Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

nucleic acids > Quiz 3 > Flashcards

Quiz 3 Flashcards

1
Q

Challenge of translation?

A

there’s no specific affinity between amino acids and DNA bases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

what is an ORF

A

ORF (open reading frame) = protein coding region(s) of mRNA and consists of contiguous, non-overlapping string of codons

  • each ORF specifies a single protein and starts and ends at internal sites within the mRNA
  • Start and Stop codons define protein-coding ORF in the mRNA
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

what are the typical Start and Stop codons for bacteria and eukaryotes?

A

Start:
Bacteria- AUG (sometimes GUG, or UUG)
Eukaryotes- AUG
Stop: UAG, UGA and UAA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

what are some differences in mRNA structures in Pro-, Eu-karyotes

A

EU:
-usually contain a single ORF (Monocistronic)
-Have Kozak sequence to recruit ribosome, increase translation efficiency (sequence= purine three bases upstream start codon and a guanine right after)
PRO:
-mRNAs often have 2 or more ORFs
-have a ribosome binding site (RBS) to recruit ribosome
-RBS called Shine-Dalgarno sequence (located 3-9 bp on the 5′ side of start codon)
-RBS is complementary to a sequence near the 3′ end of 16s rRNA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Polycistronic mRNAs?

A

Polycistronic mRNAs have multiple (poly) ORFs

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

translational coupling?

A

the phenomenon of linked translation between overlapping ORFs in prokaryotes
-because many prokaryote mRNAs are polycistronic, the start codon of the downstream ORF often overlaps the 3’ end of the upstream ORF so the ribosome that just completed translating the upstream ORF is positioned to start translating from the start codon of the downstream ORF (therefore RBS not needed to recruit the ribososome)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

describe how eukaryotes recruit the ribosome to mRNA

A

by using the 5’ cap which is located at the extreme 5’ end of the mRNA. the 5’ cap is a methylated guanine that is joined to the 5’ end by an unusual 5’ to 5’ linkage. the ribosome binds to the mRNA and moves in a 5’->3’ direction until it encounters a 5’-AUG-3’ start codon (process called scanning)
-3′ poly a promotes efficient recycling of ribosome

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

describe Transfer RNAs (tRNAs)

A

tRNAs are adaptors between mRNA codons and amino acids
Every tRNA has 5′-CCA-3′ at 3′ end where cognate amino acid is attached
-Aminoacyl tRNA synthetase enzymes attach amino acids to tRNas
-Up to 45 different tRNAs needed to bind all mRNA codons types
-Unusual bases are present in tRNA structure-> tRNA modifications to improve function!
Ψ=pseudouridine (isomer of uridine),
dihydrouridine is reduced form
->these bases are not needed for tRNA function but the cells lacking them show reduced rates of growth

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

tRNAs secondary structure?

A

All tRNAs have 1 stem, 3 stem-loops, and 4th variable loop;

1) The acceptor stem/arm (accepts the amino acid) with CCA sticking out beyond paired bases
(2) ⍦U(pseudouridine) loop: has the presence of the unusual base ⍦U in the loop. this modified base is often found within the sequence 5’-T⍦UCG-3’
(3) D loop (rich in Dihydrouridines)
(4) anticodon loop (pairs with mRNA codon), anti codon always bracketed by a purine on its 3’ end and a uricil on its 5’ end
(5) variable loop: varies in size from 3-21 bases

D and U-loops interact, so do stems to stabilize

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

describe the 2 steps for the attachment of amino acids to tRNA

A

-tRNAs are ‘charged’ by adding an amino acid to tRNA’s 3’ terminal Adenosine nucleotide via a high-energy acyl linkage. Aminoacyl tRNA synthetase ‘charges’ tRNAs in two steps:

Step 1: adenylylation (tranfer of AMP)
-the amino acid reacts with ATP to become adenylylated with the concomitant release of pyrophosphate
Activation of the Amino Acid:
aa + ATP -> aa-AMP + PPi

Step 2: tRNA charging
-the adenylylated amino acid (which remais tightly bound to the synthetase) reacts with tRNA resulting in the transfer of the aa to the 3’ end of the tRNA and release of AMP
Aminoacyl group transfer to tRNA:
aa-AMP + tRNA -> aa-tRNA + AMP

->Aminoacyl tRNA synthetase fall
into 2 classes, differing by where
amino acid attaches to tRNA
and number of enzyme subunits

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

“isoaccepting” tRNAs ?

