Mcdb 144 Lect. 1

Front 1

1.what is most condensed state

Back 1

mitotic nucleus

Front 2

1. metacentric

Back 2

centromere in middle ( humans)

Front 3

1. acrocentric

Back 3

centromere closer to one end than the other (mice)

Front 4

1. telocentric

Back 4

centromere located at the terminal end of the chromosome

Front 5

1.nucleotide vs. nucleoside

Back 5

Nucletide: a sugar + phosphate + base (A, G, C, T)

Nucleoside: sugar + base

Front 6

1.chromatin

Back 6

DNA + Histones

Front 7

1.histones

Back 7

responsible for first and most basic level of DNA packaging

-N terminal tails are modified, extends out of the core to be subject to covalent modifications that control critical aspects of chromatin structure and function
-all four histones have a characteristic fold

Front 8

1.each nucleosome

Back 8

nucleosome core particle consist of H3-H4 and H2A-H2B dimers.
both combine to form 2 sets of tetramers and
2 tetramers combine to make 1 octamer

Front 9

1.histone proteins are conserved

Back 9

most higlhy conserved proteins across species
ex. aa sequence of histone H4 in peas and cows differ by 2 aa
high conservation suggest important functional role
-mutations in histones are usually leathal (e.i if no N terminal tail then die)
-NON-LETHAL CHANGES affect GENE EXPRESSION
if change aa in folds, changes how histones assoicate with e/o, wont connect properly, might not compact tigtly
or if mutate a critical residue that needs to be modiend in N terminal tail then can cause a change in gene exression

Front 10

1.histone octamer

Back 10

are rebuilt everytime a cel divides bc when cell divides you are doubing DNA, so double content of histone to form 4histone dimers and 2tetramers and 1octamer

Front 11

1.DNA winding preference of MINOR groove inside

Back 11

A and T dinucleotide preered in minor grove inside, needs flexibitlity! A and T form 2 H bonds between DNA and nucleosome

Front 12

1.DNA winding preference of MINOR grove outside

Back 12

G-C preferred bc dont need same flixibitly as that side is away from nucleosome: have 3 H bongds

Front 13

1.Nucleosome Remodeling

Back 13

-nucleosome is not fixed ( always moving!)
-DNA winids and unwinds from each nucleosome ~ 4x/sec
-DNA remains exposed: this is req for enabling access of the DNA to enzymes such as those req for TRANSCRIPTION or REPLICATION

Front 14

1.Nuclesome sliding

Back 14

need ATP dependent chromatin remodeling complex will push the nucleosome to reveal DNA so instead of being wrapped will be in the linker dna region

Front 15

1.histone 1

Back 15

linker histone
less conserved, 1:1 ration with nucleosome cores
-changes the exit path of DNA out of the nucleosome core
-important for compaaction of nuclesomes to generate the 30 nm fiber
-if dont have Linker histones wont have basic compacton of DNA and cell will not survive

Front 16

1.zig-zag model predicted by

Back 16

X-ray chrystallography

Front 17

1.solenoid model predicted by

Back 17

cryo electron microscopy

Front 18

1.histone assembly is mediated by

Back 18

histone chaperons

Front 19

how many single stranded DNA in each chromatid?
in 2 sister chromatids ( 1 chromosome)?
in 1 homologous chromosome1.

Back 19

2 ss DNA
4 ss DNA
8 ss DNA

Front 20

what is the purpose of having exposed histone tails?

Back 20

so other proteins can bind and make DNA functional

Front 21

1.epigenetic inheritance

Back 21

cell memory that results is inherited by PROTEIN STRUCTURES rather than change in DNA sequence

Front 22

1.genetic inheritance vs epigenetic inheritance

Back 22

genetic in heritance: direct inheritance of DNA nucleotide sequence during DNA replication. If have a DNA CHANGE, then will have a PERMANENT CHANGE in both somatic and germc cells bc DNA is transmitted faithfully to somatic and germ cells
-if change happens: somatic cels changes and predispose to diseas , in germ cells will cause fetal demise

*Epigenetic inheritiance: baed on molecules bound to DNA and is LESS PERMANENT. creates a permanent state in somatic cells but in germ line ti gets erased in formation of eggs and sperm so next generation can be developed

Front 23

1.heterochromatin

Back 23

usually in centromeres and telomeres and on the edge
DNA contains few genes
high degree of compaction
-DNA regions containing genes that do not need to be expressed or highly resisant to gene expression
-euchromatin can turn to heterochromatin and therefore genes are turned off by this action, the genes are said to be silenced

Front 24

1.position effect

Back 24

differences of gene expression in which activity of the genes depends on its position relative to a nearby region of heterochromatin

Front 25

1.heterochromatin is important because..

Back 25

it is critical to prevent disease so have genses that dont want to be expressed in heterochromatin region
ex. gene glia dont want to turn on in adult brain bc if it is can cause problems with existing neruons in glia and can cause cancer

Front 26

1.spreading of heterochromatin exp with Drosophila

Back 26

hetero is usually prevented from spreading to euchrom regions by barrier
-when fleis that INHERIT certain chromosla rearrangements with no barrier , heterochrom can spread to euchrom regions with diff distances and to diff cells and then the spreading stops to make an ESTABLISHED pattern that is then inherited to make larger amonunts of clones ofprogeney that have the same neighboring genes CONDENSED into HETEROCHROM and thus INACTIVATED

Front 27

1.Chromatin can move to diff sites within the nucleus to alter GENE expression

Back 27

diff regions where chromosomes can move as they are subject to diff biochemical processes such as when their gene expression changes (transcripiton, binding to RNA proteins for transcription ect.)
-diff nuclear neighborhoods are knownw to have distnct effect on gene expression
-movemetn of chromosomes presumably reflect changes of chromatin, rna binding affinities SURROUNDING A GENE for the diff NUCLEAR NEIGHBORHOODS

Front 28

1.chromosome packing and histone state determines..

