Hall Book Ch 2

Question 1

Define Dose D0.

Answer 1

A dose of radiation that induces an average of one lethal event per cell leaving 37% of irradiated cells still viable.

Question 2

For mammalian cells, the x-ray D0 usually lies between ( ) and ( ) Gy.

Answer 2

1 and 2 Gy

Question 3

List the number and type of DNA lesions per cell detected immediately after a dose of 1 Gy of x-rays is approximately:

1. Double-strand breaks (DSBs):
2. Single-strand breaks (SSBs):
3. Base damage:
4. DNA-DNA crosslinks:

Answer 3

1. 40
2. 1,000
3. > 2,000
4. 30

Question 4

What is the most important lesions produced in chromosomes by radiation?

Answer 4

Double-strand break resulting in the cleavage of chromatin into two pieces, because the interaction of two DSBs may result in cell killing, carcinogenesis, or mutation.

Question 5

What is the yield of DSBs in the irradiated cells compared to the SSBs?

Answer 5

0.04 times that of SSBs

Question 6

DSBs are induced linearly with dose, indicating that they are formed by ( ) tracks of ionizing radiation.

Answer 6

single

Question 7

The energy from ionizing radiations is not deposited uniformly in the absorbing medium but is located along the ( ) of the charged particles set in motion - electrons in the case of x- or gamma-rays and protons and alpha particles in the case of neutrons.

Answer 7

tracks

Question 8

What are the 3 terms that radiation chemists use to describe the radiation energy deposit in the tissue?

Answer 8

Spurs, Blobs, and Short tracks

Question 9

Define the Spur (energy).

Answer 9

Spur contains up to 100 eV of energy and involves, on average, three ion pairs.

Question 10

In the case of x- or gamma rays, how much is the energy deposition events are spurs? (these rays have diameter of 4 nm, which is about twice the diameter of the DNA double helix)

Answer 10

95%

Question 11

Define Blobs (energy) in terms of the diameter, number of ion pairs involved, and the energy range.

Answer 11

Blobs are much less frequent for x- or gamma-rays with 7 nm diameter and contain on average 12 ion pairs with energy range of 100 to 500 eV.

Question 12

Because spurs and blobs have dimensions similar to the DNA double helix, what happens?

Answer 12

Multiple radical attacks occurs if they overlap the DNA helix.

Question 13

Due to the fact that both spurs and blobs have similar dimensions similar to the DNA double helix, there are multiple radical attacks which is likely to be a wide variety of complex lesions including DSBs and what term is used to describe this?

Answer 13

Locally multiply damaged site which is now replaced with the term “clustered lesion.”

Question 14

Given the size of a spur nd the diffusion distance of hydroxyl free radicals, the clustered lesion could be spread out up to ( ) base pairs.

Answer 14

20

Question 15

In the case of densely ionizing radiations, such as neutrons or alpha-particles, ( ) of blobs are produced and the damage produced, therefore, is qualitatively different from that produced by x- or gamma-rays, and it is much more ( ) for the cell to repair.

Answer 15

a greater proportion, difficult

Question 16

What are the various techniques to measure the DNA strand breaks?

Answer 16

• Sucrose gradient sedimentation
• Alkaline and neutral filter elution
• Pulsed-field gel electrophoresis (PFGE)
• Single-cell gel electrophoresis

Question 17

What is another name for single-cell gel electrophoresis?

Answer 17

Comet assay

Question 18

What technique is currently being used to measure DNA strand breaks?

Answer 18

Pulsed-field gel electrophoresis (PFGE)

Question 19

What other technique has become popular approach to visualize DNA damage through the recruitment of DNA repair proteins to sites of DNA damages?

Answer 19

Radiation induced nuclear foci

Question 20

What technique is most widely used to detect the induction and repair of DNA DSBs?

Answer 20

Pulsed-field gel electrophoresis (PFGE)

Question 21

Describe the mechanism of PFGE (pulsed-field gel electrophoresis) which is the most widely used method to detect DNA DSBs.

Answer 21

Based on the electrophoretic elution of DNA form agarose plugs within which irradiated cells have been embedded and lysed.

PFGE allows separation of DNA fragments according to size in the megabase-pair range, with the assumption that DNA DSBs are induced randomly.

Question 22

In PFGE (pulsed-field gel electrophoresis), the fraction of DNA released from the agarose plug is directly proportional to ( ).

Answer 22

Dose

Question 23

The kinetics of DNA DSB rejoining exhibits a ( ), which then decreases with repair time.

Answer 23

fast initial rate

Question 24

What is the most widely accepted description of the kinetics of DNA DSB rejoining?

Answer 24

Uses two first-order components (fast and slow) plus some fraction of residual DSBs.

Question 25

What causes the rejoining of incorrect DNA ends? What is the significance of this?

Answer 25

Originates solely from slowly rejoining DSBs is what is manifested as chromosomal damage (e.g. chromosome translocations and exchanges)

Question 26

What is the advantage of single-cell electrophoresis (Comet assay)?

Answer 26

It can detect differences in DNA damage and repair at the single-cell level.

It is particularly useful for biopsy specimens from tumors in which a relatively small number of cells can be assayed to determine DNA damage and repair.

Question 27

1. Describe the mechanism of single cell electrophoresis (comet assay).
2. What technique is applied to assess DNA SSBs and alkaline-sensitive sites?

Answer 27

1. Similar to PFGE (pulsed-field gel electrophoresis), cells are exposed to ionizing radiation, embedded in agarose, and lysed under neutral buffer conditions to quantify induction and repair of DNA DSBs.
2. Lysis is performed with an alkaline buffer.

Question 28

If the cells incurred DNA DSBs, the amount of damage is ( ) to the migration of DNA in the agarose.

As a result of the lysis and electrophoresis conditions, the fragmented DNA that migrates takes the appearance of a ( ).

Answer 28

directly proportional

comet’s tail

Question 29

Single cell electrophoresis (comet assay) is also highly ( ) and ( ) for ( ) and ( ) and to a lesser degree DNA DSBs.

Answer 29

sensitive, specific, SSBs, alkaline sensitive sites

Question 30

How can the comet assay (single cell electrophoresis) be used to detect DNA DSBs?

