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Title: 3'8 B years


1
3.8 B years
Chemotaxis
2
Mechanisms/organs that have evolved several
times, independently Eyes, Sound production,
Echo-location, Venomous sting, Flapping flight.
Evolved only once Language, Flagellar rotation
3
Julius Adler Howard Berg (biophysics) Mel Simon
(biochemistry) Sandy Parkinson (genetics)
David DeRosier (microscopy)
(crystallography)
4
A quantitative method for studying chemotaxis
5
Chemoreceptors in bacteria Julius Adler
1. What are the bacteria attracted to? 2. What
is the mechanism of attraction?
6
To serve as attractants 1.
Metabolism is not necessary. 2. Transport is not
necessary. 3. Structure important for
recognition by receptors
Taste will do. Consumption is not necessary
7
What are the bacteria attracted to? Specific
sugars amino acids
Galactose Ribose Aspartate Serine
8
Structurally related compounds compete
Fucose/galactose galactose
Aspartate/serine serine
Idea of specific receptors
9
To test if two structurally related attractants A
and B share the same receptor, a competition
capillary assay was set up. In experiment I, A
was included in the capillary at its optimum
concentration, while B was present in saturating
amounts in both the capillary and the pond.
Bacteria failed to enter the capillary. In II,
the experiment was reversed, i.e. B was in the
capillary and A in the capillary and pond. Now, a
significant accumulation of bacteria was seen in
the capillary (50 of that seen when only B was
in the capillary in the absence of A). From the
experiments in I and II, what can you conclude
about the receptors for A and B?
I. A and B likely share the same receptor. II.
B is likely detected by another receptor which
does not recognize A. An alternative explanation
is that they share the same receptor, but that B
has a higher affinity for it than A.
10
Rotation of Flagella
Samuel et al., 1999
11
F. C. Neidhardt. E. coli and Salmonella. 2nd
ed. ASM Press. 1999.
12
Two models of motility
Rotate
Flex
13
Cross-linking flagella with bivalent (and not
monovalent) antibody affects motility
c-phage bind flagella and stop motility
14
  • Three mechanical requirements of the model
  • Flagellar rotation
  • Correct direction of rotation
  • Correct pattern of grooves on filament surface

The handedness of the groove on the flagellar
filament and direction of flagellar rotation are
important for the direction of phage travel. On a
normal (left-handed - L) filament, phage will
travel down a right-handed groove when the
filament is rotating CCW. When the motors turn
CW, we know that the filament changes shape and
becomes right-handed (R). R filaments might have
tighter grooves that phage may not be able to
bind.
15
Direction of rotation is important
When CheY concentration is low (CCW), cells are
sensitive (?) When CheY concentration is high
(CW) cells become resistant ()
16
H. C. Berg. 2003. The rotary motor of bacterial
flagella. Annu. Rev. Biochem. 72 19-54.
F. C. Neidhardt. E. coli and Salmonella.
2nd ed. ASM Press. 1999.
f0, f1, and f2 are sensitive, but f5/6 and f11
are resistant
17
Tethered cells demonstrated flagellar rotation
directly (Silverman and Simon)
  • Anti-hook antibodies were used to tether cells
    either to each other or the glass slide

