L22 The Cardiorespiratory Response to Exercise Flashcards
Centre of Gravity of the Human Body
Falls at the Crest of the Ilium of a Lean male
- Centre of Gravity is Lower in woman than it is in men due to:
a) wider hips
b) tendency to put weight on there/
except: middle aged man who have a pronounced abdomen, have to lean backwards in order to keep their centre of gravity is Above their feet - if their centre of gravity isnt above their feet, then goes horizontal
Where is the centre of gravity in a foseberry flop in pole vaulting?
-
Increase in Ventilation from rest to exercise
V.O2 and V.CO2 can increase 20-30 fold from rest–> exercise
Fick principle with Ventilation
can measure ventilation using fick principle, measure O2 content at mouth and change of oxygen content around circulatory system
During steady state exercise
Same amount of oxygen being consumed be the cells
as is the net difference between what is coming in and going out again
-same for CO2 production
During Steady State exercise equations
V.O2=V.(FIo2-FEo2)
=Q.(Cao2-Cv-o2) (CO find the difference between arterial and mixed venous contents of oxygen
Rate of oxygen consumption = rate at which it is disappearing at the mouth x rate at which it is disappearing out of the blood system
Steady state= Rate O2 disappearing through ventilation as same as rate disappearing through circulation
V.= VTidal x Freq
Q.=VStroke x Freq
During Steady State exercise equations with Quantitative values
Rest= Vtidal x Freqresp (FIo2-FeO2) = Vstroke x Freqresp (Cao2-Cv-o2)
REST: 240 = 0.5 x 12 x (0.21-0.17) = 70 x 20 x (200-150)
During Exercise equations with Quantitative values
Rest= Vtidal x Freqresp (FIo2-FeO2) = Vstroke x Freqresp (Cao2-Cv-o2)
V.O2max= 5000 = 0.5 x 12 x (0.21-0.17) = 70 x 20 x (200-150) = 5Lmin-1
Profound increase in amount of O2 extracted, and the consequent lowering in the content of oxygen in venous blood
Ratio = x20 = x4 x3.5 x1.5(minor change for the oxygen extracted at level of the lung) = x2 x3 x3.5
Rate of Metabolic Energy expenditure (V.O2) proportionality
Rate of Metabolic Energy expenditure (V.O2)
-attributing totally to O2 consumption. partially true as go into lactic acid producing stage during end of long race
Proportional to
Rate of Mechanical Work. -= Power (rate at which work is done)
Rate of Metabolic Energy expenditure (V.O2) proportionality graph
Oxygen consumption(Lmin-1) as a function of Power(watts) -Oxygen Uptake values obtained via sitting on bicycleodometer (mlkg-1min-1) Sedentary= 30 Normal active= 45 Conditioned=53 Endurance athletes=85 = cross country skiers (tall large chest body build)
Efficiency Graph
Invert Oxygen consumption(Lmin-1) as a function of Power(watts) graph to have Power vs Oxygen uptake = Efficiency
=400W/ ((6L/60s)x(20kj/L)) = 0.20 = 20% efficiency
How much energy does one L of oxygen provide?
60kJ
Increased work of breathin
Deeper breaths
More frequently
Work of breath graph
Increase rate of work
V.O2 Modest Rate of O2 consumption by respiratory =Resting, Walking slowly, Working quite quickly
V.O2 increased quickly= climbing stairs, sawing, running 12kmhr-1
-remains a small part of the total o2 consumption
-but a Disproportionate increase (3% –> 10%)
What is the cause of the disproportionate increase in the work of breathing?
-
“Local” (peripheral) circulatory effects of exercise
- Decreased tissue PO2
- Decreased tissue pH
- Increased PCO2 (accompanies lactate production)
- Increased tissue T (temperature) (increases with work)
=Each causes local Vasodilation (i.e. increased local capillary blood flow and elevation of local PO2)
=Each causes a Right shift of the HbO2 relation (Oxyhaemaglobin relation)
OxyHb relation Effect of Temperature
Percent of O2 saturation of Hb as a function of PO2 mmHg
Sigmoidal relationship
-Slop inversely depends on the temperature.
Sometimes body temperature during exercise can reach extreme values (43) that wouldnt be tolerated in any other scenarios other than exercise (e.g. these temp reached in fever)
=Right shift
=reduced amount of O2 that can be onloaded at lower partial pressures of Oxygen
OxyHb relation Effect of pH
- 4 normal
7. 2= blood becomes more acidic = right shift = as lactate production develops at high rates
OxyHb relation Effect of PCO2
PCO2=40mmHg normal
Increase/right shift= PCO2 80mmHg
What is an obligatory consequence of exercise?
