Exam 2: Chp 25 and 26

Question 1

What is the trombone model in prokaryotes?

Answer 1

In order for the replisome to move as a single unit in the 5′ → 3′ direction along the leading strand, the lagging strand template must loop around

Question 2

Why DNA polymerases are called processive enzyme?

Answer 2

DNA polymerases are processive enzymes because they catalyze addition of 20 or more nucleotides to the growing DNA chain without releasing the single-stranded template.

Question 3

What promotes the processivity of Pol III? Why?

Answer 3

The sliding clamp (β subunit)

binds tightly to DNA polymerase and it keeps DNA polymerase attached to the template,keeping the Pol III holoenzyme from diffusing away.

must beinstalled around the lagging strand template about once every second

the clamp protein binds DNA polymerase and prevents this enzyme from dissociating from the template DNA strand.

a ring-shaped protein that encircles duplex DNA, binds to the DNA polymerase and tethers it to the DNA template,

Sliding DNA clamp is composed of multiple identical subunits that assemble in a shape of a “doughnut”.
The opening through the center of the sliding clamp is large enough to encircle double DNA helix and allow a layer of one or two water moleculesbetween the DNA and the proteins, making the clamp sliding along the DNA easily.
The clamp encircles the newly synthesized DNA and binds tightly to DNA polymerase associating with primer:template junction at the fork, which increases the processivity of Pol III from 10 -15 residues to more than 5000 residues

Question 4

What is The 𝛄 complex of the Pol III holoenzyme (subunit composition γ τ 2 δδ′ χψ ) is also known as the clamp loader?

Answer 4

it opens the dimeric β clamp to load it onto the DNA template in an ATP-dependent manner. Once the β clamp has been loaded onto the DNA, the Pol III core binds to the β clamp more tightly than does the γ complex, thereby displacing it and permitting processive DNA replication to occur. W

Question 5

By which enzyme are the nicks in the lagging strand sealed?

Answer 5

The nicks in the lagging strand are sealed through the action of DNA ligase.
This enzyme seals a broken phosphodiester bond

E. coli DNA ligase uses NAD+ or ATP
done through the coupled hydrolysis of either NAD + to nicotinamide mononucleotide (NMN+) + AMP, or ATP to PP i+ AMP.

Question 6

By which enzyme are the RNA primers removed and the gaps filled?

Answer 6

The β clamp, which remains around the completed Okazaki fragment, recruits DNA Pol I and DNA ligase to replace the RNA primer on the previously synthesized Okazaki fragment with DNA and seal the remaining nick.

Question 7

The structure of the E. coli replication terminus.

Answer 7

The E. coli replication terminus is a large (350-kb) region flanked by ten nearly identical nonpalindromic ∼23-bp terminator sites, TerH, TerI, TerE, TerD, and TerA on one side and TerJ, TerG, TerF, TerB, and TerC on the other

Question 8

Replication fork traveling counterclockwise passes through ___ and stops encountering ___? Backup sites?

Answer 8

TerJ, TerG, TerF, TerB, and TerC but stops on encountering either TerA, TerD, TerE, TerI, or TerH (TerD, TerE, TerI, and TerH are presumably backup sites for TerA).

Question 9

Replication fork traveling clockwise passes through ___ and stops encountering ___. Backup sites?

Answer 9

TerH, TerI, TerE, TerD, and TerA
and but halts at TerC or, failing that, TerB, TerF, TerG, or TerJ.

Question 10

The arrest of replication fork motion occurs where?

Answer 10

at Ter sites that require the action of Tus protein.Tus protein specifically binds to a Ter site, where it prevents strand displacement by DnaB helicase, thereby arresting replication fork advancement.

Tus protein, which binds to Ter sequences to prevent replication forks from moving through the termination region. The Tus protein blocks replication by forming a tight association with a specific G-C base pair in the Ter sequences.

tus, a termination protein that is found in e coli and it binds to the ter sites to create a tus-ter complex which prevents helicase from moving. this will stall the replication fork from moving in one direction but not the other.

Question 11

After replication is complete, the two daughter DNA molecules remain linked together as catenates what are the final steps of prokaryote DNA replication?

Answer 11

Termination of DNA replication occurs when the two forks meet and fuse, creating two separate double-stranded DNA molecules.

these must be separated by the action of topoisomerases II.

The resulting catenated circular dsDNAs must be separated so that each can be passed to a different daughter cell.

Question 12

How are the 2 catenane DNA molecules separated?