A

“isoaccepting” tRNAs are synthetases that recognize, attach more than 1 type of tRNA to appropriate amino acid
-however, Each aminoacyl tRNA synthetase attaches only a single amino acid to tRNAs

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

tRNA synthetase recognizes what unique structural features of cognate tRNAs?
-and why cant the anticodon itself be used for recognition?

A

-Anticodon loop and acceptor stem:
allow tRNA synthetase to identify what type tRNA is and what amino acid to add
-‘Discriminator base’ within acceptor stem encodes recognition information
-Anticodon itself can’t be used for recognition since 1 amino acid encoded by multiple codons (e.g., Ser uses 6 codons, two are completely different!)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

how does tRNA synthetases prevent mis-charging of tRNAs?

A

=tRNA synthetases use two recognition pockets

  • In addition to its catalytic pocket (for adenylylation), some aminoacyl tRNA synthetases use an editing ‘pocket’ (as a molecular sieve) to charge tRNAs with high fidelity
  • the editing pocket proofreads the product of the adenylylation reaction.
  • Amino acids are small & similar, so discrimination between some is difficult
  • Each amino acid fits into an active site pocket in the enzyme (for 1st charging step) and the amino acid forms a network of hydrogen bonds, electrostatic, and hydrophobic interactions
  • Only amino acids with a sufficient number of favourable interactions bind..
  • If incorrect amino acid-AMP is made by tRNA synthetase, it can be bound by editing pocket that will hydrolyze it (but not the correct amino acid-AMP as it would not fit in the pocket)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

why is it important to emphasize need for tRNA synthetase to use binding pockets to prevent mischarged tRNAs?

A

Ribosomes can’t discriminate between correctly and incorrectly charged tRNAs->blindly adds amino acids even if they are on the wrong tRNA
this is demonstrated experimentally by two methods:
1. Genetic demonstration
-Can mutate anticodon so it binds to the wrong codon
for that tRNA but it delivers the right amino acid to the
wrong codon
2. Biochemical demonstration
-Can test with cysteine-tRNA charged with Cys or Ala
tRNA will correctly bind with proper codon, but deliver wrong amino acid

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Ribosomes are complex ribonucleoproteins that convert mRNA info to proteins, why are they complex?

A
  • Far more complex than RNA & DNA-making machinery due to difficulty
  • 50 different proteins (or more) and at least 3 RNAs form ribosome
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Rate of translation in prokaryotes and eukaryotes?

A

-Rate of translation in prokaryotes: 2-20 amino acids/sec
-Rate of translation in eukaryotes: 2-4 A.A.s/sec (why? spatial separation…)
both much slower than Rate of DNA replication (200-1000 nt/sec )

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

what are the Ribosomes large and small subunits

A

1) Peptidyl transferase center (makes peptide bonds) -large

2) Decoding center (charged tRNAs “decode” mRNA codon sequence) -small

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

compare large and small subunits of pro and eukaryotes

A

Prokaryotes have a 50S and 30S ribosome (together is 70S-> difference due to shape of whole ribosome when individual subunits associate, binding causes compaction)
-Bacterial 50S subunit made up of many proteins and 5S, 23S RNA, 30S subunit made up of 16S RNA (remember, 16S RNA hybridizes to mRNA’s RBS site)

Eukaryotes have 80S ribosome with 60S + 40S subunits

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

polyribosomes or polysomes?

A

=mRNAs with multiple ribosomes

  • Allows more protein to be produced , can stack ribosomes 80 bases apart
  • Explains why there isn’t much mRNA in cell…
    • > most of the RNA is rRNA that produces protein
  • therefore, Ribosomes make 1 protein at a time, but can work simultaneously on a mRNA
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Ribosome cycle ?

