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GENETICS CHAP 10-12

True/False

____ 1. Homologous chromosomes are two chromosomes with identical DNA sequences.

____ 2. Recent research suggests that beneficial mutations accumulate faster when species undergo sexual reproduction rather than asexual reproduction.

____ 3. Sexual reproduction would be more advantageous than asexual reproduction for organisms living in an environment that is diverse and undergoes frequent changes.

____ 4. The separation of genes during crossing over occurs more frequently between genes that are far apart on a chromosome than for genes that are close together.

____ 5. During meiosis I, homologus chromosome pairs are separated when the centromeres split apart.

____ 6. Gregor Mendel’s research supports the idea each organism carries a pair of alleles.

____ 7. Griffith’s experiment with R and S strain Streptococcus pneumoniae showed that DNA is the genetic material of the cell.

____ 8. The central dogma means that environmental factors have no influence on the transcription and translation of genes.

____ 9. The enzymes involved in DNA replication are named for the tasks they perform.

____ 10. DNA is a structurally simpler molecule than protein.

____ 11. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____.a. homozygous recessive c. homozygous dominantb. heterozygous d. haploid

____ 12. In mink, brown fur color is dominant to silver-blue fur color. If a homozygous brown mink is mated with a silver-blue mink and 8 offspring are produced, how many would be expected to be silver-blue?a. 0 c. 6b. 3 d. 8

Figure 10-8

____ 13. In Figure 10-8, which set of chromatids illustrates the result of a single crossover of the homologous chromosomes?a. A c. Cb. B d. D

____ 14. In Figure 10-8, which set of chromatids will result if each chromatid crossed with a nonsister chromatid?a. A c. Cb. B d. D

____ 15. Crossing over would most likely occur during which stage of the cell cycle?a. when DNA is being replicatedb. when homologous chromomosomes line up in pairsc. when centromeres are separatedd. when cytokinesis begins

____ 16. Which is the best description of the events that take place during anaphase II?a. The replicated chromosomes become visible.b. Homologous chromosomes line up along the equator.c. Sister chromatids are separated and pulled to opposite sides of the cell.d. Homologous pairs are separated and pulled to opposite sides of the cell.

____ 17. A heterozygous organism is best described as which of these?a. dominant c. hybridb. genotype d. true-breeding

____ 18. In which situation are the phenotypes of F2 offspring expected to follow the ratio of 9:3:3:1.a. a monohybrid cross for 2 unlinked traitsb. a monohybrid cross for 2 closely linked traitsc. a dihybrid cross for 2 unlinked traitsd. a dihybrid cross for 2 closely linked traits

____ 19. Of the following species used in agriculture, which is most likely a polyploid?a. cow c. henb. goat d. wheat

____ 20. A geneticist crossed fruit flies to determine whether two traits are linked. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which results in the next generation would suggest these traits are linked?a. 35 normal wings, normal bristles, 28 normal wings, spineless bristles, 23 blistery wings,

normal bristles, 30 blistery wings, spineless bristlesb. 105 normal wings, normal bristles, 101 normal wings, spineless bristles, 111 blistery

wings, normal bristles, 115 blistery wings, spineless bristlesc. 198 normal wings, normal bristles, 200 normal wings, spineless bristles, 185 blistery

wings, normal bristles, 189 blistery wings, spineless bristlesd. 222 normal wings, normal bristles, 27 normal wings, spineless bristles, 22 blistery wings,

normal bristles, 228 blistery wings, spineless bristles

Figure 11-1

____ 21. Refer to Figure 11-1. If individual III-2 marries a person with the same genotype as individual I-1, what is the chance that one of their children will be afflicted with hemophilia?a. 0% c. 50%b. 25% d. 75%

____ 22. What type of inheritance pattern does the trait represented by the shaded symbols in Figure 11-1 illustrate?a. incomplete dominance c. codominanceb. multiple alleles d. sex-linked

____ 23. For the trait being followed in the pedigree, individuals II-1 and II-4 in Figure 11-1 can be classified as ____.a. homozygous dominant c. homozygous recessiveb. mutants d. carriers

____ 24. What is the relationship between individual I-1 and individual III-2 in Figure 11-1?a. grandfather-granddaughter c. great aunt-nephewb. grandmother-grandson d. mother-son

____ 25. If a female fruit fly heterozygous for red eyes (XRXr) crossed with a white-eyed male (XrY), what percent of their offspring would have white eyes?a. 0% c. 50%b. 25% d. 75%

____ 26. When roan cattle are mated, 25% of the offspring are red, 50% are roan, and 25% are white. Upon examination, it can be seen that the coat of a roan cow consists of both red and white hairs. This trait is one controlled by ____.a. multiple alleles c. sex-linked genesb. codominant alleles d. polygenic inheritance

____ 27. A cross between a white rooster and a black hen results in 100% blue Andalusian offspring. When two of these blue offspring are mated, the probable phenotypic ratio seen in their offspring would be ____.a. 100% blue c. 75% blue, 25% whiteb. 75% black, 25% white d. 25% black, 50% blue, 25% white

Figure 11-2

____ 28. According to the pedigree in Figure 11-2, how many of the offspring in the III generation show the normal trait?a. 1 c. 4b. 2 d. 5

____ 29. A phenotype that results from a dominant allele must have at least _____ dominant allele(s) present in the parent(s).a. one c. threeb. two d. four

____ 30. Examine the graph in Figure 11-3, which illustrates the frequency in types of skin pigmentation in humans. Another human trait that would show a similar inheritance pattern and frequency of distribution is ____.

