s4o3 pretest 2015-2016 multiple choice - wikispacespretest... · s4o3 pretest 2015-2016 multiple...

S4O3 Pretest 2015-2016

Multiple Choice

Identify the choice that best completes the statement or answers the question.

____ 1. What is the origin of nondisjunction for an XXX genotype that produces a phenotype of a nearly normal

female?

a. meiosis in sperm formation

b. meiosis in egg formation

c. meiosis I in sperm or egg formation

d. meiosis II in sperm or egg formation

____ 2. Nondisjunction is related to a number of serious human disorders. How does nondisjunction cause these

disorders?

a. alters the number of gametes produced

b. alters the number of zygotes produced

c. alters the chromosome structure

d. alters the chromosome number

____ 3. What occurs during the process of meiosis in humans that can lead to a child with the condition of Down

Syndrome?

a. production of a duplicate chromosome set

b. production of gametes which are diploid

c. production of gametes with one duplicate chromosome

d. production of gametes with one duplicate sex chromosome

____ 4. Which of the following could only be a result of nondisjunction during meiosis of sperm formation and not

egg formation?

a. XYY c. XXY

b. XXX d. XO

____ 5. A person has a non-normal set of sex chromosomes but is obviously female. Her cells show two Barr bodies.

Which condition accounts for these observations?

a. XXX c. XYY

b. XXY d. XO

____ 6. A series of plants, produced through vegetative propagation, have been planted at different altitudes. The

researcher observes that the higher the altitude, the shorter the plant will grow. Please make a hypothesis to

explain the results.

a. Higher altitudes affect the presence of genes that determine height.

b. Higher altitudes decrease the expression of genes that promote height.

c. The plants at lower altitudes are genetically different from plants at higher altitudes.

d. Lower altitudes cause mutations in genes that determine height.

____ 7. 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.

____ 8. What part of the chromosome might be involved with processes such as aging and cancer?

a. karyotype c. telomere

b. nondisjunction d. telophase

____ 9. Which of the following would be least likely to happen as a result of a mutation in a person's skin cells?

a. skin cancer

b. reduced functioning of the skin cell

c. no change in functioning of the skin cell

d. the person's offspring have mutated skin

____ 10. Which series is arranged in order from largest to smallest in size?

a. chromosome, nucleus, cell, DNA, nucleotide

b. cell, nucleus, chromosome, DNA, nucleotide

c. nucleotide, chromosome, cell, DNA, nucleus

d. cell, nucleotide, nucleus, DNA, chromosome

Figure 12-2

____ 11. In which part of the cell does this process shown in Figure 12-2 take place?

a. in the nucleus c. at the ribosomes

b. in food vacuoles d. on the chromosome

____ 12. Structure III in Figure 12-2 represents a(n) ____.

a. gene c. codon

b. amino acid d. DNA molecule

____ 13. The process illustrated in Figure 12-2 is called ____.

a. translation c. monoploidy

b. replication d. transcription

____ 14. Which of the structures in Figure 12-2 are composed of RNA?

a. II and IV c. I and V

b. III and IV d. III and V

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

____ 15. Applicants for the fourth job of the Help Wanted ad in Table 12-1, "Supervisor," could qualify if they were

____.

a. DNA c. tRNA

b. mRNA d. rRNA

____ 16. Applicants for the third job of the Help Wanted ad in Table 12-1, "Executive Position," could qualify if they

were ____.