Back 28

wherether DNA is found in hetero or euchrom state

Front 29

the N-terminal tails are modified by a variety of covalent modifcations

Back 29

mono, di, tri methylation (SPECIFIY WHICH ONE! BC CAN HAVE DIFF FUNCTIONS)
acetylation
phosphorylation
ubiquitylation

MODIFICATIONS ARE REVERSIBLE! modifications can be removed by other enzymes, UNLike DNA where changes are PERMANENT

Front 30

acetyl groups ADDED by

Back 30

histone acetyl transferases (HATs)

Front 31

acetyl groups are REMOVED by

Back 31

histone deacetylases (HDAC's)

Front 32

methyl groups ADDED by

Back 32

histone methyl transferases (HMT_

Front 33

methyl groups are REMOVED by

Back 33

histone methyl deametylases

Front 34

histone code hypothesis

Back 34

suggest that histones INSTRUCT the cell about transcription, DNA damage + Repair, expression, suppression

Front 35

code reader complex
leading to gene expession gene silencing or other biological functions

Back 35

a scaffold protein + protein modules that bind to specifific histone modifications on nuleosome = iguana

-code reader binds and attract other components such as protein complex with catalytic activities and aditional binding sites

Front 36

modification of H3: methylation of lysine (K) 9

Back 36

heterochromatin formation , gene silencing

Front 37

modification of H3: methylation of lysine (K) 4 and acetylation of lysine (k)9

Back 37

gene expression

Front 38

H3: phosphorylation of serine (s) 10 and acetylation of lysine (K)14

Back 38

gene expxression

Front 39

H3: methylation of lysine (k)27

Back 39

silencing of Hox genes,
X chromosome inactivation

Front 40

acetylation of lysine on N terminal

Back 40

tends to LOOSEN crhomatin structure bc adding an aceytyl gorup to lysine removes its positive charge thereby reducing the affinity of the tails for adjacent nucleosmes

Front 41

histone variants of H3

Back 41

H3.3
CENP-A

Hint 41

H3.3: transcriptional activation
CENP-A: centromere function and kinetochore assembly

Front 42

histone variants of H2A

Back 42

H2X
H2Z
macroH2A

Hint 42

H2x: DNA repair and recombination
H2Z: gene expression & chromosome segregation
macroH2A: transcirptional repression & X-chromosome inactivation ( in X chromosomes that become silent)

Front 43

when DNA is damaged what histone variant is incorporated at damaged sites?

Back 43

H2X

Front 44

code writer

Back 44

enzyme that creates a specific modification on one or more of the four histones.
-recruited by gene reulatory protein
-collaborates with code-reader to spread its mark from nuclosome to nucleosome by reader-writer complex

Front 45

reader-writer-ATPdependent chromatin remodeling

Back 45

packs nucleosomes makes them more condensed , spreads heterochromatin for long distances along the chromosome

Front 46

3 ways to stop reader-writer complex

Back 46

1. protein will bind to nuclear pore: can form a barrier that stops spread of hetero chrom. hetero is near the nuclear envelope
2. tight biding of barrier proteins to a gropu of nucleosomes to competer with heterochrom stoping the spreading of reader-writer complex
3. barrier protein binding to sequence of DNA that can erase the histone marks that are required for heterochrom spsread

Front 47

alpha-satellite DNA

Back 47

special repeat type of DNA at human centromeres
-AT rich
-171 base pair monomeric unit repeats , which are arranged in tandem arrays
these tandem arrays have CENP-A hisntones
-to be human centromere need CENP-A

Front 48

if no CENPA

Back 48

no centromere Identity

Front 49

CENP-B

Back 49

binds to CENP-B box (17bp seq) in ALPHA satellite DNA
-CENPB boxes are found in every other repeat of the 171 bp satellite DNA
-NOT a histone variant
-HELPS POSITION NUCLEOSOMES
acts with CENP-A and CENP-C to form a centromere

Front 50

have centromeres to

Back 50

to build microtubule apparatus to pull sister chromatids apart in mitosis

Front 51

the use of histone variant to

Back 51

to direct cell division machinery..help assemble proteins some of which form the kinetichore and capture the microtubules

Front 52

centromere assembly during mitosis

Back 52

cenpa is interesting in that its loaded at specific times during the cell cycle unlike other histone variants that can go in and out and loaded at any time of the cell cycle
cenp-a only loaded inG1 phase
G2 entering into meiosis: cenpa begins to be expressed ( the protein begins to be translated) cenpa present in nucleome as cell division and DNA rep is occuring

Hint 52

at replication fork nucleosomes are distributed randomely in the 2 sister chromatids at the end of s phase have 1/2 amount of cenpa protein entering into mitosis with 1/ 2 critical protin necessary to pull chromatids apart
at G1 create a full centromere.

Front 53

HJURP loading of CENP-A in late telophase/early G1

Back 53

cell incubated in culture with siRNA (preventing translation or removing RNA) of gene of interest vs. si-HJURP (removing HJURP protein)
-si-control looking if expressing a fussion protien between GFP and CENPA means wherever see GFP it is where cenpa is localized: fussion protien
-if KO HJURP end up with greenbut green but not dots so CENPA cant be loaded into nucleosomes
-this experiment proved that HJURP is necessary to load cenpa into nucleosome does it in G1 phase

Front 54

what is HJURP:

Back 54

holiday junction recognition protein
-intially ID in a screen for proteins over expessd in human lung tumor cells
-expression of HJURP is induced in response to DNA damage
-at first was thought to only be involved in DOUBLE STRAND BREAK but now know that its critial in cenpa loading

Front 55

neocentromeres

Back 55

centromere has alpha satellite sequence with cenpa and H3 that form microtubules, can alos have other regions in chromosomes with alpha satellites but dont have cenpa so cant form microtublues... so MUST have cenpa to make microtubules
-are other chromosomes that are fromed from chromosome breakage that have 2 microsatelites with cenpa, these chromosomes dont have alpha microsatelites but do have cenpa so can form microtubules

Front 56

extra centromeres?

