Molecular Biology: Chromatin And Translation

Front 1

Experiment showing that TBP alone cannot respond to Sp1 (313)

Back 1

METHOD
A composite Sp1-responsive promoter was formed that contained 6 GC-boxes and a TATA box. Run off assay was performed in which a complete transcript should be 375 nt. TFIID was mixed with mutant forms of TBP as well as with other TFs and RNApolII. Run off assay performed with labeled nt.

RESULTS
Natural TFIID supported a high level of transcription and transcription was enhaced by the TF Sp1. Transcription due to TBP was not stimulated by Sp1 in the absence of TAFIIs.

Front 2

coactivators (311)

Back 2

general TFs that also play a role in activation usually via a protein-protein interaction with an activator

(e.g., TAFII150)

Front 3

TAFII250 (313)

Back 3

a histone acetyltransferase (HAT)--adds acetyl groups to Lys on histones

also acts as a kinase

Front 4

Experiment showing that activation by Sp1 requires TAFII 110 (314)

Back 4

METHOD
Using cell extracts depleted of TFIID, they replaced the TFIID with different mixtures of coactivators and added Sp1 in increasing amounts. Primer extension assay.

RESULTS
activation of transcription was observed in each extract only if TAFII 110 was present (along with 150, 250, and TBP). 150, 250, and TBP were enough for basal transcription.

Front 5

activators (3.1)

Back 5

proteins that bind DNA (enhancers and upstream promoter elements) at specific sequences that can activate (or repress) gene expression.

Front 6

3 types of domains for activators (343)

Back 6

1. DNA-binding
2. transcription-activating
3. (only in some) ionization domains

Front 7

3 types of DNA binding domains (343)

Back 7

1. zinc binding (zinc fingers- found in TFIIIA and Sp1; zinc modules found in glucocorticoid receptor; zinc modules containing cys, found in GAL4)
2. homeodomains (~60aa)
3. bZIP and bHLH (b=basic) (e.g., MyoD protein)

Front 8

3 types of transcription-activating domains (343)

Back 8

1. acidic (GAL4)
2. glutamine-rich (Sp1)
3. proline-rich (CTF, which binds at the CCAAT box)

Front 9

Gal4 protein, general function (346)

Back 9

yeast activator that controls a set of genes responsible for metabolism of galactose

Front 10

how does GAL4 bind to upstream activating sequences? (346)

Back 10

as a dimer

cooperatively

(activators are usually dimers, tetramers)

Front 11

experiment to determine the activity of a chimeric transcription factor (352)

Back 11

METHOD
2 plasmids were introduced into yeast cells: plasmid encoding LexA-GAL4 (transcription activating domain of GAL4 and DNA-binding domain of LexA) AND a test plasmid construct (see fig 12.13). Each test plasmid had GAL1 promoter linked to lacZ reporter gene.

RESULTS
With a UAS to bind to, transcription was active and did not depend on the transcription factor because of endogenous GAL4. With no DNA target site, LexA-GAL4 could not activate because it couldn't bind to the DNA. With the lexA operator, transcription was greatly stimulated.

Thus, the transcription-activating and DNA-binding domains of GAL4 can operate independently!

Front 12

experiment that shows that GAL4 binding is cooperative (358)

Back 12

METHOD
Transcription measured by primer extension on preinitiation complexes assembled with nuclear extract and templates containing one or five GAL4-binding sites, with or without GALAH

RESULTS
5 sites allowed for cooperativity in activation of transcription by GAL4.

Front 13

advantages of activators binding as dimers (362)

Back 13

1. affinity of binding between a protein and DNA varies with the square of the free energy of binding, meaning binding dimers is much tighter than binding a monomer.

2. greater affinity between activator and DNA is needed because activators are usually found in very low concentrations.

Front 14

experiment showing how Fos and Jun bind (362)

Back 14

METHOD: Mobility shift assay
Mixed Fos, Jun, and the bZIP domains of Fos and Jun (Fos- and Jun-core) with a labeled DNA that contains an AP-1 binding site. Measured DNA-protein interaction with EMSA.