Answer 30

By changing the lysis conditions from an alkaline to a neutral pH, the comet technique can be used to measure DNA DSB repair.

Question 31

Both of these assays (comet assay and pulsed gel electrophoresis) are cell based, where DNA in cells is much more resistant to damage by radiation than would be expected from studies on free DNA.

What are the two reasons for this?

Answer 31

(1) the presence in cells of low-molecular weight scavengers that mop up some of the free radicals produced by ionizing
radiation and (2) the physical protection afforded the DNA by packaging with
proteins such as histones. Certain regions of DNA, particularly actively
transcribing genes, appear to be more sensitive to radiation, and there is some
evidence also of sequence-specific sensitivity.

Question 32

Which assay responses to ionizing radiation represents complexes of signaling and repair proteins that localize to sites of DNA strand breaks in the nucleus of a cell?

Answer 32

DNA damage-induced nuclear foci (radiation-induced foci assay)

Question 33

What are the advantages of assaying for foci formation over other techniques to
measure DNA strand breaks, which include the ease of the protocol and that it
can be carried out on both tissue sections and individual cell preparations?

Answer 33

Technically, cells/tissues are incubated with a specific antibody raised to the
signaling/repair protein of interest, and binding of the antibody is then detected
with a secondary antibody, which carries a fluorescent tag.

Fluorescence microscopy detects the location and intensity of the tag, which can then be
quantified.

Question 34

What are the two most commonly assayed proteins for foci formation?

Answer 34

γH2AX and 53BP1

Question 35

What is H2AX? Also, what happens when it’s damaged?

Answer 35

H2AX is a histone protein, which is rapidly phosphorylated in response to damage to form γH2AX.

Question 36

Staining for the unmodified histone (H2AX) gives a ( ) or () on a western blot while
γH2AX is rapidly induced on a western blot in response to stress and can be seen
to form ( ) in damaged cells

Answer 36

pan nuclear stain, unchanging band, discreet nuclear foci

Question 37

What is another protein that also becomes phosphorylated in response to stress and forms nuclear foci at the sites of DNA DSBs?

Answer 37

53BP1

Question 38

What is unique about 53BP1 compared to H2AX?

Answer 38

Antibodies to either the phosphorylated or unmodified form can be used to detect DSBs as the protein relocalizes to the damaged chromatin (i.e., it is not already part of the chromatin as is the case for H2AX).

Question 39

DNA damage-induced increases in γH2AX or phosphorylated 53BP1
can also be quantified by ( ).

Answer 39

flow cytometry

Question 40

Other proteins also form foci in response to damage are (name 4).

Answer 40

such as ataxia-telangiectasia mutated (ATM), replication protein A (RPA), RAD51, and BRCA1 (discussed in subsequent sections).

Question 41

The γH2AX or 53BP1 foci that form in a damaged cell correlate with the
presence of DSBs. Thus, the decrease of foci over time reflects the kinetics of
( ).

Answer 41

DSB repair (i.e., as the DSBs are repaired, the number of foci decreases).

Question 42

What are the two proteins involved in the repair of DNA damage by homologous recombination which is also used to detect repair defects in breast cancer biopsies.

Answer 42

BRCA1 and RAD51

Question 43

Mammalian cells have developed specialized pathways to sense, respond to, and
repair ( ).

Answer 43

base damage, SSBs, DSBs, sugar damage, and DNA–DNA crosslinks.

Question 44

Research from yeast to mammalian cells has demonstrated that the mechanisms
used to repair ionizing radiation-induced base damage are different from ( ).

Answer 44

the mechanisms used to repair DNA DSBs.

Question 45

Different repair pathways are used to repair DNA damage, depending on the stage of the ( ).

Answer 45

cell cycle

Question 46

Much of our knowledge of DNA repair is the result of studying how
mutations in individual genes result in ( ).

Answer 46

radiation hypersensitivity

Question 47

Radiationsensitive mutants identified from yeast and mammalian cells appear either to be directly involved in the repair process or to function as ( ).

Answer 47

molecular checkpoint–controlling elements

Question 48

Base damage is repaired through the ( ) illustrated in Figure 2.6.

Answer 48

base excision repair (BER) pathway

Question 49

Bases on opposite strands of DNA must be complementary: adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). U represents a ( ).

Answer 49

putative single-base mutation that is first removed by a glycosylase/DNA lyase (Fig. 2.6A).

Question 50

In base excision repair, removal of the base is followed by the removal of the sugar residue by ( ), replacement with the correct ( ), and joined by ( ).

Answer 50

apurinic endonuclease 1 (APE1)

nucleotide by DNA polymerase β

DNA ligase III–XRCC1–mediated ligation

Question 51

If more than one nucleotide is to be replaced illustrated by the putative mutation UU in Fig. 2.6B), then the ( ) performs the repair synthesis, the overhanging flap structure is removed by ( ), and DNA strands are sealed by
( ) (see Fig. 2.6B).

Answer 51

complex of replication factor C (RFC)/proliferating cell nuclear antigen (PCNA)/DNA polymerase δ/ε

the flap endonuclease 1 (FEN1)

ligase I

Question 52

Although ionizing radiation–induced base damage is efficiently repaired, defects in BER may lead to an increased mutation rate but usually do not result in cellular ( ). One exception to this is the mutation of ( ), which confers about a ( ) increase in radiation sensitivity.

Answer 52

radiosensitivity

the x-ray cross complementing factor 1 (XRCC1) gene

1.7-fold

Question 53

However, the radiation sensitivity of XRCC1 (x-ray cross complementing factor 1)-deficient cells may come from XRCC1’s potential involvement in other repair processes such as ( ).

Answer 53

SSBs

Question 54

Nucleotide excision repair (NER) removes ( ) in the DNA such as
( ).

Answer 54

bulky adducts

pyrimidine dimers

Question 55

The process of NER (nucleotide excision repair) can be subdivided into two pathways. What are they?

Answer 55

global genome repair (GGR or GG-NER) and transcription-coupled repair (TCR
or TC-NER).

Question 56

How is the process of GG-NER (Global Genome Nucleotide Excision Repair) and TCNER (Transcription Coupled Nucleotide Excision Repair) different?