Back 56

excess amount of cenpa or incorrect loading of cenpa in wrong places can create extra centromeres, this leading to hypothesis that extra centromeres can lead to cancer
-bc if had 2 centromeres DNA can get ripped and end up with translocations, floating chrom and can add itself anywhere to another chrom,
-extra centromeres possible to cause deletions by pulling aprt dna or by causing translocation mutaiton

Front 57

Y chrom

Back 57

sex chrom are not homologs
-y chrom is about 1/3 the size and recombines only at the tips
-autosomes recombine along the lenght

-females are capable of homologous recomb bc 2 XX

Front 58

understanding evo of Y chrom

Back 58

sex chromosomes began as AUTOSOMES called protoX and protoY 300 million yrs ago
-evo biologists monitored Y chrom evo by examining its genomic sequence to compare with other species
-by comparing diff species discovered structural changes in teh Y resulted i the modern form we have now

Front 59

evo of Y : 200-300 Million yrs ago

Back 59

Proto X and ProtoY recombined during meiosis (like autosomes could do homologous recomb)
-both X and Y contained Sox3 gene
-mutation in Sox3 gene resulted in conversion to SRY (maleness)
-X an Y still have RPS4 and retain similar function

Front 60

evo of Y: 150-200 milllion yrs ago

Back 60

-Y chromosome recombined with itself forming an INVERSION (rare): exrremely important, big effect
-the X and Y NO LONGER HOMOLOGS, everything in between diff, only homologs at the tips
-no longer capable of recomb
-SRY now located at top
-due to its inability to recombine proto Y accumulated mutation that werent able to be fixed bc no longer homologs with proto X so became susceptible to deletions

Front 61

evo of Y: 130-170 million yrs ago
(after marsupials)

Back 61

a SECOND INVERSION scrambled loci on teh Y Chrom
-in females (XX) recomb occured along the legth
-in males recomb was found exclusively at the ends further deletions

Front 62

evo of Y: 130-50 million yrs ago
(after mammals divereged from marsupials)

Back 62

- a bloc of autosomal genes translocated to the X and Y
-recombination occured in these regions
-Y chrom continued t odev inversions, gene order was further rearranged
-w/o recombintation to preserve the Y chromo integrity regions were deleted and the Y continued to shrink

Front 63

evo of Y: 50 million yrs ago
(after primates diverged from squirrel monkeys)

Back 63

-an AUTOSOMAL DAZ gene relocated to the Y chrom
-DAZ gene was copied MULTIPLE TIMES
-DAZ is critical for spermatogenesis
-these mulitiple compies of DAZ ensure that small deleiton in the Y do not cause INFERTILITY
-Females also have DAZ gne bc have autosomal DNA

Front 64

will Y evolve itself to extinction

Back 64

lack of Y chrom recomb resulted in what is hypotthesized to be genomic wasteland in central regions and no mechanism to repair damaged DNA
-found Y chrom regions of genome that have palindromes

Front 65

1milion year old Palindromes

Back 65

-the "GENOMIC WASTELAND" contained unique stretches of palindromic DNA
(replicated GENES in MIRROR images of e/0)
-hypothesized that palindromes allow genes that have developed mutations to recombine with normal genes for repair

Front 66

there are 8 palindromes on the Y chrom

Back 66

palindromes: like hair piins that have DNA sequences that can be read the same way either direction from 5-3' or 3-5'
- if there is damage, the chrom hsould be able to loop back to itself such that damage can be repaired off palindrome
-function like homolougous but its a mirror image unique to chrom Y

Front 67

summary of Y chrom

Back 67

Y chrom origins as autosomes with full recombination ability
-inversion event resulted in lack of recomb
-lack of recomb resulted in accumulaiton of mutations and deletions
-Y chrom began to reduce in size
-translocation of autosomal loci involved in spermatogenesis to the Y chromosome enabled critical male fertility loci to be clustered onto Y chromosome
(DAZ gene)

Front 68

3. DNA rep is semi conservative

Back 68

pink strand is the new trand generated in s-phase
-immortal strands because will remain the cell populaiton after replication
-immortal cells are instructed to remain in stem cell populaitons nothing will degrade it

Front 69

origin of replication

Back 69

repication in euky intiated in Sphase
-by end of S phase each chrom contain a sister chromatid joined with centromeres
-there are multiple origins of replication on a given chrom (this is how they are coordinated)
-given that S phase last 8 hours not all origns of rep are initated at the same time

Front 70

how does chromatin affect replication

Back 70

heterochromatin is always replicated LATE in s Phase
-therefore thought that orgins of rep associated with chromatin state
-determined this by looking at x chrom in females that is tightly compact, silenced by heterochrom and replicated late in s phase its one of last to get replicated

Front 71

DNA sequnce that serves as origin of replication contains

Back 71

1.binding site for ORC (origin recognition complex) : multisubuniti initiator protein
2. AT rich DNA sequence : easy to unwind
3. at least 1 biding site forproeites that help attract OR tor origin

Front 72

ORC
origin recognition complex

Back 72

binds to chromatin at replication origins throughout the cell cycle
-serves as foundation for assembly of pre-replicative complex includes: Cdc6 and cdt1 plus helicases (recruited by cdc6 and cdt1) in G1

Front 73

what happens if dont have Helicase

Back 73

there will be NO DNA replication because its a high energy protein complex necessary to unwind DNA

Front 74

what holds RNA poly to DNA strand?

