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Data reduction for biological assays Esbatech Course 15/12/2008

Data reduction for biological assays Esbatech Course 15/12/2008. Session 3 Gillian Raab Professor (Emeritus) of Applied Statistics Napier University Edinburgh. This session. Weighted versus unweighted fits Parallelism – definition Competitive binding assays How to test for parallelism

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Data reduction for biological assays Esbatech Course 15/12/2008

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  1. Data reduction for biological assaysEsbatech Course15/12/2008 Session 3 Gillian Raab Professor (Emeritus) of Applied Statistics Napier University Edinburgh 1

  2. This session • Weighted versus unweighted fits • Parallelism – definition • Competitive binding assays • How to test for parallelism • Short practical session 2

  3. Unweighted least squares • Assumes that the s.d. about the dose response curve is constant at all levels of Y. • For your good dose response curves this does NOT seem to be true • A weighted fit changes this assumption • Usually the s.d. is larger at higher doses 3

  4. Unweighted analysis is appropriate 4

  5. Wׁeighted analysis is appropriatewith weights 1/(fitted)^2 5

  6. Wׁeighted analysis is appropriatewith weights 1/(fitted) 6

  7. Weighting choices 7

  8. General expression for weighting • Weight proportional to 1/(fitted)^J • J=0 gives an unweighted fit • J=2 gives constant CV • J=1 gives Poisson-like weighting • Weighted sum of squares is 8

  9. How to decide on choices • Examining replicates about curve • Difficult to get a good estimate by formal means • Statistics are difficult and estimates are not very reliable unless lots of data • A simple method will be discussed in the last session • Best to have a consistent choice • Most of your well behaved assays seem to have Poisson-like errors (J=1 weight 1/Y) 9

  10. This needs a J value greater than zero Lower s.d. and larger weights at low Y A weighted fit pulls the fit closer to the points at lower doses Fit here used J=1 (Poisson-like) 10

  11. Unweighted fit to the same data The parameters almost the same When the fitting is good, little difference in estimates But the precision estimates will be different 11

  12. Precison profiles may differ 12

  13. And especially estimates of LLQ • From an unweighted fit 2254 ng/l • From a weighted fit 965 ng/l • The weighted fit gives a better estimate if the assumption about J is correct 13

  14. Weighting conclusions • Weighting seldom affects the fit much • But it does affect estimates of precision • Especially at the low end when there the non-specific response is low • And especially LLQ estimates • It is also important in checking for outliers (last session) 14

  15. Unweighted better here 15

  16. Parallelism • Interpolating on a dose response curve assumes • The biological activity that causes the response is the same for the standards and the unknown • Other factors are operating similarly for both standards and unknowns • For example, a different diluent can affect the response (Matrix effects?) so it is important to make these the same • Testing for parallelism gives a partial check on this • It needs at least 2 (better more) doses for each agent 16

  17. Dose-response curve on log (10) scale Unknown Standard 17

  18. EC50 Standard 1.0 units EC50 unknown 0.1 units Unknown Standard Relative potency Unknown To standard = 10 18

  19. If same is True everywhere On the curve Relative potency Unknown To standard = 10 Unknown Standard Distance between curves log(10) = 1 is constant So curves are parallel 19

  20. Some simple results Relative potency (RP) of unknown= (EC50 of standard) (EC50 of unknown) Log(RP) = log(EC50 of standard) - log(EC50 of unknown) = distance from unknown to standard 20

  21. Parallelism assumption • If two substances are essentially just dilutions of each other then • Response for agent 2 at concentration K1 units = Response for agent 1 at concentration RP*K1 units • RP is the relative potency • It means that RP units of agent 2 are equivalent to RP units of agent 1 • Or Agent 2 has a concentration of RP in units of agent 1 • Log (RP * k1) = log(RP) + log(k1) • So if we have a log x axis then:- • Response for agent 2 at concentration log(K1) units = Response for agent 1 at concentration log(k1) + log(RP) 21

  22. Standard Unknown 22

  23. Standard Distance from unknown to standard = Approx -0.3 Antilog(-0.3) =10^(-0.3) Approx 0.5 Unknown 23

  24. Comparative assays • Should use the parallelism assumption • It will give more precise results • And you can test for parallelism • In PRISM 4 the way to do this is via shared parameters • On the web site (changed) is a background paper about the advantages of shared parameters by the Prism 4 team 24

  25. Example of a comparative assay Fitted as two separate curves and making no use of the positive control data 25

  26. Fitting two parallel curves 26

  27. To get log(RP) directly • Rewrite the model so that the parameters are EC50 (for one) and log(RP) • the fit will be the same • But this will give a direct estimate of log(RP) • And a confidence interval for it • And we can get the estimate for RP by taking the antilog (10^value) 27

  28. To get log(RP) directly • Rewrite the model so that the parameters are EC50 (for one) and log(RP) • the fit will be the same • But this will give a direct estimate of log(RP) • And a confidence interval for it 28

  29. How to do this in Prism 4 • Open the 2 parallel curves data set from the web page • We will follow through how to do it together • Use the next few slides for your own notes as I demonstrate • My answers are on the web version (this one) 29

  30. Open the first data set • This is your analysis with two different curves • Look at the graph and the table of results • Check that this is the same data we are demonstrating and that the two slopes are the same as on the table below • How could you check if there is evidence for lack of parallelism? 30

  31. Look at the confidence intervals for the two slopes.Do they overlap? Same thing for TOP also 31

  32. Reorganise the data • Duplicate the data file (>insert>duplicate current sheet) • Copy the response for the zero dose into the columns for C and name this something like “zero dose” • Remove the responses for zero dose from the first two columns • Was it right to have it there in the first place? 32

  33. Reorganise the data • Duplicate the data file (>insert>duplicate current sheet) • Copy the response for the zero dose into the columns for C and name this something like “zero dose” • Remove the responses for zero dose from the first two columns • Was it right to have it there in the first place? • No it was pretending you had more data than you really had 33

  34. Change the model to be fitted • First transform the X scale to logs (>analyse>data manipulation>transform X values using X=log(X)) • Check the graph • OPTIONAL I would change the way the graph looks too 34

  35. Mine looks like this 35

  36. Change the model to be fitted • Now go to >analyse >non linear regression >more equations >your own equation • Enter the following <A>Y=BOTTOM+(TOP-BOTTOM)/(1+10^(SLOPE*(EC50-X))) <B>Y=BOTTOM+(TOP-BOTTOM)/(1+10^(SLOPE*(EC50+LOGRP-X))) <C>Y=BOTTOM • Give it a name. I called it “two parallel 4PL with RP” 36

  37. You are asked to set initial values for the fit • Enter these from the menu boxes Bottom *YMIN Top *YMAX EC50 *VALUE OF X AT YMID SLOPE *SIGN(YATXMAX-YATXMIN) LOGRP * enter value 0.0 • Now go to the constraints TAB • Set all parameters except LOGRP to be shared between all equations • Click OK to fit 37

  38. Go to the table of results • Notice the shared parameters with same values for all equations • Here we get • log(RP) 0.042 95% CI (0.021,0.062) • So evidence that agent is more potent than control by a factor of 10^0.042 = 1.10 • And how would you get the CI for this? 38

  39. Go to the table of results • Here we get • log(RP) 0.042 95% CI (0.021,0.062) • So evidence that agent is more potent than control by a factor of 10^0.042 = 1.10 • Just take the antilogs (10^value) of the interval for log(RP) • 10^0.021 to 10^0.062 =1.05 to 1.15 • RP is 1.01 95% CI (1.05 to 1.15 39

  40. Two different slopes vs common slopes * = incorrect on hard copy 40

  41. Things to try on your own (if time) • See if a weighted fit gives different results • Write a new model that fits two EC50s <A>Y=BOTTOM+(TOP-BOTTOM)/(1+10^(SLOPE*(EC50-X))) <B>Y=BOTTOM+(TOP-BOTTOM)/(1+10^(SLOPE*(EC502-X))) <C>Y=BOTTOM And you need to change the sharing options and initial rules • Try fitting the other data sets • Try working with the example with three curves (ambitious!) 41

  42. Estimates from fitting two curves 42

  43. Parallelism conclusions • Comparative experiments should be analysed with the parallelism assumption • This means defining shared parameters • A fit without shared parameters should be used to check parallelism from the confidence intervals • The estimated RP should be caclulated with its confidence interval • Ideally a suitable weighted fit should be used. 43

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