Dna Replication, Repair And Recombination

Front 1

The process of DNA replication requires that each of the parental DNA strands be used as a ___________________ to produce a duplicate of the opposing strand.
(a) catalyst
(b) competitor
(c) template
(d) copy

Back 1

c

Front 2

DNA replication is considered semiconservative because ____________________________.
(a) after many rounds of DNA replication, the original DNA double helix is still intact
(b) each daughter DNA molecule consists of two new strands copied from the parent DNA molecule
(c) each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand
(d) new DNA strands must be copied from a DNA template

Back 2

C
a and b are false
d is correct but not the reason that DNA replication is called semiconservative.

Front 3

The classic experiments conducted by Meselson and Stahl demonstrated that DNA replication is accomplished by employing a ________________ mechanism.
(a) continuous
(b) semiconservative
(c) dispersive
(d) conservative

Back 3

b

Front 4

If the genome of the bacterium E. coli requires about 20 minutes to replicate itself, how can the genome of the fruit fly Drosophila be replicated in only 3 minutes?
(a) The Drosophila genome is smaller than the E. coli genome.
(b) Eucaryotic DNA polymerase synthesizes DNA at a much faster rate than procaryotic DNA polymerase.
(c) The nuclear membrane keeps the Drosophila DNA concentrated in one place in the cell, which increases the rate of polymerization.
(d) Drosophila DNA contains more origins of replication than E. coli DNA.

Back 4

Choice (d) is the correct answer. Bacteria have one origin of replication, and Drosophila has many. Choice (a) is incorrect because the Drosophila genome is bigger than the E. coli genome. Choice (b) is incorrect, because eucaryotic polymerases are not faster than procaryotic polymerases.

Front 5

Meselson and Stahl grew cells in media that contained different isotopes of nitrogen (15N and 14N) so that the DNA molecules produced from these different isotopes could be distinguished by mass.

Explain how “light” DNA was separated from “heavy” DNA in the Meselson and Stahl experiments.

Back 5

DNA samples were placed into centrifuge tubes containing cesium chloride.

After high-speed centrifugation for 2 days, the heavy and light DNA products were separated by density.

Front 6

Meselson and Stahl grew cells in media that contained different isotopes of nitrogen (15N and 14N) so that the DNA molecules produced from these different isotopes could be distinguished by mass.

Describe the three existing models for DNA replication when the studies were begun and explain how one of them was ruled out definitively.

Back 6

Conservative: original paternal strands stayed together after replication. Daughter duplex made of entirely newly synthesized DNA.

Semiconservative: two DNA duplexes produced during replication each had one of the parentals and one new.

Dispersive: New DNA contained segments of parental and daughter strands.

Front 7

Meselson and Stahl grew cells in media that contained different isotopes of nitrogen (15N and 14N) so that the DNA molecules produced from these different isotopes could be distinguished by mass.

What experimental result eliminated the dispersive model of DNA replication?

Back 7

using heat to denature the DNA duplexes and then comparinging densities of the single stranded DNA.

If dispersive model had been correct, individual strands should have had an intermediate density.

However, only heavy strands and light strands were observed.

Front 8

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

When DNA is being replicated inside a cell, local heating occurs allowing the two strands to separate

Back 8

False. Strands do not need to separate for replication to occur, but this is accomplished by binding of initiator proteins at the origin of replication.

Front 9

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

DNA replication origins are typically rich in G-C base pairs.

Back 9

False - DNA replication origins are typically rich in A-T base pairs, which are held together by only two hydrogen bonds (instead of three for G-C base pairs), making it easier to separate the strands at these sites.

Front 10

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Meselson and Stahl ruled out the dispersive model for DNA replication

Back 10

True

Front 11

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

DNA replication is a bidirectional process that is initiated at multiple locations along chromosomes in eucaryotic cells.

Back 11

True

Front 12

On a DNA strand that is being synthesized, which end is growing—the 3′ end, the 5′ end, or both ends? Explain your answer.

Back 12

The 3' end. DNA polymerase can add nucleotides only to the 3'-OH end of a nucleic acid chain.

Front 13

On a DNA strand that is being used as a template, where is the copying occurring relative to the replication origin—3′ of the origin, 5′, or both?

Back 13

Both, as a result of the bidirectional nature of chromosomal replication.

Front 14

How does the total number of replication origins in bacterial cells compare with the number of origins in human cells?
(a) 1 versus 100
(b) 5 versus 500
(c) 10 versus 1000
(d) 1 versus 10,000

Back 14

D

Front 15

The chromatin structure in eucaryotic cells is much more complicated than that observed in procaryotic cells. This is thought to be the reason that DNA replication occurs much faster in procaryotes. How much faster is it?
(a) 2×
(b) 5×
(c) 10×
(d) 100×

Back 15

C

Front 16

DNA polymerase catalyzes the joining of a nucleotide to a growing DNA strand. What prevents this enzyme from catalyzing the reverse reaction?
(a) hydrolysis of PPi to Pi + Pi
(b) release of PPi from the nucleotide
(c) hybridization of the new strand to the template
(d) loss of ATP as an energy source

Back 16

A

Front 17

Which of the following statements about the newly synthesized strand of a human chromosome is true?
(a) It was synthesized from a single origin solely by continuous DNA synthesis.
(b) It was synthesized from a single origin by a mixture of continuous and discontinuous DNA synthesis.
(c) It was synthesized from multiple origins solely by discontinuous DNA synthesis.
(d) It was synthesized from multiple origins by a mixture of continuous and discontinuous DNA synthesis.

