Chapter 25: DNA Damage, Repair, & Homologous Repair

Question 1

Is DNA replication semiconservative or conservative?

Answer 1

DNA replication is semiconservative, that is, one strand of the newly synthesized DNA double helix is from parent molecule (template) and the other strand is newly synthesized and complementary to the template.

Question 2

When and by whom was the semiconservative nature of DNA replication elegantly demonstrated? What methods did they use in their experiment?

Answer 2

in 1958 by Matthew Meselson and Franklin Stahl.

Question 3

What is conservative replication?

Answer 3

in conservative replication, the parental DNA would remain intact and both strands of the daughter duplex would be newly synthesized).

Question 4

In which direction does DNA polymerase catalyze the synthesis of DNA chain?

Answer 4

DNA polymerases extend DNA strands only in 5’ to 3’ direction

Question 5

Replication Is Semidiscontinuous. Describe how this affects the strands.

Answer 5

Because DNA polymerases extend DNA strands only in 5’ to 3’ direction, the lagging strand is discontinuously synthesized in the opposite direction of the replication fork and forms Okazaki fragments.

Question 6

Okazaki fragments are resulted from? In which strand?
How are they eventually joined together?

Answer 6

semidiscontinuous replication of DNA.

Lagging strand

The Okazaki fragments are later covalently joined together by the enzyme DNA ligase.

Question 7

What is needed to initiate chain extension? How does this affect both strands?

Answer 7

RNA primers, they are made from Primases

Only one priming event is required to initiate the synthesis of the leading strand, but multiple priming events are required for lagging strand synthesis.

The RNA primers are eventually replaced with DNA.

Question 8

DNA replication occurs unidirectionally or bidirectionally in a cell?

Answer 8

DNA replication is bidirectional from the origin of replication

Question 9

Which strand in a DNA double is used as a template for the leading strand synthesis?

Answer 9

each of the two strands that make up the double helix serves as a template from which new strands are copied.

Question 10

The directions of DNA synthesis of the leading strand and lagging strand relative to the motion of replication fork.

Answer 10

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

Question 11

Where does replication occur?

Answer 11

Replication forks

Question 12

what kind of mutations are generally caused by intercalating agents?

Answer 12

Insertion/Deletion Mutations

Chemical mutagens acting as base analogs can also cause this type of mutation

Question 13

Maintaining low rates of mutation is essential for?

Answer 13

species perpetuation

Question 14

High rates of mutation affect germ line and somatic cells?

Answer 14

High rates of mutation in the germ line would destroy the species.
High rates of mutation in the somatic cells would destroy the individual (for example, cancer).

In multicellular organisms, genetic changes are usually notable only when they occur in germline cells so that the change is passed on to all the cells of the organism’s offspring. Damage to the DNA of a somatic cell, in contrast, rarely has an effect beyond that cell unless the mutation contributes to a malignant transformation (cancer).

Question 15

Genetic mutation is a driven force of ?

Answer 15

evolution, so life and biodiversity depends on a happy balance of between mutation and its repair.

Question 16

Difference between DNA and mutations?

Answer 16

although both are types of error in DNA. DNA damage is an abnormal chemical structure in DNA, while a mutation is a change in the sequence of base pairs

Question 17

DNA is susceptible to damage from what two sources?

Answer 17

DNA can also be chemically altered by agents that are naturally present in the cell or in the cell’s external environment.
DNA damage can occur naturally or via environmental factors,

Question 18

Even when damaged DNA can be mended, the restoration may be imperfect, producing a ?

Answer 18

mutation, a heritable alteration of genetic information

Question 19

What two things generate mutations? describe

Answer 19

Environmental and Chemical Agents Generate Mutations

Environmental agents such as ultraviolet light, ionizing radiation, and certain chemical agents can physically damage DNA

Question 20

What Environmental agents generate mutations?

Answer 20

Environmental agents such as ultraviolet light, ionizing radiation, X rays

UV radiation causes fusion of two pyrimidines called a thymine dimer which causes DNA replication error when the DNA polymerase ceases during replication.
Gamma radiation and X-rays cause double-strand breaks which are difficult to repair.
Ionizing radiation also damages DNA either through its direct action on the DNA molecule or indirectly by inducing the formation of free radicals,This can lead to strand breakage.

“thymine dimers formed by UV light”

Question 21

What are the two classes of DNA damage produced by chemical mutagens?

