Modeling the Disease Progression from Healthy to Overt Diabetes in ZDSD Rats

Abstract

Studying the critical transitional phase between healthy to overtly diabetic in type 2 diabetes mellitus (T2DM) is of interest, but acquiring such clinical data is impractical due to ethical concerns and would require a long study duration. A population model using Zucker diabetic Sprague–Dawley (ZDSD) rats was developed to describe this transition through altering insulin sensitivity (IS, %) as a result of accumulating excess body weight and β-cell function (BCF, %) to affect glucose-insulin homeostasis. Body weight, fasting plasma glucose (FPG), and fasting serum insulin (FSI) were collected biweekly over 24 weeks from ZDSD rats (n = 23) starting at age 7 weeks. A semi-mechanistic model previously developed with clinical data was adapted to rat data with BCF and IS estimated relative to humans. Non-linear mixed-effect model estimation was performed using NONMEM. Baseline IS and BCF were 41% compared to healthy humans. BCF was described with a non-linear rise which peaked at 14 weeks before gradually declining to a negligible level. A component for excess growth reflecting obesity was used to affect IS, and a glucose-dependent renal effect exerted a two- to sixfold increase on the elimination of glucose. A glucose-dependent weight loss effect towards the end of experiment was implemented. A semi-mechanistic model to describe the dynamics of glucose and insulin was successfully developed for a rat population, transitioning from healthy to advanced diabetes. It is also shown that weight loss can be modeled to mimic the glucotoxicity phenomenon seen in advanced hyperglycemia.

Appendix I

(a) Derivation of the Brody Growth Function as Differential Equation

The Brody growth function is formulated as

$$ WGT(t)=WG{T}_{\max }-\left(WG{T}_{\max }-WG{T}_0\right){e}^{-kt} $$

where WGT _max is the animal’s maximum weight, WGT ₀ is the animal’s birth weight, and k is the growth rate of the animal per unit of time t.

The Brody growth function is transformed to a differential equation for use in model analysis through the following proof:

Let

$$ a=WG{T}_{\max } $$

$$ b=WG{T}_0 $$

$$ WGT(t)=a-\left(a-b\right){e}^{-kt} $$

(1)

$$ \frac{dWGT}{dt}=-\left(a-b\right)\left(-k\right){e}^{-kt}=k\left(a-b\right){e}^{-kt} $$

(2)

Subtract a from both sides of (1):

$$ WGT(t)-a=-\left(a-b\right){e}^{-kt} $$

(3)

Multiply both sides of (3) by -1:

$$ -\left(WGT(t)-a\right)=\left(a-b\right){e}^{-kt} $$

(4)

Substitute (4) into (2) to obtain:

$$ \begin{array}{l}\frac{dWGT}{dt}=k\cdot -\left(WGT(t)-a\right)\hfill \\ {}\frac{dWGT}{dt}=-k\left(WGT(t)-a\right)\hfill \\ {}\frac{dWGT}{dt}=k\left(a-WGT(t)\right)\hfill \\ {}\frac{dWGT}{dt}= ka-k\cdot WGT(t)\hfill \\ {}\frac{dWGT}{dt}=k\cdot WG{T}_{max}-k\cdot WGT(t)\hfill \end{array} $$

(b) Derivation of the Natural Growth Rate of Brody Function

The Brody growth function is formulated as

$$ WGT(t)=WG{T}_{\max }-\left(WG{T}_{\max }-WG{T}_0\right){e}^{-kt} $$

where WGT _max is the animal’s maximum weight, WGT ₀ is the animal’s birth weight, and k is the growth rate of the animal per unit of time t.

We are interested in calculating the natural growth rate k, which is achieved by rearranging the Brody growth function as follows:

$$ \begin{array}{l}WGT(t)-WG{T}_{\max }=-\left(WG{T}_{\max }-WG{T}_0\right){e}^{-kt}\hfill \\ {}\frac{WGT(t)-WG{T}_{\max }}{\left(WG{T}_{\max }-WG{T}_0\right)}=-{e}^{-kt}\hfill \\ {}-\left(\frac{WGT(t)-WG{T}_{\max }}{\left(WG{T}_{\max }-WG{T}_0\right)}\right)={e}^{-kt}\hfill \\ {} ln\left(-\left(\frac{WGT(t)-WG{T}_{\max }}{\left(WG{T}_{\max }-WG{T}_0\right)}\right)\right)=-kt\hfill \\ {}k = \frac{- ln\left(-\left(\frac{WGT(t)-WG{T}_{\max }}{\left(WG{T}_{\max }-WG{T}_0\right)}\right)\right)}{t}\hfill \end{array} $$

The rats were at age 7 weeks at the start of experiment. Values of WGT_max and WGT₀ are obtained from the literature (12), and weight at 7 weeks is equal to the observed baseline weight (BLWT).

Substituting WGT _max = 840g, WGT(7) = BLWT = 220g, WGT ₀ = 6.1g, t = 7, we get:

$$ \begin{array}{l}k = \frac{- ln\left(-\left(\frac{220-840}{\left(840-6.1\right)}\right)\right)}{7}\\ {}k \approx 0.043\ {\mathrm{week}}^{-1}\end{array} $$

(c) Derivation of the Steady-State Quadratic Solution from Differential Equations of FSI and FPG

Assuming steady-state,

$$ 0=\frac{dFSI}{dt}=E{F}_B \cdot B{C}_0\cdot \left(FPG-3.5\right)\cdot Ki{n}_{FSI}-FSI\cdot Kou{t}_{FSI} $$

(1)

$$ 0=\frac{dFPG}{dt}=\frac{Ki{n}_{FPG}}{E{F}_S \cdot I{S}_0\cdot FSI}-FPG \cdot Kou{t}_{FPG}\cdot E{F}_{Urine} $$

(2)

Rearranging Eq.2 for FPG₀ (Note: Kin_FPG / Kout_FPG = KIOG and EF_Urine, EF_S, EF_B = 1 at baseline)

$$ FP{G}_0=\frac{KIOG}{I{S}_0\cdot FSI} $$

(3)

Substitute FPG from Eq.3 back to Eq. 1

$$ 0=B{C}_0 \cdot \left(\frac{KIOG}{I{S}_0\cdot FSI}-3.5\right)\cdot Ki{n}_{FSI}-FSI\cdot Kou{t}_{FSI} $$

(4)

Expand the brackets (Note: Kin_FSI/Kout_FSI = KIOI)

$$ 0=\frac{B{C}_0\cdot KIOI\cdot KIOG}{I{S}_0\cdot FSI}-3.5\cdot B{C}_0\cdot KIOI-FSI $$

(5)

Rearranging to get the quadratic equation form ax ² + bx + c = 0

$$ \begin{array}{l}0=FS{I}^2+3.5\cdot B{C}_0\cdot KIOI\cdot FSI-\frac{B{C}_0\cdot KIOI\cdot KIOG}{I{S}_0}\\ {}a = 1\\ {}b=3.5\cdot B{C}_0\cdot KIOI\\ {}c=-\frac{B{C}_0\cdot KIOI\cdot KIOG}{I{S}_0}\end{array} $$

(6)

FSI₀ can now be solved using the quadratic formula:

$$ FS{I}_0=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$

Note: only the positive root is of interest.

Substitute FSI₀ back into Eq 3 to calculate FPG₀:

$$ FP{G}_0=\frac{KIOG}{I{S}_0\cdot FS{I}_0} $$