REGULATION OF RESPIRATION lecture note 200L [Autosaved]-1-1.ppt

Respiration: Introduction
 a process by which atmosp O is delivered to the tissues for metabolism;
₂
and CO produced by the tissues is discharged into the atmosp.
₂
The LUNG
 A collection of 500 million bubbles, each 0.3 mm in diameter.
The only organ except the heart that receives the whole of
circulation
2

Respiration: function
1. Respiratory functions
i. Transport of O₂
ii. Transport of CO₂
2. Non-respiratory functions
i. metabolic function;
a. Manufacture of Surfactant
b. conversion of angiotensin I to II,
ii. filteration; remove clots and other substances
4

iii. Defence/immunity: macrophages, IgA, removal of inhaled particles
iv. Reservoir for blood: about 1L
v. Regulation of body pH: getting rid of CO₂
vi. Synthetic: phospholipids, protein synthesis (collagen, elastin), proteases
from leucocytes, carbohydrates (mucopolysaccarides of bronchial
mucous)
5

FIG. 2: Comparison of pressures (mm Hg) in the pulmonary and systemic
circulations.
7

 PROBLEM OF THE LUNG:
1.Stability of alveoli – surfactant
2.Removal of inhaled particles:
- nose filter (large particles)
- mucus (small particles) – mucous glands, goblet cells
- tiny cilia (non in the alveoli)
- macrophages (alveoli)
- leucocytes
8

Respiration: types
 External: taking O and giving out of CO
₂ ₂
 Internal: utilization of the O for cellular metabolism and the
₂
resultant production of energy, water and CO₂
9

Respiration: functional anatomy
 thoracic cage: ribs
 muscles of respiration
1. inspiration:
- Major m.: diaphragm, external intercostal
- accessory muscle: sternocleidomastoid, scalenes, serratus anterior,
levator scapulae, erectus spinae, pectoralis major and minor
2. expiration:
- passive in quiet breathing
- in strenous exercise: internal intercostal, accessory m. (abdominal recti
and posterior inferior serratus m.)
10

 air passages
◦ nasal cavity
◦ pharynx
◦ larynx
◦ trachea
◦ bronchi
◦ bronchioles
◦ terminal bronchioles
12

FIG. 4: Idealization of the human airways according to Weibel. Conducting
airways, the respiratory zone (or the transitional and respiratory zones). 13

FIG. 5: RESPIRATORY PASSAGES 14

 the LUNGS
 2 lungs
 left; 2 lobes, right; 3 lobes
 alveoli or air sacs; functional unit (about 300-500 million)
 blood supply and drainage
 capillary/alveolar membrane (blood-gas barrier)
NB: each RBC spends about 0.75 sec in the capillary network
18

capillary/alveolar membrane (blood-gas barrier)
◦ Histology of Lung Tissue
◦ Cells Types of the Alveoli
1. Type I alveolar cells
◦ – simple squamous cells where gas exchange occurs
2. Type II alveolar cells (septal cells)
– free surface has microvilli
– secrete alveolar fluid containing surfactant
3. Alveolar dust cells
– wandering macrophages remove debris
19

◦ Alveolar-Capillary Membrane
1.Respiratory membrane = 1/2 micron thick
2.Exchange of gas from alveoli to blood
- 4 Layers of membrane to cross
i. alveolar epithelial wall of type I cells
ii. alveolar epithelial basement membrane
iii. capillary basement membrane
iv. Endothelial cells of capillary
20

FIG. X: Details of Respiratory Membrane 21

 Pulmonary blood vessels:
- pulmonary artery
↓
- capillaries
↓
- pulmonary veins
22

 innervation:
◦ ANS
◦ The walls of the bronchi and bronchioles are innervated by the vagus n.
24

1. The blood-gas barrier is extremely thin with a very large area,
making it ideal for gas exchange by passive diffusion.
2. The conducting airways extend to the terminal bronchioles, with a
total volume of about 150 ml. All the gas exchange occurs in the
respiratory zone, which has a volume of about 2.5 to 3 liters.
3. Convective flow takes inspired gas to about the terminal
bronchioles; beyond this, gas movement is increasingly by diffusion in
the alveolar region.
4. The pulmonary capillaries occupy a huge area of the alveolar
wall, and a red cell spends about 0.75 second in them.
KEY CONCEPTS
26

Respiration: gas laws
1. Boyle’s law- the pressure (P) of a given mass of gas at constant temp is
inversely proportional to its volume (V). P ∞1/V
P V =P V (important in the mechanism of breathing)
₁ ₁ ₂ ₂
2. Dalton’s law of partial pressure: in a mixture of gases which do not act
chemically together, each gas exerts a PP which is the pressure it
would exert if it alone filled the containing vessel at the same T and P
27

3. Graham’s law of diffusion: the rates of diffusion (D) of two gases at the
same T and P are inversely proportional to the square roots of their
molecular weights (mw)
D √mw =D √mw
₁ ₁ ₂ ₂
4. Henry’s law: at equilibruim, the amount of gas dissolved in a given
volume of fluid at a given temp is proportional to PP of a gas in the gas
phase.
28

5. Fick’s law of diffusion: the rate of diffusion of a substance through a
membrane is directly proportional to the area (A) of the membrane, the
solubility (S) of the substance in the membrane, and the concentration
gradient (∆C) of the substance across the membrane and inversely
proportional to the thickness (t) of the membrane and the square root of
the mw of the substance.
rate of diffusion α A x S x ∆C/t x √mw
29

Respiration: mechanics of breathing
 refers to the study of factors involved in altering lung volume;
 muscular forces needed to expand the resp. system
 forces that impede expansion (resistance and elastance)
 determination of lung volume
NB: the mechanics of ventilation involve the dynamic
interaction of the lungs, chest wall and diaphragm
30

◦ Respiration occurs in two phases namely inspiration and expiration
◦ During normal quiet breathing, inspiration is the active process and
expiration is the passive process
◦ Respiratory muscles are of two types: Inspiratory muscles and Expiratory
muscles
◦ Lungs are under constant threat to collapse even in resting conditions
because of certain factors
- Elastic property of lung tissues
- Surface tension: It is the tension exerted by the fluid secreted from
alveolar epithelium on the surface of alveolar membrane
31

Alveoli collapse
 Factors preventing Alveoli collapse during Expiration;
1. surfactant
2. continuous outward pull by the chest wall
3.Intrapleural pressure
32

1. Surfactant
o present by 24th
week and secreted at about 30wks.
o a lipid surface-tension lowering agent
o Secreted by type II alveolar epithelial cells and Clara cells
o Located btw the fluid and air interface in the alveoli
◦ Surfactant that lines the epithelium of the alveoli in lungs is
known as pulmonary surfactant and it decreases the surface
tension on the alveolar membrane
◦ Surfactant is a lipoprotein complex formed by lipids especially
phospholipids, proteins and ions
33

o constituent;
 Phospholipids (about 75%)- dipalmitoylphosphatidylcholine
(DPPC)
◦ Proteins (specific surfactant protiens- SP- SPA, SPB, SPC and SPD
 other lipids (phosphatidylglycerol, neutral lipids, triglycerides
Ions- Ca₂⁺
34

o functions of surfactant
i. reduces surface tension force and hence ↑compliance in the alveoli
ii. stabilization of the alveoli
iii. opposes La place effect esp on smaller alveoli
iv. ↓ capillary filtration force
v. Plays a role in defense within the lungs against infection and
inflammation-Hydrophilic proteins SPA and SPD destroy the bacteria and
viruses by means of opsonization
NB: the magnitude of the force (P) produced by surface (T) in an alveolus
of known radius (R) is given law of Laplace
P = 2T
R
35

v. inflation of the lungs after birth
vi. Innate immunity
36

RDS
Pathology: Respiratory Distress Syndrome (RDS)
◦ Absence of surfactant in infants, causes collapse of lungs and the condition is
called respiratory distress syndrome or hyaline membrane disease
◦ It occurs in the newborn babies (especially premature) due to inadequate
formation of surfactant, resulting in an elevated alveolar surface tension
= lecithin : sphingomyelin ratio < 2
◦ Inspiratory difficulty
◦ ↑STF (surface tension force = ↓compliance
◦ Atelectesis (collapse of alveoli)
◦ pulmonary oedema (↑se capillary filtration force)
◦ This result into severe respiratory insufficiency and the infant may die
◦ Plasma levels of thyroid hormones and cortisol are low in infants with RDS
◦ Therapy of RDS includes administration of exogenous surfactant and
application of positive-end expiratory pressure (PEEP)
37