Back 17

(d). Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins. Consider a single replication origin. The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis; the fork moving in the opposite direction synthesizes a portion of the same daughter strand discontinuously as part of lagging-strand synthesis.

Front 18

You have discovered an “Exo–” mutant form of DNA polymerase in which the 3′-to-5′ exonuclease function has been destroyed but the ability to join nucleotides together is unchanged. Which of the following properties do you expect the mutant polymerase to have?
(a) It will polymerize in both the 5′-to-3′ direction and the 3′-to-5′ direction.
(b) It will polymerize more slowly than the normal Exo+ polymerase.
(c) It will fall off the template more frequently than the normal Exo+ polymerase.
(d) It will be more likely to generate mismatched base pairs.

Back 18

D

Front 19

A molecule of bacterial DNA introduced into a yeast cell is imported into the nucleus but fails to replicate the yeast DNA. Where do you think the block to replication arises? Choose the protein or protein complex below that is most probably responsible for the failure to replicate bacterial DNA. Give an explanation for your answer.
(a) primase
(b) helicase
(c) DNA polymerase
(d) initiator proteins

Back 19

Choice (d) is the correct answer. DNA from all organisms is chemically identical except for the sequence of nucleotides. The proteins listed in choices (a) to (c) can act on any DNA regardless of its sequence. In contrast, the initiator proteins recognize specific DNA sequences at the origins of replication. These sequences differ between bacteria and yeast.

Front 20

Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and is therefore not expressed. An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression. Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.

Back 20

In the absence of telomerase, the life-span of a cell and its progeny is limited.

With each round of DNA replication, the length of telomeric DNA will shrink until finally all telomeric DNA has disappeared.

Without telomeres capping the chromosome ends, the ends might be treated like breaks arising from DNA damage or crucial genetic information might be lost.

Cells whose DNA lacks telomeres will stop dividing or die.

However, if telomerase is provided to cells, they may be able to divide indefinitely because their telomeres will remain a constant length despite repeated rounds of DNA replication.

Front 21

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Primase is needed to initiate DNA replication on both the leading strand and the lagging strand.

Back 21

True

Front 22

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

The sliding camp is loaded once on each DNA strand where it remains associated until replication is complete.

Back 22

False. Although the sliding camp is only loaded once on the leading strand the lagging strand needs to unload the clamp once the polymerase reaches the RNA primer from the previous segment and then reload it where a new primer has been synthesized.

Front 23

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Telomerase is a DNA polymerase that carries its own RNA molecule to use as a primer at the end of the lagging end.

Back 23

True

Front 24

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication.

Back 24

False. Primase does not have a proofreading function, nor does it need one because the RNA primers are not a permanent part of the DNA. THe primers are removed and a DNA polymerase that does have a proofreading function fills in the remaining gaps.

Front 25

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

Primase

Back 25

3

Front 26

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

Single strand binding protein

Back 26

2

Front 27

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

sliding clamp

Back 27

3

Front 28

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

RNA primers

Back 28

3

Front 29

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

leading strand

Back 29

1

Front 30

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

lagging strand

Back 30

2

Front 31

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

Okazaki fragments

Back 31

2

Front 32

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

DNA helicase

Back 32

3

Front 33

Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (1) continuous replication, (2) discontinuous replication, or (3) both modes of replication.

DNA ligase

Back 33

2

Front 34

Although 3′-to-5′ synthesis of DNA is chemically possible, it does not occur in living systems. Why not?

Back 34

DNA synthesis from 3′ to 5′ does not allow proofreading. If the last nucleotide added was mispaired and is removed, the last nucleotide on the growing strand is a nucleotide monophosphate and the nucleotide coming in only has a hydroxyl group on the 3′ end. Thus, there is no favorable hydrolysis reaction to drive the addition of new nucleotides.

Front 35

DNA polymerases are processive, which means that they remain tightly associated with the template strand while moving rapidly and adding nucleotides to the growing daughter strand. Which piece of the replication machinery accounts for this characteristic?
(a) helicase
(b) sliding clamp
(c) single-strand binding protein
(d) primase

Back 35

B

Front 36

Researchers have isolated a mutant strain of E. coli that carries a temperature-sensitive variant of the enzyme DNA ligase. At the permissive temperature, the mutant cells grow just as well as the wild-type cells. At the nonpermissive temperature, all of the cells in the culture tube die within 2 hours. DNA from mutant cells grown at the nonpermissive temperature for 30 minutes is compared with the DNA isolated from cells grown at the permissive temperature. The results are shown in Figure Q6-22, where DNA molecules have been separated by size by means of electrophoresis (P, permissive; NP, nonpermissive). Explain the appearance of a distinct band with a size of 200 base pairs (bp) in the sample collected at the nonpermissive temperature.

Back 36

After Okazaki fragments are synthesized they must be ligated to each other so that they form one continuous strand.

At the nonpermissive temperature this doesn't happen.

The notable band at 200 base pairs is the typical size of an individual Okasaki fragment.

Front 37

Telomeres serve as caps at the ends of linear chromosomes. Which of the following is not true regarding the replication of telomeric sequences?
(a) The lagging strand telomeres are not completely replicated by DNA polymerase.
(b) Telomeres are made of repeating sequences.
(c) Additional repeated sequences are added to the template strand.
(d) The leading strand doubles back on itself to form a primer for the lagging strand.

