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Unit 3: DNA Flashcards

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1
Q

How is DNA packaged?

3.1

A
  • Eukaryotic DNA is organised on multiple chromosomes, each of which contains a linear DNA molecules.
  • DNA binds to histones that package it in the cell nucleus. Primary characteristic e.g. length of human DNA is 2 metres but must fit in a core of 5-10 micrometers)
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2
Q

Characteristics of chromosomes

3.1

A
  • Tightly packaged DNA
  • Found only during cell division
  • DNA is not used for macromolecule synthesis
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3
Q

Chromatin characteristics

3.1

A
  • Unwound DNA
  • Found throughout interphase
  • DNA is used for macromolecule synthesis
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4
Q

What is chromatin?

3.1

A

It is a complex structure of DNA and proteins and its degree on condensation depends on the activity of each chromosomal region and the cell cycle stage
*Note that there are different levels of DNA packaging

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5
Q

What are the 2 majors categories of chromatin?

3.1

A
  1. Heterochromatin: condensed chromatin structure and is inactive for transcription
  2. Euchromatin: has loose chromatin structure and is active for transcription.
  • see diagram of chromatin at interphase (pg. 6)
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6
Q

What is level 1 of DNA packaging?

3.1

A

The DNA double helix (about 147bp) coils around a core or octamer of histones ( 2 molecules of each of the histones H2A, H2B, H3 + H4) in a structure called a **nucleosome

  • The binding of H1 histone to the core nucleosome forms a chromatin subunit called the **chromatosome (166 bp, 2 turns of super helix).
  • There are about 200 bp between consecutive nucleosome, so the structure is called **”beads on a string” as it looks like a necklace under electron microscope. The linked nucleosomes give rise to an **11nm chromatin fibre.
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7
Q

Where is H1 located?

3.1

A

Where the DNA enters and leaves the nucleosome

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8
Q

Histone characteristics

3.1

A
  • basic proteins with evolutionarily highly conserved sequence
  • have between 102-135 aa
  • H2A, H2B, H3 + H4 associate to form an *octamer
  • (H3 x2) + (H4 x2) = tetramer and (H2A x2) + (H2B x2) = tetramer. Both tetramer associate together forming the protein nucleus of the nucleosome which the DNA wraps.
  • In sperm histones are replaced by *protamines allowing more packaging
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9
Q

What is level 2 of DNA packaging?

3.1

A

The 11nm fibre is rewound to form a **30 nm chromatin fiber, containing about 6 nucleosomes per turn

  • Most euchromatin at interphase is in the 30nm fibres or more condensed (60-130 nm). Actively transcribed genes are in a less condensed state.
  • The 30 nm chromatin fiber undergoes different degrees of packaging during interphase until mitosis, in which this is maximum and allows observation of chromosomes
  • 30 nm fiber forms loops linked at its base to a protein scaffold structure
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10
Q

How is the protein scaffold structure formed and how do the loops attach?
3.1

A
  • Scaffold formed by *condensins Sc1 and Sc2
  • Chromatin loops are attached to scaffold by *sequences rich in A and T called SAR (scaffold-attachment regions)
  • *Result is a 300 nm fiber
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11
Q

What is level 3 of DNA packaging?

3.1

A
  • Mitosis/ Cell division

- the 300 nm fiber is recondensed to form a *600-700 nm fiber AKA chromatid

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12
Q

What are chromosomes?

3.1

A

The degree of packaging of DNA reaches 10,000 times

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13
Q

What is the degree of packaging of DNA?

3.1

A

Level 1: 2nm fiber (dsDNA) => 11nm (necklace beads) *X6
Level 2: 11nm fiber => 30nm (solenoid structure) *X4
Level 3: 30 nm => 300 nm (hairpins) *X2,000
300nm => 600-700nm (chromatids) *X10,000

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14
Q

Which areas of the histones can be modified?

3.1

A

Histones have an amino-terminal tail so the Lys can be acetylated or methylated, and the Ser can be phosphorylated.

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15
Q

What do the histone modifications do?

3.1

A

They make up part of the histone code that can be ‘read’ by proteins during replication or expression of genetic material.

