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nucleic acids > midterm > Flashcards

midterm Flashcards

1
Q

Arthur Kornberg demonstrated DNA synthesis in cell-free extracts of bacteria and identified what?

A

Identified DNA polymerase I (unusual enzyme since it needs a template to work from)

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2
Q

what is required for new DNA synthesis

A

dNTPs, Mg2+, DNA template

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3
Q

Nucleic acids are composed oF

A

phosphates esterified to the C5′ of the sugar, 5 carbon pentose sugar, and heterocyclic base

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4
Q

where do the links of a nucleotide occur?

A

the bases attach to carbon 1 of the sugar forming a glycosidic bond. the phosphate attaches to the sugar on the 5th carbon and forms a phosphoester bond.
Nucleotides are attached to each other by linking a phosphoryl group to the 5’ hydroxyl of the sugar and another esterfied to it through the 3’ hydroxyl (phosphdiester bond)

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5
Q

what are Xeno Nucleic Acids, XNAs (xeno=alien)?

A

Artificial nucleic acids that use same bases, phosphodiester bonds as DNA but use very different nucleoside sugars

  • are invulnerable to acids, nucleases (enzymes that cleave up DNA, RNA) -> Can perhaps be used to target viral DNA
  • carries genetic info like DNA/RNA
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6
Q

Matthew Meselson & Frank Stahl demonstrated what ?

A

Demonstrated that the two strands of the double helix separate during DNA replication

-Semi-conservative DNA replication discovered by stable isotope labelling

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7
Q

3 possible models of DNA replication

A

semi-conservative(one parent strand, one new strand), conservative, and dispersive(all chopped up)

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8
Q

how did Matthew Meselson and frank stahl make their discovery?

A

-Used stable heavy isotope 15N to differentially label the
parental and daughter DNA strands
• Separated heavy from light DNA in cesium chloride density gradients that exploits minor mass difference of 15N and 14N

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9
Q

Bases can exist in 2 different forms; one predominates
what are these forms
=tautomerization

A

the N atom attached to A and C rings are in the amino form (normal state) and rarely take on the imino configuration.
the O atoms attached to G and T normally have the keto form and rarely take on the enol configuration

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10
Q

what components parts of a DNA molecule can be successfully altered?

A

Synthetic biologists created 2 more bases, ‘Z’ and ‘P’

  • DNA replicating enzymes (polymerases) can replicate Z & P both in vitro and in vivo
  • Sequencing services, cloning all work with Z and P
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11
Q

what components parts of a DNA molecule can’t be successfully altered?

A

the DNA backbone->Arsenate almost certainly cannot replace phosphate in DNA backbone

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12
Q

DNA can exist in 3 conformations

A
  1. B-DNA (left handed)
  2. A-DNA (DEHYDRATED/right handed)
  3. Z-DNA (left handed)
    * B-DNA is the major biological conformation of DNA (but Z occasionally
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13
Q

what is a propeller twist?

A

The 2 members of each base pair don’t always lie in same plane
This causes alteration in width
of major and minor grooves, so
DNA is never a perfect helix
(we just draw it like one!)
->this is dictated by local DNA sequence (like 2o structure in proteins)

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14
Q

what is the make up or chromosomes

A

Chromosomes are half DNA, half protein and result in beneficial organization

  • > (eukaryotic) chromosomes are 50% protein by mass-mostly hostones
  • can be eukaryotic, prokaryotic, even viral
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15
Q

DNA packing into chromosomes ensures what? (4)

A

Compactness
Protection
Efficient division
Organization for transcription, recombination

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16
Q

what is the difference between chromatin and chromosmoes

A

chromosome is the whole molecule, chromatin is a region

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17
Q

what are nucleosomes

A

(=DNA + histones) = they are chromosomes building blocks (“stores of negative supercoiling energy”)

  • Core of 8 proteins (Histone =H, H2A, H2B, H3, H4 – 2 of each)
  • DNA (the string) wrapped around each nucleosome (beads) 1.65X’s
DNA not on the nucleosomes is
     linker DNA (length varies by species)

DNA wrapped around nucleosome
is an invariant/fixed ~147 bp across species
nucleosome repeat length=~200bp of DNA