Back 37

D

Front 38

Sickle-cell anemia is an example of an inherited disease. Individuals with this disorder have misshapen (sickle-shaped) red blood cells caused by a change in the sequence of the β-globin gene. What is the nature of the change?
(a) chromosome loss
(b) base-pair change
(c) gene duplication
(d) base-pair insertion

Back 38

B

Front 39

Even though DNA polymerase has a proofreading function, it still introduces errors in the newly synthesized strand at a rate of 1 per 107 nucleotides. To what degree does the mismatch repair system decrease the error rate arising from DNA replication?
(a) 2-fold
(b) 5-fold
(c) 10-fold
(d) 100-fold

Back 39

D

Front 40

A mismatched base pair causes a distortion in the DNA backbone. If this were the only indication of an error in replication, the overall rate of mutation would be much higher. Explain why.

Back 40

The distortion in the backbone is insufficient information for the mismatch repair system to identify which base is incorrect and which was originally part of the chromosome when replication began.

Without additional marks that identify the difference between the newly synthesized strand and the template strand the repair would be corrected only 50% of the time by random chance.

The error rate would still be less than in a system that lacked the mismatch repair enzymes but greater than the error rate in a system that accurately identifies the newly synthesized strand.

Front 41

Beside the distortion in the DNA backbone caused by a mismatched base pair, what additional mark is there on eucaryotic DNA to indicate which strand needs to be repaired?
(a) a nick in the template strand
(b) a chemical modification of the new strand
(c) a nick in the new strand
(d) a sequence gap in the new strand

Back 41

C

Front 42

A pregnant mouse is exposed to high levels of a chemical. Many of the mice in her litter are deformed, but when they are interbred with each other, all their offspring are normal. Which two of the following statements could explain these results?
(a) In the deformed mice, somatic cells but not germ cells were mutated.
(b) The original mouse’s germ cells were mutated.
(c) In the deformed mice, germ cells but not somatic cells were mutated.
(d) The toxic chemical affects development but is not mutagenic.

Back 42

Choice (a) or (d) is correct. Choice (b) cannot account for these results because a mutation in the original mouse’s germ cells would have no effect on the fetuses she was already carrying. Neither can choice (c), because mutations in the germ cells of the fetuses while in utero would have had no effect on their development, but they might have led to mutant mice among their offspring.

Front 43

The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process. Which choice below describes the known sequence of events in this process?
(a) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by repair proteins, the gap is filled by DNA polymerase, and the strand is sealed by DNA ligase.
(b) DNA repair polymerase simultaneously removes bases ahead of it and polymerizes the correct sequence behind it as it moves along the template. DNA ligase seals the nicks in the repaired strand.
(c) DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by an exonuclease, and the gap is repaired by DNA ligase.
(d) A nick in the DNA is recognized, DNA repair proteins switch out the wrong base and insert the correct base, and DNA ligase seals the nick.

Back 43

A

Front 44

You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit. What is the most likely explanation for this?
(a) Coding sequences are repaired more efficiently.
(b) Coding sequences are replicated more accurately.
(c) Coding sequences are packaged more tightly in the chromosomes to protect them from DNA damage.
(d) Mutations in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

Back 44

D

Front 45

Sometimes chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the cytosine in the sequence TCAT is deaminated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication?
(a) TTAT
(b) TUAT
(c) TGAT
(d) TAAT

Back 45

A

Front 46

During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine that this error was not corrected and that it has no effect on the ability of the progeny to grow and reproduce.

After this original bacterium has divided once, what proportion of its progeny would you expect to contain the mutation?

Back 46

50%

DNA replication in the original bacterium will create two new DNA molecules, one of which will now carry a mismatched C-T base pair.

So, one daughter cell of that cell division will carry a completely normal DNA molecule; the other cell will have the molecule with the mutation mispaired to a correct nucleotide.

Front 47

During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine that this error was not corrected and that it has no effect on the ability of the progeny to grow and reproduce.

What proportion of its progeny would you expect to contain the mutation after three more rounds of DNA replication and cell division?

Back 47

25%

At the next round of DNA replication and cell division the bacterium carrying the mismatched C-T will produce and pass on one normal DNA molecule from the undamaged strand containing the T and one mutant DNA molecule with a fully mutant C-G base pair.

One of the four progeny of the original bacterium is mutant.

Subsequent cell division of these mutant bacteria will give rise only to mutant bacteria, whereas the other bacteria will give rise to normal bacteria.

The proportion of progeny containing the mutation will remain at 25%

Front 48

Sometimes chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the adenosine in the sequence TCAT is depurinated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication?
(a) TCGT
(b) TAT
(c) TCT
(d) TGTT

Back 48

C

Front 49

Which of the following statements is not an accurate statement about thymidine dimers?
(a) Thymidine dimers can cause the DNA replication machinery to stall.
(b) Thymidine dimers are covalent links between thymidines on opposite DNA strands.
(c) Prolonged exposure to sunlight causes thymidine dimers to form.
(d) Repair proteins recognize thymidine dimers as a distortion in the DNA backbone.

Back 49

B

Front 50

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Ionizing radiation and oxidative damage can cause DNA double strand breaks

Back 50

True

Front 51

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

After damaged DNA has been repaired, nicks in the phosphate backbone are maintained as a way to identify the strand that was repaired.

Back 51

False. The nicks are generated during DNA replication as a means of easy identification of the new strand but are sealed by DNA ligase shortly after completion of replication.