MECHANISM OF TIDAL RESPIRATION
◦ Inspiration
◦ Inspiration is an active process, normally produced by contraction of the
inspiratory muscles
◦ During tidal inspiration (quiet breathing), the diaphragm and external
intercostal muscles contract and cause increase in all the three
dimensions of thoracic cavity
◦ In tidal inspiration (quiet breathing), 70–75% of expansion of chest is
caused due to contraction of diaphragm
◦ When the diaphragm contracts following changes occur:
- The dome becomes flattened and the level of diaphragm is lowered
increasing the vertical diameter of the thoracic cavity
- During quiet breathing, the descent of diaphragm is about 1.5 cm and
during forced inspiration it increases to 7 cm
- The descent of diaphragm causes rise in the intraabdominal pressure
which is accommodated by the reciprocal relaxation of the abdominal
wall musculature. 38

◦ External intercostal muscles
- The fibres of external intercostal muscles slope downward and
forward
- They are attached close to the vertebral ends of the upper ribs
than the lower ribs
- From pivot-like joint with the vertebrae the ribs slope obliquely
downwards and forwards
- So, when the external intercostal muscles contract, the ribs are
elevated causing lateral and anteroposterior enlargement of
thoracic cavity
◦ Laryngeal muscles
- The abductor muscles of the larynx contract during inspiration
pulling the vocal cords apart
39

Expiration
◦Expiration in quiet breathing is largely a passive phenomenon and
is brought about by the:
-Elastic recoil of the lungs
-Decrease in size of the thoracic cavity due to relaxation of
diaphragm and external intercostal muscles
-An increase in the tone of muscle of the anterior abdominal wall
which forces the relaxing diaphragm upward and
-The serratus posterior inferior muscles play a minor role in pulling
down the ribs
40

MECHANISM OF FORCED RESPIRATION
◦ Forced inspiration
◦ Forced inspiration is characterized by:
- Forceful contraction of diaphragm leading to descent of diaphragm by 7–
10 cm as compared to 1–1.5 cm during quiet inspiration
- Forceful contraction of external intercostal muscles causing more
elevation of ribs leading to more increase in transverse and
anteroposterior diameter of thoracic cavity
- Contraction of accessory muscles of inspiration which cause the following
effects:
◦ Sternomastoid muscles contract and lift the sternum upwards,
◦ Anterior serrati muscles contract and lift many of the ribs upwards and
◦ Scaleni muscles contract and lift first two ribs.
41

◦ Forced expiration
◦ Forced expiration is required when respiration is increased during
exercise or in the presence of severe respiratory disease
◦ It is an active process caused as follows:
- Contraction of abdominal muscles (abdominal recti, transversus
abdominis, internal and external oblique)
◦ Compression of the abdominal contents which increases the
intra-abdominal pressure and forces the diaphragm upward
thereby reducing vertical diameter of the thoracic cavity
42

◦ Contraction of the internal intercostal muscles causes the effect
which is just opposite to that of the external intercostal muscles
◦ This is because of the leverage mechanism of the direction of the
muscle fibres which slope downward and backward creating a
longer force arm for the upper ribs
◦ Hence, their contraction tends to pull all the ribs downwards
reducing: anteroposterior diameter as well as the transverse
diameter
43

FIG. 9: RESPIRATION: AT REST 44

FIG. 10: RESPIRATION: INSPIRATION
45

A. Inspiration
 refers to ↑se in lung vol.,
 active force usually generated by the insp. M.
 lungs are passive structures that follow any movt of the chest wall
B. Expiration
 refers to decrease in lung vol, relaxation of diaphragm, elastic recoil of
the thoracic cage
 normally a passive process
47

RESPIRATORY PRESSURES
◦ Two types of pressures are exerted in the thoracic cavity and lungs during
process of respiration:
- Intrapleural pressure or intrathoracic pressure
- Intraalveolar pressure or intrapulmonary pressure
INTRAPLEURAL PRESSURE
◦ Intrapleural pressure is the pressure of fluid in the space between the
visceral pleura and parietal pleura
◦ Normal values are:
- At the end of normal inspiration: –6 mm Hg (760 – 6 = 754 mm Hg)
- At the end of normal expiration: –2 mm Hg (760 – 2 = 758 mm Hg)
- At the end of forced inspiration: –30 mm Hg
- At the end of forced inspiration with closed glottis (Müller maneuver): –70
mm Hg
- At the end of forced expiration with closed glottis (Valsalva maneuver):
+50 mm Hg.
48

◦ Throughout the respiratory cycle intrapleural pressure remains
lower than intraalveolar pressure
◦ This keeps the lungs always inflated
◦ Intrapleural pressure has two important functions:
1. It prevents the collapsing tendency of lungs
2. Because of the negative pressure in thoracic region, larger veins
and vena cava are dilated. Also, the negative pressure acts like
suction pump and pulls the venous blood from lower part of body
towards the heart against gravity. Thus, the intrapleural pressure is
responsible for venous return. So, it is called the respiratory pump
for venous return
49

INTRA-ALVEOLAR PRESSURE
◦Intra-alveolar pressure is the pressure existing in the alveoli of the lungs
◦Normally, intra-alveolar pressure is equal to the atmospheric pressure,
which is 760 mm Hg
◦It becomes negative during inspiration and positive during expiration
◦Normal values are:
-During normal inspiration: –1 mm Hg (760 – 1 = 759 mm Hg)
-During normal expiration: +1 mm Hg (760 + 1 = 761 mm Hg)
-At the end of inspiration and expiration: Equal to atmospheric pressure
(760 mm Hg)
-During forced inspiration with closed glottis (Müller maneuver): –80 mm Hg
-During forced expiration with closed glottis (Valsalva maneuver): +100
mm Hg
50

- Intraalveolar pressure causes flow of air in and out of alveoli
◦During inspiration, the intraalveolar pressure becomes negative, so
the atmospheric air enters the alveoli
◦During expiration, intraalveolar pressure becomes positive. So, air is
expelled out of alveoli.
-Intraalveolar pressure also helps in exchange of gases between
the alveolar air and the blood
◦N:B: Transpulmonary pressure is the pressure difference between
intraalveolar pressure and intrapleural pressure. It is the measure of
elastic forces in lungs, which is responsible for collapsing tendency
of lungs
51

C. Events of the respiratory cycle
 has 2 phases (I&E) that can be measured by a spirometer( measures and
record changes in lung volume)
NB: Compliance; - a measure of the distensibility (elasticity) of the lungs
and thoracic structures
- the vol change produced by a unit change of pressure
- obeys Hooke’s law (the extension of an elastic body – string - is proportional to the
applied force up to the elastic limit of the spring)
52

COMPLIANCE
◦ Distensibility or stretchability of lung and thorax is called
compliance
◦ Therefore, compliance is defined as the change in lung volume
(V) per unit change in transpulmonary pressure (P)
◦ Determination of compliance is useful as it is the measure of
stiffness of lungs
◦ Stiffer the lungs, less is the compliance
53

