1. MCB 140 – 11/8/2006 1 Question What is the chemical nature of the repressor?

2. MCB 140 – 11/8/2006 2 If you think about it … The repressor has to directly bind to a specific DNA sequence (the operator) in the E. coli genome. From a reverse-engineering perspective, the simplest way to design something that interacts with DNA sequence- specifically is to use a nucleic acid that is complementary to that DNA!

3. MCB 140 – 11/8/2006 3 In 1959, a biochemical experiment was done “proving” that the lac repressor is not a protein, and is most likely an RNA molecule In 1965, a genetic experiment was done proving that biochemical experiment entirely wrong

4. MCB 140 – 11/8/2006 4 1959: (When the z+ gene and the i+ gene arrive in an i- cytoplasm (no repressor), the z+ gene becomes active, and stays active for about 2 hours. At that point, the repressor is made, and shuts the z+ gene off. This offers an elegant opportunity to determine the biochemical nature of the repressor: add an inhibitor of protein synthesis, and let the cells spend their first two hours post-mating in that good stuff. A.B. Pardee and L.S. Prestidge BBA 36: 545. If the repressor is a protein, then inhibiting its synthesis following transfer of z+ into an i- cytoplasm should allow for constitutive synthesis of  -galactosidase!

5. MCB 140 – 11/8/2006 5 The result Inhibiting protein synthesis does not destroy the “plateau” effect:  -gal still goes off! Conclusion: “… the repressor probably is not a protein, since it was made when [protein] synthesis was inhibited. Ribonucleic acid would seem a likely candidate for the role of the repressor.” A.B. Pardee and L.S. Prestidge BBA 36: 545.

6. MCB 140 – 11/8/2006 6 A distinction between genetics and biochemistry The validity of Pardee’s conclusion is unequivocally dependent on a biochemical phenomenon (total inhibition of protein synthesis by 5-methyltryptophan) It turns out that not all protein synthesis in E. coli is inhibited by 5-me-T… A genetic experiment (=crosses between strains of different genotype) was performed to prove Pardee wrong

7. MCB 140 – 11/8/2006 7 8.28

8. MCB 140 – 11/8/2006 8 Let’s mate 1.Make an i- strain of E. coli (constitutive) 2.Make several such strains (=different kinds of mutations) 3.Mate those i- E. coli to other E. coli carrying nonsense suppressor tRNA genes 4.See what happens.

9. MCB 140 – 11/8/2006 9 “Suppression of and complementation among mutants of the regulatory gene of the lactose operon of Escherichia coli.” Bourgeois S, Cohn M, Orgel LE. JMB 14: 300 (1965) 1.Cross an i- strain with a tRNA sup strain 2.Measure activity of  -gal (+ or - lactose in the medium).

10. MCB 140 – 11/8/2006 10 In their own words “We know … that the suppressor strains used here act at the level of translation by allowing the chain-terminating codon to be read as an amino acid. We have shown suppression of i- mutations by these suppressors and conclude therefore that the i gene of the lactose operon codes for a protein.”

11. MCB 140 – 11/8/2006 11 Regulation of genes occurs via the interaction of trans- acting factors (proteins) with cis-acting sequences near the genes themselves. + stimulus +

12. MCB 140 – 11/8/2006 12 François Jacob: “If it’s true for E. coli, it must be true for E. lephant.”

13. MCB 140 – 11/8/2006 13 Budding (brewer’s and baker’s) yeast, Saccharomyces cerevisiae

14. MCB 140 – 11/8/2006 14 Yeast ferment all available sugar even in the presence of oxygen Mammals: sugarpyruvate CO 2 +H 2 O S.c.: sugarpyruvate C 2 H 5 OH+CO 2 glycolysis respiration TCA+O.P. fermentation

15. MCB 140 – 11/8/2006 15 A major evolutionary conservation of cellular response to sugar in the medium In Saccharomyces cerevisiae (budding yeast = a fungus = a eukaryote): 1.Enzymes for utilization of the sugar galactose are induced ~1000-fold by galactose. 2.These enzymes are also severely repressed by glucose in the medium. 3.Thus, for these genes to be induced fully, the medium must contain galactose and no glucose. Just like E. coli.

16. MCB 140 – 11/8/2006 16 Analogy and homology as tools in genetic investigation Animal Mandibular Arch (ventral) Mandibular Arch (dorsal) Hyoid Arch (dorsal) SharkMeckel's cartilage Palatoquadrate cartilage Hyomandibular cartiliage AmphibianArticular (bone)Quadrate (bone)Stapes MammalMalleusIncusStapes

17. MCB 140 – 11/8/2006 17

18. MCB 140 – 11/8/2006 18 First experiment (Howard Douglas, 1963) Goal: make i- yeast – that is, yeast that synthesize galactose-metabolizing enzymes constitutively. 1.Take mutant strain (gal3) that has a markedly delayed response to galactose and does not grow on it very well at all. 2.Grow on galactose – see what grows. Douglas HC, Penroy G (1963) A gene controlling inducibility of the galactose pathway enzymes in Saccharomyces. Biochim Biophys Acta 68: 155.

