Day 105 NEET PG

1041. NEET PG

Chapter: Cell Biology; Topic: Cell Organelles; Subtopic: Enzymatic Markers and Functions

Key Definitions & Concepts

Marker Enzyme: A specific enzyme localized almost exclusively to a particular organelle, allowing for its identification during cell fractionation and isolation.
Glucose-6-Phosphatase: The enzyme responsible for the final step of gluconeogenesis and glycogenolysis; located on the luminal surface of the Endoplasmic Reticulum.
Microsomes: Vesicular artifacts formed from the fragmentation of the Endoplasmic Reticulum (ER) when cells are homogenized; they contain the ER marker enzymes.
Acid Phosphatase: The classic marker enzyme for Lysosomes, functioning optimally at acidic pH to hydrolyze phosphates.
Catalase: A key antioxidant enzyme found in Peroxisomes that decomposes hydrogen peroxide into water and oxygen.
Succinate Dehydrogenase: An enzyme of the Krebs cycle and Electron Transport Chain (Complex II), serving as a marker for the Inner Mitochondrial Membrane.
5'-Nucleotidase: An integral membrane protein often used as a marker for the Plasma Membrane.
Galactosyl Transferase: An enzyme involved in glycosylation, serving as a marker for the Golgi Apparatus.
Lactate Dehydrogenase (LDH): A glycolytic enzyme located in the Cytosol (Cytoplasm), used as a marker for the soluble fraction of the cell.
Von Gierke's Disease (GSD Type I): A metabolic disorder caused by a deficiency of Glucose-6-Phosphatase, leading to hypoglycemia and hepatomegaly; directly links ER pathology to clinical disease.

Lead Question - 2016

Marker of endoplasmic reticulum?

a) Acid phosphatase

b) Glucose-6-phosphatase

c) Catalase

d) LDH

Explanation: Identifying subcellular organelles relies on specific marker enzymes. Glucose-6-phosphatase is the established marker enzyme for the Endoplasmic Reticulum (specifically utilized to identify the microsomal fraction during centrifugation). It plays a crucial role in blood glucose homeostasis by hydrolyzing glucose-6-phosphate to free glucose, primarily in the liver and kidney. Acid phosphatase is the marker for Lysosomes. Catalase is the marker for Peroxisomes. LDH (Lactate Dehydrogenase) is found in the Cytosol. Therefore, the correct answer is b) Glucose-6-phosphatase.

1. Which enzyme serves as the specific marker for the Golgi Apparatus?

a) Cytochrome oxidase

b) Galactosyl transferase

c) Urate oxidase

d) Acid phosphatase

Explanation: The Golgi apparatus functions as the processing and packaging center of the cell, heavily involved in the modification of proteins and lipids. A key modification process is glycosylation. Galactosyl transferase (or Thiamine Pyrophosphatase) is an enzyme located in the trans-Golgi cisternae that catalyzes the transfer of galactose. Due to this specific localization and function, it is widely used as the marker enzyme for the Golgi apparatus. Cytochrome oxidase marks mitochondria. Urate oxidase marks peroxisomes. Acid phosphatase marks lysosomes. Therefore, the correct answer is b) Galactosyl transferase.

2. A researcher isolates a subcellular fraction containing the Electron Transport Chain complexes. Which enzyme would confirm the presence of the Inner Mitochondrial Membrane in this fraction?

a) Monoamine Oxidase

b) Adenylate Kinase

c) Succinate Dehydrogenase

d) Citrate Synthase

Explanation: Mitochondria have distinct compartments with specific markers. The Inner Mitochondrial Membrane is the site of oxidative phosphorylation. Succinate Dehydrogenase (Complex II of the Electron Transport Chain and a Krebs cycle enzyme) is embedded within this membrane, making it the definitive marker. Monoamine Oxidase (MAO) is the marker for the Outer Mitochondrial Membrane. Adenylate Kinase marks the Intermembrane space. Citrate Synthase marks the Mitochondrial Matrix. Therefore, the correct answer is c) Succinate Dehydrogenase.

3. A 6-month-old infant presents with hepatosplenomegaly, coarse facial features, and developmental regression. The diagnosis is I-cell disease. This condition results from a failure to tag enzymes with Mannose-6-Phosphate in which organelle?

a) Endoplasmic Reticulum

b) Golgi Apparatus

c) Lysosomes

d) Mitochondria

Explanation: I-cell disease (Mucolipidosis II) is a lysosomal storage disorder caused by a sorting defect. Normally, lysosomal enzymes are tagged with a Mannose-6-Phosphate (M6P) residue, which acts as a "zip code" directing them to the lysosome. This tagging process occurs in the cis-Golgi network. In I-cell disease, the enzyme GlcNAc-phosphotransferase (located in the Golgi Apparatus) is defective. Consequently, lysosomal enzymes are secreted outside the cell instead of being sorted to lysosomes, leading to the accumulation of substrates. Therefore, the correct answer is b) Golgi Apparatus.

4. Zellweger syndrome is a peroxisomal biogenesis disorder. Which enzyme is the standard marker for the organelle absent or dysfunctional in this disease?

a) Beta-glucuronidase

b) Catalase

c) Myeloperoxidase

d) Alkaline phosphatase

Explanation: Peroxisomes are responsible for the breakdown of very-long-chain fatty acids (VLCFA) and the detoxification of hydrogen peroxide. The enzyme Catalase is highly concentrated in peroxisomes (often forming a crystalline core) and functions to decompose the hydrogen peroxide generated by oxidative reactions into water and oxygen. Thus, Catalase (and Urate Oxidase) serves as the marker for Peroxisomes. In Zellweger syndrome, peroxisomes fail to form properly (empty ghosts), leading to VLCFA accumulation. Beta-glucuronidase is lysosomal. Alkaline phosphatase is a plasma membrane enzyme. Therefore, the correct answer is b) Catalase.

5. Which enzyme is commonly used as a biochemical marker for the Plasma Membrane (Cell Membrane)?

a) 5'-Nucleotidase

b) DNA Polymerase

c) Hexokinase

d) Glutamate Dehydrogenase

Explanation: The plasma membrane separates the cell from the external environment. Marker enzymes for this structure are typically ecto-enzymes or ion pumps. 5'-Nucleotidase (and Na+/K+ ATPase) is an integral membrane protein widely accepted as the marker for the Plasma Membrane. It catalyzes the hydrolytic cleavage of phosphate from 5'-nucleotides. Elevated serum levels of 5'-Nucleotidase are also clinically used to assess cholestasis (biliary obstruction), reflecting its presence on the bile canalicular membrane of hepatocytes. DNA Polymerase is nuclear. Hexokinase is cytosolic. Glutamate Dehydrogenase is mitochondrial. Therefore, the correct answer is a) 5'-Nucleotidase.

6. In the fractionation of liver cells, which enzyme would be found in the supernatant (soluble fraction) after removing all organelles?

a) ATP synthase

b) Lactate Dehydrogenase (LDH)

c) Acid Maltase

d) HMG-CoA Reductase

Explanation: The soluble fraction of the cytoplasm is the Cytosol. Marker enzymes for the cytosol are those involved in soluble metabolic pathways like glycolysis. Lactate Dehydrogenase (LDH) is a classic cytosolic enzyme involved in anaerobic glycolysis. When cells are disrupted and centrifuged to pellet membranes and organelles, LDH remains in the liquid supernatant. ATP synthase is mitochondrial. Acid Maltase (alpha-glucosidase) is lysosomal (deficiency causes Pompe disease). HMG-CoA Reductase is located in the Smooth Endoplasmic Reticulum. Therefore, the correct answer is b) Lactate Dehydrogenase (LDH).