RESULTS
Combinations of Fos or Fos-core with Jun or Jun-core bound to DNA, but they did not bind alone. The Fos-Jun heterodimer bound better than homodimers.

Front 15

4 models that explain how activators function (363)

Back 15

1. change in topology: binding of an activator causes supercoiling which allows other TFs, pol to bind.
2. sliding: TF binds to enhancer and slides down the DNA to the promoter, where it facilitates binding of TFs, pol.
3. looping: TF binds to enhancer and loops DNA in order to bind to the promoter, where it facilitates binding of TFs, pol.
4. facilitated tracking: TF binds to the enhancer and causes a short DNA segment to loop out downstream. Increasing the size of the loop allows the factor to track along DNA until it reaches the promoter and can facilitate binding of TFs, pol. (2-step mechanism)

Front 16

catenane (364)

Back 16

physically linked plasmids --> circles linked as a chain

Front 17

how do catenanes form? (364)

Back 17

Promoter and enhancer are on opposite ends of a plasmid that contains phage attB and attP sites. In vitro recombination between these sites causes the promoter and enhancer to end up on 2 linked plasmids

Front 18

experiment with catenanes (365)

Back 18

METHOD
Injected mixtures of plasmids into oocytes and measured transcription rates with S1 mapping. Included a test (minigene=40bp) and reference plasmid (minigene=52bp). Interested in the ratio between test and reference transcription.

RESULTS
When the plasmid contained only the promoter, the reference signal was stronger than the test signal. When the enhancer is next to the promoter, the test plasmid signal is much stronger than reference. When the enhancer and promoter were opposite from each other, the enhancer still worked, but not as well. *When the enhancer and promoter were on separate plasmids in a catenane, the enhancer still worked.

Thus, enhancer does not need to be on the same DNA as the promoter, but it does need to be nearby so the two can interact.

Front 19

5 types of histones (386)

Back 19

H1, H2A, H2B, H3, H4

2A&B, 3, 4 make up the nucleosome core.

H1 binds linker DNA; important for packing, falls off easily

Front 20

features of histones (386)

Back 20

1. basic
2. contain N- and C- terminal extensions

Front 21

features of 5s rRNA genes--what transcribes them and types (395)

Back 21

1. transcribed by RNA pol III with the help of TF III A, B, C
2. oocyte-specific (98%) and somatic (2%) genes

Front 22

experiment with oocyte/somatic 5s rRNA genes (13.16)

Back 22

METHOD
1. Gently purified genes from both oocyte and somatic cells.
2.Electrophoresed histones from somatic cells (washed different ways)
3. ion-exchange chromatography

RESULTS
Initially found that both somatic and oocyte-specific genes can be transcribed in vitro if H1 is missing.

Found that when histones were added to somatic cell chromatin at one molecule per 200 bp, 5s rRNA synthesis fell dramatically and remaining transcription was all in the somatic genes.

somatic genes are transcribed in both somatic cells and oocytes because they form more stable complexes with the transcription factors.

Front 23

explain the concept of a "race" between TFs and histones when it comes to transcriptional repression/activation (397)

Back 23

Transcription is suppressed in the presence of histones because they make the DNA more compact. TFs prevent nucleosomes from forming.

So, if histones get to the control region of the DNA first, txn will be repressed. If TFs get there first, txn will be active.

Front 24

How much do core histones repress txn?

How much do core histones + H1 repress txn? (398)

Back 24

core histones: 4 fold

core histones + H1: 25-100 fold

Front 25

experiment: in vitro transcription of reconstituted chromatin (fig. 13.18)

Back 25

METHOD
1. reconstituted chromatin with plasmid DNA containing Kruppel gene +/- core histones (in varying ratios of protein to DNA) +/- polyglutamate +/- H1.
2. primer extension analysis to measure txn

RESULTS
Found that core histones can inhibit txn of the Kruppel gene up to 75% in a dose-dependent manner

Front 26

2 explanations for 75% repression of the reconstituted chromatin experiment--and experiment to show which is true (398)

Back 26

1. Nucleosomes could slow progress of all RNA pol by 75%, but not stop any of them.

2. 75% of polymerases could be blocked entirely, but 25% of promoters aren't blocked (are free of nucleosomes)

to test, they cut DNA w/RE and found txn was eliminated, meaning option 2 is correct because it means that the cut site was nucleosome-free.

