The Respiratory Center

Breathing is controlled by a neural central pattern generator called the respiratory center. The respiratory center is composed of three major groups of neurons located at the base of the brain. The dorsal respiratory group, located in the dorsal portion of the medulla, mainly causes inspiration; the ventral respiratory group can cause either inspiration or expiration, depending upon which neurons in the group are stimulated; and the pneumotaxic center, located above the medulla in the superior portion of the pons, helps control the rate and pattern of breathing.

The basic rhythm of respiration is generated mainly by the dorsal group, by emitting repetitive bursts of inspiratory action potentials. While the basic cause of these bursts is unknown, in primitive animals, neural networks have been found in which one set of neurons stimulates a second set, which in turn inhibits the first set, leading to periodic bursting activity that lasts throughout the lifetime of the animal.

Action Potential

Figure 17.14 A mutual inhibition network for the control of respiration.

The inspiratory signal is said to be a ramp signal, as it begins slowly and increases steadily for about 2 seconds, whereupon it abruptly ceases for the next 3 seconds before a new cycle begins. During normal quiet breathing, the ventral respiratory group is almost totally inactive. Expiration results primarily from elastic recoil of the lungs and thoracic cage.

In addition to neural mechanisms operating entirely within the brain, reflex signals from the periphery also help control respiration. Located in the walls of the bronchi and bronchioles throughout the lungs are stretch receptors that transmit signals to the dorsal respiratory group. Thus, when the lungs become overly inflated, the stretch receptors activate a feedback response that switches off the inspiratory ramp and stops further inspiration. This reflex is called the Hering-Breuer inflation reflex.

The real mechanism for the generation of the respiratory pattern is not known. However, a speculative, qualitative model for a neural network that can control breathing can be built using two neurons, or clumps of neurons, that inhibit each other (von Euler, 1980; Wyman, 1977), as illustrated in Fig. 17.14. (An alternate model is suggested in Exercise 12.) We suppose that there are two neurons with time-dependent outputs (their firing rates) h and I2 governed by where F\ and F2 are related to the firing rates of inhibitory and excitatory inputs. For simplicity we assume that the arrangement is symmetric, so that the time constants of the neuronal output are the same, x\ = t2 = t. We further assume that the neurons have steady excitatory inputs, E\ and E2, respectively, and that they are cross-inhibited,

so that the output from neuron 1 inhibits neuron 2, and vice versa. Thus we take F1 = F(E1 — I2) and F2 = F(E2 — Ii). The function F(x) is zero for x < 0 (so that the input and output are never negative), and a positive, increasing function of x for x > 0. Thus, we have the system of differential equations r<~t + Ii = F(Ei — I2) dI2

At this point there is no feedback from the lungs.

Equations (17.59) and 17.60) are easily studied using phase-plane analysis. There are three different possible phase portraits depending on the relative sizes of E1 and E2, two of which are shown in Figs. 17.15 and 17.16. In what follows we assume that F' > 1 for all positive arguments, although this restriction can be weakened somewhat. If E2 is much larger than E1, so that E1 < F(E2) and E2 > F(E1), then, as shown in Fig. 17.15, there is a unique stable fixed point at I2 = F(E1),I1 = 0, in which neuron 2 is firing and neuron 1 is quiescent. If E1 is much larger than E2, then the reverse is true, namely, there is a unique stable fixed point at I1 = F(E2),I2 = 0, with neuron 1 firing and neuron 2 quiescent. There is an intermediate range of parameter values when E1 and E2 are similar in size, E1 < F(E2) and E2 < F(E1), shown in Fig. 17.16, for which there are three steady states, the two on the axes, and one in the interior of the positive quadrant. The third (interior) steady state is a saddle point, and is therefore unstable.

This neural network exhibits hysteresis. Suppose we slowly modulate the parameter E1. If it is initially small (compared to E2, which is fixed at some positive level), then neuron 2 fires steadily and inhibits neuron 1. As E1 is increased, this situation remains unchanged, even when E1 and E2 are of similar size, when two stable steady solutions exist. However, when E1 becomes sufficiently large, the steady-state solution at I1 = F(E2), I2 = 0 suddenly disappears, and the variables I1, I2 move to the opposite steady state at I2 = F(E1),I1 = 0. Now if E1 is decreased, when E1 is small enough there

■ F(Ei) E2 ¡1 Figure 17.15 Phase portrait for mutual inhibition

¡1=F(E1-|2) network with E < F(E2) and E2 > F(E,).

-Figure 17.16 Phase portrait for mutual inhibition

F(El) 1 network with E, < F(E2) and E2 < F(E,).

is a reverse transition back to the steady state at /i = F(E2), 12 = 0, completing the hysteresis loop.

To use this hysteresis to control breathing, we model the diaphragm as a damped mass-spring system driven by 11, the (firing rate) output from neuron 1, the inspiratory neuron:

We model the effect of the stretch receptors by a function f (x) that is a monotone increasing function of diaphragm displacement x. The stretch receptors are assumed to excite only neuron 2, so that the output variables are governed by

We could allow stretch receptors to inhibit neuron 1 as well.

With this model, oscillation of the diaphragm is assured if the time constant t is sufficiently small. The stretch receptors act to modulate the excitatory inputs, so that as the lung expands, they excite neuron 2. With E2 + f (x) sufficiently large, neuron 1, the inspiratory neuron, is switched off. With no inspiratory input, the lung relaxes, returning f (x) toward zero and decreasing the excitation to neuron 2. This removes the inhibition to neuron 1 and allows it to fire once again. Thus, if parameters are adjusted properly, the hysteresis loop is exploited, and the inspiration-expiration cycle is established. The oscillations are robust and easily established.

