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BIPN 100 Fall 2018

Problem set #1

1. If a cell (300 mOsm) is placed in the following solutions. Designate each solution according to its osmolarity and tonicity and explain what happens to the cell. Assume that the cell is permeable to urea and impermeable to NaCl. A. 300 mOsM urea Isosmotic: the osmolarity of the cell is 300 mOsM. The same as the solution Hypotonic: Tonicity refers to the net movement of water in or out of the cell. In this case, the cell swells, because urea penetrates the cell, raising the osmolarity and causing a net flow of water into the cell. Note that the solution is isosmotic to the cell, but hypotonic to the cell. Even though the cell’s osmolarity will increase eventually due to urea entering the cell down a concentration gradient, we still call the solution isosmotic to the cell

B. 200 mOsM urea plus 300 mOsM NaCl Hyperosmotic: the solution is 500 mOsM total and the cell is 300 mOsM. Isotonic: No change in cell size at equilibrium. Initially water leaves the cell due to the higher osmolarity outside the cell (solution is 500 mOsm). Then, because there is a concentration gradient for urea, urea will enter the cell, increasing its osmolarity, and bringing some water into the cell. The nonpenetrating solute concentrations in cell and solution initially are equal, therefore there will be no net movement of water at equilibrium

C. 400 mOsM NaCl Hyperosmotic, the osmolarity of the solution is higher than that of the cell Hypertonic, the water exits due to the higher osmolarity

D. 300 mOsM NaCl Isosmotic, isotonic. No change in cell size. Sodium and chloride do not cross the membrane. Since the osmolarity of the solution and the cell are the same there is no net movement of water

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BIPN 100 Fall 2018

Problem set #1

2. Your goal is to make an isotonic intravenous solution for an emergency room patient who has lost a lot of blood. The available solutes include glucose (300 mM), NaCl (150 mM), and urea (300 mM). What solutes should she use to make up the IV solution with an osmolarity of 300 mOsm (isosmotic), making sure that it will also be isotonic? Explain your answer. Assume that the cell is permeable to urea, glucose, but not to NaCl and that the osmolarity of the cell is 300 mOsm. Osmolarity 300 mM glucose → 300 mOsm 150 mM NaCl → 300 mOsm 300 mM Urea → 300 mOsm All the solutions are isosmotic to the cell Glucose can enter the cell →  osmotic pressure () in the cell → water follows → cell gains water → solution is hypotonic to the cell Urea can enter the cell →  osmotic pressure ( ) in the cell → water follows → cell gains water → solution is hypotonic NaCl cannot enter the cell → cell does not gain or lose volume → isotonic The solution should only contain solutes that will not cross the cell membrane (ensure that there is no movement of water into the cell, isotonic), i.e., she should use NaCl, but not glucose or urea)

3. If a 10% sucrose solution is separated from a 20% sucrose solution by a membrane impermeable to sucrose, in which direction will net movement of water occur? A) from the 10% sucrose solution to the 20% sucrose solution only B) from the 20% sucrose solution to the 10% sucrose solution only C) There will be no net movement of water in this case. D) None of the answers are correct.

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BIPN 100 Fall 2018

Problem set #1

4. If a 10% sucrose solution is separated from a 20% sucrose solution by a semipermeable membrane, in which direction will net diffusion of sucrose and water take place? A) from the 10% sucrose solution to the 20% sucrose solution B) from the 20% sucrose solution to the 10% sucrose solution C) There will be no diffusion in this case. D) from the 10% sucrose solution to the 20% sucrose solution and from the 20% sucrose solution E) neither from the 10% sucrose solution to the 20% sucrose solution nor from the 20% sucrose solution to the 10% sucrose solution Since the membrane is semipermeable it only allows the solvent (water) to cross. Although there is a concentration gradient for sucrose (driving force) there is no pathway (impermeable membrane). Sucrose does not move (E) Water will move from the 10% solution to the 20% solution, from lower to higher concentration of solutes (A)

5. Indicate the relative osmolarities (hyper, hypo or iso) of the following solutions. Solution a: 2 osmolar NaCl Solution b: 1 molar NaCl Solution c: 900 milliosmolar glucose A. a is ________ osmotic to b.

D. a is ________ osmotic to c.

B. b is ________ osmotic to a.

E. b is ________ osmotic to c.

C. c is ________ osmotic to a.

F. c is ________ osmotic to b.

This is easier to answer if all three solutions are described in equivalent terms. Solution b is 2 osmolar, because of the dissociation of sodium and chloride. Solution c is 0.9 osmolar. A. iso D. hyper B. iso E. hyper C. hypo F. hypo

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BIPN 100 Fall 2018

Problem set #1

6. Red blood cells (RBC) with an internal osmolarity of 300 mOsM are placed in the following solutions. For each solution, calculate the osmolarity (before any water flows across the RBC membrane) and circle to indicate the solution’s osmolarity and tonicity (compared to the inside of the RBCs). Also circle to indicate if the RBCs will swell, shrink or not change their volume when put into the solution. Assume that the RBCs don’t lyse and that the RBC membrane is impermeable to NaCl and permeable to water and urea. A. Solution A: 100 mM NaCl Calculate the solution’s osmolarity: __________200 mOsm___________________________ The solution is:

hyperosmotic

hypoosmotic

isosmotic (circle one)

The solution is:

hypertonic

hypotonic

isotonic (circle one)

The RBC will:

swell

shrink

not change volume (circle one)

B. Solution B:400 mM urea Calculate the solution’s osmolarity: __________400 mOsm___________________________ The solution is:

hyperosmotic

hyposmotic

The solution is:

hypertonic

hypotonic

The RBC will:

swell

shrink

isosmotic(circle one) isotonic(circle one) not change volume (circle one)

7. Draw diagrams describing the Gs and Gq transduction pathways See lecture slides

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BIPN 100 Fall 2018

Problem set #1

8. A patient is suffering of high Cl- loss from the gut and severe diarrhea and dehydration In steady state there is more Cl- in the intra cellular fluid (ICF). cAMP is required to activate Cl- channels in the intestine. See the diagram shown below. A. What G protein subunit is involved? (circle one)

B. What effector is involved? (circle one)

AC

Gi

Gs

Gq

PLC

Cholera Toxin A Inactivation of GTPase → constitutively activated Gαs → hyperactivation of AC → increase in cAMP → excess Cl- loss

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BIPN 100 Fall 2018

Problem set #1

9. Based on the feedback loop answer the following questions

A. Can changes in component A ever cause the value of variable B to become smaller? Why or why not? If the value of component A becomes smaller, the value of component B also becomes smaller because the two components are connected by an arrow and a positive sign

B. Can changes in component D ever cause the value of variable E to get larger? Why or why not? If the value of component D becomes smaller, the value of component E will increase because the two components are connected by a sign-inverting arrow

C. Is there a positive feedback loop in this diagram? Briefly justify your answer. The loop that connects components A, B, C, D, E, and F is a positive feedback loop because it includes two sign-inverting arrows

D. Is there a negative feedback loop in this diagram? Briefly justify your answer. The loop that connects components B and C is a negative feedback loop because it includes one sign-inverting arrow

PLEASE NOTE: You must consider only one loop at a time. The two loops in this diagram both include components B and C, but they are two separate loops. Remember that you can’t go backwards in a loop; you must proceed only in the directions in which the arrows point

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