Front 27

experiment showing that H1 adds to inhibition of txn (presence of GAL4-VP16) (fig 13.19a)

Back 27

METHOD
1. reconstituted chromatin +/- core histones, +/- H1, and +/- GAL4-VP16 (activator)
2. primer extension assay to measure txn
3. look at ratios to determine amount of repression/anti-repression

RESULTS
Found that at moderate histone H1 levels, the presence of activators could prevent repression.

Front 28

experiment showing that H1 adds to inhibition of txn (presence of Sp1) (fig 13.19b)

Back 28

METHOD
primer extension assay, as in related experiment

RESULTS
Since Sp1 can bind to the GC boxes of the E4 promoter, it caused a 92 fold increase in activity. (Activation by Sp1 on naked DNA=2.8 fold, so 33-fold of it was due to anti-repression)

Front 29

anti-repression (399)

Back 29

prevention of repression by histones or other transcription-inhibiting factors. Antirepression is part of a typical activator's function.

Front 30

2 pieces of evidence for nucleosome-free zones (400)

where would nucleosome-free zones likely be?

Back 30

1. EM studies

2. DNase sensitive sites

nucleosome-free zones would likely be in the control regions of active genes

Front 31

experiment showing DNase I sensitive sites (nucleosome-free zones) (fig 13.22)

Back 31

METHOD
1. Use SV40 viral DNA in infected monkey cells (the DNA exists as a minichromosome)
2. Cut with DNase I
3. strip off nucleosomes from DNA
4. cut DNA w/REs
5. electrophorese, Southern blot

RESULTS
Can determine where the nucleosome-free zone is (assuming you know where the DNA control region is located) by studying RE digest.

Here, they found a great deal of 67% product and not a lot of 33%, meaning the cut-site was in the 33% product, which is where the control region is

Front 32

experiment mapping DNase hypersensitive sites in the 5'-flanking region of the globin gene (fig 13.26)

Back 32

METHOD
1. extract DNA from HEL, PUTKO, J6 cells
2. add DNase I; measure over increasing time periods
3. strip nucleosomes
4. cut with RE
5. electrophorese, Southern

RESULTS
DNase hypersensitivity was seen in the cells (HEL and PUTKO) that express globin, but not in J6 cells, which don't.

Thus, hypersensitivity corresponds to the presence of gene-specific factors that exclude nucleosomes from active genes, but not from inactive genes.

Front 33

histone code

Back 33

all of the modifications of the nucleosome (histone proteins): acetylations, phosphorylations, methylations

Front 34

how does acetylation increase gene expression? (3 ways)(405-ish)

Back 34

1. The basic N-term of H4 can interact with acidic pocket in H2A-H2B dimer
2. provides a docking site for proteins with a bromodomain (e.g., TAF250), which are HATs
3. Binding of HATs and other proteins result in more modification

Front 35

2 types of HATs (406-ish)

Back 35

1. nuclear: bind to partially-acetylated histones
H3: K9 and K14
H4: K5, 8, 16.

2. cytoplasmic: acetylate new proteins

Front 36

activity assay for HAT (fig 13.27)

Back 36

METHOD
1. electrophoresed a nuclear extract in an SDS-polyacrylamide gel containing histones, BSA, or no protein
2. silver stained gel to detect protein OR treated with radioactive acetyl-CoA (3H)
3. washed, fluorographed

RESULTS
presence of HAT activity known because the HAT would transfer labeled acetyl groups from acetyl-CoA to the histones in the gel.

