Nucleus DNA Replication & Repair

Question 1

How is DNA synthesis initiated?

Answer 1

Initiator proteins facilitate duplex opening at origins of replication and recruit helicase; this establishes a replication bubble where replicative enzymes can associate with each parent strand.

Question 2

Single-stranded binding proteins (SSBs)

Answer 2

bind with newly separated parent strands, providing a physical barrier on each strand that prevents the exposed nucleotides from interacting with free nucleotides or other single-strand polynucleotides

Question 3

topoisomerase

Answer 3

topoisomerase acts to unwind the helix ahead of the replication fork. Topoisomerase I introduces a single-strand break within the helix; this break allows the helix to rotate around the intact strand before the break is resealed. topoisomerase II grabs two portions of the same supercoiled helix. Topoisomerase II introduces a double-strand break at one location; the other portion of the helix is then passed through that break before the strand is resealed.

Question 4

Topoisomerase I

Answer 4

Topoisomerase I introduces a single-strand break within the helix; this break allows the helix to rotate around the intact strand before the break is resealed.

Question 5

Topoisomerase II

Answer 5

introduces a double-strand break at one location; the other portion of the helix is then passed through that break before the strand is resealed.

Question 6

On the lagging strand, each new stretch of DNA polymerase activity continues until the previous RNA primer is encountered, creating a polynucleotide referred to as an

Answer 6

Okazaki fragment

Question 7

In which direction do DNA polymerases construct new DNA strands?

Answer 7

DNA polymerase synthesizes new strands in a 5’ → 3’ direction by moving along the template strand in a 3’ → 5’ direction.

Question 8

Why does DNA pol III have high processivity?

Answer 8

DNA pol III has high processivity because it associates with a sliding clamp that anchors it to the template strand.

Question 9

How are Okazaki fragments processed in prokaryotes?

Answer 9

DNA polymerase replaces the RNA primer with DNA nucleotides, and ligase establishes phosphodiester bonds between fragments.

Question 10

How are some species of DNA polymerase able to proofread?

Answer 10

Certain species of DNA polymerase possess 3’ → 5’ exonuclease activity, which allows them to remove any nucleotides that are incorrectly incorporated and continue synthesis with the correct nucleotide.

Question 11

Certain species of DNA polymerase possess 3’ → 5’ exonuclease activity, which allows them to remove any nucleotides that are incorrectly incorporated and continue synthesis with the correct nucleotide.

Question 12

How are primers laid down in eukaryotes?

Answer 12

One of the subunits of DNA pol α has RNA primase activity, allowing it to initiate polymerization.

Question 13

Helicase

Answer 13

Severs hydrogen bonds within base pairs to separate the strands of a DNA helix

Question 14

Primase

Answer 14

Lays a primer (short segment of RNA) on parent strand

Question 15

DNA polymerase

Answer 15

Conducts polymerization by adding nucleotides to the 3’ of polynucleotides

Question 16

Ligase

Answer 16

Creates phosphodiester bonds between Okazaki fragments to seal gaps

Question 17

Single-stranded binding proteins

Answer 17

Bind to exposed nucleotides of separated parent strands to prevent reannealing

Question 18

Topoisomerase

Answer 18

Unwinds supercoils introduced by helicase’s activity

Question 19

DNA pol III and DNA pol δ have high processivity because they are associated with a ____ that keeps them anchored to the template strand.

Answer 19

Sliding clamp

Question 20

DNA pol III and DNA pol δ both have 3’ → 5’ _______ activity

Answer 20

exonuclease

Question 21

the only polymerase that lays down its own primer

Answer 21

DNA pol α

Question 22

DNA pol III and DNA pol δ operate on both the leading and lagging strands

Answer 22

True

Question 23

Quickly dissociating and reassociating with a template strand defines low ____

Answer 23

processivity.

Question 24

Which DNA polymerase has both 3’ → 5’ and 5’ → 3’ exonuclease activity?

Answer 24

DNA pol I uses its 5’ → 3’ exonuclease activity to remove RNA primers and its 3’ → 5’ exonuclease activity for proofreading.

