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Stages of Fetal and Adult Mammary Development Hennighausen and Robinson, Dev. Cell 1, 467-475. 2001 MCB 140 11-6-06 1 MCB 140 11-6-06 2 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). MCB 140 11-6-06 3 Hormone-free medium MCB 140 11-6-06 4 + estrogen (E2) 43 thousand deaths from breast cancer annually in the US. MCB 140 11-6-06 5 estradiol Tamoxifen (SERM) MCB 140 11-6-06 6 THE BIOLOGICAL PROBLEM Small-molecule regulation of gene expression: 1. Gene activity changes during ontogeny (not all genes are on/off at all times). 2. 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. MCB 140 11-6-06 7 MCB 140 11-6-06 8 frog thyroid hormone tadpole ecdysone pupa Breast epithelium fly estrogen Breast cancer MCB 140 11-6-06 9 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 MCB 140 11-6-06 10 An outline of the next 3 lectures ? environment regulation of DNA function Pre-1943: “just so stories” 1940s-1960s: bacteria and phage MODEL 1960s-1970s: yeast 1970s-1996: mammals 1996-curr.: the complexity of the truth MCB 140 11-6-06 11 MCB 140 11-6-06 12 The metabolism of lactose in E. coli by b-galactosidase: 16-3 MCB 140 11-6-06 13 Simple biological fact Lactose induces synthesis of the enzyme that breaks down lactose MCB 140 11-6-06 14 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? MCB 140 11-6-06 15 Enzymatic adaptation (<1950) Fact: addition of lactose to the growth medium causes a rapid and dramatic increase in bgalactosidase activity inside E. coli. Explanation (don’t laugh): 1. Protein structure is “dynamic.” 2. When an inducer (lactose) shows up, an existing protein is reorganized from small bits to become active b-galactosidase. MCB 140 11-6-06 16 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. MCB 140 11-6-06 17 lactose + inducer + b-gal ribosomes off b-gal ribosomes on MCB 140 11-6-06 18 The woefully erroneous “ribosome inducer” model of b-gal regulation by lactose was terminated by one single experiment. This experiment was purely genetic. MCB 140 11-6-06 19 PaJaMo A. Pardee F. Jacob J. Monod MCB 140 11-6-06 20 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– MCB 140 11-6-06 21 François Jacob and Jacques Monod The Nobel Prize in Physiology or Medicine 1965 MCB 140 11-6-06 22 Is inducibility dominant or recessive? Mate two bacteria, one inducible, one noninducible (constitutive), and examine the phenotype of their child. MCB 140 11-6-06 23 14-14 MCB 140 11-6-06 24 lactose + inducer + b-gal ribosomes off b-gal ribosomes on MCB 140 11-6-06 25 The expectation Hfr (male) + + z i inducible = female zi constitutive +/+/z i Simplest expectation: if the i gene 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. MCB 140 11-6-06 26 The actual data oooh, mama! MCB 140 11-6-06 27 +/+/z i MCB 140 11-6-06 28 Point #1. Inducibility is dominant. The simplest explanation to this is that the i- allele is a lossof-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?!! MCB 140 11-6-06 29 THE CONTROL EXPERIMENT = reverse the direction of the cross MCB 140 11-6-06 30 Textbook p. 555-556 Fig. 16.6 MCB 140 11-6-06 31 ♂ ♀ +/+/=z i + + z i z i + + z i zi MCB 140 11-6-06 32 ♀ ♂ + + z i zi + + z i z i In the absence of inducer: = transient burst of bgal synthesis, then the synthesis shuts off. = NO TRANSIENT BURST OF SYNTHESIS; b-gal stays off! MCB 140 11-6-06 33 Bill Watterson MCB 140 11-6-06 34 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. MCB 140 11-6-06 35 z+ i+ z- iz- i- z+ i+ ← only in an i- cytoplasm!!! MCB 140 11-6-06 36 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. MCB 140 11-6-06 37 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. MCB 140 11-6-06 38 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. MCB 140 11-6-06 39 Orgo cis-2-butene trans-2-butene MCB 140 11-6-06 40 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? MCB 140 11-6-06 41 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) lz(g) lz(BS) wt MCB 140 11-6-06 42 Cis: Trans: wt wt wt lz(g) lz(BS) lz(g) lz(BS) wt 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!!! MCB 140 11-6-06 43 Classical classical genetics Construct a strain to test a hypothesis. Morgan, Muller, etc. MCB 140 11-6-06 44 Partial diploid (Fig. 14.18) MCB 140 11-6-06 45 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. Cis: Trans: z+ i+ z+ i- z- i- z- i+ That was shown to be true. Both these strains are inducible. MCB 140 11-6-06 46 stimulus + + Regulation of genes occurs via the interaction of transacting factors (proteins) with cis-acting sequences near the genes themselves. MCB 140 11-6-06 47 François Jacob: “If it’s true for E. coli, it must be true for E. lephant.” MCB 140 11-6-06 48