1. Introduction to Biotransport Housekeeping… syllabus…
BMOLE Biomolecular Engineering Engineering in the Life Sciences Era
BMOLE – Transport
Introduction to Biotransport
Housekeeping… syllabus…
Inner life of cell video (get your Kleenexes ready… it is a tear jerker)
Dr. Corey J. Bishop Assistant Professor of Biomedical Engineering Principal Investigator of the Pharmacoengineering Laboratory: pharmacoengineering.com Dwight Look College of Engineering Texas A&M University Emerging Technologies Building Room 5016 College Station, TX
© Prof. Anthony Guiseppi-Elie; T: F:

2. 1.1
The oxygen diffusion coefficient in tissue is about 1.1e-5 cm2/s. The fluid filtration velocity is typically 1 μm/s.
A) how large of a distance between capillaries would be needed for convection to influence oxygen transport to tissues?
B) Based upon reported values for the distance between capillaries, do you think that convection is an important mechanism for oxygen transport in tissues?

3. Problems at the end of Chapter 1: ICPS
1.1:
𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s
A. how large of a distance between capillaries would be needed for convection to influence oxygen transport?

4. Problems at the end of Chapter 1: ICPS
1.1:
𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s
A. how large of a distance between capillaries would be needed for convection to influence oxygen transport?
𝑃𝑒= 𝑣𝑒𝑙∗𝐿 𝐷 𝑖𝑗 =1 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠 = 1𝑒−4 𝑐𝑚 𝑠 ∗𝐿 1.1𝑒−5 𝑐 𝑚 2 𝑠 ; L = 0.11 cm

5. Problems at the end of Chapter 1: ICPS
1.1:
𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s
B. Quantitatively show importance of convection of oxygen transport in tissues?

6. Problems at the end of Chapter 1: ICPS
1.1:
𝐷 𝑡𝑖𝑠𝑠𝑢𝑒 =1.1𝑒−5 𝑐 𝑚 2 𝑠 ; fluid filration = 1 um/s
B. Quantitatively show importance of convection of oxygen transport in tissues?
Distance between capillaries = 1e-2 cm and oxygen needs to travel half of this: so…
𝑃𝑒= 𝑣𝑒𝑙∗𝐿 𝐷 𝑖𝑗 =1 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠 = 1𝑒−4 𝑐𝑚 𝑠 ∗0.5𝑒−2 𝑐𝑚 1.1𝑒−5 𝑐 𝑚 2 𝑠 = … so convection is negligible…

7. Solubility of oxygen in plasma at 37C is 1.4e-6 mol/L/mmHg.

8. Solubility of oxygen in plasma at 37C is 1.4e-6 mol/L/mmHg.
Heme concentration in RBCs is mol/L = 4CHb

9. Solubility of oxygen in plasma at 37C is 1.4e-6 mol/L/mmHg.
Heme concentration in RBCs is mol/L = 4CHb
Hct male = 0.45; Hct female = 0.4

10. ICPS: 1.2
Objective:
Fraction of oxygen in solution vs bound to hemoglobin in arterial and venous blood (male and female)?
Assume solubility of oxygen in RBCs =solubility in serum
CO2 = HO2*(PO2 )+ 4CHbSbar*Hct
𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡

11. CO2 = HO2*(PO2 )+ 4CHbSbar*Hct
HO2 = oxygen solubility in plasma
HHb = oxygen solubility in hemoglobin
CO2 = oxygen concentration in blood
PO2 = oxygen partial pressure = 95 mmHg arterial and 38 mmHg venous
CHb = hemoglobin concentration in blood
Sbar = average fractional saturation of hemoglobin =95% art and 70% for venous
Hct = hematocrit (fraction of whole blood that is RBCs)
Female Hct: 0.4

12. 1.2
HO2 = HHb = 1.4e-6 M/PO2
CO2 = HO2*(PO2 )+ 4CHbSbar*Hct; 4CHb= M
CO2 = HO2*(PO2 =95 or 70 mmHg) M Sbar*(Hct=0.4 or 0.45)
CO2 = HO2*(PO2 =95 or 70 mmHg)+ ( M female or M men) (Sbar = 95% art or 70% venous
Genderd
where
Co2 (M) =
Ho2*PO2 (M)
P02 (mmHg) +
4CHb (M)
Sbar (fr)
Hct (f)
Female
art 95
0.0203
0.95
0.4
female
ven 38
0.70
male
0.45
PLASMA
BOUND TO HEMOGLOBIN
1.724%
98.276%
0.936%
99.064%
1.532%
98.468%
0.832%
99.168%

13. 1.3
Difference in CO2 in serum is 2.27 mL (pre- vs post-lungs) per 100 mL (70%)
And released 1.98 mL Co2 from RBCs (pre- vs post-lungs) per 100 mL
Actual total lost = 2.18 mL of Co2/100 mL
O2 goes from 38 to 100 mmHg
Arterial oxygen concentration is M vs venous at M
PV=nRT; 1 mol of gas = 22.4 L at standard thus per 100 mL we gained 5.58 mL of oxygen…
Greater propensity to gain oxygen than to lose co2.

