Physiology-membrane Channels-cell Vol Reg
.JPG "Physiology-membrane Channels-cell Vol Reg - Quiz")
Welcome to another quiz under the umbrella of physiology, where today we’ll be discussing the membrane channel – which is a family of biological membrane proteins which allow for water, ions and other solutes to move passively through the membrane. What do you know about it?
- 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 travel through, resulting in slower diffusion. The diffusion rate is directly proportional to the surface area available for diffusion and inversely proportional to the distance the substance needs to travel. Therefore, an increase in membrane thickness will hinder the movement of the non-polar substance, leading to a decrease in the rate of 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. In this method, the amount of indicator in the plasma is divided by the concentration of the indicator in the injected solution, and then multiplied by the volume of the injected solution. Since the concentration of both Evans blue dye and antipyrine in the plasma is 1 g/L, it means that the patient's blood volume is equal to the volume of the injected solution, which is 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. This means that the overall amount of fluid in the extracellular space will decrease. However, since the loss is isotonic, the concentration of solutes in the extracellular fluid will remain the same. Therefore, there will be no change in extracellular osmolarity.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). This is because glucose symporters often work in conjunction with sodium ions to transport glucose into the cell. Sodium ions create a concentration gradient by actively pumping out of the cell, and glucose is transported along with the sodium ions, utilizing the energy from the sodium gradient.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 the change in fluid volume that will occur as a result of an isotonic saline infusion. The dotted line starts at the initial fluid volume and remains flat, indicating that there is no change in fluid volume after the saline infusion. This is characteristic of an isotonic solution, where the fluid volume remains the same.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 movement of water across a membrane by osmosis is determined by the concentration of water on each side of the membrane. Osmosis is the process by which water moves from an area of lower solute concentration to an area of higher solute concentration. Therefore, the concentration of water, rather than the permeability of the membrane to solutes or the presence of ion channels or transporters, is the primary determinant for the movement of water across a membrane by 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 determine the capacity of the carrier to bind and transport the substances. The more binding sites available, the greater the capacity for transport, leading to a higher maximal rate of transport. The concentration gradient for Cl-, the transmembrane electrical gradient, and the electrical gradient for glucose are also important factors, but the number of binding sites has the greatest impact on the transport rate.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 fluid from the extracellular space, it means that there is a higher concentration of solute inside the cell compared to outside. As a result, water will move from the extracellular space into the intracellular space to balance out the concentrations. This influx of water will cause an increase in the intracellular osmolarity, leading to an increase in the concentration of solute inside the cell. Therefore, the correct answer is 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. AExplanation
The correct answer is A because when the RBC is placed in an isotonic NaCl solution, there is no net movement of water into or out of the cell, so the volume remains unchanged. At point 1, the RBC is transferred to a solution with a lower osmolarity (225 mOsm/L), causing water to enter the cell by osmosis, leading to an increase in volume. At point 2, more NaCl is added to the solution, increasing its osmolarity to 375 mOsm/L. This higher osmolarity causes water to leave the cell by osmosis, resulting in a decrease in volume. Therefore, the correct graph should show an initial constant volume, followed by an increase in volume at point 1, and then a decrease in volume at point 2.Rate this question:
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- 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 total amount of mannitol and Evans blue dye that was given to the patient (1.0 L + 0.5 L = 1.5 L) and comparing it to the plasma concentration of these substances two hours later (0.1 g/L). Since there was no loss of indicator in the urine, it can be assumed that the entire amount of mannitol and Evans blue dye remained in the plasma. Therefore, the patient's interstitial fluid volume must be equal to the volume of fluid given (1.5 L) minus the plasma volume (0.1 g/L x 1.5 L = 0.15 L), which equals 15 L.Rate 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. This is because urea has an osmotic coefficient of 0, meaning it does not contribute to the osmotic pressure. Therefore, water will move into the cells to equalize the osmotic pressure, causing them to swell.
In the NaCl solution, the cells will not change in size. This is because NaCl has an osmotic coefficient of 1, meaning it fully dissociates into ions and contributes to the osmotic pressure. The osmotic pressure inside and outside the cells will be equal, so there will be 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 extracellular fluid volume is determined by the concentration of the indicators in the plasma and the amount of indicator given to the patient. Since the concentration of inulin and antipyrine in the plasma is 1 g/L after two hours, it means that the indicators have equilibrated throughout the patient's extracellular fluid. Given that 1 L of 12 g/L inulin and 100 ml of 10 g/L antipyrine were given, it indicates that the initial amount of indicators distributed in the extracellular fluid was 12 g of inulin and 1 g of antipyrine. Therefore, 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
To calculate the plasma volume, we can use the formula: Plasma Volume = (Amount of dye in plasma / Concentration of dye in plasma) * 100.
Given that there is 0.4 mg of dye per 100 ml of plasma, and the woman's plasma volume is being asked, we can substitute these values into the formula.
Plasma Volume = (0.4 mg / 0.4 mg per 100 ml) * 100 = 100 ml.
Since 1 L = 1000 ml, the plasma volume is 100 ml / 1000 = 0.1 L or 0.1 L.
To calculate the whole blood volume, we can use the formula: Whole Blood Volume = Plasma Volume / (1 - Hematocrit ratio).
Given that the hematocrit ratio is 45%, we can substitute these values into the formula.
Whole Blood Volume = 0.1 L / (1 - 0.45) = 0.1 L / 0.55 = 0.18 L or 0.18 L.
Therefore, the woman's plasma volume is 0.1 L and her whole blood volume is 0.18 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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Mar 21, 2023Quiz Edited by
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Dec 10, 2011Quiz Created by
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