Transcript Determining the structure of Platinum Streptadine using X

Determining the Structure of Platinum
Streptidine using X-Ray Absorption
Spectroscopy
Michaëlle Mayalu
Massachusetts Institute of Technology
SULI Program: Stanford Synchrotron Radiation Laboratory, SLAC
Mentor: Serena DeBeer George
August, 15, 2007
Outline




Platinum Anti-Cancer Agents
X-Ray Absorption Spectroscopy
Data Analysis
Conclusion
Platinum Anti-Cancer
Agents
•Background
•Platinum
Streptidine
Background

Platinum complexes have been discovered to stop cell division

This has useful applications in treating cancer and considerable advances have
been made
However…..

mechanisms are that lead to the drug being taken up by the cell membrane and
integrated into the DNA are still unknown

the drug is successful in treating some types of cancers but ineffectual treating
other types

platinum drugs such as cisplatin has been found to be very toxic causing
nephrotoxity, neuropathy, ototoxicity, hematological toxicity, neuropathology and
seizures.

Consequently, the search for improved platinum drugs that treat a wider range of
cancers and display fewer toxic side effects still continues.
Platinum Streptidine



Because the structure and
coordination of the drug (especially
NH14
in solution) is essential to
H14N
understanding how the drug interacts
•Streptidine
with molecules in the body, X-ray
absorption spectroscopy (XAS) is a
H5
H1
NH11
powerful tool for determining the
H3 NH13
local structure of this newly
OH7
developed drug.
H12O
OH8
main questions to hopefully be
H11N
answered by analysis of XAS data
N
H10
measured with XAS are:
H6
H9O
1.
Is it the Pt coordinated to the
H4
H2
Streptidine?
2.
And if this is the case through which
atoms
Putative structures (obtained from
NH
elemental analysis and nuclear
H2N
NH2
NH
magnetic resonance)
HN
Cl
N. Aliaga-Alcalde and Jan Reedijk
HO
N
Pt
S
Cl
OH
OH
O
1
S
Pt
HN
HN
H2N
HN
NH
HN
HO
N
H
HO
S
OH
OH
2
N
HN
NH
H2N
PtNHNH
H2N
HO
NH
Cl
NH
NH HO
OH
OH
O
Pt
HO
S
3
NH
OH
OH
X-Ray Absorption
Spectroscopy
•What
Is It
•Measurements
•Applications
What is it?
Auger electron
Emitted photo-electron
continuum
Direct Absorption
(Transmittance)
h
2p
2s
h
fluorescent photon
1s
George DeBeer, S.
X-ray Fluorescence
What is it?
Pt-L3 Edge 11550 eV
www.chem.ucalgary/groups/farideh/xas.pdf
Measurements
X-ray
Source
Experimental Hutch
Ta Slits
Sample
I0
Mono
slits
I1
Double
Crystal
Monochromator
Detector
George DeBeer, S.
Foil
I2
Measurements: Transmittance

When a beam of monochromatic X-rays goes through matter, it loses its
intensity due to interaction with the atoms in the material. The intensity
drops exponentially with distance if the material is homogeneous, and
after transmission the intensity is:
I0=Ie-μt

Where:


I
I0
µ
t

Absorption can therefore be measured as:


incident X-ray intensity
transmitted X-ray intensity
absorption coefficient
is the thickness of the sample
A= µt=ln(I0/I)
www.ssrl.slac.stanford.edu/dichroism/xas
Transmittance
A = μt = ln (I0/I1)
t
Ion chamber
Ion chamber
μ ,absorption coefficient
t sample thickness
X-rays
I0
X-rays