A

=Ribosome cycle involves initiation, elongation, termination & disassembly
binding of mRNA and initiator tRNA to small subunit, association of large peptidyl transferase(recruit large subunit), elongation, disassembly after stop codon reached

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

describe Peptide bond synthesis of growing polypeptide chain

A

-New amino acids are attached to growing polypeptide chain at its C-terminal end (not vice versa)
-Peptides bonds are formed by transfer of the growing polypeptide chain from one tRNA to another
(Peptide bond is switched from outgoing to incoming tRNA to lengthen protein)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Two charged species of tRNAs ?

A

1) Aminoacyl-tRNA: attached at its 3’ end to
the carboxyl group of the amino acid

2) Peptidyl-tRNA: attached to the carboxyl-
terminus of the growing polypeptide

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

The ribosome has three binding sites for tRNA

A

Ribosome’s functional core is mostly rRNA and most ribosomal proteins are on ribosome’s periphery
-sites are at interface between 2 subunits

A site: to bind the aminoacylated-tRNA

P-site: to bind the peptidyl-tRNA

E-site: to bind the uncharged tRNA (E is for exit)

  • Channels through ribosome exist that allow the mRNA and growing polypeptide to enter and/or exit the ribosome
  • Ribosome holoenzyme creates active sites and tunnels for making protein
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

why is there a pronounced kink in mRNA between two codons at P and A sites?

A

-Ribosome provides kink in mRNA which facilitates tRNA entry, enzyme action
This kink opens the vacant A site codon for aminoacyl-tRNA interaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

why is ribo Channel size important?

A

Channel size only allows very limited folding of the newly made polypeptide
Channel size only allows unpaired mRNA to enter the ribosome too

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

The steps of protein synthesis?

A

Initiation
Assembly of a functional ribosome in correct place on an mRNA ready to start protein synthesis.

Elongation Cycle
correct amino acid is brought to the ribosome, is joined to the polypeptide chain, and ribosome moves 1 position along the mRNA.

Termination
when a stop codon reached, there is no amino acid to be incorporated and the newly-synthesized polypeptide is released from the ribosome.

Disassembly
a special factor binds to ribosome so it can release the mRNA and tRNA that is still bound to it, & so it can be recycled in next round of protein synthesis.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
Q

give an overview of the events of translation initiation in prokaryotes (3 steps, 4 things needed)

A

-Recruit ribosome to mRNA
-Place a charged tRNAi in the P site
-Small ribosomal subunit must align P site on start codon
Components needed:
Two ribosome subunits
mRNA
Initiator formylmet- tRNAi Met (fMet)
Three Initiation Factors

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
Q

how does small ribosomal subunit accurately dock at the right site in mRNA? why must it be accurate?

A

The 16S rRNA interacts with RBS to position the AUG in the P site.

Only small subunit needed here
(has the 16S rRNA as part of it)

Must position mRNA exactly or a
frameshift error will occur

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

what is formyl-Met-tRNAi^fMet ?

A

A special tRNA charged with a modified methionine

  • it binds directly to the prokaryotic small subunit
  • it is Initiator tRNA that base-pairs with AUG or GUG or UUG
  • Deformylase removes the formyl group during or after the synthesis
  • NH2-terminal Met, and next 1-2 amino acids commonly removed after too!
  • Only prokaryotes use formyl-Met as the first amino acid of a new protein
30
Q

3 initiation factors direct assembly of an initiation complex containing mRNAs and initiator tRNA, describe them please and thank you (Pro)

A

Initiation Factors promote correct sequential ribosomal assembly
IF1= prevents tRNA from binding to the protein of the small subunit that will become part of the A site-> blocks tRNAs from binding to “A” site (Aminoacyl)
IF2= is a GTPase aka a protein that binds/hydrolyzes GTP and it interacts with the 3 key components of the initiation machinery: the small subunit, IF1 and the charged initiator tRNA (fMet-tRNAi^fmet). This facilitates the association of fMet-tRNAi^fmet with the small subunit and prevents other charged tRNA from associating with the small subunit.
IF3= Binds to small subunit and blocks it from reassociating with the large subunit. initiation requires a free small subunit so binding of IF3 is important for new cycle of translation. it helps to dissociate the 70S ribo to its large and small parts (key role in keeping small subunit open and ready for other factor, mRNA, tRNA binding)