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18
Q

compare genome size of prokaryotes and humans and the overall genome range

A

Prokaryotes rarely have genomes > 10 mega bases (Mb)
->Prokaryote gene number can range from 382* to ~7,000
Humans have ~20,000 genes in 3.2 gigabases (Gb)

genome size is usually but not always correlated with complexity
Genomes can range from 500 to 45,000+ genes, 600kb - 120G

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19
Q

why is Sequencing of genomes increasing exponentially

A

To discover gene complement, regulation, technology application, basic research…
-DNA sequencing is quickly changing our knowledge about organisms, their genomes and metabolism, etc

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20
Q

what is the relationship of complex organisms to gene density

A

Complex organisms have decreased gene density
-Inverse relationship of complexity to gene density
(virus

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21
Q

Eukaryotic genomes are large because of; 2 things

A
  1. Genes (units of heredity) are bigger
  2. Spaces between genes, intergenic sequences, are much larger
    Proteins in eukaryotes not necessarily bigger than prokaryotes, but their protein-encoding genes are….
    only 5% of your genes encode a protein!
    (Larger genomes in complex life due to larger genes (+introns) and intergenic sequences)
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22
Q

what are intergenic sequences and what are they composed of

A

=secuence b/w genes
60% of your genome is intergenic sequences
-Formerly thought of as completely ‘junk DNA’

Intergenic sequences are either;
1) Unique DNA (25%)
2) Repetitive DNA (75%)
Intergenic DNA is a mix of some regulatory(direct & control transcription), but mostly unused DNA left over from replication errors

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23
Q

Pseudogenes are nonfunctional relics ?

A

-from when (your) ancestors were infected with viruses
-Pseudogenes arise from RT action (mRNA->DNA)
Viruses encode reverse transcriptase (RT), expressed upon infection of host to copy viral RNA genome into DNA
-Reverse transcribed genes are re-integrated into
genome and become pseudogenes (not expressed)

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24
Q

Nucleosome structure must allow it to have two opposite functions which are?

A
  1. Must be stable, small structures
    that protect DNA from damage
  2. Must allow access to information sequestered
    in DNA’s information (DNA polymerases
    must gain access to replicate DNA when
    cell divides, RNA polymerase must transcribe
    messenger RNA)
    *majority of eukaryotic DNA wrapped up in
    nucleosomes (but not all)
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25
Q

how did they find Nucleosome repeat length(200bp=core +1 linker) and DNA linker length?

A

by nuclease treatment:
DNA-digesting enzyme destroys non-protein-
bound, and thus unprotected DNA (nuclease)
Incomplete DNA digestion gives multiples
of 200 (400, 600, 800…)

200=core +1 linker, corresponds to
‘nucleosome repeat length’ (species variable)

Extended digestion cleaves off linker DNA
leaving ~147bp core DNA
(protected by histones)

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26
Q

describe histone proteins

-charge, axis, size, parts etc.

A

small, positively charged proteins (basic)
-H1 is slightly bigger than the others and is not included in the core nucleosome but instead is used as a linker protein
-Minimum 20% of the histone proteins is Arg, Lys
=octomer
have a dyad axis (2 axes of symmetry)

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27
Q

why is Minimum 20% of histone proteins Arg, Lys?

A

Lys+, Arg+ needed for salt bridges with DNA backbone phosphates-

28
Q

how is the histone core assembled?

A

DNA presence is needed for histone core assembly and is mediated by the histone-fold domain that has 3 alpha helixes.
-the core is made up of two heterodimers (H2A/H2B and H3/H4) that both have N-terminus tails that are post-translationaly modified to control structure. Histone N-terminal tails not involved in DNA binding and are cut off

29
Q

describe the Histone Octamer Assembly

A

first, the H3 and H4 histones come together to form two dimers and then these two dimers come together to form a stable tetramer which binds the DNA. Two dimers of H2A and H2B are formed and associate with the Tetramer/DNA

  • H3-H4 tetramers bind both middle and both ends of DNA
  • H2A/H2B bind the rest
30
Q

explain how/why Histone complexes have 2 planes of symmetry

A

why: specific DNA binding (-H3-H4 tetramers bind both middle and both ends of DNA, H2A/H2B bind the rest)
how:
Flat side of disc towards you (U-shaped…with loop)
Disc edge towards you (W-shaped)

31
Q

Histone DNA-wrapping relies on what?

