DNA Replication

Question 1

Structure of DNA - sugars

Answer 1

– ribose or deoxyribose
 C1’ – nitrogenous base attaches via N-glycosidic bond
 C2’ – H (DNA) or OH (RNA)
• lack of oxygen in DNA makes DNA more stable
 C3’ – OH in both
 C5’ – has 1-3 phosphate groups attached via ester linkage

Question 2

structure of DNA - nitrogenous bases

Answer 2

 Pyrimidines – 1 ring – cytosine, thymine, and uracil (RNA)
• Thymine has a methyl group whereas uracil does not
• Cytosines can have a methyl group added when next to guanine
o Modification that changes how proteins interact with DNA, influences its structure indirectly, and plays important role in regulation of gene expression
 Purines – 2 rings – adenine and guanine
 Deamination – spontaneous removal of an amine group (-NH2) from a nucleotide
• Common type of DNA damage
• Deaminated cytosine = uracil  easily detected
• Deaminated adenine = hypoxanthine  easily detected
• Deaminated methyl-cytosine = thymine  NOT easily detected and repaired

Question 3

structure of DNA - phosphate groups

Answer 3

 1-3 phosphate groups (alpha, beta, gamma) – alpha is closest to the sugar and the only one incorporated in nucleic acids; negatively charged
• Hydrolysis of other to phosphate groups provide energy

Question 4

Nucleotide

Answer 4

= a nucleoside triphosphate

nucleoside = sugar + nitrogenous base

Question 5

DNA Helix Structure

Answer 5

o Nucleotides joined by phosphodiester bonds
o Polar
 5’ PO4
 3’ OH
o H bonds between nitrogenous bases
 A-T base pairs have 2 H bonds
 G-C base pairs have 3 H bonds
o Purine always base paired with pyrmidine
o Anti-parallel strands that coil around an axis of symmetry
o Alpha-helix
 Bases lie perpendicular to axis of symmetry
 ~10 bp per turn
 Major groove = between turns
 Minor groove = between strands
o Proteins interact with DNA via the major and minor grooves
o Complimentary base pairs and antiparallel structure allows both strands to be used as templates for new DNA synthesis and allows for DNA repair

Question 6

packaging DNA into the nucleus

Answer 6

o Chromatin = DNA + proteins
• Euchromatin – less condensed; transcriptionally active
• Heterochromatin – more condensed; inactive
 Mitosis – chromatin becomes very condensed and chromosomes are visible
o 3 billion base-pairs of DNA compacted into nucleus 10,000:1
o Nucleoli – sites of ribosome synthesis
o Nucleosome – “beads on a string”
o Nucleofilaments – 30nm fibers of packed nucleosomes in a “2 start helix” consisting of 2 strands of nucleosomes stacked like coins
 Anchored via scaffolding proteins to form chromosomes

Question 7

chromatin structure and DNA replication and transcription

Answer 7

o Proteins involved in replication and transcription must get around histone proteins
o Histones have positively charged lysine amino acids that interact with negatively charged phosphate groups of DNA
o Histone acetyl transferases (HATs) – add acetyl groups
 Favors formation of euchromatin (active)
o Histone deacetylases (HDACs) – remove acetyl groups
 Favors formation of heterochromatin (inactive)
o Other types of post-translational modifications – phosphorylation, methylation, ubiquitination

Question 8

DNA replication - Cell cycle

Answer 8

 Checkpoints that regulate cell cycle progression
• G1 – are chromosomes intact after mitosis
• G2 – did all the DNA get replicated once and only once
 DNA damage results in cell cycle being halted and initiation of repair pathways
• If DNA damage cannot be repaired the cell undergoes apoptosis
 S phase – DNA synthesis phase ~6-8 hours
• Regions of genome that are actively transcribed are replicated in early S phase
• Regions of genome that are untranscribed are replicated in late S phase
 Licensing – process that ensures that all regions of DNA are replicated once and only once per cycle

Question 9

DNA replication features

Answer 9

 Semi-conservative – each of 2 daughter strands contain one strand of parental DNA and one strand of newly synthesized DNA
 Multiple origins of replication
• Bidirectional – fast and efficient – 5’  3’
• Replication forks
• Replication bubble – region of unwound DNA

Question 10

Initiation of DNA replication

Answer 10

 Each origin of replication is marked/licensed in late G1; once replicated in S phase the license is removed; ensures same area of DNA is not copied twice
 Origin Recognition Complex (ORC) assembles at origin throughout cell cycle
• Cannot initiate replication on its own
 Cdc6 and Cdt1 bind to ORC in G1 and recruits Mcm to form pre-replicative complex (preRC) – origin is now licensed
 S-Cdk phosophorylates Cdc6, Cdt1 and Mcm causing the disassembly of preRC and initiation of transcription
 Mcm has helicase activity that form replication bubble; it migrates with the replication fork