7. A 3-month-old child presents with severe hypoglycemia, lactic acidosis, and a doll-like face. Liver biopsy shows increased glycogen content. The deficient enzyme is a marker for which organelle?

a) Mitochondria

b) Lysosomes

c) Endoplasmic Reticulum

d) Cytosol

Explanation: The clinical presentation is classic for Von Gierke's Disease (Glycogen Storage Disease Type I). The defect is in Glucose-6-Phosphatase. As established, Glucose-6-Phosphatase is the marker enzyme for the Endoplasmic Reticulum. The enzyme's catalytic site faces the ER lumen, necessitating transporters (T1, T2, T3) for glucose-6-phosphate to enter and glucose to exit. A defect in the enzyme (Type Ia) or the transporter (Type Ib) impairs glucose release, causing hypoglycemia and glycogen accumulation. Lysosomal glycogen storage disease is Pompe disease (Type II). Therefore, the correct answer is c) Endoplasmic Reticulum.

8. Which enzyme serves as a marker for the Outer Mitochondrial Membrane?

a) Monoamine Oxidase (MAO)

b) Malate Dehydrogenase

c) Cytochrome c

d) Pyruvate Dehydrogenase

Explanation: The mitochondrion has two membranes. While Succinate Dehydrogenase marks the inner membrane, the Outer Mitochondrial Membrane is characterized by the presence of Monoamine Oxidase (MAO). This enzyme plays a vital role in the breakdown of neurotransmitters like serotonin and dopamine. Clinically, MAO inhibitors are used as antidepressants. Cytochrome c is in the intermembrane space (loosely attached to the inner membrane). Pyruvate dehydrogenase and Malate dehydrogenase are found in the mitochondrial matrix. Therefore, the correct answer is a) Monoamine Oxidase (MAO).

9. Acid Phosphatase is the marker for lysosomes. Which other property is unique to this organelle's interior environment?

a) High pH (Alkaline)

b) Presence of genomic DNA

c) Low pH (Acidic)

d) Double membrane structure

Explanation: Lysosomes are the digestive disposal systems of the cell. They contain hydrolytic enzymes (nucleases, proteases, lipases, phosphatases) that function optimally in an acidic environment. The marker enzyme is Acid Phosphatase. To maintain enzyme activity, the interior of the lysosome is kept at a Low pH (Acidic, around pH 4.5-5.0). This acidification is achieved by proton pumps (H+-ATPases) in the lysosomal membrane. If the lysosome breaks, the enzymes are inactive in the neutral cytosolic pH, protecting the cell from autolysis. Therefore, the correct answer is c) Low pH (Acidic).

10. The Smooth Endoplasmic Reticulum (SER) is abundant in hepatocytes involved in drug detoxification. Which enzyme system serves as a functional marker for this organelle?

a) Acid Lipase

b) Cytochrome P450 system

c) ATP synthase

d) Cathepsins

Explanation: The Endoplasmic Reticulum has two domains: Rough (RER) for protein synthesis and Smooth (SER) for lipid synthesis and detoxification. In the liver, the SER is the primary site for the biotransformation of drugs and xenobiotics. The Cytochrome P450 (CYP450) system (specifically NADPH-cytochrome P450 reductase) is embedded in the SER membrane and serves as a key functional marker. Induction of these enzymes (e.g., by phenobarbital) leads to proliferation of the SER. Acid lipase and Cathepsins are lysosomal enzymes. Therefore, the correct answer is b) Cytochrome P450 system.

1042. NEET PG

Chapter: General Physiology; Topic: Transport Across Cell Membranes; Subtopic: Mechanisms of Passive Transport (Simple Diffusion)

Key Definitions & Concepts

Simple Diffusion: The passive movement of solute molecules across a semipermeable membrane from an area of higher concentration to an area of lower concentration without the aid of transport proteins.
Fick's Law of Diffusion: States that the rate of diffusion (J) is directly proportional to the surface area (A), concentration gradient (ΔC), and lipid solubility, and inversely proportional to the thickness of the membrane (Δx).
Lipid Solubility (Partition Coefficient): A measure of how easily a substance dissolves in lipids; substances with high lipid solubility (non-polar) diffuse rapidly through the lipid bilayer.
Channel Proteins: Transmembrane proteins that form water-filled pores (e.g., aquaporins, ion channels) allowing specific hydrophilic substances to cross the membrane; distinct from simple diffusion through the lipid matrix.
Facilitated Diffusion: Passive transport mediated by carrier proteins that undergo conformational changes; unlike simple diffusion, it exhibits saturation kinetics (Tm).
Concentration Gradient: The driving force for simple diffusion; net flux stops when the concentration is equal on both sides (equilibrium).
Permeability Coefficient: A value representing the ease with which a specific molecule crosses a specific membrane; depends on lipid solubility and molecular size.
Meyer-Overton Correlation: Specifically relates the potency of general anesthetics to their lipid solubility; a clinical application of simple diffusion principles.
Diffusion Capacity (DLCO): A clinical measure of the lung's ability to transfer gas (CO) from alveoli to blood; reduced in fibrosis (thickened membrane) and emphysema (lost surface area).
Brownian Motion: The random thermal motion of molecules that underlies the physical process of diffusion.

[Image of Simple diffusion vs Facilitated diffusion graph]

Lead Question - 2016

Following is a feature of simple diffusion?

a) Against a concentration gradient

b) Easy for non-polar substance

c) More in thick membrane

d) Requires carrier protein

Explanation: Simple diffusion describes the movement of molecules directly through the lipid bilayer of the cell membrane. The core of the membrane is hydrophobic (lipid). Therefore, non-polar (lipophilic) substances like Oxygen (O2), Carbon Dioxide (CO2), nitrogen, steroid hormones, and anesthetics can dissolve in this lipid matrix and pass through easily. Diffusion is a passive process driven by thermal energy, moving substances down their concentration gradient (high to low), not against it. According to Fick's Law, the rate of diffusion is inversely proportional to membrane thickness (thicker = slower) and does not require carrier proteins (which characterizes facilitated diffusion or active transport). Therefore, the correct answer is b) Easy for non-polar substance.

1. According to Fick's Law of diffusion, the rate of net diffusion (J) is inversely proportional to which of the following parameters?

a) Concentration gradient

b) Surface area of the membrane

c) Thickness of the membrane

d) Lipid solubility of the solute

Explanation: Fick's Law mathematically describes passive diffusion: J = (D * A * ΔC) / Δx. Here, 'J' is the rate of diffusion. 'A' (Surface Area) and 'ΔC' (Concentration Gradient) are in the numerator, meaning diffusion increases as these increase. Lipid solubility also increases the diffusion coefficient (D). However, 'Δx' represents the diffusion distance or the Thickness of the membrane and is in the denominator. This means that as the membrane becomes thicker, the rate of diffusion decreases. This is clinically relevant in conditions like pulmonary fibrosis, where a thickened respiratory membrane impairs gas exchange. Therefore, the correct answer is c) Thickness of the membrane.

2. A scuba diver suffering from "the bends" (decompression sickness) has nitrogen bubbles forming in the blood. Nitrogen gas crosses cell membranes primarily by:

a) Facilitated diffusion via gas transporters

b) Simple diffusion through the lipid bilayer

c) Active transport

d) Solvent drag

Explanation: Respiratory gases such as Oxygen, Carbon Dioxide, and Nitrogen are small, non-polar, uncharged molecules. Because of their lipophilic nature, they do not require specific transporters, channels, or energy to cross cell membranes. Instead, they move via Simple diffusion through the lipid bilayer. This allows for rapid equilibration between the alveoli and pulmonary capillaries. In decompression sickness, the rapid decrease in ambient pressure causes nitrogen (which had diffused into tissues under high pressure) to come out of solution faster than it can diffuse back into the lungs to be exhaled. Therefore, the correct answer is b) Simple diffusion through the lipid bilayer.