Answer 56

The process of GG-NER is genome-wide (i.e., lesions can be
removed from DNA that encodes or does not encode for genes). In contrast, TCNER only removes lesions in the DNA strands of actively transcribed genes. When a DNA strand that is being actively transcribed becomes damaged, the
RNA polymerase can block access to the site of damage and hence prevents
DNA repair. TC-NER has evolved to prevent this blockade by RNA polymerase
by effectively removing it from the site of damage to allow the repair proteins
access.

Question 57

The mechanism of GG-NER and TC-NER differs only in the ( ); the remainder of the pathway used to repair the damage is the same for both.

Answer 57

detection of the lesion

Question 58

The essential steps in the nucleotide repair pathway are: (list 5 of them.)

Answer 58

(1) damage recognition;
(2) DNA incisions that bracket the lesion, usually between 24 and 32 nucleotides in
length;
(3) removal of the region containing the adducts;
(4) repair synthesis to fill in the gap region;
(5) DNA ligation (Fig. 2.7).

Question 59

Mutation in NER genes does not lead to ( ). However, defective NER increases ( ) to ultraviolet (UV)-induced DNA damage and anticancer agents such as alkylating agents that induce bulky adducts.

Answer 59

ionizing radiation sensitivity

sensitivity

Question 60

Germline mutations in NER (Nucleotide Excision Repair) genes lead to human DNA repair deficiency disorders such as ( ) in which patients are hypersensitive to UV light.

Answer 60

xeroderma pigmentosum

Question 61

In eukaryotic cells, the two predominant pathways for the repair of DNA DSBs
are:

Answer 61

homologous recombination repair (HRR) and nonhomologous end-joining
(NHEJ).

Question 62

Mechanistically, HRR (homologous recombination repair) requires an undamaged DNA strand to serve as a ( ), and NHEJ (nonhomologous end-joining) repairs DNA DSBs through the ( ).

Answer 62

template for repair to proceed through strand invasion

orchestration of end-to-end joining

Question 63

In lower eukaryotes such as yeast, ( ) is the predominant pathway used for repairing DNA DSBs.

Homologous recombination is an ( ) process because repair is performed
by copying information from the undamaged homologous chromatid/chromosome.

Answer 63

HRR (homologous recombination repair)

error-free

Question 64

In mammalian cells, the choice of repair for double-strand break is biased by ( ) and by ( ).

Answer 64

the phase of the cell cycle

the abundance of repetitive DNA

Question 65

HRR (homologous recombination repair) occurs primarily in the ( ) phase of the cell cycle, when an ( ) is available to act as a template, whereas NHEJ (nonhomologous end-joining) occurs in the ( ) of the cell cycle, when no such template exists (Fig. 2.8).

Answer 65

late S/G2

undamaged sister chromatid

G1 phase

Question 66

NHEJ (nonhomologous end-joining) is ( ) prone and probably accounts for many of the ( ) induced in the DNA of human cells by ionizing radiation. However, it is important to keep in mind that NHEJ and HRR are not ( ), and both have been found to be active in the ( ) of the cell cycle, indicating that other as-yet-unidentified factors, in addition to cell cycle phase, are important in determining
what repair program is used.

Answer 66

error

premutagenic lesions

mutually exclusive

late S/G2 phase

Question 67

Furthermore, two additional repair pathways in DNA Double-Strand Break Repair have been identified that can also be considered error prone. What are they?

Answer 67

Single-strand annealing (SSA) is a specialized DNA end-joining process between interspersed repetitive DNA throughout the genome and requires RAD52 and DNA ligase I.

Interestingly, although this type of repair uses a homology-based mechanism, loss of repeat sequences and associated intervening sequences is quite common.

An alternative DNA end-joining (Alt-EJ) process has also been described that involves XRCC1, DNA ligase III, and PARP. Alt-EJ is still not well understood but is active in cells
where classical NHEJ is lost.

Similar to SSA, Alt-EJ is also error prone and can lead to insertions and deletions.

Question 68

FIGURE 2.8 Illustration showing that nonhomologous recombination occurs in
the ( ) phase of the cell cycle, at which stage, there is no ( ) to use
as a template for repair. In contrast, homologous recombination occurs in the ( ) phases of the cell cycle, when there is a sister chromatid to use as a
template in repair.

Answer 68

G1, sister chromatid, S and G2

Question 69

Single-strand annealing (SSA) is a specialized DNA end-joining process between interspersed repetitive DNA throughout the genome and requires ( ).

Interestingly, although this type of repair uses a homology-based mechanism, loss of repeat sequences and associated intervening sequences is quite common.

An alternative DNA end-joining (Alt-EJ) process has also been described that involves (
). Alt-EJ is still not well understood but is active in cells
where classical NHEJ is lost.

Similar to SSA, Alt-EJ is also error prone and can lead to insertions and deletions.

Answer 69

RAD52 and DNA ligase I

XRCC1, DNA ligase III, and PARP

Question 70

The immediate response of a cell to a DNA DSB is the activation of a group of
sensors that serve both to promote DNA repair and to prevent the cell from
proceeding in the cell cycle until the break is faithfully repaired.

These sensors, ( ), are protein kinases that belong to the
phosphatidylinositol-3-kinase–related kinase (PIKK) family and are recruited to
the sites of DNA strand breaks induced by ionizing radiation.

Answer 70

ATM and Rad3-related (ATR)

Question 71

The competition for repair by HRR (homologous recombination repair) versus NHEJ (nonhomologous end-joining) is in part regulated by the protein ( ).

Functionally, ( ) promotes the processing of broken DNA ends to generate
recombinogenic single-strand DNA by regulating the activity of the
( ) protein complex (Fig. 2.9), and this resection activity is diminished by 53BP1.

Answer 71

53BP1

ATM

NBS/MRE11/Rad50

Question 72

Fig 2.9 describes nonhomologous end-joining.

In nonhomologous end-joining, DNA strand breaks are recognized
by the (        ) complex, resulting in (         ) ends. 

Homologous recombination is inhibited by the activity of ( ). The initial step of the core NHEJ pathway starts with the binding of the ends at the double-strand break by the ( ) heterodimer.

