Basic Sciences - Oxygen Transport and Consumption

Question 1

Site and use of oxygen

Answer 1

Used in mitochondria
Produces ATP via oxidative phosphorylation in Kreb’s cycle / electron transport chain chain

Question 2

Oxygen consumption calculation

Answer 2

Oxygen consumption = Minute volume x (FiO2 - FeO2)

FeO2 = fraction expired O2 (usually 16% at rest)

Question 3

Oxygen consumption at rest

Answer 3

250 ml/min

From equation:
Oxygen consumption = Minute volume x (FiO2 - FeO2)
= 5000 x (0.21 - 0.16)
= 250

Question 4

Partial pressure of oxygen at sea level

Answer 4

Approx 21 kPa

(Atmospheric partial pressure 101.3 kPa)

Question 5

Solubility of Oxygen in plasma (not haemoglobin)

Answer 5

0.23 ml/L/kPa

Very low

Question 6

Normal Hb concentration

Answer 6

130-150 g/L

Question 7

Molecular weight of Haemoglobin

Answer 7

Around 68,000

Question 8

Main regulator of haemoglobin production

Answer 8

Erythropoietin

Secreted by kidney in response to tissue O2 level

Question 9

Haemoglobin structure description

Answer 9

4 intertwined subunits:
- 2 alpha polypeptide globin chains
- 2 beta polypeptide globin chains

Each unit contains a haem group - porphyrin ring containing Fe2+ ion

Question 10

Haemoglobin structure image

Answer 10

Question 11

How many oxygen molecules bind to each haem group

Answer 11

1

Question 12

Driving factor for oxygen binding to Hb

Answer 12

Partial pressure

However relationship between PaO2 and Hb O2 binding is not proportional

Question 13

Relationship between haem and O2 binding

Answer 13

Initial O2 binding is difficult

As first O2 molecule binds to first haem group, it alters shape of Hb molecule making other binding sites more accessible

Subsequent binding of second and third O2 molecules are easier (cooperativity)

Full saturation with fourth O2 molecule is difficult as only one remaining binding site free

Question 14

Haemoglobin oxygen dissociation curve

Answer 14

Question 15

Haemoglobin oxygen dissociation curve with arterial and venous PaO2

Answer 15

Question 16

Venous blood oxygen saturation at rest

Answer 16

~ 75%

Approx 25% of available O2 extracted by tissues at rest

Question 17

Venous blood PaO2 at rest

Answer 17

~ 5.3 kPa

Question 18

P50 definition

Answer 18

PaO2 at which the Hb-O2 saturation 50%

Question 19

Usual PaO2 of P50

Answer 19

~ 3.5 kPa

Question 20

Use of P50

Answer 20

Reference point that describes position of the Hb-O2 dissociation curve and changes as the curve moves under different conditions

Question 21

3 methods to describe oxygenation

Answer 21

PaO2
SpO2
CaO2

All related but not the same

Question 22

Oxygen cascade definition

Answer 22

Three stage drop in arterial PaO2 to 13 kPa from atmospheric level of 21 kPa

Question 23

Oxygen cascade process

Answer 23

1) In the upper airway humidification occurs adding water vapour

2) In the alveoli O2 is taken up in exchange for CO2

3) In the circulation a small physiological shunt caused by bronchial circulation and thebesian veins

Question 24

Mechanism of saturation probe functioning

Answer 24

Uses different absorption characteristics of Hb and oxy-Hb for red and infrared light

Extracts only pulsatile signal to obtain arterial values

Question 25

Oxygen content of blood definition

Answer 25

Truest measure of oxygen present but difficult to directly measure

Therefore usually estimated from other values

Question 26

Oxygen content equation

Answer 26

CaO2 = Hb bound O2 + Plasma dissolved O2

Hb bound = [Hb] x 1.34 x O2 saturations

Dissolved = 0.23 x PaO2
(0.23 is dissolvability of O2 in plasma (from earlier flashcard)

Therefore:
CaO2 = (1.34 x [Hb] x saturations) + (0.23 x PaO2)

Question 27

Volume of oxygen held by 1g of Hb when fully saturated

Answer 27

1.34 ml

Hence why constant of 1.34 is included in CaO2 equation
CaO2 = (1.34 x [Hb] x sats) + (0.23 x PaO2)

Question 28

Approximate CaO2 in healthy adults

Answer 28

200 ml/L

CaO2 = (1.34 x [Hb] x sats) + 0.23 x PaO2)

= (1.34 x 150 x 0.98) + (0.23 x 13)

= 197 + 3

= 200 ml/L

Question 29

Delivery rate of O2 equation

Answer 29

DO2 = CaO2 x CO

Question 30

Usual DO2 in healthy adult

Answer 30

DO2 = CaO2 x CO

= 200 ml/L x 5 L/min

= 1000 ml/min

Question 31

Rate of oxygen consumption at rest

Answer 31

VO2 only 25% of DO2 (25% of 1000 ml/min = 250 ml/min)

Also remember VO2 equation:
VO2 = minute volume x (FiO2 - FeO2)

Question 32

Venous content of oxygen at rest

Answer 32

25% consumed at rest leaving 75% in venous blood

75% of 200 ml/L = 150 ml/L

Question 33

States which increase oxygen consumption by tissues

Answer 33

Exercise
Shivering
Pregnancy
Childhood
States that increase metabolic rate inc pyrexia and thyrotoxicosis

Question 34

Methods of physiologically providing extra oxygen for when tissue has higher oxygen consumption

Answer 34

Increasing:
- CO
- Minute ventilation
- Extraction of O2 from blood (reducing leftover venous blood CaO2)

Question 35

How much VO2 can increase during vigorous exercise

Answer 35

Up to 10x the value at rest

Question 36

Bohr effect description

Answer 36

Increase in tissue metabolism (eg exercising muscle)

More CO2 produced

Causes local acidosis

Shifts Hb-O2 curve to the right

Therefore at a given PaO2, the Hb is less saturated so gives up more O2 to the tissues

Small rise in temperature locally also contributes to the right shift of the curve

Opposite occurs at lungs with CO2 removal leading to left shift of the Hb-O2 curve, increasing affinity of Hb for O2

Question 37

When is cyanosis detected clinically

Answer 37

When ≥ 50 g/L deoxy-Hb can be seen in the skin or mucous membranes. With a [Hb] of 150 g/L this is 33% of the total, with the remaining 67% being saturated.

In practice, patients appear cyanosed when the pulse oximetry reading is much higher, around 85%.

Cyanosis is seen in capillary blood, whilst a pulse oximeter reading is based on the arterial value, which will be substantially higher because:
* PO2 is slightly higher in the artery than the capillary

• The saturation curve in the capillary has a small right shift (Bohr effect)