Resp 1 - West Ch 6 Gas Transport By The Blood

Front 1

O2 is carried through the blood in 2 forms:
1) Accounts for 0.3 ml O2/ 100 ml blood at PO2 100mm Hg
2) Accounts for 20.8 ml O2/ 100 ml blood at 100mmHg

Back 1

Dissolved O2 - inadequate

Hemoglobin bound O2 - very efficient

Front 2

One gram of pure Hb can combine with ___ml O2.
Normal blood has about __gram of Hb/100ml blood.
=> the O2 capacity is about ___ml O2/100 ml blood

Back 2

1.39ml
15gm
20.8ml

Front 3

The O2 saturation of arterial blood with PO2 of 100mmHg is about ___%
While that of mixed venous blood with a PO2 of __mmHg is about __%

Back 3

97.5%
40mmhg
75%

Front 4

If you have an anemic patient with an Hb of only 10 gm/100ml blood what is their O2 capacity?

Back 4

20.8 x 10/15 = 13.9 ml/100 ml blood
13.9 x 97.5% (O2 saturation) = 13.5 ml/100 ml blood
+ .3ml from dissolved O2

=> 13.8 ml / 100ml blood
= 13.8

Front 5

Why does a person have cyanosis with hypoxia?

Back 5

Reduced Hb is purple - low arterial O2 saturation causes cyanosis

Front 6

What happens to the following conditions in order to shift the O2 dissociation curve to the RIGHT?
Temp
PCO2
H+
DPG

Back 6

Increase of all - decreases the O2 affinity of Hb so tissues can get more O2

Front 7

What happens to the following conditions in order to shift the O2 dissociation curve LEFT?
Temp
PCO2
H+
DPG

Back 7

All go down - increases the affinity of Hb so tissues get less O2

Front 8

Small addition of CO to blood causes a shift to the ____

Back 8

Left
CO interferes with the O2 transport function of blood by combining with Hb to form COHb. CO has 240x the affinity of O2 for Hb.
At PCO of .16mmhg, 75% Hb is combined with CO.
Also interferes with the unloading of O2

Front 9

A climber reaches an altitude of 4500 m (14,800 ft) where the barometric pressure is 447 mmHg. The PO2 of moist inspired gas (in mmHg) is:
A. 47
B. 63
C. 75
D. 84
E. 98

Back 9

Answer: D.
The moist gas at body temp would have a partial pressure of oxygen of 47 mmHg.
Therefore; dry gas partial pressure would be (447-47) mmHg = 400 mmHg.
The remaining 400mmHg is 0.21% oxygen. Therefore;
PIO2 = FIO2 x 400 mmHg 0.21 x 400 = 84 mmHg

Front 10

A man with normal lungs and an arterial PCO2 of 40 mmHg takes an overdose of barbiturate that halves his alveolar ventilation but does not change his CO2 output. If his respiratory exchange ratio is 0.8, what will be his arterial PO2 (in mmHg) approximately?
A. 40
B. 50
C. 60
D. 70
E. 80

Back 10

Answer: B.
If remains constant and decreases by 50%, will double.

Front 11

In the situation described in Question 2 how much does the inspired O2 concentration (%) have to be raised to return the arterial PO2 to its original level?
A. 7
B. 11
C. 15
D. 19
E. 23

Back 11

Answer: A.
Assuming is constant, = 80 mmHg, F = 0, PB = 760 mmHg and we want to increase to 100 mmHg, relieving hypoxia. Rearrange the alveolar gas equation as follows:
Want PAO2 to become 100 mmHg therefore:
PIO2 = 100mmHg + 80/0.8 = 100 mmHg + 100 mmHg = 200 mmHg
Assume PB = 760 mmHg, PH2O = 47 mmHg

Front 12

A patient with normal lungs but a right to left shunt is found at catheterization to have oxygen concentrations in his arterial and mixed venous blood of 18 and 14 ml ?100 ml-1, respectively. If the O2 concentration of the blood leaving the pulmonary capillaries is calculated to be 20 ml ? 100 ml-1, what is his shunt as a percentage of his cardiac output? ?
A. 23
B. 33
C. 43
D. 53
E. 63

Back 12

Answer: B.

