Help me biochem whizzes!

[free99]

Assume a hemoglobin has p50 = 30 torr, what percentage of O2 is unloaded by this protein from lung (pO2 = 100 torr) to muscle (pO2 = 20 torr)?

The answer is 75%... where does that come from?

We were given the hill equation (yO2=pO2^n/(pO2^n+p50^n)) but I don't see how they got 75%... I'm totally confused, help me obi wan kenobi 😕

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[Jbrowndds]

Assume a hemoglobin has p50 = 30 torr, what percentage of O2 is unloaded by this protein from lung (pO2 = 100 torr) to muscle (pO2 = 20 torr)?

The answer is 75%... where does that come from?

We were given the hill equation (yO2=pO2^n/(pO2^n+p50^n)) but I don't see how they got 75%... I'm totally confused, help me obi wan kenobi 😕

I have not even started on pre reqs so now I'm freaked out. What IS that lol. Well I'm ready for anything.

 

[free99]

I have not even started on pre reqs so now I'm freaked out. What IS that lol. Well I'm ready for anything.

Stick with it - focus on the basics and build up... Biochem is a bit of a nightmare, but I'm working through it. Having a nearly incomprehensible asian man as the professor certainly doesn't help

 

[Jbrowndds]

Stick with it - focus on the basics and build up... Biochem is a bit of a nightmare, but I'm working through it. Having a nearly incomprehensible asian man as the professor certainly doesn't help

Math is my weakest point. But I'm sure ill find a good tutor. I'm sure someone knows the answer to your question.

 

[UCSFx2017]

Assume a hemoglobin has p50 = 30 torr, what percentage of O2 is unloaded by this protein from lung (pO2 = 100 torr) to muscle (pO2 = 20 torr)?

The answer is 75%... where does that come from?

We were given the hill equation (yO2=pO2^n/(pO2^n+p50^n)) but I don't see how they got 75%... I'm totally confused, help me obi wan kenobi 😕

The Hill equation is also expressed as:

log [theta/(1-theta)] = n log PO2 - n log P50

You are solving for PO2, which is your percent of oxygen unloaded.
​

log [theta/(1-theta)] + n log P50 = n log PO2

What is n? N is the Hill constant where n = 1 indicates no cooperation and n > 1, usually between 2.8-3.0, indicates cooperation. I'm assuming n = 1.
​

log [theta/(1-theta)] + log P50 = log PO2

Theta probably indicates bound oxygen. In this case the ~100% bound oxygen from lung becomes only ~20% bound oxygen by the time it gets to the muscles.
​

log [0.20/(1-0.20)] + log P50 = log PO2

P50 is given as 0.30.
​

log [0.20/0.80] + log (0.30) = log PO2

Simplify the left side.
​

- 0.602059991 + (- 0.522878745) = -1.124938736 = log PO2

Get rid of the log on the right side.
​

10^(-1.124938736) = 10^log PO2 = PO2
0.075 = PO2.
PO2 = 7.5%, instead of 75%.

Yet your question asks for hemoglobin which has a hill constant of 2.8. I use n = 1 thinking it was myoglobin. I'm clueless.

 

[UCSFx2017]

http://faculty.smu.edu/svik/5310/5310lectures/lect11.html

If someone figured it out, please share.

 

[free99]

The Hill equation is also expressed as:

log [theta/(1-theta)] = n log PO2 - n log P50

You are solving for PO2, which is your percent of oxygen unloaded.
​

log [theta/(1-theta)] + n log P50 = n log PO2

What is n? N is the Hill constant where n = 1 indicates no cooperation and n > 1, usually between 2.8-3.0, indicates cooperation. I'm assuming n = 1.
​

log [theta/(1-theta)] + log P50 = log PO2

Theta probably indicates bound oxygen. In this case the ~100% bound oxygen from lung becomes only ~20% bound oxygen by the time it gets to the muscles.
​

log [0.20/(1-0.20)] + log P50 = log PO2

P50 is given as 0.30.
​

log [0.20/0.80] + log (0.30) = log PO2

Simplify the left side.
​

- 0.602059991 + (- 0.522878745) = -1.124938736 = log PO2

Get rid of the log on the right side.
​

10^(-1.124938736) = 10^log PO2 = PO2
0.075 = PO2.
PO2 = 7.5%, instead of 75%.

Yet your question asks for hemoglobin which has a hill constant of 2.8. I use n = 1 thinking it was myoglobin. I'm clueless.

Great effort there - thanks for putting in the time to shed some light on the problem. Funny coincidence (maybe it's a mistake in the question?) that using a hill constant of 1 would result in an answer that is off by a factor of 10^-1...

 

[free99]

Well, I tried nearly every combo in the book to figure out where 75% came from and could not come up with that answer. I even used his answer (.75=yO2) and solved for pO2 and got something like 44.xxx... So, this one will have to be left inconclusive, I suppose - I'm stumped. Took the test today and lo and behold this exact questions popped up with different numbers. Took an educated guess based on the known values from the problem in this thread and got it right 😎. This is my last ditch effort to finding an answer to this problem out of pure curiosity at this point.

Thanks for the attempt UCSF!