Question 25

DNA pol III and DNA pol δ have 3’ → 5’ exonuclease activity for proofreading.

Answer 25

True

Question 26

DNA pol α and DNA pol ε have no exonuclease activity.

Answer 26

True

Question 27

DNA pol I replaces the RNA primer with DNA nucleotides in _____

Answer 27

prokaryotes

Question 28

DNA pol I replaces the RNA primer with DNA nucleotides in _____

Answer 28

prokaryotes

Question 29

DNA pol α initiates polymerization in _____

Answer 29

eukaryotes

Question 30

Flap endonuclease trims the displaced RNA primer in eukaryotic lagging strand processing.

Answer 30

True

Question 31

______ separates the two strands of a DNA helix.

Answer 31

Helicase

Question 32

Histone ____ binding is required for higher order folding of chromatin (as seen in heterochromatin)

Answer 32

Histone H1

Question 33

Nucleosome:

Answer 33

Basic unit of organization of DNA around a histone octamer

Question 34

Binding of histone H1 coils the nucleosomes into higher order and is called what?

Answer 34

Chromatin

Question 35

Interphase chromosome

Answer 35

Nucleosomes arranged into chromatin: for transcription, replication, repair, etc.

Question 36

Mitotic chromosome

Answer 36

Mitotic chromosome: Chromatin folded into highly condensed metaphase chromosomes for cell division

Question 37

Heterochromatin

Answer 37

Condensed chromatin Transcriptionally inactive Eg: Barr body in female cells represents inactive X chromosome, centromere, telomere Types: Constitutive heterochromatin: - Always condensed (inactive)

Question 38

Euchromatin

Answer 38

Euchromatin: - Dispersed chromatin - Transcriptionally active genes

Question 39

When does DNA Replication happen?

Answer 39

during S phase

Question 40

Telomerase

Answer 40

Eukaryotes only. A reversetranscriptase (RNA-dependent DNA polymerase) that adds DNA(TTAGGG) to 3′ ends of chromosomes to avoid loss of genetic material with every duplication.

Upregulated in progenitor cells and also often in cancer; downregulated in aging and progeria.

TelomeraseTAGsforGreatnessandGlory.

Question 41

Origin of replication

Answer 41

Particular consensus sequence in genome where DNA replication begins. May be single(prokaryotes)or multiple(eukaryotes).

AT-rich sequences (such as TATA box regions) are found in promoters and origins of replication.

Question 42

Replication fork

Answer 42

Y-shaped region along DNA template where leading and lagging strands are synthesized.

Question 43

A biologist has identified a new chemical from a fungus that has new antibacterial effects. Exposure of bacterial cells to the chemical interrupts DNA replication and the RNA remains as Okazaki fragments.

Which of the following would most likely be inhibited by the compound?

Answer 43

DNA polymerase I - This is the correct answer. DNA polymerase I in bacteria contains the 5’-3’ exonuclease that removes the primer from Okazaki fragments and replaces RNA with DNA, allowing the ligase to seal the fragments. DNA polymerase II is believed to function in DNA repair in bacteria.

TAKEAWAY: DNA polymerase I in bacteria contains the 5’-3’ exonuclease that removes the primer from Okazaki fragments and replaces RNA with DNA, allowing the ligase to seal the fragments.

Question 44

A researcher is studying DNA replication and the function of various enzymes during the process. In recreating the experiment of Meselson and Stahl, he re-establishes the necessity of one enzyme commonly used in the lagging strand while only being used once in the leading strand.

Which of the following eukaryotic enzymes satisfies the above observation?

Answer 44

Ligase – This is the correct answer. Since the lagging strand is generated away from the replication fork, it generates numerous Okazaki fragments. These fragments are joined together by DNA ligase. This is not the same for the leading strand. Since that is generated as one long strand of DNA, it uses ligase once after the initial primer is removed.

*The experiment done by Meselson and Stahl demonstrated that DNA replicated semi-conservatively, meaning that each strand in a DNA molecule serves as a template for synthesis of a new, complementary strand.

