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Nucleic acid structure

Jenna Robinson Biology English 37 cards

Cards in this deck (37)

  1. 1How can proteins interact with DNA?Positively charged protein will interact with -vely charge phosphate backbone - but need sequence specific recognition - can be done easily if strands are separated but harder when ds as the majority of chemical information is used in base pairing
    - Docking of proteins is determined by sequence dependent twist/roll/ slide then sequence dependent probing by AAs in grooves- comes from spare info in major groove - the distribution of HB donors (D), acceptors (A), methyl groups (M) and non-polar hydrogens (H) - MADA, ADAM, DAAH, HAAD for TA, AT, CG, GC
  2. 2What is B form DNA structure?B-form - idealised structure -X-ray crystallography right handed double stranded helix - antiparallel - 5'-3' - backbone formed from sugar-phosphate chain - sugar is 2'-deoxyribose - phosphodiester bond links 3'-OH with 5'-OH of adjacent sugars (draw) - bases are linked to sugar C1 by N-glycosidic bonds to N1 of pyrimidines and N9 of purines - project inwards - complementary H bonding forms base pairs (w-c only - isomorphic) - displaced from helical axis - creates major and minor grooves - bases exist primarily of amino and kept tautomers - base pairing provides stability and specificity
    Rotated by 36 degrees (28-40) per base pair
  3. 3What forces stabilise and destabilise DNA?Stacking of adjacent base pairs provide stability - entropically favoured exclusion of water- maximised by propellor twist of 16-18 degrees - bases in base pairs don't lie precisely on same plane - favours extended DNA as bending reduces stacking - stacking optimal in PU-PY bp steps - maximal in GC minimal in TA - reason why sequences with TATA easily melt
    Electrostatic repulsion between -ve back bone - destabilises - favours extended conformation - can be counted by bound protein that facilitate bending
  4. 4What structural variations are present in B form DNA?variable base-pair steps:
    - twist - rotation ber bp - 36 degrees (28-40)
    - Roll - opening along bp axis +20 to -10
    - Slide - displacement along bp axis - +2 to -1 Armstrong - TA steps can adopt high slide configuration - low stacking
    -Sequence dependent variation in dimensions of grooves - minor groove narrower in A/T rich DNA
  5. 5What are the characteristics of the major and minor grooves?- Unequal grooves doue to 120/240 degree angle between glycosidic bonds in each bp i.e. not rotationally symmetrical and base pairs are displaced from helix
    -Major groove is information rich - readily accessible as 22 A wide
    -Minor groove is information poor - AHA vs AHA and ADA vs ADA
    -
  6. 6What is the structure of Z form DNA?-In Z DNA - backbone zigzags due to syn position at Pu n9-c1 bonds - left handed - can occur with Pu-Py repeats - stabilised by Z binding proteins unclear relevance - possibly important for RNA
  7. 7What is the structure of A form DNAMajor groove in A DNA is deep and narrow - inaccesible - W-C bp RNA adopts A form (rare) - difficult to recognise often seen as foreign - 11 bp/turn
    -B in high humidity, A in low humidity
  8. 8Describe the structure of RNA?- Uracil instead of thymine - lacks 5-me but H bonding is the same
    -ribose has 2'OH group - not deoxy- additional H bonding potential - vunerable to alkaline hydrolysis
    - Lack of complementary strand - more versatile - intramolecular bp - w-c as well as non-canonical bp - as no need to adopt the regular double-helical structure - allows for more conformation e.g. the G-G N1-carbonyl-- N7 amino bp makes 4 stranded G-quadruplex structure - each G makes 4 H-bonds (draw)
    - x-ray shows many structural motifs e.g. tertiary fold
    - can be cataltic e.g. ribosome, spliceosome - mg2+ and proteins often needed
    - can bind small ligands - riboswitches
  9. 9Give two examples of RNA structural motifs1. Base triple interactions - each A can make U-A hoogsteen and A-U Watson crick (draw)
    2. Psuedoknot - base-pairing of loop with complementary sequence outside the stem enclosing the loop - draw
  10. 10How can RNA be recognised?- If not bp - all function groups available for easy recognition
    - structurally diverse RNAs can be recognised by shape
    - ds RNA is A form - inaccesible major groove - ends and distortion open up major groove to allow for recognition
  11. 11What is the basic outline for investigating mechanism of gene expression?1. clone and sequence gene - genomic if looking at pre-mRNA processing as many reg sequences not transcribed or removed by RNA processing or cDNA if looking at mRNA function - lack promoter and introns but have poly a or for expressing eukaryotic genes in E.coli - as e.coli have no splicing machinery but also can use synthetic genes - optimised with codons matching host cells tRNA population