a. DNA c. tRNA

b. mRNA d. rRNA

____ 17. A DNA segment is changed from-AATTAG- to -AAATAG-. This is a ____.

a. frameshift mutation c. insertion

b. substitution d. deletion

Figure 12-3

____ 18. What type of mutation has occurred in Figure 12-3?

a. substitution c. lethal

b. frameshift d. insertion

____ 19. What will be the result of the mutation in Figure 12-3?

a. it will have no effect on protein function

b. only one amino acid will change

c. nearly every amino acid in the protein will be changed

d. translation will not occur

____ 20. A DNA segment is changed from -AATTAGAAATAG- to -ATTAGAAATAG-. This is a ____.

a. frameshift mutation c. inversion

b. insertion d. translation

____ 21. Where would a DNA substitution probably have the smallest or least effect on the organism?

a. exon c. intron

b. histone d. operon

____ 22. 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 thymine

b. two cytosine, two adenine, two uracil

c. two adenine, two thymine, one guanine, one cytosine

d. two adenine, two guanine, two cytosine

____ 23. Which of the following sequences of processes correctly reflects the central dogma?

a. protein synthesis, transcription, translation

b. protein synthesis, translation, transcription

c. transcription, translation, protein synthesis

d. translation, transcription, protein synthesis

____ 24. 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 use

b. a chemical that cannot pass into the cell nucleus

c. a chemical that is neutralized by cytoplasm

d. a chemical that works only in the presence of histones

____ 25. 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. T

b. C d. U

____ 26. Here are two related mRNA sequences: 5'UUUAGCGAGCAU3' and 5'UUUAGCCAUAAAAAAAA3'.

How was the second sequence formed?

a. a tandem repeat mutation formed the second sequence

b. processing removed an exon and added a poly-A tail to form the second sequence

c. processing removed an intron and added a poly-A tail to form the second sequence

d. RNA polymerase matched up several incorrect nucleotide bases to the original DNA as the

second sequence formed

____ 27. A bacterium that was once able survive in a tryptophan-free environment is now unable to synthesize its

own tryptophan. The bacterium is otherwise unaffected. Where is the most likely location for the mutation

causing the change?

a. the trp repressor c. the histone

b. the trp promoter d. the RNA polymerase

____ 28. 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 promoter

b. an intron d. a repressor

Table 12-2

____ 29. 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 2

b. 1 and 3

c. 2 and 3

d. cannot tell without data on guanine and thymine

____ 30. You have a building toy set consisting of parts that can be connected together. You are going to use it to

model a piece of DNA. You have decided that each part of DNA will be represented by a different type of toy

piece. You have chosen the following four pieces so far: adenine = large red cube; guanine = large green

cube, thymine = small orange cube; cytosine = small blue cube. How many other types of pieces do you

need to represent the remaining parts both the 3-prime and the 5-prime strands of a section of DNA?

a. 0 c. 2

b. 1 d. 4

____ 31. 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. G

b. C d. T

____ 32. DNA replication of a leading strand involves which enzymes?

a. primase, polymerase c. helicase, primase, ligase

b. primase, polymerase, ligase d. helicase, primase, polymerase, ligase

____ 33. In 1974, Stanley Cohen and Herbert Boyer inserted a gene from an African clawed frog into a bacterium. The

bacterium produced the protein coded for by the inserted frog gene. The bacterium containing functional frog

DNA would be classified as a ____.

a. clone c. plasmid

b. DNA fingerprint d. transgenic organism

____ 34. A virus isolated from monkeys contains a circular double strand of DNA. The virus, called Simian Virus 40,

interests scientists because it causes cancer in laboratory animals. Using a restriction enzyme, the strand is

separated into six unequal segments, as shown in Figure 13-2. A scientist hypothesizes that the segment of

DNA causing cancer can contain no fewer than 600 base pairs. Using Figure 13-2, decide which segments of

the virus have the highest chance of containing the segment of interest. Identify in DESCENDING order,

from the HIGHEST chance to the LOWEST.

Figure 13-2

a. C, B, A c. A, B, C

b. F, E, D d. D, E, F

____ 35. What must be on either end of any genetic material that is inserted into the cleaved DNA in Figure 13-4?

Figure 13-4

a. 5'AATT3' c. 5'CCGG3'

b. 5'ATAT3' d. 5'CGCG3'

____ 36. Which DNA sequence in Figure 13-5 will be cleaved by EcoRI, which cuts AATT/TTAA?

Figure 13-5

a. A c. C

b. B d. D

Figure 13-6

____ 37. According to Figure 13-6, which are the parents of the child?

a. A c. C

b. B d. D

____ 38. According to Figure 13-6, which parents might give a false positive if only the longer DNA fragments were

analyzed?

a. A c. C

b. B d. D

____ 39. A DNA molecule containing regions from different sources is called

a. DNA ligase. c. restriction DNA.

b. recombinant DNA. d. template DNA.

