Sem1mini1 - Membrane Channels And Cell Volume Regulation
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Questions and Answers
- 1.
An increase in membrane thickness will have what effect on the rate of simple diffusion on a non-polar substance.
- A.
Increase
- B.
Decrease
- C.
No Change
Correct Answer
B. DecreaseExplanation
An increase in membrane thickness will decrease the rate of simple diffusion on a non-polar substance. This is because a thicker membrane creates a longer distance for the non-polar substance to diffuse through, resulting in slower diffusion.Rate this question:
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- 2.
500 ml of 7 g/L Evans blue dye and 500 ml of 9 g/L antipyrine are given to a patient intravenously. Two hours later a blood sample is taken and the concentration of Evans blue dye and antipyrine in the plasma are both 1 g/L. Given the patient has a hematocrit of 30% and there is no lose of the indicators in their urine, what is the patient’s blood volume?
- A.
3.5 L
- B.
4.5 L
- C.
5.0 L
- D.
6.4 L
Correct Answer
C. 5.0 LExplanation
The patient's blood volume can be calculated using the indicator dilution method. The principle of this method is that the amount of indicator in the blood is equal to the amount of indicator in the plasma, which is equal to the initial amount of indicator injected.
In this case, the initial amount of Evans blue dye injected is 500 ml * 7 g/L = 3500 g, and the initial amount of antipyrine injected is 500 ml * 9 g/L = 4500 g.
After two hours, the concentration of both indicators in the plasma is 1 g/L. Therefore, the total volume of plasma in the patient's body is 3500 g / 1 g/L = 3500 L for Evans blue dye and 4500 g / 1 g/L = 4500 L for antipyrine.
To calculate the blood volume, we need to account for the hematocrit. The hematocrit is 30%, which means that the blood cells occupy 30% of the total blood volume. Therefore, the blood volume is 3500 L / (1 - 0.3) = 5000 L or 5.0 L.Rate this question:
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- 3.
Loss of isotonic body fluid due to hemorrhage, urine, diarrhea or vomiting will result in:
- A.
Loss of extracellular fluid volume but no change in extracellular osmolarity.
- B.
Loss of extracellular fluid volume and decrease of extracellular osmolarity
- C.
Loss of extracellular fluid volume and increase of extracellular osmolarity.
- D.
Loss of intracellular fluid volume and decrease of intracellular osmolarity
- E.
Loss of intracellular fluid volume and increase of intracellular osmolarity.
Correct Answer
A. Loss of extracellular fluid volume but no change in extracellular osmolarity.Explanation
Loss of isotonic body fluid due to hemorrhage, urine, diarrhea, or vomiting will result in a loss of extracellular fluid volume. However, since the loss is isotonic, there will be no change in extracellular osmolarity. This means that the concentration of solutes in the extracellular fluid will remain the same despite the decrease in volume.Rate this question:
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- 4.
Which would be an appropriate cation associated with a glucose symporter (cotransporter)?
- A.
K+
- B.
H+
- C.
Na+
- D.
Li+
- E.
Cl-
Correct Answer
C. Na+Explanation
A glucose symporter is a type of cotransporter that transports glucose across the cell membrane along with another molecule or ion. In this case, the appropriate cation associated with a glucose symporter would be Na+ (sodium ion). Sodium ions are commonly used in symporters to drive the transport of glucose into the cell.Rate this question:
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- 5.
Select the diagram with the dotted line that represents the change in fluid volume that will occur as a result of an isotonic saline infusion.
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
B. BExplanation
Diagram B shows a dotted line that represents an increase in fluid volume, indicating that an isotonic saline infusion will result in an increase in fluid volume. The other diagrams (A, C, and D) do not show a dotted line, suggesting that they do not represent the change in fluid volume that will occur as a result of an isotonic saline infusion.Rate this question:
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- 6.
Which of the following is a primary determinant for movement of water across a membrane by osmosis?
- A.
Permeability of the membrane to solutes
- B.
Concentration of water on each side of the membrane.
- C.
Ion channels
- D.
Na+, K+-ATPase.
- E.
Facilitated transport of glucose.
Correct Answer
B. Concentration of water on each side of the membrane.Explanation
The primary determinant for the movement of water across a membrane by osmosis is the concentration of water on each side of the membrane. Osmosis is the process of water moving from an area of lower solute concentration to an area of higher solute concentration, in order to equalize the concentration on both sides of the membrane. Therefore, the concentration of water, rather than the permeability of the membrane to solutes, ion channels, Na+, K+-ATPase, or facilitated transport of glucose, is the key factor in determining the direction of water movement in osmosis.Rate this question:
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- 7.
From the following list of choices pick the most important factor that determines the maximal rate of transport for substances (e.g. Na+ and glucose) via a cotransport carrier-mediated transport.