Figure 11-3a. height c. number of fingers and toesb. blood type d. incidence of cystic fibrosis

____ 31. What phenotype is depicted in Figure 11-5?

Figure 11-5

a. O c. Ab. AB d. B

____ 32. Nondisjunction is related to a number of serious human disorders. How does nondisjunction cause these disorders? a. alters the number of gametes producedb. alters the number of zygotes producedc. alters the chromosome structured. alters the chromosome number

____ 33. A pea plant homozygous for the trait of smooth seeds is crossed with a pea plant that is homozygous for the trait of wrinkled seeds. The first generation produces seeds that are all smooth. What percent of the second-generation plants will have smooth seeds when the F1 generation is self-fertilized? a. 100% c. 50%b. 75% d. 25%

Figure 11-6This pedigree shows the transmission of a rare disease that is dehabilitating but not lethal. Carriers are not shown.

____ 34. Which type of heredity does the pedigree in Figure 11-6 demonstrate?a. autosomal recessive c. X-linked recessiveb. autosomal dominant d. X-linked dominant

____ 35. What do galactosemia and Tay-Sachs disease have in common?a. Both are conditions in which the genotype will be seen in the phenotype.b. Both are conditions characterized by respiratory failure.c. Both are conditions caused by the lack of a gene that codes for particular enzymes. d. Both are conditions caused by dominant alleles.

____ 36. The coat color in Labrador retrievers is controlled by two sets of alleles. The gene E/e determines whether the fur has pigment or not and is epistatically dominant to the gene B/b, which controls the darkness of pigment when it is there. A breeder crosses a purebred black Lab, with the genotype BBEE, and a purebred yellow Lab with the genotype bbee, producing offspring that are black. A test cross is done between these offspring and an individual with the genotype bbee. What is the expected ratio of black:chocolate:yellow?

a. 1:1:2 c. 2:1:1b. 1:2:1 d. 2:1:0

Figure 12-2

____ 37. Structure III in Figure 12-2 represents a(n) ____.a. gene c. codonb. amino acid d. DNA molecule

____ 38. The process illustrated in Figure 12-2 is called ____.a. translation c. monoploidyb. replication d. transcription

Help Wanted

Positions Available in the genetics industry. Hundreds of entry-level openings for tireless workers. No previous experience necessary. Must be able to transcribe code in a nuclear environment.

Accuracy and Speed vital for this job in the field of translation. Applicants must demonstrate skills in transporting and positioning amino acids. Salary commensurate with experience.

Executive Position available. Must be able to maintain genetic continuity through replication and control cellular activity by regulation of enzyme production. Limited number of openings. All benefits.

Supervisor of production of proteins—all shifts. Must be able to follow exact directions from double-stranded template. Travel from nucleus to the cytoplasm is additional job benefit.

Table 12-1

____ 39. Applicants for the third job of the Help Wanted ad in Table 12-1, "Executive Position," could qualify if they were ____.a. DNA c. tRNAb. mRNA d. rRNA

Figure 12-3

____ 40. What type of mutation has occurred in Figure 12-3?a. substitution c. lethalb. frameshift d. insertion

____ 41. What will be the result of the mutation in Figure 12-3?a. it will have no effect on protein functionb. only one amino acid will changec. nearly every amino acid in the protein will be changedd. translation will not occur

____ 42. A DNA segment is changed from -AATTAGAAATAG- to -ATTAGAAATAG-. This is a ____.a. frameshift mutation c. inversionb. insertion d. translation

____ 43. Where would a DNA substitution probably have the smallest or least effect on the organism?a. exon c. intronb. histone d. operon

____ 44. A particular sequence of parent DNA has four purine bases and two pyrimidine bases. According to base-pairing rules, which of the following sequences could be formed during replication?a. two cytosine, two adenine, two thymineb. two cytosine, two adenine, two uracilc. two adenine, two thymine, one guanine, one cytosined. two adenine, two guanine, two cytosine

____ 45. Which of the following sequences of processes correctly reflects the central dogma?a. protein synthesis, transcription, translationb. protein synthesis, translation, transcriptionc. transcription, translation, protein synthesisd. translation, transcription, protein synthesis

____ 46. You are a medical researcher trying to create a new antibiotic that will interfere with bacterial DNA replication without harming the eukaryotic host. You have found several chemicals that prevent DNA from unwinding and separating. Which of the following is the best chemical to use?a. a chemical that blocks uracil useb. a chemical that cannot pass into the cell nucleusc. a chemical that is neutralized by cytoplasmd. a chemical that works only in the presence of histones