This complex then recruits and activates the catalytic subunit of DNA-PK (DNA-PKcs), whose role is the juxtaposition of the two DNA ends. The DNA-PK complex then
recruits the ligase complex (XRCC4/XLF-LIGIV/PNK) that promotes the final
ligation step.

Answer 72

ATM and the MRN (Mre11-Rad50-Nbs1)

resection of the DNA

53BP1

Ku70/Ku80

Question 73

The ligation of DNA DSBs by NHEJ (nonhomologous end-joining) does not require ( ).

However, the damaged ends of DNA DSBs cannot simply be ligated together;
they must first be modified before they can be rejoined by a ligation reaction.
NHEJ can be divided into five steps:

Answer 73

sequence homology

(1) end recognition by Ku binding,
(2) recruitment of DNA-dependent protein kinase catalytic subunit (DNA-PKcs),
(3) end processing,
(4) fill-in synthesis or end bridging, and (5) ligation (see Fig. 2.9).

Question 74

NHEJ (nonhomologous end-joining) is error prone and plays an important physiologic role in generating ( ) through V(D)J rejoining. The error-prone nature of NHEJ is essential for generating ( ) and often goes undetected in mammalian cells, as errors in the noncoding DNA that composes most of the human genome has little consequence.

NHEJ is primarily found in the ( ) phase of the cell cycle, where there is no sister chromatid.

Answer 74

antibodies

antibody diversity

G1

Question 75

HRR (homologous recombination repair) provides the mammalian genome a high-fidelity mechanism of repairing ( ) (Fig. 2.10). In particular, the increased activity of this recombination pathway in late ( ) suggests that its primary function is to repair and restore the ( ) with DNA DSBs.

Compared to NHEJ, which requires no sequence homology to rejoin broken ends, HRR requires ( ) with an undamaged chromatid or chromosome (to serve as a template) for repair to occur.

Answer 75

DNA DSBs

S/G2

functionality of replication forks

physical contact

Question 76

FIGURE 2.10 shows homologous recombinational repair. The initial step in HRR is
the recognition of the lesion and processing of the double-strand DNA ends into
3′ DNA single-strand tails by the ( ) complex, which are then coated by ( ) forming a nucleoprotein filament. Then, specific HRR proteins are recruited to the nucleoprotein filaments, such as ( ).

Answer 76

MRN (Mre11-Rad50-Nbs1)

RPA

RAD51, RAD52, and BRCA1/2

Question 77

( ) is a key protein in homologous recombination as it mediates the ( ) of the homologous strand of the sister chromatid, leading to formation of ( ) junctions. The Holliday junctions are finally resolved into two DNA duplexes. See text for details.

Answer 77

RAD51, invasion, Holliday

Question 78

In HRR (homologous recombination repair), during recombination, evidence exists that ( ) phosphorylates the breast cancer tumor suppressor protein ( ), which is then recruited to the site of the DSB that has been bound by the ( ) protein complex (see Fig. 2.10).

Answer 78

ATM, BRCA1, NBS/MRE11/Rad50

Question 79

In HRR (homologous recombination repair), ( ) and perhaps other yet unidentified endonucleases resect the DNA, resulting in a ( ) that serves as a binding site for ( ).

BRCA2, which is attracted to the DSB by BRCA1, facilitates the loading of Rad51 onto RPA-coated single-strand overhangs produced by endonuclease resection.

Answer 79

MRE11, 3′ single-strand DNA, Rad51

Question 80

Rad51 protein is a homologue of the ( ) and possesses the ability to form nucleofilaments and catalyze strand exchange with the complementary strand in the undamaged chromosome.

Five additional paralogues of ( ) also bind to the RPA-coated single-stranded region and recruit ( ), which protects against exonucleolytic degradation.

To facilitate repair, ( ) uses its ATPase activity to unwind the double-stranded molecule.

Answer 80

Escherichia coli recombinase RecA

Rad51

Rad52

Rad54

Question 81

The two invading ends serve as primers for DNA synthesis, and the so-called ( ) are resolved by ( ) by ( ) over.

Answer 81

Holliday junctions

MMS4 and MUS81

noncrossing

Question 82

The Holliday junctions disengage and DNA strand pairing is followed by ( ), or by (
) of the Holliday junctions, which is followed by gap filling. The identities of the polymerase and ligase involved in these latter steps are unknown.

Because inactivation of HRR (Homologous Recombination Repair) genes results in (
) and ( ), these genes provide a critical link between HRR and chromosome stability.

Answer 82

gap filling

crossing over

radiosensitivity

genomic instability

Question 83

Dysregulated homologous recombination can also lead to cancer by ( ).

Answer 83

loss of heterozygosity (LOH)

Question 84

Crosslinks can be divided into ( ) crosslinks, which occurs on one strand of DNA, and ( ) crosslinks, which occur between two strands of DNA.

Answer 84

intrastrand

interstrand

Question 85

Crosslinks are particularly problematic during when? and why?

Answer 85

During replication and can lead to cell cycle arrest and even cell death.

Question 86

How is interstrand crosslink different from intrastrand crosslink?

Answer 86

Intrastrand crosslinks can lead to a block of DNA polymerase activity and a single strand DNA gap, which can be resolved by DNA polymerase switching to a template w/o the intrastrand crosslink or by translesion synthesis.

Interstrand crosslinks will result in a complete block to replication and prevent the unwinding of duplex DNA, leading to cell death if unrepaired.

Question 87

Interstrand crosslinks are repaired by different protein complexes depending on the phase of the cell cycle.

In G1 phase, interstrand crosslinks are removed by ( ).

Answer 87

NER (Nucleotide Excision Repair)

Question 88

Both sides of the interstrand crosslink are cleaved by ( ) on the 3’ end and ( ) on the 5’ end.

Answer 88

XPF

Fanconi-associated nuclease 1 (FAN1)

Question 89

Excision of the interstrand crosslink is followed by ( ).

Answer 89

translesion synthesis

Question 90

However, only a small percentage of interstrand crosslinks are repaired in ( ) phase, and the unrepaired lesions will present a major problem for the cell in ( ) phase.

