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Page 1: Joyce paper #1

Joyce paper #1

Page 2: Joyce paper #1

Their System

• A 157-mer RNA that has 2 64 nucleotide stretches in which G, A, and U were randomly incorporated, a central sequence between them for substrate nucleic acid to bind, and flanking sequences of some importance• Run reactions with the RNA pool and the substrate nucleic acid• The activity they were looking would join the nucleic acid substrate to

the 5’ end of the ribozyme (i.e., ligase)

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Their System

• They then ran the reaction products on a polyacrylamide gel, isolating from the gel product molecules of the correct size (the ligation products should be 191-mers)• The products were reverse transcribed from RNA to DNA, amplified by

PCR, then transcribed with T7 RNA polymerase to generate the RNA pool for the next round• By using primers that anneal to the 3’ end of the ribozyme and the 5’

end of the substrate, the only PCR products should be those in which the ligation reaction has occurred (it is possible a 191-mer could have come out of the gel that was not the desired product)

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Their System

• They did this for five rounds: round 1 reaction time was 18 hr, round 2 was 3.5 hr, round 3 was 30 min, round 4 was 30 s, and round 5 was 10 s• After the 5th round, the products were cloned, and individual clones

were tested for activity• One complication in their system was that the ligation could have

been either 2’-5’ or 3’-5’ (i.e., normal)

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Ligation type

• To determine whether each ribozyme was a 2’-5’ ligase or a 3’-5’ ligase, they incubated each with a 32P-labeled oligonucleotide substrate (which can be visualized on gels), and then digested the products with RNase T2• RNase T2 cleaves 3’-5’ bonds (only) in RNA• If the ribozyme formed a 3’-5’ bond, the product of the reaction seen on the

gel would be the substrate molecule itself• If the ribozyme formed a 2’-5’ bond, then the product would be one

nucleotide longer than the substrate• 7 clones were examined, only one had 3’-5’ ligase activity (the others were

2’-5’; previous work in similar ribozyme systems indicates 2’-5’ ligase ribozymes are much more likely)

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Improvement

• The sequence of the only 3’-5’ ligase they produced was used as a starting point for additional work• They chemically synthesized the 157-mer, but at each position there was

an 8% chance of a mutation (e.g., for position 38, 8 out of 100 molecules would have a different nucleotide than the original sequence)• Using this semi-randomized pool, 5 more rounds of selection were done

and the products cloned• 9 clones were examined, all were 3’-5’ ligases• The most active of these (designated R3) was used for subsequent work

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Trimming R3

• R3 was partially hydrolyzed (giving a pool of R3 fragments of random length) then reacted with 32P-labeled substrate nucleic acid (34 nucleotides in length)• When visualizing the products on a gel, anything larger than 34

nucleotides (the substrate size) must have been the result of the ligation reaction• So smallest band on the gel >34 nucleotides would contain the

shortest fragment of R3 that retained activity• Result was 108 nucleotides, so the minimal R3 length that retains

activity is 74 nucleotides (and its rate was 2X that of the full size R3)

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Minimal R3

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Minimal R3

• The 74-mer displayed Michaelis-Menten kinetics (can you use in vitro evolution to get an allosteric ribozyme? Do allosteric ribozymes even exist?)

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Minimal R3

• The 74-mer also retains 3’-5’ specificity

E100 is a negative control, it is a ribozyme known to have 3’-5’ ligase activityCF is a positive control, it is a ribozyme known to have 2’-5’ ligase activity

I have no idea what the 8-mer data are for

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Minimal R3

• A second experiment was run to verify that the 74-mer has 3’-5’ ligase activity and not 2’-5’, this experiment used a catalytic DNA (you read that correctly) with a 3’-5’ hydrolase activity (but it does not break 2’-5’ bonds)• The other lanes show 25-mer controls of

chemically synthesized RNAs the sequence of which spans the sequence where the ligation occurs in the ribozyme; in one, the simulated ligated product had a 2’-5’ linkage, in the other a 3’-5’ linkage

+ and – refers to the presence or absence of the catalytic DNA

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Secondary Structure

• Used chemical reagents that covalently modify adenine and/or guanine bases, treat 74-mer in absence and presence of substrate nucleic acid• After chemical treatment, use reverse transcriptase to synthesize DNA

from the 74-mer, anywhere a base is modified the RT stops• A10, A11, and A14 were modified in absence of substrate but were

strongly protected in the presence of substrate• Note these bases are not involved in base-pairing to substrate (they’re

at the opposite end of the 74-mer from the substrate binding sequence), this implies the 74-mer undergoes a conformational change when substrate binds

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Secondary StructureProtected bases when substrate is bound

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Secondary Structure

• Still couldn’t be too sure that substrate wasn’t somehow base-pairing with the protected adenines• Connected 5’ end of substrate to 3’ end of 74-mer using a 5’GAAA3’

loop, this made it much more likely the substrate was in fact base-pairing to 3’ end of 74-mer