Answer 21

1. Point mutations, in which one base pair replaces another. These are subclassified as:

(a) Transitions, in which one purine (or pyrimidine) is replaced by another.
e.g. A to G and T to C.

(b) Transversions, in which a purine is replaced by a pyrimidine or vice versa.
e.g. A to C or T and T to G or A.

2. Insertion/deletion mutations, in which one or more nucleotide pairs are inserted in or deleted from DNA. These are collectively known as indels.

Answer 22

Replication error
Mismatches can be generated and mutations can become permanent by replication.
Proofreading increases the fidelity of DNA replication by ~100-fold.
Some replication errors can escape proofreading.
The replication errors will become permanent mutations during a second round of replication.
DNA repair has to be done before the next replication occurs to prevent the formation of a permanent mutation.

Question 23

When must replication repair occur?

Answer 23

Mismatches can be generated and mutations can become permanent by replication.
Proofreading increases the fidelity of DNA replication by ~100-fold.
Some replication errors can escape proofreading.
The replication errors will become permanent mutations during a second round of replication.
DNA repair has to be done before the next replication occurs to prevent the formation of a permanent mutation.

Question 24

Chemical mutagens can cause deamination.
Deamination of cytosine changes C to which base and cause what kind of mutation after replication?

Answer 24

Nitrous Acid (HNO2), which acts as a mutagen by deamination (removal of the NH 2 group) of adenine and/or cytosine to an ether group, thus altering their base pairing.
Deaminating agents, for example nitrous acid which can cause transition mutations by converting cytosine to uracil.

Deamination of cytosine generates uracil which preferentially pairs with adenine, resulting in G:C to A:T transition.

Deamination converts adenine to hypoxanthine which pairs with cytosine rather than thymine, resulting in a A:T to G: C transition.

Deamination of guanine generates xanthine which continues to pair with cytosine but with just two hydrogen bonds.

-Point Mutation

Question 25

Another deamination, of the modified base methylcytosine, can also lead to a mutation upon replication. Some cytosines may be methylated as part of a regulatory process to inactivate certain genes in eukaryotes, or in prokaryotes as protection against restriction endonucleases. When the methylated cytosine is deaminated, it produces a?
Deamination of methylated cytosine changes it into which base?

Answer 25

When the methylated cytosine is deaminated, it produces a thymine, which changes the complementary nucleotide (upon replication) from a guanine to an adenine

Question 26

Why are methylated cytosines hotspots for C to T transition?

Answer 26

In vertebrate, methylated cytosines result in gene silencing and also are hotspots for C to T transition. This is because hydrolysis to deaminate methylated cytosine generates a natural base thymine, which can not be recognized by repair system.

Some cytosines may be methylated as part of a regulatory process to inactivate certain genes in eukaryotes, or in prokaryotes as protection against restriction endonucleases

Question 27

What is resulted from hydrolysis of the glycosidic bond in DNA molecule?

Answer 27

Alkylation of the N7 position of a purine nucleotide renders its glycosidic bond susceptible to hydrolysis, leading to loss of the base and abasic site formation.

Question 28

This process is caused by spontaneous hydrolysis of a glycosidic bond.`

Answer 28

depurination

Question 29

What kind of mutation can be caused by alkylation at the oxygen atom of guanine?

Answer 29

G:C to A:T transversion

Guanine when alkylated may be mispaired with thymine

which is one of the most common mutations found in human cancer.

Question 30

What kind of mutation can be caused by the incorporation of thymine analog 5-bromouracil?

Answer 30

Transition mutations by mispairing with guanine (G) rather than pairing with adenine (A) during replication
A:T to G:C

Question 31

Any replicational mispairing that has eluded the editing functions of the DNA polymerases may still be corrected by a process known as ?

Answer 31

mismatch repair (MMR).

Mismatch repair system: detecting and repairing mismatches, including MutS, MutL, MutH, UvrD, exonuclease, DNA polymerase, DNA ligase.

Question 32

Which protein is responsible to detect the distortion in DNA caused by mismatches?

Answer 32

MutS (mutant scan)

Question 33

The mutations that can be directly removed and the mechanisms involved.

Answer 33

Direct Methyl Group Removal
O6-methylguanine resulted from alkylation of the oxygen of carbon atom 6 of guanine base pairs with thymine, causing G:C to A:T change when damaged DNA is replicated.
A methyltransferase can directly remove the methyl group from guanine residue by transferring it to one of its own cysteine residues.

