[3dem] [ccp4bb] am I doing this right?
Marin van Heel
marin.vanheel at googlemail.com
Tue Oct 26 06:53:38 PDT 2021
Dear All,
In the context of this discussion, I suggest the readers / discussion
participants to have a look at new ideas on: Information Content of images,
SNR, DQEs, A priori Probabilities, Shannon's Channel Information
capacity,etc., expressed in: Van Heel & Schatz, arXiv 2020. (There is also
an easy-to-find Youtube lecture on the issue by "marin van heel").
Cheers,
Marin
On Tue, Oct 26, 2021 at 3:58 AM Jacob Keller <jacobpkeller at gmail.com> wrote:
> Hi All,
>
> haven't been following CCP4BB for a while, then I come back to this juicy
> Holtonian thread!
>
> Sorry for being more practical, but could you use a windowed approach:
> integrate values of the same pixel/relp combo (roxel?) over time (however
> that works with frame slicing) to estimate the error over many frames, then
> shrink the window incrementally, see whether the successive
> plotted shrinking-window values show a trend? Most likely flat for
> background? This could be used for the spots as well as background. Would
> this lose the precious temporal info?
>
> Jacob
>
> On Fri, Oct 22, 2021 at 3:25 AM Gergely Katona <gergely.katona at gu.se>
> wrote:
>
>> Hi,
>>
>>
>>
>> I have more estimates to the same problem using a multinomial data
>> distribution. I should have realized that for prediction, I do not have to
>> deal with infinite likelihood of 0 trials when observing only 0s on an
>> image. Whenever 0 photons generated by the latent process, the image is
>> automatically empty. With this simplification, I still have to hide behind
>> mathematical convenience and use Gamma prior for the latent Poisson
>> process, but observing 0 counts just increments the beta parameter by 1
>> compared to the prior belief. With equal photon capture probabilities, the
>> mean counts are about 0.01 and the std is about 0.1 with
>> rate≈Gamma(alpha=1, beta=0.1) prior . With a symmetric Dirichlet prior to
>> the capture probabilities, the means appear unchanged, but the predicted
>> stds starts high at very low concentration parameter and level off at high
>> concentration parameter. This becomes more apparent at high photon counts
>> (high alpha of Gamma distribution). The answer is different if we look at
>> the std across the detector plane or across time of a single pixel.
>>
>> Details of the calculation below:
>>
>>
>>
>>
>> https://urldefense.proofpoint.com/v2/url?u=https-3A__colab.research.google.com_drive_1NK43-5F3r1rH5lBTDS2rzIFDFNWqFfekrZ-3Fusp-3Dsharing&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=SH8-k6_ktONw_SFkmg4v6mJm95Tfznxi63jj8FQjahI&e=
>>
>>
>>
>> Best wishes,
>>
>>
>>
>> Gergely
>>
>>
>>
>> Gergely Katona, Professor, Chairman of the Chemistry Program Council
>>
>> Department of Chemistry and Molecular Biology, University of Gothenburg
>>
>> Box 462, 40530 Göteborg, Sweden
>>
>> Tel: +46-31-786-3959 / M: +46-70-912-3309 / Fax: +46-31-786-3910
>>
>> Web: https://urldefense.proofpoint.com/v2/url?u=http-3A__katonalab.eu&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=WtAab9SqD68vXGNd_F4ilEWXlIeUv4CmaJeYQuDfROQ&e= , Email: gergely.katona at gu.se
>>
>>
>>
>> *From:* CCP4 bulletin board <CCP4BB at JISCMAIL.AC.UK> *On Behalf Of *Nave,
>> Colin (DLSLtd,RAL,LSCI)
>> *Sent:* 21 October, 2021 19:21
>> *To:* CCP4BB at JISCMAIL.AC.UK
>> *Subject:* Re: [ccp4bb] am I doing this right?
>>
>>
>>
>> Congratulations to James for starting this interesting discussion.
