Question

Subject 3 has the following values: Minute ventilation = 8, liters/minute, Respiratory frequency = 4 Br/minute, PaCO2 = 20 mm Hg, PECO2 = 15 mm Hg. Identify whether each of the following is correct or incorrect.

a. VD/VT is within normal limits

b. VT is within normal limits

c. the subject is hyperventilating

d. the PaCO2 is low because the patient is breathing rapidly

e. the PaCO2 is low because the alveolar ventilation is higher than normal.

Answer 1

Questions:

Subject 3 has the following values: Minute ventilation = 8, liters/minute, Respiratory frequency = 4 Br/minute, PaCO2 = 20 mm Hg, PECO2 = 15 mm Hg. Identify whether each of the following is correct or incorrect.

a. VD/VT is within normal limits

b. VT is within normal limits

c. the subject is hyperventilating

d. the PaCO2 is low because the patient is breathing rapidly

e. the PaCO2 is low because the alveolar ventilation is higher than normal.

Answers

Given parameters

• Minute ventilation = 8, liters/minute
• Respiratory frequency = 4 Br/minute
• PaCO2 = 20 mm Hg
• PECO2 = 15 mm Hg.

a. VD/VT is within normal limits

Answer:

Correct

Rationale:

According to Bohr the ratio of the dead space (Vd) ventilation to tidal volume (Vt) is a measurable variable denoted as Vd/Vt = (PaCO₂ - PECO₂)/PaCO₂. Normal values are 0.20–0.40.

VD/VT =20-15/20

            =.25

Normal value of VD/VT is 0.20–0.40.

Patient value is 0.25.therefore,patient’s VD/VT is within normal limits.

b. VT is within normal limits

Answer:

Incorrect. VT is higher

Rationale:

Minute ventilation (VE) is the total volume of gas entering (or leaving) the lung per minute. It is equal to the tidal volume (TV) multiplied by the respiratory rate (f). Minute ventilation = VE = TV x f

Minute ventilation (or respiratory minute volume or minute volume) is the volume of gas inhaled (inhaled minute volume) or exhaled (exhaled minute volume) from a person's lungs per minute.

Minute ventilation = VE = TV x f

Minute ventilation=VT X respiratory rate

SO,

   VT = minute volume/respiratory rate

   VT =   8/4

        =2 L

Tidal volume (symbol V_T or TV) is the lung volume representing the normal volume of air displaced between normal inhalation and exhalation when extra effort is not applied.In a healthy, young human adult, tidal volume is approximately 500 mL per inspiration or 7 mL/kg of body mass.

c. The subject is hyperventilating

Answer:

Correct.

Rationale:

Hyperventilation results in reduction in arterial pressure of carbon dioxide (PaCO₂), which causes vasoconstriction, thus reducing CBF, cerebral blood volume, and, subsequently, ICP. Partial pressure of carbon dioxide (PaCO2): 38 to 42 mm Hg.

Patient value= PaCO2 = 20 mm Hg

Therefore, the subject is hyperventilating

d. The PaCO2 is low because the patient is breathing rapidly

Answer:

The statement is Correct. But in this case .the respiratory rate is 4 Br/minute. So increase in Paco2 can be because of any other physiological condition.

Rationale:

the PaCO2 is low in case of Hypocapnia.Hypocapnia or hypocapnea is known as hypocarbia.It is a state of reduced carbon dioxide in the blood. Hypocapnia usually results from deep or rapid breathing, known as hyperventilation.The main physiologic causes of hypocapnia are related to hyperventilation. Hypocapnia is sometimes induced in the treatment of medical emergencies such as intracranial hypertension and hyperkalaemia.

e. The PaCO2 is low because the alveolar ventilation is higher than normal

Answer:

Correct

Rationale:

Partial pressure of carbon dioxide (PaCO2): 38 to 42 mm Hg

Hypocapnia occurs when alveolar ventilation is excessive relative to carbon dioxide production and usually results from hyperventilation due to hypoxia, acidosis or lung disease.