Hennighausen and Robinson, Dev. Cell 1 , 467-475. 2001

1. Hennighausen and Robinson, Dev. Cell 1, 467-475. 2001

3. Note: this diagram is not entirely correct (in an entirelly incorrect sort of way), because the estrogen receptor (ER), a class I nuclear hormone receptor, is not typically bound to target genes in the absence of hormone – such behavior is characteristic of class II nuclear hormone receptors, such as the thyroid hormone receptor (TR) and the retinoic acid receptor (RAR).

4. Hormone-free medium

5. + estrogen (E2) 43 thousand deaths from breast cancer annually in the US.

6. estradiol Tamoxifen (SERM)

7. THE BIOLOGICAL PROBLEM Small-molecule regulation of gene expression: • Gene activity changes during ontogeny (not all genes are on/off at all times). • In the genomes of all living forms studied, there are genes whose transcription responds to changes in the titer of some molecule provided by the environment.

9. frog thyroid hormone tadpole ecdysone pupa fly estrogen Breast epithelium Breast cancer

10. A fact “Nothing in biology makes sense except in light of evolution.” Theodosius Dobzhansky  the established scientific fact that all life forms on Earth arose via Darwinian natural selection provides an explanatory device and a research tool of unrivalled power

11. An outline of the next 3 lectures ? environment  regulation of DNA function Pre-1943: “just so stories” 1940s-1960s: bacteria and phage  1960s-1970s: yeast  1970s-1996: mammals 1996-curr.: the complexity of the truth MODEL

13. The metabolism of lactose in E. coli by b-galactosidase: 16-3

14. Simple biological fact Lactose induces synthesis of the enzyme that breaks down lactose

15. txn: Let’s use a bit of lactose since it’s available, but focus on metabolizing glucose because, biochemically, it’s the best. Well, there’s no glucose, so the only sugar we have is lactose, and thus we have to metabolize it as rapidly as we can. There’s no lactose. Why make the enzyme and waste rNTPs and amino acids?

16. Enzymatic adaptation (<1950) Fact: addition of lactose to the growth medium causes a rapid and dramatic increase in b-galactosidase activity inside E. coli. Explanation (don’t laugh): • Protein structure is “dynamic.” • When an inducer (lactose) shows up, an existing protein is reorganized from small bits to become active b-galactosidase.

17. The inducer model of lacZ control <1958: In the cell, there are ribosomes that could be making b-galactosidase, but these ribosomes are inactive (“off”). When lactose shows up, it acts via some inducer to turn “on” those ribosomes, and then b-galactosidase is made.

18. lactose inducer b-gal ribosomes off b-gal ribosomes on + +

19. The woefully erroneous “ribosome inducer” model of b-gal regulation by lactose was terminated by one single experiment. This experiment was purely genetic.

20. PaJaMo A. Pardee F. Jacob J. Monod

21. Two types of mutations 1. Mutagenize bacteria, grow on lactose only. = cannot metabolize lactose: z +z – 2. Try to make mutant E. coli that grow on neolactose (E. coli do not normally metabolize neolactose) = cells that make b-galactosidase constitutively, i.e., that are non-inducible i +i –

22. François Jacob and Jacques Monod The Nobel Prize in Physiology or Medicine 1965

23. Is inducibility dominant or recessive? Mate two bacteria, one inducible, one non-inducible (constitutive), and examine the phenotype of their child.

24. 14-14

25. lactose inducer b-gal ribosomes off b-gal ribosomes on + +

26. The expectation Hfr (male) female z+ i+ z- i- inducible constitutive = z+/-i+/- Simplest expectation: if the igene codes for an inducer, then the constitutive synthesis seen in i-cells must be due to this inducer being somehow always active. This means that constitutive synthesis must be dominant – as soon as the b-gal gene gets into the cell with the “always active inducer,” it should go on and stay on.