Front 37

mechanism of HATs (406)

Back 37

1. bind near transcription start site
2. acetylate core histones, which neutralizes their positive charge
3. this causes loosening of histones on DNA and remodeling of chromatin
4. After chromatin remodeling, DNA is more accessible to TFs and txn is stimulated

Front 38

examples of co-activators with HAT activity(407)

Back 38

p55, Gcn5p, CBP/p300, TAFII250

Front 39

What is the effect on txn of Max-Myc?

What is the effect on txn of Max-Mad? (407)

Back 39

Max-Myc: activator

Max-Mad: repressor

Front 40

mechanism of histone deacetylases (407)

examples of repressors, corepressors, and histone deacetylases

Back 40

1. known txn repressors (Mad-Max) interact with corepressors (SIN3A), which interact with histone deacetylases (HDAC1)
2. the deacetylases remove acetyl groups from basic tails of core histones, which tightens the nucleosomes' grip on DNA
3. Thus, nucleosomes are stabilized and this causes txn repression

Front 41

experiment corraborating the histone deacetylase model (ternary complex)(fig 13.30)

Back 41

METHOD
1. transfect cells with plasmid encoding FLAG epitope alone or FLAG-HDAC2. Also add a plasmid encoding either no Mad1, Mad1, or mutant Mad1.
2. immunoprecipitate cell lysates with anti-FLAG antibody or just collect lysates.
3. electrophorese, blot, probe with anti-SIN3A or anti-Mad1

RESULTS
SIN3A coprecipitated with FLAG-HDAC2. In presence of Mad1Pro (mutant), which cannot bind to SIN3A, SIN3A did NOT coprecipitate with FLAG-HDAC2, which shows that SIN3A does not interact with HDAC2 directly, but interacts with it via Mad1.

Front 42

model and example of activation and repression by the same nuclear receptor (fig 13.31)

Back 42

Example: thyroid hormone receptor (TR), which binds as a heterodimer to RXR and then binds to thyroid hormone response element. In the absence of thyroid hormone, it acts as a repressor via the deacetylase ternary model.

In the presence of thyroid hormone, TR-RXR serves as an activator, partly due to interactions with p300/CBP, TAFII250, etc.

Front 43

4 factors that participate in chromatin remodeling, what they require, and how they work (410)

Back 43

SWI/SNF family, ISWI, NuRD, INO80

all require ATP

alter the structure of nucleosome cores to make DNA more accessible to TFs and nucleases

Front 44

chromatin immunoprecipitation (ChIP) method (fig 13.32)

Back 44

1. epitope-tag a protein
2. obtain cells in various stages of cell cycle and add formaldehyde to crosslink DNA + proteins
3. make cell extracts, immunoprecipitate with antibodies against epitope
4. to verify that complexes contain the right gene, do PCR with specific primers

Front 45

experiment showing timing of histone acetylation in chromatin of IFN-beta promoter (fig 13.33)

Back 45

METHOD
1. performed ChIP with HeLa cell extracts at various times after infection with virus, using antibodies directed against specific acetylation points and TBP.
2. PCR on immunoprecipitated chromatins with IFN-beta specific primers

RESULTS
Found pattern of histone acetylation was not random and timing of acetylation varied from position to position.

Front 46

enhanceosome (gloss)

Back 46

the complex formed by enhancers coupled to their activators

Front 47

model for the histone code at IFN-beta promoter (fig 13.35)

Back 47

1. TFs bind (enhanceosome assembles)
2. HAT (GCN5) interacts with TFs and binds
3. HAT acetylates K8(H4), K9(H3) on both dimers
4. causes protein kinase to bind
5. kinase phosphorylates S10(H3)
6. This causes GCN5 to acetylate K14(H3)
7. Causes SWI/SNF remodeling complex to bind and make chromatin accessible to TFIID
8. TFIID binds and nucleosome moves 36bp

Front 48

model for involvement of histone methylation in chromatin repression (fig 13.38)

