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BMS238 Molecular and Cell Biology > Lecture 7 > Flashcards

Lecture 7 Flashcards

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1
Q

What happens in patients with mutations in genes needed for homologous recombination

A

Homologous recombination is defective and thus these cells become critically dependant on other pathways for correct DNA repair i.e. base excision and nucleotide excision repair

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2
Q

What chemical modifications can lead to DNA damage

A

Hydrolysis, oxidation or random uncontrolled methylation

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3
Q

Draw a representation of a heteroduplex structure between two dsDNA molecules

A

See image

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4
Q

Explain the process of nucleotide excision repair

A

An excision endonuclease enzyme cleaves the single strand of the DNA containing the defect by cleaving either side of the dimer etc. This creates a polynucleotide fragment from the DNA molecule that contains the defect which is then removed by DNA helicase. DNA polymerase then extends the primer template junction created to replace the excised sequence. This is followed by resealing of the nick mediated by DNA ligase.

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5
Q

Which base is primarily affected by UV light and more prone to form dimers

A

Thymine

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6
Q

How can this defective recombination pathway seen in some cancers be harnessed as a cancer treatment

A

Combined inactivation of homologous recombination by mutations and the inhibition of base or nucleotide excision repair by anticancer drugs make cells that suffer DNA damage unable to make adequate repairs to their DNA and die. This is known as synthetic lethality.

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7
Q

What happens when cytosine is deaminated and what are the downstream effects of this

A

Converted to uracil. This will still pair with guanine but during replication the guanine will be replaced with an adenine, leading to a nucleotide substitution from CG–>TA

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8
Q

Homologous recombination is required to mediate genetic recombination and diversity created in meiosis, T or F

A

T

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9
Q

How many ways can the Double Holliday Junction be resolved

A

2

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10
Q

What is the most frequency types of DNA damage

A

Hydrolytic depurination and deamination of bases

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11
Q

Pyrimidine dimers can only occur between identical adjacent pyrimidine bases, T or F

A

F – it can be the same pyrimidine or different ones (i.e. T-C, T-T, C-C)

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12
Q

What are the products of genes encoded by the genes mutated in xeroderma pigmentosum

A

These genes encode proteins that participate in the nucleotide excision repair pathway

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13
Q

Nucleotide excision repair is only used to repair pyrimidine dimers, T or F

A

F – it is used to repair a variety of different types of DNA damage including pyrimidine dimers

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14
Q

dsDNA breaks are often caused by non-ionising radiation, T or F

A

F – they are caused by ionising radiation

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15
Q

What two methods are there of repairing dsDNA breaks

A

Non-homologous end joining, homologous recombination

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16
Q

Recall the purine bases

A

Adenine, guanine

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17
Q

What are the four main types of DNA damage

A

Deamination, depuration, pyrimidine dimers, DNA breaks

18
Q

Explain how different resolutions of the Double Holliday Junction account for whether homologous recombination occurs or not

A

If the internal strands are broken and re-joined than the same DNA strands are broken and re-joined at each junction and recombination is not achieved as each chromatid will contain the same sequence of alleles it did prior to cleavage. If external and internal strands are broken and re-joined then different DNA strands are broken and re-joined at each junction. Hence recombination is achieved. This recombination of allelic forms results in chromatids with different combinations of alleles (Ab and aB)

19
Q

Explain the process of non-homologous end joining

A

The ends of the double stranded break are rendered flush with loss of bases via degradation from the ends of the strands. These flush ends are then ligated together. However, this leads to a loss of DNA sequence due to the degradation that occurs prior to ligation.

20
Q

What are pyrimidine dimers and what causes them

A

Pyrimidine dimers are caused by the covalent linkage of benzene rings in two adjacent pyrimidine bases. This is usually caused by UV light

21
Q

What is significant about the function of the XP genes

A

Their function is tightly coupled and targeted to regions of the genome that are highly transcribed. This acts as a surveillance system to signify regions of the genome that are transcribed. RNA polymerase required for transcription is physically coupled to the DNA repair machinery

22
Q

Xeroderma pigmentosum is a disease caused by defective nucleotide excision repair machinery, what is the effect of this on patients

A

XP renders patients extremely sensitive to sunlight-induced skin cancer

23
Q

What does homologous recombination rely on

A

Using the intact DNA sequence information in the undamaged homologous chromosome

24
Q

Homologous recombination also isn’t a perfect repair mechanism, T or F

A

F – homologous recombination provides a perfect repair of dsDNA breaks and is an accurate and preferred method of repair

25
Q

Uvr genes are homologues of the XP genes found in yeast and are also transcription coupled, T or F

A

T

26
Q

Put these enzymes in the order in which they act in Base Excision repair?

a - DNA polymerase, b - DNA glycosylase, c - phosphodiesterase, d - DNA ligase, e - apurinic/apyrimidinic endonuclease.