Back 74

sliding ring
- w/0 this protein poly falls off
its is placed by clamp loader with ATP on the DNA at the primer-template junction ,
-Then DNA Polymerase is recruited to the sliding clamp

Front 75

DNA primase

Back 75

generates short 10 nucleotides of RNA primers

use ribonucleotides triphosphates to make rna primers
-are in 100-200 nucleotides on the lagging strand

Front 76

lagging strand involves:

Back 76

1. DNa priming
2, erasing
3. ligating

clamp and DNA poly extend okasaki fragment until reach next primer where fall off bc of double strand

Front 77

topoisomerase

Back 77

like a reversible nuclease that adds itself to DNA backbone covalently , breaking the phosphodiester bond in the DNA strand. since its reversible the phosphodiester bond reforms after the protein leaves

Front 78

topoisomerase 1

Back 78

produces single strand break (nick)
break in phosphodiester backbone allows DNA helix to ROTATE freely
-bc covalent linkage that joins DNA topoisomerase to DNA phosphate retains the energy of the cleaved phosphodiester bond, resealing is rapid and does not require additional energy. so no DNA ligase needed

Front 79

topoisomerase 2

Back 79

forms a covalent linkage to both strand of the DNA at THE SAME TIME making a transient double strand break in the helix
-are brought when 2 double helices cross over each other
-use AtP hydrolysis to

Front 80

polymerase proofreading
self correcting enzyme

Back 80

when have mismatched base pair at 3' OH , not effective bc polymerase cant extend strand
corrects with 3-5' EXONUCLEASE
continuing until enough nucleotides have been removed to regeneare correctly base paired 3'OH terminus for DNA synthesis

Front 81

why does poymerization occur from 5-3'

Back 81

the need for accuracy explains why its 5-3
-if 3-5 then growing 5' would proide activating triphosphate needed for covalent linkage rather than incoming mononucleotide triphophate, polymerization would not continue becasue 5' end does not hae extry 2phosphates to releae thus terminating synthesis

Front 82

histones are synthesized

and requires

Back 82

mainly in the S phase

requires histone chaperons to add histones

Front 83

telomerase

Back 83

recognize the tip of the esisting telomere (tandem repeats of short sequences gggtta)and elongates it in te 5-3 direction using an RNA template that is part of the enzyme (like reverser transcriptase by using RNA template (in enzyme) to make DNA strand
then continued by DNA poly that carries a DNA primase as one of its subunits

Front 84

5. repair mechanisms

Back 84

single nucleotide repairs
small regions of mismatch
double strand breaks

Front 85

DNA poly has 2 proofreading mech

Back 85

1. correct nucleotide has a high affinity for the moving polymerase
2. conformational change in DNA poly is required AFTER nucleotide biding but BEFORE covalent addition check DNA strand (finger tighten)

Front 86

DNA poly classic repair pathway

Back 86

fine for randomly incorporated NUCLEOTIDES

for exonucleus to fuctio, the primer stand will unclouple from template strand for exonucleus activity to occur

Front 87

what happens in catastrophic cases of DNA damage?

Back 87

classic DNA repair does not function so DNA poly STALLS when encounteres damage
-LESS ACURATE dan poly are used
-human cells have more than 10 such dna poly
-some recognize specific types of DNA damage
-others make "good guess"
*they LACK PROOFREADING ACTIVITY

Front 88

why is DNA repair so importan ?

Back 88

germ line cells dangers because permanent change that is inherited from mother to dughter
in somatic cells bit less damaging because not permanent but can still lead to diseases, cancer

Front 89

what creates/ propagates DNA damage?

Back 89

UV radiation, chemicals (ENU; creates mutaions in celss damagin nucleotides), ionizing radiation (X-ray machines) and errors in DNA repairs mechanisms

Front 90

what isotope released in atomic test in nevada

Back 90

realease of isotope I 131,
atomic cowboy sheppered cattle above ground to see what effect the test had on lilvestock, then was moved undergrown where the I 131 moved into soil

Front 91

syndromes caused by defects in DNA repair machinery

Back 91

BRCA2: breast, ovarian and prostate cancer: repair by homologous recomb

Xeroderma pigmentosum : skin cancer, uv sensitivity , neurological abnormalities: nucleotide escision reipair

fanconi anemia : leukemia, genome instability: DNA interstrand crosslink repair

Front 92

repairing single nucleotide errors
(damage that occurs spontaneously)
2 most common SINGLE NUCLEOTIDE ERRORS

Back 92

DEPURINATION: N-glycosyl linkage to deoxyribose hydorlyses:
humans lose about 5000 purines a day (A, G) left with sugar phosphate

DEAMINATION: removal of NH3 groups from bases
from NH2-->C=O

Front 93

deamination of cytosine give

Back 93

uracil : unnatural base product for DNA but natural for RNA

Front 94

deamination of thymine

Back 94

no deamination bc no amine to hydrolyze deal with already has C=O

Front 95

special case of METHYLATED C residues

Back 95

5-methyl cytosine -->deamination-->thymine (natural base)

Front 96

pyrimidine dimers

Back 96

occurs in cells exposed to UV irridation
-forms between two neighboring pyrimidines
-most common are thymididne dimers

Front 97

method by which chemical modifications produce mutations

Back 97

chemical modification: deamination
only has to happen on one strand
originally C-->deaminated -->U so doesnt really attach to G
-goes through DNA replication
top strand with mutation contains U so have a new strand with an A nucleotide
bottom strand with normal nucleotide G is paired with c : unchanged DNA

Depurinated chemical ex:
depurinated A: so no A nucleotide on strand goes through DNA replication :
top strand no base (mutated) so loss of nucleotide pair: AT bp has been deleated
bottom strand: normal where T is paired with A
* if not repaired will have 1 daughter cell with mutation and other daugher cell WT