>>
>>
>>
>> For those who are like me, nowhere near a black belt in statistics, the
>> thread has included a number of distributions. I have had to look up where
>> these apply and investigate their properties.
>>
>> As an example,
>>
>> “The Poisson distribution is used to model the # of events in the
>> future, Exponential distribution is used to predict the wait time until the
>> very first event, and Gamma distribution is used to predict the wait time
>> until the k-th event.”
>>
>> A useful calculator for distributions can be found at
>>
>> https://urldefense.proofpoint.com/v2/url?u=https-3A__keisan.casio.com_menu_system_000000000540&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=vSFYZr4wbWSz1xVDpjU0ou57-qAxsXr5zxmefXAblf8&e=
>>
>> a specific example is at
>>
>> https://urldefense.proofpoint.com/v2/url?u=https-3A__keisan.casio.com_exec_system_1180573179&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=e_uXyWeSVfcQs_ypUwBLU_b120FEMZZRNsd4OMw7MP0&e=
>>
>> where cumulative probabilities for a Poisson distribution can be found
>> given values for x and lambda.
>>
>>
>>
>> The most appropriate prior is another issue which has come up e.g. is a
>> flat prior appropriate? I can see that a different prior would be
>> appropriate for different areas of the detector (e.g. 1 pixel instead of
>> 100 pixels) but the most appropriate prior seems a bit arbitrary to me. One
>> of James’ examples was 10^5 background photons distributed among 10^6
>> pixels – what is the most appropriate prior for this case? I presume it is
>> OK to update the prior after each observation but I understand that it can
>> create difficulties if not done properly.
>>
>>
>>
>> Being able to select the prior is sometimes seen as a strength of
>> Bayesian methods. However, as a strong advocate of Bayesian methods once
>> put it, this is a bit like Achilles boasting about his heel!
>>
>>
>>
>> I hope for some agreement among the black belts. It would be good to end
>> up with some clarity about the most appropriate probability distributions
>> and priors. Also, have we got clarity about the question being asked?
>>
>>
>>
>> Thanks to all for the interesting points.
>>
>>
>>
>> Colin
>>
>> *From:* CCP4 bulletin board <CCP4BB at JISCMAIL.AC.UK> *On Behalf Of *Randy
>> John Read
>> *Sent:* 21 October 2021 13:23
>> *To:* CCP4BB at JISCMAIL.AC.UK
>> *Subject:* Re: [ccp4bb] am I doing this right?
>>
>>
>>
>> Hi Kay,
>>
>>
>>
>> No, I still think the answer should come out the same if you have good
>> reason to believe that all the 100 pixels are equally likely to receive a
>> photon (for instance because your knowledge of the geometry of the source
>> and the detector says the difference in their positions is insignificant,
>> i.e. part of your prior expectation). Unless the exact position of the spot
>> where you detect the photon is relevant, detecting 1 photon on a big pixel
>> and detecting the same photon on 1 of 100 smaller pixels covering the same
>> area are equivalent events. What should be different in the analysis, if
>> you're thinking about individual pixels, is that the expected value for a
>> photon landing on any of the pixels will be 100 times lower for each of the
>> smaller pixels than the single big pixel, so that the expected value of
>> their sum is the same. You won't get to that conclusion without having a
>> different prior probability for the two cases that reflects the 100-fold
>> lower flux through the smaller area, regardless of the total power of the
>> source.