During S phase (Fig. 2.11), the interstrand crosslink is recognized by ( ), which binds to the abnormal crosslinked structure as well as binding to the Fanconi anemia (FA) core complex containing FANCA, FANCB, FANCC, FANCE, FANCF, FANCG, and FANCL. FANCM binding to the crosslink structure protects against collapse of the replication fork.

Answer 90

G1

S

Fanconi anemia complementation group M (FANCM)

Question 91

( ) are also recruited to the crosslink site and excise the
crosslink. Replication through the excised sites is mediated by translesion
polymerases such as ( ).

After translesion synthesis, the replication fork is repaired by ( ) and the crosslink removed by ( ).

Answer 91

FAN1 and XPF/MUS81

REV1/DNA polymerase ζ

HRR

NER

Question 92

Cells with mutations in ( ) pathways are modestly sensitive to
crosslinking agents.

In contrast, individuals afflicted with the syndrome FA are ( ) to ( ) agents.

Answer 92

NER and HRR

hypersensitive

crosslinking

Question 93

Chromatin that contains ( ) is more susceptible to ( ) crosslinks, and the crosslinked proteins are usually ( ).

Answer 93

actively transcribed genes

DNA–protein

nuclear matrix proteins

Question 94

DNA–DNA interstrand crosslink repair in ( ) phase. The initial signal for DNA–DNA crosslink repair is ( ) (A).

Fork stalling results in recruitment of ( ) and the ( ).

Answer 94

S phase

stalling of the replication fork

FANCM (Fanconi Anemia Complementation Group M)

FA core complex

Question 95

Incision or unhooking of the crosslink is achieved by the action of ( ).

B: Translesion synthesis is achieved by polymerases such as ( ) and extension past the unhooked crosslink by polymerase ζ (C).

Answer 95

FAN1 and accessory proteins

Rev1 and polymerase ζ

Question 96

The resulting DNA double-strand break is repaired by ( ), and the crosslink is removed from the DNA by ( ) (E–F). This schema for crosslink repair is still a work in progress.

Answer 96

homologous recombination (D)

nucleotide excision repair

Question 97

The mismatch repair (MMR) pathway removes ( )
mismatches that occur during replication.

In addition, the MMR pathway removes base–base mismatches in ( ). See Figure 2.12 for schematic representation and an indication of the critical gene products.

Answer 97

base–base and small insertion

homologous recombination intermediates

Question 98

The process of MMR can be subdivided into four components:

Answer 98

First, the mismatch must be identified by sensors that transduce the signal of a
mismatched base pair;

second, MMR factors are recruited;

third, the newly synthesized strand harboring the mismatch is identified and the incorrect/altered nucleotides are excised;

and in the fourth stage, resynthesis and ligation of the excised DNA tract is completed.

Question 99

MMR was first characterized in E. coli by the characterization of the ( ) genes, of which homologues of these gene products have been identified and extensively characterized in both yeast and humans.

Mutations in any of the mismatch MSH, MLH, and PSM families of repair genes lead to (
) (small base insertions or deletions) and cancer, especially ( ).

Answer 99

Mut

microsatellite instability

hereditary nonpolyposis colon cancer (HNPCC)

Question 100

Mismatch repair. The initial step in the MMR pathway is the
recognition of mismatched bases through either ( )
complexes.

These recognition complexes recruit ( ), alongside the exonuclease EXO1 that catalyzes the excision step that follows.

A gap-filling step by polymerases ( ) is followed by a final ligation step.

Answer 100

Msh2-Msh6 or Msh2-Msh3

MLH1-PMS2, MLH1-PMS1, and MLH1-MLH3

δ/ε, RFC, and PCNA

Question 101

Cell killing does not correlate with ( ) but relates better to ( ).

Agents such as hydrogen peroxide produce ( ) efficiently, but very few ( ), and kill very few cells.

Answer 101

SSBs

DSBs

SSBs

DBSs

Question 102

Cells defective in DNA DSB repair exhibit hypersensitivity to killing by ionizing radiation and increased numbers of chromosome aberrations.

On the basis of evidence such as this, it is concluded that ( ) are the most
relevant lesions leading to most biologic insults from radiation, including cell
killing.

The reason for this is that DSBs can lead to ( ) aberrations that present problems at cell division.

Answer 102

DSBs

chromosomal

Question 103

The backbone of DNA is made of molecules of sugar and phosphates, which
serve as a framework to hold the bases that carry the genetic code. Attached to
each sugar molecule is a base: thymine, adenine, guanine, or cytosine. This
whole configuration is coiled tightly in a double helix.

Figure 2.13 is a highly schematized illustration of the way an organized
folding of the long DNA helix might be achieved as a closely packed series of
looped domains wound in a tight helix. The degree of packing also is illustrated
by the relative dimensions of the DNA helix and the ( ).

Answer 103

condensed metaphase chromosome

Question 104

The largest part of the life of any somatic cell is spent in ( ), during
which the nucleus, in a stained preparation, appears as a lacework of fine and
lightly stained material in a translucent and colorless material surrounded by a
membrane.

In the ( ) nucleus of most cells, one or more bodies of various
sizes and shapes, called ( ), are seen. In most cells, little more than this can
be identified with a conventional light microscope. In fact, a great deal is
happening during this time: The quantity of DNA in the nucleus doubles as each
chromosome lays down an exact replica of itself next to itself.

Answer 104

interphase, interphase, nucleoli

Question 105

When the chromosomes become visible at mitosis, they each present in duplicate. Even during interphase, there is good evidence that the chromosomes are not free to
move about within the nucleus but are restricted to ( )

The various events that occur during mitosis are divided into several phases.

The first phase of division is called ( ).

The beginning of this phase is marked by a thickening of the chromatin and condensation into light coils.

By the end of prophase, each chromosome has a lightly staining constriction known as a ( ); extending from the ( ) are the arms of the chromosome.

Answer 105

“domains.”

prophase

centromere

centromere

Question 106

Prophase ends when the chromosomes reach ( ).

With the disappearance of the nuclear membrane, the nucleoplasm and the
cytoplasm mix.

Metaphase then follows, in which two events occur simultaneously. The chromosomes move to the center of the cell (i.e., to the cell’s equator), and the spindle forms. The spindle is composed of fibers that cross the cell, linking its poles.