Shunt equation

Front 13

If a climber on the summit of Mt. Everest (barometric pressure 247 mmHg) maintains an alveolar PO2 of 34 mmHg and is in a steady state (R ? 1), his alveolar PCO2 (in mmHg) cannot be any higher than:
A. 5
B. 8
C. 10
D. 12
E. 15

Back 13

Answer: B
Dry total pressure = 247 mmHg - 47 mmHg(water vapor at body temperature) = 200 mmHg

PIO2 = 0.21 x 200 mmHg = 42 mmHg

PAO2 = PIO2 ? PACO2/R
Note: R ? 1
34 mmHg = 42mmHg ?PACO2/1
PACO2 < 8 mmHg

Front 14

A patient with severe chronic obstructive pulmonary disease, which causes marked ventilation-perfusion inequality, has an arterial PO2 of 50 mmHg and an arterial PCO2 of 40 mmHg. The PCO2 is normal despite the hypoxemia because:
A. Ventilation-perfusion inequality does not interfere with CO2 elimination.
B. Much of the CO2 is carried as bicarbonate.
C. The formation of carbonic acid is accelerated by carbonic anhydrase.
D. CO2 diffuses much faster through tissue than O2.
E. The O2 and CO2 dissociation curves have different shapes.

Back 14

Answer: E
All the statements are essentially true but only E is responsible of the normal carbon dioxide in the face of hypoxemia.

Front 15

The apex of the upright human lung compared with the base has :
A. A higher PO2.
B. A higher ventilation.
C. A lower pH in end-capillary blood.
D. A higher blood flow.
E. Smaller alveoli.

Back 15

Answer: A.
A is correct because of the high ventilation-perfusion ratio in this region. B,C and D are false see Figure 5-10. E is incorrect due to gravity and larger magnitude of the intrapleural pressure.

Front 16

If the ventilation-perfusion ratio of a lung unit is decreased by partial bronchial obstruction while the rest of the lung is unaltered, the affected lung unit will show:
A. Increased alveolar PO2 .
B. Decreased alveolar PCO2.
C. No change in alveolar PN2.
D. Rise in pH of end-capillary blood.
E. Fall in oxygen uptake.

Back 16

E. . Oxygen uptake will decrease due to a decreased driving force.
A & B, will be opposite due to the decreased ventilation-perfusion ratio. Answer C is incorrect because nitrogen concentration will have to change to compensate for the other gases, see Figure 5-10.

Front 17

A patient with lung disease who is breathing air has an arterial PO2 and PCO2 of 49 and 48 mmHg respectively, and a respiratory exchange ratio of 0.8. The approximate alveolar-arterial oxygen difference for PO2 (in mmHg) is:
A. 10
B. 20
C. 30
D. 40
E. 50

Back 17

Answer: D.

Front 18

CO2 is carried in three forms in the blood - what are they?

Back 18

Dissolved, bicarbonate, combined with proteins such as carbamino compounds

Front 19

Bicarbonate is formed in two reactions. The first by carbonic anhydrase where?
Then there is ionic dissociation of carbonic acid. How does the cell maintain electrical neutrality?

Back 19

In the red blood cell

Cl- ions move into the cell (chloride shift)

Front 20

What is the Haldane effect?

Back 20

The fact that deoxygenated blood increases its ability to carry CO2
(More reduced Hb available for H+ ions)

Front 21

The bulk of CO2 in blood is in what form?

Back 21

bicarbonate

Front 22

The CO2 dissociation curve is much more _____ than the O2 dissociation curve.
Also, the lower the saturation of HbO2, the ______ the CO2 concentration for a given PCO2. (Haldane effect)

Back 22

linear
greater

Front 23

The CO2 curve is much _____ than O2 explaining why the PO2 difference between A-mixed V blood is large and PCO2 difference is _____

Back 23

steeper

PCO2 small (about 5 mmHg)
PO2 about 60 mmHg