>>
>>
>>
>> Best wishes,
>>
>> Randy
>>
>>
>>
>> On 21 Oct 2021, at 13:03, Kay Diederichs <kay.diederichs at uni-konstanz.de>
>> wrote:
>>
>>
>>
>> Randy,
>>
>> I must admit that I am not certain about my answer, but I lean toward
>> thinking that the result (of the two thought experiments that you describe)
>> is not the same. I do agree that it makes sense that the expectation value
>> is the same, and the math that I sketched in
>> https://urldefense.proofpoint.com/v2/url?u=https-3A__www.jiscmail.ac.uk_cgi-2Dbin_wa-2Djisc.exe-3FA2-3DCCP4BB-3Bbdd31b04.2110&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=mhUwmSlF2o-ffQtdYI3zFn1XhBTscxvjXTfk5O47CjM&e=
>> actually shows this. But the variance? To me, a 100-pixel patch with all
>> zeroes is no different from sequentially observing 100 pixels, one after
>> the other. For the first of these pixels, I have no idea what the count is,
>> until I observe it. For the second, I am less surprised that it is 0
>> because I observed 0 for the first. And so on, until the 100th. For the
>> last one, my belief that I will observe a zero before I read out the pixel
>> is much higher than for the first pixel. The variance is just the inverse
>> of the amount of error (squared) that we assign to our belief in the
>> expectation value. And that amount of belief is very different. I find it
>> satisfactory that the sigma goes down with the sqrt() of the number of
>> pixels.
>>
>> Also, I don't find an error in the math of my posting of Mon, 18 Oct 2021
>> 15:00:42 +0100 . I do think that a uniform prior is not realistic, but this
>> does not seem to make much difference for the 100-pixel thought experiment.
>>
>> We could change the thought experiment in the following way - you observe
>> 99 pixels with zero counts, and 1 with 1 count. Would you still say that
>> both the big-pixel-single-observation and the 100-pixel experiment should
>> give expectation value of 2 and variance of 2? I wouldn't.
>>
>> Best wishes,
>> Kay
>>
>> On Thu, 21 Oct 2021 09:00:23 +0000, Randy John Read <rjr27 at CAM.AC.UK>
>> wrote:
>>
>> Just to be a bit clearer, I mean that the calculation of the expected
>> value and its variance should give the same answer if you're comparing one
>> pixel for a particular length of exposure with the sum obtained from either
>> a larger number of smaller pixels covering the same area for the same
>> length of exposure, or the sum from the same pixel measured for smaller
>> time slices adding up to the same total exposure.
>>
>> On 21 Oct 2021, at 09:54, Randy John Read <rjr27 at cam.ac.uk<
>> mailto:rjr27 at cam.ac.uk <rjr27 at cam.ac.uk>>> wrote:
>>
>> I would think that if this problem is being approached correctly, with
>> the right prior, it shouldn't matter whether you collect the same signal
>> distributed over 100 smaller pixels or the same pixel measured for the same
>> length of exposure but with 100 time slices; you should get the same
>> answer. So I would want to formulate the problem in a way where this
>> invariance is satisfied. I thought it was, from some of the earlier
>> descriptions of the problem, but this sounds worrying.
>>
>> I think you're trying to say the same thing here, Kay. Is that right?
>>
>> Best wishes,
>>
>> Randy
>>
>> On 21 Oct 2021, at 08:51, Kay Diederichs <kay.diederichs at uni-konstanz.de<
>> mailto:kay.diederichs at uni-konstanz.de <kay.diederichs at uni-konstanz.de>>>
>> wrote:
>>
>> Hi Ian,
>>
>> it is Iobs=0.01 and sigIobs=0.01 for one pixel, but adding 100 pixels
>> each with variance=sigIobs^2=0.0001 gives 0.01 , yielding a
>> 100-pixel-sigIobs of 0.1 - different from the 1 you get. As if repeatedly
>> observing the same count of 0 lowers the estimated error by sqrt(n), where
>> n is the number of observations (100 in this case).