◦ Compliance(c)=change in volume
change in pressure
= litre/cmH O
₂
NB: pulmonary compliance/lung compliance=0.22L/cmH O
₂
thoracic cage compliance= 0.2L/cmH O
₂
total (lungs and cage)= 0.13L/cmH O
₂
NB: In men over 60yrs, compliance is about 25% greater than in
younger men, while very little change with age appears to
occur in women.
54

 factors affecting compliance
i. lung volume
ii. posture
iii. lung disease
iv. thoracic cage disease (kyphoscoliosis)
55

The Work of breathing
◦ Work of breathing is the work done by respiratory muscles during
breathing to overcome the resistance in thorax and respiratory tract
◦ During respiratory processes, inspiration is active process and the
expiration is a passive process
◦ So, during quiet breathing, respiratory muscles perform the work only
during inspiration and not during expiration
◦ 1. compliance work - work done to expand the lungs against its elastic
forces: work done to overcome elastic resistance
CW= increase in volume/increase in intra-pleural pressure
◦ tissue resistance work – work done to overcome tissue resistance
3. airway resistance work – work done to overcome airway resistance
during movement of air onto the lungs: Resistance increases during
bronchiolar constriction.
56

PULMONARY FUNCTION TESTS
◦ Pulmonary function tests or lung function tests are useful in assessing the
functional status of the respiratory system both in physiological and
pathological conditions
◦ Lung function tests are of two types:
1.Static lung function tests
- Include static lung volumes and static lung capacities
- These tests are based on volume of air that flows into or out of lungs
- These tests do not depend upon the rate at which air flows.
2. Dynamic lung function tests
- These tests are based on time, i.e. the rate at which air flows into or out of
lungs
- Useful in determining the severity of obstructive and restrictive lung
diseases
- These tests include forced vital capacity, forced expiratory volume,
maximum ventilation volume and peak expiratory flow
57

Pulmonary volumes and capacities
1. Tidal volume (TV) {500ml}; vol of air inspired or expired per breath during
quiet breathing.
2. Inspiration Reserve Volume (IRV) {3L}; vol of air that can be inhaled
above the normal TV using maximal inspiratory effort.
(range 2000–3200 mL)
58

3. Expiratory Reserve volume (ERV) {1L}; vol of air that can be
expelled after a normal tidal expiration using maximal
expiratory effort.
4. Residual volume (RV) {1.2L}; this is the vol of air remaining in the
lungs after a maximal expiration.
• TV= 500ml IRV= 3L ERV= 1L RV= 1.2L
59

Pulmonary volumes and capacities
◦ Inspiratory capacity (IC) {3.5L}- vol of air that is inhaled maximally after a
normal expiration. Therefore, it equals the tidal volume plus inspiratory reserve
volume (TV + IRV)
◦ Functional Residual capacity (FRC) {2.2L}- vol of air that remains in the lungs
after a normal expiration, (this air ensures that blood flowing
through the lungs during expiration is oxygenated).
Includes expiratory reserve volume and residual volume
60

◦ 3. Vital capacity (VC) {4.8L}- the maximal vol. of air that can be expired
from the lungs after the deepest possible inspiration. Therefore it equals tidal
volume plus inspiratory reserve volume plus expiratory reserve volume (TV +
IRV + ERV)
◦ Estimation of VC allows assessment of maximum inspiratory and expiratory
efforts and thus gives useful information about strength of the respiratory
muscles.
4. Total lung capacity (TLC) {6L}- the total volume of air contained in the lungs
at the end of a maximal inspiration
IC= 3.5L TLC= 6L
FRC= 2.2L
VC= 4.8L
N:B: All volumes and capacities except residual volume, functional residual
capacity and total lung capacity are recorded by a spirometer
61

SUMMARY
• TLC= IRV+ERV+RV+TV
• TLC= FVC+RV
• TLC= FRC+IC
• RR (NORMAL)= 12-15cpm
64

NB:
1. the above lung vol., and capacities represents
average values for normal young adults. Female
values are 20 – 35% lower than the male values
2. age affect the values, children have lower values
than young adults (because of reduced
compliance of the lungs)
3. body size affects the values and this may account
for the differences in different race
4. values in Nigerians are lower than caucasian values
◦ N:B: Volume and capacities, which cannot be
measured by spirometry, are measured by nitrogen
washout technique or helium dilution technique or
by body plethysmograph.
65

◦ Diseases of respiratory tract are classified into two types:
1. Restrictive respiratory disease
2. Obstructive respiratory disease
◦ These two types of respiratory diseases are determined by lung
functions tests, particularly FEV
66

Anatomical & physiological dead space
 Anatomical DS {150mls}- the anatomical structures that make up
the air passages in which no gaseous exchange occurs.
 the vol of air that reaches the alveoli per breath
is the tidal vol minus the dead space
 the value can vary with age and sex
 the ADS can be increased by breathing
through a tube
68

 Physiological DS- is the anatomic DS plus the vol of areas of the
lungs that are not taking part in gaseous exchange
 non-functional areas result from poor or absent
perfusion in a well ventilated lung
 PDS can be equal to or greater than the ADS.
 in a healthy individual anatomical DS is equal to
PDS.
69

Total ventilation (VT) and alveolar
ventilation (AV)
 VT- vol of fresh air moved into the respiratory system per minute
 VT=Tidal volume x RR
 Alveolar ventilation (VA)- is the vol of fresh air that enters the alveoli
per minute i.e the amount of air utilized for gaseous exchange every
minute.
 VA= (Tidal volume- dead space) x RR
= (TV- DS)x RR
71

 VT =Tidal volume x RR
 VT =500 x 15
=7,500ml
 VA = (Tidal volume- dead space) x RR
= (TV- DS)x RR
= (500 – 150) X 15
= 350 X 15
= 5,250ml
(2000ml loss in DS)
72

◦ NB:
◦ Ventilation- the movement of air into and out of the lungs
◦ Perfusion- the flow of blood through the lungs
◦ The lower zones of the lungs more ventilated due to gravity
73

FIG., 14: Measurement of regional differences in ventilation with radioactive
xenon. 74

Partial pressure of gases
1. Inspired air: 0 : 158.0, CO : 0.3
₂ ₂
2. Alveoli: 0 : 40.0, CO : 40.0
₂ ₂
3. Arteries (left heart): 0 : 95.0, CO : 40.0
₂ ₂
4. Capillaries: 0 : 40.0 , CO : 46.0
₂ ⁻ ₂ ⁺
5. Veins (right heart): 0 : 40.0, CO : 46.0
₂ ₂
6. Expired air: 0 : 116.0, CO : 32.0
₂ ₂
75

1. Transport of O₂
 the process by which atm. O gets to the tissues for use in metabolic
₂
processes
a. mov’t of oxygen from atm. air into the alveoli
b. diffusion of O in the blood from the alveolar sac into the blood in the
₂
pulmonary capillaries
c. transport of O in the blood from the lungs to the tissues
₂
d. delivery of O from the systemic capillary blood to the tissues
₂
76

Transport of O₂
 Oxygen is transported in two forms in the blood:
 total Oxygen content= 20% (20ml/100ml of blood)
1. O dissolved in plasma= 0.3% vol
₂
2. O carried in combination with Hb in the red blood cell
₂
(concentration/saturation)= 19.7% vol
77

1. O dissolved in plasma
₂
◦ plasma is a poor carrier of O . At a PO of 100mmHg, 100ml of
₂ ₂
plasma can carry only 0.3ml of O (poor solubility of oxygen in
₂
water content of plasma)
1. O diffuses from a region of high press in the alveolar sacs
₂
(100mmHg) into region of low press in the plasma (40mmg)
2. at PO 100mmHg in plasma (0.3ml/100ml), O is expose to RBC Hb,
₂ ₂
hence it diffuses from the plasma to the Hb.
3. Still, transport of oxygen in this form becomes important during the
conditions like muscular exercise to meet the excess demand of
oxygen by the tissues.
78