18
c Phage care about motor rotation but not
bacterial motility. Justify
Polymorphic forms that have straight flagellar
filaments (all L or f0 all R or f11) do not
generate thrust to propel the bacterium, even
though they rotate. These mutants are still
susceptible to the phage, even though the
bacteria are non-motile.
19
Tracking and Gradient Sensing
20
Howard Berg with Tracker, Boulder, CO, 72
21
From Berg Brown, 72 see Berg book
22
How can such a small organism detect the
concentration differences necessary to sense a
gradient in space?
23
The Gradient-Sensing Mechanism
An instantaneous spatial comparison
2 micron
Head
Tail
A temporal comparison
24
Temporal Gradient Apparatus
instantaneous spatial comparison- behave as if
in the uniform environment.
temporal comparison- behave as if they are in
a gradient.
25
Control (CiCf)
12 sec
Negative (Ci1 mM, Cf0.24 mM)
5 min
Positive (Ci0, Cf0.76 mM)
Because the bacteria behave as if they are in a
gradient, even though Cf is uniform, the data
support a temporal sensing mechanism
26
Adler
a
b
c
d
e
Correspondence between CCW and runs and CW and
tumbles
27
Macnab Koshland (1972) found that cells
suddenly exposed to a positive step of serine (0
to 0.8mM) swam smoothly (without tumbling) for up
to 5 minutes. Cells exposed to a negative step (1
to 0.24 mM) tumbled continuously for about 12
sec. The recovery time after the addition of
repellents is also 12-15 sec. Thus, removal of
attractant acts like a repellent signal. What is
happening during the recovery period? From what
you have learned about the signaling pathway,
explain why the length of the two recovery times
may be different.
Recovery time is the time it takes for the
receptors to return to their pre-stimulus
signaling state. This involves
methylation/demethylation of the receptors
catalysed by CheR/CheB-P respectively.
Methylation is slower and de-methylation.
Attractants inhibit CheA and hence CheB-P levels
drop. They also change the receptor conformation
so that the methylatable glutamates are more
exposed to CheR. Since methylation is slow,
recovery from attractants is slow. The repellent
response stimulates CheA and CheB-P levels rise.
Demethylation is fast, hence recovery times are
faster.
28
Structure and Signaling Movements of
Chemoreceptors
Hughson, A.G and Hazelbauer, G.L. 1996. Detecting
conformational change of transmembrane signaling
in a bacterial chemoreceptor by measuring effects
on disulfie cross-linking in vivo. Proc. Natl.
Acad. Sci. USA. 93 11546-11551.
29
Milburn et al
  • The Asp-binding fragment of Tar was overproduced
    and crystallized.
  • Crystallization was facilitated by introduction
    of a Cys at position 36.
  • Structural differences between the apo and
    aspartate complex of the receptor (transducer)
    were examined.

30
Symmetry generates two ligand binding sites at
the dimer interface
Binding stoichiometry believed to be one
aspartate per receptor subunit, but only one of
the two ligands were observable
31
Comparison of /- aspartate structures reveals
-individual subunits remain largely unaffected by
aspartate binding, including the positions of a1
and a4
-when apo and complex are superimposed upon one
another, a small shift is indicated by the
positions of the subunits
- 4 degree rotation b/w subunits about a pivot
axis
32
Hughson and Hazelbauer paper
  • Designed to understand mechanism of movement of
    chemoreceptors during transmembrane signaling.
  • Assays were performed using intact cells.
  • Measure the effects of ligand occupancy on the
    rates of oxidative cross-linking between
    cysteines introduced into the Trg chemoreceptor

33
Immunoblot using a anti-Trg serum, in the absence
or presence() of saturating amounts of
ribose. Samples were taken before and at
various times after the addition of an oxidizer.
Monomeric Trg
Results are from cells containing Trg with a
cysteine at position 42, 42 and 203, or 38 and
202.
34
Thin lines indicate no substantial effect on the
rate of cross-linking, here seen between the two
different subunits.
35
If TM1 does not move, TM2 must. Data consistent
with a downward motion
36
Crosslinking of TM1-TM2 with cysteine-substituted
Tar receptors was monitored 0, 5 and 15 minutes
after uniform addition (i.e. no gradient) of 10
mM aspartate to E. coli cells grown in the
absence of aspartate. If increase in a specific
crosslink is diagnostic of TM1-TM2 movement, at
which of these time points, if any, do you expect
to see the highest level of crosslinking?
5 minutes. They are not signaling at 0 min, and
will have adapted by 15 minutes.
37
  • What is the contribution of individual subunits?

Signaling by the E. coli aspartate chemoreceptor
Tar with a single cytoplasmic domain per
dimer Tatsuno et al, 1996
Attractant signaling by an aspartate
chemoreceptor dimer with a single cytoplasmic
domain Gardina Manson, 1996
38
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39
Problem
If receptors are homodimers, how does one
construct a heterodimer?
Complementing mutations to produce heterodimers
40
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41
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42
Conclusions
  • Both domains of Tar are not necessary for
    signaling to Asp.
  • Signaling may be mediated by interaction between
    dimers

43
Tatsuno et al decided to test Gardina Mansons
conclusions, by using an additional R73K mutation
in their experiments. With R73K included, what
arrangement of subunits in A and B will allow
signaling?
  • Only one of the two subunits can carry the R73K
    mutation. They would
  • have to therefore express two plasmids, one
    carrying A19K-A198E and the
  • other with A19K-A198E-R73K.
  • B. They would have to introduce the A198E
    mutation in a separate subunit and
  • co-express with the A19K-R73K mutant to suppress
    the A198E defect.