____ 47. This is a template DNA sequence: 3'AATCGC5'. This is a partially-completed mRNA strand transcribed from the DNA template: 3'GCGA5'. What is the next nucleotide that RNA polymerase will attach?a. A c. Tb. C d. U

____ 48. Using DNA sequencing, you discover that a bacterium has experienced a deletion mutation that removed three nucleotides. The bacterium appears completely unaffected in all its functions. Where is the mostly likely location for the mutation?a. an exon c. a promoterb. an intron d. a repressor

Table 12-2

____ 49. Three samples of DNA contain the percentages of nitrogenous bases listed in Table 12-2. According to Chargaff’s law, which two samples probably belong to the same species?a. 1 and 2b. 1 and 3c. 2 and 3d. cannot tell without data on guanine and thymine

____ 50. The template strand of a piece of DNA being replicated reads: 5'-ATAGGCCGT-3'. A partially synthesized Okazaki fragment is 5'CCTA3'. If the next fragment is four bases long, what is its first base?a. A c. Gb. C d. T

Completion

51. An individual with the genotype Aa is ____________________ for the trait.

52. Genetic recombination occurs when chromosomes undergo ____________________.

53. Gregor Mendel is best known for his research on ____________________ plants.

54. Aneupoidy, the condition of having an abnormal number of chromosomes, can occur during meiosis II, during which ____________________ fail to separate.

55. ____________________, found at the end of chromosomes, prevent the chromosomes from fusing into rings.

56. A karyotype arranges chromosomes in the order of their number, which is also in ____________________ order of size.

57. Analyzing the karyotype of a child before it is born has the benefit of alerting a mother to the presence of ____________________ abnormalities.

58. A normal human karyotype displays ____________________ pairs of autosomes.

59. In the spring, many trees produce flowers and regrow the leaves that were lost during the fall. This is an example of a change in ____________________.

Figure 11-7

60. The abnormality of the karyotype shown in Figure 11-7 is ____________________.

61. The karyotype in Figure 11-7 has a total of ____________________ chromosomes.

62. This person to whom the karyotype in Figure 11-7 belongs has a total of ____________________ autosomes.

63. A woman’s father developed Huntington’s disease at age 40. Her probability of also developing the disease as she ages is ____________________%.

64. A bacterium that synthesizes an enzyme that neutralizes an environmental toxin would probably use a(n) ___________________ operon to regulate production of the enzyme.

Help Wanted

Positions Available in the genetics industry. Hundreds of entry-level openings for tireless workers. No previous experience necessary. Must be able to transcribe code in a nuclear environment.

Accuracy and Speed vital for this job in the field of translation. Applicants must demonstrate skills in transporting and positioning amino acids. Salary commensurate with experience.

Executive Position available. Must be able to maintain genetic continuity through replication and control cellular activity by regulation of enzyme production. Limited number of openings. All benefits.

Supervisor of production of proteins—all shifts. Must be able to follow exact directions from double-stranded template. Travel from nucleus to the cytoplasm is additional job benefit.

Table 12-1

65. The first position listed in Table 12-1 would be filled by ____________________.

66. Prokaryotic and eukaryotic chromosomes are similar because they are both made of DNA and ____________________.

67. Similar enzymes interact with the leading and lagging strands in DNA replication. Enzymes involved with the RNA primer are more active on the lagging strand, as is the enzyme ____________________.

68. You find a salamander that has an extra eye in the place where its left ear should be. This unusual appearance is most likely due to a mutation in a(n) ____________________ gene.

69. During DNA replication, single-stranded binding proteins associate with DNA after the ____________________ enzyme has acted on the DNA.

70. In the early 20th century, some scientists were reluctant to accept the conclusion that DNA could be the genetic material because it is so ____________________.

Short Answer

71. Describe how genetic recombination through segregation and crossing over can lead to variations in the offspring.

72. Approximately one out of every 20 persons of European descent in the United States carries an allele for a hereditary disorder known as cystic fibrosis (CF), but only one in every 2000 babies born to parents in this group in the United States is afflicted with the disorder. Individuals with two alleles for cystic fibrosis produce large amounts of mucus that accumulate in the lungs, liver, and pancreas. This mucus clogs important ducts in these organs and causes extensive damage. Why is there such a difference between the number of individuals who carry the allele and the number actually born with the disease?

73. Explain why it was important that Mendel crossed “true breeding” varieties of peas when he performed his crosses?

74. A bacterial cell tolerates low levels of molecule X in the environment. High levels of X, however, are lethal unless the bacterium is producing a set of protective enzymes. These enzymes can be transported out of the cell and incorporated by other nearby bacteria. What type of gene regulation is most effective at protecting the bacterium? Justify your answer.