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16
Q
  • *What are histone acetylation or methylation associated with?
    3. 1
A
  • Acetylation: associated with transcriptional activation. Acetylated nucleosomes associate with more transcriptionally active chromosomal regions.
  • Methylation: associated with both active and repressed chromatin
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17
Q

Chromosomes

3.1

A
  • *Consequence of nuclear DNA packaging together with various proteins in their maximum degree of condensation
  • During metaphase both copies of replicated DNA are held by *the centromere
  • Each copy is called a *chromatid
  • 2 chromatids= a metaphase chromosome (*sister chromatids)
  • *the centromere divides the chromatids to 2 arms, can be same or different length
  • Ends of the chromatids are *telomeres
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18
Q
  • How are chromosomes classified based on centromere position?
    3. 1
A
  1. ) Metacentric: centromere is located approx. in centre, chromosome divided into 2 identical arms
  2. ) Submeracentric: centromere stays displaced to one side, dividing chromosome into short and long arm
  3. ) Submetacentric with satellite sones: same as 2.), but with chromosome segment separate from rest due to 2° constriction
  4. ) Acrocentric: centromere is near one end, making p-arm v.small
  5. ) Telocentric: centromere at extreme, making p-arm practically non-existent

See diagram on pg. 17

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19
Q

Anatomy of chromosome

3.1

A
  • Centromere: point where sister chromatids join together
  • P: short arm; upward
  • !: long arm; downward
  • Telomere: tips of chromosome
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20
Q

What is the difference between centromeres and telomeres?

3.1

A

They are made of *different specific DNA sequences
- in yeast cent. Are short (125 bp). In mammals they are made of hundreds of Kb of repetitive DNA, which is sometimes chromosome specific

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21
Q

What is a centromere?

3.1

A
  • Specialised region of the chromosome
  • Function: correct distribution of duplicated chromosomes to daughter cells during mitosis
  • *sister chromatid association sites
  • *binding site for the mitotic spindle
  • at end of mitosis nuclear membranes are re established and chromosomes are decondensed
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22
Q

What are telomeres?

3.1

A
  • made up of specialised structures of DNA and proteins. Sequences are similar, presenting repeats of a single DNA sequence containing groups of G residues on 1 strand.
  • Function:
    1. *maintain structural integrity of chromosomes (without they cannot fuse with each other or degrade)
    2. *position chromosome in nucleus
    3. *ensure complete DNA replication
  • *In humans: TTAGGG, is repeated in tandem to a variable length between 3-20 Kb
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23
Q
  • Structure of a telomere

3. 1

A
  • Telomeric DNA forms a loop on itself to form a circular structure with *shelterin (protein complex) that protects the ends of chromosomes against degradation
  • *telomerase replicates telomeric DNA (DNA polymerase cannot replicate chromosomal endings)
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24
Q

What is a genome?

3.2

A

Set of genetic material of an organism

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25
Q

What is a gene?

3.2

A

DNA (structural and regulatory) required to encode a gene product (mature RNA or protein)

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26
Q

What is extragenic DNA?

3.2

A

DNA that does not encode any protein. Research suggests they are not devoid of functionality, such as regulation of gene expression

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27
Q

Length of haploid nuclear human genome

3.2

A

3200 Mb

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28
Q

What is an exon?

3.2

A

A fragment of messenger RNA that survives the assembly processes (splicing) to become part of the mature messenger RNA. They make up the coding region and the untranslated transcribes regions that flank the coding region.

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29
Q

What is an intron?

3.2

A

The non-coding sequence of a gene which is initially transcribes into messenger RNA, but is lost during splicing and is therefore not present in mature mRNA

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30
Q

What is alternative splicing?

3.2

A
  • Generation of different mature RNA transcribe from the same gene, obtained by deleting one or more exons in the splicing process during *RNA maturation.
  • Alternative splicing occurs in RNA which codes for many proteins so that by alternating the exons used, a set of related proteins can be generated that are differentials expressed throughout development or in different tissues.
  • *Alternative splicing allows 21,000 human protein Golding genes to specify nearly 85,000 different proteins
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31
Q

Types of sequences in our genome?

3.2

A

There are 21,000 genes (exons, introns, regulatory sequences)
- Regulatory sequences: promotors, silencers, enhancers
Extragenic DNA:
- DNA which is transcribed to non coding RNA
- Repetitive DNA sequences:
1. Tandem repeats: satellite DNA
2. Sparse repeats: LINES, SINES, LTR, transposing
- Gene duplication and pseudogenes

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32
Q

What is the ENCODE Consortium?

3.2

A
  • *An ongoing collaboration funded by the National Human Genome Research Institute with the goal of building a comprehensive parts list of the functional elements in the human genome
  • *ENCODE investigators employ a variety of assays and methods to identify functional elements

assay = an investigative procedure for qualitatively assessing or quantitatively measuring the presence, amount, or functional activity of a target entity.