Front 52

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Depurination of DNA is a rare event that is caused by ultraviolet irradiation.

Back 52

False. Depurination occurs constantly in our cells through spontaneous hyrolysis of the bond linking the DNA base to the deoxyribose sugar.

Front 53

Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

Nonhomologous end joining is a mechanism that ensures that DNA double-strand breaks are repaired with a high degree of fidelity to the original DNA sequence.

Back 53

False. Homologous recombination can repair double strand breaks without any change in DNA sequence.

Nonhomologous end joining always involves loss of genetic information because the ends are degraded by nucleases before they can be ligated back together.

Front 54

Several members of the same family were diagnosed with the same kind of cancer when they were unusually young. Which one of the following is the most likely explanation for this phenomenon? It is possible that the individuals with the cancer have _______________________.
(a) inherited a cancer-causing gene that suffered a mutstion in an ancestor’s somatic cells
(b) inherited a mutation in a gene required for DNA synthesis
(c) inherited a mutation in a gene required for mismatch repair
(d) inherited a mutation in a gene required for the synthesis of purine nucleotides

Back 54

Choice (c) is the correct answer. In fact, affected individuals in some families with a history of early-onset colon cancer have been found to carry mutations in mismatch repair genes. Mutations arising in somatic cells are not inherited, so choice (a) is incorrect. A defect in DNA synthesis or nucleotide biosynthesis would probably be lethal, so choices (b) and (d) are incorrect.

Front 55

Homologous recombination is an important mechanism in which organisms use a “back-up” copy of the DNA as a template to fix double-strand breaks without loss of genetic information. Which of the following is not necessary for homologous recombination to occur?
(a) 3′ DNA strand overhangs
(b) 5′ DNA strand overhangs
(c) a long stretch of sequence similarity
(d) nucleases

Back 55

B

Front 56

In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes. During which process does swapping occur?
(a) DNA replication
(b) DNA repair
(c) meiosis
(d) transposition

Back 56

C

Front 57

The events listed below are all necessary for homologous recombination to occur properly:
A. Holliday junction cut and ligated
B. strand invasion
C. DNA synthesis
D. DNA ligation
E. double-strand break
F. nucleases create uneven strands

Which of the following is the correct order of events during homologous recombination?
(a) E, B, F, D, C, A
(b) B, E, F, D, C, A
(c) C, E, F, B, D, A
(d) E, F, B, C, D, A

Back 57

D

Front 58

Consider the copy of chromosome 3 that you received from your mother. Is it identical to the chromosome 3 that she received from her mother (her maternal chromosome) or identical to the chromosome 3 she received from her father (her paternal chromosome), or neither? Explain.

Back 58

Neither. The copy of chromosome 3 you received from your mother is a hybrid of the ones she received from her mother and her father.

Front 59

Mobile genetic elements are sometimes called “jumping genes,” because they move from place to place throughout the genome. The exact mechanism by which they achieve this mobility depends on the genes contained within the mobile element. Which of the following mobile genetic elements carry both a transposase gene and a reverse transcriptase gene?
(a) L1
(b) B1
(c) Alu
(d) Tn3

Back 59

A

Front 60

Which of the following is true of a retrovirus but not of the Alu retrotransposon?
(a) It requires cellular enzymes to make copies.
(b) It can be inserted into the genome.
(c) It can be excised and moved to a new location in the genome.
(d) It encodes its own reverse transcriptase.

Back 60

D

Front 61

Which of the following DNA sequences is not commonly carried on mobile genetic elements? You may choose more than one option.
(a) transposase gene
(b) Holliday junction
(c) recognition site for transposase
(d) antibiotic resistance gene

Back 61

(b). A Holliday junction is not a sequence but a structural intermediate in homologous recombination. Choices (a), (c), and (d) are all sequences that can be found in mobile genetic elements.

Front 62

HIV is a human retrovirus that integrates into the host cell’s genome and will eventually replicate, produce viral proteins, and ultimately escape the host cell. Which of the following proteins is not encoded in the HIV genome?
(a) reverse transcriptase
(b) envelope protein
(c) RNA polymerase
(d) capsid protein

Back 62

C

Front 63

Some retrotransposons and retroviruses integrate preferentially into regions of the chromosome that are packaged in euchromatin and are also located outside the coding regions of genes that contain information for making a protein. Why might these mobile genetic elements have evolved this strategy?

Back 63

The most evolutionarily successful mobile genetic elements are those that are best at reproducing themselves.

To increase the number of copies of a particular element, the element must:
1. not kill it's host
2. maximize its ability to continue reproducing.

If an element inserts into the coding region of a gene, it might disable the gene and thereby confer a selective disadvanatage in the reproductive or survival of its host.

Thus elements that devised a way to avoid insertion into coding regions were probably better able to increase their copy number throughout the human population.

If an element inserts into a heterchromatic region of a chromosome, its genes may not be expressed and therefore it may become immobile.

Elements that devised a way to direct insertion into euchromatin would be more likely to maintain mobility and thereby increase their copy number over time.

Front 64

Restriction enzymes are frequently a _____ compelx. Each subunit is responsible for the cleavage of the ____ bond on one face of the helix.

