I. Ron: - Seminar Minutes of 10-21-74 A.Discussion of the survival of linkage of the markers‘in Bacillus globizii DNA after Ry endonuclease digestion. 1. 2. 4. Implications of a.-f.; The recipient was a hybrid strain between B, globizgii and B.s. The donor was B. globigii DNA. Using the hybrid strain 62 as recipient, His and Tyro cotransfer were tested after Ry digestion. a. It was only 19% of normal (low), if Tyr was the first selection. and His, the second by replica plating. b. If His, was the primary selection and Tyr the second, then cotransfer was about 807%. There was thought to be a cut near Hisy somewhere. A hybrid strain 38-1 that is tryp tyr has been prepared, for use as recipient in transformation. The same competent celis were used for all these experiments. el a. After transformation, selecting for Tryp first and then for cotransfer of Tryp-Tyro 3 67% cotransfer. b. Selecting for Tyro first and then for cotransfer of Tryp-Tyro -) 65% cotransfer. This is the same result as seen in 8.s. coetransfer? Cotransfers were done by replica plating and by use of plates lacking both amino acids; results were the same. ec. After R; digestion of the DNA used in transformation, selection for Tryp first and then for cotransfer of Tryp-Tyro ———% the cotransfer was lowered to 23%, d. After Ry digestion of the DNA, selection for Tyro first and then for cotransfer of Tryp-Tyro —=— the cotransfer showed only 18% Linkage. Linkage of Tryp and Tyro markers is preserved but is reduced by Ry digestion to about 1/3 its previous value. e. Using strain 6-2 as the recipient, selection for His_first and then for cotransfer of His-Tyro ——» a high co-transfer of 75-80% before Rr digestion, & the same value after Rr digestion of the DNA. f. Again using 6-2 as a recipient, selectinn for Tyro first and then for cotransfer of His-Tyro —™» a cotransfer of 887, before Ry digestion, but a cotransfer of only 18% after Rr digestion of the DNA. of His-Tyro a. Some wkerl polarityg(there is a difference in co-transfer depending upon which marker is first selected) effect possibly, though this should not be so, since TrypsTyr is symmetrical in crossing behavior (co-transfer behavior is the same regardless which marker ‘5 selected first). b. Do these results represent a disparity in the recombinant types or, most likely, a disparity in the way in which the selection was done? ¢. One is probably running into artifacts during the transformation and selection procedures. d. Something is strange about the His marker.After Ry digestion, polarity is seen in the His coetransfer, (a difference in co-transfer depending upon which marker is first selected). The His-Tyro # Tyro-His co-transfer frequency. His transformants are more than Tyr transformants, using Ry treated DNA not separated by electrophoresis. 10-21-74, p.2 5. Refer to the relative frequencies of transformation by the three modes of selection: a. Selection for one marker -+ (plate lacks one amino acid) b. Selection for other marker +- (plate lacks other amino acid) e¢. Selection for both markers ++ (plate lacks both amino acids) selection for 2nd marker -- “+ +- ss #4#44_both markers transformed ection for lst marker % OF co-transfer = ++ = ineidence of co-transfer 7 (+-) + (++) incid. of single transfer + of cotransf % co-transfer = ++ = =«sincidence of co-transfer : (-+) » ( ++) incid. of single transfer + of cotransfe Another value to measure is ++ * (+-) + (++) + (++) This has been done for Tryp-Tyr and no difference was obtained, i.e., the frequency of Tryp-Tyr double transformants was the same on +-, -+, and ++ plates. This has not yet been done for His-Tyr. Polyphosphate has an augmenting effect on transformation. There is a several-fold increase in transformation, comparable to the helper effect with increase in DNA concentration (high: DNA conc. . increase in co-transfer => This may be a way to increase the competence of cells. . Continuation of the search for markers in B. subtilis DNA that surt#ive Rr endonuclease treatment (i.e., that are resistant to Ry) 1. 2. 3. 4. 3. 5 markers have survived Ry treatment: marker % survival tryp-tyro 5 his 10 uracil 7 methionine 60 adenine 8 lysine and glycine are completely destroyed after Ry treatment. Other markers gave strange effects. 