Endocrinology Math Problem #3

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FamilyGuy0395

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Problem: You’ve discovered a new enzyme (called "enzyme X" for reference) which travels in the blood and temporarily alters neuronal signaling when you need it most. The normal concentration of enzyme X in the blood is 0.12 mg/dL. If a man weighs 80 kg, what is the quantity of enzyme X in his blood (assume that 8% of his body weight is blood with a density (or specific gravity) of 1)? Your answer must be given in micrograms (μg).

I believe the solution is 7,680 μg (or 7.68 × 103 μg), but I'd love if someone could confirm I did the problem correctly. I have my work at the ready in case it's needed.
 
Your answer's almost right, but there's a units error. I assume your working went something along these lines, first calculating the weight of the man's blood

\(\displaystyle 8\% \text{ of } 80 \: kg = 6.4 \: kg\)

Then we're assuming blood has a density of 1, so:

\(\displaystyle 6.4 \: kg \times \frac{1 \: m^3}{1 \: kg} = 6.4 \: m^3\)

Then convert to volume:

\(\displaystyle 6.4 \: m^3 \times \frac{10^6 \: cm^3}{1 \: m^3} \times \frac{1 \: mL}{1 \: cm^3} \times \frac{0.01 \: dL}{1 \: mL} = 64000 \: dL\)

And finally, find the concentration of Enzyme X:

\(\displaystyle 64000 \: dL \times\frac{0.12 \: mg}{1 \: dL} = 7680 \: mg\)

But that's the answer in milligrams, not micrograms. One more conversion is needed to finish up.
 
Hmm, I'm confused. Let me write out my work and tell me where I went wrong.

[Enzyme X] = 0.12 mg/dL = 1.2 mg/L = 1,200 μg/L

mperson = 80 kg

mblood = 8% of 80 kg = 0.08 × 80 kg = 6.4 kg

specific gravity = ρblood/ρH2O
1 = ρblood/(1 g/mL)
ρblood = 1 g/mL = 1 kg/L

ρblood = mblood/Vblood
1 kg/L = 6.4 kg/Vblood
Vblood = 6.4 L

[Enzyme X] × Vblood = 1,200 μg/L × 6.4 L = 7,680 μg

Sorry for the formatting being pretty pedestrian, I don't understand how to format math on this site.
 
ksdhart2 said:
Your answer's almost right, but there's a units error. I assume your working went something along these lines, first calculating the weight of the man's blood

\(\displaystyle 8\% \text{ of } 80 \: kg = 6.4 \: kg\)

Then we're assuming blood has a density of 1, so:

\(\displaystyle 6.4 \: kg \times \frac{1 \: m^3}{1 \: kg} = 6.4 \: m^3\)

Then convert to volume:

\(\displaystyle 6.4 \: m^3 \times \frac{10^6 \: cm^3}{1 \: m^3} \times \frac{1 \: mL}{1 \: cm^3} \times \frac{0.01 \: dL}{1 \: mL} = 64000 \: dL\)

And finally, find the concentration of Enzyme X:

\(\displaystyle 64000 \: dL \times\frac{0.12 \: mg}{1 \: dL} = 7680 \: mg\)

But that's the answer in milligrams, not micrograms. One more conversion is needed to finish up.
Click to expand...

\(\displaystyle 6.4 \: kg \times \frac{1 \: m^3}{1 \: kg} = 6.4 \: m^3\) is way too much blood :)
 
Oh, shoot, yes. Units are confusing. Sorry about that. :( The multiplying factor for density should be either:

\(\displaystyle \frac{1 \: cm^3}{1 \: g}\) or \(\displaystyle \frac{1 \: m^3}{1000 \: kg}\)

And that means the original answer would be in micrograms, and therefore correct.

I should have noticed this before. I never even thought about what my result meant - of course it's absurd for someone to have 6400 liters of blood! Honestly, it's pushing it to have 6.4 liters, as the revised math indicates...
 
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