27. The actual data oooh, mama!

28. z+/-i+/-

29. Point #1. Inducibility is dominant. The simplest explanation to this is that the i- allele is a loss-of-function allele (a hypomoph or a null). Why on Earth, then, is a cell that is genotypically i-, which means it has a loss-of-function mutation in the “inducer” gene have a “constutitively on” phenotype?! Its inducer is broken! It should have a “constitutively off” phenotype!! Point #2. Why on Earth does the dominance of inducibility take 2 hours to show up?!!

30. THE CONTROL EXPERIMENT = reverse the direction of the cross

31. Textbook p. 555-556 Fig. 16.6

32. z+ i+ z- i- z- i-  z+ i+ ♀ ♂ = z+/-i+/-

33. z+ i+ z- i- z- i-  z+ i+ ♀ ♂ In the absence of inducer: = transient burst of b-gal synthesis, then the synthesis shuts off. = NO TRANSIENT BURST OF SYNTHESIS; b-gal stays off!

34.  Bill Watterson

35. Oh, and one additional fact During bacterial mating, the only thing transferred from one cell to another is the DNA. The cytoplasm is not transferred.

36. z+ i+ z- i- z- i-  z+ i+ ← only in an i- cytoplasm!!!

37. What was thought • That inducibility was due to an inducer – i.e., that induction of lactose somehow affects a molecule that activates the gene • So what PJM showed – because z+ when transferred into an i- cytoplasm IMMEDIATELY BECAME INDUCIBLE, was that enzyme synthesis is the default state – it requires nothing; i.e., that i is a repressor.

38. In their own words   “Up until now, it had seemed reasonable to postulate that constitutive mutants synthesized an endogenous inducer which was absent in inducible cells. The results described here suggest an exactly opposite hypothesis. The facts can be explained by the supposition that the i gene determines (via an enzyme intermediate) the synthesis, not of an inducer, but of a "repressor" which blocks the synthesis of b-galactosidase, and the exogenous inducer displaces this repressor and restores enzyme synthesis. With the i- allele, present in an inactive form in the constitutives, the repressor is not formed, and b-galactosidase is synthesized, the exogenous inducer therefore being without effect. This hypothesis, although at first surprising, is in agreement with many other facts. It has been known for many years that the synthesis of certain constitutive enzymes is inhibited by "repressors", exogenous or endogenous.” Pardee AB, Jacob F, Monod J. (1959) Sur l’expression et le role des alleles `inductible` et `constitutif` dans la synthese de la b-galactosidase chez les zygotes d’ Escherichia coli. Comptes rendus des Academie des Sciences 246: 3125-3127.

39. Firm proof If the repressor is real, and is a soluble molecule, then i+ should be dominant over i- regardless of whether it is in cis or in trans to the b-gal gene.

40. Orgo cis-2-butene trans-2-butene

41. The cis-trans test(aka complementation test – p. 107) Edward Lewis (NP 1995) Are two different mutations that appear to affect the same trait in the SAME gene or in DIFFERENT genes?

42. The cis-trans test, 1949:lozenge (M. Greene) Two different recessive mutants, both with the same phenotype (small eyes and fused facets). Are they mutations in the same gene? Make two different fly lines and compare their phenotypes. Cis: Trans: wt wt wt lz(g) lz(BS) wt lz(BS) lz(g)

43. Cis: Trans: wt wt wt lz(g) lz(BS) wt lz(BS) lz(g) This is a control experiment. The flies will be wild-type regardless of whether BS and g are in the same gene or not. If flies are normal, then mutations are in different genes. If the phenotype is still mutant, then BS and g must be in the same gene!!!

44. Classical classical genetics Construct a strain to test a hypothesis. Morgan, Muller, etc.

45. Partial diploid (Fig. 14.18)

46. Firm proof If the repressor is real, and is a soluble molecule, then i+ should be dominant over i- regardless of whether it is in cis or in trans to the b-gal gene. That was shown to be true. Both these strains are inducible. Cis: Trans: z+ i- z+ i+ z- i+ z- i-

47. stimulus + + Regulation of genes occurs via the interaction of trans-acting factors (proteins) with cis-acting sequences near the genes themselves.

48. François Jacob: “If it’s true for E. coli, it must be true for E. lephant.”