Question 11

movementof the replciation fork requires several proteins

Answer 11

 Helix must be unwound in order to separate the strands; supercoiling that results has to be relieved
 Helicase enzymes (including Mcm) move along one strand of DNA and change conformation to bind to double strand; using ATP they separate the two strands and return to original conformation
 Single strand binding proteins (including RPA) bind to single stranded regions and prevent the two strands from re-annealing
 Topoisomerases (type 1) – make reversible nicks in DNA ahead of replication forks, pass unbroken strands through the gap, and then reseal the gap
 Topoisomerases (type 2) – make double strand breaks in the DNA, allow uncoiling, and then re-ligate
• Clinical Scenario – Topoisomerases – targets for chemotherapy for cancer
• Clinical Scenario – Base analogs substitute for normal bases and inhibit DNA replication (often used for anti-virals or to treat cancer)

Question 12

DNA synthesis by DNA polymerase

Answer 12

 DNA polymerase creates a phosphodiester bond between 5’ PO4 of incoming nucleotide and 3’ OH of previous nucleotide
 DNA syntheses occurs in 5’ to 3’ direction but moves along the template 3’ to 5’
 First 3’ OH is provided by an RNA primer made by primase (an RNA polymerase)
 DNA polymerase has 3’5’ exonuclease activity which allows it to proofread by excising the last nucleotide added if it doesn’t base pair correctly with template
 Leading strand is copied continuously from origin of replication in same direction as replication fork
 Lagging strand is copied discontinuously in small fragments beginning at replication fork and moving toward origin; fragments are called Okasaki fragments
 RNA primers are removed by RNA hydrolases (RNases)
 DNA ligase uses ATP to make the final phosphodiester bond to seal nicks and join the fragments together
 PCNA (proliferating cell nuclear antigen) is a sliding clamp that increases the processivity of DNA polymerase; also involved in DNA repair, chromatin remodeling, and cell cycle regulation
 Multiple (atleast 14) DNA polymerases

Question 13

maintenance of epigenetic signals

Answer 13

o Epigenetics – study of heritable changes in gene expression that occur without changes in the primary DNA sequence
o Epigenetic Signals – histone modifications and DNA methylation
 Patterns are maintained during DNA replication
 Influence gene expression through effects on chromatin structure by regulating accessibility of the DNA to transcription factors
 Most forms of epigenetic inheritance are not passed to the next generation because they are erased during the production of germ cells

Question 14

patterns of DNA methylation

Answer 14

 On cytosines next to guanines on the same strand
 Associated with heritable gene inactivation (inhibits transcription)
 Important for development, X chromosome inactivation, imprinting
 Necessary to maintain chromosomal stability by keeping repetitive sequences in non-coding region in a repressed state
 Many CG sequences have mutated to TG via deamination throughout evolution
 CG islands (rich CG areas), often at 5’ end of ‘housekeeping genese’ that are constitutively expressed, are normally NOT methylated
 Since methylation in 5’ regulatory regions of genes suppresses gene expression, patterns of DNA methylation represent a type of epigenetic information that must be passed on to each of two daughter cells
 Same patterns of methylation are maintained during DNA replication by DNA maintenance methylase that recognizes hemi-methylated sites and methylates the other, newly synthesized strand

Question 15

histone code

Answer 15

 Histone tails often modified by acetylation, methylation, phosphorylation, ubiquitination
 Modification = epigenetic signal that helps regulate gene expression
 Histones are distributed during replication so that they leave gaps
 New histone proteins bind the gaps and are modified by reader-writer remodeling complexes to match the ones already bound

Question 16

problem of end replication

Answer 16

– because of removal of the RNA primer from the end of the lagging strand, the 3’ end of the parental DNA cannot be replicated completely
 Telomeres protect the ends of chromosomes
• Many copies of 6 base pair DNA sequence, plus proteins that bind
• 3’ overhang folds back and provides structure for proteins to bind
• Functions to protect ends of chromosomes from degradation and fusion
• Functions to distinguish ends of intact chromosomes from broken ones
• Telomeric DNA is lost with every replication of DNA, fewer proteins can bind; and risk increases that telomeres will not be able to protect ends of chromosomes
• Loss of telomeric DNA triggers cell cycle checkpoint mechanisms and causes the cell to stop growing permanently (cells are said to be senescent)

Question 17

telomerase and mechanism of activity

Answer 17

• Length of telomeres is maintained in germ line cells and some stem cells by telomerase, a ribonucleoprotein that carries its own RNA template complimentary to the G-rich strand of the telomere
o Enzyme is NOT detectable in normal somatic cells
o Found in germline cells, stem cells and >85% of cancer cells
• Expression associated with immortalization of cell lines
• Uses RNA template to extend 3’ end of parental strand (reverse transcriptase)
• Experimental cancer vaccines are being developed that target telomerase as tumor-specific antigen

Question 18

mitochondrial DNA

Answer 18

o Some mitochondrial proteins are transcribed and translated in the mitochondria but most are transcribed in the nucleus and translated in the cytosol before the protein is imported to mitochondria
o Mitochondrial DNA is more economical than nuclear DNA
 Every base pair is used to make a functional product whereas nuclear genome has a lot of repetitive DNA
o Mitochondrial DNA has higher mutation rate than nuclear DNA (5-10 times)
 Partly due to its close proximity to sites of generation of oxygen radicals but also may be due to less efficient repair mechanisms
o Accumulated mutations in mitochondria DNA have been suggested to contributed to decreased efficiency of oxidative metabolism with increased age and overall aging process