3. Which graphical representation best characterizes the relationship between the rate of transport and substrate concentration in simple diffusion?

a) Sigmoidal curve

b) Rectangular hyperbola (saturation kinetics)

c) Linear relationship

d) Exponential decay

Explanation: This is a key distinguishing feature between simple and carrier-mediated transport. In simple diffusion, the rate of transport is directly proportional to the concentration gradient. As long as the concentration difference increases, the rate of diffusion increases without limit. This produces a straight Linear relationship on a graph (non-saturable). In contrast, carrier-mediated transport (like facilitated diffusion) relies on a limited number of protein carriers. Once all carriers are occupied (Vmax), the rate plateaus, showing saturation kinetics (rectangular hyperbola). Therefore, the correct answer is c) Linear relationship.

4. A premature infant develops Respiratory Distress Syndrome (RDS) due to surfactant deficiency. This leads to alveolar collapse (atelectasis). Physiologically, this hypoxia is primarily caused by a reduction in which factor of Fick's Law?

a) Membrane thickness

b) Diffusion coefficient

c) Surface area

d) Concentration gradient

Explanation: Fick's Law states diffusion is proportional to Surface Area (A). The lungs normally have a massive surface area for gas exchange provided by millions of open alveoli. In Neonatal RDS, the lack of surfactant leads to high surface tension and the collapse of alveoli (atelectasis). This collapse drastically reduces the total functional Surface area available for oxygen diffusion from the air into the blood. While the membrane might thicken later due to damage, the primary immediate physical deficit causing hypoxia is the loss of surface area. Therefore, the correct answer is c) Surface area.

5. Which of the following substances has the highest permeability coefficient across a pure phospholipid bilayer?

a) Glucose

b) Sodium ion (Na+)

c) Water

d) Steroid hormones (e.g., Cortisol)

Explanation: Permeability depends on size, charge, and lipid solubility. Ions like Sodium (Na+), despite being small, are highly charged and surrounded by a hydration shell, making them essentially impermeable to the hydrophobic lipid core without channels. Glucose is large and polar, requiring transporters (GLUTs). Water is small and polar; it can diffuse slowly but relies mostly on aquaporins. Steroid hormones are lipids (cholesterol derivatives). Being non-polar and lipophilic, they have the highest partition coefficient and diffuse most readily directly through the bilayer to reach intracellular receptors. Therefore, the correct answer is d) Steroid hormones (e.g., Cortisol).

6. The "Partition Coefficient" of a solute is a measure of its:

a) Molecular weight

b) Water solubility relative to its size

c) Lipid solubility relative to water solubility

d) Electrical charge density

Explanation: The Partition Coefficient (K) is a physicochemical property that predicts how easily a substance can cross a biological membrane via simple diffusion. It is defined as the ratio of a substance's solubility in oil (lipid) to its solubility in water. A high partition coefficient means the substance is more soluble in lipids than in water (lipophilic). Since the cell membrane core is lipid, substances with a high partition coefficient (like general anesthetics) enter cells very rapidly. This concept is central to the Meyer-Overton rule in anesthesiology. Therefore, the correct answer is c) Lipid solubility relative to water solubility.

7. A patient with severe pneumonia has fluid accumulation in the alveoli (pulmonary edema). This impairs oxygenation because the presence of fluid essentially increases the:

a) Membrane surface area

b) Diffusion distance (path length)

c) Solubility of Oxygen

d) Partial pressure of Oxygen in alveoli

Explanation: Gas exchange involves diffusion through the respiratory membrane (alveolar epithelium, interstitium, capillary endothelium). In pulmonary edema, fluid accumulates in the interstitial space and within the alveoli. Oxygen molecules must now diffuse not only through the tissue barriers but also through a layer of fluid to reach the red blood cells. This effectively increases the Diffusion distance (path length) or membrane thickness (Δx in Fick's Law). Since diffusion time is proportional to the square of the distance, even small increases in fluid thickness cause significant hypoxemia. Therefore, the correct answer is b) Diffusion distance (path length).

8. Which variable determines the direction of net simple diffusion of an uncharged solute?

a) The chemical concentration gradient only

b) The electrical gradient only

c) The electrochemical gradient

d) ATP availability

Explanation: For charged particles (ions), movement is dictated by both the chemical gradient (concentration difference) and the electrical gradient (potential difference), collectively known as the electrochemical gradient. However, for uncharged solutes (like glucose, urea, O2), the electrical potential across the membrane exerts no force on the molecule. Therefore, the driving force for net diffusion is governed solely by the Chemical concentration gradient (difference in concentration) across the membrane. Diffusion continues until the concentrations on both sides are equal. Therefore, the correct answer is a) The chemical concentration gradient only.

9. Unlike simple diffusion, facilitated diffusion is characterized by:

a) Transport against the concentration gradient

b) Non-specific transport of any molecule

c) Requirement for metabolic energy (ATP)

d) Competitive inhibition and stereospecificity

Explanation: Both simple and facilitated diffusion are passive (downhill) processes requiring no ATP. However, facilitated diffusion uses specific carrier proteins (like GLUT4 for glucose). Because it involves binding to a protein, it shares properties with enzymes: 1) Stereospecificity (e.g., transporting D-glucose but not L-glucose), 2) Saturation (Tm), and 3) Competitive inhibition (structurally similar molecules can block the transporter). Simple diffusion, being a physical process through the lipid bilayer, does not display stereospecificity or competitive inhibition in the classical sense. Therefore, the correct answer is d) Competitive inhibition and stereospecificity.

10. In a laboratory setting, increasing the temperature of a solution will have what effect on the rate of simple diffusion?

a) It will decrease linearly

b) It will increase

c) It will remain unchanged

d) It will stop completely

Explanation: Diffusion is driven by the random thermal motion (Brownian motion) of molecules. The kinetic energy of molecules is directly related to temperature. As temperature increases, the kinetic energy of the solute molecules increases, causing them to move and collide more frequently and vigorously. This results in an Increase in the rate of diffusion. Conversely, cooling a solution slows down molecular motion and the rate of diffusion. This is a fundamental physical principle applicable to biological systems as well. Therefore, the correct answer is b) It will increase.

1043. NEET PG

Chapter: General Physiology; Topic: Transport Across Cell Membranes; Subtopic: Modes of Membrane Transport

Key Definitions & Concepts

Simple Diffusion: Passive movement of substances (like gases, water, lipophilic drugs) down a concentration gradient directly through the lipid bilayer; generally considered the most common mechanism overall.
Primary Active Transport: Transport against a gradient utilizing direct energy from ATP hydrolysis (e.g., Na+/K+ ATPase, Ca2+ ATPase).
Secondary Active Transport: Transport driven by the energy stored in the electrochemical gradient of another molecule (usually Na+), not direct ATP usage.
Symport (Cotransport): A type of secondary active transport where two substances move in the same direction (e.g., SGLT1 moving Na+ and Glucose into the cell).
Antiport (Counter-transport): Secondary active transport where substances move in opposite directions (e.g., Na+/Ca2+ exchanger).
Facilitated Diffusion: Passive transport aided by carrier proteins or channels, exhibiting saturation kinetics (e.g., GLUT transporters).
Pinocytosis: "Cell drinking"; a form of endocytosis where the cell engulfs extracellular fluid and dissolved solutes non-specifically.
Phagocytosis: "Cell eating"; engulfment of large particles (bacteria, debris) by specialized cells like macrophages.
Receptor-Mediated Endocytosis: Highly specific uptake of ligands (like LDL or Transferrin) involving clathrin-coated pits.
Aquaporins: Specialized channel proteins that facilitate the rapid passive diffusion of water molecules.