33
Q

What are the types of non coding RNA?

3.2

A
  1. tRNA
  2. rRNA
  3. miRNA (micro RNA)
  4. Long ncRNAs, lncRNA (long non-coding RNAs)
  5. snRNA
  6. snoRNA
34
Q

What is the role of tRNA and rRNA?

3.2

A

They play fundamental roles in protein synthesis

35
Q

What is the role of miRNA?

3.2

A
  • Its is a double stranded RNA with approx. 22 nucleotides.
  • 1 strand associates with RISC (RNA-inducing silencing complex) and the miRNA directs the RISC to the complementary mRNAs where they *inhibit translation or stimulate mRNA degradation.
  • Estimates that miRNA can target up to 100 different mRNAs
  • have role in various developmental processes, regulation of cell proliferation + survival, their abnormal expression is associated with heart disease and cancer although functions are not fully understood.
36
Q

What is the role of long ncRNA/lncRNA?

3.2

A
  • non-coding RNA more than 200 nucleotides long
  • important *regulators of gene expression in eukaryotes
  • more than 50,000 in humans e.g. Xist RNA, 17Kb that blocks transcription of 1 our 2 X chromosomes in females
37
Q

See pg. 34 for Diagram

A

GO GO!

38
Q
  • What are the types of repetitive DNA sequences?

3. 2

A
  • Single sequence repeats

- Sparse repeats

39
Q
  • What are single sequence repeats?

3. 2

A
  • Tandems made of thousands of copies of short sequences (1-500 nt).
  • Not transcribed but important for structure of chromosome
  • Represent 40% of total genomic DNA in Drosophili and approx. 10% in humans
     This is satellite DNA because when separating genomic DNA by density gradient centrifugation, it appears as “satellite” bands of the main band
  • AT rich sequences are less dense
  • GC rich sequences are more dense
40
Q
  • What are sparse repeats?

3. 2

A

Very important contributor to genome size (45% of human genome)

   * SINE (short interspersed nuclear elements): 100-300 bp. 1.5 million of these sequences, *13% of human genome
   * LINE (long interspersed nuclear elements): 4-6kb 850,000 of these sequences, *21% of human genome
  • Both are transposable elements. These are retrotransposons - transposition is by reverse transcription (RT).
  • LINEs can autonomously transpose since it codes for reverse transcriptase and an endonuclease, both of which are used by SINEs.
    • LTR transposing: retroviruses like sequences integreated into human genome through evolution. Flanked by long terminal direct repeats (LTR). 2-10 kb. 450,000 of these sequences, 8% of human genome. Also RT
    • DNA transposing: move through genome being copied and reinserted as DNA sequences. 80-3000bp. 300,000 of these sequences, 3% of human genome
41
Q

What is gene duplication?

3.2

A
  • The duplication of a region of DNA containing at least 1 gene.
  • The 2 genes that result from this usually code for the same protein.
  • Groups of genes with a similar sequence are called *gene families, formed by duplication of ancestral genetic material, and then divergence through mutation
  • Genome-wide duplications can also occur especially in plants e.g. A. thaliana
42
Q

What are pseudogenes?

3.2

A
  • Functional but inactive copies of genes, often due to accumulation on mutations
43
Q

What is the mechanism that allows DNA to duplicate?

3.3

A
  • Semi-conservative replication: each chain serves as a template for a new chain.

The experiment of Meselson and Stahl (1958) with isotope N15 and density centrifugation showed that replication is semi conservative.

44
Q

What are the types of DNA polymerase used for?

3.3

A
  • *Both eukaryote and prokaryotes contain different DNA polymerases with different roles in DNA replication + repair
    • *Bacteria: DNA polymerase III is the main polymerase responsible for replication.
    • *Eukaryotes: DNA polymerases α, δ + ε function in nuclear DNA replication, and DNA polymerase γ is responsible for replication of mitochondrial DNA.
45
Q

What fundamental properties do all DNA polymerases share?