2. O carried in combination with Hb in the red blood cell
₂
i. Normal [Hb]= 14.5g/dl
ii. Each gram of Hb is capable of carrying 1.34ml of O at full saturation
₂
iii. Oxygen combines with the iron in heme part of hemoglobin (4 atoms of
iron)
iv. Normal hemoglobin content in blood is about 15 g%.
2. Since oxygen carrying capacity of hemoglobin is 1.34 mL/g, blood with 15
g% of hemoglobin should carry 20.1 mL% of oxygen, i.e. 20.1 mL of oxygen
in 100 mL of blood.
3. But, blood with 15 g% of hemoglobin carries only 19 mL% of oxygen, i.e. 19
mL of oxygen is carried by 100 mL of blood
4. Oxygen carrying capacity of blood is only 19 mL% because the
hemoglobin is not fully saturated with oxygen. It is saturated only for about
95%.
79

iv. Each molecule of Hb (which contains 4 unit of Hb) combines with four
molecule of O :-
₂
Hb +O →Hb O (slower)
₄ ₂ ₄ ₂
Hb +O →Hb O
₄ ₂ ₄ ₄
Hb +O →Hb O
₄ ₂ ₄ ₆
Hb +O →Hb O
₄ ₂ ₄ ₈
NB: → a reverse reaction, takes 0.01sec, a self catalytic rxn,
deoxygenation of Hb O is very fast
₄ ₈
◦ Saturation is the state or condition when hemoglobin is unable to hold or
carry any more oxygen.
◦ When all the four sites on haemoglobin are occupied by O2, then that
molecule of haemoglobin is 100% saturated.
◦ Saturation of hemoglobin with oxygen depends upon partial pressure of
oxygen.
80

Oxygen-Haemoglobin dissociation curve
1. the curve shows the degree of saturation of Hb at different
oxygen tension
2. enable us to know the additional O taken up by Hb when
₂
PO is increased by a known amount
₂
3. enable us to know how much O is given-up by Hb when O
₂ ₂
tension falls by a known amount
4. It explains hemoglobin’s affinity for oxygen.
5. Oxygen-hemoglobin dissociation curve is shifted to left or right
by various factors:
6. Shift to left indicates acceptance (association) of oxygen by
hemoglobin
7. Shift to right indicates dissociation of oxygen from hemoglobin.
81

FIG. 15: OXYHAEMOGLOBIN DISSOCIATION CURVE
82

Factors for shift to the right- release of O₂
i. Decrease in partial pressure of oxygen
ii. ↑se temp
iii. ↑se PCO₂
iv. ↑se acidity (fall in PH)
v. ↑se in 2,3 diphosphoglycerate (2,3-DPG)
Bohr Effect
◦ Bohr effect is the effect by which presence of carbon dioxide
decreases the affinity of hemoglobin for oxygen.
◦ Bohr effect was postulated by Christian Bohr in 1904.
◦ In the tissues, due to continuous metabolic activities in tissues, PCO is
₂
very high and the partial pressure of oxygen is low.
83

◦ Due to this pressure gradient, carbon dioxide enters the blood
and oxygen is released from the blood to the tissues
◦ Presence of carbon dioxide decreases the affinity of hemoglobin
for oxygen and enhances further release of oxygen to the tissues
and oxygen-dissociation curve is shifted to right
◦ All the factors, which shift the oxygen-dissociation curve to right
enhance the Bohr effect.
84

Factors for shift to the left- withholding of O₂
i. ↓se temp
ii. ↓PCO₂
iii. reduced acidity (↑se PH)
iv. ↓se 2,3-DPG
v. presence of foetal Hb
85

1. Oxygen transport in the blood
 through the pulmonary circulation (CO=5L/min), same as that in systemic
circulation
 blood takes about 1sec to traverse the pulmonary circulation which is
adequate for full oxygenation to occur
86

2. Oxygen delivery to the tissues
 arterial blood with a PO of 97mmHg diffuses to the tissues with a
₂
PO of 40mmHg
₂
 diffusion continues until the PO in the blood leaving the tissues
₂
(venous blood) is equal to that of the tissues i.e 40mmHg
Thus, the diffusion of oxygen from blood to tissues is 5 mL/100 mL
of blood.
87

CO transport
₂
1. CO is produced continuously in the body tissues (from cellular
₂
respiration)
2. the CO must be transported from the tissues to the lungs where it is
₂
expired
3. the transport of CO involves:
₂
a. diffusion of CO from the tissues (PCO , 46mmHg) into the blood
₂ ₂
b. transport of CO in the blood (PCO , 40mmHg) to the lungs
₂ ₂
c. diffusion of CO from the pulmonary capillaries into the alveoli
₂
d. movement of CO from the alveoli into the atmospheric air
₂
88

Transport of CO in blood
₂
 3 forms
1.as dissolved CO (10%)
₂
2. in combination with plasma proteins (NH group) and Hb-carbamino
₂
compound (30%)
3. as bicarbonate (60%)
CO + H O H CO → H + HCO
₂ ₂ ₂ ₃ ⁺ ₃⁻
↑
CA
NB: H CO (
₂ ₃ carbonic acid)
HCO (
₃⁻ bicarbonate)
89

NB:
 CA- carbonic anhydrase (found in RBC, speeds up rxn about 500x compared
with that in plasma)
 H is buffered by Hb and phosphate
⁺
 H CO (carbonic acid)
₂ ₃
 HCO (bicarbonate)- moves out of the Red cell and replaced by Cl from
₃⁻ ⁻
plasma (Cl shift or the Hamburger effect)
⁻
90

 the Hamburger effect:
◦ Chloride shift or Hamburger phenomenon is the exchange of a chloride ion for a
bicarbonate ion across RBC membrane.
◦ maintain electrical neutrality
◦ the HCO3− diffuses out of the RBCs into the plasma, the inside of the cells become
less negatively charged.
◦ Because the RBC membrane is relatively impermeable to cations, so in order to
neutralize this effect, negatively charged chloride ions (Cl−) diffuse from the plasma
into the RBCs to replace the HCO3−.
◦ When the negatively charged bicarbonate ions move out of RBC into the plasma,
the negatively charged chloride ions move into the RBC in order to maintain the
electrolyte equilibrium (ionic balance).
◦ This process is mediated by Band 3, a major ion exchange membrane protein.
◦ it results in a lower [Cl ] (2% lower in venous blood than in arterial blood)
⁻
◦ water follows the Cl into the RBC so that RBC in venous blood is about 3% bigger
⁻
than in arterial blood
◦ Bicarbonate ions combine with sodium ions in the plasma and form sodium
bicarbonate.
91

FIG. 16: Scheme of the uptake of CO2 and liberation of O2 in systemic capillaries.
92

CO dissociation curve
₂
◦ Carbon dioxide dissociation curve is the curve that demonstrates the
relationship between the partial pressure of carbon dioxide and the
quantity of carbon dioxide that combines with blood.
◦ The volumes of CO in 100 volume of blood is plotted against PCO
₂ ₂
Factors affecting the rate of CO transport
₂
i. Rate of tissue metabolism (determines rate of CO production)
₂
ii.Rate of blood flow
iii.Degree of deoxygenation of Hb at the tissue level
iv.Rate / depth of pulmonary ventilation
93

Haldane Effect
◦Haldane effect is the effect by which combination of oxygen with
hemoglobin displaces carbon dioxide from hemoglobin.
◦Excess of oxygen content in blood causes shift of the carbon dioxide
dissociation curve to right.
◦Due to the combination with oxygen, hemoglobin becomes strongly
acidic and causes displacement of carbon dioxide from hemoglobin in
two ways:
-Highly acidic hemoglobin has low tendency to combine with carbon
dioxide
- Because of the acidity, hydrogen ions are released in excess which bind
with bicarbonate ions to form carbonic acid.
◦Haldane effect is essential for:
1. Release of carbon dioxide from blood into the alveoli of lungs
2. Uptake of oxygen by the blood.
94