A

the hydrogen bonds formed between the negative phosphodiester DNA backbone and the positive histone sidechains

  • only few bonds made between DNA bases and histone side chains, all non-sequence specific
  • histones highly basic sequence overpowers DNAs neg backbone and bends it on its inside edge
32
Q

how do Histone amino-terminal tails stabilize DNA wrapping around the octamer?
aka how do they stick out?

A

H2B and H3 tails stick out between DNA coils, poking out through minor grooves
H4 and H2A stick out above and below the DNA
*This directs DNA to wrap around the histone like a string
around the grooves of a screw, causing left-hand
DNA wrapping about the histone

33
Q

what do prokaryotes use in place of nucleosomes to maintain negative supercoiling (Negative supercoiling favours DNA unwinding)

A

use special topisomerase (gyrase, ATP) in place of nucleosomes to maintain negative supercoiling
-Hyperthermophilic prokaryotes avoid unwanted negative
supercoiling with ‘reverse gyrase’ – promotes positive supercoiling

34
Q

Two types of chromatin? how do they differ?

A

1) Euchromatin – stains poorly, more open structure

2) Heterochromatin – stains densely, more compact (associated with key chromosomal regions such as centromere, telomere)
->explains lower gene expression
(and vice versa)
*Euchromatin and heterochromatin differ in density, function, gene expression

35
Q

what is the role of Histone H1

A

it binds linker DNA to the rest of the nucleosome and pulls it closer protecting it from nuclease digestion and increases the length of DNA that is bound around the histone octomer
-causes “zig-zag” in DNA

36
Q

what is the “Solenoid model”

first model of 30 nm fibers

A

Beyond nucleosomes, there is 30 nm fibers
– solenoid model is superhelix (6 nucleosomes/turn)
-model supported by helical pitch of 11nm, same as nucleosome diameter
-nucleosome discs face each other, linker DNA in middle but not in axis

37
Q

explain the zigzag model

second model of 30 nm fibers

A
  • folds back on itself, no nucleosome disc stacking like the solenoid model…
  • model supported by experiments where H1 is added
  • linker DNA in middle and does go through axis
  • supported by X-ray structure, biophysical expts. (spring-like)
  • model favoured by longer linker DNA, but linker DNA length is species-dependent…both models may be correct
38
Q

why are Histone amino-terminal tails needed
for 30nm fiber formation?
(Without tails, no fiber formation)

A

Histone’s NH3-terminal tails implicated in stabilizing nucleosomes
-Histone amino-tails stick out from core and tails can interact with adjacent nucleosomes
-Positively charged H4 tail and negatively charged H2A tail
region are important for fiber formation
-These regions in ‘tails’ are conserved across species, have no other roles unlike the rest of the amino-terminal tail

39
Q

Nucleosomal, 30 nm fiber compaction of DNA

insufficient to cram it all in nucleus…. what is the futher compaction?

A

-A nuclear (chromosome) scaffold proposed for
loops of the 30 nm fiber that are 40-90kb long
-Structural Maintenance of Chromsome (SMC)
protein and Topoisomerase II involved

40
Q

Histone variants alter nucleosome function

what can H2A and H3 be replaced with and what is the purpose?

A
  1. H2A can be replaced with H2A.X which can be phosphorylated when there is a chromosomal double-strand break next to it which then serves to act as a signal to enzymes to repair DNA break
  2. H3 can be replaced with CENP-A when near the centromere because it is incorporated into kinetochore to attach mitotic spindle
    - CENP-A differs from H3 as it has a longer amino terminal tail which serves to recruit kinetochore
41
Q

what are the three ways that Nucleosome-remodelling complexes facilitate DNA movement on histones

A

Nucleosome stability governed by
“nucleosome-remodelling proteins”

1) Nucleosome can change location by
‘sliding’ along DNA strand (all can do this)

2) Some of these proteins swap/transfer
histone octamer to another strand

3) H2A & H2B dimer exchange occurs for
H2A.X/H2B in the nucleosome core

42
Q

how do Nucleosome-remodeling protein complexes unwind & restore nucleosomes?

A

DNA starts off being bound within the nucleosome and is inaccessible but then remodelling protein complex A is recruited and nucleosome slides down DNA so that DNA binding proteins can come in and do their job. once finished, remodelling protein complex B takes of the binding proteins and the DNA is exposed but then goes back into its protected bound fotmation

43
Q

how do nucleosome-remodelling complex move?

what is required for them to work?