31
Q

Summary of translation initiation in prokaryotes

A

The 2 Initation Factors (I.F.s) bind at or near tRNA binding sites (A, P, E sites)

  1. IF3 promotes the dissociation of the apo-ribosome…
  2. IF1 binds near the A site, blocking it
    (want 1st tRNA in Peptidyl site, not Aminoacyl site!)
  3. IF2 (+GTP) assists the fMet-tRNAfMet binding to P site

The 30S pre-initiation complex is formed & ready for mRNA binding
(The order of mRNA or fMet-tRNAi binding is uncertain…)

4. mRNA binds and positions the start
     codon correctly
5. Conformational changes occur in the
       small subunit, causing IF3 loss
6. 50S subunit can now bind and stimulates
     the GTPase activity of IF2 (->IF2-GDP has
      less affinity for ribosome….)
7. IF2(GDP) and IF1 are released

->The 70S initiation complex is now formed

32
Q

how does Eukaryotic protein synthesis differ from pro

A
  1. Initiator tRNA always binds small subunit before mRNA does
  2. The first Met in eukaryotes is not formylated (for nuclear genes…)
  3. A separate set of accessory proteins prepares mRNA before it is bound by small subunit
  4. Ribosome + tRNA scans the mRNA for first start codon
33
Q

what are the Eukaryotic initiation factors and describe their role in ribosome assembly

A

eIF1A and eIF3 dissociate the ribosome large subunit

eIF3, eIF5, and eIF1 act as 1 functional unit (1-3-5)

Initiator tRNA is escorted to the small subunit by the 3-subunit GTP-binding protein eIF2 (the complex between the initiator tRNA and eIF2 is called the ternary complex)

  • > eIF2(GTP)•Met-tRNAi Met binds the small subunit and positions the tRNAi in the P site…
  • eIF3 is large and interacts with every member of the 43S PIC
  • > 43S preinitiation complex formed
34
Q

how is Eukaryotic ribosomes recruited to mRNA by 5′ Cap

A

mRNA cap is recognized by eIF4 subunits:

eIF4E • eIF4G • eIF4A (in that order, E-G-A)

eIF4B binds –A, activates RNA helicase activity of A to get rid of unwanted secondary structure

Translation can be regulated by proteins interacting with the E subunit

  • > Capped mRNA proteins bind & unwind mRNA before preinitiation complex made
  • > now 48S preinitiation complex
35
Q

how and why do translation initiation factors hold eukaryotic mRNAs in circles

A
  • eIF4G protein that binds to 5′ cap area interacts with the proteins that coat the poly-A tail
  • This allows small subunit to run on a circular mRNA track to minimize the time needed to find a new mRNA
  • Maximizes amount of protein made in short time (translation efficiency can be maximized)
  • it may also enhance reinitiation of the recently terminated ribosome
  • also enhances eIF4E binding to mRNA cap and large subunit recruitment
36
Q

how does start codon binding set off cascade leading to eIF loss & large subunit gain

A

identification of start AUG codon by the small ribosomal subunits. 48S preinitiation complex scans from 5′ capped end to 3′ and uses ATP in helicase activity.

  • Proper base pairing of start codon with tRNAi-Met causes conformational change of the 48S complex and results in release of eIF1 and change in conformation of eIF5. This leads to eIF2 hydrolyzing its associated GTP and in its GDP-bound state, no longer binds the initiator tRNA and is released along with eIF5
  • loss of eIF2 causes eIF5B-GTP binding and recruits large 60S subunit (this is possible bc eIF1 and eIF5 are released)
  • binding of the large subunit causes eIF5B to hydrolyze GTP, causing loss of remaining eIFs (eIF1A and 5B-GDP)
37
Q

mechanism of elongation is highly conserved in prokaryotes & eukaryotes, what are the three key events that must occur for the correct addition of each amino acid?