Answer 12

Topoisomerase II (topoisomerase IV) separates the two circular DNA by breaking one double-stranded DNA molecule and passing a second double-stranded DNA through this break
at the replication termination site.

Question 13

Type II topoisomerases purpose?

Answer 13

Type II topoisomerases are essential for resolving topologically entwined double-stranded DNA.

separates the two circular DNA by breaking one double-stranded DNA molecule and
pass a second double-stranded DNA through this break.

Question 14

Prokaryote DNA replication has nearly perfect fidelity. (only one mismatch occurs per 10^8 to 10^10 base pairs) Why can organisms keep high fidelity of replication?

Answer 14

1. Cells maintains balanced levels of dNTPs
2. Only when the incoming dNTP base pairs with the template, the DNA polymerase undergoes conformational changes (catalytically inactive open conformation to closed active conformation) to catalyze the addition of the dNTP to the growing end of DNA chain
3. The proofreading functions of DNA polymerases (3’→5’exonuclease activity) detect and eliminate the occasional errors made by their polymerase function
4. DNA repair systems detect and repair errors in the newly synthesized
   DNA as well as any damage that may occur after its synthesis

The fidelity of a DNA polymerase refers to its ability to accurately replicate a template.

The high fidelity of DNA replication is achieved by the regulation of dNTP levels, by the low error rate of the polymerase reaction, by the requirement for RNA primers, by 3′ → 5′ proofreading, and by DNA repair mechanisms.

Question 15

What are the 3 main polymerases involve din replicating eukaryotic nuclear DNA and what are their activities?

Answer 15

polymerases 𝛂, 𝛅, and 𝛆
All these DNA polymerases catalyze DNA synthesis in the 5’→3’ direction.

(pol 𝛂),replicates DNA by extending a primer in the 5′ → 3′ direction under the direction of a ssDNA template. This enzyme tightly associates with primase indicating that it is involved in initiating DNA replication. It has no exonuclease activity (3’→5’) and therefore cannot proofread its polymerization product.

(pol 𝛅), does not associate with a primase and contains a 3′ → 5′ exonuclease active site. (it can replicate the entire length of a template DNA), but only when it is in complex with a sliding-clamp protein named proliferating cell nuclear antigen (PCNA)

(Pol 𝛆)d has a 3′ → 5′ exonuclease activity, doesn’t need PCNA

Question 16

By which eukaryotic DNA polymerase is the leading and lagging strand likely to be synthesized?

Answer 16

It was proposed that DNA polymerase ε catalyzes the synthesis of the leading strand (may also contribute to lagging strand synthesis) because of its high processivity without PCNA and essential for replication, whereas DNA polymerase α and δ cooperate to synthesize the lagging strand (δ may also participate in leading strand synthesis).

DNA polymerase ε is probably the leading strand replicase.

Question 17

Pol δ in complex with what is required for lagging strand DNA synthesis?

Answer 17

Pol δ in complex with PCNA is required for lagging strand DNA synthesis.

Question 18

What is the PCNA protein and it is structurally similar with what protein?

Answer 18

in complex with DNA polymerase δ to replicate entire length of template DNA. it forms a trimeric ring with almost identical structure (and presumably function) as the E. coli β2 sliding clamp

dentical to the sliding clamp (β subunit) of the Pol III holoenzyme and increases the processivity of DNA polymerases

Question 19

Processivites of eukaryotic DNA polymerases?

Answer 19

Pol α has low processivity, involved in initiation of replication.

Pol δ, only high processivity/infinite/unlimited when it is in complex with a PCNA protein.

Pol ε has has high processivity in the absence of PCNA

Question 20

DNA polymerase γ function?

Answer 20

occurs exclusively in the mitochondrion, where it presumably replicates the mitochondrial genome.

Question 21

What is reverse transcriptase (RT) and What is the template of reverse transcriptase?

Answer 21

Reverse transcriptase (RT) is an essential enzyme of retroviruses, which are RNA-containing eukaryotic viruses such as human immunodeficiency virus (HIV, the virus causing AIDS).
RT catalyzes DNA synthesis in the 5’→3’ direction from RNA template.
The viral RT uses its RNA as a template to synthesize a complementary DNA strand in a host cell, yielding an RNA-DNA hybrid helix.

Question 22

What kind of RNA can be destroyed by RNaseH?