Loop added here

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Secondary Structure

• Now when ligation occurs, product should be a circular RNA; that is in fact what they observed, also noting dimers were formed as well as circular dimers (why?)• Now, to spice things up a bit, note that the 74-mer itself (nor any of

its precursors) contains no C at all, but the substrates used have contained them• Are the cytosine bases in the substrate important for catalysis?• One way to find out: get rid of them

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Getting Rid of C

• Cs in substrate were replaced with Us to form G•U “wobble” base pairs, result was an inactive ribozyme• Next step was to replace the Gs in the ribozyme with As to regain Watson-

Crick base pairs, also inactivated ribozyme• By trying various combinations of changing Cs to Us and Gs to As, was

determined that the critical issue was that having a U in place of c5 in the substrate abolished activity, the other G•C base pairs can be safely changed to A•U• Since a U replacing c5 in the substrate was not acceptable, could still get rid

of the C by putting the A in its place (instead of the U) and then changing the ribozyme sequence so U base pairs with the A in the substrate: full activity!

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Getting Rid of C

• New C-free form of substrate had a different sequence that necessitated changing the 74-mer to a 77-mer (in all of the excitement they changed the substrate over to an all-RNA substrate that would base pair correctly with the 3’ end of the ribozyme, but was 3 nucleotides longer; so the complementary 3 nucleotides were added to the 3’ end of the ribozyme)• 74-mer: kcat=0.016 min-1, Km=0.2 mM• C-free 77-mer system: kcat=0.013 min-1, Km=6.2 mM

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Getting Rid of C

• The higher Km for the 77-mer may be due to the replacement of the stronger G•C base pairs with the weaker A•U base pairs (note the 77-mer system had more total base pairs between substrate and ribozyme); they also speculate that the binding may also be affected by as-yet undetected differences in 3D structure of the 74-mer and the 77-mer• They kept referring to Fig. 3 as evidence that the 77-mer retained the

3’-5’ specificity, but it doesn’t look like Fig. 3 has anything to do with the 77-mer

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Putting C Back In (make up your minds)

• Since the ribozyme itself had never contained C, they wondered how well it would tolerate inclusion of C• The 74-mer sequence was chemically synthesized as DNA such that

each position had a 1% chance of having a C• After synthesis, the sequences were randomized even more through

hypermutagenic PCR followed by transcription to give RNA sequences (effectively randomly incorporating C into each position of the 74-mer)• 6 rounds of in vitro evolution resulted in all clones examined having

the same sequence

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R3C Ligase

• The resulting ribozyme was designated as the R3C ribozyme

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R3C Ligase

• Notice that it is now a 73-mer (position 7 was lost somewhere along the way; stuff happens you know, they’re very busy people and sometimes things fall through the cracks), 11 other mutations were also noted• R3C ligase: kcat=0.32 min-1, Km=0.4 mM, remains a 3’-5’ ligase• Previous slide also shows results of protection experiments

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R3C Ligase

• Including C is thought to give improved catalysis because R3C resulted from (among other changes) 1 A•U and 4 G•U base pairs being replaced by stronger G•C base pairs (perhaps making structure more stable)

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Bulges and Stems

Note movement of P4-P5 bulge when C is included, also note the unpaired UIf the U is removed, activity decreases 64-foldIf the entire P4-P5 stem is mutated into a continuous helix (remove bulge and U), result is no activity

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Bulges and Stems

• To pick this apart, they started with R3-derived sequence and added in only the stem-stabilizing mutations (squares), little change in activity• When they started with R3 sequence

and added in only the bulge-remodeling mutations (circles), result was much less activity• Combining only 2 stem-stabilizing

mutations with the bulge-remodeling mutations resulted in more activity than starting R3 sequence Changing length and/or sequence of the P3

and P5 stems had no effect on activity, implies the loops at the ends are unimportant

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Break It Up

• If the loops at the ends of P3 and P5 are dispensable, then breaking the P3 loop (and extending the P3 stem 4 base pairs) results in a catalytic system consisting of the regular 12-mer substrate, an 18-mer “substrate” (really the 5’ end of R3C), and a 58-mer catalytic piece• For this system: kcat=0.2 min-1,

Km18-mer=0.1 mM (when 12-mer is

saturating)

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Break It Up

• This “fragmented” system will provide ideas for subsequent papers in which they attempt to develop a self-replicating ribozyme• Note that in this context it is not a self-replicating RNA polymerase,

but R3C will provide the basis for self-replicating RNA systems• Putting the fragmented R3C into the context of prebiotic reactions,

fragmented R3C will assemble the 12-mer substrate and the 18-mer substrate into a 30-mer product• If the catalytic core emerges from the prebiotic soup by chance, then

12-mer and 18-mer that result from chance will be converted to product


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