Tracer Kinetic Modeling: Basics and Concepts

Abstract

In nuclear medicine all studies are dynamic. This may seem like a controversial statement as most studies performed in nuclear medicine departments consist of a single static scan. However, even in this case, the temporal dimension plays an important role in terms of the time point for the scan after administration of the tracer as well as its length.

This chapter is dedicated to the memory of Prof. Lyn S Pilowsky.

Appendices

Appendix A – Compartmental Models

Expressions for the impulse response functions for the 1-TC and 2-TC models are derived below. L{·} represents the Laplace transform, Laplace-domain functions are identified with a tilde, and s is a complex Laplace domain variable.

1.1 The 1-TC Model

$$ \begin{array}{l} \frac{\text{d}}{{{\hbox{d}}t}}{C_T}(t) = {K_1}{C_p}(t) - {k_2}^{\prime \prime }{C_T}(t) \\ \hskip 4pc\Leftrightarrow \\ L\left\{ {\frac{\text{d}}{{{\hbox{d}}t}}{C_T}(t)} \right\} = L\left\{ {{K_1}{C_p}(t) - {k_2}^{\prime \prime }{C_T}(t)} \right\} \\ \hskip 4pc\Leftrightarrow \\ s{{\hat{C}}_T}(s) - {C_T}(0) = {K_1}{{\hat{C}}_p}(s) - {k_2}^{\prime \prime }{{\hat{C}}_T}(s) \\ \hskip 4pc\Rightarrow \\\end{array} $$

With initial condition, C _T (0)=0:

$$ {\hat{C}_T}(s) = \frac{{{K_1}}}{{s + {k_2}^{\prime \prime }}}{\hat{C}_p}(s) $$

$$ \begin{array}{l} \hskip 4pc\Leftrightarrow \\ {C_T}(t) = {K_1}{{\hbox{e}}^{ - {k_2}^{\prime \prime }t}} \otimes {C_p}(t) \\ \hskip 4pc\Leftrightarrow \\\end{array} $$

Impulse response function:

$$ {H_1}(t) = {K_1}{{\hbox{e}}^{ - {k_2}^{\prime \prime }t}} $$

1.2 The 2-TC Model

$$ \begin{array}{l} \left\{ \begin{array}{l} \frac{\text{d}}{{{\hbox{d}}t}}{C_{ND}}(t) = {K_1}{C_p}(t) - \left( {{k_2}' + {k_3}'} \right){C_{ND}}(t) + {k_4}{C_S}(t) \\ \frac{\text{d}}{{{\hbox{d}}t}}{C_S}(t) = {k_3}'{C_{ND}}(t) - {k_4}{C_S}(t) \\\end{array} \right. \\ \hskip -9pc\Leftrightarrow \\\left\{ \begin{array}{l} s{{\hat{C}}_{ND}}(s) - {C_{ND}}(0) = {K_1}{{\hat{C}}_p}(s) - ({k_2}' + {k_3}'){{\hat{C}}_{ND}}(s) + {k_4}{{\hat{C}}_S}(s) \\ s{{\hat{C}}_S}(s) - {C_S}(0) = {k_3}'{{\hat{C}}_{ND}}(s) - {k_4}{{\hat{C}}_S}(s) \\\end{array} \right. \\\end{array} $$

With initial conditions, C _ND (0) = C _S (0) = 0:

$$ \begin{array}{l} \hskip -8pc\Rightarrow \\\left\{ \begin{array}{l} \left( {s + {k_2}^\prime + {k_3}^\prime } \right){{\hat{C}}_{ND}}(s) = {K_1}{{\hat{C}}_p}(s) + {k_4}{{\hat{C}}_S}(s) \\ \left( {s + {k_4}} \right){{\hat{C}}_S}(s) = {k_3}'{{\hat{C}}_{ND}}(s) \\\end{array} \right. \\\end{array} $$