A

cyclic process:

  1. an aminoacyl-tRNA is loaded into the A site ribosome as dictated by the A-site codon
  2. A peptide bond is formed (the peptidyl transferase reaction) between aminoacyl-tRNA in A-site and peptide chain that is attached to the peptidyl-tRNA in the P-site
  3. the peptidyl-tRNA•mRNA is translocated in the ribosome to the P-site so the ribosome is ready for another cycle of codon recognition and peptide bond formation
    * Precise translocation is difficult, helper proteins are needed
38
Q

describe the role of elongation factor EF-Tu

A

EF-Tu = elongation factor thermo unstable
-Aminoacyl-tRNA is delivered to
A-site by elongation factor EF-Tu
-EF-Tu (GTP) binds with an aminoacyl-tRNA and brings it to the ribosome
-interaction between EF-Tu and its binding site on large subunit triggers its GTPase activity and uses GTP to
catalyze drop-off reaction

39
Q

describe the state of the ribosome sites at the start of each elongation cycle:

A
  • the A (aminoacyl) site on the ribosome is empty
  • the P (peptidyl) site contains a peptidyl-tRNA
  • the E (exit) site contains an uncharged (empty) tRNA…
40
Q

how does EF-Tu help to ensure that the correct

aminoacyl-tRNA is in place? (5)

A

1) Only EF-Tu(GTP) can bind to the aminoacyl end of the aa•tRNA (wont bind if not associated with GTP)

2) It prevents the aminoacyl end of charged tRNA
from entering peptidyl transferase centre early (it masks the aa and prevents peptide bond formation)

3) The codon-anticodon pairing is verified
4) GTPase is activated by the ‘factor binding centre’
5) EF-Tu(GDP) is released from tRNA as it has less affinity for it

41
Q

what are the ribosome’s 3 mechanisms to select against incorrect aminoacyl-tRNAs?

A
  1. two adjacent adenine residues in the 16S rRNA component (located within the A-site of small subunit) form h bonds with the minor groove of each correct base pair between the anti-codon and first two bases of codon in A-site (less affinity for incorrectly matched tRNAs
  2. GTPase is not activated if EF-Tu & tRNA is not positioned correctly by base pairing with mRNA (single mismatch of codon-anticodon base pairs can alter EF-Tu position and prevent it from interacting with factor binding center)
  3. to participate successfully in the peptidyl-transferase reaction, the tRNA must rotate into the peptidyl transferase center in the large subunit in process called accommodation. only a correctly paired codon-anticodon can withstand the strain of the rotation and incorrect tRNA will dissociate from ribosome before entering peptidyl transferase reaction
42
Q

what are the three events of translocation?

A

The P-site tRNA must move to the E site
since it is now “empty”

2) The A-site peptidyl tRNA must move to the P site
since it now has a peptidyl-bond

3) The mRNA must move by three nucleotides to
expose the next codon

43
Q

How does Peptide bond formation and the elongation factor EF-G drive translocation of tRNAs & mRNA?

A

After peptidyl bond made, tRNAs and mRNA all need
to move to the next site to continue translation
->EF-G protein needed for this, formerly known as “translocase”, use GTP energy to ‘kick’ tRNAs 1 site over
->tRNA base pairing to mRNA drags it along too!
1. EF-G translocase mimics a tRNA to displace the tRNA
bound to the A site
2. EF-G bound to GTP can slip into A site and then use GTP energy to act like a molecular “flicker” -> flicks tRNAs one site over
3. EF-G with GTP bound is activated by entering the A site,
stimulates the GTP hydrolysis to GDP

44
Q

describe how both EF-Tu-GDP and EF-G-GDP exchange GDP for GTP prior to participating in a new round of elongation

A

-EF-G has a lower affinity for GDP than GTP so it easily loses GDP spontaneously and also spontaneously binds new GTP
-For EF-Tu, a GTP-exchange factor EF-Ts is required for the GDP-GTP exchange….
*A cycle of peptide bond formation consumes two molecules of GTP and one molecule of ATP
( 1 ATP for adenylylation, 2 GTP for tRNAs)

45
Q

describe briefly the termination of translation

A
  • release factors (RF) terminate translation in response to stop codons
  • Release factors activate the hydrolysis of polypeptide from peptidyl-tRNA
46
Q

describe both classes of RFs and how they differ for pro and eukaryotes

A

Class I RF: recognizes stop codon, break protein off tRNA

prokaryotes: RF1 (UAG, UAA); RF2 (UGA, UAA)
eukaryotes: eRF1 (UAG; UGA; UAA)

Class II RF: stimulate dissociation of class I RF from ribosome (regulated by GTP binding and hydrolysis)

prokaryotes: RF3
eukaryotes: eRF3
* Eukaryotic and prokaryotic release factors almost identical in number, function

47
Q

how do short regions of class I release factors recognize stop codons and trigger release of the peptidyl chain?

