Question

4. Given the following data: Mrs. Blanche Dubois is 52 years old, lives in Leaksville, is 62 inches tall, and has a white cat named Fluffy. Mr. Homer Jones of Ruffin is her brother. While assessing her history and physical, you note that she has a blood pressure of 125/75 mm Hg, a heart rate of 105/min., a respiratory rate of 17/min., and has diabetes. Her lungs are clear and normal (there is no alveolar dead space present). What is her approximate: Normal predicted resting tidal volume? Anatomic dead space? Minute ventilation (VE)? Alveolar ventilation in one minute (VA)? Pulse pressure? Mean Arterial Pressure?

Answer 1

* Normal predicted resting tidal volume = approximately 500 mL in one inspiration or 7 mL/kg in normal adult

Ideal body weight(IBW)= 45.5kg + 2.3( height in inches -60)

Substituting the values ,IBW= 45.5 +2.3×(62-60) =50.1~ 50kg.

Since body weight is not given ,IBW=50kg

Blanche's normal resting tidal volume = 7ml/kg ,

for 50 kg ,tidal volume = ~350ml.

*Anatomic dead space : It is volume of air in the trachea, bronchus, bronchioles and the terminal bronchioles. In normal person anatomic dead space is approximately 2 mL/kg or 150 ml

Her anatomic dead space = 2ml×50= ~ 100ml

* Minute ventilation : It is volume of gas inhaled and exhaled in 1min.

Minute ventilation (VE)= tidal volume × respiratory rate

VE= 350ml ×17 breathe/min = 5950ml/min or 5.95 L/Min

*Alveolar ventilation = ([tidal volume -dead space] × respiratory rate)

([350 -100] ×17) =250×17

= 4250 mL/min

*Pulse pressure : It is the difference between systolic and diastolic blood pressure.

Systolic pressure =125

Diastolic pressure = 75

  125 - 75 = 50 mmHg

*Mean arterial pressure

= (systolic blood pressure +[2× diastolic blood pressure ])÷3

=(125 + [2×75])÷3

=275÷3

=91.7 mmHg.