    2. Develop assay system
    3. Identify cis acting sequences - seq on DNA or RNA that act as binding sites - find consensus sequence - investigate mutations
    4. Identify trans-acting factors - proteins or RNA that bind to cis sequences
  12. 12Compare and contrast in vivo and in vitro assay?In vivo - physiological conditions, little control over variables, difficult to monitor intermediate, can be unphysiological if test gene is too abundant
    In vitro - precisie control over conditions - can purify trans-factors, but often in efficient and unphysiological as not all components present at right level
    In practice a combination of both are used
  13. 13Compare and contrast one gene at a time vs genome-wide analysesGlobal analyses e.g. microarrays and deep sequencing - general insights that one gene at a time methods can't - hypotheses made from in silico analyse of genome, transciptome or proteome then tested using tradition in vivo/vitro approaches
  14. 14Describe the basics of transcription- two DNA strands - coding/sense strand and template strand
    - Promoter specifies TSS - consists of cis elements usually upstream of TSS - recognised by RNA pol
    - main phases: initiation, elongation and termination
    -5' -3' synthesis
    -NTP substrates - NMP incorporated, PPI released
    - No primer
    - ~40 nt/s at 37 degrees
    - error rate 1 in 10,00
    -90-95% of mRNAs start with A or G
  15. 15What is the NAD+ cap used for?At some promoters with +1A NAD+ can be incorporated by RNAP - contains ADP-ribose nicotinamide - only occurs at some promoters - NAD+ cap stabilises mRNA
    - means some mRNAs have NAD+ at 5' end instead of normal triphosphate
    - in eukaryotes cap added by other enzymes either post or co-transcriptionally
    - in prokayrotes - NAD competes with ATP for incorporation as it has adenosine diphosphate
    - With the cap mRNA much moe stable
  16. 16What is the structure of RNA polymerase?Single core RNAP synthesises all RNA - alpha 2 for assembly - only N terminal domain visible ins structure - don't contact the B units - the CTD connected to NTD by a flexible linker and can bind DNA
    beta and beta prime form large cleft with active site at base - cleft is flanked by mobile pincers/jaws which enclose 20 bp of downstream DNA in elongating complex
    all 3= core enyzyme - one its own efficient elongation but multiple start sites
    sigma confers promoter recognition - less abundant then core-enzyme - accurate initiation, holoenzymes includes all and more
    -Overall structure v similar to eukaryotic pol II despite low degree of sequence identity and smaller number of subunits - sequence identity most around active site
  17. 17How are promoters characterised and identified?- Generate consensus sequence - by aligning sequences - sequence logo used to illustrate conservation - height indicates the degrees of conservation
    Mutations in promoter affect quantity but not sequence of mRNA - UP or DOWN mutations - for strong consensus matching promoter, mutations more likely to be DOWN
    KMnO4 reacts preferentially with unpaired thymine bases - oxidising the C5-C6 double bond - adding hydroxyl groups to C5 and C6 - subsequent alkali treatment cleaves the phosphodiester backbone at modified position - so sensitivity to KMnO4 is useful for monitoring where DNA becomes unwound e.g. during transcription initiation
  18. 18What is electrophoretic mobility shift assays?'gel-shift' assay - does protein interact with DNA fragment
    1. 32P end labelled DNA
    2. Incubate with pure DNA-binding protein or cell extract
    3. On a non-denaturing gel- run electrophoresis - then do autoradiography (measures level of decay of radioactive substances) or phoshorimager - run against free DNA
    4. To get sequence specificity, compare to mutant DNA unlabelled - should compete if W/T resulting in more free DNA as W/T titrated in - but, won't compete if bad mutant - no change in free vs complexed labelled DNA (see book for pics)
    or could just use labelled probes with specific mutation then compare free/complexed ratio
    - Supershift assay - can test the binding protein using antibodies - DNA-protein-antibody complex migrates even slower than the DNA-protein complex - causes super shift
  19. 19How does footprinting work?Asks at which positions protein interacts with DNA
    32P labelled DNA, incubate with protein (also without for negative control)