A

-works like inchworm
-These proteins require ATP for action
-Move in a directional manner a.k.a.
nucleosome ‘translocation’
-Bind to histone octamer and move DNA relative to octamer by pulling DNA over octamer
-cheaper to move short parts of DNA a bit at a time rather than entire DNA strand at once

44
Q

how do different Nucleosome-remodelling complexes compare?

A

structurally dissimilar

  • Variable subunit number, all slide on DNA, only 2 swap DNA strands
  • Different histone binding domains, but similar ATP hydrolysis subunit
  • Some nucleosome-remodelling complexes subunits target transcription factor proteins to specific chromosomal regions
  • *Subunit structure, histone-binding domains, & DNA targets differ
45
Q

Nucleosomes don’t bind specific DNA sequences but why can it be good if they are placed in certain spots?

A

because their location can expose promotor DNA sequence on linker region for regulatory purpose

46
Q

what is one way that nucleosomes can be placed to expose promoter sequence on linker region?

A

2 ways to do this - #1 is DNA binding proteins
-some can act like bookends, preventing
nucleosomes from binding to a small
stretch of DNA as there is insufficient space b/w the binding proteins for nucleosome wrapping
-Some proteins bind to target DNA, then
facilitate nucleosome assembly next to it

47
Q

Second way to promote positioning of nucleosomes?

A

=Nucleosomes can be positioned by DNA sequence by binding to A-T rich minor groove
(Broad overall DNA sequence can be used)
-A-T rich sequences are preferentially bound due to more minor grooves that form those multiple H-bonds
*done by the nucleosomes of yeast

48
Q

name the 4 main Post-translational modifications of histones and how modification can alter DNA accessibility

A

acetylation (lys, neutralize +ve)
methylation (lys and arg)
phosphorylation (ser and thr, -ve)
ubiquitination (lys)
*Modifications are (mostly) reversible
-modifications alter the charge and therefore the DNA binding properties of individual histones
-marks histones for recognition by other proteins
-Inhibits ability to form zigzag/solenoids
(amino-tails can’t bind other amino-tails)

49
Q

what is the Histone code hypothesis

A

states that all the histone post-translational modifications associated with any gene is a code dictating the expression of that gene
(Sum of histone post-translational modifications encodes gene expression status)
*Lysine acetylation and methylation are of primary importance

50
Q

why are Lysine acetylation and methylation are of primary importance for dictating the expression of a gene

A

Lys Acetylation usually leads to Activation (with some exceptions)
• Lys methylation can activate or repress (depending on the site)

51
Q 
Histone modifications encode specificity in what domains, what proteins bind them?
theses domains (4) recognize and bind modified amino-tails
A

bromodomains: recognize/binds acetylated tails
chromodomains: recognize/binds methylated regions
PHD finger domains: recognize/binds methylated tails
SANT domains: binds unmodified tails
*These proteins only bind very specific modified sequences, not all of them

52
Q

Duplicated chromosome means all associated proteins need to be replicated so how are Histone modifications passed to daughter chromosomes after DNA duplication?

A

nucleosomes break down into histone heterodimers at the replication fork. H3/H4 dimer stays attached to the daughter chromos while H2A/H2B dimer is free(soluble)
-H2A/H2B dimers and H3/H4 tetramers passed on
are either all old or all new histones (about every other H3/H4 dimer on a strand is old)

53
Q

what is the role of histone chaperones

A

assembly of nucleosomes is not spontaneous and need help from histone chaperones which prevent unproductive interactions between DNA and histones

54
Q

describe the role for these two histone chaperones:
CAF-1 = Chromatin Assembly Factor
PCNA = Proliferating Cell Nuclear Antigen

A

CAF-1 is activated only during replication and is a chaperone for H3/H4
PCNA is a ring-shaped sliding clamp protein that identifies DNA for CAF-1 (tells it that the DNA is replicating). It attaches DNA polymerase and also releases from DNA, but stays on the DNA itself
-the PCNA-CAF-1 complex assembles H3/H4 tetramer into nucleosome to protect naked DNA

55
Q

Idea that template was protein ended

when proven that;

A

1) DNA polymerase only needs Mg, dNTPs
+ DNA (not protein) template to replicate

2) Messelson-Stahl experiments showed
semiconservative replication –separation
of complementary strands

56
Q

what are the Substrate requirements for DNA polymerase

A

mg2+, dNTPs, DNA template, primer with a free 3’ OH group

57
Q

how does the 3′ hydoxyl of the DNA primer attach the next dNTP?