This oscillation can be externally controlled. For example, by increasing E2, the cycle can be stopped after expiration, whereas by increasing E1 the inhibition of the stretch receptors can be overridden and inspiration lengthened (as in, take a deep breath). Decreasing E1 shortens the inspiration time and can stop breathing altogether.

In Fig. 17.17 is shown a plot of the two inhibitory variables 11 and 12 (shown dashed) plotted as functions of time. Parameter values for this simulation were t = 1.0,m =

T~dt + Ii = F(Ei -12), dl , _, ^ T—2 + I2 = F(E2 - Il + f (x)).

0 20 40 60 80 100

Time

Figure 17.17 Neural output variables I1 and I2 (dashed) shown as functions of time.

0.5, p = 5.0,k = 1.0,Ei = 0.5,E2 = 0.3. The function F was specified as F(x) = for positive x and zero otherwise, and the stretch response curve was taken to be f (x) = x3/(1 + x3).

17.5 EXERCiSES

1. Give a "proper" mathematical derivation of (17.14) by introducing appropriate dimension-less parameters. What dimensionless parameter must be small for this approximation to be valid?

Dco2

2. (a) Develop a model of carbon dioxide and oxygen transport that includes the oxyhe-

moglobin buffering reaction and the effect of free hydrogen ions on the concentration of bicarbonate. Does the inclusion of proton exchange improve or hinder the rate at which oxygen and carbon dioxide are transported into or out of the blood?

(b) Estimate the overall effect of this exchange by assuming that the pH of pulmonary venous blood is about 0.04 lower than that of arterial blood.

3. Develop a model of CO2 transport that accounts for its competitive binding with Hb. What is the effect of this binding on total carbon dioxide and oxygen flux?

4. Construct a simple model for the total oxygen and total carbon monoxide in the blood.

Assume that the circulatory system is a well-mixed container and that oxygen is removed by metabolism, while oxygen is added and carbon monoxide eliminated during transport through the lungs. Use the models of Section 17.1.5 to determine reasonable transfer rate functions. Estimate the parameters of the model and use numerical computations to de termine the half-clearance times for elimination of carbon monoxide at different oxygen levels. How well does your model fit the experimental data shown in Table 17.3?

Table 17.3 Experimental half-clearance times for elimination of carbon monoxide from the blood (Pace et al., 1950; also see Exercise 4).

O2 in atm

Half-clearance time (min)

0.21

249

1.0

47

2.5

22

Table 17.4 Alveolar gas concentration and oxygen saturation at different altitudes. The last column shows the alveolar PO2 when breathing pure oxygen at atmospheric pressure. At this pressure, O2 saturation is 100%. (Guyton and Hall, 1996, Table 43-1, p. 550.)

Barometric

PO2

Alveolar

O2

Alveolar

Pressure

in air

PO2

Saturation

PO2

(in air)

(in air)

(in oxygen)

Altitude (ft)

(mm Hg)

(mm Hg)

(mm Hg)

(%)

(mm Hg)

0

760

159

104

97

673

10,000

523

110

67

90

436

20,000

349

73

40

73

262

30,000

226

47

18

24

139

5. Suppose the respiratory exchange rate is fixed. Show that there is a linear relationship between the alveolar carbon dioxide and oxygen partial pressures.

6. (a) Assume that regulatory mechanisms maintain the arterial oxygen partial pressure at

40 mm Hg and the ventilation-perfusion ratio at 1. Find the alveolar PO2 and the oxygen saturation, leaving the alveolus as a function of atmospheric PO2.

(b) Data are shown in Table 17.4 for breathing normal air or breathing pure oxygen. What assumption from part 6a is apparently wrong? From the data, determine the arterial oxygen partial pressure.

7. Devise a different model in which metabolism and ventilation are held fixed. How do the alveolar PO2 and O2 saturation vary as a function of atmospheric pressure?

8. Using data from Table 17.4, estimate the altitude at which incoming alveolar blood has zero PO2, at normal metabolism.

9. Determine the red blood cell count (concentration of hemoglobin) that is necessary to maintain constant venous oxygen partial pressure as a function of altitude at fixed metabolism.

10. Find the rate of carbon monoxide clearance as a function of external PO2, with fixed metabolism and ventilation.

11. (a) Determine the structure of stable steady solutions of equations (17.59-17.60) in the

(E1,E2) parameter plane using F(x) = for positive x and zero otherwise.

(b) Numerically simulate the system of equations (17.61-17.63) using the parameters in the text. Plot the function E2 + f (x) as a function of time in the above parameter plane to see how hysteresis is exploited by this system.

12. Consider the following as a possible model for the respiratory center. Two neural FitzHugh-Nagumo oscillators have inhibitory synaptic inputs, so that dvi r, . , .

for i = 1, 2. The synaptic input si is some neurotransmitter that is released when the opposite neuron fires:

and the amplitude of the release Xj decreases gradually when the neuron is firing, via dxi

(a) Simulate this neural network with f (v,w) = 1.35v(1 - |v2) - w, wOT(v) = tanh(5v), F(v) = 1(1 + tanh 10v), and with parameters tv = 5, ve = -2, as = 0.025, jis = 0.002, ax = 0.001, jx = 0.01, gs = 0.19.

(b) Give an approximate analysis of the fast and slow phase portraits for these equations to explain how the network works.

(c) How does this bursting oscillator compare with those discussed in Chapter 6?

(d) What features of this model make it a good model for the control of the respiratory system and what features are not so good?

CHAPTER 18

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