A

RF1 structure very similar to tRNA structure,
RF1s seem to have a peptide anticodon
-RF1s have strictly conserved GGQ (Gly Gly Gln)
-GGQ: involved in polypeptide hydrolysis(essential for polypeptide release); located close to peptidyl-transferase center
-SPF: peptide anticodon; interacts with stop codon

48
Q

explain how Class 2 RF evicts Class 1 RF from ribosome including GDP-GTP exchange and GTP hydrolysis

A

RF-3 has a higher affinity to GDP than to GTP
-Class I release factor binds to A site
• SPF peptide anticodon pairs with stop codon
• GGQ is positioned in peptidyl transferase site
• Hydrolysis occurs • Polypeptide dissociates
-RF3•GDP binds to a ribosome containing a Class I RF
• RF3 exchanges GDP for GTP (kind of like EF-Ts does for EF-Tu.. class 1 factor stimulates this change)
• Class I factor is displaced caused by RF3-GTP’s new shape
• GTP is hydrolyzed • RF3•GDP dissociates (has low affinity for ribosome on its own)

49
Q

whats the role of the ribosome recycling factor (RRF)?

A

after the polypeptide, RF1 (RF2), RF3 are released, the RRF mimics a tRNA to stimulate the release of tRNA and mRNA from a terminated ribosome

  • RRF and EF-G use GTP hydrolysis to push out the tRNA and mRNA (P site tRNA leaves directly from P site, not E site)
  • RRF and EF-G then lost, allowing IF3 to bind to small subunit and displace the large subunit
  • This process effectively recycles ribosome
50
Q

what is C.R.I.S.P.R? where are they found?

A

= Clustered Regularly Interspaced Short Palindromic Repeats

  • CRISPR regions are structured, recognizable DNA sequences in non-eukaryotes
  • unusual but characteristically organized sequence in the genomes of several bacteria
  • CRISPR DNA sequences are key part of adaptive immunity in non-eukaryotes
  • in ~40% of bacteria, 90% of Archaeons
  • Number of CRISPR clusters ranges from 1 per genome (usually) but up to 20 in more unusual cases, but up to 400 in extreme example
51
Q

describe the structure of CRISPR

A

consists of repeated sequences (each ~30bp long and highly conserved) that are interleaved with spacer sequences that are similar length but have highly divergent sequence
-has a leader sequence (includes a promoter)
at one end that is usually A-T rich ~500bp in length

52
Q

Where do CRISPR Spacer DNA Sequences Originate From?

A
  • Divergent spacer sequences’ source was originally unknown
  • Bioinformatics revealed spacer sequence DNA was from bacteriophages (viruses that target bacteria) and plasmids
  • CRISPR system now known to be defence system against foreign (viral) DNA as it was discovered that a significant number of spacer regions were identical to regions of known phage or plasmids
  • > Foreign DNA is captured and turned against itself by integrating it into genome
53
Q

How to prove that CRISPR system is used for defence system against foreign (viral) DNA? (3)

A
  1. Bacteria infected with virus gain new DNA from infecting viruses (bacteria incorporated spacer sequences derived from that phage)
  2. Can add in new viral DNA to make “naïve” bacteria virus-resistant
  3. Can delete existing spacer (viral) DNA, bacteria are now virus-susceptible
54
Q

what are Cas proteins

A

CRISPR sequences are partnered with accessory proteins called CRISPR-Associated (Cas) proteins
- required for integration/defence against foreign DNA
-Cas and other proteins are part of set, but not all
are present in all genomes
-Cas1 “integrase” and Cas2 RNase present in all systems, needed for defence, acquisition
*Other Cas, Cmr, Cse, Csx proteins used, but not essential in all cases
->Other Cas proteins involved in expression, processing of existing spacers or antiviral defence

55
Q

how are Spacer Sequences acquired from Infecting Viruses?

