Hi mashhour-ga, I'll answer these one by one, with some explanation
(and the steps, where math is involved) for each:
1. C
We start off with 41 L Total Body Water, then add 2L of fluid, giving
43 L total. The amount of "stuff" dissolved or suspended in the water
doesn't change the result, since the problem states that we should
assume "no losses," and that the system (person) is at equilibrium.
2. C
First, calculate how many milli-Osmoles are present to start with:
mOsm [Tot] = 41 L * 273 mOsm/L = 11,193 mOsm.
Next, we're told that we add 2L of 1.8% saline. That's a
concentration of 273*2 mOsm/L (2 times normal, which is 273mOsm/L).
So, the number of mOsm added is 273*2 mOsm/L * 2L = 1092 mOsm.
Now, the total number of mOsm after adding the 2L of hypertonic saline
is (11,193 + 1092) mOsm = 12,285 mOsm. This is in a total volume of
43 L (see #1), so the final concentration is 12,285 mOsm / 43L = 286
mOsm/L. The closest answer is 288 mOsm/L, so there may be an error in
the choices (typo?).
3. D
When the hypertonic saline is added to the extracellular (vascular)
space, the initial concentration of solute outside the cells will be
greater than that inside the cells. This will drive water out of the
cells, toward the higher NaCl concentration, until the concentration
is the same inside and outside the cells, which is equilibrium for
this system (same amount of water moving out as in).
Before adding the saline, we have how many mOsmoles in the
extracellular space?
mOsm [ECV, initial] = ECV * (concentration of solute in ECV)
= 13 L * 273 mOsm / L = 3549
mOsmoles.
We then add 1092 mOsm (see #2) to this space for a total of (3549 +
1092) mOsm = 4641 mOsm.
Inside the cells, we have 28L * 273 mOsm / L = 7644 mOsmoles.
Now, water moves (not the particles) so that the final concentration
inside equals the final concentration outside. Mathematically, this
is written as follows:
(particles inside)/ICV = (particles outside)/ECV
Also, ICV + ECV = TBW. (So ICV = TBW - ECV)
Substituting the numbers gives the following:
7644 mOsm / (TBW - ECV) = 4641 mOsm / ECV
7644 mOsm / (43 L - ECV) = 4641 mOsm. / ECV
We need to solve this equation for ECV:
7644 mOsm (ECV) = (4641 mOsm) (43L - ECV)
12,285 mOsm (ECV) = 199,563 mOsm * L
ECV = 16.2 L
Although no answer given matches exactly again, the closest (best)
answer is D.
Another way to do this problem is to take the final concentration from
#2 (286 mOsm/L), and, knowing that this is the concentration both in
the ECV and the ICV, use the number of mOsmoles to calculate the
required volumes to get this concentration inside and outside the
cells:
ECV = 4641mOsm / (286 mOsm/L) = 16.2 L
4. C
Since the ECV = 13.6L, and the TBW=43L, the ICV must be the difference
of the two:
ICV = TBW-ECV = 43L - 16.2L = 26.8L
Or, using the same strategy as in #3, ICV = 7644 mOsm / (286 mOsm/L) =
26.7 L.
The closest answer to either of these is choice C.
5. C
First off, the patient's pH of 7.15 is lower than normal (7.40), so we
know they have some type of acidosis, which eliminates choices B and
D. Next, look at the PaCO2. This is 30mmHg, which is lower than one
would expect if the person had a blood pH of 7.40. At pH 7.40, the
PaCO2 would be around 40mmHg (a good trick is that it equals the last
two digits of the pH when respiratory compensation is working). So, a
low PaCO2, by itself, would *raise* the pH, so this is not a
respiratory acidosis. In respiratory acidosis, the PaCO2 is too high
(the person's not breathing fast enough). What's left? The acidosis
must be of metabolic origin, and the PaCO2 is low because the body is
trying to compensate for the metabolic effect by making the person
breath faster (lower CO2, which is an acid in solution, make blood
more alkaline).
6. E
The (anatomic+alveolar) dead space volume (VD) is the volume of the
portion of the respiratory tract that is not involved in air exchange
(conduction only), e.g., the trachea, the bronchi, and the
non-respiratory bronchioles. This space can be measured using the
Fowler nitrogen washout method (see link below), eliminating choice D.
Choice A is the minute ventilation (volume per minute). Choice B is
the Residual Volume. The VD is usually much less than the tidal
volume (typically around 160mL).
7. C
See the link below for a definition of pneumothorax. Pneumonia is an
infection. Pleurisy is an irritation of the lining of the lung.
Tachypnea is rapid breathing. A pneumotachograph (misspelled in the
question) is the device used to measure flow-volume curves.
8. C
Diffusing capacity is the rate at which O2 crosses the alveolar /
capillary interface, also called DLCO. It is measured in (mL O2) /
minute / mmHg.
Some links:
Info on the movement of water when adding hypertonic saline and how
this results in thirst:
http://soma.npa.uiuc.edu/labs/greenough/statements/rswain/hormones/032296.html
Arterial Blood Gas practice:
http://www.rnceus.com/abgs/abgmethod.html
Discussion of measurement of Dead Space using Fowler method:
http://human.physiol.arizona.edu/sched/respiration/morgan44/morgan.l44.html
Definition of pneumothorax:
http://www.dictionary.com/search?q=pneumothorax
A discussion of lung diffusing capacity:
http://www.healthsci.utas.edu.au/medicine/teaching/physiol/student_projects/2001/resppres/Lung%20Diffusing.htm
Conditions that affect DLCO:
http://www.fpnotebook.com/LUN93.htm
The Family Practice Notebook is a great general resource for answering
these types of question:
http://www.fpnotebook.com/
I hope this was helpful. Let me know if you require any
clarification.
-welte-ga |
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