Once the chromosomes are stabilized at the equator of the cell, their ( ) divide, and metaphase is complete.

Answer 106

maximal condensation and the nuclear membrane disappears, as do any nucleoli

centromeres

Question 107

The phase that follows, anaphase, is characterized by a movement of the
chromosomes on the ( ). The chromosomes appear to be pulled
toward the poles of the cell by fibers attached to the centromeres. The arms,
particularly the long arms, tend to trail behind.

Anaphase is followed by the last phase of mitosis, ( ). In this phase,
the chromosomes, congregated at the poles of the cell, begin to uncoil.

The nuclear membrane reappears, as do the nucleoli, and as the phase progresses, the
chromosome coils unwind until the nucleus regains the appearance characteristic
of interphase.

Answer 107

spindle to the poles

telophase

Question 108

Telomeres cap and protect the terminal ends of chromosomes. The name
telomere literally means “end part.” Mammalian telomeres consist of long arrays
of ( ) repeats that range in total length anywhere from ( )
kilobases.

Each time a normal somatic cell divides, telomeric DNA is lost from
the lagging strand because DNA polymerase cannot synthesize new DNA in the
absence of an RNA primer. Successive divisions lead to progressive shortening,
and after 40 to 60 divisions, the telomeres in human cells are shortened
dramatically, so that vital DNA sequences begin to be lost. At this point, the cell
cannot divide further and undergoes ( ).

Answer 108

TTAGGG, 1.5 to 150, senescence (irreversible arrest of proliferation)

Question 109

Telomere length has been described as the “molecular clock” or
generational clock because it shortens with age in somatic tissue cells during
adult life. Stem cells in self-renewing tissues, and cancer cells in particular,
avoid this problem of aging by activating the enzyme telomerase.

Telomerase is a ( ) that includes the complementary sequence to the TTAGGG repeats and so continually rebuilds the chromosome ends to offset the degradation that occurs with each division.

Answer 109

reverse transcriptase

Question 110

In tissue culture, immortalization of cells—that is, the process whereby cells
pass through a “crisis” and continue to be able to divide beyond the normal limit —is associated with ( ).

Virtually, all human tumor cell lines and approximately 90% of human
cancer biopsy specimens exhibit telomerase activity.

By contrast, normal human somatic tissues, other than stem cells, do not ( ) of this enzyme.

Both immortalization and carcinogenesis are associated with ( ) expression.

Answer 110

telomere stabilization and telomerase activity

possess detectable levels

telomerase

Question 111

In the traditional study of chromosome aberrations, the effects of ionizing
radiations are described in terms of their appearance when a preparation is made
at the first ( ) after exposure to radiation. This is the time when the
structure of the chromosomes can be discerned.

The study of radiation damage in mammalian cell chromosomes is hampered
by the large number of mammalian chromosomes per cell and by their small
size.

Most mammalian cells currently available for experimental purposes have a
diploid complement of 40 or more chromosomes. There are exceptions, such as
the Chinese hamster, with 22 chromosomes, and various marsupials, such as the
rat kangaroo and woolly opossum, which have chromosome complements of 12
and 14, respectively.

Many plant cells, however, contain fewer and generally
much larger chromosomes; consequently, much information on chromosomal
radiation damage accrued from studies with plant cells.

Answer 111

metaphase

Question 112

If cells are irradiated with x-rays, ( ) are produced in the chromosomes.
The broken ends appear to be “sticky” because of unpaired bases and can rejoin
with any other sticky end. It would appear, however, that a broken end cannot
join with a normal, unbroken chromosome.

Once breaks are produced, different fragments may behave in various ways:

Answer 112

DSBs

1. The breaks may restitute, that is, rejoin in their original configuration. In this
   case, of course, nothing amiss is visible at the next mitosis.
2. The breaks may fail to rejoin and give rise to an aberration, which is scored
   as a deletion at the next mitosis.
3. Broken ends may reassort and rejoin other broken ends to give rise to
   chromosomes that appear to be grossly distorted if viewed at the following
   mitosis.

Question 113

The aberrations seen at metaphase are of two classes: ( ) aberrations and ( ) aberrations.

Answer 113

chromosome, chromatid

Question 114

Chromosome aberrations result if a cell is irradiated early in ( ), before the chromosome material has been duplicated. In this case, the radiation-induced break is in a ( ); during the DNA synthetic phase that follows, this strand of chromatin lays down an identical strand next to itself and replicates the break that has been produced by the radiation.

This leads to a chromosome aberration visible at the ( ) because there is an identical break in the corresponding points of a pair of chromatin strands.

Answer 114

interphase, single strand of chromatin, next mitosis

Question 115

If, on the other hand, the dose of radiation is given ( ), after the DNA material has (
) and the chromosomes consist of ( ) of chromatin, then the aberrations produced are called ( ) aberrations.

Answer 115

later in interphase, doubled, two strands, chromatid

Question 116

In regions removed from the ( ), chromatid arms may be fairly well ( ), and it is reasonable to suppose that the radiation might break ( ) chromatid without breaking its ( ) chromatid, or at least not in the same place.

Answer 116

centromere, separated, one, sister

Question 117

A break that occurs in a ( ) chromatid arm after chromosome replication and leaves the opposite arm of the same chromosome ( ) leads to chromatid aberrations.

Answer 117

single, undamaged

Question 118

Many types of chromosomal aberrations and rearrangements are possible, but an
exhaustive analysis is beyond the scope of this book.

Three types of aberrations that are lethal to the cell are described, followed by two common rearrangements that are consistent with cell viability but are frequently involved in carcinogenesis.

The three lethal aberrations are the ( ); the ( ), which are
( ) aberrations; and the ( ), which is a ( )
aberration.

All three represent gross distortions and are clearly visible.

Many other aberrations are possible but are not described here.

Answer 118

dicentric, ring, chromosome, anaphase bridge, chromatid

Question 119

The formation of a dicentric is illustrated in diagrammatic form in Figure
2.14A.

This aberration involves an interchange between ( ) chromosomes.

If a break is produced in each chromosome early in interphase and the sticky ends are close to one another, they may rejoin as shown. This bizarre interchange is replicated during the DNA synthetic phase, and the result is a grossly distorted chromosome with ( ) centromeres (hence, dicentric).

