Exam 4: chapter 15-18

Question 1

If you already know that the infectious agent was either bacterial or viral, which
method(s) listed above would allow you to distinguish between these two possibilities?
A) I
B) II
C) II or III
D) IV
E) either II or IV

Answer 1

C) II or III

Question 2

2) In electron micrographs of HSV infection, it can be seen that the intact virus initially reacts with cell surface proteoglycans, then with specific receptors. This is later followed by viral capsids docking with nuclear pores. Afterward, the capsids go from being full to being “empty.” Which of the following best fits these observations?
A) Viral capsids are needed for the cell to become infected; only the capsids enter the nucleus.
B) The viral envelope is not required for infectivity, since the envelope does not enter the nucleus.
C) Only the genetic material of the virus is involved in the cell’s infectivity and is injected like the genome of a phage.
D) The viral envelope mediates entry into the cell, the capsid mediates entry into the nuclear membrane, and the genome is all that enters the nucleus.
E) The viral capsid mediates entry into the cell, and only the genomic DNA enters the nucleus, where it may or may not replicate.

Answer 2

D) The viral envelope mediates entry into the cell, the capsid mediates entry into the nuclear membrane, and the genome is all that enters the nucleus.

Question 3

3) Which of the following characteristics, structures, or processes is common to bacteria
and viruses?
A) metabolism
B) ribosomes
C) genetic material composed of nucleic acid
D) cell division
E) independent existence

Answer 3

C) genetic material composed of nucleic acid

Question 4

4) A bacterium is infected with an experimentally constructed bacteriophage composed
of the T2 phage protein coat and T4 phage DNA. The new phages produced would
have ________.
A) T2 protein and T4 DNA
B) T2 protein and T2 DNA
C) T4 protein and T4 DNA
D) T4 protein and T2 DNA

Answer 4

C) T4 protein and T4 DNA

Question 5

5) Viruses ________.
A) manufacture their own ATP, proteins, and nucleic acids
B) use the host cell to copy themselves and make viral proteins
C) use the host cell to copy themselves and then synthesize their own proteins
D) metabolize food and produce their own ATP

Answer 5

B) use the host cell to copy themselves and make viral proteins

Question 6

6) In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts?
A) DNA contains sulfur, whereas protein does not.
B) DNA contains phosphorus, whereas protein does not.
C) DNA contains nitrogen, whereas protein does not.
D) DNA contains purines, whereas protein includes pyrimidines.
E) RNA includes ribose, whereas DNA includes deoxyribose sugars.

Answer 6

B) DNA contains phosphorus, whereas protein does not.

Question 7

7) In an analysis of the nucleotide composition of DNA, which of the following will be found?
A) A = C
B) A = G and C = T
C) A + C = G + T
D) G + C = T + A

Answer 7

C) A + C = G + T

Question 8

8) For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogen’s. Thus, labeling the nitrogen’s would provide a stronger signal than labeling the phosphates. Why won’t this experiment work?
A) There is no radioactive isotope of nitrogen.
B) Radioactive nitrogen has a half-life of 100,000 years, and the material would be too
dangerous for too long.
C) Although there are more nitrogen’s in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive.
D) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity
would not distinguish between DNA and proteins.

Answer 8

D) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity
would not distinguish between DNA and proteins.

Question 9

9) In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules?
A) A = G
B) A + G = C + T
C) A + T = G + T
D) A = C
E) G = T

Answer 9

B) A + G = C + T

Question 10

10) Of the following, which is the most current description of a gene?
A) a unit of heredity that causes formation of a phenotypic characteristic
B) a DNA subunit that codes for a single complete protein
C) a DNA sequence that is expressed to form a functional product: either RNA or polypeptide
D) a DNA—RNA sequence combination that results in an enzymatic product
E) a discrete unit of hereditary information that consists of a sequence of amino acids

Answer 10

C) a DNA sequence that is expressed to form a functional product: either RNA or polypeptide

Question 11

11) In the polymerization of DNA, a phosphodiester bond is formed between a phosphate group of the nucleotide being added and ________ of the last nucleotide in the polymer.
A) the 5’ phosphate
B) C6
C) the 3’ OH
D) a nitrogen from the nitrogen-containing base

Answer 11

C) the 3’ OH

Question 12

12) Within a double-stranded DNA molecule, adenine forms hydrogen bonds with
thymine and cytosine forms hydrogen bonds with guanine. This arrangement ________.
A) allows variable width of the double helix
B) permits complementary base pairing
C) determines the tertiary structure of a DNA molecule
D) determines the type of protein produced

