Cellular Engineering - Principles of Biomedical Engineering - page 266

Y

⎛ ⎞

[]

[ ]

(7.13) ln = nLnL ln − ln

⎜ ⎟ 50

⎝ ⎠

1 − Y

The exponent, n , is the slope of the line. Plotting saturation versus oxygen pres-
sure (or concentration), apparent cooperativity can be determined from the slope.
For Hb the slope is 2.8, that is, Hb is partially cooperative as the 4 subunits only
partially cooperate to behave like 2.8 completely cooperative subunits.

EXAMPLE 7.2

Suppose you get the following data for a type of hemoglobin that you have isolated from
a new animal species:

(a) What is the Hill coefficient? What is the P 50 ?

(b) Describe the cooperativity and affinity of this new hemoglobin as compared to

normal human hemoglobin.

Y%
(HbO 2 /Hb)

pO 2
(mmHg)

25.1 39.8

6.3 15.8

1.6 6.3

0.4 2.5

Solution: Using the given data calculate the X- and Y-axes in (7.13).

Y

⎛ ⎞

pO 2
(mmHg) Y

Y

ln

⎜ ⎟

Y%

⎝

1 Y

⎠

ln(pO 2 ) −

1 Y

−

25.1 39.8 0.251 0.3351

− 1.093

3.684

6.3 15.8 0.063 0.0672 − 2.700 2.760

1.6 6.3 0.016 0.0163 − 4.119 1.841

0.4 2.5 0.004 0.004 − 5.517 0.916

(a) Plot ln(Y/(1 − Y) and ln(pO 2 ) and fit linear regression lines and obtain the slopes

and intercepts. From the slope, Hill coefficient is 1.593.

From the intercept, n*ln(p 50 ) = 7.022.

Hence, P 50 = 82.12 mmHg

(b) The cooperativity of the new hemoglobin is positive but less than the normal

hemoglobin.