>>
>> best wishes,
>> Kay
>>
>> On Wed, 20 Oct 2021 13:08:33 +0100, Ian Tickle <ianjt05 at GMAIL.COM<
>> mailto:ianjt05 at GMAIL.COM <ianjt05 at GMAIL.COM>>> wrote:
>>
>> Hi Kay
>>
>> Can I just confirm that your result Iobs=0.01 sigIobs=0.01 is the estimate
>> of the true average intensity *per pixel* for a patch of 100 pixels? So
>> then the total count for all 100 pixels is 1 with variance also 1, or in
>> general for k observed counts in the patch, expectation = variance = k+1
>> for the total count, irrespective of the number of pixels? If so then
>> that
>> agrees with my own conclusion. It makes sense because Iobs=0.01
>> sigIobs=0.01 cannot come from a Poisson process (which obviously requires
>> expectation = variance = an integer), whereas the total count does come
>> from a Poisson process.
>>
>> The difference from my approach is that you seem to have come at it via
>> the
>> individual pixel counts whereas I came straight from the Agostini result
>> applied to the whole patch. The number of pixels seems to me to be
>> irrelevant for the whole patch since the design of the detector, assuming
>> it's an ideal detector with DQE = 1 surely cannot change the photon flux
>> coming from the source: all ideal detectors whatever their pixel layout
>> must give the same result. The number of pixels is then only relevant if
>> one needs to know the average intensity per pixel, i.e. the total and s.d.
>> divided by the number of pixels. Note the pixels here need not even
>> correspond to the hardware pixels, they can be any arbitrary subdivision
>> of
>> the detector surface.
>>
>> Best wishes
>>
>> -- Ian
>>
>>
>> On Tue, 19 Oct 2021 at 12:39, Kay Diederichs <
>> kay.diederichs at uni-konstanz.de<mailto:kay.diederichs at uni-konstanz.de
>> <kay.diederichs at uni-konstanz.de>>>
>> wrote:
>>
>> James,
>>
>> I am saying that my answer to "what is the expectation and variance if I
>> observe a 10x10 patch of pixels with zero
>> counts?" is Iobs=0.01 sigIobs=0.01 (and Iobs=sigIobs=1 if there is only
>> one pixel) IF the uniform prior applies. I agree with Gergely and others
>> that this prior (with its high expectation value and variance) appears
>> unrealistic.
>>
>> In your posting of Sat, 16 Oct 2021 12:00:30 -0700 you make a calculation
>> of Ppix that appears like a more suitable expectation value of a prior to
>> me. A suitable prior might then be 1/Ppix * e^(-l/Ppix) (Agostini §7.7.1).
>> The Bayesian argument is IIUC that the prior plays a minor role if you do
>> repeated measurements of the same value, because you use the posterior of
>> the first measurement as the prior for the second, and so on. What this
>> means is that your Ppix must play the role of a scale factor if you
>> consider the 100-pixel experiment.
>> However, for the 1-pixel experiment, having a more suitable prior should
>> be more important.
>>
>> best,
>> Kay
>>
>>
>>
>>
>> On Mon, 18 Oct 2021 12:40:45 -0700, James Holton <jmholton at LBL.GOV<
>> mailto:jmholton at LBL.GOV <jmholton at LBL.GOV>>> wrote:
>>
>> Thank you very much for this Kay!
>>
>> So, to summarize, you are saying the answer to my question "what is the
>> expectation and variance if I observe a 10x10 patch of pixels with zero
>> counts?" is:
>> Iobs = 0.01
>> sigIobs = 0.01 (defining sigIobs = sqrt(variance(Iobs)))
>>
>> And for the one-pixel case:
>> Iobs = 1
>> sigIobs = 1
>>
>> but in both cases the distribution is NOT Gaussian, but rather
>> exponential. And that means adding variances may not be the way to
>> propagate error.
>>
>> Is that right?
>>
>> -James Holton
>> MAD Scientist
>>
>>
>>
>> On 10/18/2021 7:00 AM, Kay Diederichs wrote:
>> Hi James,
>>
>> I'm a bit behind ...
>>
>> My answer about the basic question ("a patch of 100 pixels each with
>> zero counts - what is the variance?") you ask is the following:
>>
>> 1) we all know the Poisson PDF (Probability Distribution Function)
>> P(k|l) = l^k*e^(-l)/k! (where k stands for for an integer >=0 and l is
>> lambda) which tells us the probability of observing k counts if we know l.