Diffusion: Fick´s law
◦ This states: that the rate of transfer of a gas through a sheet of tissue
is proportional to the tissue area and the difference in gas partial
pressure between the two sides, and inversely proportional to the
tissue thickness.
◦ the rate of transfer is proportional to a diffusion constant, which
depends on the properties of the tissue and the particular gas.
◦ The constant is proportional to the solubility of the gas and inversely
proportional to the square root of the molecular weight.
◦ CO₂ diffuses about 20 times more rapidly than does O₂ through tissue
sheets because it has a much higher solubility but not a very different
molecular weight.
95

FIG. 17: Diffusion through a tissue sheet
Area – 50 - 100 m²
Thickness – 0.3μm
V (gas) = DL * (P1 – P2)
DL = Diffusing capacity of the lung (area, thickness,
diffusion properties of the sheet and the gas concerned
96

CONTROL OF BREATHING
◦ A- Introduction
◦ B- Neural control of breathing
1- Medullary centers
2- Pontine centers
3- Central chemoreceptors (chemical control)
◦ C- Peripheral chemoreceptors
1- Carotid bodies
2- Aortic bodies
◦ D- Other respiratory reflexes/ non chemical influences on respiration
98

A- INTRODUCTION
The chief function of the lung is to exchange O₂ and CO₂ between blood
and gas
 spontaneous respiration is produced by rhythmic discharge of motor
neurons that innervate the respiratory muscles
 the rhythmic discharges from the brain are regulated by alteration in
arterial PO₂, PCO₂ and [H⁺] ion
 this remarkable regulation of gas exchange is made possible because the
level of ventilation is so carefully controlled
99

FIG. 18: Showing the 3 basic elements of respiratory control system
100

B- Neural control of breathing
◦ Two separate neural mechanisms regulate respiration:-
1- voluntary control
2- automatic control
◦ The voluntary system is located in the cerebral cortex and sends impulses
to the respiratory motor neurons via corticospinal tracts
◦ The automatic system is located in the pons and medulla and the efferent
output is located in the white matter of the spinal cord between the
lateral and ventral corticospinal tracts
101

◦ The nerve fibres mediating inspiration converge on the phrenic motor
neurons located in the ventral horns from C3 to C5 and external intercostal
motor neurons in the ventral horns throughout the thoracic cord
◦ The fibers concerned with expiration converge primarily on the internal
intercostal motor neurons in the thoracic cord
102

◦ 1- Medullary centers
 consist of two group of neurons:-
a. ventral respiratory group (VRG)
b. dorsal respiratory group (DRG)
(another group of cells in the ventrolateral region known as the pre-
Bottzinger complex also appears to be essential for the generation of the
respiratory rhythm)
103

◦ These centers are influenced by activity in :-
- pons
- reticular activating system (RAS)
- cerebral cortex and
- afferent activity in the vagus, glossopharyngeal and somatic
nerves
104

◦ A- DRG
- lie within and near the tractus solitarius
- primarily an inspiratory cell, contains I-neurons
-afferent input comes mainly from vagus and
glossopharyngeal n.
- activity influenced by low PO₂, high PCO₂, low PH, changes
in lung volume, RAS
- efferents go to the contralateral phrenic nerve, intercostal
neurons and to VRG
◦ All the neurons of dorsal respiratory group are inspiratory neurons
and generate inspiratory ramp by the virtue of their autorhythmic
property
◦ Dorsal group of neurons are responsible for basic rhythm of
respiration
105

◦ B- VRG
- a long column of neurons that extends through the nucleus
ambiguus and the nucleus retroambiguus in the ventrolateral medulla
- contains both I and E neurons
- comprises of the upper motor neurons of the vagus and the
nerves to the accessory muscles of respiration
◦ The VRG neurons normally remain totally inactive during quiet breathing.
◦ The VRG neurons become active during inspiration (role of I-neurons) as
well as expiration (role of E-neurons) during forceful respiration.
◦ Thus, VRG operates when high levels of pulmonary ventilation is
required, for example, during exercise.
106

◦ 2- Pontine centers (areas of the brainstem)
They modify the activities of medullary respiratory centers
 group into two:-
a- the pneumotaxic centre
b- the apneustic centre
107

a- the pneumotaxic centre
- lies in the upper part of the pons
- control the medullary respiratory centers, particularly the dorsal
group neurons.
- inhibits apneustic center
- inhibit or shortens inspiration leading to a slower and more
rapid respiratory rate, thus it regulate inspiration volume and
secondarily respiratory rate
- it fine tunes respiratory rhythm
108

b- the apneustic center
- lies in the lower part of the pons
- has an excitatory effect on the inspiratory area of the medulla,
hence it increases the duration of inspiration
- Apneustic center increases depth of inspiration by acting directly
on dorsal group neurons.
- inhibited by impulses carried in the vagus n. and also by the
activity in the pneumotaxic center
- loss of inhibitory activity on the apneustic center results in
prolonged periods of inspiration termed “apneusis”
109

◦ Chemoreceptors are the sensory nerve endings, which give response to
changes in chemical constituents of blood
◦ The chemical factors regulating respiration are pCO2, pO2 and pH of blood
◦ 3- Central chemoreceptors (chemical control)
- these are cells that lie just beneath the ventral surface of the medulla
and respond to [H⁺] ion in the cerebrospinal fluid (CSF) and the surrounding
interstitial fluid
Central chemoreceptors are connected with respiratory centers, particularly
the dorsal respiratory group
a- charged ions do not readily cross the endothelium of the blood vessels
in the brain (the blood brain barrier)
b- however, CO₂ does cross this barrier, since it is a small uncharged
molecule
110

Central chemoreceptors (chemical control)
- an increase in CO₂ stimulates the CCR after hydration to carbonic acid
and dissociation into Hydrogen and bicarbonate ions. Liberated H⁺ ion
stimulate the chemoreceptors
- the CCRs are surrounded by brain ECF and responds to changes in its [H⁺]
ion
- the composition of the ECF is governed by the CSF, local blood flow and
local metabolism
- The central chemoreceptors regulate the respiration from minute-to-minute
- Their stimulation leads to an increase in rate and depth of respiration
- It is important to note that about 70–80% of the resting respiratory drive is
due to the stimulatory effect of CO2 on the central chemoreceptors
111

FIG. 19: Showing the effect of BBB on the transport of charged ions
112

FIG. 20: Showing the effect of BBB on the transport of charged ions 113

◦ C- Peripheral chemoreceptors
- located in the carotid bodies at the bifurcation of the common
carotid arteries and in the aortic bodies above and below the aortic
arch
- Hypoxia is the most potent stimulant for peripheral chemoreceptors.
◦ - Due to the presence of oxygen sensitive potassium channels in
the glomus cells of peripheral chemoreceptors.
◦ 1- the carotid bodies
-contain glomus cells of two types, type I and type II cells
- they are rich in fenestrated sinusoidal capillaries
114

FIG 21: Showing location of carotid and aortic bodies
115

Peripheral chemoreceptors
- type I cells:-
- contain dopamine
- they are in close apposition to endings of the afferent carotid sinus
nerve
- type II cells:-
- they are glia-like and surrounds 4 or 6 type I cells
- functions probably as a support
◦ Hypoxia causes closure of oxygen sensitive potassium channels and
prevents potassium efflux.
◦ This leads to depolarization of glomus cells (receptor potential) and
generation of action potentials in nerve ending.
◦ These impulses pass through aortic and Hering nerves and excite the
dorsal group of neurons
◦ Dorsal group of neurons in turn, send excitatory impulses to respiratory
muscles, resulting in increased ventilation.
◦ This provides enough oxygen and rectifies the lack of oxygen.
116

