Afferent bladder nerve activity in the rat: a mechanism for starting and stopping voiding contractions

Abstract

The objective of this work was to study the relation between afferent bladder nerve activity and bladder mechanics and the mechanisms that initiate and terminate bladder contractions. Bladder nerve activity, pressure and volume were recorded during the micturition cycle in the rat. The highest correlation was found between afferent nerve activity and stress (pressure×volume). Afferent nerve activity depended linearly on stress within 6%, and both slope and offset were independent of the bladder-filling rate. The levels of afferent bladder nerve activity at the onset and cessation of efferent firing to the bladder were highly reproducible with coefficients of variation of ≤17%. We propose a model in which afferent activity is proportional to bladder wall stress, and bladder contraction is initiated when afferent activity exceeds a threshold due to an increasing pressure and volume. The contraction continues until afferent activity drops below a threshold again as a result of a decreasing volume.

Appendix

Calculation of bladder wall tension (T) and stress (σ).

The bladder is approximated to a sphere with radius R (Fig. 9):

Fig. 9

Calculation of bladder wall tension and stress. The bladder is approximated to a sphere with radius R. The force that drives the two semi-spheres apart is balanced by the force keeping them together: F_tension. Tension is calculated as F_tension/(circumference of cross section) and stress as F_tension/(area of cross section)

Tension

The ring area, A_ring, can be calculated as

$$ A_{{ring}} = 2\pi \cdot R \cdot \cos \alpha \cdot R \cdot d\alpha = 2\pi R^{2} \cdot \cos \alpha \cdot d\alpha $$

(A1)

The force on this ring, F_ring equals P·A_ring. P is bladder pressure.

The vertical component, F_ring,_up, equals F_ring·sinα. Thus:

$$ F_{{ring,up}} = 2\pi \cdot R^{2} \cdot P \cdot \cos \alpha \cdot \sin \alpha \cdot d\alpha $$

(A2)

Integration yields the force that drives the two semispheres apart, F_up:

$$ F_{{up}} \begin{array}{*{20}c} {{ = {\int\limits_0^{\pi /2} {\pi \cdot R^{2} } }P \cdot 2\sin \alpha \cos \alpha \cdot d\alpha }} \\ {{\left. { = \pi \cdot R^{2} P\sin ^{2} \alpha } \right|^{{\pi /2}}_{0} }} \\ {{ = \pi \cdot R^{2} \cdot P}} \\ \end{array} $$

(A3)

The force that keeps both semispheres together, F_tension, equals F_up. Wall tension is defined as F_tension divided by the circumference of the cross section:

$$ T = \frac{{\pi \cdot R^{2} P}} {{2\pi \cdot R}} = P \cdot R/2 $$

(A4)

Stress

Stress (σ) is defined as F_tension divided by the area of the cross section. If d is the wall thickness, this area equals π(R+d)²−πR²=2πRd+πd². Assuming that d<<R yields:

$$ \sigma _{{thinwall}} = \frac{{\pi \cdot R^{2} P}} {{2\pi \cdot Rd}} = \frac{{PR}} {{2d}} $$

(A5)

However, when the bladder is almost empty this approximation is not valid.

With a non-negligible wall thickness, the tangential stress in the wall at radius r can be calculated [41]:

$$ \sigma {\left( r \right)} = \frac{{P_{{in}} r^{3}_{{in}} - P_{{out}} r^{3}_{{out}} }} {{r^{3}_{{out}} - r^{3}_{{in}} }} + \frac{{r^{3}_{{in}} r^{3}_{{out}} }} {{2r^{3} }} \cdot \frac{{P_{{in}} - P_{{out}} }} {{r^{3}_{{out}} - r^{3}_{{in}} }} $$

(A6)

$$ \sigma {\left( r \right)} = \frac{{P_{{in}} r^{3}_{{in}} }} {{2r^{3} }} \cdot \frac{{r^{3}_{{out}} + 2r^{3} }} {{r^{3}_{{out}} - r^{3}_{{in}} }} - \frac{{P_{{out}} r^{3}_{{out}} }} {{2r^{3} }} \cdot \frac{{r^{3}_{{in}} + 2r^{3} }} {{r^{3}_{{out}} - r^{3}_{{in}} }} $$

(A7)

With P_in=pressure inside sphere; P_out=pressure outside sphere; r_in=inner radius; r_out=R=outer radius.