A
  • Invading foreign DNA is cleaved by Cas2 RNase, part of which becomes new proto-spacer(= sequence in virus that will become new spacer) between two repeat sequences
  • Cas1 integrase will integrate this new protospacer into host DNA
  • proto-spacer incorporated at the proximal end near the leader sequence
56
Q

what is PAM?

A

Protospacer-Adjacent Motif

-it is a sequence that is close/adjacent to the protospacer and is part of the Repeat

57
Q

describe CRISPR transcription in relation to E coli. (Role of cascade/CasE, RNA “handles”, character of crRNAs?)

A

CRISPR is transcribed as a single Long RNA which is then processed into shorter RNA species that target the destruction of invading DNA or RNA

  • Leader sequence contains promoter for driving CRISPR transcription which contains spacers, repeats
  • the promotor generates a single RNA transcript called the pre-crRNA
  • CasE processes the pre-crispr RNA (pre-crRNA) and the resulting short crRNAs=length of 1 spacer + 1 repeat
  • E coli CRISPR associated with 8 cas genes, 5 of which form a complex called Cascade (has one subunit that is implicated in the processing of the long transcript into the individual short crRNAs)
  • these crRNAs remain bound to the Cascade complex and direct it to the invading DNA genomes of foreign DNA
  • each crRNA includes repeats that are termed the 5’ and 3’ handles, RNA “handles” serve as binding sites for proteins that form the Cascade complex targeting the foreign DNA
  • “guide” RNA binds complementary strand in target DNA sequence (spacer ensures specificity)
  • Cas9 nuclease cleaves both of DNA helix’s 2 strands
58
Q

Piwi-Associated RNAs?

A

CRISPR is mimicked in animals by Piwi-Associated RNAs

  • Animals use Piwi-associated RNAs (piRNAs) to target transposons to suppress them (defend against detrimental RNA)
  • Transposon remnants are 45% of our genome, kept suppressed by histone modification, formation of heterochromatin
  • Transposons (“jumping genes”) targeted by piRNAs presumably made from transposons
  • Failure to destroy transposons leads to spontaneous mutagenesis
59
Q

how is CRISPR/Cas9 exploited to generate different mutations?

A
  • > CRISPR/Cas9 now used to create knock-out (NHEJ) or knock-in(donor DNA) mutations and allows scientists/doctors to break/fix genes
  • > Guide RNA (crRNA) matches targeted DNA and allows Cas9 to cut both strands
  • > Double-strand breaks are made and prompts DNA damage response (Recall highly mutagenic NHEJ)
  • > If donor DNA is present or supplied, new DNA can be inserted into site
60
Q

how do mutated Cas9 nucleases increase possibilities of genetic engineering?

A

-Cas9 mutant proteins exploited to repress/activate target genes
-Cas9 protein can be modified/disabled so DNA is
bound using guide RNA but not cut
-Transcription Repressor/Activator proteins fused to Cas9 alter gene expression in controlled fashion
-Cas9 mutant proteins can be improved to increase specificity ->Cas9n ‘nickase’ mutant cuts only once-Requires 2 different Cas9/guide RNAs

61
Q

New CRISPR Prime Editing ?

A

-greatly improves prospects for genetic surgery
-Prime editing reduces “off-target” hits, doesn’t use DSB repair or need to supply donor DNA with desired edits
=Catalytically broken Cas9 is fused to reverse transcriptase and Prime guide RNA encodes targeting sequence which encodes desired edits
-Original opposite complementary strand nicked, identified for “repair” by removal and re-synthesis to match new engineered strand
-New method can fix insertions/deletions, all 12 types of mutations ->“In principle, (method) could correct 89% of known pathogenic genetic variants”

62
Q

what three parts of RNA are needed for splicing?

