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NBME 21 Answers

nbme21/Block 4/Question#25

An 80-year-old woman cannot concentrate her urine ...


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 +11  upvote downvote
submitted by flashvoyger(11),

The question is asking you how much water must the woman take in to maintain the same osmolality. This woman takes in 450 mOsm of solute per day. This is a unit of measurement -- think of it like grams.

For her to keep the same osmolality she must excrete 450 mOsm per day. The only way for her to excrete the solute is via the kidneys. The only way for her kidneys to excrete 450 mOsm is if they excrete 1 liter of water also. This is the max concentration that her kidneys can produce. (Her kidneys are not “powerful enough” to make her urine any more concentrated than that.)

This woman is also losing another liter of water to feces, sweating and respiration. This is the “insensible water loss”. That means if she losing 1 liter of water to sweat, respiration and feces per day plus 1 liter of water to urine (because she needs to dissolve her solute in something!), for her blood to stay the same osmolality, she must replace the water she lost thus must, at minimum, drink 2 liters of water per day.

jean_young2019  great explanation!Thank you! +  

 +8  upvote downvote
submitted by yotsubato(282),

This question is stupid. Water intake for a healthy individual is 2.0 L a day.

She’s intaking 450 mOsm per day so she needs to excrete 450 mOsm per day to maintain equilibrium. You can’t just excrete mOsm’s by themselves -- they have to be dissolved in some amount of water.

Let’s say you excrete 450 mOsm with 500 mL of water -- that means your kidneys are concentrating urine to:

450 mOsm ÷ 500 mL = 900 mOsm/L

But the maximum this lady’s kidneys can concentrate urine to is 450 mOsm/L, so she has to excrete more water to get it that dilute. That amount of water is 1 L, because 450 mOsm/1 L = 450 mOsm/L.

Now there’s nothing stopping her from excreting the 450 mOsms in an even more dilute urine -- for example if she drank an extra L of water one day, the kidneys could get rid of that extra L with the same amount of 450 mOsm by diluting the urine to 450 mOsm ÷ 2 L = 225 mOsm/L. But the question asks for the minimum amount of water -- which is 1 L by the kidneys (+ 1 L from the other stuff for a total of 2 L).

Water(1) 1Kg water=1L water 450mOsm/kg water=450mOsm/L water Min H2O needed to dissolve 450 solute by kidney is 1L water Water(2) Other (feces et al.)=1L Total: water(1)+ water(2)=2L

To maintain plasma osmolality -> Need to exactly replace all the fluids lost in the day

She cannot concentrate urine above 450 mOsm/kg , so the minimum amount of water required to be excreted by kidneys is 1 (to excrete the 450 mOsm she accumulates per day). The minimum excretion water required is necessarily at max concentration; if you were to say produce diluter urine, say 225mOsm/kg, this would require 2 L of water. The question wants the minimum possible water volume, so we assume she’s concentrating to the max.

1 L losses from kidney + 900 mL insensible + 100 mL in sweat and feces = 2L losses -> need to ingest 2 L of water to replace.