Back 64

multisubunit; phosphodiester

Front 65

Bases are ___ attached to ___ of sugar backbone

Back 65

covalently; C1

Front 66

A single nucleic acid strand is a polymer of ____ linked by ____ bonds

Back 66

nucleotides; phosphodiester

Front 67

The 5' carbon of incoming nucleotide is linked to the

Back 67

3' carbon of existing polymer

Front 68

Chagaff's Rule

Back 68

molar ratios for bases
adenine = thymine
guanine = cytosine

Front 69

Two stand of DNA would in a helix with bases on the ____ and sugar phosphate backbone on the ___. Strands are antiparallel - __ to __

Back 69

inside; outside; 5' to 3'

Front 70

Life is not created de novo: a cell must come from a

Back 70

cell

Front 71

Part of the process of creating new cell is

Back 71

duplication of genetic information

Front 72

Semiconservative replication

Back 72

each daughter duplex contains one parental nucleic acid strand and one newly synthesized strand.

Front 73

Conservative replication

Back 73

daughter duplex formed from two newly synthesized strands and parent duplex is conserved

Front 74

Delbruck dispersive replication

Back 74

backbone breaks and rejoins at multiple positions resulting in each strand containing a mixture of parental and newly synthesized fragments.

Front 75

Meselson Stahl Experiment

Back 75

Grew cells in heavy medium and then transfer to light medium for 20' to allow replication. Result was only intermediate weight band.

This result eliminates the conservative model which would have had both a heavy and a light band.

Front 76

20'

Back 76

one round of replication in bacteria

Front 77

How did scientists distinguish between semiconservative vs. dispersive models?

Back 77

Repeated the experiment but heated the DNA to break the H bonds and separate strands before centrifugation.

Result one heavy; one light = semiconservative.

Front 78

Replication fork

Back 78

the region on the parental DNA double helix where replication is occurring

Front 79

DNA acts as a ____ for its own ____

Back 79

template; duplication

Front 80

In E. coli, one round of replication takes about 20-30 minutes. In humans, takes about 8 hours. What is the rate difference from?

Back 80

due to higher order chromatin organization in eukaryotes.

Front 81

Replication begins at specific chromosomal sites:

Back 81

replication origin

Front 82

Human genome must have about ____ origins of replication

Back 82

10,000

Front 83

Replication is ____ with two replication forks per replication bubble. The forks move in ___ directions

Back 83

bidirectional; opposite

Front 84

Restriction enzymes can ___ the phosphodiester bonds between adjacent nucleotides.

Back 84

cleave

Front 85

Ligase can create a phosphodiester bond to

Back 85

repair the linkage between nucleotides.

Front 86

DNA ligase ___ ___ restriction fragments

Back 86

covalently links

Front 87

Catalyzing the formation of 3' - 5' phosphodiester bonds requires

Back 87

ATP

Front 88

Any cloned DNA segment can be reinsterted into cells and

Back 88

tested for biological activity

Front 89

A DNA fragment is linked to a vector through

Back 89

phosphodiester bonds

Front 90

Vector with ORI can

Back 90

replicate in the host cell

Front 91

Plasmid

Back 91

Circular double stranded DNA molecule separate from the cell's chromosome

Front 92

To be maintained in a cell, plasmids require

Back 92

three DNA regions

Front 93

Three DNA regions required by plasmids

Back 93

1. Replication origin
2. Selectable marker drug
3. polylinker

Front 94

polylinker

Back 94

multiple cloning site - a region in which exogenous DNA gragments can be inserted

Front 95

Plasmid DNA is duplicated before

Back 95

every cell division

Front 96

Restriction fragments are readily inserted into plasmid vectors using a

Back 96

polylinker - mutiple cloning site

Front 97

In the presence of DNA ligase, DNA fragments with compatible ends (produced by same RE) will be

Back 97

inserted into the plasmid.

Front 98

Plasmid cloning permits

Back 98

isolation of DNA fragments from complex mixtures

Front 99

Each colony arises from a ___ ___ that has been transformed with a ___ ___

Back 99

single cell; single plasmid

Front 100

The initial fragment of DNA is replicated in a ___ of ___ into a larger number of ___ ___ (___)

Back 100

colony; cells; identical copies; clone

Front 101

A colony contains about

Back 101

a million cells

Front 102

Transformation

Back 102

uptake and expression of foreign DNA

Front 103

Make cells competent for taking up DNA by using ___ ____.

Back 103

divalent cations

Front 104

Eukaryotes do not contain an easily identifiable

Back 104

orgin (DNA sequence)

Front 105

When screening for origins in yeast, need:

Back 105

plasmid to survive on selective media but will also need ORI to replicate

Front 106

Using microarrays to find replication origins:

Back 106

1. Allow replication to occur in different tubes for different time intervals.

2. For each tube, create DNA fragments and fluorescently label all fragments. Hybridize with plates that have DNA spots of known sequences.

3. Sequences at origins will replicate first and have twice as much DNA as other regions faster.

Front 107

DNA polymerase synthesizes DNA ___ to ___ and requires an ___ ____

Back 107

5' to 3'; RNA primer

Front 108

Multienzyme complexes that contain DNA polymerases perform

Back 108

asymmetric replication

Front 109

The leading strand is synthesized:

Back 109

continuously and requires only one RNA primer.

Front 110

The lagging strand is synthesized

Back 110

discontinuously

Front 111

Both strands are synthesized in the

Back 111

5' to 3' direction

Front 112

DNA polymerase requires a ____ to start synthesis. The leading strand requires only a ___ ___ at the start of replications

Back 112

primer; single primer

Front 113

On the lagging strand, DNA primase synthesizes a short __ ___ of about __ ___. These primers are elongated by the polymerase to begin each ___ fragment.