Front 98

Base Excision repair

Back 98

ideal for single nucleotide mismatch: depurination and deamination bc removing single nucleotide
-uses enzymes: DNA GLYCOSYLASES
-recognizes altered bases in DNA and catalyzes its hydrolytic removal from the surgar
recognition is based upon a "FLIPPING-OUT" mechanism
-then the catalyzed base is recognized by DNA ENDONUCLEASE by missing tooth to remove sugar phosphate
-DNA POLY repairs the gap with DNA LIGASE

Front 99

Nucleotide excision repair

Back 99

used to remove large changes in structure of DNA such as pyrimimdine dimers or BULKY LESIONS
-multienzyme complex scans DNA for distortions
cleaves strand on sides of leasion, DNA HELICASE necessary to damaged ssdna and requires DNA poly and DNA ligase

Front 100

Transcription-coupled repair

Back 100

linking RNA poly to repair DNA damage
-RNA poly wills tall on DNA lesions and use coupling proteins to direct repair machinery to these sites
-works with BER, NER to direct repair immediately to the cells most important DNA sequnces , like those expressed when the damage occurs
if RNA becomes permanently stalled at sites of DNA damage that lie in important genes can have Cockayne's syndrome

Front 101

Cockayne's Syndrome

Back 101

homozygous recessive disorder
-children age very quickly
-growth retardation, skeltal abnormalities, severe sensitivity to light
-bc rna ploly become permanently stallled at sites of DNA damage in important genes
- therefore important genes are not transcribed

Front 102

repair of double strand breaks

Back 102

occurs when both strands cleaved
cuased by ionizing radiation (I131, replication error, oxidizing agents cell metabolites)
-if not repairedcan result in loss of entire chromosomal regions and genes which are not segregated in mitosis
-Non homologous END JOINING
-homologous recombination

Front 103

non-homologous end joining

Back 103

accidental ds break; so loss of nucleotides due todegradation from ends, (consequence: have deletion of DNA sequence where break occured); have an end joining
- is common repair mech used by cells , its primary mech in some cells of the body

Front 104

end recoginign by

Back 104

ku heterodimers , that will recruit additonal protiens that process dna end with limited repair synthesis and ligation
-repair has suffered a deletion of nucleotides
quick dirty method

Front 105

DNA Ligase IV

Back 105

acts with XRCC1 protein in NHEJ to COVALENTLY link 2 strands of DNA

-involves 2 compies of the XRCC! and DNA ligase IV to bind to DNA with nicks and broken ends

Front 106

homologous recombination

Back 106

use sister chromatid for repair:
have accidental break , loss of nucleotides due to degraditon from edns , damage repaired ACCURATELY using info from sister chromatid

Front 107

DNA damage and the cell cycle;

Back 107

HOMOGOUS RECOM DOESNT JUST REGULATE in sister chromatids from s phas e it can occur off the homoogus chrom in any phase of the cell cycle depding of what

Front 108

rapid localization of rpair proteins to double strand breaks

Back 108

1. DAPI: dye that recognized DNA , 1 nucleus within 1 cell ; investigators put grids over cell , the grids protected nucleus from damage (CONTROL no damage)
2. BrdU : synthetic nucleoside analogous to Thymidine used to detect replicatio of cells; black where girds are, not labeled with BrdU; BrdU incorporated so see green , incorporated in ares that are damaged, doesnt tell you what type of damage bc incorporated by dNA poly just tells you that poly is functioning; funciton in BER, NER< NHEJ, HR
3. Mre11 (double strand repair protein) part of HR machine ; majoritiy of reapir is due to HR can tell bc Mre11 complex is recruited tospecifically to double strand breaks that go to homologous recomb

Front 109

6. homologous recomb ;

Back 109

the exchange of genetic material btwn homologous DNA sequences
-thre major bio uses of homolous recomb:
-ds breaks,
-meiosis
-repair stalled rep fork

one major exp use of HR:
generatio nof Null mutant mice

Hint 109

-

Front 110

non Productive

Back 110

nonpairing interactions , regions not homologous, gene A may pair with Gene C

Front 111

productive :

Back 111

pairing interactions, gene b with gene b, so go through nucleation and zippering

Front 112

repairing at Replication Fork Error
SINGLE STRAND INVASION

Back 112

have nicks created by topoisomerase, these error canoccur frequently bc are in teh active DNA replication and are creating nicks to unwind DNA
-if nick not repaired results in rep fork break
-so sister chromatid terminates too short
-the other okazaki fragment hasnt been created
-so needs to go through homologous recombination to do that need EXONUCLEASE to digest the 5' strand of DNAto creat 3' overhand of newly synthesized DNA
-the 3' overhang will invade the other strand of the other sister chromatid that was forming the lagging strand
-invades into the double helix (open helix)
-have 3' invasion
with a breakage of the newly formed lso that the top strand now has a small piece of the old laggin stand
and have additonal DNA synthesis that will continue the generation of the leading strand and the remainder of the older lagging strand
can continue generating gen through okazking and leading

Front 113

RAD 51 (RecA) protein

Back 113

necessary for HR
-acts after EXONUCLEASE before starnd invastion
-exonuclease allow binding of proteins that recognize ssDNA
--ssDNA bidng proteins
--Rad52 (binds ssDDBP)-recruits Rad 51
--Rad 51( RecA)

Front 114

RAD 51

Back 114

binds tightly in cooperative cluster to ssDNA ; direct the invasion
-like SSBP, RecA/rad51 bids tightly in long cooperative cluster to form NUCLEOPROTEIN FILAMENT
-B/c each RecA has more than one biding sitee can hold single strand a double helix together
-allows for DNA synpsis between the double helix and homogous region of single stranded DNA
-forms heteroduplex: region of DNa double helix formed by the pairing of 2 dna strands that ere from diff dna molecules