14. 1.4 L = 1um thick; s.s. = 0.33 seconds
Is oxygen diffusion across the alveolus a significant factor in the time required for hemoglobin to oxygenate as it traverses the capillary? A.k.a is this diffusion-limited?
L2/Dij = diffusion time = (1e-4)2 cm2/(2e-5cm2/s)
= s (vs 0.33 seconds);
Not diffusion limited.

15. 1.5 LSA = lateral surface area… order Order Diam (mm) Length #
Circ. (mm)
LSA/length *mm^2)
total surface area (cm^2)
area of each
vol of each
tot vol (cm^3)
cum sa
cum vol 11 1
0.024
0.116
300358
E-05 10 2
0.044
0.262
97519 9 3
0.073
0.433
31662 8 4
0.122
0.81
9736 7 5
0.192
1.51
2925 6
0.352
2.72
774
0.533
4.6
202
0.875
8.19 49
1.519
14.26 12
2.486
11.87
5.08 25

16. 1.6
Essentially same as 1.5

17. 1.7 Renal artery flow rate = 1.25 L/min
A) Water fraction across glomerulus?
B) renal vein flow rate?
C) [Na+] leaving glomerulus and in renal vein

18. Relevant Eqn Mass balance RPF = renal plasma flow
CarteryRPFartery –CveinRPFvein = CurineQbladder
In – out = what was diverted
RPFartery =RPFvein + Qbladder
Water content for male and female = 55% and 60%
Per day: CO = 4-8 L/min; vs 990 L/day male and 1080 L/day for female for kidney filtration
A) So Males = % and females %
B) from table excretion rate of h20 = 1.44 L/day
So 1.25 L/min (from problem) –1.44 L/day/1440 min/day=1.249L/min

19. C) CarteryRPFartery –CveinRPFvein = CurineQbladder
[Na+ vein]/time = [Na+ artery]/time – [Na+ excreted]/time
From table: Filtered: 25,200 mmol/day *180 L/day = 140 mM = Cartery
From table: excreted: 150 mmol/day
CarteryRPFartery –CveinRPFvein = CurineQbladder
140mM*1.25 L/min – Cvein* L/min = Curine* L/min
Curine = 150 mmol/day / 1.44 L/day (table) = mM
Cvein = mM; slight increase in [Na+] due to loss of water volume

20. 1.8 Permeability = 5e-9 cm/s = km Deff in the tissue= 1e-10 cm2/s
Thickness of membrane = 150 um = 150 e-4 cm = 1.5 e-2 cm
A) Biot # = ratio of relative resistances from membrane to tissue = kmL/Deff
If Biot >> 1, then tissue-limited or tissue is providing the greatest resistance to transport
If Biot << 1, then membrane-limited or membrane is providing the greatest resistance to transport
Biot # = (5e-9 cm/s)*(1.5e-6 cm)/(1e-10 cm2/s) = 0.75
Similar transport resistances in membrane compared to tissue.

21. 1.9 VdotOxygen=Q(Cv-Ca) = oxygen consumption rate
Q = pulmonary blood flow
Cv or a = venous or arterial oxygen concentrations
HO2 = 1.33e-6 M/mmHg
HHb = 1.5e-6 M/mmHg
𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡
M/mmHg
mmHg
Fraction
Fraction
Fraction
mmHg
M/mmHg
𝑃 𝑜2 = 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 = 𝑆 𝑏𝑎𝑟

22. 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 2.6 1+ 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 2.6 = 𝑆 𝑏𝑎𝑟

23. 1.9 continued
𝑆 𝑏𝑎𝑟 𝑓𝑜𝑟 𝑣𝑒𝑛𝑜𝑢𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝑠𝑎𝑚𝑒 𝑓𝑜𝑟 𝑣𝑒𝑛𝑜𝑢𝑠 𝑑𝑢𝑟𝑖𝑛𝑔 𝑟𝑒𝑠𝑡 𝑜𝑟 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒) =
𝑃 𝑜2 = 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 = assuming P02 is 100 mmHg, 0.971
𝑆 𝑏𝑎𝑟 𝑓𝑜𝑟 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑟𝑒𝑠𝑡 = 𝑃 𝑜2 =40 𝑚𝑚𝐻𝑔 𝑃 𝑃 𝑜2 𝑃 =
𝑆 𝑏𝑎𝑟 𝑓𝑜𝑟 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑟𝑖𝑛𝑔 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒 = 𝑃 𝑜2 =15 𝑚𝑚𝐻𝑔 𝑃 𝑃 𝑜2 𝑃 =