____ 40. Which outcome is possible using genetic engineering, but not using selective breeding?

a. A sheep with wool longer than wool produced by any other sheep.

b. Corn that produces one large corn cob per plant.

c. A bacterium that produces human insulin.

d. A hairless variety of cow.

____ 41. In which of these processes do scientists use restriction enzymes?

a. genetic engineering c. inbreeding

b. hybridization d. selective breeding

____ 42. What is the term used to describe the complete genetic information of a cell or organism?

a. clone c. haplotype

b. genome d. nucleotide

____ 43. The regions of DNA that are unique to each individual are the

a. nucleotide regions. c. non-coding regions.

b. phosphate regions. d. protein-coding regions.

____ 44. Why have viruses been used in gene therapy in humans?

a. They inject DNA into cells.

b. They are passed from one person to another.

c. They are disease-causing factors.

d. They contain restriction endonucleases.

____ 45. Regions of linked variations in the genome that can be associated with human diseases are known as

a. haplotypes. c. coding regions.

b. plasmids. d. non-coding regions.

____ 46. Researchers can determine the difference between normal and disease states of the human body by analyzing

the structure and function of

a. large numbers of phosphate groups. c. single phosphate groups.

b. large numbers of proteins. d. single proteins.

____ 47. Santa Gertrudis cattle were developed by mating shorthorn beef cattle, who produce high quality beef, with

heat- and insect-resistant Brahman cattle from India. The result of this cross are cattle that are resistant to

heat and insects and also produce high-quality beef. This process is an example of

a. cloning. c. hybridization.

b. genetic engineering. d. inbreeding.

____ 48. A genetically engineered organism that contains a gene from another organism is called a

a. bacterial organism. c. genetic organism.

b. cloned organism. d. transgenic organism.

Matching

Match each item with the correct statement below.

a. restriction enzymes g. haplotypes

b. genetic engineering h. gene therapy

c. transgenic organism i. gel electrophoresis

d. genome j. test cross

e. selective breeding k. clone

f. hybrid

____ 49. A method of determining an exact genotype.

____ 50. Inserting the DNA of one organism into the DNA of another organism.

____ 51. The total DNA present in the nucleus of each cell.

____ 52. Technique that separates DNA fragments by size.

____ 53. A genetically engineered organism that contains a gene from another organism.

____ 54. A technique aimed at correcting mutated genes that cause human diseases.

____ 55. Genetic variations that are closely linked together.

____ 56. An identical copy of an organism.

Problem

57. In Figure 12-5, use the letter P to label all of the phosphate groups. Use an S to label all the sugar molecules.

For labeling the nitrogen bases, use a T for thymine and a C for cytosine. Guanine and adenine have been

filled in for you. Circle and label a codon. Circle and label a nucleotide.

Figure 12-5

58. Review the results of Hershey and Chase’s experiment with radioactive labeling of bacteriophages. Describe

what conclusions the researchers might have drawn if no viral replication occurred in the bacteria infected by

viruses labeled with 32

P. What follow-up experiment could be done to test that conclusion?

59. Review the prokaryotic trp operon. If you were going to create a mutation that would cause the least amount

of damage to the bacterium, what type of mutation would you choose? Would you locate it in the repressor,

the promoter, or somewhere else? Give reasons for your answer.

Agrobacterium tumefaciens is a bacterium that causes crown gall disease, a tumorous growth on the growing

tip of certain plants. The bacterium is able to enter a plant through small cuts in the outer cell layer. When

Agrobacterium enters a plant cell, a DNA sequence from the bacterium integrates into the plant's DNA. This

new section of DNA causes the plant's cell to reproduce quickly to form a tumor and to synthesize a food

molecule needed by the bacterium. A critical bit of information that scientists have learned about the process

is that the tumor-causing information is carried on a large plasmid that is separate from the bacterium's main

chromosome. During the infection process, the DNA on the plasmid that codes for food production and rapid

reproduction leaves the plasmid, moves into the plant cell nucleus, and integrates with one of the plant cell's

chromosomes. Thus, when the plant cell reproduces, it passes along the bacterium's genetic information,

which has been incorporated into the plant genome.