Problem

75. In guinea pigs, the allele for rough coat (R) is dominant to the allele for smooth coat (r), and the allele for black fur (B) is dominant to the allele for white fur (b). If two guinea pigs that are heterozygous for rough, black fur (RrBb) are mated, what are the possible phenotypes and what is the frequency of each? Show your work in a Punnett square, Figure 10-4.

Figure 10-4

Some biology students wanted to determine whether a pair of brown mice purchased at a pet store was homozygous dominant or heterozygous for fur color. They let the mice mate and examined the offspring. Six mice were born. All six had brown fur.Some of the students felt that this was enough evidence to prove that the mice were homozygous for brown fur color. Other students did not, so another experiment was planned.

76. Describe the next experiment the students could conduct to determine whether the parent mice are homozygous brown or heterozygous. Explain your answer.

77. Hypophosphatemia, an X-linked dominant disorder, causes muscle weakness due to low phosphate levels. A man with this disorder plans to marry a woman who does not display symptoms of the disorder. Describe the pattern of inheritance and determine how many of their male children will have this disorder.

Essay

78. Imagine that you are a geneticist working with a couple who are expecting a child. The mother is 41 and the father’s brother has cystic fibrosis. Please describe the risks and benefits of performing amniocentesis and give them information that will help them make a decision as to whether or not they will do fetal testing.

79. Imagine that several years ago one of your parents began to exhibit a decline in mental functions and coordination in movement. The condition has recently been diagnosed as Huntington’s disease. Would you want to be tested for the Huntington’s allele? Why or why not? In what ways would this information affect your life if you did test positive?

80. Consider the function of homeobox genes, described on page 344. They were first discovered and are best known in fruit flies (Drosophila), though similar genes exist in many other organisms. How useful do you think it is to use Drosophila to investigate the genetics of other eukaryotes? Give some reasons for your opinion.

GENETICS CHAP 10-12Answer Section

TRUE/FALSE

1. ANS: FHomologous chromosomes contain the same genes as one another, but the DNA sequences vary. One homolog comes from an individual’s father, while the other comes from the mother.

PTS: 1 DIF: Bloom's Level B REF: 270NAT: LS_2a TOP: 10-3

2. ANS: TSexual reproduction allows for more variety in offspring, allowing natural selection to play a stronger role in favoring those individuals with beneficial traits.

PTS: 1 DIF: Bloom's Level B REF: 276NAT: LS_3a TOP: 10-3

3. ANS: TSexual reproduction allows for greater variety in offspring. Thus, in a variable and diverse environment, there is a greater chance of success for organisms that reproduce sexually as compared to organisms that reproduce asexually.

PTS: 1 DIF: Bloom's Level C REF: 276NAT: LS_3a TOP: 10-3

4. ANS: TThe farther apart two loci are, the more likely it is that they will be separated by a crossover.

PTS: 1 DIF: Bloom's Level B REF: 284NAT: LS_2b TOP: 10-8

5. ANS: FHomologous chromosomes separate, but centromeres do not.

PTS: 1 DIF: Bloom's Level A REF: 270–275NAT: LS_2b TOP: 10-2

6. ANS: THe concluded the genes come in pairs, such as round versus smooth seed, or green versus yellow seed.

PTS: 1 DIF: Bloom's Level A REF: 278NAT: LS_2b TOP: 10-4

7. ANS: FGriffith showed only that live R strain bacteria became disease-causing in the presence of heat-killed S strain bacteria.

PTS: 1 DIF: Bloom's Level F REF: 326NAT: LS_2a TOP: 12-1

8. ANS: FThe central dogma summarizes the production sequence from DNA to RNA to protein. Environmental factors influence gene transcription and translation through a variety of regulatory mechanisms.

PTS: 1 DIF: Bloom's Level E REF: 342–345NAT: LS_1d TOP: 12-8

9. ANS: TThe enzymes are given names that reflect their tasks: helicase unwinds the helix, primase adds a primer, etc.

PTS: 1 DIF: Bloom's Level D REF: 333–334NAT: LS_2a TOP: 12-4

10. ANS: TDNA is made up of only 4 nucleotide bases, but protein is a chain that can contain up to twenty different types of amino acids folded into a complex shape.

PTS: 1 DIF: Bloom's Level D REF: 329NAT: LS_2a TOP: 12-2

MULTIPLE CHOICE

11. ANS: AThe most likely scenario is that the white mouse displays the recessive trait. If this is the case, then the white mouse must be homozygous. If the white mouse were either homozygous dominant or heterozygous, then it would likely produce white offspring when mated with a brown mouse.

FeedbackA Correct.B If the white mouse is heterozygous, then this would mean white is dominant and its

brown mate is homozygous recessive. Such a cross would yield about half white and brown offspring.

C If the white mouse were homozygous dominant, then its offspring would have to be white.

D Sperm and eggs are haploid. Individual mice are not.