5 markers are currently being tested. Aro, Cannot show its presence by transformation after R, treatment, but can do so before such digestion; it is not destroyed but its activity is greatly decreased after R; treatmant. Bands are 2-3 mm wide, but the gel is being cut in 1 mm slices. a. The Tyro marker location (determined by biologicallactivity of transformation) is at mm 23 from the origin. b. His marker is at mm 39 but has transformants throughtout the spectrum = background. c. Aro, is not found; there is no transformant for phenylalanine after two days incubation of plates. d. Methionine is found at mm 32 —— > 99+ 7% enrichment. C. One l. 10-21-74, p.3 7. Using whole DNA, get the enrichment ratio. After R and before gel separation calculate, for example, His and Meth* Meth? His* After gel separation, calculate the same frequencies in the respective peaks ( in the His peak His*, should now be much higher and in the Meth peak, Meth* should Mee be much higher). His? ' The ratio of the ratios after separation and before separation = enrichment = the ratio after gel electrophoresis separation the ratio before gel electrophoresis separation 8. For Tryp. there is almost 100% separation of its biological activity. In Tryp-Tyr cotransfer there were 50 colonies/.05 ml from the peak. His} has a 99% separation. There were 1,200 colonies/.05 ml from the peak of the His, band gel slice (count of colonies = absolute # of transformants). Less than 1% His transformant colonies were elsewhere. Background of His is about 2-10/.05 ml. Rr treatment acts at specific sites. Shearing of DNA cuts randomly, Since it is the specific sites that are now being observed, the R may be destroying more of the activity of these specific sites than would sheering. rough test for a rough degree of purification of markers. Adenine survives to a lower frequency than the other markers; it is not present in the first 2.5 em. The fractionation is good. fragment Look for size mutants - an increase in size of DNAacarrying a specific marker, caused by a mutation at an Ry site resulting in that cut not being cut. Are there any colonies at the big end of the spectrum where not expected , or any shift; are there any colonies seen that could be the results of Ry site mutations (a change in a nucleotide so that the enzyme does not recognize that particular site). Mutation may account for His, marker behavior; the His) may be in a hot spot. Twice consecutively a major His peak and a minor His peak were found; the minor peak (less than 1 % of total) was found a few m from tryp-tyr peak. Extract the DNA from the expected mutants & pool the DNA. Look for the location of his), by locating the peak in gel electrophoresis. See if the His); peak can be moved over to where the minor peak is, due to a mutation in the R, site near His;. Pick one colony —— subculture (it may be a mutant). Pool 39 colonies ——¥» subculture (this provides a chance of finding a mutant). His, has almost no reversion. Methionine seems very stable; it doesn't appear to have any revertants (the recipient negative marker stays negative and doesn't revert to positive). 10. ll. 12, 13. 10-21-74, p.4 Rationale for looking for mutants: It's a mapping tool; it corroborates the general picture of what is going on & in so doing, vertfies the model. Any structural mutants that can be used to map the DNA have many uses, Pool the R, treated and gel-separated DNA over several intervals. For example, pool big peak His), then pool minor peak His; (which could be a major peak if a size mutant were found). What fraetion of His (a short piece) will hybridize with the big band His? large pieces: ----.+--.----.-... Y mt tspocceeseen- -H a new element that appears only in the : mutant H ( Histidine) A hypothetical example: A mutant is found that has histidine activity in a heavier band than is normally found on gels. Its DNA could then be isolated, Ry treated, & gel separated to get "pure" big mutant His molecules (H’). One could then hybridize the mutant "big His " with the nonmutant "small His". Only a portion of the "big His " will hybridize with the "small His" since this is a specific type of aggregation. When observed under the electron microscope the proportion of double stranded to Single stranded DNA in the hybrid molecules can provide a measure of the His length between the end of the doublesstranded section & the next Rr gut - the end of the single-strand H , and also a measure of the nfs length. If a size mutant were found and its DNA isolated, Ry treated & separated small pieces e on gels, there would be a change in the height of the H band; also, the band for the small piece H would be missing or smaller. The DNA missing in the small piece should be in the big band. The absolute amount of DNA in the hybrid H can be measured. How many fragments form one band? Results may be confused by the ectopic problem. Currently a way to measure the purity of the piece is being sought. Cut the gel into 1. mm pieces. Bake counts of each band. How many counts per band, & how many per piece? This is a crude measure of purity. Band % Counts of H2-DNA Tryp-Tyr 3.7 Methionine 4.2 His 4.4 Uracil 3 It is estimated that there are about 5 different fragments of DNA per band. Most of the DNA is in 3 cm; the rest is :pread over 5 ecm. Try to check purity using Specific Activity (biological activity/physical activity = BubTPPSSCTOAn ts ATEACOUBESATAR ; ey, -Rebanding AS YST2y*SeoPhaBeRRTEnSE uy, Tryp is still the heaviest band - the first seen with biological activity; it is probabby not yet pure. What is the E.M. length? 