    - digest madly for single hit conditions using DNase (sequence non-specific), OH' radicals (non-specific) or methylation of N7 G and N3 A by DMS a (permeable to cell so can be used in vivo - but do in vivo can't used radio labelled probe) - heat treatment depurinates Me bases, alkali cleaves backbone at depurated sites
    - on high resolution denature gel (6M urea) to GE and autoradiography -
  20. 20What is modification interference?Asks at which positions prior chemical modifications prevent proteins binding
    Use 32P labelled DNA - only one strand - modify DNA - average of 1 mod per strand e.g. using ehtylnitrosourea (ENU) modifies phosphate or DMS methylates purines - incubate DNA with protein - separate bound from unbound, purify DNA and cleave at modified sites (e.g. for ENU incubate at high pH)- run on denaturing PAGE - missing bands are where modification prevents protein binding
  21. 21What is chromatin immunoprecipitation?1. Treat live cells with formaldehyde - crosslinks amino or amino groups within 2A via -CH2 bridge - protein-protein and protein-nucleic acid crosslinks
    2. Lyse cells, sonicate to break DNA to ~200 bp
    3. Use specific antibody to immunoprecipitate DNA bound to protein of interest
    4. Heat treat to remove crosslinks - purify DNA and analyse using qPCR for DNA regions of interest - or deep sequence to globally profile binding to DNA
    Chip is a low resolution technique ~+/- 100bp - but can be applied genome wife
  22. 22What are the bacterial core promoter elements?1. -10/ pribnow box: TATAATG - recognised by sigma region 2 - down mutation inhibit promoter mutations
    2. -35 BOX - TTGACA - recognised by sigma region 4 by helix-turn-helix motif- down mutations inhibit initial binding
    3. Up elements at highly active rRNA promoters, ~20 bp A/T rich element upstream of -35 - consensus AWTNA- recognised by C-terminal domain of alpha subunits - can be displaced by 10-20 bp upstream and remain active - binds via minor groove - preferred narrowed minor groove of A/T rich DNA -
    - 10 and -35 box are both symmetric
    - Space between the boxes is conserved - in >95% of E.coli - it's between 16-19 bp
  23. 23What is the structure of the sigma subunit and why it important- Core enzyme, binds DNA non-specifically, without sigma you get non-specific transcription - initiating from multiple locations, with sigma non-specific binding is reduced by 10^3-10^4
    Promoter affinity increased by 10^3 - accurate initiation of transcription
    Domains 2 and 4 contain helix-turn-helix motifs -recognisises -10 and -35 boxes - but sigma alone does not bind DNA
    Sigma has 1.1 section - negatively charged DNA mimic - in free sigma contacts region 4 - stops binding to DNA - but in holoenzymes it occupies downstream DNA binding cleft - DNA can only bind to holoenzymes if it can displace the sigma 1.1 - needs to be specific to do this
    region 3.2 occupies RNA exit channel
  24. 24How does initiation occur?Holoenzyme binds promoter - reversible - B and B' pincer surround downstream DNA channel, active site located at base of pincers
    -
    - DNA melts to form 14 nt bubble and ( detected by KMnO4 sensitivity - promoted by neg. supercoiling) - separate tunnels for template and non-templated DNA - isomerisation of the protein - irreversible
    Binding of NTP - first one has low affinity -H-bond interaction but no stacking - second one binds - formation of phosphpdiester bonds -
    - Multiple cycles of abortive elimination can occur - extension for 2-9 nt - bubble expands to up to 23 nt -
    - If the extension gets beyond 9 nt, the transcription bubble contracts from the 5' end - marking successful transition to elongation - the RNA:DNA duplex forms interactions with active site channel at 9bp - more stability needed for elongation
    then sigma unit released - elongation needs to be completely sequence non-specific - 3.2 region of sigma leaves RNA exit channel - instead ssRNA interacts with exit, sigma regions 4,2 and 3 break contacts with B and B' which isn't compatible with elongation
    -
  25. 25How can transcription generally be regulated?- Classes of genes can be co-ordinately controlled by switching sigma factors - 7 sigma subunits in E.coli
    E.g. sigma 32 deals with heat shock where you gets hurt term general switch off of expression - instead switches on expression of protective proteins like chaperone which helps with the correct folding of proteins
    - Inducers and repressors are allosteric regulators of activator and repressor proteins - regulatory proteins have 2 conformations - one is stabilised by inducer or co-repressor, one conformation binds DNA with high affinity - often bind as dimers to symmetric binding sites - contrast to RNAP
