Exercise 6 Dose linearity and dose proportionality by l50J1n6c


									      Exercise 6
Dose linearity and dose
          Objectives of the exercise

• To learn what is dose linearity vs. dose

• To document dose proportionality using ANOVA

• To test dose linearity/proportionality by linear

• To test and estimate the degree of dose
  proportionality using a power model and a
  bioequivalence approach
        Linearity: an overview

• In drug development, it is essential to
  determine whether the disposition a new
  drug are linear or nonlinear
      Linearity and stationary
• Basic PK parameters (F%, Cl, Vss,…) are
  usually independent of the dose (linearity)
  and repeated (continuous) administrations

• Otherwise they are
  – Dose dependent (non-linearity PK)
  – Time dependent (non-stationary PK)
       Why is it important for a drug
       company to recognize dose
          dependent kinetics?
• Drugs which behave non-linearly are difficult
  to use in clinics, specially if the therapeutic
  window is narrow (e.g.: phenytoin)
  – drug monitoring

    • Thus, EU guidelines required to
            document linearity
  Why is it important for a drug company to
   recognize dose dependent kinetics?

• Drug development is often stopped if non-linearity
  is observed for the usual therapeutic
  concentration range
 • Non linearity is often observed in toxicokinetics
! (higher doses are tested)
    – sophisticated data analysis
    – data interpretation: need to know whether non-linearity
      exists or not
    Linearity and order of a reaction

• Order 1 : linearity
  – In a linear system all processes
    (absorption, distribution, … are governed
    by a first order reaction

• Order < 1 : Michaelis-Menten
• Order zero : perfusion, implant
         Dose proportionality
• For a linear pharmacokinetic system, measures
  of exposure, such as maximal concentration
  (Cmax) or area under the curve from 0 to infinity
  (AUC), are proportional to the dose.
                 AUC  Dose
• This can be expressed mathematically as:

             AUC    Dose
Assessment of dose proportionality

1. Analysis of variance (ANOVA) on PK
   response normalized (divided) by dose

2. Linear regression (simple linear model
   or model with a quadratic component)

3. Power model
 ANOVA for dose proportionality

    AUCdose1 AUCdose 2 AUCdose 3
H0:                            ....
     Dose1    Dose2     Dose3

  If H0 not rejected, no evidence against DP
              Linear regression
• The classical approach to test DP is first to fit the
  PK dependent variable (AUC, Cmax…) to a
  quadratic polynomial of the form:

    Y    1 ( Dose )   2 ( Dose )                  2

    Where the hypothesis is whether beta2 and alpha equal or not zero.
       Dose non-proportionality is declared if either parameter is
                    significantly different from zero.
 • If only beta2 is not significantly different from 0, the
   simple linear regression is accepted.

        Y     (Dose)  
•where alpha is tested for zero equality.
   • If alpha equals zero, then Eq. 1 holds and dose
   proportionality is declared.
   •If alpha does not equal zero, then dose linearity (which is
   distinct from dose proportionality) is declared.
  Limits of the classical regression

• The main drawback of this regression approach
  is the lack of a measure that can quantify DP;
  also when the quadratic term is significant or
  when the intercept is significant but close to
  zero, we are unable to estimate the magnitude
  of departure from DP.
• This point is addressed with the power model
Power model
         Power model and dose
          proportionality (DP)
• An empirical relationship between AUC
  and dose (or C) is the following power
       Y  Exp ( )( Dose ) Exp ( )

In this model, the exponent (beta), i.e. the slope
is a measure of DP.
            Power model and dose
             proportionality (DP)
• Taking the LN-transformation leads to a linear equation
  and the usual linear regression can then applied to this

    log e (Y )     log e (dose)  
•where beta, the slope, measures the proportionality
between Dose and Y.
•If beta=0, it implies that the response is independent from
the dose
         •If beta=1, DP can be declared.
   Power model and DP:
A bioequivalence approach
       Power model and DP:
issue associated to the classical H0

• If an imprecise study lead to large
  confidence intervals around β,
  – You cannot reject the classical H0 and you
    can conclude to a DP but that is in fact
The test problem for BE
Bioequivalence : the test problem

 From a regulatory point of view the
 producer risk of erroneously rejecting
 bioequivalence is of no importance

 The primary concern is the protection of
 the patient (consumer risk) against the
 acceptance of BE if it does not hold true
Bioequivalence : the test problem
   Classical test of null hypothesis (I)