44
The phosphotransfer reaction of 2-component
systems
(CheA)
(CheY/CheB)
(CheZ)
45
Hess, J. F. et al., 1988. Phosphorylation of
three proteins in the signaling pathway of
bacterial  chemotaxis. Cell 53 79-87.
46
Why do the authors conclude that the order of
phoshate transfer is from A to Y to water?
47
Conclusions
  • CheA becomes autophosphorylated in the presence
    of ATP
  • CheAP can transfer its phosphate to CheB or CheY
  • CheZ accelerates the hydrolysis of CheYP

48
Molecular Structure of CheAD289
  • Its a dimer.
  • There are 3 distinct domains
  • Kinase domain (residues 355-540)
  • Dimerization domain (residues 290-354)
  • Regulatory domain (residues 541-671)
  • The two subunits in the dimer are not perfectly
    symmetrical

49
  • In which of the following tubes will CheY be
    maximally
  • phosphorylated and why?
  • 1) CheA CheY ATP
  • 2) CheA CheW CheY ATP
  • 3) CheA CheW Methylated Receptors CheY ATP
  • 4) CheA CheW Methylated Receptors CheZ
    CheYATP.

In 3, because the kinase activity of CheA is
maximal in ternary complexes with receptors and
CheW, and methylated receptors stimulate CheA.
CheZ would dephosphorylate CheY.
50
Regulation by phosphorylation CheB CheY    
Anand, G. S. et al., 1998. Activation of
methylesterase CheB evidence of a dual role for
the regulatory domain. Biochemistry 37
14038-14047.
Zhu et al., 1997. Crystal structures of CheY
mutants Y106W and T87I/Y106W CheY activation
correlates with movement of residue 106. J. Biol.
Chem. 272 5000-5006.
51
CheB is a two-domain protein
What is the nature of stimulation by
phosphorylation?
Anand et al.
52
C h e B D 5 6 N
  • Asp56 is the site of phosphorylation
  • Substituted Asp(D) with Asn(N)

Mutants that regained activity mapped
predominantlyin linker region.
53
C o n c l u s i o n s
  • 1. N-terminal domain plays a dual role in
    regulation of methylesterase activity
  • Phosphorylation both relieves inhibition as
    well as stimulates methylesterase activity of the
    C-terminal domain.
  • 2. Stimulation of methylesterase activity by
    mutations in the linker region suggest this must
    propagate the conformation chage or move during
    CheB activation by phosphorylation

54
Model for CheB activation
  • We draw these arrows because if can measure rates
    in the forward and reverse directions and get
    real numbers, we can begin to precisely model the
    chemotaxis response

55
CheY is a single domain protein
  • A) Asp 57 is phosphorylated
  • B) Phosphorylation controls movement of Tyr 106
  • C) Rotomeric nature of Tyr 106 controls switch
    activation

56
  • Structure of CheY indicates that 106 a highly
    conserved residue in the CheY superfamily - is a
    rotamer, found bothinside and outside
  • In the structure, Y106 contacts T87 when it is
    inside
  • Structure of T87I indicates that Y106 is
    outside. This mutant is non-chemotactic
  • Hypothesis movement of the side chain of residue
    106 modulates the activation state of CheY.

57
Alternate models for the physical mechanism of
throwing the switch
The deterministic model a sequential model of
binding and switching
58
The Stochastic model binding of CheY-P to the
switch merely increase the probability of a CW
rotation
59
CheBc
CheB D56N
In this experiment, why does CheBc, which has
unregulated methylesterase activity, first show
an increase and then a decrease in chemotaxis?
The initial increase in activity reflects
restoration of chemotaxis in a CheB- background
with increasing levels of CheBc expression.
Overexpression causes decrease in chemotaxis
because the unregulated methylesterase activity
prevents the system from adapting.
60
Phosphorylation activates CheY to cause CW motor
rotation. According to Zhu et al., two CheY
mutants - Y106W and T87I - can both still
be phosphorylated, but are non-chemotactic. The
former is hyperactive (causes CW rotation) and
the latter is inactive (CCW rotation). a) Why is
Y106W non-chemotactic if it can cause CW
rotation? b) Why is phosphorylation not enough to
activate T87I?
a) Chemotaxis requires the ability to switch
between CW and CCW states. The two mutants are
effectively locked in one or the other state.
b) Zhu et al propose that the rotameric state
of Y106 controls CheY activity, based on their
observation that in wild-type CheY, this residue
rotates freely between inside and outside
positions, in the active CheY mutant Y106W this
residue is inside, while in the inactive mutant
T87I it is outside. Phosphorylation is unable to
overide the steric hindrance caused by the T87I
mutation which forces Y106 outside.
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