[Image of Cell membrane transport mechanisms summary]

Lead Question - 2016

Most common mechanism for transport into the cell?

a) Diffusion

b) Primary active transport

c) Antiport

d) Cotransport

Explanation: When considering the sheer volume of molecules moving across cell membranes generally (including water, oxygen, carbon dioxide, nitrogen, and lipophilic substances), the most common mechanism is Diffusion (specifically Simple Diffusion). This process is passive, requires no energy, and follows the concentration gradient. In the context of pharmacology, the vast majority of drugs are absorbed and enter cells via passive diffusion (following Fick's Law). While nutrient uptake (glucose, amino acids) often relies on carrier-mediated transport (cotransport), diffusion remains the fundamental and most prevalent baseline process for cellular life and gas exchange. Therefore, the correct answer is a) Diffusion.

1. The Na+/K+ ATPase pump maintains the transmembrane potential by moving ions against their gradients. For every ATP molecule hydrolyzed, what is the stoichiometry of ion movement?

a) 3 Na+ in, 2 K+ out

b) 2 Na+ out, 3 K+ in

c) 3 Na+ out, 2 K+ in

d) 1 Na+ out, 1 K+ in

Explanation: The Na+/K+ ATPase is the quintessential example of primary active transport. It is ubiquitous in animal cells. Its function is to maintain the low intracellular sodium and high intracellular potassium concentrations. The pump operates by hydrolyzing one molecule of ATP to pump 3 Sodium ions (Na+) OUT of the cell and 2 Potassium ions (K+) INTO the cell. Because 3 positive charges leave and only 2 enter, the pump is electrogenic, contributing slightly (about -4mV) to the negative resting membrane potential. Therefore, the correct answer is c) 3 Na+ out, 2 K+ in.

2. A diabetic patient is prescribed an SGLT2 inhibitor (flozin) to lower blood glucose. This drug targets a transporter in the proximal renal tubule. The physiological mechanism of glucose reabsorption via SGLT2 is an example of:

a) Facilitated Diffusion

b) Primary Active Transport

c) Secondary Active Transport (Symport)

d) Simple Diffusion

Explanation: Glucose is reabsorbed from the renal tubule lumen against its concentration gradient. To achieve this uphill transport, the cell utilizes the energy stored in the inward Sodium gradient (established by the Na+/K+ pump). The Sodium-Glucose Linked Transporter (SGLT) binds both Na+ and Glucose and transports them together into the cell. Since both move in the same direction, this is Symport (or Cotransport). Because the energy comes from the ion gradient rather than direct ATP hydrolysis at the transporter, it is classified as Secondary Active Transport. Therefore, the correct answer is c) Secondary Active Transport (Symport).

3. Insulin stimulates glucose uptake in skeletal muscle and adipose tissue by inducing the translocation of which transporter to the cell membrane?

a) GLUT1

b) GLUT2

c) GLUT4

d) SGLT1

Explanation: Glucose transporters (GLUTs) mediate facilitated diffusion of glucose. GLUT4 is unique because it is the insulin-sensitive transporter. In the basal state, GLUT4 is sequestered in intracellular vesicles. Upon insulin binding to its receptor, a signaling cascade causes these vesicles to fuse with the plasma membrane, inserting GLUT4 and increasing glucose uptake by 10-20 fold. This is the mechanism for postprandial glucose clearance. GLUT1 is ubiquitous (blood-brain barrier). GLUT2 is in the liver/pancreas (glucose sensor). SGLT1 is in the intestine. Therefore, the correct answer is c) GLUT4.

4. In cardiac muscle cells, Calcium extrusion during relaxation is partially achieved by the Na+/Ca2+ Exchanger (NCX). This transporter moves 3 Na+ in and 1 Ca2+ out. This is an example of:

a) Uniport

b) Antiport (Counter-transport)

c) Primary Active Transport

d) Passive Diffusion

Explanation: The Na+/Ca2+ Exchanger (NCX) is crucial for preventing calcium overload in cardiac myocytes. It uses the energy of the Na+ gradient (sodium wanting to enter the cell) to push Calcium out of the cell against its massive concentration gradient. Since the two species move in opposite directions (Na+ in, Ca2+ out), this mechanism is defined as Antiport or Counter-transport. It is a form of secondary active transport. The drug Digitalis inhibits the Na+/K+ pump, accumulating intracellular Na+, which weakens the NCX gradient, keeping Ca2+ inside to increase contractility. Therefore, the correct answer is b) Antiport (Counter-transport).

5. A patient with familial hypercholesterolemia has extremely high levels of LDL cholesterol due to a defect in cellular uptake. The mechanism by which cells normally internalize LDL particles is:

a) Pinocytosis

b) Phagocytosis

c) Receptor-Mediated Endocytosis

d) Simple Diffusion

Explanation: Low-Density Lipoprotein (LDL) is too large to pass through channels or carriers. Its uptake is highly regulated. The LDL particle binds to specific LDL Receptors concentrated in Clathrin-coated pits on the cell membrane. This binding triggers the invagination of the membrane to form a coated vesicle, internalizing the LDL-Receptor complex. This specific, high-affinity process is Receptor-Mediated Endocytosis. Defects in the LDL receptor or the protein apolipoprotein B-100 impair this binding, leading to elevated serum cholesterol. Pinocytosis is non-specific "drinking." Therefore, the correct answer is c) Receptor-Mediated Endocytosis.

6. The MDR1 (Multi-Drug Resistance) protein, often overexpressed in cancer cells, pumps chemotherapeutic drugs out of the cell. Structurally and functionally, this transporter belongs to the:

a) ABC Transporter superfamily

b) Solute Carrier (SLC) family

c) Ion Channel family

d) Aquaporin family

Explanation: P-glycoprotein (MDR1) is an efflux pump that removes toxins and drugs from cells. It functions by binding ATP and using the energy of hydrolysis to pump substrates against their gradient. This ATP-dependence classifies it as a Primary Active Transporter. Structurally, it contains ATP-Binding Cassettes, placing it in the ABC Transporter superfamily. The Cystic Fibrosis Transmembrane Conductance Regulator (CFTR) is another famous member of this family (though it functions as a chloride channel, it is structurally an ABC protein). Therefore, the correct answer is a) ABC Transporter superfamily.

7. Which transport mechanism exhibits the property of Transport Maximum (Tm), where the rate of transport plateaus at high substrate concentrations?

a) Simple Diffusion through lipid bilayer

b) Simple Diffusion through channels

c) Carrier-Mediated Transport

d) Osmosis

Explanation: Saturation kinetics is a hallmark of processes requiring a specific binding site. In Carrier-Mediated Transport (both Facilitated Diffusion and Active Transport), there are a finite number of transporter proteins in the membrane. As substrate concentration increases, the rate of transport increases until all binding sites are occupied (saturated). At this point, the rate reaches a maximum velocity (Vmax) or Transport Maximum (Tm) and cannot increase further. Simple diffusion does not saturate; its rate is linear with concentration. Therefore, the correct answer is c) Carrier-Mediated Transport.

8. Omeprazole is used to treat peptic ulcers by inhibiting the H+/K+ ATPase in gastric parietal cells. This pump is responsible for the secretion of acid. What type of transport does this pump perform?

a) Secondary Active Transport

b) Primary Active Transport

c) Facilitated Diffusion

d) Passive leak channel

Explanation: The proton pump (H+/K+ ATPase) in the stomach must concentrate H+ ions in the lumen by a factor of over a million compared to the intracellular space. This massive gradient requires significant energy input. The pump hydrolyzes ATP directly to drive H+ out into the lumen and K+ into the cell (electroneutral 1:1 exchange). Since it uses ATP directly, it is a Primary Active Transporter. Proton Pump Inhibitors (PPIs) like Omeprazole irreversibly bind to and inhibit this pump. Therefore, the correct answer is b) Primary Active Transport.