Back 113

short RNA primer; 10 nucleotides; Okazaki

Front 114

Primase makes a short RNA primer from

Back 114

DNA template

Front 115

On the lagging strand, after synthesizing a DNA fragment, a ____ removes the ___ ___, a ___ ___ fills in the missing DNA and a ___ joins the ____ fragments.

Back 115

nuclease; RNA primer; repair polymerase; ligase; Okazaki

Front 116

DNA polymerase has an important proofreading activity:

Back 116

3' to 5' exonuclease

Front 117

THe polymerase

Back 117

checks each nucleotide for correct base pairing to the template strand before the next is added

Front 118

Separate sites within the polymerase are responsible for the ___ ___ activities.

Back 118

two enzymatic

Front 119

The polymerase undergoes a ___ ___ when an incorrect match is detected. This allows the mismatched nucleotide to be at the ___ for the ___ of the ____ nucleotide

Back 119

conformational change; E site; removal; incorrect

Front 120

In addition to the polymerases, complex contains a helicase and single-stranded DNA _______

Back 120

binding proteins (SSBs)

Front 121

The DNA helix must be opened up ahead of the replication fork so the

Back 121

deoxynucleotide triphosphates can form base pairs with the template strand.

Front 122

DNA helicase separates the strands exposing the

Back 122

template

Front 123

Single-stranded DNA binding proteins bind to the

Back 123

exposed DNA strands and keep them destabilized (unwound)

Front 124

The leading and lagging strands are synthesized

Back 124

concurrently

Front 125

Telomerase elongates the

Back 125

lagging strand

Front 126

Problems arise when the DNA replication reaches the ends of the chromosome because the lagging strand needs to have a primer that would be beyond the end of the chromosome. Otherwise a cell would loose a little bit of the end DNA every time it replicated the DNA. The cell's solution

Back 126

involves having DNA sequence repeats at the end of the chromosomes. These specific sequences attract the enzyme telomerase which contains a RNA template as part of the enzyme complex.

Front 127

Telomerase elongates the

Back 127

template strand

Front 128

Long term survival of a species is ___ by genetic changes

Back 128

enhanced

Front 129

The survival of an individual depends on

Back 129

genetic stability

Front 130

Mutation

Back 130

A permanent change in DNA

Front 131

Sickle cell anemia is an example of a disease caused by a

Back 131

single nucleotide change

Front 132

DNA probes can be used to detect

Back 132

single nucleotide changes

Front 133

How can the cell differentiate the new strand from the template strand?

Back 133

Cells mark the newly synthesize strand to ensure that they repair the correct strand

Front 134

How do procaryotes differentiate the new strand from the template?

Back 134

mthylation

Front 135

How do eukaryotes differentiate the new strand from teh template?

Back 135

nicks in the sugar phosphate backbone.

Front 136

Two basic types of cells in mammals

Back 136

germ and somatic

Front 137

Germ cells

Back 137

precurosr cells that give rise to gametes

Front 138

Somatic cells

Back 138

all other cells

Front 139

Mutations in germ cells

Back 139

will be passed on to the offspring

Front 140

somatic cell mutations

Back 140

will only effect the individual and not their offspring

Front 141

Nucleotide changes in somatic cells

Back 141

give rise to variant cells that grow unchecked. Uncontrolled cell proliferation is cancer

Front 142

Cancer usually requires an accumulation of

Back 142

multiple mutations

Front 143

Types of DNA damage

Back 143

DNA replication errors
Spontaneous DNA damage
Induced DNA damage

Front 144

Spontaneous DNA damage

Back 144

depurination and deamination

Front 145

Induced DNA damage

Back 145

UV radiation; chemicals

Front 146

Consequences of unrepaired deamination

Back 146

can change the codon and lead to amino acid substitution or a nonsense mutation - creates a truncated protein

Front 147

Consequences of unrepaired depurination will lead to a

Back 147

nucleotide deletion and shift how the remaining infomration will be read during translation

Front 148

Consequence of thymine dimers

Back 148

they stall the replication machinery

Front 149

DNA mutations can cause different possible protein outcomes:

Back 149

nonsense mutations, missense mutations, silent mutations and neutral mutations

Front 150

nonsense mutations

Back 150

convert a codon to a stop codon - truncated protein - no activity

Front 151

missense mutations

Back 151

result in an amino acid substitution that alters protein function

Front 152

silent mutations

Back 152

codes for the same amino acid

Front 153

Neutral mutations

Back 153

results in a different amino acid - does not affect protein function

Front 154

Excision repair is used when only

Back 154

one strand of the helix has mutations

Front 155

Types of damage corrected by excision repair

Back 155

mismatch repair pathways
Spontaneous DNA damage
Induced DNA damage

Front 156

Recognition and excision step:

Back 156

multiple enzymes are needed to recognize and remove each type of damage

Front 157

Excision repair: After specific enzyme recognizes damage, three steps:

Back 157

1. nuclease breaks covalent bond
(the nuclease is specific for type of damage)
2. Repair polymerase binds 3' OH and adds new nucleotides
3. Ligase seals the nick

Front 158

Excision repair: The repair polymerase binds 3' OH end of the DNA strand and adds new nucleotides in the ____ direction and has ___ abilities. This enzyme is different than the DNA replication polymerase but the same repair polymerase can be used for