Question 45

Polymerase beta

Answer 45

Polymerase beta functions to repair errors in DNA replication, particularly in nucleotide excision repair. After establishing the template strand, it repairs the daughter strand which could be from the leading or lagging strand. As such it has equal utilization by both strands of the DNA after replication is complete.

Question 46

Polymerase gamma

Answer 46

Strictly used for mitochondria, polymerase gamma is utilized for replication and proofreading.

Question 47

Single stranded binding protein

Answer 47

This protein functions to keep the two strands from joining back together after the helicase enzyme separates the double stranded DNA molecule.

Question 48

Topoisomerase

Answer 48

Topoisomerase – This is not the correct answer. Topoisomerase functions to stabilize the supercoiling that occurs as the DNA double helix is unwound. As such it is utilized by either strand of the DNA.

Question 49

A 27-year-old male presents to the ER with the flu despite having been vaccinated that year. The physician explains that his longstanding history of HIV infection has probably played a role in making him more susceptible for various infections the body usually could protect against. HIV is a unique virus in that it utilizes a reverse transcriptase that has RNA-dependent DNA activity. Although the virus carries its genetic material in the form of ssRNA, it utilizes its reverse transcriptase to synthesize viral DNA following infection of a host cell.

Which of the following enzymes in DNA replication functions in a similar capacity?

Answer 49

Telomerase – This is the correct answer. Telomerase is a reverse transcriptase or RNA-dependent DNA polymerase. As such it adds DNA nucleotides to the 3’ end of chromosomes in order to avoid the loss of genetic material that occurs with every duplication. In humans specifically, the telomeric DNA is made up of numerous tandem repeats of the sequence TTAGGG.

TAKEAWAY: Telomerase is an essential enzyme utilized by only eukaryotes to prevent the loss of genetic material with every replicative cycle. It does this by utilizing its RNA-dependent DNA polymerase activity to add DNA to the 3’ end of the chromosomes.

Question 50

Polymerase alpha

Answer 50

Eukaryotic equivalent to the primase, adding an initial RNA sequence to allow for strand elongation. It does not contain RNA-dependent DNA activity.

Question 51

Topoisomerase 2

Answer 51

Topoisomerase 2 – This is not the correct answer. An enzyme that utilizes ATP to relieve both negative and positive supercoils by making transient breaks in both DNA strands. It does not contain RNA-dependent DNA activity.

Question 52

DNA replication in eukaryotic cells produce _____ discontinuously from the single lagging strand of double-stranded DNA; DNA replication of the leading strand is continuous and does not involve the production of _______. Defects in the formation of ______ are thought to be common sources of the trinucleotide repeat expansion mutations causing more than 40 heritable neuromuscular and neurodegenerative conditions such as Huntington’s disease. In summary, the mutation rate is elevated in _______ because the lagging strand is replicated discontinuously; whereas, the leading strand is synthesized continuously. The more complex answer is that DNA polymerase alpha (Pol-alpha) is the primase and it lacks 3’ exonuclease activity for proofreading errors.

Answer 52

Okazaki fragments

Question 53

Homologous recombination is a

Answer 53

Homologous recombination is a double-strand break repair mechanism based on using a complementary strand of DNA as a template

Question 54

Base excision repair is a

Answer 54

Base excision repair is a single-strand break repair mechanism based on using a complementary strand of DNA as a template;

Question 55

Nonhomologous end-joining is

Answer 55

Nonhomologous end-joining is the DNA repair mechanism that is most susceptible to frequent mutation errors due to nucleotide insertions and deletions.

The ability of mitosis and environmental factors (e.g., ionizing radiation, carcinogens, mutagens, toxic drugs, THC-marijuana smoking, etc.) to induce targeted DNA double-strand breaks (DSBs). The subsequent repair of chromosomal DSBs by the cell can be classified into two categories of repair pathways: non-homologous end joining (NHEJ) and homology-directed repair (HDR). At its core, NHEJ-break ends can be ligated without a homologous template, whereas HDR-breaks requires a template to guide repair. NHEJ is a very efficient repair mechanism that is most active in the cell. It is also susceptible to frequent mutation errors due to nucleotide insertions and deletions (indels). HDR is considered the dominant mechanism for precise DSB repair, but suffers from low efficiency as it requires higher sequence similarity between the severed and intact donor strands of DNA. There are fewer errors or chances of mutations if the DNA template used during repair is identical to the original undamaged DNA sequence.

