CVRS TO CPP L2

Question 1

Question 1
Why does cardiac output (heart rate & stroke volume) and minute ventilation (tidal volume & respiratory frequency) increase when we undertake aerobic exercise (e.g. jogging)?

Answer 1

Necessary for meeting the heightened metabolic demands of the body during physical activity.
Coordinated response aimed at enhancing oxygen delivery to tissues and removing carbon dioxide efficiently.

Question 2

Cardiac output (CO) : volume of blood the heart pumps per minute

CO = HR × SV

Answer 2

Heart Rate (HR):
During exercise, HR increases due to sympathetic activation, mainly
Pump more blood to meet the increased oxygen demand of muscles

Stroke Volume (SV):
Increase with exercise due to enhanced venous return (the volume of blood returning to the heart).
Affected by increased blood volume, improved cardiac contractility, and reduced systemic vascular resistance contribute to this rise

The combination of a higher HR and SV leads to a significant increase in cardiac output

Question 3

Minute ventilation
Total volume of air inhaled or exhaled from the lungs per minute

MV = Tidal Volume (VT​) × Respiratory Frequency (f)

Answer 3

Tidal Volume (VT​)
During aerobic exercise, VT​ increases as the body requires more oxygen (deeper breaths that enhance lung inflation and gas exchange efficiency

Respiratory Frequency (f):
As VT​ increases, so does the frequency of breaths per minute during vigorous activities

This increase is driven by both metabolic demands and chemoreceptor feedback responding to elevated carbon dioxide levels produced during exercise.

Question 4

Under which of the following conditions would you get the most flow through a capillary bed with a resistance of 10 resistance units?

Pa = pressure at the arteriolar end of the capillary bed
Pv = pressure at the venous end of the capillary bed

A.Pa = 35mmHg, Pv = 15mmHg
B.Pa = 70mmHg, Pv = 35mmHg
C.Pa = 120mmHg, Pv = 90mmHg
D.Pa = 215mmHg, Pv = 200mmHg

Answer 4

Q = ΔP / R
Pa-Pv / 10
B

Question 5

At which points in the respiratory cycle is airflow zero?

A.At Functional residual capacity (FRC): the volume remaining in the lungs after a normal, passive exhalation
B.At the end of expiration
C.At the end of inspiration
D.Mid-inspiration
E.Mid-expiration

Answer 5

At FRC (Functional Residual Capacity):
FRC is the volume of air remaining in the lungs after a normal expiration. At this point, airflow is zero because there is no net movement of air in or out of the lungs.
The forces that promote lung expansion (inspiratory muscles) are balanced by those that promote lung collapse (elastic recoil). Therefore, there is no change in lung volume or pressure.

At the end of expiration:
At this point, when all available air has been exhaled from the lungs, airflow is zero.
The diaphragm and intercostal muscles are relaxed, and there is no active force promoting air movement.

At the end of inspiration:
At this point, when all available air has been inhaled into the lungs, airflow is also zero.
The inspiratory muscles have reached their maximum contraction, and there is no active force promoting further air movement

Question 6

With a constant driving pressure how much would blood flow and air flow increase if the radius of a blood vessel or airway doubled?

A.2-times
B.4-times
C.8-times
D.16-times
E.32-times

Answer 6

Flow rate (Q) through a cylindrical tube is directly proportional to the 4th power of its radius (r) when other factors such as pressure and viscosity remain constant.
Q ∝ r4

Therefore, if the radius doubles, the flow rate increases by a factor of (2)4 = 16

Question 7

In which of the following situations will inspired PO2 (PIO2) be highest (assume PH2O at body temperature is 6.25kPa and that atmospheric pressure at sea level is 100kPa)?

A.Breathing 100% oxygen at an altitude of 11,000m
B.Breathing aeroplane cabin air at cruising altitude of a commercial jet with a cabin pressure of 80kPa
C.Breathing room air at sea level
D.Breathing room air at sea level in a hyperbaric chamber with twice the normal level of atmospheric pressure

Answer 7

PIO2 = FIO2 × (Pb - PH2O)
FiO2 is the fraction of inspired oxygen (for room air, it’s approximately 0.21; 1.0 for 100% O2).
Pb is the barometric pressure.
PH2O is the partial pressure of water vapor (assumed to be 6.25 kPa at body temperature).

A. Breathing 100% oxygen at an altitude of 11,000m:
FiO2 = 1.0 (100% oxygen)
Pb = 26.54 kPa (barometric pressure at 11,000m altitude, approximately)
PH2O = 6.25 kPa
PIO2 = 1.0 × (26.54 - 6.25) = 20.29 kPa

B. Breathing airplane cabin air at a cruising altitude of a commercial jet with a cabin pressure of 80 kPa:
FiO2 = 0.21 (assuming it’s room air)
Pb = 80 kPa (cabin pressure)
PH2O = 6.25 kPa
PIO2 = 0.21 × (80 - 6.25) = 15.49 kPa

C. Breathing room air at sea level:
FiO2 = 0.21 (room air)
Pb = 100 kPa (sea level atmospheric pressure)
PH2O = 6.25 kPa
PIO2 = 0.21 × (100 - 6.25) = 19.69 kPa

D. Breathing room air at sea level in a hyperbaric chamber with twice the normal level of atmospheric pressure:
FiO2 = 0.21 (room air)
Pb = 200 kPa (twice the normal atmospheric pressure)
PH2O = 6.25 kPa
PIO2 = 0.21 × (200 - 6.25) = 40.69 kPa

Option D