24. Gender
where
state
Co2 (M) =
Ho2*PO2 (M)
P02 (mmHg)
(1-Hct) +
Hct (f)
(4CHb (M)
Sbar
Hhb*PO2
Po2)
Female
art
rest
5.32E-05
40.00
0.6
0.4
0.0203
0.754
6.00E-05
exercise
2.00E-05
15.00
0.193
2.25E-05
female
ven
1.33E-04
100.00
0.971
1.50E-04
male
0.55
0.45
=1.33e-6 M/mmHg
=1.5e-6 M/mmHg
VdotOxygen Q*
(Cv --
Ca)
Rest
5.8
Exercise 25
Male
moles/min
L/min M

25. 1.10 A) determine rate of oxygen removal from lung venules given:
Inspired air is 21% oxygen; alveolar oxygen partial pressure is 105 mmHg; blood flow rate of 5 L/min; respiration of 10 breaths per minutes/ and 0.56 L per breath; alveolar volume and dead volume is 0.15 L and T=37C; R = L*atm/mol/K and 1atm = 760 mmHg
𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 =𝑄( 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 )

26. Cinhaled=Pinhaled/RT and Calveolar=Palveolar/RT
𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 =𝑄( 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 )
PV = nRT; P/RT = n/V = C
Cinhaled=Pinhaled/RT and Calveolar=Palveolar/RT
Cinhaled=(Pinhaled=0.21*1 atm)/(RT= L*at/mol/K*310K)= M
Calveolar=(Palveolar=105/760 atm)/RT = M
Males: 0.56 L/breath and females: 0.45 L/breath
Dead volume males: 0.19L and female: 0.14 L
Males: 10 breaths/min (Inhaled vol – dead volume)
Males = 3.7 L/min and females = 3.1 L/min
𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 = moles of oxygen/min for males and moles of oxygen/min for females
Knowing , 22,400 L of oxygen per mole then 236 mL of oxygen/min for males and 197 mL of oxygen/min for females

27. 1.10B
What is 𝑉 𝑑𝑜𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 =𝑄( 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 )
given that you know, during exercise, blood flow rate can be 25 L/min, and rises to 30 breaths per minute and oxygen demand increases to 4 L/min. Partial pressure of oxygen in the pulmonary artery declines to 20 mmHg but remains at 100 mmHg in the pulmonary vein. Given that 4CHb is M; Hct for females is 0.4 and for males it is HO2 = 1.33e-6 M/mmHg; HHb = 1.5e-6 M/mmHg
𝐶 𝑂2 = 𝐻 𝑂2 𝑃 𝑂2 1−𝐻𝑐𝑡 + 4 𝐶 𝐻𝑏 𝑆 𝑏𝑎𝑟 + 𝐻 𝐻𝑏 𝑃 𝑂2 𝐻𝑐𝑡
𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 𝑃 50 =26 𝑚𝑚𝐻𝑔 = 𝑆 𝑏𝑎𝑟 = for f/m vein (0.971) artery (0.336)
note the pulmonary artery during rest is ~0.754.
Hence 𝐶 𝑂2 is:
Gender
where
exc
Co2 (M) =
Ho2 (M/mmHg)*PO2
P02 (mmHg)
(1-Hct) +
Hct (f)
(4CHb (M)
Sbar
Hhb*PO2
Po2)
Female
art
Exc
2.66E-05
20.00
0.6
0.4
0.0203
3.00E-05
female
ven
1.33E-04
100.00
1.50E-04
male
0.55
0.45
=1.33e-6 M/mmHg
=1.5e-6 M/mmHg
𝑉 𝑑𝑜 𝑡 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 = 𝑄 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 = 4 𝐿/𝑚𝑖𝑛 𝐶 𝑣𝑒𝑛𝑜𝑢𝑠 − 𝐶 𝑎𝑟𝑡𝑒𝑟𝑖𝑎𝑙 𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 ; V dot oxygen = V dot inhalation (𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 )
Vdotinhaled = Q*
(Cv --
Ca)/(Cinhaled-Calveolar)
C=n/V = P/RT
Female 25 /
Male
L/min M

28. 1.11 Stroke volume Peripheral Resistance Powder of Left side of heart
Athlete
Rest and exercise
Sedentary Person

29. CO = HR*SV Peripheral Resistance = mean arterial pressure/CO
Athlete:
Raise from 5 L/min to 25 L/min
Raise from 60 bpm to 105 bpm
Raise arterial pressure from 100 to 130 mmHg
Sedentary Person:
Raise from 72 bpm to 125 bpm
Raise arterial pressure from 100 to 150 mmHg
exercise
CO=
HR* SV
Athlete 25
105 SP
125
0.2
rest 5 60 72
Periph resistance =
mean art pressure/ CO
5.2
130 6
150 20
100