Back 158

5'-3'; proofreading; all types of excision repairs

Front 159

The ligase used in excision repair is the same type used to _____

Back 159

seal the nicks between the Okazaki fragments during DNA replication

Front 160

Two types of DNA repair for when double strand damage occurs

Back 160

nonhomologous end-joining
homologous end-joining

Front 161

Repair of double stranded DNA breaks by

Back 161

nonhomologous end-joining

Front 162

Net result of nonhomologous end-joining

Back 162

double strand break repaired with deletion of nucleotides at repair site

Front 163

Homologous recombination: Mechanism to repair double strand breaks cause by

Back 163

ionizing radiation, oxidation

Front 164

Mobile genetic elements are another source of

Back 164

DNA variation

Front 165

Mobile genetic elements

Back 165

transposons and transposase

Front 166

Transposons contain a

Back 166

transposase gene which encodes for an enzyme that catalyzes the movement of the element

Front 167

Transposons are classified according to the

Back 167

mechanism by which they move.
1. DNA only transposons
(both cut and paste mechanism and replicative transposition)
2. retrotransposons
(move through an RNA intermediate)

Front 168

Below is the sequence from the 3′ end of an mRNA.

5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′

If you were told that this sequence contains the stop codon for the protein encoded by this mRNA, what is the anticodon on the tRNA in the P-site of the ribosome when release factor binds to the A-site?
(a) 5′-CCA-3′
(b) 5′-CCG-3′
(c) 5′-UGG-3′
(d) 5′-UUA-3′

Back 168

(a). The stop codon (UAA) is underlined in the mRNA sequence below; this is the only stop codon on this piece of mRNA. The codon (UGG) preceding the stop codon will be binding to a tRNA in the P-site of the ribosome when release factor binds to the A-site. The anticodon of the tRNA will bind to the codon UGG and will be CCA.

5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′

Front 169

One strand of a section of DNA isolated from the bacterium E. coli reads:

5′-GTAGCCTACCCATAGG-3′

A. Suppose that an mRNA is transcribed from this DNA using the complementary strand as a template. What will be the sequence of the mRNA in this region (make sure you label the 5′ and 3′ ends of the mRNA)?

Back 169

5′-GUAGCCUACCCAUAGG-3′

Front 170

One strand of a section of DNA isolated from the bacterium E. coli reads:

5′-GTAGCCTACCCATAGG-3′

How many different peptides could potentially be made from this sequence of RNA, assuming that translation initiates upstream of this sequence?

Back 170

Two. (There are three potential reading frames for each RNA. In this case, they are
GUA GCC UAC CCA UAG …
UAG CCU ACC CAU AGG …
AGC CUA CCC AUA GG? … .
The center one cannot be used in this case, because UAG is a stop codon.)

Front 171

One strand of a section of DNA isolated from the bacterium E. coli reads:

5′-GTAGCCTACCCATAGG-3′

What are these peptides? (Give your answer using the one-letter amino acid code.)

Back 171

VAYP
SLPIG
Note: PTHR will not be a peptide because it is preceded by a stop codon.

Front 172

A strain of yeast translates mRNA into protein inaccurately. Individual molecules of a particular protein isolated from this yeast have variations in the first 11 amino acids compared with the sequence of the same protein isolated from normal yeast cells, as listed in Figure Q7-36. What is the most likely cause of this variation in protein sequence?

(a) a mutation in the DNA coding for the protein
(b) a mutation in the anticodon of the isoleucine tRNA (tRNAIle)
(c) a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different amino acids
(d) a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules

Back 172

(c). A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between amino acids would allow an assortment of amino acids to be attached to the tRNAIle. These assorted aminoacyl-tRNAs would then base-pair with the isoleucine codon and cause a variety of substitutions at positions normally occupied by isoleucine. A mutation in the gene encoding the protein would cause only a single variant protein to be made (choice (a)). A mutation in the anticodon loop of tRNAIle (choice (b)) or a mutation in the isoleucine-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules (choice (d)) would cause the substitution of isoleucine for some other amino acid (which is the opposite of what is observed).

Front 173

Which of the following statements is true?
(a) Ribosomes are large RNA structures composed solely of rRNA.
(b) Ribosomes are synthesized entirely in the cytoplasm.
(c) rRNA contains the catalytic activity that joins amino acids together.
(d) A ribosome binds one tRNA at a time.

Back 173

(c). Ribosomes contain proteins as well as rRNA (choice (a)). rRNA is synthesized in the nucleus, and ribosomes are partly assembled in the nucleus (choice (b)). A ribosome must be able to bind two tRNAs at any one time (choice (d)).

Front 174

A poison added to an in vitro translation mixture containing mRNA molecules with the sequence 5′-AUGAAAAAAAAAAAAUAA-3′ has the following effect: the only product made is a Met-Lys dipeptide that remains attached to the ribosome. What is the most likely way in which the poison acts to inhibit protein synthesis?
(a) It inhibits peptidyl transferase activity.
(b) It inhibits movement of the small subunit relative to the large subunit.
(c) It inhibits release factor.
(d) It mimics release factor.

Back 174

(b). Choice (a) would prevent all peptide bond formation. Choice (c) would have no affect on translation until the stop codon was reached. Choice (d) would be likely to result in a mixture of polypeptides of various lengths; a poison mimicking a release factor could conceivably cause only Met-Lys to be made, but this dipeptide would not remain bound to the ribosome.

Front 175

In eucaryotes, but not in procaryotes, ribosomes find the start site of translation by ____________________________.
(a) binding directly to a ribosome-binding site preceding the initiation codon
(b) scanning along the mRNA from the 5′ end
(c) recognizing an AUG codon as the start of translation
(d) binding an initiator tRNA

Back 175

(b). Choice (a) is true only for procaryotes. Choices (c) and (d) are true for both procaryotes and eucaryotes.