Answer 22

The RNA strand of the DNA-RNA hybrid helix made by Viral RT

The RNA strand in the hybrid is then degraded by an RNase H, an RNase that recognize RNA in an RNA-DNA hybrid. The remaining DNA strand acts as a template for the synthesis of its complementary strand, yielding a double-stranded DNA that is then integrated into a host cell genome.

Question 23

RT has been a particularly useful tool in genetic engineering because it can transcribe mRNAs to?

Answer 23

to complementary strands of DNA (cDNA)

Question 24

Additional Enzymes Participate in Eukaryotic DNA Replication
What is MCM function?
What is RPA?
What is RNase H1?
What is FEN1?

Answer 24

MCM: heterohexameric helicase

RPA: replication protein A, a SSB protein

RNase H1: removes RNA primers

FEN1: flap endonuclease-1, removing the last residue of RNA primer and few nts that is synthesized by Pol α, functioning as a proofreading endonuclease for Pol α

Question 25

Eukaryotic chromosome contains single or multiple replication origins?

Answer 25

Multiple

Question 26

Meaning of origin of replication?

Answer 26

the specific sites at which DNA unwinding and initiation of replication occur.

Question 27

The definition of replicon.

Answer 27

all the DNA replicated from a particular origin. The single chromosome in E. coli has only one origin of replication, the entire chromosome is a single replicon. multiple origins exist in each eukaryotic chromosome and divide it into multiple replicons.
the entire region of DNA that is independently replicated from a single origin of replication

Question 28

What is the end replication problem during the replication of eukaryotic linear chromosome? how is it solved?

Answer 28

Goal: Finishing replication in linear chromosome ends known as telomeres

Problem:
-The leading strand can be extended to the extreme 5’ terminus of the template. However, theLagging strand can not be extended to the extreme end because after removal of RNA primer, DNA polymerase can not synthesized a complementary strand without a primer bound to the end providing 3’-OH.
In other words, The ends of linear chromosomes present a problem for the replication machinery. Specifically, DNA polymerase cannot synthesize the extreme 5′ end of the lagging strand.

-Consequently, in the absence of a mechanism for completing the lagging strand, DNA molecules would be shortened at both ends by the length of an RNA primer with each round of the replication. This would eventually lead to the loss of essential genetic information at the ends of the chromosome.

Solution:
Telomerase solves the end replication problem by extending the 3-end of the chromosome.

Question 29

The definition of telomere.

Answer 29

The ends of eukaryotic chromosomes,
Telomeres are generally composed of head-to-tail repeats of a TG-rich DNA sequence. For example, human telomeres consist of many 5’-TTAGGG-3’ head-to-tail repeats.
Telomerase solves the end replication problem by extending the 3-end of the chromosome.

Question 30

The components of telomerase and their roles?

Answer 30

Telomerase Extends Chromosome Ends
telomeric DNA is synthesized and maintained by an enzyme named telomerase, which is a ribonucleoprotein (a complex of protein and RNA).

Telomerases are ribonucleoproteins including both protein and RNA components-protein component has DNA polymerase activity extending the 3’-end of its DNA substrate, RNA component serves as the template for adding the telomeric sequence to the 3’ terminus at the end of the chromosome.

The telomerase enzyme attaches to the end of the chromosome; complementary bases to the RNA template are added on the 3′ end of the DNA strand.

The ends of the chromosomes pose a problem as polymerase is unable to extend them without a primer. Telomerase, an enzyme with an inbuilt RNA template, extends the ends by copying the RNA template and extending one end of the chromosome. DNA polymerase can then extend the DNA using the primer. In this way, the ends of the chromosomes are protected.
How does telomerase work? The enzyme binds to a special RNA molecule that contains a sequence complementary to the telomeric repeat. It extends (adds nucleotides to) the overhanging strand of the telomere DNA using this complementary RNA as a template. When the overhang is long enough, a matching strand can be made by the normal DNA replication machinery (that is, using an RNA primer and DNA polymerase), producing double-stranded DNA.

Question 31

If DNA damage is not corrected, what may happen? Even when damaged DNA can be mended, the restoration may be imperfect, producing a?

Answer 31

may alter the nucleotide sequences of genes.
mutation, a heritable alteration of genetic information

Question 32

What environmental and chemical agents generate mutations?

Answer 32

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

Question 33

The DNA damage produced by chemical mutagens, substances that induce mutations, falls into two major classes:

Answer 33

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.pyrimidine-to-pyrimidine and purine-to-purine substitution, e.g. A to G and T to C.