Front 115

Branch migration

Back 115

once invation , strand exchange can move in branch migration , wher an unpaired region of one othe siggle strand displaces a paired region of the other ss stand moving the branch point w/o chaing the toal number od dna bp
1. can either go spontaneou branch migration where makes little net progress, but can move either direction
2. Helicase mediated branch migration expend energy ATP to move branch point thousands of base pairs at a particular direction

Front 116

HR can slo repair dsDNA breaks

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recombination occur afte cell has replicated its DNA and sister chromatid can be used as template
-double strand break caused by radiation , exonuclease degreades 5' ends: bind ssb proteins, Rad52 and Rad51 (hod together invading and origan strand to facilitate to go to branch migreaion) then have synthesis and migragon ; moving using bottom strand ALWAYS USE HELICASE

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EX of UNIDERECTIONAL BRANCH MIGRATION

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Is ejected from duplex and reconnects with original strands , DNA synthesis provied 3' and ligase

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wh is meiosis improtant

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one set of chrom is obtain from each parent ( 22 autosomes and 1 sex chrom)
during meisois the homologs recognize e/o and separte ito cells
during pairing homologs exchange DNA to create diversity

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diversity

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have haploid cell with pair of homologs go through chrom duplication and meiosis an dcome out with 4 cells with chromosmes that have crossed over ..all come from 1 chrom but rearranged

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meiosis overview

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1 round of replication
2 rounds of cell division

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meitotic prophase 1

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leptotene
zygotene
pachytene
diplotene

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leptotene

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condensation

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zygotene

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synapsis

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pachytene

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recombination

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diplotene

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chiasmata (preparing to separate apart)

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synaptonemal complex (SC )

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complex req to facilitate homologous recomb of meiotic cells (this is not req for DNA damage)

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steps to generating SC

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1. begins ith formation of lateral elements to form proteinaceous coure with siter chromatids of one homogours paoir (Leptotene)

2. during zygotene and pachytene the homogous pairs boudn by their lateral elements aight along thier lenght by transvers fibers for pairs to synaps w eo
3. a central element forms in the middle btwn the lateral elements of the homologours pair through the transverse fibers
4 this proces enables pairng and synapsis

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meiosis homologous recomb. occruing

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only in prophase 1 meisois 1
have SC that hold chromosome togher to allow homologous recomb evnets to occur

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mammalian protiens req for SC

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SCP1- transverse filament
SCP2 -lateral elements
SCP3-lateral elements

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using homologous recomb as a tool to understand the importance of scp1 and scp3

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use immunoflurescence to study celluar locaizaiton of these protiens
-know have nuclei bc dapi stain in early leptotene ; begin to have the production of SCP3 but its not localized and not loades
Leptotene: protein beings to accumualte
zygotene: scp3 is coating the homogous chrom; chmosome not fully condensed yet and homogous chrom havent found e/o not paired with e/0 but have spc3 coating on them
Pachytene: have chromo that have synapse with e/0 looks like fewer strands bc pair w e0 cant dsitingusih homolous strands fully synapsed
diplotene: homogous chrom are stargin to move aprt from eo , homolgous recm occurs at pahytne during diplotene pull apsrt recobm will be resoved

after diplotene: SCP complex breaks down and is no longer observed

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scp3 gene is required for SC assembly , chrom synapsis and male fertiltiy

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scp3 mutatn do not have synaptonemal complexes : spermatocytes fromt the testis staine with silver nitrate to detect protien on condensing chrmo
-WT: squggly line pachytene : can see thembc the dna is coated inprotein of spc3 complex taking upthe silver nitrate
in mutatn: is partially looking suqiggly not fully so telling us tath dna is dd from wt so theere msut be somehting wrong w the assemply of the sc proteins

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scp3 mutanat cells

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A. zygotene: chrom not condensed
B. pachytene: chromo paired and less centromeres
c -/-

scpe protein is staine d in greeen
centromere is stained in red
in murine spematocytes we would expect 20 centromeres signals if chromosome were undergoing synapsis B
the mutants have double the number of centromeres indidcation faiure of synapsis
so dont have SCP3
no scp3= no synapsis

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scp1 ( transerse filament)

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part of transverse filaments of SC
-parallel coil-coil homodimer with c-terminal embedded in in lateral elements of opposing homologous chrom
-N term overlap in the central element created by pairing of 2 homologous chrom
-in yeast loss of scp1 does not have disrupts pairing but cross over is reduced
-in yeast loss of scp1 does not disrupt pairing but cross over is reduced bc cross over is mediated by scp1

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scp1 mutants

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a/b: high powered micrograph of permatocytes hmoogous chrom

a: have 2 strands in WT
AE: axial lateral ( dark black line
CE: central element gray stuff inmiddle where SCP1 is

B: mutant
AE: axial but missing CE
so if dont ahve SCp1 in mice are missing the normal structre of the central element i the synaptomenal protien
KL : nuclei that are stained with REC 8 (protein found in axial lateral) and SYcp3 markers for axolateral element; begining to et the production of protien both look norma lin leptotene , tells us that the production of scp3 is not affected in absece of scp1 , so if delet scp1 does not affect production of scp3. if it was woulnt see anyscp3 in leptone stage

Q: homologour chorom syanps w eo
R : mutant
same color yellow sc in oth so scp3 must be able to colocolaie with rec8 and form axial elements

R: homologous chrom found e .o but not able to undergo synapses so wont undero homogous recomb

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double strand break programmed by

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spo11 : uniq to meiotic homoglogous recomb bc it is meidaing the double srand breakage on one of the sister chromatids

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spo11

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recruited with mre11 nuclease comples (also from general recomb) make 1 chrom cut and then spo11 is dissociated
-exonuclease esposes sibble sranded 3' bc eats 5
-have RAD51 hlp invade
-DNA synthesis for branchpoing formation

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why do programmed double strand break not occur inevery cell in mammals?