There also are two fragments that have no centromere ( ), which
will therefore be lost at a subsequent mitosis.

Answer 119

two separate, two, acentric fragment

Question 120

The appearance at metaphase is shown in the bottom panel of Figure 2.14A. An example of a dicentric and fragment in a metaphase human cell is shown in Figure 2.15B; Figure 2.15A shows a normal metaphase for comparison.

Question 121

The steps in the formation of a dicentric by irradiation of ( ) chromosomes. A break is produced in each of ( ) chromosomes.

The “sticky” ends may join incorrectly to form an ( ) between the two chromosomes. Replication then occurs in the DNA synthetic period. One chromosome has two centromeres: a dicentric.

The acentric fragment will also ( ), and both will be lost at a subsequent mitosis
because, lacking a ( ), they will not go to either pole at anaphase.

Answer 121

prereplication (i.e., G1), two separate, interchange, replicate, centromere

Question 122

B: The steps in the formation of a ring by irradiation of a ( )
chromosome. A break occurs in each arm of the ( ) chromosome.

The sticky ends rejoin ( ) to form a ring and an acentric fragment.

Replication then occurs.

Answer 122

prereplication (i.e., G1), same, incorrectly

Question 123

C: The steps in the formation of an anaphase bridge by irradiation of a
( ) chromosome. Breaks occur in ( ) of the ( ) chromosome. Incorrect rejoining of the sticky ends then occurs in a sister union.

At the next anaphase, the acentric fragment will be lost, one centromere
of the dicentric will go to each pole, and the chromatid will be stretched between
the poles. Separation of the progeny cells is ( ) possible; this aberration is likely
to be lethal. (Courtesy of Dr. Charles Geard.)

Answer 123

postreplication (i.e., G2), each chromatid, same, not

Question 124

The formation of a ring is illustrated in diagrammatic form in Figure 2.14B.
A break is induced by radiation in ( ) arm of a ( ) chromatid early in the
cell cycle. The sticky ends may rejoin to form a ring and a fragment. Later in the
cycle, during the DNA synthetic phase, the chromosome replicates.

The ultimate appearance at metaphase is shown in the lower panel of Figure 2.14B.

The fragments have no centromere and probably will be lost at mitosis because they
will not be pulled to either pole of the cell. An example of a ring chromosome in
a human cell at metaphase is illustrated in Figure 2.15C.

An anaphase bridge may be produced in various ways. As illustrated in
Figure 2.14C and Figure 2.16, it results from breaks that occur late in the cell
cycle ( ).

Breaks may occur in ( ), and the sticky ends may rejoin incorrectly to form a sister union. At anaphase, when the two sets of chromosomes move to opposite poles, the section of chromatin between the two centromeres is stretched across the cell between the poles, hindering the separation into two new progeny cells, as illustrated in Figure 2.14C and Figure 2.16B. The two fragments may join as shown, but because there is no centromere, the joined fragments will probably be lost at the first mitosis. This type of aberration occurs in human cells and is essentially always lethal. It is hard to demonstrate because preparations of human chromosomes usually are made by accumulating cells at metaphase, and the bridge is only evident at anaphase.

Figure 2.16 is an anaphase preparation of Tradescantia paludosa, a plant used extensively for cytogenetic studies because of the small number of large chromosomes. The anaphase bridge is seen clearly as the replicate sets of chromosomes move to opposite poles of the cell.

Answer 124

each, single, (in G2) after the chromosomes have replicated, both chromatids of the same chromosome

Question 125

Two important types of chromosomal changes that are not lethal to the cell
are ( ) and ( ).

Answer 125

symmetric translocations, small deletions

Question 126

The formation of a symmetric translocation is illustrated in Figure 2.17A. It involves a break in (            ) chromosomes, with the broken ends being exchanged
between the two chromosomes as illustrated. An aberration of this type is
difficult to see in a conventional preparation but is easy to observe with the
technique of (            ), or chromosome painting, as it commonly is called. Probes are available for every human chromosome that makes them fluorescent in a different color. 

Exchange of material between two different chromosomes then is readily observable (Fig. 2.18). Translocations are associated with several human malignancies caused by the activation of an oncogene; Burkitt lymphoma and certain types of leukemia are examples.

Answer 126

two prereplication (G1), fluorescent in situ hybridization (FISH)

Question 127

FIGURE 2.17 A: Formation of a symmetric translocation. Radiation produces
breaks in (            ) chromosomes. The broken pieces are exchanged between the two chromosomes, and the “sticky” ends rejoin. This aberration is not (           ) to the cell. 

There are examples in which an exchange aberration of this type leads to the activation of an oncogene. See Chapter 10 on radiation carcinogenesis.

B: Diagram of a deletion. Radiation produces two breaks in the same arm of the same chromosome. What actually happens is illustrated more clearly in Figure 2.18.

Answer 127

two different prereplication, necessarily lethal

Question 128

The other type of nonlethal chromosomal change is a ( ). This is illustrated in Figure 2.17B and may result from two breaks in the same arm of the same chromosome, leading to the loss of the genetic information between the two breaks. The actual sequence of events in the formation of a deletion is easier to understand from Figure 2.19, which shows an interphase chromosome. It is a simple matter to imagine how two breaks may
isolate a loop of DNA—an acentric ring—which is lost at a subsequent mitosis.

Deletions may be associated with ( ) if the lost genetic material
includes a ( ). This is discussed further in Chapter 10 on
radiation carcinogenesis.

Answer 128

small interstitial deletion, carcinogenesis, tumor suppressor gene

Question 129

FIGURE 2.19 Formation of a deletion by ionizing radiation in an interphase
chromosome. It is easy to imagine how two breaks may occur (by a single or two
different charged particles) in such a way as to isolate a loop of DNA. The
“sticky” ends rejoin, and the deletion is lost at a subsequent mitosis because it
has no centromere.

Answer 129

This loss of DNA may include the loss of a suppressor gene
and lead to a malignant change. See Chapter 10 on radiation carcinogenesis.
The interaction between breaks in different chromosomes is by no means
random. There is great heterogeneity in the sites at which deletions and
exchanges between different chromosomes occur; for example, chromosome 8 is
particularly sensitive to exchanges. As mentioned previously, each chromosome
is restricted to a domain, and most interactions occur at the edges of domains,
which probably involves the nuclear matrix. Active chromosomes are therefore
those with the biggest surface area to their domains.