$$ \begin{array}{l} \hskip -9pc\Leftrightarrow \\\left\{ \begin{array}{l} {{\hat{C}}_{ND}}(s) = {\left( {s + {k_2}^\prime + {k_3}^\prime - \frac{{{k_3}^\prime {k_4}}}{{s + {k_4}}}} \right)^{ - 1}}{K_1}{{\hat{C}}_p}(s) \\ {{\hat{C}}_S}(s) = \frac{{{k_3}^\prime }}{{s + {k_4}}}{{\hat{C}}_{ND}}(s) \\\end{array} \right. \\\end{array} $$

$$ \begin{array}{l} \hskip 7pc\Rightarrow \\{{\hat{C}}_T}(s) \equiv {{\hat{C}}_{ND}}(s) + {{\hat{C}}_S}(s) = \left( {1 + \frac{{{k_3}'}}{{s + {k_4}}}} \right){{\hat{C}}_{ND}}(s) \\ = \left( {\frac{{s + {k_3}' + {k_4}}}{{s + {k_4}}}} \right)\left( {\frac{{s + {k_4}}}{{(s + {k_2}' + {k_3}')(s + {k_4}) - {k_3}'{k_4}}}} \right){K_1}{{\hat{C}}_p}(s) \\\end{array} $$

$$ \ = \left( {\frac{{s + {k_3}^\prime + {k_4}}}{{{s^2} + \left( {{k_2}^\prime + {k_3}^\prime + {k_4}} \right)s + {k_2}^\prime {k_4}}}} \right){K_1}{\hat{C}_p}(s) $$

Find poles:

$$ \begin{array}{l} {s^2} + ({k_2}' + {k_3}' + {k_4})s + {k_2}'{k_4} = 0 \\ \hskip 8pc\Leftrightarrow \\ s = - \frac{1}{2}\left( {{k_2}' + {k_3}' + {k_4} \mp \sqrt {{{{({k_2}' + {k_3}' + {k_4})}^2} - 4{k_2}'{k_4}}} } \right) \equiv - {\theta_{1,2}} \\ \hskip 7pc\sim \;\sim \;\sim \\\end{array} $$

Partial fraction expansion:

$$ \begin{array}{l}\frac{{{\phi_1}}}{{s + {\theta_1}}} + \frac{{{\phi_2}}}{{s + {\theta_2}}} = {K_1}\frac{{s + {k_3}' + {k_4}}}{{{s^2} + ({k_2}' + {k_3}' + {k_4})s + {k_2}'{k_4}}} \\\end{array}$$

$$ \begin{array}{l} \hskip -5pc\Leftrightarrow \\\left\{ \begin{array}{l} {\phi_1} = {K_1}\frac{{{k_3}' + {k_4} - {\theta_1}}}{{{\theta_2} - {\theta_1}}} \\{\phi_2} = - {K_1}\frac{{{k_3}' + {k_4} - {\theta_2}}}{{{\theta_2} - {\theta_1}}} \\\end{array} \right. \\\end{array} $$

$$ \sim \;\sim \;\sim \quad\quad \\{(16.27)} + {(16.28)} \\\Rightarrow \\ $$

$$ {\hat{C}_T}(s) = \left( {\frac{{{\phi_1}}}{{s + {\theta_1}}} + \frac{{{\phi_2}}}{{s + {\theta_2}}}} \right){\hat{C}_p}(s) $$

$$ \begin{array}{l} \hskip -6pc\Leftrightarrow \\{C_T}(t) = \left( {{\phi_1}{e^{ - {\theta_1}t}} + {\phi_2}{e^{ - {\theta_2}t}}} \right) \otimes {C_p}(t) \\ \hskip -6pc\Leftrightarrow \\\end{array} $$

Impulse response function:

$$ {H_2}(t) = {\phi_1}{e^{ - {\theta_1}t}} + {\phi_2}{e^{ - {\theta_2}t}} $$

Appendix B – Outcome Measures

2.1 Volume of Distribution

For the 1-TC model we obtain:

$$ \frac{\text{d}}{{{\hbox{d}}t}}{C_T}(t) = {K_1}{C_p}(t) - {k_2}^{\prime \prime }{C_T}(t) = 0 \\\Rightarrow \\{V_T} \equiv {\left[ {\frac{{{C_T}(t)}}{{{C_p}(t)}}} \right]_{eq}} = \frac{{{K_1}}}{{{k_2}^{\prime \prime }}} \\ $$

