Answer:
Explanation:
Given that:
P₅₀ = 20 torr in muscle
The partial pressure of O2 in lungs Po₂ = 100 torr
The hill coefficient [tex]n_h[/tex] = 3
According to the Hill coefficient;
[tex]\dfrac{\Upsilon}{1-\Upsilon}= (\dfrac{PO_2}{P_{50}})^n[/tex]
here;
[tex]\Upsilon[/tex] = fraction of haemoglobin monomers
[tex]\dfrac{\Upsilon}{1-\Upsilon}= (\dfrac{100}{20})^3[/tex]
[tex]\dfrac{\Upsilon}{1-\Upsilon}[/tex] = 125 θ lungs
However for θ muscle :
[tex](\dfrac{PO_2}{P_{50}})^n=( \dfrac{20}{20})^3[/tex]
= 125 (θ lungs) - 1 (θ muscle)
= 124
B.
if nh =1
[tex]\mathtt {\theta \ lungs} = \dfrac{\Upsilon}{1-\Upsilon}= (\dfrac{100}{20})^1[/tex]
[tex]\mathtt {\theta \ lungs} =5[/tex]
[tex]\mathtt {\theta \ muscle} = \dfrac{20}{20}^1[/tex]
(θ muscle) = 1
(θ lungs) - (θ muscle) = 5 - 1
(θ lungs) - (θ muscle) = 4
A) The efficiency of hemoglobin at O₂ delivery = 124
B) Efficiency of O₂ delivery will change by : 4
Given data:
Partial pressure of O₂ in lungs ( PO₂ ) = 100 torr
Partial pressure of O₂ in muscle = ( P₅₀ ) = 20 torr
Hill coefficient nh = 3
r = fraction of hemoglobin monomers
A) Determine the efficiency of hemoglobin at O₂ delivery
Given that Hill coefficient = [tex]\frac{r}{1-r} = (\frac{PO_{2} }{P_{50} } )^{n}[/tex]
= [tex]\frac{r}{1-r} = (\frac{100}{20})^{3}[/tex] = 125 ∅ Lungs
For Muscle :
[tex](\frac{20}{20} )^{3}[/tex] = 1 ∅ muscle
Therefore the efficiency of hemoglobin at 0₂ delivery
= ( ∅ lungs - ∅ muscle )
= ( 125 - 1 ) = 124
B ) Determine the change in efficiency of O2 delivery
when hill coefficient ( nh ) = 1
for lungs :
[tex]\frac{r}{1-r} = (\frac{100}{20})^{1}[/tex] = 5
for muscle :
[tex](\frac{20}{20} )^{1}[/tex] = 1
Therefore change on efficiency = ( 5 - 1 ) = 4
Hence we can conclude that The efficiency of hemoglobin at O₂ delivery = 124 and Efficiency of O₂ delivery will change by 4
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