Question 56

The 5′–3′ exonuclease activity of DNA polymerase moves ahead, and is used to remove RNA primers from newly synthesized ___

Answer 56

DNA

Question 57

A research investigator is studying the molecular components of tumorigenesis. Analysis of a cell line that rapidly transforms into a tumor cell line demonstrated an increased mutation rate within the cell. Further analysis indicated that there was a mutation in the DNA polymerase enzyme that synthesizes the leading strand.

This inactivating mutation is likely to be in which of the following activities of this DNA polymerase?

Answer 57

3′–5′ exonuclease activity – This is the correct answer. DNA polymerase has an error checking capability which enables it to remove the mispaired base before proceeding with the next base insertion. This is due to the 3′–5′ exonuclease activity of DNA polymerase by which, prior to adding the next nucleotide to the growing DNA chain, the base put into place in the previous step is examined for correct base-pairing properties. If it is incorrect, the enzyme goes “backwards” and removes the incorrect base, then replaces it with the correct base.

TAKEAWAY: DNA polymerase sometimes makes mistakes when inserting bases into a newly synthesized strand and base-pairing with the template strand. However, mistakes do occur at a frequency of about one in a million bases synthesized. DNA polymerase and it’s 3′–5′ exonuclease activity helps to correct these mistakes.

Question 58

Glucose + glucose =

Answer 58

Maltose

Question 59

Glucose +galactose=

Answer 59

Lactose

Question 60

Glucose + fructose=

Answer 60

Sucrose

Question 61

Genomic instability

Answer 61

Genomic instability refers to increased propensity for mutations, including DNA base changes and structural alterations, in the genomic DNA. People with deficient DNA repair mechanisms, or those with increased exposure to sources of genetic damage, will have increased genomic instability.

Question 62

Exogenous sources involve influences from our physical environment

Answer 62

(eg, UV light, ionizing radiation, or carcinogens).

Question 63

Endogenous sources

Answer 63

involve unintended consequences of metabolic processes (eg, oxidation, nitrosylation, or hydrolysis of DNA strands). Errors in DNA replication are another endogenous source of DNA lesions.

Question 64

How can pyrimidine dimers be created and why are they significant?

Answer 64

UV light creates pyrimidine dimers. If left uncorrected, they will interfere with DNA replication and transcription.

Question 65

A ______ is an alteration in a single nucleotide.

Answer 65

point mutation

Question 66

Introns are flanked by specific sequences (usually 5’-GU… AG-3’) that serve as splicing sites to be used by spliceosomes for identification. If the “end sequence” (-AG), technically known as the splicing acceptor site, is altered, the spliceosome will scan for the next “end sequence” in the pre-mRNA. This will result in an entire exon being removed from the final mRNA.

Answer 66

True

Question 67

When a single nucleotide or any nontriplet DNA segment is deleted or inserted within the coding region of a gene, a _______ usually occurs.

Answer 67

frameshift mutation

Question 68

How are point mutations in coding regions of DNA characterized?

Answer 68

Point mutations are characterized based on how they alter the codon sequence in mRNA.

Question 69

What is direct repair?

Answer 69

Direct repair attempts to correct damaged nucleotides without severing the polynucleotide’s phosphodiesterase backbone.

Question 70

What are some common types of lesions BER corrects?

Answer 70

Deamination, depurination, and uracil incorporation into DNA.

Question 71

Deficiency of NER activity results in a disorder called xeroderma pigmentosum. Individuals with xeroderma pigmentosum have extreme photosensitivity, developing severe sunburn from only a few minutes of sunlight exposure. Because they are unable to correct UV damage, individuals who lack NER have a significantly increased risk of skin cancer. There is a variant form of xeroderma pigmentosum that results from a deficiency of a specific type of DNA polymerase. DNA polymerase η (pol η) is a unique polymerase in that it can still accurately replicate DNA in the presence of pyrimidine dimers. Pol η isn’t a repair mechanism as much as a means of circumventing the structural defects caused by UV light.