Right shift of the OxyHb curve
Alveolar capillary interface equation (increase DLco)
V.O2= DL(PAo2 - Pc'o2) Lung-->Circulation= Increase of DL =Vo2/DeltaT = V.O2 = Do2 x A x (Solubility x (PAo2 - Pco2))/d Do2= Physical property of oxygen SolubilityO2= Physico-chemical property of oxygen and blood Can change (to increase Dl)= A (increase area) and d ( decrease diffusion distance) Dlo2 = Do2 x Solubility x A/D -Increase DLco diffusing capacity by Opening up more pulmonary capillaries =Increase SA and decrease diffusion distance d where capillaries hadnt been diffused
Alveolar Capillary interface (Decrease of Pc’O2) with exercise
Decrease in the Partial Pressure of Oxygen at the end of the pulmonary capillary (Decrease of Pc’O2)
Alveolar Capillary interface (Decrease of Pc’O2) with exercise
Decrease in the Partial Pressure of Oxygen at the end of the pulmonary capillary (Decrease of Pc’O2)
- want to have it as high as possible
- Effect of cardiac output on the Transit time of Erythrocytes through the pulmonary capillaries
Respiratory responses to Exercise- Transit time
Effect of Cardiac Output on the Transit Time of erythrocytes/RBC through the pulmonary capillaries
-takes time to equilibriate the Partial pressure in air (Pair) with the Partial pressure in passing Hb (Phb)
-enter and exit time to be oxygenated
Transit time= function of Heart rate
-Increase HR = More blood + faster = Smaller Transit time per RBC
-HR=70 –> RBC 0.5sec to transit capillary and be fully O2
-HR=200 –> RBC
Effect of Transit Time on Pulmonary end-capillary PO2 (Pc’O2)
End capillary = about to send blood back to Left heart as maximally oxygenated as can be under the circumstances
1. Normal/Rest= Transit time0.4s =
Enter capillary= Po2 40mmHg (rel. deoxygenated)
Exit capillary= Pc’o2 100mmHg (fully oxygenated)
2. Exercise: Transit time 0.2s
Exit capillary= Pc’o2 90mmHg
-less time in capillary to be oxygenated = leaves capillary relatively less oxygenated (decreased saturation by 10-15mmHg)
Is the effect of Effect of 0.2s Transit Time on Pulmonary end-capillary PO2 (Pc’O2) decrease of 10-15mmHg Large?
Graph: Oxygen content as a function of Oxygen partial pressure
Normal= 200 mL/L Oxygen content. 100mmHg Oxygen Partial pressure
–> Decrease in Partial Pressure
Abnormal/Exercise=
a) 97% saturated/195mL/L Oxygen content. 95mmHg Oxygen Partial pressure
b) BUTTT recall the Right shift of the HbOxygen relation:
-previously modest (content) change has now doubled in size (worsened by right shift) for the same O2 partial pressure
- 94% saturated/185mL/L Oxygen content. SAME 95mmHg Oxygen Partial pressure
O2 Content as a function fo a) PO2(mmHg) and b) CO (Lmin-1)
HbO2 relation under normal conditions (37 degrees. ph 7.4)
=Onload 200mLL-1 (content) of O2 when the partial pressure is 100mmHg in the lung (will be 100 if young and healthy)
Arbituary O2 consumption= V.O2= 3.6 Lmin-1 = achieved at CO of 30 Lmin-1 (increase by 5-6x)
-Shaded area: indicate total area of 3.6Lmin-1 = insync with the decrease in O2 content of (200–>80) 120 mLL-1 = 120 x CO = 120x30Lmin-1=3.6Lmin-1
(multiply Vertical distance by horizontal distance to get rate of O2 consumption)
(Rate of O2 consumption = V.O2 = Delta Content x CO)
What is the unit for CO
L min-1
Interrelation between (Q.) Blood Flow and the Right Shift of the HbO2 relation
HbO2 relation under exercise conditions (40 degrees. ph 6.9.) Right shift
=Can off Load more O2 at a give Partial Pressure
Content: (increased vertical distance)
a) Cannot load on as much O2 (start 200–>175 content)
b) Can take O2 content down to a lower value (31content)
But can REGAIN the SAME V.O2 3.6Lmin-1 Oxygen consumption with a..
CO:
Reduced CO 25Lmin-1
(SAME V.O2 Rate of O2 consumption, therefore Same Area (just changing length of sides))
Written explanation for Interrelation between (Q.) Blood Flow and the Right Shift of the HbO2 relation
Right shift of the OxyHb curve
=Can off Load more O 2 at a give Partial Pressure (175-31 > 200-80)
=additional advantage of
=this more than compensates the O2 content onload has decreased
Overall:
Reduced CO Cardiac Output during Exercise
-Same Rate of O2 Consumption (V.O2)
-at a Reduced Blood Flow (Q.)
-With no loss of Pv’O2 (no partial pressure change)
-The partial pressure that is left over = can be used to drive O2 out of the capillaries, across the cell membrane and to the inspiring mitochondria
Logic behind Pv’ falling much lower
-Because if you decreased O2 content even more
=decreased CO
=still have enough Po2 to drive O2 to mitochondria (as mitochondria can use O2 right down to 1mmHg)
Why doesnt Pv’O2 fall much lower?
Tissues are getting blood that arent muscle tissues
-Cardiovascular system has a limitation to the amount of blood that it can pump
-still has to keep other tissues happy (Brain(steady), skin(cool down, Limbs(arms turn red w. vasodilation))
=compromise b/w
1) Need to lose heat
2) Desire to keep exercising
Note: overall all dependant on the environmental conditions how these are met
Why does blood need to be pumped to your skin to cool down?
Efficiency=20% = Work
80%=Inefficient = Heat
-must get rid of the heat, because if the body temperature rises too high = death
(Efficiency
=400W/ ((6L/60s)x(20kj/L)) = 0.20 = 20% efficiency)
Methods of getting rid of body heat
- Radiation (temperature difference between yourself and the environment)
- Evaporation (perspiring)
-cannot get rid of heat via perspiration if High humidity
Overall:
-Slave to the environmental conditions as to how much exercise you can do
