Question

Need cardiac output and respiratory physiology wiz

Part 2 BONUS 10 free points to apply to any assignment TIME LIMIT 30 Minutes (time yourself) Open book, notes, and a collaboration with ONE class mate is allowed. Nancy has a resting tidal volume of 470 ml/breath, Vo of 120 ml/breath, and a respiration frequency of 14 breath/min. The O2 content of her inspired air is 20.94% and of the expired air at rest is 16-44%. Her arterial blood is 95% saturated and her venous blood 75%. Ignore the O2 dissolve in plasma. Her Hb concentration was 18 g/dl blood, (1 gram of Hb carries 1.34 ml of O2 when fully saturated). If her stroke volume is 95 ml/beat, What is her heart rate at rest ? 20 pts Nancy's SBP was 110 and DBP 30. What is her TPR at rest. 5 pts Stop ! Think ! Lay out your equations! Figure out what you need to know and think how you can calculate it. Now solve the problem. SHOW ALL YOUR WORK -KEEP TRACK OF UNITS IN ALL CALCULATIONS

Answer 1

We know that Cardiac output (Q)= Heart rate(HR) x stroke volume (S)

____________________________________________________________________________________________________

Fick’s principle states that the rate of diffusion is proportional to the difference in concentration.

So, the volume of oxygen consumed per unit time is proportional to the difference in oxygen content between arterial and venous blood

we can write oxygen consumption (V_O2 ) = cardiac output(Q) x [ arterial oxygen concentration(C_a) - venous oxygen concentration (C_v) ]

So ,   V_O2 = Q( C_a - C_v )

_____________________________________________________________________________________________________

Now from question we know the arterial oxygen content, which is

Oxygen carried by hemoglobin + plasma dissolved oxygen

= (1.34 x Hb x hemoglobin saturation of oxygen) + (0) [ according question dissolved oxygen to be discarded]

= ( 1.34 x 18 x 95/100 ) +0            [ in question it is given that in arterial blood Hb is 95% saturated ]

= 22.914 ml oxygen in 100 mL arterial blood = 0.2291 ml O2 / ml arterial blood (C_a)

Same way in oxygen content in vein venous blood:

( 1.34 x 18 x 75/100 ) + 0

= 18.09 ml oxygen in 100 mL blood of vein = 0.1809 ml o2 / 1 ml venous blood (C_v)

____________________________________________________________________________________________________

We know (V_D) is physiological dead space which does not takes part in the oxygen diffusion.

So amount of inhaled air that takes part in the gaseous exchange :

tidal volume(V_t) - Physiological dead space (V_D)

= 470 - 120 ml / breath

= 350 ml / breath

or ( 350 x 14) = 4900 ml /minute

____________________________________________________________________________________________________

Now we need to calculate oxygen consumption from inhaled and exhaled oxygen percentage given in the question

So oxygen consumption ( V_O2):

Oxygen in inspired air - Oxygen exhaled air

(4900 * 20.94/100) - (1900 * 16.44/100 )

= 4900 ( 0.045)

=220.5 ml / min

____________________________________________________________________________________________________

So now we can write that :

V_O2 = Q ( C_a - C_v )

220.5 = Q ( C_a - C_v )

220.5 = Q ( 0.2291 - 0.1809 )

Q = 220.5 / 0.0482 = 4574.7 mL 4575 ml

null

So, we can write:

Heart rate(HR) x Stroke volume(S) = 4575

or HR = 4575 / S = 4575 / 95 = 48.15 $\approx$ 48 beats / minute

_____________________________________________________________________________________________________

Part 2:

We know

TPR ( Total peripheral Resistance ) = ( Mean arterial blood pressure - Central venous pressure ) / cardiac output

Now in resting stste central venous pressure is near 0

So we can write TPR = Mean arterial blood pressure (MAP) / cardiac output( Q)

Now we need to determine MAP

Generally

MAP = Diastolic blood pressure (DP) + 1/3 pulse pressure

        = DP + 1/3 ( Systolic pressure[SP] - Diastolic pressure[DP] ) We know [ pulse pressure = (SP) - (DP) ]

       = DP + 1/3 ( SP - DP)

      = 30 + 1/3( 110 - 30)

      = 30 + 80/3

      = 170/3 = 56.67 mmHg

So TRP = 56.67 / 4575 mmHg.min. ml^-1 $\approx$ 0.0124 mmHg.min.mL^-1