PTS: 1 DIF: Bloom's Level D REF: 277–281NAT: LS_2b TOP: 10-6

12. ANS: AThe brown mink will always donate a dominant allele to its offspring, which means all the offspring will have the dominant phenotype. None of the offspring would be silver-blue, which is the recessive color.

FeedbackA Well done.B Remember, the brown mink is homozygous.C Remember, the brown mink is homozygous.D Brown is dominant to silver-blue.


13. ANS: AA single crossover will cause B and b to change positions relative to A and a on just two chromatids. The other two chromatids will be unaffected, one would remain AB and the other aa. The chromatids that experience the crossover will contain Ab and aB.

FeedbackA This is correct.B This shows the effect of zero crossovers.C This shows the effect of two crossovers.D There should be 2 copies of B.

PTS: 1 DIF: Bloom's Level E REF: 283–284NAT: LS_2b TOP: 10-8

14. ANS: CThis would result in recombination on all the chromatids, producing two Ab chromatids and two aB chromatids.

FeedbackA This is the result of one crossover.B Try again.C Correct.D There must be two B alleles.


15. ANS: BCrossing over refers to the exchange of DNA, which occurs when homologous chromosomes form pairs.

FeedbackA Check page 272.B Well done.C Crossing over occurs before this.D Try again.


16. ANS: CAnaphase is the phase in which chromosomes are separated. Homologous pairs are separated during anaphase I. Sister chromatids are separated during anaphase II.

FeedbackA This is prophase I.B This is metaphase I.C This is correct.D This is anaphase I.

PTS: 1 DIF: Bloom's Level B REF: 272–274NAT: LS_2b TOP: 10-2

17. ANS: CA heterozygous (Aa) individual is produced by a cross of two different true-breeding (AA and aa) individuals.Thus the heterozgous individual is a hybrid.

Feedback

A This description refers to a trait or allele.B This description refers to the genetic makeup of an individual.C Correct.D This refers to a homozygous individual.


18. ANS: CThis ratio is expected in the F2 generation of a dihybrid cross when independent assortment takes place. This is the case when two traits are unlinked, or found on different chromosomes.

FeedbackA In this case, a 3 to 1 ratio is expected.B Try again.C Correct.D If the traits are linked the results will be skewed from this ratio.

PTS: 1 DIF: Bloom's Level D REF: 282NAT: LS_2b TOP: 10-6

19. ANS: DPlants are often polyploids, while animals are not.

FeedbackA Animals are not usually polyploids.B Polyploid animals are not usually robust.C Check page 284.D Correct.

PTS: 1 DIF: Bloom's Level B REF: 284NAT: LS_2c TOP: 10-9

20. ANS: DLinkage is evident if two of the phenotypic combinations occur at a much greater rate than the other two.

FeedbackA The frequencies of all groups are about equal.B This result supports independent assortment.C Try again.D Well done.


21. ANS: A

FeedbackA You are right.B You are on the right track.C Please consider the pattern in the pedigree again.D Please refer to page 299.

PTS: 1 DIF: Bloom's Level E REF: 299NAT: LS_2b TOP: 11-1

22. ANS: D

FeedbackA You are on the right track.B Please consider again all the factors.C Please refer to pages 307-308.D You are correct.

PTS: 1 DIF: Bloom's Level D REF: 299–300 | 307–308NAT: LS_2b TOP: 11-5

23. ANS: D

FeedbackA Please consider again the reference to the individual, not the allele.B Please consider the factors again.C Please refer to pages 299-300.D You are correct.


24. ANS: B

FeedbackA You are on the right track.B You are right.C Please consider again the symbols in creating a pedigree.D Please refer to page 299.

PTS: 1 DIF: Bloom's Level B REF: 299TOP: 11-3

25. ANS: C

FeedbackA Please consider all the factors.B You are on the right track.C You are correct.D Please refer to pages 296-300.

PTS: 1 DIF: Bloom's Level C REF: 296–300NAT: LS_2b TOP: 11-1

26. ANS: B

FeedbackA You are on the right track.B You are right!C Please consider again the inheritance pattern.D You are heading in the right direction.

PTS: 1 DIF: Bloom's Level C REF: 299–300NAT: LS_2b TOP: 11-4

27. ANS: D

FeedbackA Please consider again the results of mating two from the second generation.B You are on the right track.C Please refer to pages 302-303.D You are correct.


28. ANS: C

FeedbackA Are you looking at the right family?B Are these unaffected individuals?C You are correct.D Only consider the offspring.


29. ANS: A

FeedbackA You are correct!.B Think about the effect of a dominant allele on the phenotype.C Try working a cross.D Review the concept of a dominant allele.


30. ANS: A

FeedbackA You are correct.B Please review the inheritance pattern of blood types.C Please consider again the frequency of people in your class who have the same number

of fingers and toes.D Please consider the rarity of this disease in the population.

PTS: 1 DIF: Bloom's Level D REF: 309TOP: 11-4

31. ANS: D

FeedbackA Please refer to page 304.B That's a red blood cell in the diagram.

C Please review page 204 again.D You are correct!