19. MCB 140 – 11/8/2006 19 What grew 1.One would expect revertants of the gal3 mutation. Those didn’t show up. 2.What did show up was true i- cells – yeast that synthesized the GAL enzymes constitutively (that’s why they grew). 3.They made i+ / i- cells – they were inducible.

20. MCB 140 – 11/8/2006 20 Conclusion (ta-daaa!) “The inducibility in gene in yeast fits the description proposed by Monod and Jacob for regulator genes. … By analogy with the lactose system in E. coli, the galactose- inducible state in yeast corresponds to the production of a repressor, due to i+, while the constitutive state, due to i-, represents a failure to repression.”

21. MCB 140 – 11/8/2006 21 Nothin’ but net Indeed, the i+ gene is now called GAL80. Its product, Gal80p, is a repressor of GAL genes.

22. MCB 140 – 11/8/2006 22 But “We might have expected, by further analogy with the systems in E. coli, to find mutations linked to the structural galactose genes and to be expressed as cis dominant constitutives, but these have not yet been detected in our material.” In other words, they wanted to find “operator” mutations, and didn’t find them.

23. MCB 140 – 11/8/2006 23 Screen for gal cells H. Douglas, D. Hawthorne (1964): 1. Take wt haploid yeast. 2. Zap them with UV light. 3. Replica-plate to find those that are gal. 4. 1:1000 are mutant. Douglas HC, Hawthorne D (1964) Enzymatic expression and genetic linkage of genes controlling lactose utilization in Saccharomyces. Genetics 49: 837.

24. MCB 140 – 11/8/2006 24 7.5a

25. MCB 140 – 11/8/2006 25 A.8

26. MCB 140 – 11/8/2006 26 Initial analysis of mutants 1.Cross mutant to wt. 2.Sporulate. 3.Dissect the tetrad (ascus) 4.Confirm that 2 spores are GAL and 2 are gal.

27. MCB 140 – 11/8/2006 27 What came out Large number of mutants in different genes: GAL1six mutants GAL2five mutants GAL3six mutants GAL4two mutants GAL5nine mutants GAL7four mutants GAL10one mutant

28. MCB 140 – 11/8/2006 28 “When genes are linked, PDs exceed NPDs” 5.18

29. MCB 140 – 11/8/2006 29 Linkage analysis

30. MCB 140 – 11/8/2006 30 Conclusions The following genes are very closely linked: GAL1, GAL7, and GAL10 The GAL4 gene is not linked to those three

31. MCB 140 – 11/8/2006 31 What does what

32. MCB 140 – 11/8/2006 32 The fun part of this complete breakfast The gal4 mutation is unlinked to the enzyme genes, and yet the GAL4 gene product is required for their synthesis.

33. MCB 140 – 11/8/2006 33 Ta-daaa!! “The closely linked genes GAL1, GAL7, and GAL10 [code for] the galactose pathway enzymes, galactokinase, transferase, and epimerase. The mutation gal4 blocks the synthesis of these enzymes, but unlike the phenotypically similar mutation O o in E. coli is complementable and is not linked to the genes whose expression it controls.”

34. MCB 140 – 11/8/2006 34 Summary ctd. “GAL4 apparently produces a cytoplasmic product required for the expression of GAL1, GAL7, and GAL10. The role of the repressor might be to prevent the synthesis or the activity of this product.”

35. MCB 140 – 11/8/2006 35 But still “In considering a problem of an operator locus in yeast, it is of interest that [we] have been unable to find a constitutive mutant analogous to an O C type…” = cis dominant mutation that leads to constitutive galactose enzyme synthesis “find mutation in operator that does not bind repressor…”

36. MCB 140 – 11/8/2006 36 1966: got it! Complex mutagenesis screen in a diploid strain for constitutive mutants – exactly analogous to such screen in E. coli, where a diploid was used to prevent isolation of mutants in the repressor gene itself. Got one dominant mutation that was very tightly linked to the GAL4 gene. Douglas HC, Hawthorne D (1966) Regulation of genes controlling synthesis of the galactose pathway enzymes in yeast. Genetics 54: 911.