(b) Transversions, in which a purine is replaced by a pyrimidine or vice versa.pyrimidine-to-purine and purine-to-pyrimidine substitution, 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.

Question 34

What kind of treatments can cause thymine dimer?

Answer 34

UV radiation (200-300 nm) promotes the formation of a cyclobutyl ring between adjacent thymine residues on the same DNA strand to form an intrastrand thymine dimer

UV radiation causes fusion of two pyrimidines that occupy the adjacent positions on the same polynucleiotide chain. In the case of thymine, it is called thymine dime.Pyrimidine fusion in DNA strand are incapable of base-pairing and cause the DNA polymerase to cease during replication.

Question 35

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

Answer 35

Nitrous Acid (HNO2), a delaminating agent 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.

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

Nitrous acid oxidatively deaminates aromatic primary amines, thus can deaminate bases in DNA molecules.

Question 36

What kind of damage do Gamma radiation and X-rays cause?

Answer 36

double-strand breaks which are difficult to repair.

Question 37

Deamination of methylated cytosine changes it into which base?

Answer 37

In vertebrate, methylated cytosines result in gene silencing and also are hotspots for C to T transition.

transition mutation

Gene silencing is a method used to inactivate a specific gene to prevent its expression. In this way, genes are not entirely removed but are inactivated to reduce their expression by about 70%.

Question 38

Why are methylated cytosines hotspots for C to T transition?

Answer 38

This is because hydrolysis to deaminate methylated cytosine generates a natural base thymine, which can not be recognized by repair system.

(1) the higher rate of deamination of 5meC compared to C generates more T⋅G than uracil⋅G (U⋅G) mismatches, and (2) inefficient repair of T⋅G mismatches by the very short-patch (VSP) repair system compared to the repair of U⋅G mismatches by the uracil-DNA glycosylase (Ung).

Methylated cytosines deaminate at higher rates than unmethylated cytosines and the lesions they produce are repaired less efficiently. As a result, methylated cytosines are mutational hotspots.

Question 39

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

What causes the hydrolysis of the glycosidic bond in DNA?

Answer 39

producing an abasic site in the DNA./loss of a base

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.

DNA also undergoes depurination by spontaneous hydrolysis of the N-glycosol linkage, producing an abasic site in the DNA.

Question 40

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

Answer 40

Alkylation of guanine at the O6 atom is a highly mutagenic DNA lesion because it alters the coding specificity of the base causing G:C to A:T transversion mutations

O6-methylguanine generated by alkylation often mispairs with thymine, resulting in the change of a G:C base pair into an A:T base pair during DNA replication.

Transversion: G mispairs with T which changes G:C base pair to A:T base pair

Question 41

Oxidation of guanine catalyzed by reactive oxygen species generates ?

Answer 41

7,8-dihydro-8-oxoguanine or oxoG which can base pair with adenine as well as with cytosine, resulting in G:C to A:T transversion which is one of the most common mutations found in human cancer.

Question 42

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

Answer 42

T:C to A:G

Transversion

Question 43

what kind of mutations are generally caused by intercalating agents?

What are base analogs?

Answer 43

Insertion and deletion mutations

Base analog: structural similar to proper bases but differ in ways that make them treacherous to the cell.
Intercalating agent: flat molecules containing several polycyclic rings that bind to that equally flat purine or pyrimidine bases of DNA, just as the bases bind or stack with each other in the double helix. Intercalating agents cause insertion or deletion.

nsertion/deletion mutations (indels) may arise from the treatment of DNA with intercalating agents such as acridine orange or proflavin (Section 24-3B). The distance between two consecutive base pairs is roughly doubled by the intercalation of such a molecule between them. The replication of the distorted DNA occasionally results in the insertion or deletion of one or more nucleotides in the newly synthesized polynucleotide.

Question 44

Chemical mutagens acting as ___can cause insertion/deletion.

Answer 44

base analogs and intercalating agents

Question 45

If the chemical to be tested has high mutagenicity, what kind of result is to be expected in the Ames test?

Answer 45

Many mutagens are also carcinogens. Ames test, devised by Bruce Ames, is a rapid and effective bacterial assay for carcinogenicity that is based on the high correlation between carcinogenesis and mutagenesis.

Question 46

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

Answer 46

mismatch repair (MMR).