Answer 12

B) permits complementary base pairing

Question 13

13) Who performed classic experiments that supported the semiconservative model of DNA replication?
A) Watson and Crick
B) Meselson and Stahl
C) Hershey and Chase
D) Franklin and Wilkins

Answer 13

B) Meselson and Stahl

Question 14

14) Semiconservative replication involves a template. What is the template?
A) single-stranded binding proteins
B) DNA polymerase
C) one strand of the DNA molecule
D) an RNA molecule

Answer 14

C) one strand of the DNA molecule

Question 15

15) DNA is synthesized through a process known as ________.
A) semiconservative replication
B) conservative replication
C) translation
D) transcription

Answer 15

A) semiconservative replication

Question 16

16) Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base?
A) One of the daughter cells, but not the other, would have radioactive DNA.
B) Neither of the two daughter cells would be radioactive.
C) All four bases of the DNA would be radioactive.
D) Radioactive thymine would pair with nonradioactive guanine.
E) DNA in both daughter cells would be radioactive.

Answer 16

E) DNA in both daughter cells would be radioactive.

Question 17

17) E. coli cells grown on 15N medium are transferred to 14N medium and allowed to
grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?
A) one high-density and one low-density band
B) one intermediate-density band
C) one high-density and one intermediate-density band
D) one low-density and one intermediate-density band
E) one low-density band

Answer 17

D) one low-density and one intermediate-density band

Question 18

18) In the figure associated with this question, which of the three types of viruses shown
would you expect to include a capsid(s)?
A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III

Answer 18

E) I, II, and III

Question 19

19) DNA contains the template needed to copy itself, but it has no catalytic activity in cells. What catalyzes the formation of phosphodiester bonds between adjacent nucleotides in the DNA polymer being formed?
A) ribozymes
B) DNA polymerase
C) ATP
D) deoxyribonucleotide triphosphates

Answer 19

B) DNA polymerase

Question 20

20) Refer to the figure associated with this question. Which structure is responsible for stabilizing DNA in its single-stranded form?
A) A
B) B
C) C
D) D

Answer 20

C) C

Question 21

21) What provides the energy for the polymerization reactions in DNA synthesis?
A) ATP
B) DNA polymerase
C) breaking the hydrogen bonds between complementary DNA strands
D) the deoxyribonucleotide triphosphate substrates

Answer 21

D) the deoxyribonucleotide triphosphate substrates

Question 22

22) Refer to the figure associated with this question. What bases will be added to the primer as DNA replication proceeds? The bases should appear in the new strand in the order that they will be added starting at the 3’ end of the primer.
A) C, A, G, C, A, G, A
B) T, C, T, G, C, T, G
C) A, G, A, C, G, A, C
D) G, T, C, G, T, C, T

Answer 22

C) A, G, A, C, G, A, C

Question 23

23) Identify the lagging strand during duplication of DNA starting from a double helix in the
accompanying figure.
A) a
B) b
C) c
D) d

Answer 23

C) c

Question 24

24) Put the following steps of DNA replication in chronological order.
1. Single-stranded binding proteins attach to DNA strands.
2. Hydrogen bonds between base pairs of antiparallel strands are broken.
3. Primase binds to the site of origin.
4. DNA polymerase binds to the template strand.
5. An RNA primer is created.
A) 1, 2, 3, 4, 5
B) 2, 1, 3, 5, 4
C) 3, 2, 1, 5, 4
D) 3, 1, 2, 4, 5

Answer 24

B) 2, 1, 3, 5, 4

Question 25

25) In the accompanying figure, which is the template strand?
A) a
B) b
C) c
D) d

Answer 25

A) a

Question 26

26) What is a major difference between eukaryotic DNA replication and prokaryotic DNA
replication?
A) Prokaryotic replication does not require a primer.
B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have multiple origins of replication.
C) DNA replication in prokaryotic cells is conservative. DNA replication in eukaryotic cells is semiconservative.
D) DNA polymerases of prokaryotes can add nucleotides to both 3’ and 5’ ends of DNA strands; those of eukaryotes function only in the 5’ → 3’ direction.

Answer 26

B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have multiple origins of replication.

