BIOLOGY 306 EXAM 1 - Answers to Essays
42. When a person receives intravenous fluids to help increase blood volume, why is it important for the
fluid to be isotonic? What would happen if a hypotonic or hypertonic solution was administered? (3
points)
The intravenous solution must be isotonic to prevent a net movement of water into or out of the red
blood cells which could interfere with normal RBC functions.
A hypotonic solution causes water to move into cells. This would cause the red blood cells to swell
and potentially undergo hemolysis, thus destroying the RBCs (and therefore severely impacting gas
transport).
A hypertonic solution causes water to move out of cells. This would cause the red blood cells to
undergo crenation, also leading to cell death (same final outcome as hemolysis - impaired gas
transport).
43. NADH and FADH2 both carry electrons to the electron transport chain to produce ATP. How many
ATP molecules are produced from one molecule of NADH? From one molecule of FADH 2? Explain
the difference. You may use diagrams if desired, but please label and explain thoroughly. (4 points)
NADH = three ATP; FADH2 = two ATP.
NADH delivers its electrons to FMN - ahead of all three proton pumps located in the electron transport
chain. As the electrons move down the chain, all three pumps are turned on and result in the movement
of 6 protons (H+) into the intermembrane space. When these protons move down their concentration
gradient through the ATP synthase complex, each pair of protons will facilitate the production of one
ATP molecule. Thus, the six protons will result in three ATP produced.
FADH2 delivers its electrons to CoQ - after the first proton pump located in the electron transport
chain. As the electrons move down the chain, only two pumps are turned on and result in the
movement of 4 protons (H+) into the intermembrane space. When these protons move down their
concentration gradient through the ATP synthase complex, each pair of protons will facilitate the
production of one ATP molecule. Thus, the four protons will result in two ATP produced.
47. What effect would you expect a 2.9 M solution of fructose to have if injected directly into the
bloodstream of a patient? Explain your answer. (You can assume the following information - fructose
does not dissociate in solution, there is no fructose inside the blood cells, and the cell membrane is not
freely permeable to the fructose molecules) (3 points)
(Note: Although a concentration gradient exists for fructose (2.9 M outside the cell, 0 inside the cell),
fructose is not able to enter through the cell membrane. Therefore, there will be no solute movement to
influence any water movement, and osmosis will occur only based on the solute concentrations on either
side of the membrane. To compare solute concentrations, you must first convert the M solution to
mOsm.)
2.9 mole/L x 1000 = 2.9 mmol x 1 (dissociation constant of fructose) = 2900 mOsm/L.
This is then compared to the "normal" osmolarity of blood - 300 mOsm/L
Since the solution of fructose outside the cell has a much higher total solute concentration
(hyperosmotic) than inside the cell, one is able to predict that water will move out of the cell, causing
crenation of the red blood cells (thus, the fructose solution is also hypertonic).
Bonus Question - (2 points)
From laboratory experiment - When red blood cells were placed in an isosmotic solution of urea, the
mixture became translucent very quickly, and turned quite red. When red blood cells were placed in an
isosmotic solution of glucose, the mixture remained cloudy and showed no noticeable change in
appearance. What was the difference between these two solutions, and what happened to the red blood cells
in each solution?
Both solutions were isosmotic to the cells (i.e. there was no concentration gradient for water based on the
solute concentrations). However, urea is a small, lipid-soluble molecule that is able to cross the cell
membrane freely. As it diffused down its concentration gradient into the cell, water followed to try to
maintain equilibrium. Thus, the isosmotic urea solution was hypotonic, causing water to move in, and
eventually causing hemolysis of the RBCs.
Glucose, on the other hand, is not able to freely cross the lipid bilayer of the cell membrane. Since there
was no solute movement into the cell, there also was no net water movement in either direction across the
cell membrane. Thus, the isosmotic glucose solution was also isotonic, and the RBCs were unaffected by
this solution.