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Transcript
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