With V=bladder volume and V_t=tissue volume:

$$ V_{t} = \raise0.5ex\hbox{$\scriptstyle 4$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 3$}\pi {\left( {r^{3}_{{out}} - r^{3}_{{in}} } \right)} \Leftrightarrow {\left( {r^{3}_{{out}} - r^{3}_{{in}} } \right)} = \frac{{3V_{t} }} {{4\pi }} $$

(A8)

$$ V = \raise0.5ex\hbox{$\scriptstyle 4$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 3$}\pi r^{3}_{{in}} \Leftrightarrow r^{3}_{{in}} = \frac{{3V}} {{4\pi }} $$

(A9)

with (A.7), (A.8) and (A.9):

$$ \sigma {\left( {r_{{in}} } \right)} = \frac{{3P_{{in}} }} {{2V_{t} }} \cdot {\left( {V + \frac{{V_{t} }} {3}} \right)} - \frac{{3P_{{out}} }} {{2V_{t} }}{\left( {V + V_{t} } \right)} $$

(A10)

$$ \underline{{\sigma {\left( {r_{{out}} } \right)} = \frac{{3P_{{in}} }} {{2V_{t} }} \cdot {\left( V \right)} - \frac{{3P_{{out}} }} {{2V_{t} }}{\left( {V + \raise0.5ex\hbox{$\scriptstyle 2$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 3$}V_{t} } \right)}_{{}} }} _{{\; + }} $$

(A11)

$$ \sigma {\left( {r_{{out}} } \right)} = \sigma {\left( {r_{{in}} } \right)} - \raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}P_{{in}} + \raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}P_{{out}} = \sigma {\left( {r_{{in}} } \right)} - \raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}P $$

(A12)

Mean stress in bladder wall:

$$ \sigma = {\left[ {\sigma {\left( {r_{{out}} } \right)} + \sigma {\left( {r_{{in}} } \right)}} \right]}/2 $$

(A13)

from (A.12) and (A.13):

$$ \begin{array}{*{20}l} {\sigma \hfill} & { = \hfill} & {{\sigma {\left( {r_{{in}} } \right)} - \raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}P} \hfill} \\ {{} \hfill} & { = \hfill} & {{\frac{3} {{2V_{t} }}{\left( {P_{{in}} - P_{{out}} } \right)}V + \frac{1} {2}P_{{in}} - \frac{3} {2}P_{{out}} - \frac{1} {4}P} \hfill} \\ \end{array} $$

(A14)

$$ \begin{array}{*{20}c} {{}} & { = } & {{\frac{3} {{2V_{t} }} \cdot PV + \frac{1} {4}P - p_{{out}} }} \\ \end{array} $$

(A15)

Where P=P_in−P_out, and P_out is considered a constant. The term ¹/₄P becomes negligible if (3PV/2V_t)/0.25P>10, i.e. if 6 V/V_t>10.

In rats V_t averages 0.2 cm³. Thus, the pressure term is negligible if V>0.33 cm³ and should be accounted for only at low bladder volumes.

If nerve activity is proportional to stress then:

$$ \begin{array}{*{20}c} {{{\text{NA}} = a \cdot {\text{stress}} + b = a{\left( {3V/2V_{t} + \raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}} \right)} \cdot P - {\left[ {a \cdot P_{{out}} + b} \right]} = a{\left( {3V/2V_{t} - \raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}} \right)} \cdot P + {b}\ifmmode{'}\else$'$\fi}} \\ {{{\text{with }}{b}\ifmmode{'}\else$'$\fi = - a \cdot P_{{out}} - b}} \\ \end{array} $$

(A16)

Or under the above assumption:

$$ \begin{array}{*{20}c} {{NA = a{\left( {3/2V_{t} } \right)} \cdot PV + {b}\ifmmode{'}\else$'$\fi = {a}\ifmmode{'}\else$'$\fi \cdot PV + b}} \\ {{{\text{with }}{a}\ifmmode{'}\else$'$\fi = a{\left( {3/2V_{t} } \right)}}} \\ \end{array} $$

(A17)