A

the free 3′ hydoxyl of the primer attacks the alpha-phosphoryl in and Sn2 reaction and forms a phosphdiester bond
-Keeps DNA replication going in 5′-> 3′ direction
-Coupled pyrophosphatase reaction provides free energy
to drive reaction to completion, essentially irreversible

58
Q

describe why DNA polymerase has one active site, why is it unusual?

A

Single active site for synthesis - unusual since 4 dNTPs as substrates

  • Exploits geometry of A=T and G=C base pairs
  • this ensures the bases can form proper bonds
  • “kinetic selectivity” (no base pair=no fast catalysis)
  • DNA pol can use 4 dNTPs, but slows rate when inappropriate bases try to pair
59
Q

how does DNA polymerase distinguish dNTP from rNTP

A

Rate of addition for rNTPs is 1000 X’s less than dNTPs even though they are 10x more plentiful

  • Why? Active site of DNA pol too small to take in the extra 2ʹ OH (dNTP and rNTP differ by one hydroxyl)
  • Discriminator residues in DNA polymerase interact with dNTP ribose sugar ring – this is blocked by extra 2ʹ hydroxyl (-OH ) of rNTPs
60
Q

Three domains of DNA pol are nicknamed what?

A

DNA pol tertiary structure conveniently resembles a hand, each part has a role

  1. fingers- bind incoming dNTP, folds over/closes on correct bp and brings in catalytic Arg, Lys, Mg2
  2. palm-1^o catalytic active site, binds Mg2+, “proofreads”
  3. thumb-binds the DNA as it exits the enzyme, and is important in processivity
    - Newly made DNA in palm, between thumb & index finger is 3’OH end
    - Template hangs off at sharp angle below pinkie finger
61
Q

why does DNA Pol requires 2 divalent metals for activity? what is their jobs?

A

Palm area requires metals (Mg2+, or Zn2+) to alter chemical environment (Coordinates 3ʹ OH of template and dNTP)

  • first metal reduces affinity for 3’OH to hydrogen and creates a 3’O- that is ready to attack an alpha phosphoryl
  • second metal interacts with dNTPs triphosphate to reduce negativity
62
Q

how does palm domain proof reading work

A

-Palm domain monitors new base-pairing by interacting with minor groove and new nucleotide’s base pair by H-bonding
-If wrong dNTP added, palm
doesn’t close – recruit proof-
reading enzyme to remove
mis-matched base pair

63
Q

role of DNA pol thumb domain

A

thumb domain holds DNA
-Thumb domain is non-catalytic
-interacts with newly synthesized DNA
• maintains correct position of primer and active site
• maintains strong association between DNA Pol and substrate

64
Q

Role of DNA pol finger domain

A

binds to incoming dNTP, folds over if correct pairing made
-bends template so that only one base is in the active site,
all others are excluded so no incorrect pairing
-Fingers open, DNA shifts by 1bp

65
Q

Processivity = ?

A

= avg # of bp added each time primer:template is bound

*DNA pol can add 1000 dNTP/bindingn->Speed due to processivity

66
Q

Rate-limiting step of DNA replication?

A

Rate-limiting step is time taken by DNA Pol to bind, find primer:template junction (about 1s, dNTPs added in ms)

67
Q

how does DNA Pol proof-reading remove frequent errors?

A

mediated thru exonuclease activity that clips off wrong bp and 3 before

  • ‘flickering’ of nucleotide tautomers causes wrong dNTP incorporation (Observed error rate in DNA (1/ 1010bp)
  • dNTP misincorporation slows down synthesis due to changes in 3’OH geometry
    1. Primer:template junction destabilized, several bp of newly made strand unpair
    2. Exonuclease site as 10X higher affinity for unwound DNA than polymerase does
    3. Mismatched bp cleaved off, polymerase reasserts binding, synthesis continues

nucleic acids (5 decks)

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