Sample

+

     
I




Ionizing radiation (i.e. X-rays) creates ion pairs in the gas and the sweeping
voltage results in a current flow.
George DeBeer, S.
Measurements: Fluorescence
Fluorescence
detector
Soller Slits
filter
X-ray
sample
George DeBeer, S.
Applications
Absorption Coefficient (mu)
X-Ray Absorption Spectrum (edge + EXAFS)
EXAFS (extended x-ray
absorption fine structure)
Pre-edge
and Edge
(XANES)
Energy
XAS or XAFS
George DeBeer, S.
Applications
Extended X-Ray absorption fine structure
constructive
interference
results in a maximum
George DeBeer, S.
destructive interference
results in a minimum
Data Analysis
Processing Data
Analyzing Results
Fitting Data
Processing Data
Getting the EXAFS
1.4
Raw
1.2
Raw data: This is the way that the XAS
transmission mode spectrum looks, right off
the beam line.
1.0
0.8
0.6
0.4
6800
7000
7200
7400
eV
7600
7800
8000
1.4
Pre-edge subtraction: A procedure performed
to subtract the total absorption from the
absorption of the edge in interest.
1.2
Raw
1.0
0.8
0.6
0.4
6800
7000
7200
7400
eV
7600
7800
8000
Normal
1.5
Spline: A method for removing the atomic
background from the absorption curve (i.e. the
absorption due to the photoabsorber alone, with
out any neighboring atoms) .
1.0
0.5
0.0
6800
7000
7200
7400
7600
7800
8000
George DeBeer, S.
Processing Data
Getting Information from EXAFS Data
6
15.0
4
A Fourier transform
allows you to
visualize the radial
distribution of atoms.
FT Magnitude
EXAFS *k
3
2
0
-2
10.0
5.0
-4
-6
4
6
8
k (Å-1)
10
12
0.0
0
1
2
Note:
k is the photoelectron
wave number.
EXAFS data are kweighted to enhance
oscillations at high-k.
3
Cu-N
Cu-S
Cu-C2/C5 (SS)
Cu-N-C2/C5 (MS)
Cu-C3/N4 (SS)
Cu-N-C3/N4 (MS)
George DeBeer, S.
4
5
6
data + fit
EXAFS *k
k= (2m(E-E0)/ћ2)1/2
3
R (Å)
4
6
8
k (Å-1)
10
12
EXAFS data is really a sum
of sine waves. The goal of
fitting data is to deconvolute
the total signal into its
components.
Analyzing Results
Pt-Std Edges Solid and Solution
1.60E+00
1.40E+00

As can be seen, the edges of
the solid and solution data
begin at different energies

Because the solid begins at a
lower energy, it is more reduced

This suggests that the solid
structure is surrounded by
heavier atoms
1.20E+00
Absorbance
1.00E+00
8.00E-01
solid
6.00E-01
soln
4.00E-01
2.00E-01
0.00E+00
11540
11545
11550
11555
11560
11565
-2.00E-01
Energy(eV)
11570
11575
11580
Analyzing Results
1.00E+01
K3 EXAFS and Corresponding Fourier
Transforms
8.00E+00
6.00E+00
X(k)*k^3
4.00E+00
2.00E+00

amplitude of the EXAFS slightly
decreases from solid to soln

this indicate a decrease in
coordination number or may also
be indicative of lighter atoms
present in solution.

intensity of FT greatly decreases
from solid to solution.

Similar to the decrease in EXAFS
amplitude, the decrease in peak
intensity indicates a decrease in
coordination or lighter atoms
ligated to the Platinum in solution.
0.00E+00
-2.00E+00
-4.00E+00
-6.00E+00
-8.00E+00
Energy(eV)
Transform Magnitide
2. 50E +00
2. 00E +00
1. 50E +00
1. 00E +00
5. 00E -01
0. 00E +00
0. 00E +00
1. 00E +00
2. 00E +00
3. 00E +00
4. 00E +00
R(A)
5. 00E +00
6. 00E +00
7. 00E +00
8. 00E +00
Fitting Data
Initial Model
Calculation of initial
distance & disorder
parameters.
Optimization of
distance & disorder
parameters.
Are fit parameters
reasonable & is fit
quality good?
No