PTS: 1 DIF: Bloom's Level A REF: 304TOP: 11-4

32. ANS: DNondisjunction occurs during cell division in which the homologous chromosomes fail to separate properly in meiosis I or in which the sister chromatids fail to separate in meiosis II. This results in gametes with an inaccurate number of chromosomes.

FeedbackA Please review the process of meosis.B Please review the formation of gametes.C You are on the right track.D You are correct!

PTS: 1 DIF: Bloom's Level C REF: 312–314NAT: LS_2c TOP: 11-9

33. ANS: BSmooth is the dominant allele. The first generation would be heterozygous, Rr. In the second generation, 75% would have the genotype Rr or RR, or smooth seeds, whereas 25% would have the recessive genotype, rr, and be wrinkled.

FeedbackA Did you consider all the factors?B You are correct.C Please refer to page 301.D Please review dominant and recessive inheritance patterns.

PTS: 1 DIF: Bloom's Level D REF: 301NAT: LS_2b TOP: 11-1

34. ANS: CBecause only the males demonstrate this disease, it shows that it is a recessive disease associated with the X chromosome.

FeedbackA What is the pattern of those who exhibit the trait?B Please refer to pages 307-308.C You are correct.D You are on the right track.

PTS: 1 DIF: Bloom's Level E REF: 299–300 | 307–308NAT: LS_2b TOP: 11-1

35. ANS: CIn the case of galactosemia, there is a lack of the gene that codes for the enzyme that breaks down galactose. In the case of Tay-Sachs disease, there is a lack of the gene that codes for an enzyme that breaks down fatty substances.

Feedback

A Please review recessive and dominant disorders.B Please refer to page 297.C You are correct!D Please review the characteristics of these disorders.

PTS: 1 DIF: Bloom's Level C REF: 296–298NAT: LS_2c TOP: 11-2

36. ANS: AThe genotypes resulting from the test cross and their phenotypes are:BbEe BlackbbEe YellowBbee Chocolatebbee Yellow

FeedbackA That's right!B Did you attribute the phenotypes correctly?C Check your crosses.D Have a look on page 305 to review.

PTS: 1 DIF: Bloom's Level F REF: 302–305NAT: LS_2b TOP: 11-4

37. ANS: BA tRNA molecule carries a specific amino acid corresponding to the anticodon of that tRNA molecule.

FeedbackA What is the definition of a gene?B That's correct!C You're on the right track, but the anticodon is located somewhere else on this molecule.D DNA does not have uracil as one of its bases.

PTS: 1 DIF: Bloom's Level C REF: 338NAT: LS_1c TOP: 12-6

38. ANS: ATranslation is the process by which the mRNA “template” is used to form polypeptides.

FeedbackA That's correct!B Page 334 shows replication.C Check the definition of monoploidy.D Is any DNA involved in the pictured process?

PTS: 1 DIF: Bloom's Level B REF: 338NAT: LS_1c TOP: 12-6

39. ANS: ADNA is the genetic material that replicates and is passed along when a cell divides. DNA controls the production of enzymes and other proteins.

Feedback

A That's correct!B Does mRNA replicate?C tRNA carries amino acids.D Check page 333 for clues.

PTS: 1 DIF: Bloom's Level D REF: 326–333NAT: LS_1c TOP: 12-1

40. ANS: BDeleting a nucleotide causes a frameshift mutation, since the codons following the deletion will code for different amino acids than the original sequence.

FeedbackA Check the definition of a point mutation on page 345.B That's correct!C How do you know what effects the mutation has on the whole organism?D A protein is simply a polypeptide, or chain of amino acids.

PTS: 1 DIF: Bloom's Level A REF: 346NAT: LS_2a | LS_2c TOP: 12-11

41. ANS: CSince the reading “frame” is shifted one place, all amino acids after the deletion will be different from the original sequence.

FeedbackA Protein function is determined by the sequence of amino acids in the protein.B Look closely at the codons that follow the mutation.C That's correct!D Do you see a stop codon?

PTS: 1 DIF: Bloom's Level B REF: 346NAT: LS_2a | LS_2c TOP: 12-11

42. ANS: AThe deletion of adenine from the beginning of the sequence causes a frameshift mutation.

FeedbackA That's correct!B Is the number of nucleotides the same in the original and mutated sequences?C An inversion reverses the order of nucleotides.D Review the definition of translation on page 338.

PTS: 1 DIF: Bloom's Level B REF: 345–346NAT: LS_2a | LS_2c TOP: 12-11

43. ANS: CIntrons, or intervening sequences, are removed from mRNA before it leaves the nucleus, so they are not used in enzyme synthesis.

FeedbackA You're on the right track.B Histones are proteins.

C That's correct!D What does an operon do?

PTS: 1 DIF: Bloom's Level E REF: 337NAT: LS_2a | LS_2c TOP: 12-7

44. ANS: AAccording to base-pairing rules, the number of purine bases formed must match the number of pyrimidine bases in the parent sequence.