>> The PDF is normalized: SUM_over_k (P(k|l)) is 1 when k=0...infinity is 1.
>> 2) you don't know before the experiment what l is, and you assume it is
>> some number x with 0<=x<=xmax (the xmax limit can be calculated by looking
>> at the physics of the experiment; it is finite and less than the overload
>> value of the pixel, otherwise you should do a different experiment). Since
>> you don't know that number, all the x values are equally likely - you use
>> a
>> uniform prior.
>> 3) what is the PDF P(l|k) of l if we observe k counts? That can be
>> found with Bayes theorem, and it turns out that (due to the uniform prior)
>> the right hand side of the formula looks the same as in 1) : P(l|k) =
>> l^k*e^(-l)/k! (again, the ! stands for the factorial, it is not a semantic
>> exclamation mark). This is eqs. 7.42 and 7.43 in Agostini "Bayesian
>> Reasoning in Data Analysis".
>> 3a) side note: if we calculate the expectation value for l, by
>> multiplying with l and integrating over l from 0 to infinity, we obtain
>> E(P(l|k))=k+1, and similarly for the variance (Agostini eqs 7.45 and 7.46)
>> 4) for k=0 (zero counts observed in a single pixel), this reduces to
>> P(l|0)=e^(-l) for a single observation (pixel). (this is basic math; see
>> also §7.4.1 of Agostini.
>> 5) since we have 100 independent pixels, we must multiply the
>> individual PDFs to get the overall PDF f, and also normalize to make the
>> integral over that PDF to be 1: the result is f(l|all 100 pixels are
>> 0)=n*e^(-n*l). (basic math). A more Bayesian procedure would be to realize
>> that the posterior PDF P(l|0)=e^(-l) of the first pixel should be used as
>> the prior for the second pixel, and so forth until the 100th pixel. This
>> has the same result f(l|all 100 pixels are 0)=n*e^(-n*l) (Agostini §
>> 7.7.2)!
>> 6) the expectation value INTEGRAL_0_to_infinity over l*n*e^(-n*l) dl is
>> 1/n . This is 1 if n=1 as we know from 3a), and 1/100 for 100 pixels with
>> 0 counts.
>> 7) the variance is then INTEGRAL_0_to_infinity over
>> (l-1/n)^2*n*e^(-n*l) dl . This is 1/n^2
>>
>> I find these results quite satisfactory. Please note that they deviate
>> from the MLE result: expectation value=0, variance=0 . The problem appears
>> to be that a Maximum Likelihood Estimator may give wrong results for small
>> n; something that I've read a couple of times but which appears not to be
>> universally known/taught. Clearly, the result in 6) and 7) for large n
>> converges towards 0, as it should be.
>> What this also means is that one should really work out the PDF instead
>> of just adding expectation values and variances (and arriving at 100 if
>> all
>> 100 pixels have zero counts) because it is contradictory to use a uniform
>> prior for all the pixels if OTOH these agree perfectly in being 0!
>>
>> What this means for zero-dose extrapolation I have not thought about.
>> At least it prevents infinite weights!
>>
>> Best,
>> Kay
>>
>>
>>
>>
>>
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>> <https://urldefense.proofpoint.com/v2/url?u=http-3A__www.jiscmail.ac.uk-253chttp-3A_www.jiscmail.ac.uk&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=oVaZIXgE-SRmGu-I-Kk_0QXQXjrXb00KyOtVPm6W5Qc&e= >>, terms &
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>> <https://urldefense.proofpoint.com/v2/url?u=http-3A__www.jiscmail.ac.uk_CCP4BB-253chttp-3A_www.jiscmail.ac.uk_CCP4BB&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=dApcI6QRnXBfn3ACXaiG8RSNp72HqWMCxO4uIAzbToE&e= >>, a
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>> <https://urldefense.proofpoint.com/v2/url?u=http-3A__www.jiscmail.ac.uk-253chttp-3A_www.jiscmail.ac.uk&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=oVaZIXgE-SRmGu-I-Kk_0QXQXjrXb00KyOtVPm6W5Qc&e= >>, terms &
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>>
>> ------
>> Randy J. Read
>> Department of Haematology, University of Cambridge
>> Cambridge Institute for Medical Research Tel: + 44 1223 336500
>> The Keith Peters Building Fax: + 44 1223
>> 336827
>> Hills Road E-mail:
>> rjr27 at cam.ac.uk<mailto:rjr27 at cam.ac.uk>
>> Cambridge CB2 0XY, U.K.