FIG 22: Showing types of carotid bodies cell type
117

FIG 23: Showing carotid bodies cell type
118

Peripheral chemoreceptors
- the PCR responds to ;-
- low PO₂
- increased PCO₂ and
- increased [H+] ion or low PH in the arterial blood
NB: The response of the PCRs to arterial PCO₂ is less important than that of
the CCRs. However, their response is more rapid and they may be useful in
matching ventilation to abrupt changes in PCO₂.
◦ They regulate the respiration from breath to breath and their stimulation
increases the rate and depth of respiration.
119

Assignment 1
◦ discuss the mechanism in which the carotid bodies cells execute
their functions.
120

Other respiratory reflexes / non-chemical influences on respiration
• D. Other respiratory reflexes / non-chemical influences on
respiration
1.Lung receptors
a. pulmonary stretch receptors
- receptors in the airways and lungs are innervated by
myelinated and unmyelinated vagal fibers. The unmyelinated
fibers are C fibers
121

- the receptors innervated by the myelinated fibers are divided
into:-
> slowly adapting receptors and
> rapidly adapting receptors
- the unmyelinated C fibers are divided into:-
> pulmonary C fibers and
> bronchial C fibers
NB: pulmonary stretch receptors are also known as slowly
adapting pulm. Stretch recept
122

◦ the main reflex effect of stimulating these receptors is a slowing of
respiratory frequency due to an increase in expiratory time. This is known
as the Hearing-Breuer inflation reflex (H-B reflex).
◦ A protective reflex that restricts inspiration and prevents overstretching of
lung tissues.
◦ The H-B inflation reflex is an increase in the duration of expiration
produced by steady lung inflation and the H-B deflation reflex is a
decrease in the duration of expiration produced by marked deflation of
the lung
◦ NB: H-B inflation reflex [Increase expiration=steady lung inflation]
H-B deflation reflex [decrease expiration= lung deflation]
◦ However, Hering-Breuer reflex does not operate during quiet breathing. It
operates, only when the tidal volume increases beyond 1,000 mL.
123

2- Irritant receptors:-
- located in the large airways and are stimulated by smoke,
noxious gases, and particles
- these receptors initiate reflexes via vagus nerve (myelinated
rapidly adapting receptors) like coughing, bronchoconstriction, mucus
secretion and hyperapnea (forced respiration)
3- J-Receptors
- these are endings of non-myelinated C fibers that are close to
pulmonary vessels, hence the term “juxtacapillary” or J is used
- they are stimulated by distention of the pulmonary vessels, pulmonary
congestion, pulmonary oedema, and also by intravenous or intracardiac
administration of chemicals such as capsaicin
124

4- Bronchial C fibers
- these are supplied by bronchial circulation rather than the
pulmonary circulation
- they respond quickly to chemicals injected into the bronchial
circulation
- the reflex responses to stimulation include rapid shallow breathing,
bronchoconstriction and mucus secretion
5- Chest wall receptors
- detect force generated by the respiratory muscles during breathing
- in case of an excessive force on the chest these receptors gives rise
to the sensation of dyspnoea (shortness of breath)
125

◦ E- other receptors
1- Nose and upper airway receptors
- receptors respond to mechanical and chemical stimulation
- responses include sneezing, coughing, and bronchoconstriction
> coughing begins with a deep inspiration followed by a force
expiration against a close glottis that is suddenly opened
> sneezing is a similar expiratory effort with a continously open glottis
(these above reflexes help expel irritants and keep the airways clear)
2- Joint and muscle receptors
- impulses in afferent pathways from proprioceptors in muscles,
tendons and joints stimulate the inspiratory neurons especially during the
early part of exercise
126

3- Pain and temperature
- stimulation of many afferents nerves can bring about changes in
ventilation
- pain often causes a period of apnea followed by hyperventilation
- heating the skin may result in hyperventilation
◦ Further reading:
 exercise and respiration
 regulation of acid-base balance
 respiration at high altitude and deep sea diving
127

EXERCISE AND RESPIRATION
◦ Exercise - a condition that puts an increased energy demand on the
body.
◦ Response – increase in the supply of oxygen and nutrients to the tissues
and increased removal of metabolic waste (e.g., CO₂)
◦ Above response is achieve by:
1.↑sed pulmonary ventilation (↑se in the rate and depth of respiration) brought
about by stimulation of joint proprioceptors.
2.↑sed cardiac output [↑sed blood supply to the lungs (via pulmonary
circulation) and the tissues – via systemic circulation]
3.↑sed release of oxygen to the tissues (↑sed blood flow, ↑sed gradient for
oxygen diffusion from the capillaries to the tissues and dilatation of several
capillaries which lead to ↑sed oxygen delivery to the tissues).
128

EXERCISE AND RESPIRATION cont.,
◦ Sequence of ventilation in exercise
i. At onset of exercise – abrupt ↑se in ventilation, followed by a further,
more gradual increase
- due to psychic stimuli, afferent impulses from proprioceptors in
muscles, tendons, and joints (others includes chemoreceptors, body
temperature e.t.c)
ii. An abrupt ↓se in ventilation when exercise ceases, followed by a more
gradual decrease to pre-exercise level.
129

REGULATION OF ACID BASE BALANCE (pH REGULATION)
i. pH is a measure of the degree of acidity or alkalinity of a solution
ii. It is the -log of the hydrogen ion concentration of a solution
₁₀
iii. Physiological body pH = 7.4 (slightly alkaline) – maintained within narrow
limits.
iv. Enzymes function within the narrow limit of this normal body pH.
v. Factors tilting the body to acidity:
• Production of CO₂ from metabolic processes
• Metabolism of Sulphur-containing amino acids (cystine, cysteine) to
sulphate
• Metabolism of phosphorus to phosphate
• Production of lactic acid and pyruvic acid (exercise)
• Production of acetoacetic acid and beta hydroxybutyric acid (poorly
control Diabetes- ketones)
130

vi. Factors tilting the body to alkalinity:
•Hyperventilation
•Ingestion of large quantity of vegetables
•Vomiting of contents of stomach
vii. The process of H⁺ regulation involves 3 steps:
a.Chemical buffering by the extracellular and intracellular buffers (blood)
b.Control of CO₂ level in the blood by alteration in the rate of alveolar
ventilation (lungs) – excretion of C0₂
c.Control of the bicarbonate concentration in the blood by changes in
renal H⁺ excretion (kidney) – renal excretion of H⁺
131

◦ The pH equation (Henderson-Hasselbalch equation:
pH = pK + log [Base]
[Acid]
◦ Where K is the dissociation constant.
o To maintain a constant body pH, the ratio of [Base]:[Acid] must be
maintained constant
◦ NB: The lung excretes over 10,000 mEq of carbonic acid per day,
compared with less than 100 mEq of fixed acids by the Kidney.
132

BUFFERS
◦ A buffer system is a solution which will accept hydrogen or hydroxyl ions
without an appreciable change in pH.
− it is made up of weak base and its salt with a strong acid or weak acid and its salt with
a strong base.
− body buffers are mainly weak acids and they are able to take up or release H⁺ (to
minimise changes in [H⁺].
− e.g., weak acids: - H₂CO₃, H₂PO₄
133

o The main buffers in the body are:
a.Bicarbonate / Carbonic acid buffer system (HCO₃⁻ / H₂CO₃ buffer).
b.Disodium monohydrogen phosphate / monosodium dihydrogen
phosphate buffer (Na₂HPO₄/NaH₂PO₄ buffer).
c.Protein B minus / Protein B plus buffer (Pr/Pr⁻ buffer).
d.Haemoglobin buffer (Hb⁻/Hb⁺ buffer)
134

a. Bicarbonate /Carbonic acid buffer system
1. CO₂ + H₂ O H₂CO₃
↑
CA (carbonic anhydrase)
– speeds up the reaction X5,000
– found in RBC, ion secreting tissues (renal tubular
epithelium, stomach epithelium)
2. CO₂ + H₂ O H₂CO₃ → H⁺ + HCO₃⁻
3. The pH resulting from the solution of CO2 in blood and the consequent
dissociation of carbonic acid is given by the Henderson-Hasselbalch
equation.
2. It is derived as follows. In the equation: see next slides;
135

