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+--- Thread: NBMe 3 section qn 4 - rtpcr (/showthread.php?tid=552938)
NBMe 3 section qn 4 - rtpcr - ArchivalUser - 12-05-2010
Cud someone please explain that
thx
0 - ArchivalUser - 12-05-2010
can u post number properly as nbme form number; block number ;and q number.
0 - ArchivalUser - 12-05-2010
NBME =3
Section = 1
Qn = 4
0 - ArchivalUser - 12-05-2010
Hi
This question in fact is three questions.
First you should know the normal osmolarity of a cell, in this example RBC, which is
300 mOsm/L.
Second you should know the direction of water diffusion (osmosis) which occurs
down the gradient of water concentration. In other words, water diffuces from a
compartment with few solutes to a compartment with higher solutes. Here be careful
not to be confused with water concentration vs. solute concentration.
Third, you should know the units. mM/L vs. mOsm/L. Osmole is concentration
of all particles. For example, 100 mM/L of NaCl = 200 mOsm/L NaCl or 100 mM/L CaCl2 = 300 mOsm/L CaCl2.
Now with this information you can answer the question: the graph shows that volume of RBC
has doubled. It means that it has been put in a solution with a higher concentration of water
and less solutes. So the osmolarity of the solution should be less than 300 of the RBC.
In first glance, it seems there are two options 75 and 150. But these are in mM and 150 mM of NaCl is really 300 mOsm and this is not the correct answer. Therefore, you have only
one option which is 75 mM = 150 mOsm and it make sense regarding the above facts.
Hope to be useful