60. Why is the above information about how Agrobacterium causes crown gall disease important to scientists

hoping to produce transgenic plants?

61. What could be used to cut open an Agrobacterium plasmid and insert a gene that would increase the rate of

conversion of atmospheric nitrogen into nitrates?

62. Illustrate and label what the plasmid might look like with the desired gene inserted.

63. What benefits to agriculture could stem from scientists being able to engineer plants genetically?

Essay

64. 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.

65. 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.

66. Imagine a cell that synthesizes both the leading and lagging DNA strands continuously. How would the

replication process have to be different from the process that occurs in prokaryotes and eukaryotes? Draw

and label a picture or describe the process in words.

67. Transgenic bacteria are used not only for research, but also for medical and agricultural purposes.

Transgenic animals are used in laboratories for biological research, and transgenic bacteria produce insulin,

other hormones, and vaccines. Many species of plants have been genetically engineered to be more resistant

to insect or viral pests. Some scientists are concerned about transgenic plants that have built-in pest and

disease resistance from toxin-producing genes engineered directly into the DNA of the crop, as well as the

weed killers that are poisonous to humans and are sprayed over herbicide-resistant crops. They are also

concerned about the antibiotic-resistant genes that are being incorporated into genetically engineered

organisms.

What is your assessment of the effects of genetic engineering on the future of human health?

68. Genetic engineering often results in the production of unusual traits in organisms, such as fluorescent

mosquitoes. Explain the purpose of this trait that is apparently functionless?

69. Polymerase chain reaction was invented by discovering a way to overcome a major technical problem through

the use of the DNA polymerase enzyme from a thermophilic (heat-loving) bacteria. Describe the nature of the

major technical problem that needed to be overcome?

S4O3 Pretest 2015-2016

Answer Section

MULTIPLE CHOICE

1. ANS: B

Nondisjunction during meiosis of egg formation would produce gametes that are XX and O. The male gamete

would give one X, thus forming XXX.

Feedback

A What genotypes of sperm would this produce? B You are correct! C You are on the right track. D Consider the possible source of two of the X chromosomes.

PTS: 1 DIF: Bloom's Level F REF: 312–314

NAT: LS_2c TOP: 11-9

2. ANS: D

Nondisjunction 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.

Feedback

A 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: 312–314


3. ANS: C

Nondisjunction of chromosome 21 occurs during meiosis I, resulting in gametes with two copies of that

chromosome.

Feedback

A You are on the right track. B This would produce a polyploid zygote. C You are correct. D You are on the right track.



4. ANS: A

The genetic cause of this syndrome is paternal non-disjunction during meiosis that results in YY sperm.

Feedback

A You are right! B Work through the possibilities in each parent. C You are on the right track.

D Work through the possibilities in each parent.



5. ANS: A

Two of the X chromosomes are inactivated, forming Barr bodies, to maintain the proper gene dosage.

Feedback

A You are right! B Did you consider all the factors? C Please refer to pages 305-308. D You are on the right track.

PTS: 1 DIF: Bloom's Level E REF: 305–308


6. ANS: B

All plants are genetically equivalent because they were produced through vegetative propagation. Therefore,

the phenotype is affected by the environment.

Feedback

A You are on the right track. B You are correct. C Please refer to page 309. D Did you consider all the factors?

PTS: 1 DIF: Bloom's Level D REF: 309

NAT: IS_1a TOP: 11-6

7. ANS: C

In 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.



8. ANS: C

Telomeres are protective caps found on chromosomes that might be involved with processes such as aging

and cancer.

Feedback

A Did you consider all the factors? B Please refer to page 311. C You are correct! D You are on the right track.

PTS: 1 DIF: Bloom's Level B REF: 311

TOP: 11-8

9. ANS: D

Skin cells are somatic cells, so the mutation is not passed on to the person’s offspring.

Feedback

A Are you picking the least likely event? B Think about the difference between body cells and germ cells. C Did you consider all types of mutations? D That's correct!



10. ANS: B

Nucleotides are the subunits of nucleic acids like DNA. DNA coils around histone proteins to form

chromosomes, which are contained in the nucleus of a cell.