Question 27

27) In the accompanying figure, which structure causes the two strands of DNA to
separate?
A) A
B) B
C) C
D) D

Answer 27

B) B

Question 28

28) At a specific area of a chromosome, the following sequence of nucleotides is
present where the chain opens to form a replication fork:
3’ C C T A G G C T G C A A T C C 5’
An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence?
A) 5’ G C C T A G G 3’
B) 3’ G C C T A G G 5’
C) 5’ A C G T T A G G 3’
D) 5’ A C G U U A G G 3’
E) 5’ G C C U A G G 3’

Answer 28

D) 5’ A C G U U A G G 3’

Question 29

29) Polytene chromosomes of Drosophila salivary glands each consist of multiple
identical DNA strands that are aligned in parallel arrays. How could these arise?
A) replication followed by mitosis
B) replication without separation
C) meiosis followed by mitosis
D) fertilization by multiple sperm
E) special association with histone proteins

Answer 29

B) replication without separation

Question 30

30) In E. coli, what is the function of DNA polymerase III?
A) to unwind the DNA helix during replication
B) to seal together the broken ends of DNA strands
C) to add nucleotides to the 3’ end of a growing DNA strand
D) to degrade damaged DNA molecules
E) to rejoin the two DNA strands (one new and one old) after replication

Answer 30

C) to add nucleotides to the 3’ end of a growing DNA strand

Question 31

31) The leading and the lagging strands differ in that ________.
A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.
B) the leading strand is synthesized by adding nucleotides to the 3’ end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5’ end.
C) the lagging strand is synthesized continuously, whereas the leading strand is
synthesized in short fragments that are ultimately stitched together.
D) the leading strand is synthesized at twice the rate of the lagging strand.

Answer 31

A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction

Question 32

32) What is the role of DNA ligase in the elongation of the lagging strand during DNA
replication?
A) It synthesizes RNA nucleotides to make a primer.
B) It catalyzes the lengthening of telomeres.
C) It joins Okazaki fragments together.
D) It unwinds the parental double helix.
E) It stabilizes the unwound parental DNA.

Answer 32

C) It joins Okazaki fragments together.

Question 33

33) Which of the following help(s) to hold the DNA strands apart while they are being replicated?
A) primase
B) ligase
C) DNA polymerase
D) single-strand DNA binding proteins
E) nuclease

Answer 33

D) single-strand DNA binding proteins

Question 34

34) Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following?
A) the evolution of telomerase enzyme
B) DNA polymerase that cannot replicate the leading strand template to its 5’ end
C) gaps left at the 5’ end of the lagging strand
D) gaps left at the 3’ end of the lagging strand because of the need for a primer
E) the “no ends” of a circular chromosome

Answer 34

C) gaps left at the 5’ end of the lagging strand

Question 35

35) How does the enzyme telomerase meet the challenge of replicating the ends of linear chromosomes?
A) It adds a single 5’ cap structure that resists degradation by nucleases.
B) It causes specific double-strand DNA breaks that result in blunt ends on both
strands.
C) It causes linear ends of the newly replicated DNA to circularize.
D) It catalyzes the lengthening of telomeres, compensating for the shortening that could
occur during replication without telomerase activity.
E) It adds numerous GC pairs, which resist hydrolysis and maintain chromosome
integrity.

Answer 35

D) It catalyzes the lengthening of telomeres, compensating for the shortening that could
occur during replication without telomerase activity.

Question 36

36) The DNA of telomeres has been highly conserved throughout the evolution of
eukaryotes. This most likely reflects ________.
A) the inactivity of this region of DNA
B) the low frequency of mutations occurring in this DNA
C) continued evolution of telomeres
D) that new mutations in telomeres have been advantageous
E) a critical function of telomeres

Answer 36

E) a critical function of telomeres

Question 37

37) Which of the following would you expect of a eukaryote lacking telomerase?
A) a high probability of somatic cells becoming cancerous
B) an inability to produce Okazaki fragments
C) an inability to repair thymine dimers
D) a reduction in chromosome length in gametes
E) high sensitivity to sunlight

Answer 37

D) a reduction in chromosome length in gametes

Question 38

38) What is a telomere?
A) the mechanism that holds two sister chromatids together
B) DNA replication during telophase
C) the site of origin of DNA replication
D) the ends of linear chromosomes

Answer 38

D) the ends of linear chromosomes

Question 39

39) Telomere shortening puts a limit on the number of times a cell can divide. Research has shown that telomerase can extend the life span of cultured human cells. How might adding telomerase affect cellular aging?
A) Telomerase will speed up the rate of cell proliferation.
B) Telomerase eliminates telomere shortening and retards aging.
C) Telomerase shortens telomeres, which delays cellular aging.
D) Telomerase would have no effect on cellular aging.

Answer 39

B) Telomerase eliminates telomere shortening and retards aging.