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bc its regulated in the cell type ina spceific way
only testis have a signal , have organ specifid an dcell ltyp specififc epression of spo11
only express in meiotic cell i time of meiosis
ovary is negative bc femals preform fetal ife meiosis occuring during fela ife / fetal germcesll so time specific vs males happens as addults

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spo11 req for meiotic chrom synapsis

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in meiosis w/o spo11 cant complee but if dont form ds break wont get recuritment of rad51 bc spo wont be breaking ds

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spo 11 mutants arest in zygotene

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c: WT rad 51 able to be recurited to the zygotene chrom bc of sp11 ds breads

d: mutatnt in meiosis with no spo11 nto able to complete break aos wont have recruitent of rad 51
missing rad 51 bc delel spo 11

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resolution of meitoic exchange

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1. non cross over
2. cross over ( unique only to meiosis)

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non cross over

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gets ejected then repiired
if nicks occur across branchpoint will have NCO . will have a lot of heteroduplex in that area

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cross over

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creats 2 branchpoing
"double hoiday junction, not ejected remain in the ds dna reconnect to orignal strand , must nick strands rather than eject them

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cross over interference

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90% OF SP11 EVENT ARE RESOVED IN NON CROSS OVERS : FORMATION OF HETERODUPLEX
crossovers ar uniformly distributed alon gthe chrom
-if 1 of the sp11 ds break is forming a CO will inhibit cross over in neighboring region
-for most orgnanisms roughty 2 CO per chrom occur during meiosis

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heteroduplexes

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regardeless of whter a meitoic synapsis result in a cross over or NCO the machinery leaves behind a hetroduplex
-the hetroduplex can tolerate a small amount of mismatch which wil be repaired , bc not all chom are identical each other they

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gene conversion

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when heteroduplex sequences are not always same
-mismatch repair excise portion of blue repaired by NUCLEOTIDE EXCISION REPAIR , dna synthesis fills in with DNA ligase

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mobile genetic elements

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specialized segments of dna that can move from one position to another in 1 GENOME not talking about movement in diff cells
-ALL cells euky and proky contain mobile genetic elements
-aprox 40 of human genome can be traced to mobile geneti elements
--most are no longer active due to mutations

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transpositoin:

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when dont have mutaion is mechanisms by which mobile genetic elements call transpososn move
-require: specific enzyme usually encoded by tranpooson itseld (TRANSPOSASE)

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TRANSPOSASE

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act as DNA sequences at the end of th etransposn in the given genome
-modestly selective in choosing the target
bc no requirement for homolog btwn the ends of the transpososn and the targe sequence

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transposition n tranposase

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frequncy of transpositoin can be measure in bacteri , can be m

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actual mechanisms for nonrival retrotranposons

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L1 element in chromosome,
goes through L1 RNA synthesis with poly A tail ( completley begining) unless goe throughnext step
-need reverse ranscriptase and endo nuclease both must be present! (enzyme dont have to come from the transposomes themself . both forma complex that bind to L1 RNA
-endonuclease acts by nicking targent dna to release 3'OH from target dna and is used as primer to prme reverse transcriptase so direct synthesis using rna template and ts isdoen throug reverse transcriptase then somehow integrate new synthesized dna to have 2 new dana on target

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is presence of endo nuclease required?

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long intespresed nulear elements or LINES L1
-contian 2 open reading frams (ORF0 which encode all reirement for transpositioin
-short interspered nuclear elemetn (sine ) alu element dont have ORF , so amplfiy by pirtian LINE orf bc all come from same class
-LINES and SinES make up about half of human genome so are very importatn

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transposons as tools for gene therapy

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if wan tto perform gene therapy need stable delivery of dNA ,, so can use tranposons that are not viursues and get high efficenity of integration of DNa into genome
-Sleeping beauty (SB) transposon was first describe for use in human cells
--transposase enxyme was delivereed in trans

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sleeping beauty transposon system

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DNA only tranposon
part of TC1/ mariner

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tci/mariner transposons

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have been isolated in small nubmer work in vets cells
-has inveted repeats with
--single tranposase
--or 2 tranosase bidng sites

-intergreate into TA dinucleotide, in a manner that is considered random

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what makes a transposon

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inverted repeat

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using SB lung didrecteed gene therphay

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a; dobule inverted repeaes, tranpososn, but no tranposase
b. sb tranposase but no inverted repeates so not a tranposon ; in this state cna be deivered in trans with a
c: hybrid of the two delivered in cis
when adding a dn b SB tranpose will bind inot the inverted double repeats and could uct tranposon out of the plasmsid , woudl insert tranposon into the mamalin targe DNA which is our goal

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inject vector a

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vecor a has lucierase was injected tomice in tail the plamsid will get stuck incapillary bed and will enter cells of the lungs

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analysis of luciferease:
big quesiton is it less efficient to deliver by trans than it is by cis?