3.3

A
  • Only synthesise DNA in the 5’ to 3’ direction, adding a dNTP (deoxynucleotide triphosphates) to the 3’OH group of a growing chain
  • They can add a new deoxyribonucleotide only to an existing strand (primer) that is H bonded to the template strand => can’t direct de novo synthesis by catalysing polymerisation of free dNTPs

De novo= can’t make from nothing, in this case they need primer

46
Q
  • Classification of DNA polymerase in prokaryotes

3. 3

A
  • DNA polymerase I: participates in DNA replication + repair, and base exonuclease activity (deleting nucleotides) in both directions
  • DNA polymerase II: participates in DNA repair and its exonuclease activity is in the 3’ to 5’ sense.
  • DNA polymerase III: participates in DNA replication + revision and has exonuclease activity is in the 3’ to 5’ sense.
47
Q
  • Classification of DNA polymerase in eukaryotes

3. 3

A
  • DNA polymerase γ: participates in replication of mitochondrial DNA, only found in mitochondria.
  • DNA polymerase α, δ + ε: most active in mitosis, suggests their main function is with production of DNA copies
  • DNA polymerase β: most active in cells not dividing, so main function associated with DNA repair.

-Types γ, δ + ε have 3’ to 5’ exonuclease activity

48
Q

How does DNA replication begin?

3.3

A
  • DNA molecule opens like zipper when H bonds between complementary bases are broken at certain points: *the origins of replication
  • Initiator proteins recognise specific nt sequences and facilitate attachment of other proteins that allow separation of the DNA strands, *forming 2 replication forks
  • Many enzymes and proteins are involved in replication, *forming the replication complex/replisome. They are homologous in eukaryotes and rachea, but differ in bacteria
49
Q

Origin of replication in prokaryotes

3.3

A

1 single origin.
In E.coli it’s composed of 245bp.
- binding of *initiator protein to specific sequence within the origin. It begins to unwind the DNA and recruits the other proteins involved.
- *Helicase along with *single stranded DNA binding proteins continue to unwind and present the DNA
- *2 replication forks form that move in opposite directions along circular chromosome

50
Q

Origin of replication in eukaryotes

3.3

A

*Multiple origins needed to replicate long chromosomes in a few hours (reasonable)
- E.coli genome (4 X 10^6 bp) replicates in 30 mins
- mamallin genomes (3 X 10^9 bp) from 1 origin, would take 3 weeks/30,000 mins
Problem is worse as mammalian DNA replication is X10 slower than E.coli due to DNA packaging in chromatin

51
Q
  • What is autonomous replication sequence (ARS)?

3. 3

A
  • Part of the origin of replication in S.cerevisiae genome.
  • Contains 4 regions; ACS, B1, B2 + B3
  • ACS domain is highly conserved - any mutation nullifies origin function
  • Mutations in B1, B2 + B3 decrease, but don’t stop the origin functioning
  • *ORC (origin recognition complex) binds to ACS andB1. Then recruits additional proteins, such as MCM DNA ligase, to origin
52
Q

What is the initiation process for origin of replication in higher eukaryotes?
3.3

A
  • Not known
  • Seems determined by chromatin structure characteristics rather than specific sequences like ACS of ARS
  • Origin of replication in mammalian cells separated by intervals of 50-300 kb, so human genome = 30,000 origins of replication
53
Q
  • What is a replication fork?

3. 3

A

Place of the DNA molecule in replication where the parental DNA strands are separated and 2 new daughter strands are synthesised

54
Q

What is the problem with replication forks and solution?

3.3

A
  • Problem: complementary strands of DNA run antiparallel, so continuous synthesis of new strands in the replication fork would require one strand to be synthesised in 5’-3’ direction while other in 3’-5’. DNA polymerase only goes in 5’-3’ direction
  • Solution : *Okazaki fragments (1-3 kb)
55
Q
  • How do Okazaki fragments work?

3. 3

A
  • DNA polymerase needs a primer to provide a 3’OH group.
  • unlike DNA polymerase, RNA can be synthesised de novo. *Primase: synthesises short RNA fragments (3-10 nt) complementary to the template lagged strand in replication fork. Okazaki fragments are synthesised by extension of these by DNA polymerase.
  • RNA primers then removed and replaced by DNA
    - *Prokaryotes: by DNA polymerase I
    - *Eukaryotes: RNAse H degrades the RNA strand then DNA polymerase δ fills gaps
  • Resulting DNA fragments joined by *DNA ligase
  • *Sliding clamps: associate with prokaryotic polymerase III or eukaryotic polymerase δ to position them in the primer and associated template binding site while replication occurs.
  • *Clamp-loaders: bind sliding clamps to DNA at primer and template binding site. Then released and DNA polymerase binds to sliding clamps (help polymerase stay in place)
  • *SIngle stranded DNA-binding proteins: stabilise uncoiled DNA strand, keeping extended so it can be copied by polymerase
  • *Topoisomerases: reduce stress from supercoiling of DNA
56
Q

Model of replication fork of E.coli

3.3

A

*Helicase, *primase and *2 DNA polymerase III molecules carry out coordinated synthesis of both strands of DNA (leading and lagging). Both polymerases form *complex with clamp-loading protein which is *bound to helicase at replication fork. *Topoisomerase acts as swivel at head of hairpin. *Polymerase I with Ligase removes the RNA primers and binds Okazaki fragments behind hairpin.