1. H₂CO₃ H⁺ + HCO₃⁻ …………………………………………………… i
2. the law of the mass action gives the dissociation constant of carbonic acid K’A
as
[H⁺] x [HCO₃⁻] ………………………………………………………….... ii
[H₂CO₃]
3. Because the concentration of carbonic acid is proportional to the concentration
of dissolved carbon dioxide, we can change the constant and write
KA = [H⁺] x [HCO₃⁻] …………………………………………… iii
[CO₂]
4. Taking logarithms,
log KA = log [H⁺] + log [HCO₃⁻] …………………………… iv
[CO₂]
5. Whence
- log [H⁺] = - log KA + log [HCO₃⁻] ……………………………. v
[CO₂]
6. Because pH is the negative logarithm, 136

pH = pKA + log [HCO₃⁻] ……………………………………. vi
[CO₂]
7. Because CO2 obeys Henry’s law, the CO2 concentration (in mmol.l⁻1
) can be
replaced by (Pco2 × 0.03). The equation then becomes
pH = pKA + log [HCO₃⁻] ……………………………………. vii
0.03Pco ₂
8. The value of pKA is 6.1, and the normal HCO₃⁻ concentration in arterial blood
is 24 mmol.l⁻1
. Substituting gives
pH = 6.1 + log 24 ……………………………………. viii
0.03 x 40
= 6.1 + log 20
= 6.1 + 1.3
Therefore,
pH = 7.4 137

◦ Bicarbonate buffer system is present in ECF (plasma).
◦ It consists of carbonic acid (H2CO3) which is a weak acid and the HCO3–,
which is a weak base (salt form of NaHCO3)
◦ Bicarbonate buffer system prevents the fall of pH in a fluid to which a strong
acid like hydrochloric acid (HCl) is added.
HCl + NaHCO3 → H2CO3 + NaCl
↓
CO2 + H2O
◦ Bicarbonate buffer system also prevents the increase in pH in a fluid to
which a strong base like sodium hydroxide (NaOH) is added.
◦ NaOH + H2CO3 → NaHCO3 + H2O
H+ and HCO3–.
◦ NaHCO3 is a weak base and it prevents the increase in pH by the strong
NaOH
◦ 138

TAKE HOME:
1. as long as the ratio of bicarbonate concentration to (Pco2 × 0.03) remains equal
to 20, the pH will remain at 7.4.
2. The bicarbonate concentration is determined chiefly by the kidney and the PCO₂
by the lung.
3. Bicarbonate (HCO₃⁻ ) buffer pK = 6.1
4. The proteins account for roughly 1/6th
of the buffering capacity of the blood.
139

Na₂HPO₄/NaH₂PO₄ buffer system
1. The plasma contains Na HPO and NaH PO in the ratio of 4:1
₂ ₄ ₂ ₄
2. The buffer system can remove H⁺ and OH⁻ from a solution, shown: -
a. HCl + Na HPO = NaCl + NaH PO
₂ ₄ ₂ ₄
strong weak strong weak
acid salt salt acid
b. NaOH + NaH PO =
₂ ₄ Na HPO + H
₂ ₄ ₂O
Strong weak weak neutral
base acid base fluid
3. From the pH equation:
pH = pK + log Na HPO
₂ ₄
NaH PO
₂ ₄
4. The pK of the phosphate buffer system is 6.8.
140

5. The phosphate buffer system is less effective extracellular fluid buffer
than the HCO₃⁻ buffer, because.-
a. … of its low concentration in the ECF
b. … NaH₂PO₄ is non-volatile and cannot be expelled by the lungs.
6. The phosphate buffering system is more important in the urine than in
the blood.
◦This system consists of a weak acid, the dihydrogen phosphate (H2PO4 –in
the form of sodium dihydrogen phosphate (NaH2PO4) and the base,
hydrogen phosphate (HPO4 – in the form of disodium hydrogen
phosphate (Na2HPO4).
141

c. Protein buffering system
1. Protein can be positively or negatively charged depending on the pH of
the surrounding medium.
2. As a result of this property, it can act as H⁺ acceptor and donor.
3. Proteins and their component amino acids have the acidic
COOH(carboxyl) group and the basic – NH₂ (ammonium) group
4. The – NH₂ group can accept H to become NH (protein in an acidic
⁺ ₃
environment).
5. In alkaline solution, the – COOH group acts (as an acid)and donates H⁺
and ionizes as - COO⁻
6. The anion - COO gives the molecule a predominant negative charge,
⁻
thus in alkaline solutions, protein is negatively charged (Pr )
⁻ 142

7. The behaviour of protein in acid, neutral and alkaline solution is as
follows.
NH₃⁺ NH₃⁺ NH₂
| | |
Protein Protein Protein
| | |
In acid COO⁻ COO
solutions In neutral solutions In alkaline solutions
(H⁺HPr) (H⁺Pr⁻) (Pr⁻)
143

Haemoglobin buffer system
1. Haemoglobin molecule contains 38 histidine units making it a good
buffer
2. The histidine molecule contains a = N group which behaves like an - NH₂
group.
3. It accepts H⁺ and becomes = NH⁺ (therefore a base that can neutralize acids)
4. Using the ionization of histidine, Hb exists as: Hb and HbH⁺ OH⁻
5. The ionization of histidine depends on the oxygenation degree of the Hb molecule.
6. When oxygen is released in the tissues, more HbH⁺ and OH⁻ are formed and this will
be available to neutralize the carbonic acid formed from the CO₂ entering the RBC.
144

d. Buffers in pH regulation
1. The buffer are the first line of action in the regulation body pH.
2. They help to mop-up excess hydrogen or hydroxyl ions with very little
change in over-all pH.
3. After the mop-up action, the sites at which the excess H⁺ are removed
from the body are the lungs and the kidneys.
4. The role of the lungs in pH regulation
a. by expelling CO₂ from the body, the body is expelling carbonic acid
b. It has estimated that the CO₂ produced in the body of an adult man in
24 hours is equivalent to about 15 litres of normal HCl
c. ……thus, if CO₂ is not expelled via the lungs, a lot of acid soon
accumulates in the body leading respiratory acidosis.
d. ……..conversely, if excess CO₂ is expelled via the lungs, as may occur in
hyperventilation, then the body fluid becomes alkaline. This is
respiratory alkalosis
145

Role of the kidneys in pH control
◦ The kidneys help in keeping pH constant by:
a.Reabsorption of HCO₃⁻
b.Excretion of H⁺
c.Formation of ammonia, which acts as a urinary buffer and ensures that H⁺,
once in the tubular lumen, does not diffuse back into the cells but is
excreted as NH₄⁺, having a combined with ammonia
NH₃ + H⁺ → NH₄⁺
a.When there is acidosis, there is 100% reabsorption of HCO₃⁻ in the kidneys
and lot of H⁺ is excreted
147

Role of the kidneys in pH control
e. Production of NH3 is increased several – fold to buffer the H⁺ secreted
into the tubular lumen to ensure its excretion
f. When there is alkalosis, the kidneys excrete excess HCO3- and
conserves H⁺
g. The H⁺ from non-volatile acids like H₂SO₄, H₂PO₄ and various organic
acids are excreted in the kidneys.
148

HYPOXIA
◦ Hypoxia is oxygen lack at the tissue level.
◦ This occurs when there is reduced or insufficient oxygen supply to the
tissues.
◦ Hypoxia can occur because of any one or more of the following defects:
- Decreased oxygen carrying capacity of the blood
- Decreased rate of blood flow to the tissue
- Decreased utilization of oxygen by the tissue cells
- Decreased oxygen tension (pO2) of the arterial blood
◦ 4 Types of hypoxia
1.Anaemic hypoxia
2.Stagnant hypoxia
3.Histotoxic hypoxia
4.Hypoxic hypoxia (this leads to arterial hypoxia) 149