Question 130

Chromosomal aberrations in ( ) have been used widely as
biomarkers of radiation exposure. In blood samples obtained for cytogenetic
evaluation within a few days to a few weeks after total body irradiation, the
frequency of ( ) in the lymphocytes reflects the dose received.

Lymphocytes in the blood sample are stimulated to divide with a mitogen such as phytohemagglutinin and are arrested at metaphase, and the incidence of rings and dicentrics is scored. The dose can be estimated by comparison with in vitro cultures exposed to known doses. Figure 2.20 shows a ( ) curve for aberrations in human lymphocytes produced by γ-rays.

The data are fitted by a linear-quadratic relationship, as would be expected,
because rings and dicentrics result from the interaction of two chromosome
breaks, as previously described.

The ( ) component is a consequence of the two breaks resulting from a ( ) charged particle.

If the two breaks result from ( ) charged particles, the probability of an interaction is a ( ) function of dose. This also is illustrated for the formation of a dicentric in Figure 2.20.

Answer 130

peripheral lymphocytes, asymmetric aberrations (dicentrics and rings), dose-response

linear, single, different, quadratic

Question 131

FIGURE 2.20 The frequency of chromosomal aberrations (dicentrics and rings)
is a linear-quadratic function of dose because the aberrations are the
consequence of the interaction of two separate breaks.

At low doses, both breaks may be caused by the ( ) electron; the probability of an exchange aberration is proportional to dose (D).

At higher doses, the two breaks are more likely to be caused by ( ) electrons. The probability of an exchange aberration is proportional to the ( ) of the dose (D2).

Answer 131

same, separate, square

Question 132

If a sufficient number of metaphases are scored, cytogenetic evaluations in
cultured lymphocytes readily can detect a recent total body exposure of as low as
( ) Gy in the exposed person.

Such studies are useful in distinguishing between “real” and “suspected” exposures, particularly in those instances involving “black film badges” or in potential accidents in which it is not certain whether individuals who were at risk for exposure actually received a dose of radiation.

Answer 132

0.25

Question 133

Mature T lymphocytes have a finite life span of about ( ) days and are
eliminated slowly from the peripheral lymphocyte pool. Consequently, the yield
of dicentrics observed in peripheral lymphocytes ( ) in the months and
years after a radiation exposure.

During in vivo exposures to ionizing radiation, ( ) aberrations are
induced not only in mature lymphocytes but also in lymphocyte progenitors in
marrow, nodes, or other organs.

Answer 133

1,500, declines, chromosome

Question 134

The stem cells that sustain asymmetric aberrations ( ) die in attempting a subsequent mitosis, but those that sustain a symmetric nonlethal aberration (such as a translocation) survive and pass on the aberration to their progeny.

Consequently, dicentrics are referred to as ( ) aberrations because their number (
) with time after irradiation.

Symmetric translocations, by contrast, are referred to as “stable” aberrations because they ( ) for many years.

Either type of aberration can be used to estimate dose soon after irradiation, but if many years have elapsed, scoring dicentrics ( ) the dose, and only stable aberrations such as translocations give an accurate picture.

Until recently, translocations were much more difficult to observe than dicentrics, but now, the technique of FISH makes the scoring of such symmetric aberrations a relatively simple matter. The frequency of translocations assessed in this way correlates with total body dose in exposed individuals even after more than 50 years, as was shown in a study of the survivors of the atomic bomb attacks on Hiroshima and Nagasaki.

Answer 134

such as dicentrics, “unstable,” declines, persist, underestimates

Question 135

Ionizing radiation induces ( ).

Answer 135

base damage, SSBs, DSBs, and DNA protein crosslinks

Question 136

The cell has evolved an intricate series of sensors and pathways to respond to
each type of radiation-induced damage.

DNA (          ), the most lethal form of ionizing radiation–induced damage, is
repaired by (            ) in the (          ) of the cell cycle and (        ) recombination (mainly) in the (        ) phase of the cell cycle.

Answer 136

DSBs, nonhomologous recombination, G1 phase, homologous, S/G2

Question 137

Defective ( ) recombination leads to ( ) aberrations,
immune deficiency, and ionizing radiation sensitivity.

Defective ( ) recombination leads to ( ) and ( )
aberrations, decreased proliferation, and ionizing radiation sensitivity.

Answer 137

nonhomologous, chromosome, homologous, chromatid, chromosome

Question 138

There is good reason to believe that ( ) rather than SSBs lead to important
biologic end points including cell death, carcinogenesis, and mutation.

Answer 138

DSBs

Question 139

Radiation-induced breakage and ( ) in ( ) may lead to ( ) aberrations.

Answer 139

incorrect rejoining, prereplication (G1) chromosomes, chromosome

Question 140

Radiation-induced breakage and incorrect rejoining in ( ) may lead to ( ) aberrations.

Answer 140

postreplication (late S or G2) chromosomes, chromatid

Question 141

Lethal aberrations include ( ).

( ) translocations and ( ) deletions are nonlethal.

Answer 141

dicentrics, rings, and anaphase bridges

Symmetric, small

Question 142

There is a good correlation between cells killed and cells with ( ).

Answer 142

asymmetric exchange aberrations (i.e., dicentrics or rings)

Question 143

The incidence of most radiation-induced aberrations is a (                       )
function of dose.

Answer 143

linear-quadratic

Question 144

Scoring aberrations in lymphocytes from peripheral blood may be used to
estimate total body doses in humans accidentally irradiated. The lowest single
dose that can be detected readily is ( ) Gy.

Answer 144

0.25

Question 145

Dicentrics are ( ) aberrations; they are lethal to the cell and are not passed on to ( ).

Consequently, the incidence of dicentrics ( ) slowly with time after exposure.

Answer 145

“unstable”

progeny

declines

Question 146

Translocations are (      ) aberrations; they persist for many years because
they are not (       ) to the cell and are passed on to the progeny.

Answer 146

“stable”

lethal