Feedback

A Chromosomes are in the nucleus. B That's correct! C Check page 329. D What is a chromosome made of?

PTS: 1 DIF: Bloom's Level B REF: 329 | 332


11. ANS: C

The process of translation takes place at the ribosomes.

Feedback

A This process happens after messenger RNA leaves the nucleus. B Where does messenger RNA go when it leaves the nucleus? C That's correct! D Chromosomes are made up of DNA and proteins.

PTS: 1 DIF: Bloom's Level C REF: 338


12. ANS: B

A tRNA molecule carries a specific amino acid corresponding to the anticodon of that tRNA molecule.

Feedback

A 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.



13. ANS: A

Translation is the process by which the mRNA “template” is used to form polypeptides.

Feedback

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



14. ANS: A

The mRNA “template” and the tRNA molecule are the only RNA structures. I is an anticodon composed of

nucleotide bases, III is a polypeptide, and V is an amino acid.

Feedback

A That's correct! B Only one of these is made of RNA. C What is an anticodon? D These are related to each other, but they are not made of RNA.

PTS: 1 DIF: Bloom's Level C REF: 336 | 338


15. ANS: B

mRNA is formed from the template strand of DNA, and it carries the “code” from the nucleus to the

ribosomes.

Feedback

A Does DNA leave the nucleus? B That's correct! C tRNA carries amino acids. D Look at page 336.



16. ANS: A

DNA 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: 326–333


17. ANS: B

The number and identities of the bases in the two sequences are the same except for the substitution of an

adenine molecule for a thymine.

Feedback

A A frameshift mutation changes the number of bases in the sequence. B That's correct! C Do the two sequences contain the same number of bases? D Check page 346 for the definition of a deletion.

PTS: 1 DIF: Bloom's Level A REF: 345

NAT: LS_2a | LS_2c TOP: 12-11

18. ANS: B

Deleting a nucleotide causes a frameshift mutation, since the codons following the deletion will code for

different amino acids than the original sequence.

Feedback

A 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.



19. ANS: C

Since the reading “frame” is shifted one place, all amino acids after the deletion will be different from the

original sequence.

Feedback

A 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?



20. ANS: A

The deletion of adenine from the beginning of the sequence causes a frameshift mutation.

Feedback

A 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: 345–346


21. ANS: C

Introns, or intervening sequences, are removed from mRNA before it leaves the nucleus, so they are not used

in enzyme synthesis.

Feedback

A 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: 337


22. ANS: A

According to base-pairing rules, the number of purine bases formed must match the number of pyrimidine

bases in the parent sequence.

Feedback

A 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 | 333

NAT: LS_2a TOP: 12-2

23. ANS: C

The central dogma states that DNA is transcribed to mRNA, which is translated to a sequence of amino acids

that form a polypeptide.

Feedback

A 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: 336–339


24. ANS: B

The only chemical that will prevent bacterial replication without interfering with eukaryotic replication is one

that cannot enter the eukaryotic nucleus.

Feedback

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


NAT: IS_1e TOP: 12-3

25. ANS: D

The mRNA strand is being transcribed from the DNA in a right-to-left order. The corresponding base pair to

adenine is uracil in RNA.

Feedback

A 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!


NAT: LS_1c | LS_2a TOP: 12-7

26. ANS: C

RNA processing removes introns and adds a poly-A tail.

Feedback

A A tandem repeat is a set of copied codons. B Check the definition of exons on page 337. C That's correct! D Are the nucleotides in the second sequence unrelated to those in the first sequence?



27. ANS: B

The promoter is the most likely location for the mutation, because if it cannot operate, transcription of the

tryptophan-synthesizing enzymes does not take place. The trp repressor plays a role only in the presence of

tryptophan.

Feedback

A What would happen if the repressor molecule didn't work? B That's correct! C Histones are part of eukaryotic cells. D If RNA polymerase were affected, the bacterium wouldn't survive at all.


NAT: LS_1d | LS_2c TOP: 12-9

28. ANS: B

Introns, 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.

Feedback

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


NAT: LS_1d | LS_2c TOP: 12-7

29. ANS: B

Though the numbers do not match exactly, 1 and 3 have similar percentages of adenine and cytosine.