>> www-structmed.cimr.cam.ac.uk<https://urldefense.proofpoint.com/v2/url?u=http-3A__www-2Dstructmed.cimr.cam.ac.uk_&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=Q5RQqk_mEgty2jTWeZ9gjyxt4x-14kd9kdlPHdVorfQ&e= >
>>
>>
>> ________________________________
>>
>> To unsubscribe from the CCP4BB list, click the following link:
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>>
>> ------
>> Randy J. Read
>> Department of Haematology, University of Cambridge
>> Cambridge Institute for Medical Research Tel: + 44 1223 336500
>> The Keith Peters Building Fax: + 44 1223
>> 336827
>> Hills Road E-mail:
>> rjr27 at cam.ac.uk<mailto:rjr27 at cam.ac.uk>
>> Cambridge CB2 0XY, U.K.
>> www-structmed.cimr.cam.ac.uk<https://urldefense.proofpoint.com/v2/url?u=http-3A__www-2Dstructmed.cimr.cam.ac.uk&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=dI0L9lfv53zODnTFcku_TD-m-H9GOZ4Rt8jwNG9qBT8&e= >
>>
>>
>> ########################################################################
>>
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>> This message was issued to members of https://urldefense.proofpoint.com/v2/url?u=http-3A__www.jiscmail.ac.uk_CCP4BB&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=4ZqZviVbXRmXJlCgDTPlxXimFO82w7SxBRK-q8jWWCY&e= , a
>> mailing list hosted by https://urldefense.proofpoint.com/v2/url?u=http-3A__www.jiscmail.ac.uk&d=DwIFaQ&c=-35OiAkTchMrZOngvJPOeA&r=L7-zyQ-04fFCMRqzLIOnx7H0exGZHwIQe_wMPuY600I&m=p3Ogfv3NVAsovV0OsEJMfULaG3pKbZdrI6QFq9IAVtw&s=CBmOdALwXx5WZVcS3k2sbL2JOjWwckr7m3yaegw6hSE&e= , terms & conditions are
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>>
>>
>>
>>
>>
>> ------
>>
>> Randy J. Read
>>
>> Department of Haematology, University of Cambridge
>>
>> Cambridge Institute for Medical Research Tel: + 44 1223 336500
>>
>> The Keith Peters Building Fax: + 44 1223
>> 336827
>>
>> Hills Road E-mail:
>> rjr27 at cam.ac.uk <rjr27 at cam.ac.uk>
>>
>> Cambridge CB2 0XY, U.K.
>> www-structmed.cimr.cam.ac.uk
>>
>>
>>
>>
>> ------------------------------
>>
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>>
>> --
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>
>
> --
>
> +++++++++++++++++++++++++++++++++++++++++++++++++
>
> Jacob Pearson Keller
>
> Assistant Professor
>
> Department of Pharmacology and Molecular Therapeutics
>
> Uniformed Services University
>
> 4301 Jones Bridge Road
>
> Bethesda MD 20814
>
> jacob.keller at usuhs.edu; jacobpkeller at gmail.com
>
> Cell: (301)592-7004
>
> +++++++++++++++++++++++++++++++++++++++++++++++++
>
> ------------------------------
>
> To unsubscribe from the CCP4BB list, click the following link:
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