30. What is the Power during rest/exercise for athlete/sedentary person?
𝑃𝑜𝑤𝑒𝑟=𝑤𝑜𝑟𝑘∗𝑏𝑝𝑠=𝑏𝑝𝑠 𝑃 𝑎 𝑑𝑉 =𝑏𝑝𝑠 𝑃 𝑎 𝑑𝑒𝑙𝑡𝑎𝑉 = [J/s]
bps = beats/sec
Mean arterial pressure is assumed to be constant
And in Pa, not mmHg
V = ventricular volume
Athlete:
Raise from 5 L/min to 25 L/min
Raise from 60 bpm to 105 bpm
Raise arterial pressure from 100 to 130 mmHg
Sedentary Person:
Raise from 72 bpm to 125 bpm
Raise arterial pressure from 100 to 150 mmHg
exercise
CO=
HR* SV
SV m^3
arterial pressure
Power (use m^3 not L and use Pa not mmHg)
Athlete 25
105 SP
125
0.2
0.0002
rest 5 60
E-05 72
E-05
1 mmHg = Pa = N/m2 1 L = m3
WHAT?!?!?!
Problem 1.11 is 1.11?!
That is very exciting!

31. 1.12
Body adapts to reduced barometric pressure to extract sufficient oxygen and do work.
@3650 m the barometric pressure = 485 mmHg
For an oxygen pressure drop of 30 mmHg, determine oxygen uptake rate for a respiration of 20 breaths per min.
Estimate oxygen saturation in venous blood if hct rises to 0.6 and the partial pressure of oxygen blood is at a partial pressure equal to 98% of the alveolar level.

32. 1.12
Body adapts to reduced barometric pressure to extract sufficient oxygen and do work.
@3650 m the barometric pressure = 485 mmHg
Typically the partial pressure of oxygen is mmHg so mmHg *485/760 = mmHg and the alveolar pressure would thus be mmHg with a pressure drop of 30 mmHg.
C = P/RT = C=
101.85
mmHg
1/760
mmHg/atm = M
L atm/(mol*K)
310 K
71.85

33. V dot oxygen = V dot inhalation (𝐶 𝑖𝑛ℎ𝑎𝑙𝑒𝑑 − 𝐶 𝑎𝑙𝑣𝑒𝑜𝑙𝑎𝑟 )
V dot inhalation =f(Vinhalation −Vdead) = 20 bpm*(0.56L-0.19L) = 7.4L/min
Vdot oxygen =
7.4
L/min*
(Cinhaled - Calveolar)
( M M)
mol/min
Assumed to be the same

34. 1.12
Estimate oxygen saturation in venous blood if hct rises to 0.6 and the partial pressure of oxygen blood is at a partial pressure equal to 98% of the alveolar level The venous blood oxygen saturation will be 0.98 of the alveolar so mmHg *0.98 = mmHg and so oxygen saturation is
𝑃 𝑜2 = 𝑃 50 =26 𝑚𝑚𝐻𝑔 𝑃 𝑜2 = 𝑃 50 =26 𝑚𝑚𝐻𝑔 = 𝑆 𝑏𝑎𝑟 =0.9302

35. Basal metabolic rate of resting individual = 1650 kcal/day
Fraction of metabolic energy used to pump blood: recalling problem 1.11
resting indivudal expends
1650
kcal/day
how is kcal = Joule
4.184 kJ/kcal
6903.6
kJ/day
287.65
kJ/hr
kJ/min
kJ/s
J/s
at rest
1.11
J/s expended for heart so
of energy is expended on pumping heart

36. Filtration rate mM
Excretion rate
[]urine/[]plasma
water
180
L/day
1.44
Na+
25200
mmol/day
140
150
transport out of urine K+
720 4
100
Into urine
glucose
800
0.5
Out of urine
Urea
933
467

37. 1.15 Low mw sugar for assessing Kidney function (i.v. and
Check blood and urine conc
In time. Inulin is not
reabsorbed
𝐶 𝑖𝑛𝑢𝑙𝑖𝑛 𝑖𝑛 𝑝𝑙𝑎𝑠𝑚𝑎 𝐺𝐹𝑅= 𝐶 𝑖𝑛𝑢𝑙𝑖𝑛𝑒 𝑖𝑛 𝑢𝑟𝑖𝑛𝑒 𝑄 𝑢𝑟𝑖𝑛𝑒
Plasma
0.001
g/mL
GFR =
0.125
1.44
L/day
180 1
mL/min
7.5
L/hr
L/min
125