Photoreactivation
UV irradiation causes pyrimidine dimer (thymine dimer showed).
DNA photolyase captures energy from light and uses it to break the covalent bonds linking adjacent pyrimidine.

Question 34

Which protein nicks the newly synthesized strand at the location of mismatches, and how does this protein recognize newly synthesized strand?

Answer 34

MutH binds at hemimethylated sites, is activated by interaction with MutL and MutS, and selectively nicks the unmethylated daughter strand. Therefore, only the mismatched base-pair in the newly synthesized strand will be removed.

Question 35

Damaged bases that cannot be directly repaired may be removed and replaced in a process, discovered by Tomas Lindahl, known as?

Answer 35

base excision repair (BER).
This pathway, as its name implies, begins with removal of the damaged base.

Cells contain a variety of DNA glycosylases that each cleave the glycosidic bond of a corresponding type of altered nucleotide, leaving a deoxyribose residue with no attached base (Fig. 25-34). Such apurinic or apyrimidinic sites (AP or abasic sites) also result from the occasional spontaneous hydrolysis of glycosidic bonds.

Question 36

The enzymes of BER, which correct the most frequent type of DNA damage, include ?

Answer 36

a glycosylase that recognizes 8-oxoguanine and the enzyme uracil–DNA glycosylase (UDG), which excises uracil residues. The latter arise from cytosine deamination as well as the occasional misincorporation of uracil instead of thymine into DNA

Question 37

Describe Hydrolysis of the Glycosidic Bond by Glycosylase and the Enzymes that are involved in the repair of AP site.

Answer 37

A specific glycosylase recognizes and removes the uracil from the backbone by hydrolyzing the glycosidic bond to leave an AP site (apurinic or apyrimidinic).
The resulting abasic sugar is removed from the DNA backbone by endo- and exo-nucleolytic cleavage.
A repair DNA polymerase and DNA ligase fill the single nucleotide gap.
Base excision repair and nucleotide excision repair are two principle repair system.

Question 38

Describe Nucleotide Excision Repair: UvrABC Repair System in E. coli

Answer 38

In E. coli, NER is carried out in an ATP-dependent process through the actions of the UvrA, UvrB, and UvrC proteins
Nucleotide excision repair enzymes recognize distortions of DNA helix caused by DNA damage, e.g. thymine dimer.
-Repair system includes four major proteins: UvrA, UvrB, UvrC, and UvrD.
-UvrA and UvrB scan DNA to identify a distortion.
-UvrA exits from the complex and UvrB melts DNA to create a single-stranded bubble around the lesion.
-UvrC is recruited to the complex by UvrB to create two incisions: one on the 5’ side and one on the 3’ side.
-DNA helicase UvrD releases the single-stranded fragment.
-The gap is filled by the action of DNA polymerase and DNA ligase.

Question 39

What is the SOS response system

Answer 39

“SOS response system” refers to the mechanism in which an organism initiates the production of activator protein (RecA), which results in the dissociation/or destruction of LexA repressor a repressor of SOS response genes (e.g. translesion polymerase, UvrA, UvrB, RecB, RecC, RecD) which activates the SOS inducer proteins
(e.g. transletion polymerases)Finally, Error-prone repair of damaged DNA

SOS repair refers to the DNA repair system, which uses RecA regulatory protein to inhibit the repressor’s activity and activate the SOS inducer genes to recover the DNA damage. But in mutated DNA, the inactivation of LexA repressor becomes necessary to induce the expression of SOS genes.

Question 40

During SOS response, binding of RecA onto single-stranded DNA activates or represses RecA activity?

Answer 40

activates

Question 41

RecA is a dual function protein on which two cellular processes?

Answer 41

RecA is a dual function protein on both SOS response and recombination)

Question 42

LexA is a activator or repressor of SOS response genes?

Answer 42

repressor

Question 43

What is Homologous Recombination

Answer 43

physical exchange of DNA sequences between identical or nearly identical regions of DNA molecules. I Crossing-over between two homologous chromosome during meiosis is one of the results of homologous recombination.
In HR, DNA strands cross over ti exchange places. Hr can damage replication forks and double strand breaks
Involves many proteins
Homologous recombination is a type of genetic recombination in which nucleotide sequences are exchanged between two similar or identical molecules of DNA

Question 44

Although two DNA molecules have high degree of similarity or identical, the DNA molecules can have small regions of sequence difference, known as?

Answer 44

alleles of the same gene.

