6/28/01

Lecture #4 - Catabolite repression and tryptophan operon

Cis-acting and trans-acting

trans acting mutations are normally in proteins (diffusable products)
cis acting mutations in sequences (DNA or RNA)

2 classes of mutation in lac

constitutive mutations lead to genes that are always on
uninducible mutations lead to genes that are always off

Uninducible mutations in lac

1) Promoter mutations in the &endash;35 or &endash;10 consensus sequences

RNA polymerase can't recognize the promoter, lowering level of transcription
"down" mutation
Cis-acting, only affects genes linked to the promoter

2) lac I^S : "super" repressor, repressor is always bound to the operator, always preventing transcription.

There are two possible causes for lac I^S mutations:
1. repressor can't bind to the inducer
2. repressor can bind the inducer but can't undergo the conformation change required to get the repressor to fall off of the operator
I^S / I⁺ merodiploid --> uninducible phenotype, the super repressor mutation (I^S) is dominant to wild type. Once I^S is bound, it will always be bound so the wild-type repressor will not cause the I^S repressor to fall off the operator.
note: (I⁺ is wild type)

3) lac I^{tight binding}

repressor can bind inducer and can undergo a conformational change
repressor can't be released from operator
I^tb/ I⁺ merodiploid --> uninducible phenotype, I^tb dominant to I⁺
It is important to note that I^tb is different from I^S because the mutations in these genes affect different domains of the lac repressor protein.

Constitutive Mutations in lac

1) O^C &endash; an operator mutation which can't bind to the repressor so genes can't be turned off. O^C mutations leads to a constitutive phenotype and genes are always on.

2) Lac I^- : non-functional repressor mutation that has catastrophic effects on the protein's function

I^-/ I⁺ merodiploid --> inducible phenotype; I^- is recessive to wild type

3) Lac I^D

the repressor can't bind to DNA butŠ
the repressor is still able to form tetramers
I^D/ I⁺ merodiploid --> constitutive phenotype, I^D dominant to wild type
Wild-type copy doesn't rescue the phenotype, why not?

Lac I repressor functions as a tetramer.

If all four subunits of the tetramer are wild type, then the repressor is functional. A function repressor is produced in I⁺/ I⁺ cells.

If one I^D monomer is present in the tetramer, then the repressor is "poisoned" and the repressor tetramer does not function. This happens in I^D/ I⁺ cells.

The I^D/ I⁺ result is an example of a "dominant-negative" mutation because the activity of the wild type allele is reduced by the product of the I^D allele.

Figure 10.9

This figure maps the mutations in the lacI gene responsible for different functions
The lac I^S mutations are inducer binding mutations that can be mapped over a large area (between residues 62-300)
Mutants that can't make tetramers, also known as oligomerization mutants, are mapped towards the carboxy terminal end of the lac repressor protein between amino acids 225 and 300.
Lac I^D mutations are at the amino-terminal end of the lacI protein, defining the DNA binding domain. Lac I^D mutants have a constitutive phenotype and do not bind to DNA.
Lac I^tb mutations are also in the DNA binding domain but Lac I^tb mutants have an uninducible phenotype and bind extra-strongly to DNA.

Figure 10.10

This figure describes the biochemical experiments used to identify the interaction of the lac repressor with the lac operator
There is an axis of symmetry at +11 bp and the sequence is palindromic (meaning that it contains inverted repeats)
The palindromic lac operator consists of 2 half sites. Each half site:
a.) contacts 1 monomer of the lac I protein

b.) 2 subunits of lac I repressor contact DNA at any one time
But why if only 2 subunits of lac I repressor contact DNA at one time is the lac I repressor a tetramer? Because auxillary operators allow tetramers to contact more than 1 operator at a time.
Many DNA binding proteins function as dimers binding to operators (in prokaryotes) or response elements (in eukaryotes) that have palindromic organization &endash; two half sites.

Back to Figure 10.10 &endash; looking at the 3 experiments

O^C mutations are red and clustered near the axis of symmetry.

DNAseI footprinted region is &endash;5 to +21.