9. The "Chloride Shift" in Red Blood Cells involves the exchange of Bicarbonate (HCO3-) leaving the cell for Chloride (Cl-) entering the cell. This anion exchange is mediated by Band 3 protein via:

a) Facilitated Diffusion (Antiport)

b) Active Transport

c) Simple Diffusion

d) Pinocytosis

Explanation: The Band 3 protein (AE1) is an Anion Exchanger. It facilitates the movement of HCO3- and Cl- across the RBC membrane. This movement is passive; it does not consume ATP. The direction is driven by the concentration gradients of the ions (which change between tissues and lungs). However, because it involves a carrier protein moving two ions in opposite directions, it is mechanistically an Antiport. Since it is passive (no energy), it is technically a form of Facilitated Diffusion acting as an exchanger. In standard classification, it's often grouped under Antiport/Exchangers. Therefore, the correct answer is a) Facilitated Diffusion (Antiport).

10. In nerve axons, the rapid depolarization phase of the action potential is mediated by the influx of Sodium. Through which type of membrane protein does this Sodium enter?

a) Voltage-Gated Ion Channel

b) Ligand-Gated Ion Channel

c) Leak Channel

d) Na+/K+ Pump

Explanation: Ion channels provide a watery pore for ions to diffuse passively down their electrochemical gradients. They are gated to control flow. During an action potential, the membrane potential reaches a threshold, triggering the opening of Voltage-Gated Sodium Channels. Na+ rushes into the cell (down its gradient), causing depolarization. This is simple diffusion through a channel protein. Ligand-gated channels open in response to neurotransmitters (like ACh). Leak channels are always open (responsible for resting potential). The pump builds the gradient but doesn't create the spike. Therefore, the correct answer is a) Voltage-Gated Ion Channel.

1044. NEET PG

Chapter: General Physiology; Topic: Transport Across Cell Membranes; Subtopic: Modes of Membrane Transport

Key Definitions & Concepts

Simple Diffusion: Passive movement of substances (like gases, water, lipophilic drugs) down a concentration gradient directly through the lipid bilayer; generally considered the most common mechanism overall.
Primary Active Transport: Transport against a gradient utilizing direct energy from ATP hydrolysis (e.g., Na+/K+ ATPase, Ca2+ ATPase).
Secondary Active Transport: Transport driven by the energy stored in the electrochemical gradient of another molecule (usually Na+), not direct ATP usage.
Symport (Cotransport): A type of secondary active transport where two substances move in the same direction (e.g., SGLT1 moving Na+ and Glucose into the cell).
Antiport (Counter-transport): Secondary active transport where substances move in opposite directions (e.g., Na+/Ca2+ exchanger).
Facilitated Diffusion: Passive transport aided by carrier proteins or channels, exhibiting saturation kinetics (e.g., GLUT transporters).
Pinocytosis: "Cell drinking"; a form of endocytosis where the cell engulfs extracellular fluid and dissolved solutes non-specifically.
Phagocytosis: "Cell eating"; engulfment of large particles (bacteria, debris) by specialized cells like macrophages.
Receptor-Mediated Endocytosis: Highly specific uptake of ligands (like LDL or Transferrin) involving clathrin-coated pits.
Aquaporins: Specialized channel proteins that facilitate the rapid passive diffusion of water molecules.

[Image of Cell membrane transport mechanisms summary]

Lead Question - 2016

Most common mechanism for transport into the cell?

a) Diffusion

b) Primary active transport

c) Antiport

d) Cotransport

1. The Na+/K+ ATPase pump maintains the transmembrane potential by moving ions against their gradients. For every ATP molecule hydrolyzed, what is the stoichiometry of ion movement?

a) 3 Na+ in, 2 K+ out

b) 2 Na+ out, 3 K+ in

c) 3 Na+ out, 2 K+ in

d) 1 Na+ out, 1 K+ in

a) Facilitated Diffusion

b) Primary Active Transport

c) Secondary Active Transport (Symport)

d) Simple Diffusion

3. Insulin stimulates glucose uptake in skeletal muscle and adipose tissue by inducing the translocation of which transporter to the cell membrane?

a) GLUT1

b) GLUT2

c) GLUT4

d) SGLT1

4. In cardiac muscle cells, Calcium extrusion during relaxation is partially achieved by the Na+/Ca2+ Exchanger (NCX). This transporter moves 3 Na+ in and 1 Ca2+ out. This is an example of:

a) Uniport

b) Antiport (Counter-transport)

c) Primary Active Transport

d) Passive Diffusion

5. A patient with familial hypercholesterolemia has extremely high levels of LDL cholesterol due to a defect in cellular uptake. The mechanism by which cells normally internalize LDL particles is:

a) Pinocytosis

b) Phagocytosis

c) Receptor-Mediated Endocytosis

d) Simple Diffusion

6. The MDR1 (Multi-Drug Resistance) protein, often overexpressed in cancer cells, pumps chemotherapeutic drugs out of the cell. Structurally and functionally, this transporter belongs to the:

a) ABC Transporter superfamily

b) Solute Carrier (SLC) family

c) Ion Channel family

d) Aquaporin family

7. Which transport mechanism exhibits the property of Transport Maximum (Tm), where the rate of transport plateaus at high substrate concentrations?

a) Simple Diffusion through lipid bilayer

b) Simple Diffusion through channels

c) Carrier-Mediated Transport

d) Osmosis

a) Secondary Active Transport

b) Primary Active Transport

c) Facilitated Diffusion

d) Passive leak channel

9. The "Chloride Shift" in Red Blood Cells involves the exchange of Bicarbonate (HCO3-) leaving the cell for Chloride (Cl-) entering the cell. This anion exchange is mediated by Band 3 protein via:

a) Facilitated Diffusion (Antiport)

b) Active Transport

c) Simple Diffusion

d) Pinocytosis

10. In nerve axons, the rapid depolarization phase of the action potential is mediated by the influx of Sodium. Through which type of membrane protein does this Sodium enter?

a) Voltage-Gated Ion Channel

b) Ligand-Gated Ion Channel

c) Leak Channel

d) Na+/K+ Pump

1045. NEET PG

Chapter: General Physiology; Topic: Nerve-Muscle Physiology; Subtopic: Resting Membrane Potential and Nernst Equation

Key Definitions & Concepts

Nernst Equation: Used to calculate the equilibrium potential for a single ion. Formula: E = -61 x log(Cin/Cout) at 37°C.
Equilibrium Potential (E): The membrane potential at which the electrical gradient exactly balances the chemical concentration gradient for a specific ion, resulting in no net flow.
Potassium (K+): The primary ion determining the Resting Membrane Potential (RMP). High concentration inside (ICF ~140-150 mEq/L), low outside (ECF ~4-5 mEq/L).
Sodium (Na+): High concentration outside (ECF ~140-150 mEq/L), low inside (ICF ~10-14 mEq/L). Equilibrium potential is roughly +60 mV.
Resting Membrane Potential (RMP): The baseline potential difference across the membrane, typically -70 to -90 mV in neurons/muscle, largely due to K+ leak channels.
Goldman-Hodgkin-Katz Equation: Calculates membrane potential taking into account the permeability and concentration of multiple ions (K+, Na+, Cl-).
Gibbs-Donnan Effect: The unequal distribution of permeant charged ions caused by the presence of non-permeant charged ions (like proteins) on one side of the membrane.
Na+-K+ ATPase: An electrogenic pump that maintains the ion gradients by moving 3 Na+ out and 2 K+ in against their gradients.
Depolarization: The membrane potential becomes less negative (moves towards 0 or positive), often driven by Na+ influx.
Hyperpolarization: The membrane potential becomes more negative than RMP, often driven by K+ efflux or Cl- influx.