George DeBeer, S.
Yes
Compare with
other good fits
to determine
best fit.
Solid Pt-Std
Best Fit: 4 Pt-Cl at 2.30 Å and 1 Pt-N at 2.02 Å
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4 Pt-Cl at 2.30 Å
1.60E+00
Normalized error
0.567 F/(No. pts)
10 10
8 8
6
4
Normalized
error: 1.20E+00
0.547F/(No.
1.00E+00
pts)
4
2
2
EXAFS
0
00
2
0
-2
2
Cl
4
6
4
8
6
10
8
12
14
Cl
10
12
14
16
16
EXAFS
FIT
FIT
-2
-4
-4
-6
-6
-8
8.00E-01
-8
-10
Pt
k(A^-1)
-10
3.00E+00
3.00E+00
6.00E-01
k(A^-1)
2.50E+00
2.50E+00
4.00E-01
2.00E+00
2.00E-01
0.00E+00
11540
-2.00E-01
FT Magnitude
FT Magnitude
Absorbance

EXAFS*k^3
EXAFS*k^3
1.40E+00
6
2.00E+00
Cl
Cl
N
FT
1.50E+00
FT FIT
FT
1.50E+00
FT FIT
1.00E+00
11545
1.00E+00
11550
11555
11560
11565
11570
11575
5.00E-01
5.00E-01
Energy(eV)
0.00E+00
0.00E+00

0.00E+00
5.00E-01
1.00E+00
1.50E+00
2.00E+00
2.50E+00
3.00E+00
3.50E+00
1.50E+00
2.00E+00
2.50E+00
3.00E+00
3.50E+00
4.00E+00
Comparison of K2PtCl
R(A) 4 and C8H18N6O7Cl4Pt
0.00E+00
5.00E-01
1.00E+00
R(A)
4.00E+00
11580
Solution Pt-Std
4 Pt-Cl/S at 2.30 Å 1 Pt-N at 2.03 Å

8
Normalized error:
0.359F/(No. pts)
6
EXAFS*k^3
4
2
EXAFS
FIT
0
0
2
4
6
8
10
12
14
-2
-4
-6
k(A^-1))
1.80E+00
1.60E+00
1.40E+00
FT Magnitude

1.20E+00
1.00E+00
FT
FT FIT
8.00E-01
6.00E-01
4.00E-01
2.00E-01
0.00E+00
0.00E+00
5.00E-01
1.00E+00
1.50E+00
2.00E+00
R(A)
2.50E+00
3.00E+00
3.50E+00
4.00E+00
Solution Pt-Std

Best Fit: 3 Pt-Cl/S at 2.30 Å and 2 Pt-N at 2.08 Å
8
6
EXAFS*k^3
Normalized error:
0.363 F/(No. pts)
4
Cl/S
2
Cl/S
0
0
EXAFS
Cl/S
2
4
6
8
10
12
FIT
14
-2
-4
-6
Pt
k(A^-1)
1.80E+00
1.60E+00
1.40E+00
FT Magnitude

1.20E+00
N
N
1.00E+00
FT
FT FIT
8.00E-01
6.00E-01
4.00E-01
2.00E-01
0.00E+00
0.00E+00
5.00E-01
1.00E+00
1.50E+00
2.00E+00
R(A)
2.50E+00
3.00E+00
3.50E+00
4.00E+00
Conclusion
Conclusion

It can be concluded that Platinum is in fact coordinated to the
streptidine through a bond with nitrogen.

In the solid…

the streptidine is coordinated to the Pt by one nitrogen at 2.02 Å

In the solution…

the streptidine could be coordinated to 1-2 nitrogens, although
with 2 nitrogens at 2.08 Å is more likely.

This is shown by analysis and fitting of the EXAFS data of the solid
and solution Pt-Std and comparing the edges and EXAFS of the Pt-Std
solid and solution

Future work should include the determination of how many chlorines
and sulfurs are ligated in the solution structure.
Acknowledgments

Special Thanks to:


My Mentor Serena Debeer George
U. S. Department of Energy, Office of Science for
giving me the opportunity to participate in the
SULI
References

Serena DeBeer George, Introduction to Extended X-ray
Absorption Fine Structure (EXAFS) Spectroscopy
and its Applications (Power Point)

XAS Short Course for Structural Molecular Biology Applications

www.chem.ucalgary/groups/farideh/xas.pdf

www.ssrl.slac.stanford.edu/dichroism/xas

N. Aliaga-Alcalde and Jan Reedijk, University of Leiden, the
Netherlands, unpublished results, Pt-Std (Platinum-Streptidine
complexes)