((16.31a))

for the 2-TC model:

$$ \left\{ \begin{array}{l} \frac{\text{d}}{{{\hbox{d}}t}}{C_{ND}}(t) = {K_1}{C_p}(t) - \left( {{k_2}^\prime + {k_3}^\prime } \right){C_{ND}}(t) + {k_4}{C_S}(t) = 0 \\\frac{\text{d}}{{{\hbox{d}}t}}{C_S}(t) = {k_3}^\prime {C_{ND}}(t) - {k_4}{C_S}(t) = 0 \\\end{array} \right. $$

$$ \begin{array}{l} \hskip -3pc\Rightarrow \\\left\{ \begin{array}{l} {\left[ {\frac{{{C_{ND}}(t)}}{{{C_p}(t)}}} \right]_{eq}} = \frac{{{K_1}}}{{{k_2}^\prime }} \\{\left[ {\frac{{{C_S}(t)}}{{{C_{ND}}(t)}}} \right]_{eq}} = \frac{{{k_3}^\prime }}{{{k_4}}} \\\end{array} \right. \\\end{array} $$

((16.31b))

$$ \begin{array}{l} \hskip -1pc\Rightarrow \\{V_T} \equiv {\left[ {\frac{{{C_{ND}}(t) + {C_S}(t)}}{{{C_p}(t)}}} \right]_{eq}} = {\left[ {\frac{{{C_{ND}}(t)\left( {1 + {k_3}'/{k_4}} \right)}}{{{C_p}(t)}}} \right]_{eq}} \cr = \frac{{{K_1}}}{{{k_2}'}}\left( {1 + \frac{{{k_3}'}}{{{k_4}}}} \right) \\\end{array} $$

and similarly for the 3-TC model:

$$ {V_T} \equiv {\left[ {\frac{{{C_F}(t) + {C_{NS}}(t) + {C_S}(t)}}{{{C_p}(t)}}} \right]_{eq}} = \frac{{{K_1}}}{{{k_2}}}\left( {1 + \frac{{{k_3}}}{{{k_4}}} + \frac{{{k_5}}}{{{k_6}}}} \right) $$

((16.31c))

2.2 Binding Potential

Expressions for calculating BP _F from different model parameters can be derived as shown below, using the identities k ₃ = k _on B _avail, k ₄ = k _off, k ₃′ = f _ND k ₃, k ₂′ = f _ND k ₂, and K ₁/k ₂ = f _p (from f _p [C _p ] _eq = [C _F ] _eq and K ₁[C _p ] _eq = k ₂[C _F ] _eq ):

$$ BP\equiv \frac{{{B_{\max }}}}{{{K_D}}} \geq B{P_F} \equiv \frac{{{B_{\rm{avail}}}}}{{{K_D}}} = \frac{{{B_{\rm{avail}}}{k_{\rm{on}}}}}{{{k_{\rm{off}}}}} = \frac{{{k_3}}}{{{k_4}}} \\={(1/f_{ND})(k_{3}\prime/k_4)=} \frac{1}{{{f_p}}}\frac{{{K_1}{k_3}\prime }}{{{k_2}\prime {k_4}}} \\ $$

Appendix C – Reference Tissue Models

3.1 The Simplified Reference Tissue Model

$$ \left\{ \frac{\text{d}}{{{\hbox{d}}t}}{C_T}(t) = {K_1}{C_p}(t) - {k_2}^{\prime \prime }{C_T}(t) \hfill \\\frac{\text{d}}{{{\hbox{d}}t}}{C_R}(t) = {}^R{K_1}{C_p}(t) - {}^R{k_2}^\prime {C_R}(t) \hfill \\ \right. \\\Leftrightarrow \\ $$