Lead Question - 2016

ECF concentration of 1C. is 150 meq/L and ICF concentration of le is 5 meq/L. What is the equilibrium potential for K+ is?

a) +60 mV

b) -60 mV

c) -90 mV

d) +90 mV

Explanation: The question asks for the equilibrium potential of Potassium (K+). Standard physiological values for K+ are high intracellularly (~150 mEq/L) and low extracellularly (~5 mEq/L). The text "ECF concentration of 1C...150" and "ICF...5" likely contains typos ("1C" for ICF, "le" for ECF, or swapped values), as the question specifically asks for K+. Applying the Nernst Equation: E = -61 x log (Cin / Cout). Using standard K+ values (Cin=150, Cout=5): E = -61 x log(150/5) = -61 x log(30) = -61 x 1.477 ≈ -90 mV. If the numbers 150 (outside) and 5 (inside) were strictly used for a cation, the result would be +90 mV (resembling Na+ gradients), but since the question asks for K+, the negative potential is the physiological answer. Therefore, the correct answer is c) -90 mV.

1. Which ion has the highest permeability across the resting neuronal cell membrane, thereby contributing most to the Resting Membrane Potential?

a) Sodium (Na+)

b) Chloride (Cl-)

c) Potassium (K+)

d) Calcium (Ca2+)

Explanation: The Resting Membrane Potential (RMP) is determined by the ion to which the membrane is most permeable. In the resting state, the cell membrane contains numerous "leak channels" that are open. The majority of these are Potassium (K+) leak channels. Because the membrane is much more permeable to K+ than to Na+ or other ions (roughly 50-100 times more), K+ flows out of the cell down its concentration gradient, carrying positive charge out and leaving the inside negative. This drives the RMP close to the Nernst equilibrium potential of Potassium (-90 mV). Therefore, the correct answer is c) Potassium (K+).

2. If the extracellular concentration of Potassium [K+]out is acutely increased (hyperkalemia), what happens to the Resting Membrane Potential?

a) It becomes more negative (Hyperpolarization)

b) It becomes less negative (Depolarization)

c) It remains unchanged

d) It reaches the Sodium equilibrium potential

Explanation: The RMP is sensitive to the gradient of Potassium. According to the Nernst equation (E = -61 log [Cin/Cout]), increasing the denominator ([K+]out) makes the ratio smaller, the log value smaller, and thus the equilibrium potential less negative (closer to zero). Physiologically, a high external K+ reduces the concentration gradient for K+ to leave the cell. Less K+ leaving means less positive charge is removed, making the inside of the cell less negative compared to the outside. This shift towards a more positive potential is called Depolarization, making excitable cells more prone to firing initially. Therefore, the correct answer is b) It becomes less negative (Depolarization).

3. The equilibrium potential for Chloride (Cl-) in most neurons is approximately -70 mV. If the Resting Membrane Potential is also -70 mV, what is the net flow of Chloride ions?

a) Net influx

b) Net efflux

c) No net flow

d) Active transport out of the cell

Explanation: The "Equilibrium Potential" (Nernst potential) is defined as the voltage at which the electrical driving force exactly balances the chemical concentration driving force. If the membrane potential (Vm) is exactly equal to the equilibrium potential for Chloride (E-Cl), the electrochemical forces acting on Chloride are equal and opposite. Consequently, there is No net flow of the ion across the membrane, even if channels are open. This state of equilibrium is why inhibitory postsynaptic potentials (IPSPs) mediated by Cl- (like GABA-A receptors) often "clamp" the membrane potential at -70 mV. Therefore, the correct answer is c) No net flow.

4. During the upstroke of the nerve action potential, the membrane potential transiently approaches the equilibrium potential of which ion?

a) Potassium (+90 mV)

b) Sodium (+60 mV)

c) Chloride (-70 mV)

d) Calcium (+130 mV)

Explanation: During the depolarization phase (upstroke) of an action potential, Voltage-Gated Sodium Channels open, dramatically increasing the membrane's permeability to Sodium (Na+). According to the Goldman equation, the membrane potential shifts towards the equilibrium potential of the most permeant ion. Since Na+ permeability (P-Na) becomes dominant, the potential rushes towards the Sodium equilibrium potential (E-Na), which is approximately +60 mV. It usually peaks around +30 to +40 mV due to the simultaneous/delayed opening of K+ channels and inactivation of Na+ channels preventing it from fully reaching +60 mV. Therefore, the correct answer is b) Sodium (+60 mV).

5. Which factor has a value of approximately -61 mV in the Nernst equation at body temperature (37°C)?

a) RT/zF

b) 2.303 RT/zF

c) Faraday's Constant

d) Gas Constant

Explanation: The full Nernst equation is E = -(RT/zF) * ln(Cin/Cout). To convert the natural logarithm (ln) to the base-10 logarithm (log), we multiply by 2.303. The constants are: Gas constant (R), Temperature (T in Kelvin), Valency (z), and Faraday's constant (F). At body temperature (37°C or 310 K), the combined factor 2.303 RT/F equals approximately 61.5 mV. For a monovalent ion (z=1), this is simplified to -61 mV (or +61 mV depending on the log ratio inversion). This constant is essential for quick calculations. Therefore, the correct answer is b) 2.303 RT/zF.

6. The sodium-potassium pump (Na+-K+ ATPase) is described as "electrogenic" because it:

a) Moves equal amounts of charge in both directions

b) Creates a net positive charge inside the cell

c) Moves 3 positive charges out for every 2 positive charges in

d) Consumes ATP to generate an action potential

Explanation: The Na+-K+ pump moves 3 Sodium ions (Na+) out of the cell and 2 Potassium ions (K+) into the cell for every cycle using 1 ATP. Both ions are positively charged. By pumping 3 cations out and only bringing 2 cations in, there is a net loss of 1 positive charge from the intracellular space. This net movement of charge creates a small electrical potential (making the inside more negative), contributing about -4 mV directly to the Resting Membrane Potential. This capability to generate a potential difference makes the pump Electrogenic. Therefore, the correct answer is c) Moves 3 positive charges out for every 2 positive charges in.

7. Calculate the equilibrium potential for Calcium (Ca2+) if [Ca2+]out = 2 mM and [Ca2+]in = 0.0002 mM. (Use the factor 61).

a) +122 mV

b) -122 mV

c) +61 mV

d) +244 mV

Explanation: Using the Nernst Equation: E = -61/z * log(Cin/Cout). Calcium is a divalent cation, so z = +2. The factor becomes -61/2 = -30.5. Concentration ratio: Cin/Cout = 0.0002 / 2 = 0.0001 (which is 10^-4). Log(10^-4) = -4. Calculation: E = -30.5 * (-4) = +122 mV. Alternatively, E = +61/z * log(Cout/Cin) = +30.5 * log(10000) = 30.5 * 4 = +122 mV. Calcium has a very large gradient driving it into the cell, resulting in a highly positive equilibrium potential. Therefore, the correct answer is a) +122 mV.