14. 15. 16. 17. 18. 19, 20. 21. 10-21-74, p.5 Do denaturation mapping, an analytical technique, to see how multi-modal the DNA of the band is (i-e., how many different fragments are in one band). Start wihh the whole DNA after R treatment. Then use the electrophoresis fractionated DNA for comparison. Use the differential melting point (Gan's procedure). Heat, cool rapidly (that which is denatured remains so & that which is not remains so), pass through a hydroxyapatite column to separate the fragments with different degrees of denaturation (A-T rich fragments denature at a different temperature than G-C rich fragments). Clean renaturability of the material after Ry corroborates its homogeneity in a given band. A good way to get rid of nécks is to melt and renature. That DNA which comes back to the original size is the zood DNA. Use CsCl fractionation & calculate the enrichment ratidg; a given CsCl band achieves a given enrichment. Make the gel electrophoresis a preparative method for use on a large scale, The methionine marker (smaller than the Tryp marker) survives well, It is in the middle and has the least edge effect (those markers at the edge are not as efficiently transformed as those more centrally Situated on the chromosone). Check on secondary Ry sites. Size and closeness to the edge may be criteria for survivability. There may be something special about the ends of R treated DNA so that fragments from Ry treated DNA may slide back nto the chromosone differently. To destroy ends - treat with exonuclease I. Does the ectopic insertion depend on having the kind of ends obtained after Ry treatment? Can the resolution be increased any further: To get still better Separation consider the best conditions for doing the electrophoresis, Are these optimized? a. Tonic strength of buffer (0.1M Tris + borate —% pH 8.7 is used; alteration of pH may cause gel shrinkage) . b. % agarose gel (0.7 ~ 0.73 is being used; 1.6 - no bands, 1.3 - begin to see bands, 1 and 0.7 - good bands, below 0.7 the gels shrink) ¢. Voitage (20 volts is used. A lower voltage gives less heating effects. Slower movement may provide better separation.) d. Temperature (room temperature js used. Working With very thin gels, the electrophoresis can be done at & in the cold room. There would be a better temperature cnntrol in the cold & this might give better reproducibility.) €- Solutes - try adding other solutes to influence the texturdof the gel and of the DNA. 22, 10-21-74, p.6 Set up conditions for electrophoresis such that the bands stop and whatever is smaller starts traveling faster. Use step-gels; at different % agaroses different DNA fragments are sorted out, & different % agaroses can be layered one on top of the other. D. For Publication: 1. Hone in ont level of fractionation rebanding leave the mutation work for later 2 Discuss: resolution measurement of resolution degree of fractionation specific activity number of markers 3. Include information about: &. sucrose gradient (1) to show effect of Ry treatment (2) to calculate mass b. compare bands obtained from €1) DNA Ry treated (bands obtained & activity increases) (2) sheared DNA (no bands obtained & activity decreases) €. computer data The computer simulation data agrees with the experimental data. At first it wasn't random because of the random # generator used, but now another random # generator is employed & it is more random. d. table of markers tested & the survival rate of the different markers III. Hela: Crossing Experiments 1. Seven different strains were tried. 2. Six showed no growth of colonies, + 3. Bes. SB 863 S* Tyr™ Aro” Tryp” Lys, gave no growth alone on appropriate plates, nor did SD 8 Tyr+-Arog Tryp Lys- Cys” alone. (SD8 didn't revert at all; SD 1 reverts alot.) 4. When $3863 and SD8 were cross streaked on a plate, colonies were obtained When colonies occur, either $8853 receives the Tyr Azoy Tryp from S98, or SD8 receives the Lys from $B863, in recombination, 5. SD8 is the result of an R, transformation: it may have a plasmid. 6. Repeat the experiment as it was done (grow the bacteria together on the plate for 48 hours, cross-streaked). See if the result is repeatable & mot a matter of contamination, but of crossing (recombination). 7. If repeatable, then mix the two kinds of bacteria together just before plating the mixture. 8. Isolate the colonies & find out their nutritional requirements. 9. Repeat with Streptomycin on the plate to find out which bac. is donating the markers. 10. Repeat with and without Cys on the plate. 11. Is DNA transfer involved? Add DNAase to the mixture. B.C.