37. MCB 140 – 11/8/2006 37 - galactose GAL7 GAL10GAL1 GAL4 Gal80p

38. MCB 140 – 11/8/2006 38 + galactose GAL7 GAL10GAL1 GAL4 Gal4p Gal80p

39. MCB 140 – 11/8/2006 39 - galactose in gal81 strain GAL7 GAL10GAL1 GAL4 Gal4p Gal80p gal81

40. MCB 140 – 11/8/2006 40 wrong (correct between 1966-1978)

41. MCB 140 – 11/8/2006 41 Gal4p is synthesized at all times, irrespective of the presence of galactose Yasuji Oshima (Osaka)

42. MCB 140 – 11/8/2006 42  Tokyo Kokuritsu Hakubutsukan (Ueno)

43. MCB 140 – 11/8/2006 43 sumi-e

44. MCB 140 – 11/8/2006 44 gal4-4 – a ts allele of GAL4 GAL at 25  and gal at 35  1.Take wt and gal4-4 cells. 2.Grow at 35  in glucose 3.Move them to 25  4.Immediately add galactose 5.Measure galactokinase. 6.Compare wt and gal4-4. Matsumoto et al. J. Bacteriol. 134: 446 (1978). gal4-4 (35  25  ) gal4-4 (25  ) wild-type

45. MCB 140 – 11/8/2006 45 Epistasis (1982-84) A gal80 cell synthesizes galactose utilization enzymes constitutively. A gal4 cell does not synthesize those enzymes under any condition. A double-mutant gal80 gal4 cell has the same phenotype as a gal4 cell – no enzyme synthesis. GAL4 is epistatic to GAL80. Gal4p acts downstream of Gal80p. A “superrepressor” allelic form of GAL80 (GAL80 S ) does not respond to galactose. Overexpression of Gal4p can overcome its effect! That is, GAL4 GAL80 S is uninducible, but GAL4 high copy GAL80S is inducible. Johnston and Hopper PNAS 79: 6971 (1982). Torchia et al. MCB 4: 1521 (1984).

46. MCB 140 – 11/8/2006 46 SUPPRESSION: A given mutation (A) has a discrete phenotype that is not normal, i.e. not wild-type. The presence of the second mutation (B, the suppressor mutation) causes the AB double mutant to display a phenotype that is normal or near-normal. Thus, a suppressor mutation rescues or restores or repairs, in whole or in part, the defect caused by the first mutation. Examples: A nonsense mutation in a gene can be suppressed by a mutation in a tRNA gene in which the anticodon has been mutated to read the nonsense codon (informational suppressor). A temperature-sensitive mutation that destabilizes a protein can be corrected by a compensating mutation in another gene product that binds to and acts in a complex with the first gene product (extragenic suppressor). Overproduction of a transcription factor can overcome the need for a protein kinase in the pathway that is normally needed to activate that transcription factor (dosage suppressor). Loss of a repressor of a gene can compensate for absence of the positive regulator normally needed to turn on that gene (bypass suppressor). EPISTASIS: Two mutations (A and B) each have discrete phenotypes that are readily distinguishable from each other, and neither are normal (both are non-wild-type). If the phenotype of the AB double mutant resembles that of one of the two single mutants and not the other, that mutation is said to be epistatic over the other. Thus, if the AB double mutant looks like the A mutant alone, mutation A is said to be epistatic over mutation B; conversely, if the AB double mutant looks like the B mutant alone, mutation B is said to be epistatic over mutation A. Examples: In yeast, the ade3 mutation blocks purine biosynthesis early in the pathway, whereas the ade2 mutation blocks the pathway later. Both require Ade in the medium to grow. The ade2 mutant has a white colony color, but the ade2 mutant has a pink-to-red colony color (because the metabolic intermediate that accumulates in the ade2 mutant polymerizes to form a pigment). The ade2 ade3 double mutant still requires Ade to grow, but displays a white colony color. Thus, the ade3 mutation is said to be epistatic to the ade2 mutation (which make sense because ade3 blocks production of the intermediate that would otherwise accumulate in the ade2 mutant). In flies, apterous (ap) mutations block wing formation, whereas curled wing (cw) causes a dramatic change in wing morphology. Neither have normal wings. A ap cw double mutant has a wingless phenotype. Thus, ap is epistatic to cw. Thus, in the pathway for construction of a wing, ap functions before cw. Prof. Jeremy Thorner, UCB

47. MCB 140 – 11/8/2006 47 Also “described” on pp. 592-593, under the amusing heading “The yeast GAL system is another complex regulatory mechanism.”

48. MCB 140 – 11/8/2006 48 How genes respond to environmental stimuli

49. MCB 140 – 11/8/2006 49 More from Dr. Jacob “I have always been convinced that the same principles operating in bacteria are also operating in higher organisms with added complexity. The question therefore is to understand what kind of complexity is involved and how it is generated.”

50. MCB 140 – 11/8/2006 50 Next time The answer to Dr. Jacob’s question.

51. MCB 140 – 11/8/2006 51 Further reading 1.T. Brock The Emergence of Bacterial Genetics (CSHL Press 1990) 2.M. Ptashne, A. Gann Genes and Signals (ibid) 3.H. Judson The eighth day of creation (Simon and Schuster 1979) 4.S. Brenner My Life in Science (BMC Press 2001)