Feedback

A Keep trying. B That's correct! C Which bases pair up in DNA?

D Check page 329 for Chargaff's law.



30. ANS: C

The only pieces still needed represent a sugar and a phosphate—the order of those pieces determines the

direction of the strand.

Feedback

A A nucleotide is made of one phosphate, one sugar, and one base. B A sugar and a phosphate group are two different molecules. C That's correct! D What is the difference between the 3-prime strand and the 5-prime strand?



31. ANS: A

The next fragment will begin from the 3' end of the parent strand.

Feedback

A 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.



32. ANS: D

The leading strand is formed continuously, but since replication can originate at multiple points along the

chromosome, all four enzymes are needed.

Feedback

A How does the DNA unwind? B Check page 334 for hints. C You're on the right track. D That's correct!



33. ANS: D

A transgenic organism contains a gene inserted from another organism.

Feedback

A Check the definition of a clone. B Refer to page 370 for more information. C Refer to page 370 for more information. D Correct!


TOP: 13-4

34. ANS: C

Answer C lists the base pairs in order from Highest to Lowest: A = 1768, B = 1169,

C = 1100.

Feedback

A Refer to pages 364–366 for more information. B You're close! C Correct! D Refer to pages 364–366 for more information.

PTS: 1 DIF: Bloom's Level E REF: 364 | 375–377


35. ANS: A

5'AATT3' are the bases that will match up with the bases on the cleaved ends.

Feedback

A Correct! B Refer to page 364 for more information. C Check the sequence of bases. D Refer to page 364 for more information.



36. ANS: C

Sequence C is the only one that has the base pairs AATT/TTAA.

Feedback

A Check the sequences again. B Refer to page 364 for more information. C Correct! D You're close!



37. ANS: C

The bands in columns E and F in answer C most closely match the bands in the

child’s profile.

Feedback

A You're close! B Refer to pages 373–374 for more information. C Correct! D Refer to pages 373–374 for more information.

PTS: 1 DIF: Bloom's Level D REF: 373–374

TOP: 13-8

38. ANS: A

The longer bands, though an incomplete analysis, are similar to those of the child.

Feedback

A Correct! B Check the bands again. C Refer to pages 373–374 for more information. D Refer to pages 373–374 for more information.


TOP: 13-8

39. ANS: B

Recombinant DNA contains DNA from different sources.

Feedback

A Refer to page 366 for more information. B That's correct! C Refer to page 366 for more information. D This is described on page 368.


TOP: 13-4

40. ANS: C

Genetic engineering is able to produce organisms that express genes belonging to another species.

Feedback

A Look back to page 363. B This could occur through selective breeding. C That's correct! D This is possible through selective breeding.


TOP: 13-5

41. ANS: A

Restriction enzymes cut DNA into fragments during genetic engineering.

Feedback

A That's correct! B Refer to page 361 for a definition of this term. C Refer to page 361 for a definition of this term. D Refer to page 360 for a definition of this term.


TOP: 13-5

42. ANS: B

A genome is the complete genetic information in a cell.

Feedback

A Check to see how a clone is defined. B That's correct! C Refer to page 372 for more information.

D Refer to page 372 for more information.



43. ANS: C

Unlike the protein-coding regions of DNA that are almost identical among individuals, the long stretches of

noncoding regions of DNA are unique to each individual.

Feedback

A Refer to page 373 for more information. B Refer to page 373 for more information. C That's correct! D You're close!


TOP: 13-7

44. ANS: A

Viruses are a way of getting a desired piece of DNA into cells.

Feedback

A Excellent! B This trait doesn't help with gene therapy. C Refer to page 378 for more information. D These molecules are made by bacteria.


TOP: 13-6

45. ANS: A

Haplotypes are regions of linked variations in the genome that can be associated with human diseases.

Feedback

A That's correct! B Refer to page 378 for more information. C You're close! D Refer to page 378 for more information.


TOP: 13-9

46. ANS: B

A broad analysis of hundreds or thousands of proteins can define both normal and disease states of the human

body.

Feedback

A Refer to page 379 for more information. B That's correct! C Refer to page 379 for more information. D You're close!