1) Chemical footprinting experiment: purines protected by repressor against methylation

If protein (repressor) is closely bound to DNA, methylation and cleavage are prevented
Protection in both strands was observed near the axis of symmetry

2) "Hypersensitive sites" are purines where methylation is enhanced by a repressor

If protein is bound to DNA, a region is protected from digestion by DNAseI
Some bands are darker than control bands suggesting that DNAseI cuts more often at the site than in control DNA &endash; those sites are favored for DNAseI digestion
When DNA binding proteins bind to DNA, they sometimes bend the DNA making the DNA more susceptible to DNAseI cleavage
These more susceptible sights (i.e. hypersensitive sites) indicate changes in the secondary and tertiary structure of DNA

3) chemical crosslinking assay indicates thymines that can be crosslinked to repressor

Bind protein to and add crosslinking chemical
This crosslinking chemical causes covalent bond formation between thymine and the protein if they are in close proximity to each other.
Some T's in the operator can be crosslinked to the repressor (if use a crosslinking reagent specific for T's)

Figure 10.20

The catabolite activator protein (abbreviated CAP or CRP) is involved in positive regulation of lac to get a high rate of lac expression. The lac repressor is involved in negative regulation but the lac promoter is a poor match to the consensus sequence at &endash;10 and &endash;35, thus the promoter is weak. To get a high level of expression of lac, you need positive activation mediated by the catabolite activator protein (abbreviated CAP or CRP).
CAP/CRP is an inactive activator until the inducer, cAMP, binds to CAP/CRP, yielding an active activator aiding RNA polymerase in binding to the lac promoter.

Phenomenology of glucose repression in E. coli

E. coli prefers to grow on glucose, a monosaccharide
If you give E. coli glucose and lactose (a disaccharide), then must make ß-gal to break down lactose. E. coli first uses the glucose and then uses the lactose.
In the 1950s, Jacob and Monod observed growth of E. coli on glucose and lactose.
They grew E. coli in lactose and glucose and measured the OD₆₀₀ (optical density = #cells/ml) and the ß-gal activity (lacZ activity) on the y-axis vs. time on the x-axis
Their measurements showed that cells density increased linearly at first, and during this time the cells used glucose. Then, the optical density did not change during the lag time when the cell transcribed/translated the genes of the lac operon. Next, cell density increased linearly as lactose was being utilized as a C source.
As soon as cell density began to increase as lactose was first being used as a sugar source, lacZ activity increased rapidly.
This experiment illustrates the glucose repression effect.

What is the molecular explanation for the glucose repression effect?

DOGMA: the following observations were WRONG! Why?

As [glucose] increases, [cAMP] decreases
As [glucose] decreases, [cAMP] increases

The glucose repression phenomenon is real but not via the [cAMP] difference

Instead, glucose repression phenomenon is mediated by lactose uptake in cell:

If [glucose] increases, then lactose can't get into the cell. LacY encodes permease transports lactose into the cell.
If [glucose] increases, lac permease activity decreases
If [glucose] decreases, inhibition of the lac permease does not occur and lactose can be transported into the cell

Positions of CAP and operators varies in different operons

CAP/cAMP is not specific for lac but also serves as a postive activator for the arabinose and galactose operons as well as many other operons in E. coli.

In the lac promoter:

CAP binding site overlaps with RNA polymerase binding site at around &endash;50
CAP makes a protein-protein interaction with the a-subunit of RNA polymerase
CAP stimulates transcription by helping RNA polymerase to bind to the promoter (i.e. increases the rate of closed complex formation)
In negative regulation at lac, recall that the open complex formation is prevented because the lac repressor binds to operator and the operator overlaps with the transcription start site.

In the gal promoter:

Operator overlaps with the &endash;35 consensus sequence and with the RNA polymerase binding domain. Binding of the gal repressor blocks closed complex formation.
CAP mediates positive regulation even though it overlaps with RNA polymerase binding domain.
CAP protein binds to a different face of DNA than does the RNA polymerase and makes protein-protein contacts with RNA polymerase and CAP holds RNA polymerase at promoter leading to increased levels of transcription (i.e. positive regulation).

Trp Operon

2 types of control:

1) Transcriptional

transcriptional control is a negative repressible system
trpR encodes for trp repressor
trp repressor is activated by corepressor trp.

2) Post-transcriptional

attenuation is post-transcriptional control, which is the regulated transcription termination in trp leader

Negative repressible regulation

trpR is inactive by itself
corepressor trp binds to trpR and forms functional repressor which represses transcription

Figure 10.39

The control region of the trp operon consists of the promoter, operator, leader, and attenuator.
The structural genes are trpE, trpD, trpC, trpB, etc.
The leader peptide coding region (open reading frame [ORF]) begins 26bp into the RNA
The leader controls expression of genes in trp operon

trp leader

the 14 amino acid protein is encoded in the leader but the protein has no function
2 consecutive trp residues are present in leader which is odd because trp is rare; trp is encoded 1.1% of the time
secondary structure in RNA of trp leader is important - a GC rich hairpin and subsequent U-rich single strand form a rho independent transcription terminator just 3' to the stop codon for the trp leader, but upstrem of the trp structural genes