1. Anaemic hypoxia
i. Oxygen carrying power of the blood is decreased
- The oxygen tension (PO2) in arterial blood is normal
- Blood flow is normal or may even be elevated
- ….but the total oxygen carried by the blood is inadequate for the
needs of the tissues
ii. Can be caused by anaemia of all types, in which blood [Hb] is below
normal. Decreased number of RBCs
iii. Combination of Hb with CO to form carboxyhaemoglobin
i. Formation of altered hemoglobin (Hb converted to methaemoglobin
after poisoning with ferricyanides)
150

2. Stagnant hypoxia
◦Caused by sluggish blood flow to the tissue i.e decreased velocity of
blood flow. Hence, adequate oxygen is not brought tissues.
i.Can be general or localized:
- General stagnant hypoxia is found in shock, haemorrhage or congestive
heart failure
-Localized stagnant hypoxia occurs in embolism, thrombosis, or localized
vasoconstriction
3. Histotoxic hypoxia
i.The amount of oxygen delivered to the tissue is adequate, but because
of the action of a toxic agent, e.g., Cyanide or sulfide, the cells of the
tissue cannot make use of the oxygen supplied to them.
◦These poisonous substances destroy the cellular oxidative enzymes and
there is a complete paralysis of cytochrome oxidase system.
151

4. Hypoxic hypoxia
• This is due to a reduction in the PO₂ of arterial blood, hence it is sometimes
called arterial hypoxia.
• Characterized by reduced oxygen tension in arterial blood.
• Hypoxic hypoxia can be caused by:
i. Inspiration of air that has a low PO₂ (high altitude, inhalation of artificial
gas mixture whose PO₂ is lower than that of atmospheric air)
ii.Decreased pulmonary ventilation (obstruction of an airway, paralysis or
weakness or respiratory muscles, pneumothorax, depression of respiratory
centres by drugs or other agents)
iii.Insufficient oxygenation in abnormal lungs (fibrosis of the lungs,
ventilation/perfusion imbalance, constriction of the alveoli with fluid as in
pneumonia, pulmonary oedema and drowning)
iv.Arterio-venous shunts resulting in venous blood bypassing the lungs and
not being oxygenated
152

RESPIRATION AT HIGH ALTITUDE
 The higher one goes into high altitude, the lower the amount of air
available in the environment.
→ e.g, the PO in inspired at sea-level is 159 mmHg and alveolar
₂ PO 104
₂
mmHg, while at an altitude of about 10,000 metres, PO in inspired air 47
₂
mmHg and alveolar PO is 21 mmHg.
₂
↓
 There is also a corresponding decrease in the total barometric pressure
↓
 hypoxia stimulates the peripheral chemoreceptors (carotid body)
↓
 leading to an increase in pulmonary ventilation
↓
 hyperventilation → lots of CO expelled →
₂ respiratory alkalosis
153

HIGH ALTITUDE: Acclimatization
1. ↑se in pulmonary ventilation
2. ↑se in red blood cell count (hypoxia stimulate EPO)
3. ↑se 2, 3 – DPG level which ↑ses O delivery to the tissues.
₂
4. Excretion of alkaline urine to correct the alkalosis
5. An ↑se in the number of mitochondria in the cells and
6. ↑sed vascularity of the tissues
154

HIGH ALTITUDE: Cyanosis
1. Bluish discolouration of the skin and mucous membrane.
2. ….due to presence of a large quantity of deoxygenated Hb in the
blood.
3. Sites where cyanosis can be observed:
 Skin (not seen in black Africans)
 Neonates (Africans or non-Africans)
 Tongue
 Nail bed
 Buccal mucosa in black Africans
4. Cyanosis becomes noticeable when the arterial blood contains 5g or
more of deoxygenated Hb per 100ml of blood.
155

DEEP SEA DIVING
1. Beneath the sea, the pressure ↑ses⁺
 for every 10 metres of depth in sea-water, pressure ↑ses by 1 atmosphere
2. To prevent lung collapse breathed air must be supplied under high pressure
(hyperbaric air / hyperbarism – the act of breathing hyperbaric air)
3. The gases normally breathes are Nitrogen, Oxygen and CO₂ (N and O when breathe
under high pressure can cause serious physiological effect, CO has no effect)
156

DEEP SEA DIVING
4. The increased pressure causes a lot of N and O to dissolve in the body fluids
and in the tissues.
5. As the diver descends deeper into the sea:
 the ↑sed pressure causes compression of the gases being inspired, leading
to a decrease in volume and an increase in pressure according Boyle´s law
P ∞1/V P V =P V
₁ ₁ ₂ ₂
157

DEEP SEA DIVING: Nitrogen narcosis
1. At sea level N has no effect on body functions
2. At 120 ft beneath sea level for ˃ 1hr – mild symptoms of narcosis
appears (diver becomes jovial, just happy)
3. At 150 to 200 ft - becomes drowsy
4. At 200 to 250 ft – becomes very week and too clumsy to perform
saddled work
5. Depths greater than 250 ft – becomes useless and cannot
perform any function
158

DEEP SEA DIVING: Decompression sickness
1. Bends, Caisson disease, Diver´s paralysis
2. If diver stays beneath the sea for a long period → N will dissolve in his
body
3. If the diver suddenly comes back to the surface of the sea → N forms
gas bubbles in the body fluids (ICF, ECF) →→ AIR EMBOLISM
4. ……DECOMPRESSION SICKNESS
5. …. pain in joints (reason for calling it ´bends´) and muscle of the legs
or arms
159

DEEP SEA DIVING: Decompression sickness
6. …. dizziness, paralysis, collapse and unconsciousness
7. …. massive pulmonary embolism → shortness of breath ----- pulmonary
oedema = DEATH
8. REMEDY: ascend gradually over 2 – 5 hrs or 6 hrs.
9. NB: - a. slow ascent over 1 hr will cause elimination of 70% of the
dissolved Nitrogen
- 90% will be eliminated if ascent is carried out in 6 hrs.
b. pressurized tank (use to treat decompression sickness)
c. use of helium instead of Nitrogen
d. SCUBA apparatus (Self-contained Underwater Breathing
Apparatus)
160

Oxygen toxicity at high pressures
1. When O is breathe at high pressure, the alveolar and tissue PO2
are markedly ↑sed.
2. O at a high pressure of 4 atm (PO2 of 3040 mmHg) will cause:
- seizures
- coma in most people within 30 mins
3. Other symptoms of acute poisoning include:
- nausea
- irritability
- disturbance of vision
- muscle twitching
- dizziness
- disorientation
161

Oxygen toxicity at high pressures
4. Molecular oxygen toxicity – caused by O free radicals:
- superoxide
- hydrogen peroxides
5. Enzymes (peroxidases, catalases, superoxide dismutases (SOD) in the
tissue rapidly remove these free radicals
6. When the Hb-oxygen buffering mechanism fails – free radicals then
cause damage to cell membrane, cellular enzymes leads to….
7. derangement in cellular metabolic processes
8. Nervous tissues are most vulnerable to the effect of these radicals
9. As a result most of the acute lethal effects of acute oxygen toxicity are
related to brain dysfunction.
162

F- References
1- Review of Medical Physiology, William F. Ganong, MD
2- The National Medical Series for independent Study, 2nd
edition, physiology, John Bullock, Joseph Boyle III, Michael B.
Wang
3- Respiratory physiology, The essentials, 9th
edition, John B.
West
4-
163

“ lifting heavy rocks is no sign of strength ; (real) strength lies in
controlling the wrath when one has been over-whelmed by it”.
164

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