See diagram pg. 56 if struggling to visualise

57
Q

How do topoisomerases work?

3.3

A
  • Once DNA is unwound, DNA at head of replication fork is forcibly rotated in the opposite direction - the circular molecules are wound on themselves (PROBLEM)
  • Topoisomerases solve problem by catalysing the reversible breaking and joining of DNA strands. Transient breaks are introduced by this enzyme to act as as twirling links, allowing DNA strands to rotate freely on one another.

See diagram on pg. 55 if struggling to visualise

58
Q

DNA maintenance

3.4

A
  • Accuracy of DNA replication is crípticas for cell reproduction
  • Error frequency: less than 1 incorrect base for every 10^9 incorporated nucleotides
59
Q

What mechanisms allow DNA polymerase to accurately replicate DNA?

A
  • Helps select correct base (unsure how)
  • *Double reading activity: exonuclease activity of polymerase III and Polymerase δ + ε allows to hydrolyse DNA in the 3’-5’ direction, contrary to DNA synthesis. This allows an incorrect base to be cleaved at the end of the growing DNA chain
60
Q

What is telomerase?

3.4

A
  • DNA polymerase is involved in telomere formation, with the rare peculiarity that its only *capable of synthesising oligonucleotides with the telomeric sequence
  • Enzyme contains a 159-nucleotide RNA oligonucleotide, essential for providing replication template of the telomere sequence (so its a type of reverse transcriptase)
61
Q
  • What are the 2 forms of spontaneous DNA damage?

3. 5

A
  1. Deamination of adenine, cytosine + guanine
  2. Depurination: due to breaking of bond between purine bases and deoxyribose, leaving an apurinic site (AP) in DNA

See diagram on pg. 64 for visualisation

62
Q
  • How does Radiation and Chemical Induced DNA damage ocurr?

3. 5

A
  • UV light induces formation of *pyrimidine dimmers: 2 adjacent pyrimidines linked by a cyclobutan in ring structure
  • *Alkylation at various positions on DNA bases
  • Carcinogens react with DNA bases, causing *addition of bulky chemical groups to DNA molecule
63
Q

What are the DNA repair mechanisms?

A
  1. Direct reversal of the chemical reaction responsible for damaged DNA
  2. Removal of damaged bases followed by replacement with newly synthesised DNA:
    • Base excision repair
    • Nucleotide excision repair
    • Mismatch repair
    • Translesion DNA synthesis
    • Double strand break repair
64
Q

What does direct reversal of the chemical reaction responsible for damaged DNA offer?

A

Offers a specific pathway to repair some types of DNA damage:

 - Pyrimidine dimmers from UV light exposure
 - Alkylated guanine residues modified by methyl and ethyl groups at the O^6 position of the purine ring

Mammals lack this repair mechanism

65
Q

How are pyrimidine dimmers repaired?

3.5

A

Direct repair by photoreactivation

66
Q

What do pyrimidine dimers do to the DNA molecule?

3.5

A

They distort the structure of the DNA chain and block transcription or replication, so that their repair is related to the cell’s ability to survive radiation

67
Q
  • How does Direct repair by photoreactivation work?

3. 5

A
  • Energy from visible light splits the bonds formed in the cyclobutane ring
  • Common in prokaryotic and eukaryotic cells e.g E.coli, yeast
68
Q

*How is O^6-methylguanine repaired?

A

O^6-methylguanine methyltransferase transfers the methyl group to a cysteine residue at the enzyme activation site

69
Q
  • How does Base excision repair work?

3. 5

A
    • A single damaged base is removed from the DNA molecule and replaced with the correct one
      Example: DNA containing uracil can arise from 2 mechanisms
      1. Occasionally imcorporated in place of thymine during DNA synthesis
      2. Formed by deamination of cytosine. This is of greater biological importance as it alters the normal pattern of complementary bases => represents a mutagenic event
  • Cleavage and exchange of uracil is catalysed by * various enzymes: DNA glycosylase, AP endonuclease, deoxyribosephosphodiesterase, DNA polymerase + ligase
70
Q
  • How does Nucleotide excision repair work?