Back 48

1. nucleosomes become methylated on K9(H3)
2. this recruits HP1, which binds to a methylated K9 on one nucleosome and recuits a histone methyl transferase to methylate K9 on a neighboring nucleosome.
3. propagates from one nucleosome to the next

Front 49

structure of mRNA cap (472)

Back 49

a methyl group on the N7 position in guanine (7'mG)

the methyl group contributes a positive charge

Front 50

4 functions of 7'mG cap (474)

Back 50

1. protect from degradation
2. enhance translation
3. export to cytoplasm
4. enhance splicing

Front 51

experiment showing the effect of capping on mRNA stability (fig 15.6)

Back 51

METHOD
1. use radiolabeled capped, blocked, and uncapped mRNA
2. inject in oocytes, wait 8 hours
3. isolate RNA
4. glycerol gradient

RESULTS
Capped and blocked much more stable over time than uncapped.

When caps and blocks were removed, the RNA was still degraded

Front 52

experiment showing the effect of capping/poly-A on translatability (475)

Back 52

METHOD
1. introduced luciferase mRNA +/- cap, +/- polyA into tobacco cells.

RESULTS
Poly(A) provides a 21x boost, but capping provides a 297x boost.

Front 53

experiment showing the effect of capping on transport of mRNA (475)

Back 53

METHOD
Capping occurs at pol II promoter, but not at pol III promoter.
1. created a construct in which the U1 snRNA gene was under control of the pol III promoter
2. construct + oocyte nuclei + 5sRNA gene + radiolabeled nuctide, low amounts of alpha-amanitin so no pol II would transcribe

RESULTS
Virtually all uncapped U1 snRNA made by pol III remained in the nucleus

Front 54

experiment verifying that pol III constructs remained uncapped and in the nucleus (fig 15.7b)

Back 54

METHOD
1. immunoprecipitate labeled RNAs with anti-Sm or anti-TMG antibody.
2. electrophoresed RNAs from supernatant and immunoprecipitate

RESULTS
Found that only the wild type snRNA made by pol II precipitated--none of the pol III products were in the pellet. Thus, none of the pol III products appear to have spent any time in the cytoplasm.

Front 55

experiment showing size of polyA region (fig 15.8)

Back 55

METHOD
1. isolate radiolabeled RNAs from the nucleus and cytoplasm
2. digest everything but A's using RNase A and RNase T1.
3. electrophorese, collect fractions, determine radioactivity

RESULTS
the polyAs electrophoresed more slowly than the 5s marker, making them about 200nt long.

Front 56

how do we know that polyA goes to the 3'-end of the mRNA? (478)

Back 56

if polyA was located interiorally, RNase digestion would yield polyA with a phosphate at the 3' end and base hydrolysis would give only AMP. In reality, RNase digestion yields polyA with an unphosphorylated adenosine at the 3'-end and base hydrolysis gives AMP + adenosine.

Front 57

how do we know that polyA is added post-transcriptionally? (478)

Back 57

we know polyA is done post-transcript. because there are no huge polyT sequences at the end of genes AND actinomycin D does not inhibit polyadenylation, but it does inhibit transcription.

Front 58

how are polyA's added? (478)

Back 58

posttranscriptionally by polyA polymerase

Front 59

experiment that shows polyA helps in protection of mRNA (fig 15.10)

Back 59

METHOD
1. injected globin mRNA +/- polyA into oocytes and measured rate of globin synthesis (via radioactivity) over 2 days

RESULTS
After 6 hours, the mRNA without polyA could no longer support translation whereas the mRNA with polyA could.

Later experiments contradicted this.

Front 60

experiment showing the effect of polyA on translatability (fig 15.11a)

Back 60

METHOD
1. incubated mRNAs +/- caps, +/- polyA with 35S-methionine in cell extracts
2. allow 30 min for protein synthesis, electrophorese and measure with fluorography

RESULTS
polyA enhaced translatability of both capped and uncapped mRNAs.

