dna replication and repair

Question 1

unwinding of the two strands , exposed bases , and strict Watsons-crick base pairing ( becoming the complementary strand as the double strand of dna unwinds )

Answer 1

semiconservative replication ( aka one parental strand and one newly seznsthesied )

Question 2

dna replication requirement:

Answer 2

1- single strand template
2- deoxyribonucleotide triphosphate ( dNTPs of A,G,C,T ) + Mg +2
3- replisome: nucleotide protein complex that co-ordinates the replication activities numerous enzymes and proteins
4- primer w/ a free 3’ end hydroxyl group

Question 3

—- is the separation of 2 complimentary strands that occurs in the origin of replication

Answer 3

initiation

Question 4

true or false:
1- specific points where dna replication begins consensus sequence - short AT poor region
2- eukaryotes have one site of replication

Answer 4

1- false: rich
2- false: mutible site

Question 5

from the origin 2 —- move outwards in — direction.
Active systhesis requires the assembly of — at the origin of replication

Answer 5

• replication forks
• opposite
• replisomes

Question 6

what are the roles of protein in replisomes?

Answer 6

1st step : unwinding proteins
- dna helicase ( separates the double strands into a single strand to allow the dna to be copied )
- single strand binding protein SSB ( wraps the single strand of dna )
- topoisomerase ( which prevents dna from getting tangled , can cut the backbone and reform it )
2nd step : enzymes to replicate
- primase ( sytheises the short rna sequence that is complementary to the single strand of dna that serves as a template )
- dna polymerase ( sythezies the dna , works in one direction , only adds to the 3’ , copies info from the template strand to make the daughter strand )

Question 7

in the dna polymerase it uses — strand of the dna as a template , it reads its template from —– and it make new dna from —-. It aligns and adds nucleotides alone the ss template which specifics the —- of the new chain aka —— , catalysis formation of —–

Answer 7

• single
• 3-5
  -5’ - 3’
• sequence
• Watson crick base pairing
• phosphor bonds
  (DNA)n + dNTP —> (DNA)n+1 + pyrophosphate ( PPi ) this reaction is driven by the subsequent hydrolysis of PPi

Question 8

dna polyemrase is highly — < or equal to 1000 base/second such as: —– and —– , and dna polymerase has — activity to prevent errors such as:

Answer 8

• highly processive as
  1- PCNA proliferating cell nuclear antigen
  2- sliding camp role: it encircles the dna template and keep the dna polymerase closely associated to template as it rapidly move along , the catalyses bond formation joining < or equal to 1000base / second
• it has proofreading activity as:
  1- substrate specificity
  2- proofreading

Question 9

for the dna polymerase substrate activity its active sure can bind to all —- snf catalysis occurs when the — one is bound . The dNTP base pairs with template while the enzyme is in — and catalytically — form . the enzyme conformational change with the correct W-C pair leading to —-

Answer 9

• all four dNTP types
• correct
• open
• inactive
• active enzyme

Question 10

3’-5’ exonuclease activity acts as —- in reverse direction as it removes the nucleotide at the — end of the new strand that are mismatched

Answer 10

• proofreading error correction activities
• 3’ end

Question 11

the leading strand is synthesised — by the — travelling the replication form since the dna polymerase makes new copies
from 5’-3’. While the —- is sytheised discountoisly piece by piece which is what we call —-

Answer 11

• continuously
• dna polymerase
• lagging strand
• semidiscointous replciaropn

Question 12

the lagging strand is sytheises piece by piece as the primase make a new — at the regular intervals , dna polymerase replicates the template from the primer producing a new strand in —- direction , dna is blocked by proxmity of the next primer and the result is a dna strand of 1,000bp —— , primer is — , — filled and — joined
( check slide 14 for pic)

Answer 12

• primer
• 5-3
• Okazaki fragments
• removed
• gaps
• backbone

Question 13

in eukaryotic dna polymerase the human have multible enzymes and the main ones are:

Answer 13

poly alpha ,delta , and epsilon

Question 14

—- is involved in initiating the replication and is associated tightly w/ primase to make a —-

Answer 14

poly alpha , poly alpha / primase complex that is 7-10 rna + 15dNTPs

Question 15

poly alpha replicated the dna by —– 5’-3’ and it has —- and the processive is —-

Answer 15

• extending the primer
• no exonulceas activity aka no proofreading
• moderate

Question 16

poly delta and epsilon are not associated with — and they replicate the dna by extending the primer 5-3 and their processive is — , they have — in complex with —- and they have —- activity

Answer 16

• primase
• high
• unlimited in complex w/ PCNA
• 3-5 exonuclease activity

Question 17

— is responsible for the leading stranding synthesis
—- is responsible for the lagging strand sythesis

Answer 17

• poly epsilon
• poly delta

Question 18

the two enzymes required for the primer removal are:

Answer 18

1- Rnase H1
2- Flap endoculease 1 ( FEN1 )
( check slide 17)

Question 19

— removed most of the rna leaving one 5- ribonuclide adjacent to the dna
—- removes the 5- ribonucleotide which has the endonuclease activity and poly alpha lacks proofreading , and this has endoculeas activity but the mismatch is up to — from 5- end of annealed dna

Answer 19

• Rnase H1
• FEN1
• 15 bp

Question 20

eukaryotes lack — sequence , dna replication proceeds until each replication fork — with the fork from the adjacent replicon
the problem w/ replicating the 2 ends of linear dna strands called —- , the continuous synthesis on the leading strand can proceed to very tip of the template. but what happenes to the extreme end of the lagging strand????
- primer is removed but the — available for the dna polymerase to add nucleotide so the dna can be replicated at the end of the strand ( that’s for the leading strand unlike the lagging strand it doesnt have that )