From (16.25):

$$ \left\{ {{\hat{C}}_T}(s) = \frac{{{K_1}}}{{s + {k_2}^{\prime \prime }}}{{\hat{C}}_p}(s) \hfill \\{{\hat{C}}_R}(s) = \frac{{{}^R{K_1}}}{{s + {}^R{k_2}^\prime }}{{\hat{C}}_p}(s) \hfill \\ \right. \\\Rightarrow \\ $$

$$ \left( {{\hbox{with}}\;{R_1} = \frac{{{K_1}}}{{{}^R{K_1}}}} \right): $$

$$ {{\hat{C}}_T}(s)= {R_1}\frac{{s + {}^R{k_2}^\prime }}{{s + {k_2}^{\prime \prime }}}{{\hat{C}}_R}(s) \\= {R_1}{{\hat{C}}_R}(s) + {R_1}\frac{{{}^R{k_2}^\prime - {k_2}^{\prime \prime }}}{{s + {k_2}^{\prime \prime }}}{{\hat{C}}_R}(s) \\\Rightarrow \\{C_T}(t)= {R_1}{C_R}(t) + {R_1}\left( {{}^R{k_2}^\prime - {k_2}^{\prime \prime }} \right){{\hbox{e}}^{ - {k_2}^{\prime \prime }t}} \otimes {C_R}(t) \\\Leftrightarrow \\ $$

Impulse response function:

$$ {H_{1R}}(t) = {R_1}\delta (t) + {R_1}\left( {{}^R{k_2}^\prime - {k_2}^{\prime \prime }} \right){{\hbox{e}}^{ - {k_2}^{\prime \prime }t}} $$

where δ(t) is the Dirac delta-function.

3.2 The Full Reference Tissue Model

$$ \left\{ \begin{array}{l} \frac{\text{d}}{{{\hbox{d}}t}}{C_{ND}}(t) = {K_1}{C_p}(t) - \left( {{k_2}' + {k_3}'} \right){C_{ND}}(t) + {k_4}{C_S}(t) \\ \frac{\text{d}}{{{\hbox{d}}t}}{C_S}(t) = {k_3}'{C_{ND}}(t) - {k_4}{C_S}(t) \\ \frac{\text{d}}{{{\hbox{d}}t}}{C_R}(t) = {}^R{K_1}{C_p}(t) - {}^R{k_2}'{C_R}(t) \\\end{array} \right. $$

From (16.25) and (16.29):

$$ \begin{array}{l} \left\{ \begin{array}{l} {{\hat{C}}_T}(s) = \left( {\frac{{{\phi_1}}}{{s + {\theta_1}}} + \frac{{{\phi_2}}}{{s + {\theta_2}}}} \right){{\hat{C}}_p}(s) \\{{\hat{C}}_R}(s) = \frac{{{}^R{K_1}}}{{s + {}^R{k_2}'}}{{\hat{C}}_p}(s) \\\end{array} \right. \cr \hskip 5pc\Rightarrow \\{{\hat{C}}_T}(s) = \left( {\frac{{{\phi_1}}}{{s + {\theta_1}}} + \frac{{{\phi_2}}}{{s + {\theta_2}}}} \right)\frac{{s + {}^R{k_2}'}}{{{}^R{K_1}}}{{\hat{C}}_R}(s) \cr \hskip 5pc\Leftrightarrow \\\end{array} $$

$$ \left( {{\hbox{with}}\;{R_1} = \frac{{{K_1}}}{{{}^R{K_1}}}} \right): $$

$$ \begin{array}{l} \left\{ \begin{array}{l} {{\hat{C}}_T}(s) = \left( {\frac{{{\phi_1}}}{{s + {\theta_1}}} + \frac{{{\phi_2}}}{{s + {\theta_2}}}} \right){{\hat{C}}_p}(s) \\{{\hat{C}}_R}(s) = \frac{{{}^R{K_1}}}{{s + {}^R{k_2}'}}{{\hat{C}}_p}(s) \\\end{array} \right. \cr \hskip 5pc\Rightarrow \\{{\hat{C}}_T}(s) = \left( {\frac{{{\phi_1}}}{{s + {\theta_1}}} + \frac{{{\phi_2}}}{{s + {\theta_2}}}} \right)\frac{{s + {}^R{k_2}'}}{{{}^R{K_1}}}{{\hat{C}}_R}(s) \cr \hskip 5pc\Leftrightarrow \\\end{array} $$