Front 61

polyA-binding protein I (PAB I) (479)

Back 61

binds to eukaryotic mRNA during translation and boosts translation efficiency

polyA- mRNA cannot be transcribed as efficiently because it cannot bind PAB I

Front 62

experiment showing that polyA did NOT offer protection to the mRNA (fig 15.11b)

Back 62

METHOD
1. labeled mRNAs with 32P and incubated them in cell extracts
2. electrophoresed, autoradiographed, quantified with densometry

RESULTS
PolyA made no difference in stability of mRNA over time.

Front 63

experiment showing the effect of polyadenylation on recruitment of mRNA to polysomes (480)

Back 63

METHOD
1. mixed 32P- and 3H- labeled polyA+/- with cell extract
2. separated polysomes from monosomes using sucrose gradient

RESULTS
polyA+ mRNA is clearly better at associating with polysomes

enhance translatability by recruiting mRNAs to polysomes

Front 64

experiment showing the effect of the size of polyA on recruitment of mRNA to polysomes (fig 15.12b)

Back 64

the greatest increase of mRNA association with polysomes is seen when the polyA grows from 5 to 30nt, after which point there is a gradual increase

Front 65

model for polyadenylation (fig 15.26--thanks a lot, Janine...)

Back 65

1. CPSF, CstF, and CF I and II assemble on the pre-mRNA, guided by AAUAAA and GU/U motifs
2. cleavage occurs, stimulated by the CTP of RNA pol II
3. CstF and CF I, II leave the complex and polyA polymerase enters
4. PAP, aided by CPSF, initiates polyA synthesis, yielding an oligo 10nt long.
5. PAB II enters the complex and allows rapid extension of the oligoA to polyA.
6. complex dissociates

Front 66

in what order does cleavage of mRNA and polyadenylation occur? (481)

Back 66

cotranscriptionally

Front 67

2 hypotheses for synthesis and polyadenylation of mRNA (481)

Back 67

1. transcription terminates immediately downstream of a polyA site and THEN polyadenylation occurs.

2. Transcription continues to end of region and then the primary transcript is clipped and polydenylated at each clipping site.

#2 is correct

Front 68

experiment showing that transcription of eukaryotic genes extends beyond the polyA site (fig 15.15)

Back 68

METHOD
1. isolated nuclei and incubated them with [32P]UTP to label run-on RNA.
2. hybridized this labeled RNA to DNA frags of varying sizes

RESULTS
They observed just as much hybridization to frags lying over 500bp downstream of the polyA site as to frags within the gene. Thus, transcription continues at least 500bp downstream of polyA site.

Front 69

experiment showing the importance of AAUAAA sequence to polyadenylation (fig 15.16)

Back 69

METHOD
1. created recombinant viruses with polyA sites (AAUAAA) 240 bp apart (and one mutant each in which one of these sites was knocked out).
2. performed S1 analysis with a probe that yields a 680nt signal if upstream polyA site works and 920nt signal if downstream site works.

RESULTS
Deletion of an AAUAAA prevents polyadenylation at that site.

Front 70

3 polyadenylation sites (483)

Back 70

1. AAUAAA site: 20nt upstream of polyA site
2. GU-rich motif: 23-24bp after AAUAAA site
3. U-rich motif: directly follows GU-rich site.

Front 71

relationship between RNA pol II and polyadenylation (485)

Back 71

a phosphorylated CTD of RPB1 is required for polyadenylation

Front 72

tRNA charging, general (546)

Back 72

the process of binding amino acids covalently to tRNAs: one of the pre-initiation events of translation

Front 73

where and how is an amino acid attached to the tRNA? (546)

Back 73

attached by an ester bond between the carboxyl group of the amino acid and the 2' or 3' hydroxyl group of the terminal adenosine of tRNA

Front 74

2 steps of tRNA charging (546)

Back 74

1. aminoacyl-tRNA synthetase couples an amino acid to AMP to form aminoacyl-AMP + PP.
2. aminoacyl-tRNA synthetase replaces the AMP in the aminoacyl-AMP with tRNA to form an aminoacyl-tRNA + AMP.