  26. 26How does the lac operon work?- Regulatory and structural genes organised into operon to allow for coordinated response
    - Has 2 promoters - a weak promoter that transcribes LacI mRNA - encoding the lac repressor, the regulated lac promoter - Only switched on when lactose is available and when glucose is low (glucose easier substrate to use)
    - Low basal level of transcription mean there is enough LacY product - permease to allow lac uptake when available - also a side reaction of B-galactosidase creates allolactose which is the inducer that binds to the lac repressor and weakens binding - but activation only persists when lactose available
    - Also have artificial inducers e.g. IPTG - not maetabolised by b-galactosidase but bind repressor
    - The regulated promoter has suboptimal -10 to -35 elements - maximal levels of transcription require an activator - the CAP protein - binds with cAMP and then DNA to promoter RNAP binding
    - The 'operator' - binding site for lac repressor (palindromic) - overlaps the transcription start site -
    - Binding site for CAP protein is upstream of -35 box, but not overlapping
  27. 27What is the structure and binding of the lac repressor?- Repressor is tetramer of 38 kDA subunits dimer of dimers - recognises operator by helix-turn-helix motifs with one dimer docking into the major groove
    - Additional binding at +410 and -80 - at least one required for full repression - i.e. tetramer binds two palindromic sites - looping DNA
    -Only ~10 repressor tetramers/cell
    - Affinity for non-specific DNA high - 99.9% of repressor is bound - accelerates binding to specific sequences by 100-1000 fold - restricts the effective volume protein need to diffuse - smaller volume to search for operator - facilitated transfer, by sliding along DNA or by inter segment transfer (helped by tetramer structure
  28. 28What is CAP and how does it help with regulation?- Dimer of 22 kDA subunits - cAMP bound form binds symmetric site upstream of RNAP
    - DNA binding via helix- turn-helix motif
    - 7 AA positioning or stabilisation helix makes non-specific contacts with back bone
    - 9 AA recognition helix rocks into major groove - sequence-specific recognition by van der Waals and H-bonding e.g. between Arg 180, Glutamate 181 and Arg 185 with GC base pair
    - CAP binding is necessary but not sufficient for activation -
    - Bound CAP interacts with C-terminal domain of alpha subunit - promotes RNAP binding - x-ray structures compared bound and unbound CAP shows no conformational changes - suggests CAP is only used to recruit RNAP
    - CAP helps compensate for suboptimal -35 and -10 box
    - Also acts on other promoters e.g. Gal, ARA
  29. 29Why do we use micrococcal nuclease to find out about nucleosome positioning?- Need to pack 2m of DNA into a 6 micro metre nucleus >1000 fold compaction
    - Light microscopy shows 2 types
    1. dense-staining heterochromatin - transcriptionally silent - interphase chromatin - 30nm fiber - released with high ionic strength
    2. Poorly staining, open, euchromatin - expressed genes
    - beads on a s ring = 10nm fiber
  30. 30How are nucleosome positioned and how do we know this?MNase - sequence independent DNase - bound proteins protect against digestion
    - With extensive digestion - then separate out protein and run of gel - 147 bo associated with nucleosome core particle - invariant across eukaryotes
    - With limited digestion, 200 bp ladder - core nucleosome + linker DNA - variable linker can be 20-60 bp e.g. 160 for cortical neutrons, but 200 for comical glia
  31. 31What is the structure of nucleosomes?Histone octamer makes up core - H3.H4 tetramer - interacts with ends of DNA and 60 bp in middle + 2x H2A.H2B. dimer - interacts with 30 bp flanking middle
    core histone fold domain contains 3 alpha helices - 1 long helix flanked by 2 shorter helices
    - N -terminal enriched in Lys/Arg - numerous post-translational modification eg. ser-p, lys-acetylation - arg,lys methylation - alters charge can act as binding site for proteins
    - DNA wraps 1.65 times round octamer -left-handed helix, n-terminal not visible by X-ray structure - but emerge around nucleosome - positive charges help stabilise DNA wrapping
    - 14 sites of DNA-protein contact -each time minor groove approaches core, ~33 H-bonds to p-backbone, 7H-bond to bases in minor groove
  32. 32How are nucleosome positioned?Contacts of nucleosomes generally sequence-independent, but some sequence dependent positioning e.g. AT rich DNA binds more readily towards minor groove and make minor groove narrower - 5bp periodicity of AT and GC rich sequences favours positioned nucleosome, also sequence specific DNA-binding proteins can maintain specific nucleosome-free areas