Question 40

40) Which of the following cells have reduced or very little active telomerase activity?
A) most normal somatic cells
B) most normal germ cells
C) most cancer cells
D) None of the above choices is correct.

Answer 40

A) most normal somatic cells

Question 41

41) Telomere shortening is a problem in which types of cells?
A) only prokaryotic cells
B) only eukaryotic cells
C) cells in prokaryotes and eukaryotes

Answer 41

B) only eukaryotic cells

Question 42

42) What appears to be a dark side to telomerase activity with regard to human health?
A) Telomerase is active in most cancer cells.
B) Telomerase is inhibited by p53.
C) p53 inhibits telomerase.
D) There are more chromosomal ends than can be repaired by telomerase.
E) Telomerase activity is only seen in somatic cells; therefore, chromosome shortening is likely in gametic chromosomes.

Answer 42

A) Telomerase is active in most cancer cells.

Question 43

43) DNA replication is highly accurate. It results in about one mistake per billion nucleotides. For the human genome, how often would errors occur?
A) on average, once or twice in the lifetime of an individual
B) on average, six times each time the entire genome of a cell is replicated
C) on average, once every six cell divisions
D) on average, once in a lifetime in 10 percent of the population

Answer 43

B) on average, six times each time the entire genome of a cell is replicated

Question 44

44) In a healthy cell, the rate of DNA repair is equal to the rate of DNA mutation. When the rate of repair lags behind the rate of mutation, what is a possible fate of the cell?
A) The cell can be transformed to a cancerous cell.
B) RNA may be used instead of DNA as inheritance material.
C) The cell will become embryonic.
D) DNA synthesis will continue by a new mechanism.

Answer 44

A) The cell can be transformed to a cancerous cell.

Question 45

45) Researchers found E. coli that had mutation rates 100 times higher than normal.
Which of the following is the most likely cause of these results?
A) The single-stranded binding proteins were malfunctioning.
B) There were one or more mismatches in the RNA primer.
C) The proofreading mechanism of DNA polymerase was not working properly.
D) The DNA polymerase was unable to add bases to the 3’ end of the growing nucleic
acid chain.

Answer 45

C) The proofreading mechanism of DNA polymerase was not working properly.

Question 45

46) The epsilon (ε) subunit of DNA polymerase III of E. coli has exonuclease activity.
How does it function in the proofreading process? The epsilon subunit ________.
A) removes a mismatched nucleotide
B) excises a segment of DNA around the mismatched base
C) can recognize which strand is the template or parent strand and which is the new
strand of DNA.
D) adds nucleotide triphosphates to the 3’ end of the growing DNA strand

Answer 45

A) removes a mismatched nucleotide

Question 46

47) Recent studies have shown that xeroderma pigmentosum (an error in the nucleotide excision repair process) can result from mutations in one of seven genes. What can you infer from this finding?
A) There are seven genes that produce the same protein.
B) These seven genes are the most easily damaged by ultraviolet light.
C) There are several proteins involved in the nucleotide excision repair process.
D) These mutations have resulted from translocation of gene segments.

Answer 46

C) There are several proteins involved in the nucleotide excision repair process.

Question 47

48) In humans, xeroderma pigmentosum (XP) is a disorder of the nucleotide excision repair mechanism. These individuals are unable to repair DNA damage caused by ultraviolet light. Which of the following are the most prominent types of DNA lesions in individuals suffering from xeroderma pigmentosum?
A) mismatch errors
B) telomere shortening
C) methylation of purines
D) thymine dimers

Answer 47

D) thymine dimers

Question 48

49) Which one of the following is LEAST likely to cause mutations in DNA?
A) aflatoxins that are found in moldy grains
B) hydroxyl radicals formed as by-products of aerobic respiration
C) ultraviolet radiation from sunlight
D) light from an incandescent bulb

Answer 48

D) light from an incandescent bulb

Question 49

50) Which of the following would be LEAST likely to cause DNA damage to an individual suffering from xeroderma pigmentosum (XP)?
A) direct sunlight
B) tanning beds
C) incandescent lightbulbs
D) reflected sunlight

Answer 49

C) incandescent lightbulbs

Question 50

51) Given the damage caused by UV radiation, the kind of gene affected in those with
XP is one whose product is involved with ________.
A) mending of double-strand breaks in the DNA backbone
B) breakage of cross-strand covalent bonds
C) the ability to excise single-strand damage and replace it
D) the removal of double-strand damaged areas
E) causing affected skin cells to undergo apoptosis

Answer 50

C) the ability to excise single-strand damage and replace it