    H 0 : T - R =         or T = R

    H 1 : T - R          or T  R

  T and R : population mean for test and
      reference formulation respectively

Decision on the BE cannot be based on the
  classical null hypothesis
Classical statistical hypothesis:

      F%               Ref               Test
                       n=1000            n=1000


Statistically different for p  0.05 but actually
therapeutically equivalent
 Classical statistical problem : the

   F%                Ref                       Test
 100                 n=3                       n=3



Not statistically different for p  0.05 but
actually not therapeutically equivalent
Bioequivalence : the test problem
    Classical test of null hypothesis

• Can be totally misleading

• Acceptance of B.E. despite clinically relevant
difference between R and T formulation

• Rejection of B.E. despite clinically irrelevant
difference between R and T
Bioequivalence : the test problem
    Classical test of null hypothesis

  Use of the classical null hypothesis would
  encourage poor trials, with few subjects,
  under uncontrolled conditions to answer
  an irrelevant question
 Bioequivalence: the test problem
• The appropriate hypothesis

          H01                                   H02
       (Ref -test)                           (Ref -test)
                     q1                 q2        inequivalent

                          (Ref -test)
Observation          q1                 q2        equivalent
 Bioequivalence: the test problem
• The appropriate hypothesis

                       q1                 q2
           H01              (Ref -test)
                    5%                    5%
two unilateral "t" tests
  Can we reject H01?                 Can we also reject H02?

                 YES                       YES
 Two unilateral t test and a 90%

• From an operational point of view to
  perform 2 unilateral t-tests or to compute
  the 90% CI (of the ratio) lead to exactly
  the same conclusion.
   Decision procedures for the power model

 DP not                                     DP not
accepted                                   accepted
           1 the 90 % CI of the slope2
                 DP accepted       +125%
Power model: construction of a 90% CI
• If Y(h) and Y(l) denote the value of the dependent
  variable, like Cmax, at the highest (h) and lowest (l) dose
  tested, respectively, and the drug is dose proportional

                 Y ( h) h
                         Ratio
                 Y (l ) l
where Ratio is a constant called the maximal dose
 Dose proportionality is declared if the ratio of
     geometric means Y(h)/Y(l) equals Ratio
      Construction of a 90% CI
• The a priori acceptable confidence interval (CI)
  for the SLOPE (see Smith et al for explanation)
  is given by the following relationship:

          Ln (0.8)                       Ln (1.25 )
  1                    slope  1 
     Ln (dose _ ratio)               Ln (dose _ ratio)

  Here 0.8 and 1.25 are the critical a priori values suggested by
  regulatory authorities for any bioequivalence problem after a
                     data log transformation.
A working example
            Fist analysis:
        an ANOVA to test H0

• Conclusion of the ANOVA: in the present
  experiment, there was no evidence
  against the null hypothesis of BPA dose
  proportionality for BPA doses ranging from
  2.3 and 100000 µg/kg”
    Linear regression analysis
•   Unweighted vs. weighted simple
    linear regression
Power model: raw data
                    Power model







     0      2          4            6          8      10         12

Observed Y and Predicted Y for the power (linear log-log ) model with
data corresponding to doses ranging from 2 to 100 000µg/kg (log-log
scale) ; visual inspection of figure 7 gives apparent good fitting.
                  Power model








       0      2            4              6          8          10      12

X vs. weighted (w=1) residual Y for a log-log linear power model
with data corresponding to doses ranging from 2 to 100 000µg/kg;
inspection of figure 8 indicates appropriate scatter of residuals (no
bias, homoscedasticity)
The univariate CI for the SLOPE (0.9026-1.030) as computed by WinNonLin is
     a 95% CI computed with the critical ‘t’ value for 20 ddl i.e. t=2.086.
 To compute a 90% CI i.e. (1-2*alpha) 100%, the critical “t” for 20 ddl is 1.725
  and the shortest 90% CI of the SLOPE is 0.9137-1.019; this is the classical
           shortest interval computed for a bioequivalence problem.
 a priori confidence interval for BPA
               dose ratio

              Ln (0.8)                       Ln (1.25 )
    1                      slope  1 
         Ln (dose _ ratio)               Ln (dose _ ratio)

• the a priori confidence interval for this BPA dose
  ratio was 0.9794-1.0206 it can be concluded that
  both the 95 and the 90% CI for the SLOPE were
  not totally included in this a priori regulatory CI
  and then the BPA dose proportionality cannot be
  accepted (proved) for this range of BPA doses;

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