Front 75

30s initiation complex (549)

Back 75

30s ribosomal subunit + mRNA + aminoacyl-tRNA + initiation factors

Front 76

2 types of methionine tRNAs (550)

how did they separate the two types?

Back 76

1. tRNA mMet
2. tRNA fMet (contains formyl group): must be charged with the formyl group when ON the tRNA

separated these two types by countercurrent distribution

Front 77

experiment to determine tRNA codon specificity(550)

Back 77

METHOD
1. make labeled aminoacyl-tRNA in vitro
2. make specific trinucleotides
3. add ribosomes and see if tRNA binds

RESULTS
tRNA-mMet responded to the AUG codon and tRNA-fMet responsed to AUG, GUG, and UUG. The latter two can also serve as initiation codons 8% and 1% of the time, respectively. Here, both GUG and UUG code for methionine.

Front 78

where do the two methionine tRNAs incorporate their methionines into the protein product? (550)

Back 78

tRNA-fMet incorporates methionines only in the first position of the polypeptide (often excised later).

tRNA-mMet methionines were incorporated primarily in the interior of the protein product.

Front 79

why doesn't fMet show up in mature protein? (551)

Back 79

1. formyl group is later removed
2. 10 amino acids are removed???

Front 80

positive strand phages (552)

Back 80

their genomes are also their mRNAs.

examples: R17, f2, MS2

contain 3 genes: A protein, coat protein, replicase

Front 81

experiment with R17 and ribosomal subunits (552)

Back 81

If 30s subunit came from E.coli (R17's natural environment), R17 coat protein could be translated. If it came from B. stearothermophilus, translation didn't occur.

Dissociated the 30s subunit into its RNA and protein components. 2 components stood out: S12 (ribosomal protein) and 16s rRNA. If these components came from E. coli, translation of coat gene was active. If either was from the other, translation didn't occur.

Front 82

Shine Delgarno sequence (or RBS-ribosome binding sequence) (552)

Back 82

They noticed that binding sites contained a site just upstream of the initiation codon that was complementary to the 3' end of the 16s rRNA.

Noted that the 16s of E. coli was more complementary to the coat gene than the 16s of Bacillus, implying that they had found the ribosome binding site.

*Found that wherever 3+ base pairs were possible between 16s and sequence upstream of codon, ribosome binding occurred.

Front 83

translation coupling (Phil's notes)

Back 83

when the translation of one gene relies on another gene being translated (as with the coat protein and replicase genes)

Front 84

experiment with colicin E3 in support of Shine Delgarno (553)

Back 84

METHOD
1. bound E. coli ribosomes to R17 A protein gene's initiation region
2. treated complexes with Rnase called colicin E3, which cuts near the 3' end of 16s rRNA.
3. fingerprinted, S1 mapped the RNA and found a ds RNA frag.

RESULTS
1 strand of the ds RNA was an oligo from the A protein gene initiation site and the other was an oligo from the 3' end of 16s rRNA.

Front 85

what is the small subunit of a prokaryotic ribosome made up of? (546)

the large subunit?

Back 85

small: 16s rRNA + protein = 30s

large: 23s + 5s + protein = 70s

Front 86

experiment that demonstrated ribosome subunit exchange (fig 17.4)

Back 86

METHOD
1. made heavy ribosomes labeled with 3H-U and light ribosomes labeled with 14C-U.
2. subjected ribosomes to sucrose gradient, collected fractions, detected radioisotopes. These steps allowed them to observe sedimentation behavior of the ribosomes.
3. Then, they added 3H-labeled heavy ribosomes to media with light ribosomes and extracted them.

RESULTS
The 3H-labeled ribosomes ended up sedimenting intermediately (between light and heavy), showing that there was subunit exchange occurring.

Front 87

experiment showing the effect of IF3 in dissociating ribosomes (fig 17.5)

Back 87

METHOD
1. mixed E. coli ribosomes with (a) no factors, (b) IF1, or (c) IF3 and separated intact ribosomes from subunits with centrifugation.
2. assayed concentration of ribosomes or subunits by measuring absorbance at 254nm.