Answer 20

• termination
• collides
• tolemeres
• OH

Question 21

3’ end of each chromosome —- of the tandem repeats which is — in humans

Answer 21

• 1,000
• TTAGGG

Question 22

telemetric dna synthesised and maintained by — = —–
the – acts as a template for the sytheiss of the dna and adds tandem repeat to the — end

Answer 22

• telomerase = ribonucleoprotein ie. rna + protein
• rna
• 3’

Question 23

the normal tolmerase activity is —- the cells as in unicellular eukaryotes and in humans as ——- production

Answer 23

1- dividing
2- germ line and gametes production as sperm

Question 24

during the development as cells divide and differentiate the telomerase function —–

Answer 24

declines = tolermease shortens

Question 25

telomerase contain approx –kb hexamer repeat after 100s of cell division :
the chromosome end will be — and gene will be —- which causes the dna to be —- causing the cells to stop dividing and tier g0 or apoptosis

Answer 25

• 16kb
• damaged
• deleted
• damaged

Question 26

—— maintain chromosomal stability and prevent chromosomal degradation

Answer 26

tolemere

Question 27

Some cells have the ability to reverse telomere shortening by expressing —–

Answer 27

tolemerase

Question 28

—– is an RNA-dependent DNA polymerase, meaning an enzyme that can make DNA using RNA as a template.

Answer 28

tolemerase

Question 29

—- shorten each time the cell replicates until they can no longer divide which leads to —- , however the enzyme — takes place to restore the cell division however it can cause —-

Answer 29

• telomere
• aging
• tolemrase
• cancer

Question 30

the absence of —- allows normal senescent of somatic cells aka aging , while the abnormal activity ( enhanced activity ) results in —-

Answer 30

telomerase
uncontrolled division leading to cancer , diagnosis tool potential target for treatment

Question 31

true or false:
the proofreading activity of dna polymerase reduces the error rate by 1 in 10 power 4 - 10 power 5 to 1 in 10 power 7 base replicated

Answer 31

true

Question 32

dna is constantly being damaged by :

Answer 32

radiation , high energy radiation , chemicals

Question 33

the damage is the uv light which allows the fuse of adjacent pyrimidines

Answer 33

radiation

Question 34

the damage causes the double strand to break

Answer 34

high energy radiation

Question 35

the damage affects the nitrous acid by delaminating the amines

Answer 35

chemicals eg. c –> uricle , a—> hypoxanthine

Question 36

most damage is repaired by the —- while the remains can cause: —— and —- in the dna sequence
these damages can be:

Answer 36

• cells
• mutation and changes
• insertion , deletion and base substitution

Question 37

—– occurs shortly after replication , replaces mistamche bases or loops up to 4bp in the dna
defects in humans may cause —-

Answer 37

mistmatch repair MMR , high cancer incidents

Question 38

methylation aka the discrimination between parental and daughter strand is an example in

Answer 38

mismatch repair

Question 39

HNPC ( heredity non polyposis cancer ) 70% occurs in —– and —- genes and produce a —- protein and this occurs in — repair

Answer 39

• MLH1 + MLH2
• MutL proteins
• mistmatch

Question 40

—- replaces bases lost through chemical process such as deamination and depurination

Answer 40

base excision repair
( includes: dna glycolase , ap endocunxlease and exonuclease )(dna polymerase and ligase for repair)

Question 41

—- identifies and removes damaged base leaving apurinic or apyridindic sites
—- cuts the backbone
—- removes the sugar + several adjacent bases

Answer 41

dna glycosylase
AP endonuclease
exonuclease

Question 42

—- responds to helix distortion as pyrnimidne dimers , replaces regions of damaged dna of up to 30 bases in length , defence against 2 important carcinogens ( tabacco smoke and sunlight )

Answer 42

nucleotide excision repair
( read:Humans: 16 proteins
Mutations affecting different proteins in the
pathway identified in two disorders:
Cockayne Syndrome microcephaly, premature
aging, sensitivity to sunlight, developmental
delays, shortened lifespan.
Xeroderma Pigmentosum)

Question 43

—– is a rare human skin disease –
Autosomal recessive
Deficiency in nucleotide excision repair;
lack of enzymes necessary for repair of
DNA damage induced by ultraviolet
(UV) radiation (Thymine dimers)
Symptoms:
- extreme sensitivity to light
- skin cancer
- frequent secondary tumours & associated cancer-related
death (< 30 yrs. of age)

Answer 43

Xeroderma pigmentosum ( XP)

Question 44

mechanism for repairing the double strand break - nonhomologous end-joining NHEJ :

Answer 44

The “Ku protein” = a broken DNA
sensor, recognises double stranded
breaks.
The Ku protein holds both strands of
broken DNA leaving the ends
accessible to: nucleases
polymerases
ligases
Ends of broken DNA aligned,
trimmed or filled and strands ligated
Generates mutations
Deficiency - associated with
predisposition to cancer and
immunodeficiency syndromes

Question 45

mechanism for repairing the double strand break:
( recombination or homologous repair )

Answer 45

Uses enzymes and proteins that perform
genetic recombination between homologous
chromosomes during meiosis
Uses DNA sequence information in
homologous chromosome to correct the break
During S phase, sister chromatid is physically
close, providing a homology donor for repair
Non-mutagenic
Important in humans :
Defects in proteins BRCA1 and BRCA2
incidence of breast, ovarian, prostate,
pancreatic cancers
Mutation in genes coding for BRCA1 and BRCA2
= 80% lifetime risk