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24 hr: a has high expression in lugs but could not be the resullt of transposition bc no tranposase enxym just plasmid A. so tells us that plasmid could integreat into cells oflungs and have an expressionof luciferase of the plasmid but the plamid itself is not inegrated into the genome as the consequnce of tranposition and know this because same expression in allother combintaion of trans and cis
look at 2 months and see diff:
little signal in A suggesting got inegrated into dna inot lungs and that still afer 2 months albe to get production of luciferse that is reaable above control
more expression in trans and cis: but dont see to be any diff from eithough . tranposase is present

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sb tranposon sumary

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-non viral method of integrating DNA itno host genome
- SB can be tranposed in cis or trans
SB is subject ot over expression inhibition bc needs to be controlled
-amount of acargo btwn invertdd repeats nees to be considerd intherapeutic strategies

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mrna

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messenger RNA that code for proteins
only make 3-5% of tal RNa in mmamalian cells

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transcription requires RNA poly

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RNA ploy 1
RNA ply 2 transcription of nucleotide regions
RNA poly 3

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RNA poly

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catalyzes the phosphodiester bond that link ribonucleiotides togeterh
uwind NDA ahead of its active site
-hsort wind of DNA RNA duples during transcription
-does not require are primer
-makes mistake X104 nucleotide unike dna poly hat has good proofreading this has modest proof reading

-fi make mistake snot a big consequ3nce bc its not a permanent change

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many rna poly can act on teh same geene

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if dna moves from left to right gets small to bic bc sgrand is getting onger
black dots are the rna polymeras e machines showing that tere are many transcriptionl machines thar are boudn making ribosomal rna

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if rna poly moves from Left to Right

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bottom stand is used as template

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if rna poly moves from right to left

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top strand is used at emplate

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how does rna poly know how to start and stop

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euky and proky initiate and temicnate differently

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proky

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simpler
only one type of rna plymerase
uses DNA as guide to syntheize rna
-detachable sigma factor associates with the core enzyme and assists in locaizing to sites of inintaion
-rna poly+ sigma factor = holoenzyme
sigma recognizes specific sequences thatr are in bacterial gnome
sigma 70

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transcription initnaio in bacteria is driven by sequence

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there are certain consensus sequences in DNa that sigma recognizes (TATA)
2 hexamer consensus sequ are important there is 17 nucleotide space that s important

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rna poy recongnizes consensus sequence in

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IN dsDNA not ssDNA
and have opeing of double helix that is NOT Atp dependent
(no sigma?)

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abortive RNA transcripton

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rna poly wil l make a little then will stop, start ston so slow when there is the initiation of rna synthesis

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sigma factor ineraction with DNA weakes

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transcription proceed s rapidly once sigma has sissociated and RNA poly undergoes variou strucutrusl changes
as Rna begins o move awasy from stranscripton start site the binding of rna poy to sigma weaekes bc sigmama binds to consensus seq so polky pulls awa from sigma and transcripton continues

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RNA poly catalyzes RNA synthesis until reaches terminaor signaml

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termination signals arse a string of AT nucleotide pairs and 2 fold symmetric sequnces . rna generates UA and hU A have low afinity for DNA than ct bc of hydrongs the rna makes a hiar phin that trmitaes

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euky initon

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3 rna poly
rna pol 2 transcribes the majority of dna
ran2 requires accesory proteins, calle general transcription factors
-help position euky rna poly onto protomer, pull double helix aprt , release rna poly from promoter for elongation
EQUIVALENTS OF SIGMA FACTOR FOR BACTERIA!!!!

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overview of euky transpcripton and TFII

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-tat box before the start of transcription
-TFII with TBP that recognize Tata box : TFII create a change in the are of DNA cause a bluge so have a eay secognition seques for recruitn other TF to make general trancrion complex
TFIID recruits TFIIB
then will recrui Hm E and RAN poly II with F to form a general transcription complex

TFIIH is essenstion for trancriton! if dotn have it wot have transcription bc unwind DNA and releases RNa poly from mrototer

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TFIIH hydrolyzess ATP to unwind helix

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phosphoryltes RNA pol II
-the activity of TFIIH will result in disasembly of most of TF from th inination complex
-have recruitment ofnucleotides and begins transcription for machienery are gone

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TFIID

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has TBP recognizes Tata Box
also has FAF subunit: recognizes other DNA sequces near the trascription start point , regulate dna binding by TBP

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TFIIB

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positions rna poly at start of trancription

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TFIIF

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stabilized RNa poly interaction to TBP and TFIIB
recruits TFIIH and TFIIE

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TFIIE

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attracts and regulates TFIIH

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TFIIH

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unwinds DNa at transcrition start , phhophyraltes ser5 of Rna poly , and releaseses RNA from promoter

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BRE

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TFIIB

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TATA

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TBP

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INR

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TFIID

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DPE

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TFIID

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transcription facots

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attracct poly ii and activites tanscritiption

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mediator comples

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bids to transcription facotrs together with RNA polII and TFIIs

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chromatin modifying enzymes

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chromatin remodeling coplexes and histon acetylases to allow greater accessibiltiy to DNA

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transcriptoanal activator and mediators

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activator proeitns are are far from tat box in that fold to attack to getran ll transcription factor
help rna poly, transcritio factor and mediator all to assemble at the promoter , so attract atp dependent chromatin remodeling complexes and histone acetylases

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elongation:

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rna poly must be released from large complex req proeolysis ofn activaor proein and phosph of rna pol ii at th c termianl domain

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elongation factor

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both proky and euky requiri elongatio factors
-decrease changes of poly 2 from dissociating during elongation
associate shortly after ininitnaon
-in euky eongatio factors are aided by ATP dependent chromatin remdeling complexes
-elongation factors can dissociate H2a and 2b hisones if required to aid trancription but the mechisns isnt known

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rna cappin

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at 5' end after prodced abot 25 nt
uses modified guanine nucleotied
invoes 3 enzymes :
-: phophatase: removes phosthat from 5 end of nascent rna
-guanyl transferease adds GMP usisng REVERSE LINKAGE ( 5'-5')
-methyl transferase: adds methyl gorup to guanisisnen
all these enzyme ar on poly itself

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CTD ( C terminal domaine)

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this carrie pre mRNA as rna poly 2 transcribes
-consist of 52 tandem reeats of 7 aa
-each repeat contains 2 serienes (2 and 5yh)
-capping protiens bind to rna pol 2 tail when it is phoph on ser% of the hetad repeat