$$ {\hat{C}_T}(s) = {R_1}{\hat{C}_R}(s) + \left( {\frac{{{\rho_1}}}{{s + {\theta_1}}} + \frac{{{\rho_2}}}{{s + {\theta_2}}}} \right){\hat{C}_R}(s) $$

where

$$ \left\{ \openup2pt{\rho_1} = {R_1}\frac{{\left( {{k_3}\prime + {k_4} - {\theta_1}} \right)\left( {{}^R{k_2}\prime - {\theta_1}} \right)}}{{{\theta_2} - {\theta_1}}} \hfill \\{\rho_2} = - {R_1}\frac{{\left( {{k_3}\prime + {k_4} - {\theta_2}} \right)\left( {{}^R{k_2}\prime - {\theta_2}} \right)}}{{{\theta_2} - {\theta_1}}} \hfill \\ \right. $$

$$ \begin{array}{l} \hskip -9pc\sim \;\sim \;\sim \\ \hskip -8.2pc(34) \\ \hskip -8.1pc\Rightarrow \\{C_T}(t) = {R_1}{C_R}(t) + \left( {{\rho_1}{{\hbox{e}}^{ - {\theta_1}t}} + {\rho_2}{{\hbox{e}}^{ - {\theta_2}t}}} \right) \otimes {C_R}(t) \\ \hskip -8.2pc\Leftrightarrow \\\end{array} $$

Impulse response function:

$$ {H_{2R}}(t) = {R_1}\delta (t) + {\rho_1}{{\hbox{e}}^{ - {\theta_1}t}} + {\rho_2}{{\hbox{e}}^{ - {\theta_2}t}} $$

Appendix D – Logan Graphical Analysis

4.1 1-TC Model

$$ \frac{\text{d}}{{{\hbox{d}}t}}{C_T}(t)= {K_1}{C_p}(t) - {k_2}^{\prime \prime }{C_T}(t) \\\Leftrightarrow \\{C_T}(t) - {C_T}(0)= {K_1}\int\limits_0^t {{C_p}\left( \tau \right){\hbox{d}}\tau } - {k_2}^{\prime \prime }\int\limits_0^t {{C_T}\left( \tau \right){\hbox{d}}\tau } \\ $$

with C _T (0)=0:

$$ \begin{array}{l} \hskip 1pc\Leftrightarrow \\\frac{{\int_0^t {{C_T}\left( \tau \right){\text{d}}\tau } }}{{{C_T}(t)}} = \frac{{{K_1}}}{{{k_2}^{\prime \prime }}}\frac{{\int_0^t {{C_p}\left( \tau \right){\text{d}}\tau } }}{{{C_T}(t)}} - \frac{1}{{{k_2}^{\prime \prime }}} \\ \hskip 1pc\Rightarrow \\\end{array} $$

$$ \frac{{\int_0^t {{C_T}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} = {V_T}\frac{{\int_0^t {{C_p}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} + const. $$

4.2 2-TC Model

$$ \begin{array}{l} \left\{ \begin{array}{l} \frac{\text{d}}{{{\hbox{d}}t}}{C_{ND}}(t) = {K_1}{C_p}(t) - \left( {{k_2}' + {k_3}'} \right){C_{ND}}(t) + {k_4}{C_S}(t) \\ \frac{\text{d}}{{{\hbox{d}}t}}{C_S}(t) = {k_3}'{C_{ND}}(t) - {k_4}{C_S}(t) \\\end{array} \right. \\ \hskip 8pc\Leftrightarrow \\ \left\{ \begin{array}{l} {C_{ND}}(t) - {C_{ND}}(0) = {K_1}\int\limits_0^t {{C_p}(\tau ){\hbox{d}}\tau } - \left( {{k_2}' + {k_3}'} \right)\cr \qquad\qquad\qquad\qquad\times\int\limits_0^t {{C_{ND}}(\tau ){\hbox{d}}\tau } + {k_4}\int\limits_0^t {{C_S}(\tau ){\hbox{d}}\tau } \\ {C_S}(t) - {C_S}(0) = {k_3}'\int\limits_0^t {{C_{ND}}(\tau ){\hbox{d}}\tau } - {k_4}\int\limits_0^t {{C_S}(\tau ){\hbox{d}}\tau } \\\end{array} \right. \\\end{array} $$