FeedbackA That's correct!B Does uracil occur in DNA?C Count the number of purine bases.D Review base pairing rules on page 329.

PTS: 1 DIF: Bloom's Level E REF: 329–330 | 333NAT: LS_2a TOP: 12-2

45. ANS: CThe central dogma states that DNA is transcribed to mRNA, which is translated to a sequence of amino acids that form a polypeptide.

FeedbackA The central dogma starts with DNA.B Check page 336.C That's correct!D What is the difference between transcription and translation?

PTS: 1 DIF: Bloom's Level B REF: 336–339NAT: LS_1c TOP: 12-8

46. ANS: BThe only chemical that will prevent bacterial replication without interfering with eukaryotic replication is one that cannot enter the eukaryotic nucleus.

FeedbackA Does DNA contain uracil?B That's correct!C Both bacteria and eukaryotic cells have cytoplasm.D Bacterial chromosomes don't have histones.

PTS: 1 DIF: Bloom's Level E REF: 335NAT: IS_1e TOP: 12-3

47. ANS: DThe mRNA strand is being transcribed from the DNA in a right-to-left order. The corresponding base pair to adenine is uracil in RNA.

FeedbackA Is the mRNA being formed from left-to-right or from right-to-left?B Review base pairing on page 329.C You're on the right track.

D That's correct!

PTS: 1 DIF: Bloom's Level D REF: 337NAT: LS_1c | LS_2a TOP: 12-7

48. ANS: BIntrons, or intervening sequences, get processed out of the mRNA before it leaves the nucleus, so removal of an intron would probably have little effect on bacterial functions such as enzyme synthesis.

FeedbackA You're on the right track.B That's correct!C The promoter is involved in gene regulation.D What is a deletion?

PTS: 1 DIF: Bloom's Level E REF: 337NAT: LS_1d | LS_2c TOP: 12-7

49. ANS: BThough the numbers do not match exactly, 1 and 3 have similar percentages of adenine and cytosine.

FeedbackA Keep trying.B That's correct!C Which bases pair up in DNA?D Check page 329 for Chargaff's law.


50. ANS: AThe next fragment will begin from the 3' end of the parent strand.

FeedbackA That's correct.B Check page 334 for hints.C Is this a leading strand or a lagging strand?D Try matching the fragment to the template strand.


COMPLETION

51. ANS: heterozygous

PTS: 1 DIF: Bloom's Level A REF: 279NAT: LS_2b TOP: 10-5

52. ANS: crossing over


53. ANS: pea


54. ANS: sister chromatids


55. ANS: Telomeres

PTS: 1 DIF: Bloom's Level C REF: 311TOP: 11-8

56. ANS: decreasing


57. ANS: chromosome


58. ANS: 22


59. ANS: phenotype

PTS: 1 DIF: Bloom's Level B REF: 309TOP: 11-6

60. ANS: trisomy 21

PTS: 1 DIF: Bloom's Level D REF: 311NAT: LS_2c TOP: 11-7

61. ANS: 47


62. ANS: 45


63. ANS: 50

PTS: 1 DIF: Bloom's Level C REF: 298NAT: LS_2b TOP: 11-2

64. ANS:inducibleThe bacterium would not need to produce the enzyme unless the environmental toxin were present, so the situation is analogous to the lac operon example in the text.

PTS: 1 DIF: Bloom's Level C REF: 342–343

NAT: LS_1d TOP: 12-965. ANS:

RNA polymeraseThis molecule transcribes mRNA from the DNA template.

PTS: 1 DIF: Bloom's Level D REF: 337NAT: LS_1c TOP: 12-7

66. ANS:proteinsThey do not have the same proteins, but both types of chromosomes are formed of DNA coiled around organizing proteins.


67. ANS:DNA ligaseLigase has more sugar-phosphate backbone gaps to seal on the lagging strand.

PTS: 1 DIF: Bloom's Level D REF: 333–334NAT: LS_2a TOP: 12-4

68. ANS:homeobox or hoxThese genes are responsible for major body organization in many eukaryotes.

PTS: 1 DIF: Bloom's Level C REF: 344NAT: LS_1d | LS_2c TOP: 12-10

69. ANS:helicaseBnding proteins keep the DNA separate after they have been unwound and unzipped.

PTS: 1 DIF: Bloom's Level C REF: 333NAT: LS_2a TOP: 12-4

70. ANS:simpleEven after Avery’s experiments, scientists still thought complex proteins might be the genetic material.

PTS: 1 DIF: Bloom's Level F REF: 327TOP: 12-1

SHORT ANSWER

71. ANS:Independent segregation of homologous chromosomes during gamete formation allows for a random assortment of alleles in the sex cells. This allows the members of each pair of alleles to recombine in new ways in the offspring. Crossing over leads to new gene combinations when homologous chromosomes exchange alleles during meiosis. Thus, an allele may be paired with a trait with which it was not previously paired.