Recall that transcription and translation are coupled (i.e occur simultaneously) in bacteria

Handout with Explanations of Observations

For quick reference, I will assign an arbitrary number / letter to each genotype:

Number / letter
Genotype
Level structural gene expression

*a
TrpR⁺, plus trp

1 (maximal repression)

*b
TrpR^-, plus trp

70 (depression WITH attenuation)

*c
TrpR^-, minus trp

700 (depression, NO attenuation)

1
TrpR^-, plus trp, D LD102 (trp leader deleted)

550

2a
TrpR^-, plus trp, trpT (trp tRNA)

500

2b
TrpR^-, plus trp, trpS (trp tRNA synthase)

500

3
TrpR^-, minus trp, Met à Ile missense mutation in leader

70

Explanations

700, 550, and 500 all essentially mean a high level of gene expression and are basically equivalent

*a &endash; trp is the corepressor, so an active repressor will stop structural gene expression completely

*b &endash; the repressor doesn't function and we observe the "transcriptional effect"

*c &endash; "post transcriptional" effect observed

compare *b and *c: when the cells are starved for trp, then a 10X higher level of expression is observed

the 70X *b genotype is "attenuated" compared to the 700X level

attenuation leads to lower levels of gene expression
notice that the post-transcriptional effect does not required trpR

For 1:

deleting trp leader causes attenuator to be deleted, rho independent terminator is removed
deleting the trp leader mimics starving for trp

For 2:

tRNA synthetase puts amino acids on tRNA, charging the tRNA
attenuation is sensitive to levels of charged trp-tRNAs, NOT to the levels of trp itself

For 3:

ribosome can't translate the leader because the initiating methionine codon was mutated
translation of trp leader is necessary to overcome attenuation

A variety of secondary structures are possible for the trp leader, see Fig. 10.41

Conformation 1: Region 1 and 2 pair together AND Regions 3 and 4 pair together

Regions 3 and 4 of trp leader pair (G = -20 kcal) to form a terminator hairpin or stem loop which is part of a rho-independent terminator. Regions 1 and 2 pair as well (G = &endash;11.2).

Conformation 2: Region 2 pairs with region 3 (G = &endash;11.7). If region 2 pairs with region 3, then no rho-dependent termination occurs

Based on the G values, conformation 1 is more favorable than conformation 2 so RNA preferentially folds into conformation 1 in the absence of intervention.

Figure 10.42

If no trp present:

700 fold effect is observed (see *c)

Regions 2 and 3 pair

If no trp present, there is not enough trp to formed charged trp tRNAs and the ribosome waits at the 2 trp codons in region 1 for charged trp tRNAs.

The ribosome covers up the region 1 so region 1 can not base pair with region 2.
Region 2 and 3 become base paired before region 4 has been transcribed.
Thus regions 3 and 4 can not base pair and form the rho-independent terminator.

If trp present

Cells aren't starved for trp so charged trp tRNAs are present
Ribosomes can insert trp in protein sequence when it encounters the two UGG trp codons
The ribosome continues translating until it reaches the UGA stop codon between regions 1 and 2. At the point, the ribosome disrupts 2-3 pairing so that 3-4 pairing forms the terminator hairpin. RNA polymerase terminates at the attenuator. See *b.

Back to the chart with more explanations

For *b - ribosome translates up to stop codon in between regions 1 and 2, making region 3 available to base pair with region 4 and RNA polymerase terminates at the attenuator.

For 1 - no rho independent terminator forms because the entire leader (regions 1-4) were deleted

For 2 - the ribosomes pause at trp codons because no charged trp tRNAs are available

this is the equivalent of starving cells for trp (see *c)
regions 2 and 3 base pair because region 1 is covered up by stalled ribosome
transcription continues and region 4 is eventually transcribed even though ribosome is stalled. Once region 4 is transcribed, region 3 and 4 can base pair but it's too late for rho-independent termination to stop further transcription of trp structural genes.

For 3 - translation of trp leader is prevented but transcription of trp leader still occurs

In the absence of translation (i.e. presence of ribosomes on leader RNA), regions 3 and 4 base pair together, and rho independent terminator forms.

In the trp operon spacing leads to regulation of the operon

1) spacing of trp residues is crucial

2) stem-loops are crucial

Timing is also important:

1) coupling of transcription and translation is key

2) secondary structure of leader of RNA is important because it causes RNA polymerase to pause. Ribosomes follow closely behind RNA polymerase.