Front 176

Which of the following statements about procaryotic mRNA molecules is false?
(a) A single procaryotic mRNA molecule can be translated into several proteins.
(b) Ribosomes must bind to the 5′ cap before initiating translation.
(c) mRNAs are not polyadenylated.
(d) Ribosomes can start translating an mRNA molecule before transcription is complete.

Back 176

(b). Bacterial mRNAs do not have 5′ caps. Instead, ribosome-binding sites upstream of the start codon tell the ribosome where to begin searching for the start of translation.

Front 177

A mutation in the tRNA for the amino acid lysine results in the anticodon sequence 5′-UAU-3′ (instead of 5′-UUU-3′). Which of the following aberrations in protein synthesis might this tRNA cause? (Refer to the codon table provided above Q7-29.)
(a) read-through of stop codons
(b) substitution of lysine for isoleucine
(c) substitution of lysine for tyrosine
(d) substitution of lysine for phenylalanine

Back 177

(b). The mutant tRNALys will be able to pair with the codon 5′-AUA-3′, which codes for isoleucine.

Front 178

(a) The protein binds to the small ribosomal subunit and increases the rate of initiation of translation.
(b) The protein binds to sequences in the 5′ region of the mRNA and inhibits the rate of initiation of translation.
(c) The protein binds to the large ribosomal subunit and slows down elongation of the polypeptide chain.
(d) The protein binds to sequences in the 3′ region of the mRNA and prevents termination of translation.

Back 178

b

Front 179

The concentration of a particular protein X in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed. You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that levels of X mRNA in the mutant cells are normal. Which of the following mutations in the gene for X could explain these results?
(a) the introduction of a stop codon that truncates protein X at the fourth amino acid
(b) a change of the first ATG codon to CCA
(c) the deletion of a sequence that encodes sites at which ubiquitin can be attached to the protein
(d) a change at a splice site that prevents splicing of the RNA

Back 179

(c). The decrease in level of protein X in the normal cell is most probably due to protein degradation, because levels of mRNA remain constant. The inability of the mutant cell to divide could be due to a mutation that inhibits protein degradation. This would be achieved by the removal of sites for attachment of ubiquitin, which targets proteins for destruction. Choices (a), (b), and (d) would probably not produce the result described, because without the production of a functional protein X the mutant cells could not grow in size.

Front 180

Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the __________________ subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the __________________ subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the __________________-site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the __________________-site by forming base pairs with the exposed codon in the mRNA. The __________________ enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the __________________ called release factor. Eventually, most proteins will be degraded by a

Back 180

large; small; P; peptidyl transferase; protein; proteasome

Front 181

Which of the following methods is not used by cells to regulate the amount of a protein in the cell?
(a) Genes can be transcribed into mRNA with different efficiencies.
(b) Many ribosomes can bind to a single mRNA molecule.
(c) Proteins can be tagged with ubiquitin, marking them for degradation.
(d) Nuclear pore complexes can regulate the speed at which newly synthesized proteins are exported from the nucleus into the cytoplasm.

Back 181

(d). Proteins are synthesized in the cytoplasm and therefore newly synthesized proteins would not be exported from the nucleus into the cytoplasm.

Front 182

Which of the following statements about the proteasome is false?
(a) Ubiquitin is a small protein that is covalently attached to proteins to mark them for delivery to the proteasome.
(b) Proteases reside in the central cylinder of a proteasome.
(c) Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold.
(d) The protein stoppers that surround the central cylinder of the proteasome use the energy from ATP hydrolysis to move proteins into the proteasome inner chamber.

Back 182

(c). Once proteins are sent to the proteasome, proteases degrade them. Chaperone proteins provide a place for misfolded proteins to attempt to refold.

Front 183

Which of the following molecules is thought to have arisen first during evolution?
(a) protein
(b) DNA
(c) RNA
(d) All came to be at the same time.

Back 183

(c). Because RNA is known to catalyze reactions within the cell, because the components of RNA are thought to be more readily formed in the conditions on primitive Earth, and because RNA can contain genetic information, it is the most likely of the three molecules to have arisen first in evolution.

Front 184

According to current thinking, the minimum requirement for life to have originated on Earth was the formation of a _______________.
(a) molecule that could provide a template for the production of a complementary molecule
(b) double-stranded DNA helix
(c) molecule that could direct protein synthesis
(d) molecule that could catalyze its own replication

Back 184

(d). Choice (a) is incorrect in that, although this may have been a step in self-replication, it would not by itself be sufficient. Choices (b) and (c) are incorrect, as these stages in the evolution of the cell must have succeeded the formation of the first self-replicating molecules.

Front 185

Ribozymes catalyze which of the following reactions?
(a) DNA synthesis
(b) transcription
(c) RNA splicing
(d) protein hydrolysis

Back 185

C

Front 186

You are studying a disease that is caused by a virus, but when you purify the virus particles and analyze them you find they contain no trace of DNA. Which of the following molecules are likely to contain the genetic information of the virus?
(a) high-energy phosphate groups
(b) RNA
(c) lipids
(d) carbohydrates

Back 186

B

Front 187

Give a reason why DNA makes a better material than RNA for the storage of genetic information, and explain your answer.

Back 187

DNA is double-stranded and therefore the complementary strand provides a template from which damage can be repaired accurately.