    -Active genes are not nucleosome free, but nucleosomes are repositioned during activation - not a flat line more peaks and troughs
    -Strong positioning of nucleosome at active transcription start site - found by genome-wide mapping of nucleosome positions by deep sequencing of micrococcal nuclease treated chromatin from resting and activated T cells - compared to positioning of RNA pol which was high at TSS and in expressed genes
    -Nucleosomes in expressed genes show peak at +40 and a low density at the TSS - this positioning arises from their exclusion from these reasons when the transcription machinery binds at the promoter
  33. 33What does DNase I sensitivity tell us about gene expression?- take DNA digest with DNase -then digest with REs - restriction enzymes have highly sequence specific cutting - southern blot to find bands using gene-specific probes
    - Showed expression of genes was ~5x more rapid e.g. b-globin in adult erythroid cells (expressed) or ovalbumin in oviduct (expressed) vs the reverse environments
    could also do reverse in
    -Not due to fewer nucleosomes - but could represent chromatin de-condensation e.g. 30nm to 10nm transition
    DNase hypersensitive sites i.e. nucleosome free regions can also be identified - cleaved 10-20 times more rapidly:
    - Do a light DNase I digestion, purify DNA then cut with RE - southern blot and probe with radio labelled fragment that hybridises to one end of the target restriction fragment - autorad/phosphorimager - will see bands where transcription factors would be pound - poised for promoters, enhancers etc
    - Bands consist of fragments cut at one end by restriction enzyme and the other end by DNase - from the length you can find out how far away it is from restriction site and identify it (see book for pics)
  34. 34What is the structure of the linker histone and how do they ?- Larger than core histone (21 vs 11-15 kDa)
    - N and C terminal tails have +ve charge
    - Addition of H1 to core nucleosome increases protection of linker DNA from micrococcal nuclease - also shows more ordered zig-zag arrangement of nucleosomes
    - In vitro, high ionic strength 150 mM NaCl causes 10 nm fibre to condense to 30 nm - 6 nucleosomes per turn
    - In vivo, 30 nm tail requires H1 and N-term tails of core histones
    - Two models of packing - one start=solenoid (H1 goes across edge) or two start= zig-zag the (seen at low ionic ) - H1 goes across central axis (see book for pics but they're pretty bad) - which forms depends on length of DNA
    - compaction in 30nm fibre ~40 fold - further levels of compaction involve chromatin loops - attached to proteinaceous nuclear scaffold
    -DNA in 30nm fibre not accessible to proteins
  35. 35How can chromatin compaction be regulated?- Loss (even more than core histones) or modification of H1 -depletion of H1 near active promoters is observed by ChIP
    - Acetylation of lysine residues in N-terminal tails of core histones by Histone acetyl transferases - change from charge to neutral - loosens chromatin packaging - ac-Lys also recognised by bromodomain - containing transcription regulators - some HATS (histone acetyl transferases) themselves have bromodomains - positive feedback loops - not all bromodomains recognise all Ac-Lys residues - flanking AAs are important
    - Conversely, some transcriptional repressor recruit histone deacetylases
    - Methylation leaves charge unchanged, but Me-Lys is recognised by chromodomain proteins - frequently repressive
    - Suggestions of a histone code - 26 residues in the N-terminal tails of histones can be modified - different combination result in different activation states of chromatin - code is more complex than just acetylation activates, methylation represses
  36. 36Compare the two ways of repairing a nuclease-induced double strand break1. NHEJ - non-homologous end joining - imprecise can do deletions or insertions of various lengths
    2. HDR - homologous directed repair - can introduce specific point mutations or insert sequences by recombination of the target locus with exogenously supplied DNA donor templates - high fidelity, therefore preferential
  37. 37How can zinc fingers be used in genome engineering?- Zinc finger nucleases - take a zinc finger binding protein and fuse it to a nuclease -

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Biology (Lecture notes ) Flashcards on Nucleic acid structure , created by Jenna Robinson on 04/04/2017.
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