A

Two splice sites for each intron, a 5’ (donor) splice site and a 3’ (acceptor) splice site at each end of intron (Most conserved sequence is GU for 5’ splice site and AG for 3’ splice site)
The third part needed is the branch point sites (A) followed by a stretch of pyrimidines 11

63
Q

describe how an Intron is removed in a “lasso” (lariat) form as flanking exons joined

A

Nucleophilic attack by 2′ OH
group of branch site Adenine
on the 5′ phosphoryl group
at beginning of the intron

The 3′ OH group at 3′ end of
     upstream exon performs a 
     nucleophilic attack on the
     phosphoryl group at Guanine’s
     5′ end of downstream exon

The reaction products are an intron with a lariat structure,
plus a correctly spliced RNA-> 2 bonds broken and 2 made…. no ATP needed
-The spliced out intron is rapidly destroyed to prevent reverse reaction
*Sequential transesterification events allows intron self-splicing, self-ligation

64
Q

what is a Spliceosome? what is it made of

A

=huge ribonucleoprotein with transient snRNPs doing splicing

  • > large complex, Made of 150 proteins + 5 RNAs and requires ATP for function
  • > The 5 small nuclear RNAs (snRNAs) called U1, U2, U4, U5, U6
  • > snRNPs = small nuclear RiboNucleoprotein Proteins = snRNAs bound to proteins
65
Q

what are the several functions of snRNPs when bound to spliceosome off and on? (3)

A
  • recognize 5′ splice site and branch point site
  • bring 5′ splice site and branch point site together
  • catalyse the RNA cleavage and joining reactions
66
Q

how does RNA-RNA hybrids form between the

U# snRNAs and pre-mRNA by base pairing

A
  1. U1 binds 5′ splice site early, preparing it for U6 later
  2. U2 snRNA binds branch site
  3. U2 (+branch site) binds U6 (+5′ splice site) to bring the lariat together
  4. BBP (Branch point Binding Protein) recognize polypyrimidine stretch but then displaced by U2 snRNA that binds
  5. Helicases using ATP unwind RNA-RNA complexes, drive molecular rearrangements
  6. A rearrangement occurs in the complex U4 & U6 along with U5 (tri-RNP particle) & joins the Complex
  7. U2AF lost (both subunits)
  8. U4 and U6 snRNAs are paired
  9. U6 replaces U1 at the 5’ splice site and U1 leaves the Complex
  10. U4 is released from the complex
  11. U6 now interacts with U2 to form the active site
  12. 5′ splice site and the ‘A’ branch point brought
    close together to get “C(ut) complex”
  13. The 5′ splice site and the 3′ splice sites are brought together by U5 and the second transesterification occurs
67
Q

describe the assembly in the spliceosome

A

U1 snRNP binds to 5’ splice site
U2AF subunit binds to 3’ splice site (‘35’ part, ‘65’ part binds to poly Y part of branch point)
U2 binds to the branch site and displaces BBP (bulge helps)

68
Q

“self-splicing” introns?

A

pre-mRNAs that don’t require spliceosome (have ribozyme instead)

  • Catalytic mechanism is always transesterification
  • Splicesomes may have evolved from Group II self-splicing introns-> these have same structure as spliceosomes’s snRNP (RNA part)
69
Q

how are Group I introns different from nuclear pre-mRNA and group II

A

Group I introns release a linear intron rather than a lariat

  • smaller than Group II
  • uses free G nucleotide or nucleoside
  • > Use G-site branch site
70
Q

Two ways to ensure accuracy of splicing?

A

1) Transfer of splicing factors from RNA polymerase CTD to the 5′ splice site
- >Reduces exon skipping
2) Ensure that only splice sites next to exons are selected:
- >SR proteins (Serine-aRginine rich) bind to Exonic Splicing Enhancers (ESE)
- >Recruit components of splicing machinery (U2AF proteins to 3′ splice site, U1 snRNP to 5′ splice site)

71
Q

trans-splicing?

A

Exons from different RNA molecules can be fused
together
-Alternative splicing is where an exon may be joined to another exon that isn’t necessarily the nextmost one
-> May want exon skipping
-Same chemistry as before, but no larait formed -> “Y”-shaped RNA spliced out instead (for completely different mRNA)

72
Q

Minor spliceosome ?

A

AT-AC spliceosomes =recognizes distinct 5′, 3′ splice sites but uses same process as major s. (may be an intermediate between group II introns and full-blown major spliceosome

-Most eukaryotes use spliceosome to process
pre-mRNA, but a ‘minor’ form exists too
-U1 and U2 can be replaced by U11 and U12
(AU – AC at 5′ and 3′ sites)

nucleic acids (5 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions