ryraman2661
☆    

India,
2024-03-20 05:51
(23 d 10:16 ago)

Posting: # 23911
Views: 522
 

 ICH M10 - Method validation - Anchoring points-reg [Regulatives / Guidelines]

Dear All,

I have a query with reference to the anchoring point(s) , (LLOQ and ULOQ) in method validation of normal, molecules.

As per the recent M-10 guideline, the criteria for accepting calibration curve is minimum 75 % of non-zero standards should pass. There is no mentioning of anchoring points.

In Precision and Accuracy (P & A) experiments.

We do 8 points calibrators including LLOQ and ULOQ and 6 replicates of each level QC. If one anchor point, ULOQ calibrator not meeting, and other 4 QCs are within the accepted calibrators, can we accept?.


Members can share their opinion
Ohlbe
★★★

France,
2024-03-20 11:50
(23 d 04:18 ago)

@ ryraman2661
Posting: # 23913
Views: 441
 

 ICH M10 - Method validation - Anchoring points-reg

Dear ryraman2661,

❝ I have a query with reference to the anchoring point(s), (LLOQ and ULOQ) in method validation of normal, molecules.


Please check the definition of anchor points in section 9, glossary: Spiked samples set at concentrations below the LLOQ or above the ULOQ of the calibration curve and analysed to improve curve fitting in LBAs.

So by definition: LLOQ and ULOQ samples are not anchor points, and anchor points are only to be used for ligand-binding assays, not chromatographic methods.

❝ As per the recent M-10 guideline, the criteria for accepting calibration curve is minimum 75 % of non-zero standards should pass. There is no mentioning of anchoring points.


There is, for LBA, in section 4.2.3: At least 75% of the calibration standards excluding anchor points, and a minimum of 6 concentration levels of calibration standards, including the LLOQ and ULOQ, should meet the above criteria. The anchor points do not require acceptance criteria since they are beyond the quantifiable range of the curve.

❝ In Precision and Accuracy (P & A) experiments.


❝ We do 8 points calibrators including LLOQ and ULOQ and 6 replicates of each level QC. If one anchor point, ULOQ calibrator not meeting, and other 4 QCs are within the accepted calibrators, can we accept?.


If you're using a chromatographic method, no, according to section 3.2.4: For accuracy and precision runs, if all replicates of the LLOQ or the ULOQ calibration standard in a run are rejected, then the run should be rejected, the possible source of the failure should be determined and the method revised, if necessary. If the next validation run also fails, then the method should be revised before restarting validation.

In ligand-binding assays, the LLOQ and ULOQ samples can't be excluded from the curve according to section 4.2.3, as mentioned previously.

Regards
Ohlbe
Helmut
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Vienna, Austria,
2024-03-21 18:42
(21 d 21:25 ago)

@ ryraman2661
Posting: # 23915
Views: 390
 

 LBAs: Anchor points and model fitting

Hi ryraman,

adding to Ohlbe’s post: Anchor points may – or may not! – stabilize the nonlinear model fit in LBAs. It is up to you to decide – based on the A&P of back-calculated concentrations (of calibrators and QCs) – which model performs better and has to be used in method validation and analysis of samples.

An example of a 4-parameter logistic model$$y=D+\frac{A-D}{1+\left(\frac{x}{C}\right)^B}$$

[image]

Initial guesses, parameter bounds (constraints):
  type  A   B  C   D
 start 15 0.5 50  90
 lower  0 0.1  5  50
 upper 50 1.0 80 100

Model with anchor points: AIC = 84.48761
Formula: obs ~ logistic(A, B, C, D, x)

Parameters:
  Estimate Std. Error t value Pr(>|t|)   
A  0.03825    0.95628   0.040 0.968493   
B  0.50353    0.03443  14.623 3.84e-12 ***
C 48.93165   10.79076   4.535 0.000202 ***
D 99.98011    5.34034  18.722 3.80e-14 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.251 on 20 degrees of freedom

(In)accuracy (%) = A
        x      xhat        A
   0.1000   0.09949  -0.5076
   0.1000   0.10650   6.4670
   0.1000   0.09218  -7.8210
   0.3981   0.38360  -3.6350
   0.3981   0.36040  -9.4730
   0.3981   0.39200  -1.5320
   1.5850   1.80900  14.1400
   1.5850   1.82400  15.0800
   1.5850   1.60100   1.0010
   6.3100   6.19500  -1.8170
   6.3100   5.45600 -13.5300
   6.3100   6.25800  -0.8301
  25.1200  23.44000  -6.7040
  25.1200  23.15000  -7.8620
  25.1200  29.92000  19.1200
 100.0000 123.10000  23.0600
 100.0000  94.70000  -5.3050
 100.0000  88.64000 -11.3600

(Im)precision (%) = P
        x mean (xhat)       P
   0.1000     0.09938 -0.6205
   0.3981     0.37870 -4.8800
   1.5850     1.74500 10.0700
   6.3100     5.97000 -5.3920
  25.1200    25.50000  1.5190
 100.0000   102.10000  2.1350

Model without anchor points: AIC = 62.06841
Formula: obs0 ~ logistic(A, B, C, D, x0)

Parameters:
   Estimate Std. Error t value Pr(>|t|)   
A   0.16548    1.82282   0.091    0.929   
B   0.50623    0.07225   7.006 6.19e-06 ***
C  49.06767   29.33499   1.673    0.117   
D 100.00000   15.92206   6.281 2.02e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.165 on 14 degrees of freedom

(In)accuracy (%) = A
        x      xhat        A
   0.1000   0.09704  -2.9620
   0.1000   0.10400   4.0080
   0.1000   0.08974 -10.2600
   0.3981   0.38230  -3.9640
   0.3981   0.35890  -9.8370
   0.3981   0.39070  -1.8490
   1.5850   1.81600  14.6000
   1.5850   1.83200  15.5500
   1.5850   1.60700   1.3970
   6.3100   6.21800  -1.4610
   6.3100   5.47700 -13.2000
   6.3100   6.28000  -0.4722
  25.1200  23.43000  -6.7150
  25.1200  23.14000  -7.8690
  25.1200  29.89000  19.0000
 100.0000 122.20000  22.1700
 100.0000  94.13000  -5.8720
 100.0000  88.14000 -11.8600

(Im)precision (%) = P
        x mean (xhat)      P
   0.1000      -3.073 -3.073
   0.3981      -5.216 -5.216
   1.5850      10.520 10.520
   6.3100      -5.044 -5.044
  25.1200       1.471  1.471
 100.0000       1.480  1.480

Here the model without anchor points performed better based on the AIC. Not so sure if based on the A&P. If you run the script numerous times, you will find cases where it is the other way around or the AIC agree about the ‘best’ model.

Homework: Given the small value of the parameter A and its large p-value in both models, one could remove it and try the simplified model:$$y=D+\frac{D}{1+\left(\frac{x}{C}\right)^B}$$
[image] simulation script

LLOQ   <- 0.1
ULOQ   <- 100
CV     <- 0.05 # 5% multiplicative error
reps   <- 3    # Replicates at each level
calibs <- 6    # Levels of calibrators
# Parameters of the 4-parameter logistic function for simulations
A      <- 0    # Zero concentration response
B      <- 0.5  # Slope
C      <- 50   # Inflection point
D      <- 100  # Infinite concentration response
# Calibrators equally spaced in log-scale

calib  <- sort(rep(signif(exp(seq(log(LLOQ), log(ULOQ), length.out = calibs)), 4), reps))
first  <- unique(head(calib, 2 * reps))
last   <- unique(tail(calib, 2 * reps))
anchor <- signif(c(LLOQ / (first[2] / first[1]), ULOQ * (last[2] / last[1])), 4)
x      <- sort(c(rep(anchor, reps), calib)) # Add anchor points
anchs  <- which(x %in% anchor)
# Mandatory parameter guesses (start), lower and upper bounds
guess  <- data.frame(type = c("start", "lower", "upper"),
                     A = c(15, 0, 50), B = c(0.5, 0.1, 1),
                     C = c(50, 5, 80), D = c(90, 50, 100))
logistic <- function(A, B, C, D, x) {       # Response
  D + (A - D) / (1 + (x / C)^B)
}
back <- function(x, A, B, C, D, obs) {      # For uniroot()
  logistic(A, B, C, D, x) - obs
}
y      <- logistic(A, B, C, D, x)           # Theoretical responses
# Add error
obs    <- rlnorm(n = length(y), meanlog = log(y) - 0.5 * log(CV^2 + 1),
                 sdlog = sqrt(log(CV^2 + 1)))
# Nonlinear least-squares model with anchor points
# Note: Only with algorithm = "port" bounds (parameter constraints) can be specified

m1     <- nls(obs ~ logistic(A, B, C, D, x), algorithm = "port",
                    start = as.list(guess[guess$type == "start", 2:5]),
                    lower = guess[guess$type == "lower", 2:5],
                    upper = guess[guess$type == "upper", 2:5],
                    control = list(tol = 1e-08))
x0     <- x[-anchs]   # Remove anchor points
obs0   <- obs[-anchs] # and their observations
# Nonlinear least-squares model without anchor points
m0     <- nls(obs0 ~ logistic(A, B, C, D, x0), algorithm = "port",
                     start = as.list(guess[guess$type == "start", 2:5]),
                     lower = guess[guess$type == "lower", 2:5],
                     upper = guess[guess$type == "upper", 2:5],
                     control = list(tol = 1e-08))
res1   <- res0 <- data.frame(x = x0, y = obs0, xhat = NA_real_, A = NA_real_)
for (j in seq_along(x0)) { # Back-calculate x and (in)accuracy
                           # No analytical solution for x, we need a numeric one
  res1$xhat[j] <- uniroot(back, interval = range(x), extendInt = "yes",
                          tol = 1e-8, A = coef(m1)[["A"]], B = coef(m1)[["B"]],
                          C = coef(m1)[["C"]], D = coef(m1)[["D"]],
                          obs = res1$y[j])$root
  res1$A[j] <- 100 * (res1$xhat[j] - res1$x[j]) / res1$x[j]
  res0$xhat[j] <- uniroot(back, interval = range(x), extendInt = "yes",
                          tol = 1e-8, A = coef(m0)[["A"]], B = coef(m0)[["B"]],
                          C = coef(m0)[["C"]], D = coef(m0)[["D"]],
                          obs = res0$y[j])$root
  res0$A[j] <- 100 * (res0$xhat[j] - res0$x[j]) / res0$x[j]
}
P1     <- P0 <- data.frame(x = unique(x0), xhat = NA_real_, P = NA_real_)
for (j in 1:nrow(P1)) {     # (Im)precision
  P1$xhat[j] <- mean(res1$xhat[res1$x == P1$x[j]])
  P1$P[j]    <- mean(res1$A[res1$x == P1$x[j]])
  P0$xhat[j] <- mean(res0$A[res0$x == P0$x[j]])
  P0$P[j]    <- mean(res0$A[res1$x == P0$x[j]])
}
res1$y       <- res0$y <- NULL # No more needed
names(P1)[2] <- names(P0)[2] <- "mean (xhat)"
plot(x, obs, ylim = c(0, max(obs)), type = "n", log = "x", axes = FALSE,
     xlab = expression(bolditalic(x)), ylab = expression(bolditalic(y)))
x.ax   <- axTicks(1, log = TRUE)
grid(nx = NA, ny = NULL)
abline(v = x.ax, lty = 3, col = "lightgrey")
abline(v = c(LLOQ, ULOQ), lty = 2)
axis(1, at = x.ax)
axis(2, las = 1)
axis(3, at = c(LLOQ, ULOQ), labels = c("LLOQ", "ULOQ"))
box(lwd = 2)
points(x,  obs,  pch = 21, cex = 1.25, col = "blue", bg = "#0000FF80")
points(x0, obs0, pch = 21, cex = 1.25, col = "red",  bg = "#FF000080")
new.x  <- exp(seq(log(min(x)),  log(max(x)),  length.out = 501))
new.x0 <- exp(seq(log(min(x0)), log(max(x0)), length.out = 501))
lines(new.x,  predict(m1, list(x  = new.x)),  lwd = 3, col = "blue")
lines(new.x0, predict(m0, list(x0 = new.x0)), lwd = 3, col = "red")
cat("Initial guesses, parameter bounds (constraints):\n"); print(guess, row.names = FALSE)
cat("Model with anchor points: AIC =", AIC(m1));    summary(m1)
cat("(In)accuracy (%) = A\n");  print(signif(res1, 4), row.names = FALSE)
cat("(Im)precision (%) = P\n"); print(signif(P1, 4),   row.names = FALSE)
cat("Model without anchor points: AIC =", AIC(m0)); summary(m0)
cat("(In)accuracy (%) = A\n");  print(signif(res0, 4), row.names = FALSE)
cat("(Im)precision (%) = P\n"); print(signif(P0, 4),   row.names = FALSE)


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ryraman2661
☆    

India,
2024-03-25 13:28
(18 d 02:39 ago)

@ Helmut
Posting: # 23921
Views: 244
 

 LBAs: Anchor points and model fitting

Thanks for you all for your inputs
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