3. 5

A
  • Oligonucleotides containing lesions are removed.
    E.g UV induced pyrimidine dimers, bulky groups from interactions of carcinogens
  • Lesion is recognised by a protein (UvrA in E.coli) which causes the binding of other nuclease proteins (UvrB + UvrC) and a *helicase to unzip the DNA strands.
  • DNA containing the lesion is cleaved by *nucleases.
  • Gap is filled by *DNA polymerase and joined by *ligase.
71
Q

What is Xeroderma pigmentosum (XP)?

3.5

A
  • Condition causing skin and tissue that covers the eye to be extremely sensitive to UV light
  • People with this cannot be in contact with sunlight: confers on them a *congenital deficiency in an endonuclease that *prevents DNA repair in skin cells mutated by UV light.
72
Q
  • How does Transcription-coupled repair work?

3. 5

A
  • Repair specifically to lesions in *genes that are actively transcribed
  • If lesion occurs in DNA strand being transcribed, RNA polymerase stops.
  • Arrested RNA polymerase is recognised by protein/s (in E.coli, transcriptional repair coupling factor; in mammals, CSA + CSB) that *displace RNA polymerase and *recruit different proteins to clear region affected and repair befor seen.
73
Q

What is Cockayne syndrome?

3.5

A
  • A rare hereditary disease with an autosomal recessive pattern of inheritance, caused by mutation in 2 genes; ERCC6 (75% of cases) and ERCC8 (CKN1) (25% of cases). They code of CSA and CSB proteins respectively.
  • Mutation of proteins causes alterations in genes. Of those genes, that ones transcribed are not quickly or efficiently repaired as normal cells.
74
Q
  • How does mismatch repair work?

3. 5

A
  • Recognises non-complementary bases incorporated during DNA replication that haven’t been eliminated by double reading of DNA polymerase.
  • Enzymers of this system identify and cleave specific base from newly replicated strand.
    • E. coli: *MutH endonuclease breaks unmethylated DNA strand; *MutL + MutS with an exonuclease + helicase cleave the DNA between the “gap” in strand and noncomplementarity. Space filled by *DNA polymerase + ligase
    • Mammals: MSH + MLH homologous to MutS + MutL
75
Q

What is Hereditary nonpolyposis colorectal cancer (HNPCC)?

3.5

A
  • one of most common inherited diseases, accounts for 15% of colorectal cancers in US
  • Mutations in MSH and MLH responsible
  • Defects in genes result in high frequency of mutations in other cellular genes, with a corresponding high probability that these mutations could trigger cancer
  • Affected individuals experience higher incidence of other types los cancer
76
Q
  • How does Translesion DNA synthesis work?

3. 5

A

If other repair systems fail, replication is blocked at point of damage.

  • However cells possesses specialised DNA polymerases capable of replicating through a damaged DNA site (polymerases II, IV + V in E.coli)
  • Specialised polymerases replace normal that has been stopped by the damage. They continue synthesis at site of damage, then are replaced again by the normal polymerase
  • Specialised DNA polymerases show low fidelity when copying undamaged DNA and lack double reading activity, so error rates are X100-X10,000 higher than normal polymerases
77
Q

Why is the need for double strand break repair dangerous and what are the mechanisms?
3.5

A
  • represent dangerous variant of DNA damage: large segments of chromosomes + genes lose if fragmentation isn’t repaired
  • Mechanisms: Non-homologous end joining (NHEJ) + homologous recombination (HR)
78
Q
  • How does non-homologous end joining work?

3. 5

A
  • 2 broken ends of chromosomes are put back together
  • Problem: “messy” and often result in loss or addition of nucleotides at cleavage site; tends to produce mutation but better than alternative
79
Q
  • How does homologous recombination work?

3. 5

A
  • double strand breaks repair by recombination with an uninjured homologous chromosome
  • Both strands of damaged molecule digested by nucleases at break site, in 3’-5’ direction
  • 3’ hanging ends invade other parent molecule by homologous base pairing
  • Gaps filled by repair synthesis and sealed by ligation, producing cross-stranded intermediate.
  • Division and ligation of crossed strands cause recombinant molecules

Biology (11 decks)

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