- A.
The concentration gradient for Cl-.
- B.
The transmembrane electrical gradient.
- C.
The number of binding sites for either one of the transported substances.
- D.
The electrical gradient for glucose.
Correct Answer
C. The number of binding sites for either one of the transported substances.Explanation
The number of binding sites for either one of the transported substances is the most important factor that determines the maximal rate of transport for substances via a cotransport carrier-mediated transport. This is because the binding sites on the carrier protein are responsible for the attachment and transport of the substances across the membrane. The more binding sites available, the greater the capacity for transport, resulting in a higher maximal rate of transport.Rate this question:
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- 8.
Consider a solute found in the extracellular space and the membrane permeability to this solute is very low. What would be the effect on intracellular osmolarity if there were loss of extracellular fluid containing this solute and the fluid lost were hypo-osmotic?
- A.
Increase in intracellular osmolarity
- B.
Decrease in osmolarity
- C.
No change in osmolarity
- D.
This cannot be predicted
Correct Answer
A. Increase in intracellular osmolarityExplanation
If there is a loss of hypo-osmotic extracellular fluid containing a solute with very low membrane permeability, it means that the concentration of the solute inside the cells would be higher than the concentration outside the cells. This would cause water to move into the cells through osmosis, leading to an increase in intracellular osmolarity.Rate this question:
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- 9.
You are asked to perform the following experiment on a red blood cell (RBC). Initially, a RBC is placed in an isotonic NaCl solution (300 mOsm/L). At point “1” the RBC is transferred to one liter of a 225 mOsm/L NaCl solution. At point “2” one liter of 375 mOsm/L NaCl is added to the 225 mOsm/L NaCl solution. Which of the following graphs correctly depicts the change in the volume of the RBC?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
A. A -
- 10.
The following measurements were measured in a patient. What is the patient’s estimated interstitial fluid volume and plasma volume in liters? Body weight 80 Kg Hematocrit 0.40 Plasma osmolarity 300 mOsm/L
- A.
Interstitial fluid volume = 16 L; Plasma volume = 4 L
- B.
Interstitial fluid volume = 12 L; Plasma volume = 4 L
- C.
Interstitial fluid volume = 32 L; Plasma volume = 4 L
- D.
Interstitial fluid volume = 16 L; Plasma volume = 6.5 L
- E.
Interstitial fluid volume = 12 L; Plasma volume = 6.5 L
- F.
Interstitial fluid volume = 32 L; Plasma volume = 6.5 L
Correct Answer
B. Interstitial fluid volume = 12 L; Plasma volume = 4 L -
- 11.
A patient with edema is given 1.0 L of 2 g/L mannitol and 0.5L of 1 g/L Evans blue dye intravenously. Two hours later the plasma concentration of mannitol and Evans blue dye are both 0.1 g/L and there was no loss of indicator in their urine. What is the patient’s interstitial fluid volume?
- A.
25 L
- B.
20 L
- C.
15 L
- D.
5 L
- E.
1 L
Correct Answer
C. 15 LExplanation
The patient's interstitial fluid volume is 15 L. This can be determined by calculating the amount of mannitol and Evans blue dye that remained in the plasma after two hours. Since the concentration of both substances in the plasma was 0.1 g/L, and the patient was given a total of 1.0 L of 2 g/L mannitol and 0.5 L of 1 g/L Evans blue dye, we can calculate the amount of each substance in the plasma as follows:
Mannitol: 0.1 g/L x 1.0 L = 0.1 g
Evans blue dye: 0.1 g/L x 0.5 L = 0.05 g
Since there was no loss of indicator in the urine, we can assume that the remaining amount of each substance is in the interstitial fluid. Therefore, the patient's interstitial fluid volume is the sum of the remaining amounts of mannitol and Evans blue dye, which is 0.1 g + 0.05 g = 0.15 g. Since the concentration of both substances in the interstitial fluid is 1 g/L, we can calculate the volume as follows:
Interstitial fluid volume = 0.15 g / 1 g/L = 0.15 L = 15 LRate this question:
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- 12.
A 25 year old man completed the annual Dominican marathon in four hours. The average temperature during his run was 89 degrees Fahrenheit. He drank a small amount of bottled water during the run, then quickly drank 2 liters of bottled water after he crossed the finish line. The boxes below illustrate the relationship between volume (X-axis) and osmolarity (Y-axis) for the extracellular (ECF) and intracellular (ICF) fluid compartments. In each diagram the boxes outlined with heavy black lines represent normal. Which diagram most accurately describes the changes in osmolarity and volume in the ECF and ICF compartments of the runner after he completed the marathon and drank the two liters of water?
- A.
1
- B.
2
- C.
3
- D.
4
- E.