RESULTS
Only IF3 caused a significant dissociation, as evidenced by a greater amount of subunits as opposed to whole ribosomes.

Front 88

function of IF1 (549)

function of IF3

Back 88

IF1 promotes dissociation

IF3 binds to the 30s subunit and prevents reassociation.

Front 89

experiment showing the ribosome-dependent GTPase activity of IF2 (558)

Back 89

METHOD
1. measured release of labeled inorganic phosphate from gamma-32-P GTP in the presence of IF2, ribosomes, or IF2 + ribosomes.

RESULTS
Only together could ribosomes and IF2 hydrolyze the GTP.

Front 90

experiment showing that GTP hydrolysis is required for IF2 to leave the ribosome (fig 17.14)

Back 90

METHOD
1. mixed labeled P labeled IF2 and 3H labeled fMet-tRNA with 30s subunits to from 30s initiation complexes.
2. then, added 50s ribosomal subunits in the presence of either GDPCP or GTP
3. analyzed via sucrose gradient ultracentrifugation

RESULTS
With GDPCP, the 70s complex did not break up. With GTP, dissociation of IF2 occured while fMet-tRNA remained bound to the 70s complex. Thus, GTP hydrolysis is necessary for IF2 release.

Also, more fMet-tRNA bound in the presence of GTP, hinting at IF2's catalytic ability.

Front 91

steps of prokaryotic translation initiation (559)

Back 91

1. dissociation of 70s ribosome into 50s and 30s subunits, under the influence of IF1.
2. Binding of IF3 to the 30s subunit, which prevents reassociation with the 50s subunit.
3. Binding of IF1, IF2, and GTP alongside IF3 on 30s.
4. Binding of mRNA and fMet-tRNA to form the 30s initiation complex.
5. Binding of 50s subunit, which causes the loss of IF1 and 3.
6. Dissociation of IF2 from the complex with simultaneous hydrolysis of GTP. The produce is 70s iniation complex, which is ready to begin elongation.

Front 92

experiment giving evidence that an amber mutation was a stop codon--fusion (fig 18.31)

Back 92

Made a deletion in the rIIA gene so that it was fused to rIIB. In this fusion gene, they saw B activity, but no A. When they made an amber mutation in A, they didn't see activity in A or B, unless the construct was placed inside of a suppressor strain.

This lead them to believe that the amber mutation was a stop codon.

Front 93

experiment giving direct evidence that an amber mutation is a stop codon--N term frags (fig 18.32)

Back 93

Mutated phage head genes with amber mutations throughout and digested the protein products. Found they were all N-terminal fragments.

Thus, amber mutation is a stop codon.

Front 94

how do suppressor strains work (fig 18.35)

Back 94

In a suppressor strain, an anticodon for, say, tyr has been mutated so it is complementary to the, say, amber mutation. Thus, as the mRNA is being translated and the UAG codon comes up, a tyr is going to be attached to the protein product instead of translation termination occurring.

Front 95

specificity of release factors (618)

Back 95

RF1: UAA, UAG (AG of stop codon binds to Pro-Ala-Thr)
RF2: UAA, UGA (GA of stop codon binds to Ser-Pro-Phe)

Front 96

Nirenberg's assay for release factors (fig 18.37)

Back 96

He labeled the fMet of tRNA and loaded the ribosome with the AUG codon and the tRNA-fMet. Then he added a termination sequence + a RF and observed if the fMet was released.

Front 97

how does a tmRNA work? (goodness, this thing is sweet!) (621)

Back 97

the tRNA-like domain is charged with an alanine, binds to the ribosome's A site and via peptidyl transferase, donates the alanine to the stalled polypeptide.

the mRNA part contains an ORF that causes the ribosome to stop translating the incomplete mRNA and start translating the ORF of the tmRNA. trans-translation

Front 98

ribosome recycling factor (RFF) (625)

Back 98

protein factors that resemble tRNAs that can release ribosomes from natural mRNAs.

an essential process