TOP: 13-9

47. ANS: C

Crossing parent organisms with different forms of a trait to produce offspring with specific traits results in

hybrids.

Feedback

A Refer to page 367 for more information. B Refer to page 363 for more information. C That's correct! D Refer to page 361 for more information.

PTS: 1 DIF: Bloom's Level E REF: 361 and 363

TOP: 13-5

48. ANS: D

A genetically engineered organism that contains a gene from another organism is called a transgenic

organism.

Feedback

A Refer to page 370 for more information. B What is the definition of a clone? C You're close! D That's correct!


TOP: 13-4

MATCHING

49. ANS: J PTS: 1 DIF: Bloom's Level B

REF: 362 NAT: LS_2b TOP: 13-3

50. ANS: B PTS: 1 DIF: Bloom's Level B

REF: 363 TOP: 13-4

51. ANS: D PTS: 1 DIF: Bloom's Level A

REF: 364 NAT: LS_1c TOP: 13-4

52. ANS: I PTS: 1 DIF: Bloom's Level A

REF: 365 TOP: 13-4

53. ANS: C PTS: 1 DIF: Bloom's Level B

REF: 370 TOP: 13-5

54. ANS: H PTS: 1 DIF: Bloom's Level C

REF: 378 TOP: 13-6

55. ANS: G PTS: 1 DIF: Bloom's Level B

REF: 378 TOP: 13-6

56. ANS: K PTS: 1 DIF: Bloom's Level A

REF: 367 TOP: 13-4

PROBLEM

57. ANS:

See Solution 12-6.

PTS: 1 DIF: Bloom's Level C REF: 329–330 | 338

NAT: LS_1c | LS_2a TOP: 12-2

58. ANS:

This result would have weakened the experiment. The researchers could have concluded that the labeling

process interfered with viral replication. A follow-up experiment would compare bacteria infected with

unlabeled viruses to bacteria infected with 32

P-labeled viruses.


NAT: IS_1a TOP: 12-1

59. ANS:

A point mutation (substitution) can have deleterious effects, but it has the possibility of being neutral, unlike a

frameshift, duplication, or tandem repeat. Students may choose any location, but must include a reasonable

argument.


NAT: LS_2c | IS_1e TOP: 12-9

60. ANS:

Answers may vary. This knowledge is important because plant cells are surrounded by a thick cell wall that

makes the introduction of foreign DNA difficult. Agrobacterium offers a way of successfully placing foreign

DNA in a plant cell.



61. ANS:

Restriction enzymes could be used to cut open a plasmid so that the desired gene could be inserted.


TOP: 13-4

62. ANS:

Illustrations may vary.

See Solution 13-1.

Solution 13-1


TOP: 13-4

63. ANS:

Answers may vary. They might be able to engineer plants that require less fertilizer, produce more protein, are

resistant to disease, grow in less favorable environments, and are a more nutritious food source.


TOP: 13-6

ESSAY

64. 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: 311

NAT: LS_2b TOP: 11-10

65. 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.



66. ANS:

Answers should include enzymes that add DNA nucleotides and strict base-pairing that allows identical

copies to be made. Answers should also mention that DNA needs to be unwound to be replicated, and that

the addition of nucleotides is directional.



67. ANS:

Answers will vary, but should include the benefits of bacterial production of hormones and vaccines, the

dangers of toxins spreading from crops to human food supplies, and the chance that antibiotic-resistant genes

could lead to the development of new antibiotic-resistant diseases.


TOP: 13-6

68. ANS:

The gene giving rise to fluorescence is used as a marker in a plasmid to show the researcher when another

gene in the plasmid, which does have a specific function, has been successfully integrated into the cells of the

organism being engineered. Another example of a marker gene is the antibiotic resistance genes used in

plasmids to genetically engineer bacteria.


TOP: 13-4

69. ANS:

For DNA to be copied the two strands must be separated so that a complementary strand can be added to the

template. The strands are separated by heating, but the temperature needed to separate the strands would

damage the DNA polymerase needed to synthesize the new DNA. The polymerase from the thermophilic

bacteria overcomes this problem because it is able to stay functional at the higher temperatures.


TOP: 13-4

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