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

Question 47

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

Answer 47

MutS (mutant scan)

Question 48

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

Answer 48

MutH

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.

can distinguish the two DNA strands because newly replicated DNA remains hemimethylated until methyltransferases such as Dam methyltransferase have had sufficient time to methylate the daughter strand

One possibility is that a newly synthesized daughter strand is identified by its as yet unsealed nicks between Okazaki fragments.

Question 49

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

Answer 49

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 50

Damaged bases that cannot be directly repaired may be removed and replaced in a process, discovered by Tomas Lindahl, known as? What does it require?How does it function?

Answer 50

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

A specific glycosylase recognizes and removes the uracil from the backbone by hydrolyzing the glycosidic bond to leave an AP site (apurinic or apyrimidinic)./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
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.

Question 51

Enzymes that are involved in the repair of AP site.

Answer 51

The deoxyribose residue is then cleaved on one side by an AP endonuclease, the deoxyribose and several adjacent residues are removed by the action of a cellular exonuclease (possibly associated with a DNA polymerase), and the gap is filled in and sealed by a DNA polymerase and DNA ligase.

Question 52

The pathway that recognize distortions of DNA helix caused by DNA damage, e.g. thymine dimer.rather than by the recognition of any particular group? What does this pathway require to function?

Answer 52

Nucleotide Excision Repair

UvrABC Repair System in E. coli
UvrA, UvrB, and UvrC proteins

ATP-dependent process

Question 53

The process of UvrABC repair and the function of each components.

Answer 53

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 54

What is the SOS response system

Answer 54

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.

1. DNA damage (e.g. thymine dimer) causes single-stranded DNA
2. RecA protein binds to single-stranded DNA and is activated (RecA is a dual function protein on both SOS response and recombination)
3. Proteolytic destruction of LexA, a repressor of SOS response genes (e.g. translesion polymerase, UvrA, UvrB, RecB, RecC, RecD)
4. Activation of SOS response genes (e.g. transletion polymerases)
5. Error-prone repair of damaged DNA

It is a fail-safe mechanism that allows the replication machinery to bypass the site of damage, such as thymine dimer (escaped from repair system).

Question 55

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

Answer 55

activates

Question 56

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

Answer 56

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

Question 57

LexA is a activator or repressor of SOS response genes?

Answer 57

repressor

Question 58

What is homologous recombination? What can it repair

Answer 58

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

When foreign DNA can be installed in a host’s chromosome through recombination

nucleotide sequences are exchanged between two similar or identical molecules of DNA

DNA strands cross over to exchange places.

Recombination can repair damaged replication forks and double-strand breaks.

Question 59

key steps of HR

Answer 59

The corresponding strands of two aligned homologous DNA duplexes are nicked, and the nicked strands cross over to pair with the nearly complementary strands on the homologous duplex, thereby forming a segment of heterologous DNA, after which the nicks are sealed (Fig. 25-40a–e). The crossover point is a four-stranded structure known as a Holliday junction. The crossover point can move in either direction, often thousands of nucleotides, in a process known as branch migration

Question 60

The Holliday junction can be “resolved” into two duplex DNAs in two equally probable ways/Two ways to resolve the Holliday junction?

Answer 60

1. Cleavage of the strands that did not cross over exchanges the ends of the original duplexes to form, after nick sealing, the traditional recombinant DNA molecule (right branch of Fig. 25-40j– l).
2. Cleavage of the strands that crossed over exchanges a pair of homologous single-stranded segments (left branch of Fig. 25-40j – l).
3. Cut in the two DNA strands that were not broken (did not cross over) during initiation reaction generates splice products or called crossover product (right branch of l).
4. Cut in the two DNA strands that were broken (crossed over) to initiate recombination results in the exchanges of homologous single-stranded segments or called non-crossover products (left branch of l).

Question 61

Homologous recombination requires several protein activities known as?

Answer 61

RecBCD, RecA, RuvAB, and RuvC.

Question 62

RecA promotes which step in the homologous recombination?

Answer 62

RecA Protein Assembles on Single-Stranded DNA and Promotes Strand Invasion
The RecA protein promotes pairing between a single-stranded DNA molecule and a homologous sequence in another DNA molecule,

(1) A ssDNA binds to RecA to form an initiation complex. (2) dsDNA binds to the initiation complex so as to transiently form a three-stranded helix.RecA

rotates the bases of the aligned homologous strands to effect strand exchange in an ATP-driven process.

search for sequence matches between two DNA molecules and the generation of the regions of base pairing

Question 63

RecBCD function?

Answer 63

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.
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.

Question 64

RuvAB complex promotes which step in the homologous recombination?