with C _ND (0)=C _S (0)=0:

$$ \begin{array}{l} \hskip 6.8pc\Leftrightarrow \\ \left\{ \begin{array}{l} {C_T}(t) = {K_1}\int\limits_0^t {{C_p}(\tau ){\hbox{d}}\tau } - {k_2}'\int\limits_0^t {{C_{ND}}(\tau ){\hbox{d}}\tau } \\ {C_S}(t) = {k_3}'\int\limits_0^t {{C_{ND}}(\tau ){\hbox{d}}\tau } - {k_4}\int\limits_0^t {{C_S}(\tau ){\hbox{d}}\tau } \\\end{array} \right. \\\end{array} $$

$$ \begin{array}{l} \hskip 6.8pc\Leftrightarrow \\ \left\{ \begin{array}{l} \frac{{\int\limits_0^t {{C_{ND}}(\tau )d\tau } }}{{{C_T}(t)}} = \frac{{{K_1}}}{{{k_2}'}}\frac{{\int\limits_0^t {{C_p}(\tau )d\tau } }}{{{C_T}(t)}} - \frac{1}{{{k_2}'}} \\ \frac{{\int\limits_0^t {{C_S}(\tau )d\tau } }}{{{C_T}(t)}} = \frac{{{k_3}'}}{{{k_4}}}\frac{{\int\limits_0^t {{C_{ND}}(\tau )d\tau } }}{{{C_T}(t)}} - \frac{1}{{{k_4}}}\frac{{{C_S}(t)}}{{{C_T}(t)}} \\\end{array} \right. \\\end{array} $$

$$ \begin{array}{l} \hskip 5pc\Rightarrow \\\frac{{\int\limits_0^t {{C_T}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} \equiv \frac{{\int\limits_0^t {{C_{ND}}(\tau ) + {C_S}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} = \frac{{{K_1}}}{{{k_2}'}}\frac{{\int\limits_0^t {{C_p}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} \cr \quad - \frac{1}{{{k_2}'}} + \frac{{{k_3}'}}{{{k_4}}}\left( {\frac{{{K_1}}}{{{k_2}'}}\frac{{\int\limits_0^t {{C_p}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} - \frac{1}{{{k_2}'}}} \right) \cr \quad- \frac{1}{{{k_4}}}\frac{{{C_S}(t)}}{{{C_T}(t)}} \\ = \frac{{{K_1}}}{{{k_2}'}}\left( {1 + \frac{{{k_3}'}}{{{k_4}}}} \right)\frac{{\int\limits_0^t {{C_p}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} - \frac{1}{{{k_2}'}}\left( {1 + \frac{{{k_3}'}}{{{k_4}}}} \right) \cr \quad- \frac{1}{{{k_4}}}\frac{{{C_S}(t)}}{{{C_T}(t)}} \\\end{array} $$

$$ \begin{array}{l} \hskip -7.5pc\Leftrightarrow \\\frac{{\int\limits_0^t {{C_T}\left( \tau \right){\text{d}}\tau } }}{{{C_T}(t)}} = {V_T}\frac{{\int\limits_0^t {{C_p}\left( \tau \right){\text{d}}\tau } }}{{{C_T}(t)}} - \frac{1}{{{k_2}^{\prime \prime }}} - \frac{1}{{{k_4}}}\frac{{{C_S}(t)}}{{{C_T}(t)}} \\ \hskip -7.5pc\Rightarrow \\\end{array} $$

with C _S (t)/C _T (t)=constant (pseudoequilibrium):

$$ \frac{{\int\limits_0^t {{C_T}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} = {V_T}\frac{{\int\limits_0^t {{C_p}(\tau ){\text{d}}\tau } }}{{{C_T}(t)}} + {\hbox{const}}. $$