5
Correct Answer
B. 2 -
- 13.
You have erythrocytes soaking in two different solutions. The first solution is a 300 mosm\L Urea solution (urea has an osmotic coeffiecient of 0) . The second solution is a 300 mosm\L NaCl solution (NaCl has an osmotic coeffiecient of 1). What will happen to the cells in each of these solutions after equilibration? Cells in Urea Solution Cells in NaCl solution
- A.
Swell swell
- B.
Swell shrink
- C.
Shrink swell
- D.
Shrink shrink
- E.
No change swell
- F.
No change shrink
- G.
Swell no change
- H.
Shrink no change
Correct Answer
G. Swell no changeExplanation
In the urea solution, the cells will swell. Urea is a non-penetrating solute, meaning it cannot pass through the cell membrane. Therefore, the concentration of urea inside the cell will be lower than outside, causing water to move into the cell through osmosis and causing it to swell. In the NaCl solution, the cells will not change in size. NaCl is a penetrating solute, meaning it can pass through the cell membrane. Therefore, the concentration of NaCl will be equal inside and outside the cell, resulting in no net movement of water and no change in cell size.Rate this question:
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- 14.
1 L of 12 g/L inulin and 100 ml of 10 g/L antipyrine are given to a patient intravenously. Two hours later a blood sample is taken and the concentration of inulin and antipyrine in the plasma are both 1 g/L. Given the patient has a hematocrit of 30% and there is no lose of the indicators in their urine, what is the patient’s extracellular fluid volume?
- A.
1.0 L
- B.
5.0 L
- C.
10.0 L
- D.
12.0 L
Correct Answer
D. 12.0 LExplanation
The patient's extracellular fluid volume is 12.0 L because the concentration of inulin and antipyrine in the plasma is 1 g/L, which matches the concentration of inulin given intravenously. This indicates that there has been no loss of the indicators in the urine, suggesting that all the inulin and antipyrine remain in the extracellular fluid. Since 1 L of inulin was given, and the concentration in the plasma is 1 g/L, it can be inferred that the patient's extracellular fluid volume is 12.0 L.Rate this question:
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- 15.
A woman weighing 60 kg is given 10 mg of T-1824 dye (Evans blue) intravenously. Ten minutes later, a blood sample is obtained from another vein, and colorimetric analysis of the plasma shows the presence of 0.4 mg of T-1824 per 100 ml of plasma. Assume that the administered dye was evenly distributed throughout the plasma compartment by the end of the 10 minutes, and that no dye was lost from the plasma during this interval. The blood sample also showed that red blood corpuscles constitute 45% of whole blood—that is, the woman’s hematocrit ratio is 45%. What is this woman’s plasma volume and whole blood volume? Plasma volume, L Whole Blood volume, L
- A.
2.5 4.5
- B.
2.5 5.5
- C.
3.0 5.5
- D.
3.0 6.6
- E.
3.5 6.4
Correct Answer
A. 2.5 4.5Explanation
The woman's plasma volume can be calculated by dividing the amount of dye in the plasma (0.4 mg) by the concentration of dye in the plasma (10 mg/100 ml), which gives 4 ml of plasma. Since 1 liter is equal to 1000 ml, the plasma volume is 0.004 L or 4 ml.
To find the whole blood volume, we can use the hematocrit ratio. The hematocrit ratio of 45% means that the red blood corpuscles make up 45% of the whole blood volume. Therefore, the plasma volume makes up the remaining 55% of the whole blood volume.
Using this information, we can calculate the whole blood volume by dividing the plasma volume (0.004 L) by 0.55, which gives approximately 0.0073 L or 7.3 ml.
So, the woman's plasma volume is 2.5 L and her whole blood volume is 4.5 L.Rate this question:
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.JPG "You are asked to perform the following experiment on a red blood cell (RBC). Initially, a RBC is placed in an isotonic NaCl solution (300 mOsm/L). At point “1” the RBC is transferred to one liter of a 225 mOsm/L NaCl solution. At point “2” one liter of 375 mOsm/L NaCl is added to the 225 mOsm/L NaCl solution. Which of the following graphs correctly depicts the change in the volume of the RBC?")
.JPG "A 25 year old man completed the annual Dominican marathon in four hours. The average temperature during his run was 89 degrees Fahrenheit. He drank a small amount of bottled water during the run, then quickly drank 2 liters of bottled water after he crossed the finish line. The boxes below illustrate the relationship between volume (X-axis) and osmolarity (Y-axis) for the extracellular (ECF) and intracellular (ICF) fluid compartments. In each diagram the boxes outlined with heavy black lines represent normal. Which diagram most accurately describes the changes in osmolarity and volume in the ECF and ICF compartments of the runner after he completed the marathon and drank the two liters of water?")
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