Answer 71

True

Question 72

How does NER differ from BER?

Answer 72

NER can recognize a broader range of single-strand lesions and removes an oligonucleotide instead of excising a single nucleotide.

Question 73

Defective mismatch repair is a key etiology of Lynch syndrome, formerly called hereditary nonpolyposis colorectal cancer (HNPCC). Individuals with Lynch syndrome have significantly increased risk of cancer within the gastrointestinal tract.

Answer 73

True

Question 74

Homologous recombination (HR) uses a homologous chromosome as a template to process the severed strands. When cellular machinery is able to detect two homologous double-strand DNA strands, it will attempt to reconnect the complementary strands without the loss of any nucleotide sequences.

Answer 74

True

Question 75

Individuals with BRCA1 mutations experience defective HR and as a result are at elevated risk of developing _____ and _____

Answer 75

breast and ovarian cancers.

Question 76

_______ is a mechanism that uses a special ligase complex to directly fuse the ends of two DNA fragments. When HR isn’t possible, due to the degree of damage sustained by the DNA or the current status of the cell cycle, the higher priority for the cell is to use NHEJ to reconnect the severed strands, even at the risk of jeopardizing some genetic integrity.

Answer 77

Nonhomologous end joining (NHEJ)

Question 78

Why is HR preferred over NHEJ?

Answer 78

Because NHEJ connects any two severed strands it can find (ie, the strands don’t have to be homologous), it is error-prone and can result in mutations.

Question 79

How do HR and NHEJ differ?

Answer 79

NHEJ is error-prone; HR restores nucleotide sequence. HR primarily occurs during S and G2 phases; NHEJ can occur in any phase.

Question 80

Point mutations in coding regions (silence, missense, and nonsense mutations) are characterized by their influence on the codon sequence of processed mRNA and therefore protein structure and function.

Answer 80

True

Question 81

Point mutations in coding regions (silence, missense, and nonsense mutations) are characterized by their influence on the codon sequence of processed mRNA and therefore protein structure and function.

Answer 81

True

Question 82

_______ result from an insertion or deletion of nucleotides that distorts the entire downstream codons.

Answer 82

Frameshift mutations

Question 83

A patient is found to have inherited a defective protein involved in homologous recombination repair of double-strand breaks. What pathology is the patient at increased risk for developing later in life?

Answer 83

The correct answer is breast or ovarian cancer. The patient has inherited a BRCA1 mutation, an etiology of defective homologous recombination repair, predisposing her to breast and ovarian cancer.

Question 84

________ is a condition associated with defective nonhomologous end joining.

Answer 84

Ataxia telangiectasia

Question 85

Single-nucleotide polymorphisms in nicotinic acetylcholine receptor genes may predispose to the development of lung cancer

Answer 85

True

Question 86

_______ is associated with defective mismatch repair.

Answer 86

Lynch syndrome

Question 87

_________ is associated with defective nucleotide excision repair.

Answer 87

Xeroderma pigmentosum

Question 88

______ is an error-prone repair mechanism because it does not require homology to correct double-stranded breaks.

Answer 88

NHEJ

Question 89

Researchers investigating epidermal basal cell dysplasia are quantifying the length of noncoding, short repetitive regions of DNA at the ends of linear chromosomes. Which of the following best describes the function of these sequences?

Answer 89

Noncoding, repetitive nucleotide sequences (eg, TTAGGG) found on the ends of linear chromosomes most closely describes telomere, which maintain chromosomal integrity. Telomeres do this by protecting the more important coding regions of chromosomes from damage or loss of genetic material. Each cycle of DNA replication leads to the loss of some genetic material, because DNA is synthesized in a 5’ to 3’ direction and requires a primer to get started. Thus there will always be a small region of DNA covered by the primer on the lagging strand that will not be replicated, since there is no room for a primer to precede it and initiate replication (depicted in the image).