TRUE

Section:  11.6

50) G-quartets are G-rich single-stranded tails that loop back on themselves forming G-G double stranded sections. Such looping is involved in aligning chromosomes for homologous recombination.

Answer:  FALSE

Section:  11.8