8. In the Goldman-Hodgkin-Katz equation, the contribution of Chloride (Cl-) is entered differently from Na+ and K+ because:

a) Chloride is divalent

b) Chloride has a negative valence (anion)

c) Chloride is not permeable

d) Chloride is actively transported into the cell

Explanation: The Goldman equation calculates Vm based on permeability (P) and concentrations. For cations (Na+, K+), the term is log([C]out / [C]in). However, Chloride is an Anion (negative valence, z = -1). Due to the properties of logarithms (log(A/B) = -log(B/A)), the negative charge allows the term to be inverted to maintain the positive sign of the equation's structure. Thus, for Chloride, the concentrations are inverted: [Cl-]in is in the numerator and [Cl-]out is in the denominator. This accounts for the valence difference. Therefore, the correct answer is b) Chloride has a negative valence (anion).

9. The absolute refractory period of a neuron is primarily due to the:

a) Opening of Voltage-gated K+ channels

b) Inactivation of Voltage-gated Na+ channels

c) Hyperpolarization of the membrane

d) Closure of leak channels

Explanation: During the absolute refractory period, no new action potential can be generated, regardless of the stimulus strength. This occurs during the depolarization and early repolarization phases. The molecular basis is the state of the Voltage-gated Sodium Channels. After opening to cause depolarization, these channels enter an Inactivated state (ball-and-chain mechanism closes the pore). They cannot reopen until the membrane repolarizes and the channels reset to the "closed but ready" state. This prevents back-propagation and limits firing frequency. Hyperpolarization causes the relative refractory period. Therefore, the correct answer is b) Inactivation of Voltage-gated Na+ channels.

10. Which transport mechanism is responsible for establishing the high concentration of Potassium inside the cell?

a) Simple Diffusion

b) Secondary Active Transport

c) Primary Active Transport

d) Facilitated Diffusion

Explanation: While K+ leak channels allow K+ to exit (creating the RMP), the high intracellular concentration of K+ is established and maintained against its concentration gradient. Moving ions "uphill" (from low extracellular to high intracellular concentration) requires metabolic energy. The Na+-K+ ATPase pump performs this function by using ATP hydrolysis to actively pump K+ into the cell (and Na+ out). This is Primary Active Transport. Without this pump, the gradients would dissipate via diffusion, and the cell would lose its excitability. Therefore, the correct answer is c) Primary Active Transport.

1046. NEET PG

Chapter: General Physiology; Topic: Membrane Potentials; Subtopic: The Nernst Equation and Equilibrium Potentials

Key Definitions & Concepts

Nernst Equation: A mathematical relationship used to calculate the equilibrium potential for a single ion based on its concentration gradient across a membrane.
Equilibrium Potential (E): The electrical potential difference that exactly balances the concentration gradient for a specific ion, resulting in no net flux.
Valence (z): The electrical charge of an ion (e.g., +1 for Na+, -1 for Cl-, +2 for Ca2+); a critical variable in the Nernst equation.
Concentration Gradient: The difference in ion concentration between the intracellular and extracellular fluid (Cin/Cout); the driving force for diffusion.
Electrochemical Gradient: The net driving force acting on an ion, combining both the electrical potential difference and the concentration difference.
Driving Force: Mathematically defined as (Vm - Eion); determines the magnitude and direction of current flow when channels are open.
Reversal Potential: Another term for Equilibrium Potential; the voltage at which the current direction reverses.
Temperature (T): The Nernst potential is directly proportional to absolute temperature; typically calculated at 37°C (310 K) in physiology.
Faraday's Constant (F): Represents the magnitude of electric charge per mole of electrons.
Non-ionic Solutes: Substances like glucose or urea that lack charge (z=0); the Nernst equation cannot be applied to them as they do not generate diffusion potentials.

[Image of Nernst equation formula]

Lead Question - 2016

Nernnst equation related to equilibrium potential does not depend upon?

a) Concentration gradient

b) Electric gradient

c) Non-ionic solution

d) Concentration of ions in two solution

Explanation: The Nernst equation ($E = -61/z \times \log([C]_{in}/[C]_{out})$) describes the condition of equilibrium for charged particles (ions). It calculates the Electrical gradient (voltage) required to balance a specific Concentration gradient (ratio of ions in two solutions). The variables strictly required are Temperature, Gas Constant, Faraday's Constant, Valence ($z$), and the Concentrations inside and outside. If a solute is Non-ionic (uncharged, $z=0$), the denominator in the equation becomes zero (or the concept becomes physically meaningless in this context), as uncharged molecules diffuse solely based on concentration and do not generate an opposing electrical potential. Therefore, the correct answer is c) Non-ionic solution.

1. In the Nernst equation, the equilibrium potential is inversely proportional to which property of the ion?

a) Lipid solubility

b) Valence (Charge)

c) Molecular weight

d) Extracellular concentration

Explanation: The simplified Nernst equation at 37°C is $E = -61.5/z \times \log([C]_{in}/[C]_{out})$. The variable '$z$' represents the valence (charge) of the ion. Because '$z$' is in the denominator of the constant factor, the magnitude of the equilibrium potential is inversely proportional to the valence. For example, a divalent ion like Calcium ($z=+2$) will have a smaller Nernst slope factor ($\approx 30.7$) compared to a monovalent ion like Sodium ($z=+1$, factor $\approx 61.5$) for the same concentration ratio magnitude. Molecular weight and lipid solubility are not variables in the Nernst equation. Therefore, the correct answer is b) Valence (Charge).

2. A patient with severe burns develops acute hyperkalemia (high extracellular K+). According to the Nernst prediction, how does this affect the Equilibrium Potential for Potassium ($E_K$)?

a) $E_K$ becomes more negative (Hyperpolarized)

b) $E_K$ becomes less negative (Depolarized)

c) $E_K$ becomes positive

d) $E_K$ remains unchanged

Explanation: Under normal conditions, $[K^+]_{in}$ is high ($\approx 140$) and $[K^+]_{out}$ is low ($\approx 4$), resulting in an $E_K$ of about -90 mV. In hyperkalemia, $[K^+]_{out}$ increases. This decreases the concentration ratio ($[K^+]_{in}/[K^+]_{out}$). Mathematically, as the ratio approaches 1, the logarithm approaches 0. Consequently, the equilibrium potential moves closer to 0 mV (i.e., it becomes less negative or depolarized). This shift in $E_K$ (and consequently the Resting Membrane Potential) moves the cell closer to the firing threshold, increasing excitability and risk of cardiac arrhythmias. Therefore, the correct answer is b) $E_K$ becomes less negative (Depolarized).

3. Which factor is NOT considered in the Nernst Equation but IS included in the Goldman-Hodgkin-Katz (GHK) equation?

a) Temperature

b) Membrane Permeability

c) Ion Concentration

d) Ion Valence

Explanation: The Nernst equation calculates the potential for a single ion assuming the membrane is fully permeable to it (or at equilibrium). However, real cell membranes are permeable to multiple ions simultaneously (Na+, K+, Cl-) to varying degrees. The Goldman-Hodgkin-Katz (GHK) equation calculates the actual Resting Membrane Potential by taking into account the Membrane Permeability ($P$) of each ion in addition to their concentrations. For example, the resting membrane is dominated by K+ because $P_K$ is much higher than $P_{Na}$. Nernst assumes ideal, single-ion conditions. Therefore, the correct answer is b) Membrane Permeability.

4. If the concentration of Sodium inside the cell is 14 mM and outside is 140 mM, the Nernst potential for Sodium (log 10 = 1) is approximately:

a) -61 mV

b) +90 mV

c) +61 mV

d) 0 mV

Explanation: Using the Nernst equation: $E_{Na} = -61/z \times \log([C]_{in}/[C]_{out})$. $z$ for Sodium is +1. Ratio is $14/140 = 0.1$ (or $1/10$). $\log(0.1) = -1$. $E_{Na} = -61 \times (-1) = +61 mV$. Alternatively, using the "61 log (Out/In)" form: $61 \times \log(140/14) = 61 \times \log(10) = 61 \times 1 = +61$ mV. Sodium has a strong electrochemical drive to enter the cell, creating a positive equilibrium potential. Therefore, the correct answer is c) +61 mV.