PTS: 1 DIF: Bloom's Level E REF: 283NAT: LS_2b TOP: 10-7

72. ANS:For an individual to inherit cystic fibrosis, two people, each with the recessive gene, must meet and mate. The odds of that are 1/20 1/20 or 1/400. Then, using a Punnett square, it can be determined that there is only a 25% chance that a sperm with the recessive allele will fertilize an egg with the recessive allele—hence the relatively low frequency of children with cystic fibrosis.

PTS: 1 DIF: Bloom's Level F REF: 280NAT: LS_2b TOP: 10-6

73. ANS:

Mendel wanted to make sure there were no “hidden” traits in the parental generation. This means the parents were homozygous.

PTS: 1 DIF: Bloom's Level E REF: 277NAT: IS_1a TOP: 10-4

74. ANS:Either a repressible or an inducible operon could be effective; student answers must include justification. An operon induced by X would allow protective enzymes to be made as soon as X was detected. An operon repressed by the presence of protective enzymes synthesized by other bacteria would also protect the cell.

PTS: 1 DIF: Bloom's Level F REF: 342–343NAT: LS_1d TOP: 12-9

PROBLEM

75. ANS:The phenotypic ratio is 9 rough, black fur: 3 rough, white fur: 3 smooth, black fur: 1 smooth, white fur. See Solution 10-1.

Solution 10-1

PTS: 1 DIF: Bloom's Level C REF: 282NAT: LS_2b TOP: 10-6

76. ANS:There are a number of ways students could respond correctly to this problem.a) The parent mice could be permitted to mate several more times and produce large numbers of offspring.

The larger F1 population would increase the likelihood of the recessive phenotype being expressed.b) The students could allow several of the F1 mice to interbreed. If the parents are heterozygous, about 50%

of their offspring should be heterozygous and with the larger numbers, there would be a greater chance of the recessive phenotype showing in the F3 population.

c) The parent mice each could be mated with a homozygous recessive mouse. In that way, if a parent were heterozygous, there would be a 50% chance of the offspring showing the recessive trait. There are other possible correct responses. All of the responses should in some way indicate the need for more offspring because heredity operates according to the laws of probability.

Both mice could be homozygous brown, and that is why the recessive allele does not segregate out and appear in the offspring. BUT, only one mating and six offspring are not enough to prove this mathematically. One of the pair of mice could be heterozygous and the other homozygous brown and again, the recessive trait would not be seen in the offspring. Another possibility is that both mice are heterozygous. There would be only a 25% chance that the recessive alleles would segregate out and combine during fertilization. Six offspring may not be a large enough sample mathematically to reasonably expect the 25% chance of white mice to be expressed.

PTS: 1 DIF: Bloom's Level F REF: 282NAT: IS_1a TOP: 10-6

77. ANS:Because this is X-linked dominant, all of the daughters will have this disorder because each daughter will receive the X chromosome from the father and the X chromosome from the mother will not mask the effects. None of the male children will exhibit the disorder, because they receive the normal X chromosome from their mother.

PTS: 1 DIF: Bloom's Level F REF: 296–300 | 307–308NAT: LS_2b TOP: 11-5

ESSAY

78. ANS:Amniocentesis is a method of fetal testing that has both benefits and risks for the fetus and the mother. Some of the risks associated with amniocentesis are discomfort for the expectant mother, slight risk of infection, and even the risk of miscarriage. The procedure can be of benefit because chromosome abnormalities and other genetic defects can be detected. It is not always necessary to perform fetal testing, given the risks. Because the mother is older, her risks of having a child with Down syndrome is increased. This condition could be detected with amniocentesis. Because there is no treatment for Down syndrome, the parents could be put into an ethical dilemma. On the one hand they can be more prepared, but on the other, the condition can not be changed. Because cystic fibrosis is a recessive genetic disorder, the father will be a carrier. If there has not been a demonstration of that phenotype on the side of the mother, the chances are lower that she will also be a carrier, but it is still possible. Therefore, the pedigree of the mother will need to be taken into consideration, but because the gene is recessive, it is possible for members of her pedigree to be carriers and not know it.

PTS: 1 DIF: Bloom's Level F REF: 311NAT: LS_2b TOP: 11-10

79. ANS:Answers will vary. Given that this is a dominant genetic disease, the presence of a the allele will result in its expression in the phenotype. Thus, there is at least a 50% chance of having the disease. Because there is no cure, it would be a dilemma to have this information. Possible complications include matters such as health insurance and the possibility that employers might use the information against an employee. These are questions society will need to answer as well. The personal effects of such information will vary from student to student.

PTS: 1 DIF: Bloom's Level E REF: 296–298NAT: IS_1e TOP: 11-2

80. ANS:Student answers should use evidence to support the point of view. Good answers will include characteristics not specifically mentioned in the text; e.g., that fruit flies are small and can be raised in a laboratory, or that genes in flies may work differently than genes in other organisms.

PTS: 1 DIF: Bloom's Level F REF: 344NAT: IS_1e TOP: 12-10

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