Question 45

Key steps of homologous recombination

Answer 45

1. Alignment of two homologous DNA molecules
2. Introduction of breaks in the DNA. The breaks may occur in one DNA strand or both DNA strands.
   3.Formation of initial short regions of base pairing between the two recombining DNA molecules. A single stranded DNA from one parental DNA base pairs with its complementary strand in the homologous DNA duplex. This step is called strand invasion. Strand invasion results in crossing DNA strands. This crossing structure is called a Holliday junction.
3. Movement of the Holliday junction known as branch migration.
4. Cleavage of the Holliday junction. The process of cutting the DNA strands within the Holliday junction to regenerate two separate duplex DNA molecules is called resolution.

Question 46

Two ways to resolve the Holliday Junction

Answer 46

1. Cut in 2 unbroken DNA strands (did not cross over), makes a crossover/splice product
2. Cut in 2 broken DNA strands (crossed over) to initiate recombination results

Question 47

Homologous recombination requires which four proteins?

Answer 47

RecBCD, RecA, RuvAB, and RuvC.

Question 48

Homologous recombination requires which four proteins?

Answer 48

RecBCD, RecA, RuvAB, and RuvC.

Question 49

RecB, RecC, and RecD proteins. These proteins are products of ?

Answer 49

SOS genes

Question 50

RecBCD function?

Answer 50

RecBCD Initiates Recombination by Making a Single Strand Available, also facilitates RecA loading onto ssDNA

RecBCD unwinds duplex DNA in a ATP-driven process and nicks the resulting single-stranded DNA, generating single-stranded DNA tail in the homologous recombination site.

RecBCD enzymes help load the RecA, a strand-exchange protein onto the single-stranded DNA ends.

The RecBCD enzyme is both a helicase that unwinds, or separates the strands of DNA, and a nuclease that makes single-stranded nicks in DNA.
If the DNA contains a Chi site, RecBCD nicks the DNA and loads RecA protein onto the newly generated 3′ end
Once encounters the Chi site, it stops cleaving 3’-end, but continues cleaving 5’-end thus producing a 3’-end single-strand DNA

Question 51

RecA promotes which step in the homologous recombination?

Answer 51

is a strand-exchange protein.
The RecA protein promotes pairing between a single-stranded DNA molecule and a homologous sequence in another DNA molecule

These proteins involves both the search for sequence matches between two DNA molecules and the generation of the regions of base pairing.
RecA has two distinct DNA-binding sites: a primary site bound by ssDNA and a secondary site to be occupied by dsDNA.

assembly of a presynaptic filament of RecA on the ssDNA of the broken chromosome, which in turn uses the ssDNA sequence to search for a homologous region in the dsDNA genome.
The RecA monomer contains two DNA binding sites in the large central domain, one for binding ssDNA, and the other for binding duplex DNA.

Question 52

RuvAB complex promotes which step in the homologous recombination? RuvA and RuvB functions?

Answer 52

RuvAB Complex Recognizes Holliday Junction and Promotes Branch Migration
RuvABC Mediates Branch Migration and the Resolution of the Holliday Junction.
RuvA is a specific Holliday junction DNA-binding protein that recognizes the structure of the DNA junction, regardless of its DNA sequence.

RuvA recuits RuvB to the Holliday junction.
RuvB is a hexameric ATPase that provides the energy to drive the movement of the Holliday junction by exchange of base pairs.

Question 53

RuvA binding to the Holliday junction belongs to sequence-specific or nonsequence specific protein-DNA interaction?

Answer 53

nonsequence specific

Question 54

What is the function of RuvC?

Answer 54

RuvC Cleaves the Specific DNA Strands at the Holliday Junction to Finish Recombination
The final stage of homologous recombination is the resolution of the Holliday junction into its two homologous dsDNAs. This process is carried out by RuvC, RuvA–Holliday junction complex to cleave oppositely located strands at the Holliday junction. The resulting single-strand nicks are then sealed by DNA ligase.

Question 55

How can DNA Can Be Repaired by Recombination

Answer 55

Recombination Repair Reconstitutes Double-Strand Breaks.
Homologous Recombination repairs damaged replication fork
-homology-directed repair (HDR) or homologous end-joining of DSB

Question 56

CRISPR-Cas9?

Answer 56

a System for Editing and Regulating Genomes

Question 57

What are Transposons

Answer 57

Transposons are genetic elements that move within a genome, often by mechanisms involving their replication.

Question 58

What is the difference between homologous recombination and transposition?

Answer 58

Transposition does not require homology between donor and recipient DNA