A

b, e, c, a, d

27
Q

What is the results of the formation of pyrimidine dimers

A

This arrests DNA replication or can cause mis-reading of the DNA sequence by DNA polymerase

28
Q

What is meant when we refer to non-homologous end joining being quick and dirty

A

Non-homologous end joining is a far from optimal repair mechanism. It is carried out quickly but in itself, can induce mutations

29
Q

Recall the pyrimidine bases

A

Thymine, cytosine, uracil

30
Q

Explain the process of depurination

A

The carbon to nitrogen bond between the carbon position one in the deoxyribose sugar and the nitrogen in the purine ring is hydrolysed. This releases the base and results in the loss of a base pair, nucleotide deletion, in one of the daughter DNA molecules produced during replication

31
Q

Why are dsDNA breaks especially dangerous

A

Large fragments of chromosomes can be lost

32
Q

Explain how a Double Holliday Junction is created during genetic recombination in meiosis

A

Two chromatids exist with different alleles for two genes (AB and ab). Spo11 endonuclease makes an initial cleavage in chromatid to create a dsDNA break. Mre11 then carries out a 5’ resection to create a targeted dsDNA break with a 3’ overhang in the DNA molecule. RecA then mediates strand invasion by this strand on the other chromosome creating a heterocomplex/triple helix structure. DNA polymerase then synthesises across the gap between the end of the 3’ overhang from the invading strand and the 3’ end of the other chromosome. The 3’ end is then relegated with the original strand to create a double holliday junction whereby the invading strand from one helix is forced into interactions with the complimentary strand in the helix of the other chromatid and vice versa.

33
Q

Put these enzymes in the order in which they act in Nucleotide Excision repair?

a - DNA Helicase, b - Excision endonuclease, c - DNA polymerase, d - DNA ligase

A

b, a, c, d

34
Q

Give an example of a gene involved in homologous recombination that when mutated can cause cancer

A

BCRA2 – causing breast, ovarian and prostate cancer

35
Q

Explain how homologous recombination is mediated

A

Firstly, the 5’ ends are resected by exonuclease to create single strands that can be used to prime DNA synthesis when annealed to a template strand from the complimentary chromosome. The DNA-binding protein RecA promotes strand invasion of the undamaged template molecule by one strand from the damaged DNA molecule that acts as a primer. This forces complimentary binding with the same sequence in the undamaged sister chromatid. This forms a heteroduplex structure between the dsDNA helix of the sister chromatid and the single stranded sequence from the damage DNA molecule. This facilitates templated DNA synthesis of one strand by DNA polymerase. DNA polymerase synthesises across the damaged region by reading information out of the undamaged sister chromatid. The newly synthesised DNA then dissociates from its template and re-anneals to its original partner strand allowing second strand synthesis and formation of a pair of staggered singles stranded nicks. These nicks are then ligated by DNA ligase

36
Q

What is the name of the protein that promotes strand invasion of the damaged strand to the undamaged during homologous recombination?

A

RecA

37
Q

What can make DNA molecules susceptible to dsDNA breaks

A

If the DNA has already suffered a ssDNA break due to base excision repair or nucleotide excision repair. This may weaken the DNA structure

38
Q

Explain how deamination of cytosine is achieved

A

The NH2 group attached to the purine ring of cytosine at position 4 is replaced with a carbonyl group

39
Q

Which DNA damage(s) is base excision repair used for

A

Deamination and depurination

40
Q

Explain the process of DNA excision repair

A

The damage base is first removed by DNA glycosylase which cleaves off the inappropriate base(s). The deoxyribose sugar is then removed by an enzyme known as apurinic/apyrimidinic endonuclease. Lastly the phosphate is removed by the phosphodiesterase enzyme. This leads to a ssDNA break in the polynucleotide chain. This gap is then recognised by DNA polymerase which then extends over this gap which acts as a primer template junction. Finally, DNA ligase uses ATP hydrolysis to provide the energy required to seal the nick

41
Q

What are the three main methods of DNA repair

A

Base excision repair, nucleotide excision repair, homologous recombination

42
Q

How many different genes have been identified as being responsible for xeroderma pigmentosum, give some examples

A

7 different genes including XPA, XPC, XPD, XPF and XPG

BMS238 Molecular and Cell Biology (31 decks)

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