Answer 64

RuvAB Complex Recognizes Holliday Junction and Promotes Branch Migration
RuvA forms a tetramer.
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 65

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

Answer 65

non sequence specific

Question 66

What is the function of RuvC?

Answer 66

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

RuvC recognizes the Holliday junction and specifically nicks two of the homologous DNA strands that have same polarity, generating 5’ phosphate and 3’ OH group ready for DNA ligase.

Question 67

Homologous Recombination repairs?

Answer 67

damaged replication fork homology-directed repair (HDR) or homologous end-joining of DSB Homologous

Recombination Repair Reconstitutes Double-Strand Breaks. We have seen that double-strand breaks (DSBs) in DNA can be rejoined, often mutagenically, by nonhomologous end-joining (NHEJ; Section 25-5E). DSBs may also be nonmutagenically repaired through a recombination repair process known as homology-directed repair (HDR; also called homologous end-joining), which occurs via two Holliday junctions (Fig. 25-50):

Question 68

CRISPR-Cas9?

Answer 68

a System for Editing and Regulating Genomes

The CRISPR–Cas system enables many prokaryotes to defend themselves from previously encountered bacteriophages. CRISPR–Cas9 technology has permitted the facile editing and regulation of genomes.

Question 69

What are Transposons

Answer 69

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

Question 70

What is the difference between homologous recombination and transposition?

Answer 70

Transposition does not require homology between donor and recipient DNA

Transposition Rearranges Segments of DNA and thus lead to a more dynamic genome structure.

Question 71

Structure of IS element.

Answer 71

The simplest transposons are named insertion sequences or IS elements, which generally comprises a so-called transposase gene flanked by short inverted terminal repeats. IS element is flanked by a directly repeated segment of host DNA. Direct repeat of host DNA flanking IS element suggests that an IS element is inserted in the host DNA at a staggered cut that is later filled in. The length of the target sequence (~5-9 bp), but not its sequence, is the characteristic of the IS element.

Question 72

. Most eukaryotic transposons are

Answer 72

retrotransposons.

Question 73

What are the three major types of RNA?

Answer 73

Ribosomal RNA (rRNA): structural constituents of ribosome

Transfer RNA (tRNA): small compact RNAs that deliver amino acids to the ribosomes for assembly into proteins

Messenger RNA (mRNA): RNA that directs protein synthesis

Question 74

Which type of RNA directs protein synthesis?

Answer 74

mRNa

Question 75

Difference between transcription and DNA replication.

Answer 75

Transcription: the process by which nucleiotide sequence information is transferred from DNA to RNA

Transcription is performed by RNA pol and replication is performed by DNA pol

New strand is made from ribonucleotides rather than deoxyribonucleotides
RNA polymerase does not need a primer
RNA product does not remain base-paired to the template DNA strand.
Transcription is less accurate than replication

Transcription selectively copies only certain parts of the genome and make copies from one to hundreds and thousands. Replication must copy the entire genome and do so once every cell cycle

Question 76

The definition of structural gene.

Answer 76

Protein-coding genes are called structural gene.

and are transcribed individually
in eukaryotes
A gene consists of coding (transcribed into RNA) and non-coding sequences as
well as regulatory sequences.

Question 77

What is a Operon?

Answer 77

A operon typically contains genes with related function.

multiple genes that are transcribed together

Most protein-coding genes (called structural genes) in eukaryotes are transcribed individually. In prokaryotic genomes, however, genes are frequently arranged in tandem along a single DNA strand so that they can be transcribed together.

Question 78

The components of core RNA polymerase.

Answer 78

α 2ββ ′ωσ

Contains a holoenzyme, sigma factor, and subunits

Question 79

Function of σ factor in RNA polymerase holoenzyme.

Answer 79

converts the core enzyme into the form that initiates transcription only at promoter.

dissociates from the core enzyme, α 2ββ ′ω, which carries out the actual polymerization process.

Recognizes a gene’s promoter and is released once chain initiation has occurred.

Question 80

Which strand in a gene is sense (antisense) strand?

Answer 80

Only one DNA strand serves as a template during transcription

Sense= coding because it’s the same,The other DNA strand, the nontemplate strand, which has the same sequence as the transcribed RNA

antisense= noncoding because its complementary

Question 81

What sequences in a gene are coding/noncoding sequences?

Answer 81

Antisense= noncoding= template (complementary) strand of DNA

Sense= coding= nontemplate (same) DNA strand