5. The "Driving Force" for an ion to move across the membrane is mathematically defined as:

a) $Vm - E_{ion}$

b) $E_{ion} - Vm$

c) Permeability $\times$ Concentration

d) Nernst Potential $\times$ Valence

Explanation: The direction and magnitude of ionic current (I) flow through an open channel depend on the Driving Force. This is defined as the difference between the actual Membrane Potential ($Vm$) and the ion's Equilibrium Potential ($E_{ion}$). Formula: Driving Force = $Vm - E_{ion}$. If $Vm$ equals $E_{ion}$, the driving force is zero and there is no net current (equilibrium). If $Vm$ is -70 mV and $E_{Na}$ is +60 mV, the driving force is -130 mV, representing a very strong force pulling Na+ into the cell. Therefore, the correct answer is a) $Vm - E_{ion}$.

6. A patient with renal failure presents with hypocalcemia (low extracellular Calcium). How does this affect neuronal excitability, and what is the mechanism?

a) Decreased excitability due to hyperpolarized Nernst potential

b) Increased excitability due to lowering of the threshold potential

c) No change in excitability

d) Increased excitability due to Na+ pump inhibition

Explanation: This is a high-yield clinical correlation. While the Nernst potential for Ca2+ changes, the primary effect of Hypocalcemia is on the Voltage-Gated Sodium Channels. Extracellular calcium normally binds to the outer surface of Na+ channels, stabilizing them and making them harder to open. When extracellular Ca2+ is low, this stabilizing effect is lost. Consequently, the threshold voltage required to open Na+ channels becomes more negative (closer to the resting potential). This lowers the threshold for firing, leading to Increased excitability and tetany (Chvostek's/Trousseau's signs). Therefore, the correct answer is b) Increased excitability due to lowering of the threshold potential.

7. If the membrane potential is clamped at +61 mV, and the equilibrium potential for Sodium is +61 mV, what is the net flux of Sodium ions?

a) Net influx

b) Net efflux

c) Zero net flux

d) Dependent on ATP availability

Explanation: This scenario describes the definition of the Nernst/Equilibrium potential. At +61 mV, the electrical repulsion of positive sodium ions from inside the positive cell exactly balances the chemical force driving sodium down its concentration gradient into the cell. Because the electrical and chemical forces are equal and opposite, the system is in equilibrium. Although individual ions may move, there is Zero net flux of sodium. This potential is also called the "Reversal Potential" because movement would reverse direction if the voltage went higher or lower. Therefore, the correct answer is c) Zero net flux.

8. What effect does increasing the temperature of the solution have on the Nernst Equilibrium Potential?

a) It decreases the potential magnitude

b) It increases the potential magnitude

c) It has no effect

d) It reverses the polarity

Explanation: The Nernst equation includes the term $RT/zF$, where $T$ is the absolute temperature in Kelvin. Since $T$ is in the numerator, the magnitude of the equilibrium potential is Directly proportional to the temperature. As temperature increases, the kinetic energy of the ions increases, leading to a stronger diffusion tendency. Consequently, a larger electrical potential is required to oppose this increased diffusion force and maintain equilibrium. Thus, the potential magnitude Increases. Therefore, the correct answer is b) It increases the potential magnitude.

9. The resting membrane potential of a neuron (-70 mV) is closest to the Nernst potential of Potassium (-90 mV) rather than Sodium (+60 mV) because:

a) Potassium has the highest concentration gradient

b) The membrane has high resting permeability to Potassium

c) The Sodium-Potassium pump moves more Potassium

d) Potassium is a divalent ion

Explanation: According to the GHK equation, the membrane potential is a weighted average of the equilibrium potentials of all permeant ions, weighted by their permeabilities. In a resting neuron, the membrane possesses numerous "leak channels" that are highly selective for Potassium. Thus, the resting membrane Permeability to Potassium ($P_K$) is about 50-100 times higher than that for Sodium ($P_{Na}$). This high permeability "pulls" the resting membrane potential towards $E_K$. If Na+ channels open (increasing $P_{Na}$), the potential moves towards $E_{Na}$. Therefore, the correct answer is b) The membrane has high resting permeability to Potassium.

10. If the concentration gradient of a monovalent cation is reversed (i.e., higher concentration inside than outside becomes higher outside than inside), the new Nernst potential will:

a) Change in magnitude only

b) Change in sign (polarity) only

c) Become zero

d) Remain exactly the same

Explanation: The Nernst equation depends on the log of the concentration ratio ($C_{in}/C_{out}$). If the gradient is reversed (e.g., swapping the values of $C_{in}$ and $C_{out}$), the ratio becomes the reciprocal (inverted). The logarithm of a reciprocal ($\log(1/x)$) is equal to the negative logarithm ($-\log(x)$). Therefore, reversing the gradient simply reverses the sign of the calculated voltage. For example, if high inside gives -90 mV (like K+), high outside (with same ratio) would give +90 mV. The magnitude remains the same (assuming the ratio value is the same), but the Polarity reverses. Therefore, the correct answer is b) Change in sign (polarity) only.

1047. NEET PG

Chapter: General Physiology; Topic: Body Fluids and Membranes; Subtopic: Gibbs-Donnan Equilibrium

Key Definitions & Concepts

Gibbs-Donnan Equilibrium: The behavior of charged particles near a semi-permeable membrane that typically fails to distribute evenly across the two sides due to the presence of a non-diffusible charged substance (usually protein).
Non-Diffusible Anions: Large intracellular proteins ($Pr^-$) and organic phosphates that cannot cross the cell membrane; they carry a net negative charge at physiological pH.
Donnan Product Rule: At equilibrium, the product of the concentrations of diffusible cations and anions on one side equals the product on the other side ($[K^+]_{in} \times [Cl^-]_{in} = [K^+]_{out} \times [Cl^-]_{out}$).
Donnan Excess: The phenomenon where the total concentration of diffusible ions is greater in the compartment containing the non-diffusible ion (ICF) than in the compartment without it (ECF).
Osmotic Effect: Because total ions are higher in the ICF due to the Donnan effect, water tends to enter the cell; the Na+/K+ pump acts as a functional "anti-Donnan" mechanism to prevent cell swelling.
Intracellular pH: Due to the Donnan distribution, $H^+$ ions are slightly more concentrated inside the cell, making the ICF slightly more acidic than the ECF.
Chloride Shift: An example of passive anionic redistribution (Hamburger phenomenon) that respects Donnan principles across the RBC membrane.
Resting Membrane Potential: The Donnan effect contributes to the negative charge inside the cell, although the K+ leak channels and Na/K pump are the primary determinants.
Electroneutrality: Despite the unequal distribution of individual ions, each compartment (ICF and ECF) remains electrically neutral macroscopically.
Colloid Osmotic Pressure: In capillaries, plasma proteins create a Donnan effect that contributes to the oncotic pressure keeping fluid in the vessels.

[Image of Gibbs Donnan Equilibrium diagram]

Lead Question - 2016

Due to Donnan-Gibbs effect?

a) Concentration of K. is greater in ECF

b) Concentration of cl is greater in ECF

c) Total ions are more in ICF

d) All are true

Explanation: The Gibbs-Donnan effect occurs because non-diffusible anions (proteins) are trapped inside the cell (ICF). To maintain electrical neutrality, these proteins attract diffusible cations ($K^+$) into the cell and repel diffusible anions ($Cl^-$) out of the cell. Mathematically, to satisfy the equilibrium product ($[Cat]_{in} \times [An]_{in} = [Cat]_{out} \times [An]_{out}$), the sum of the concentrations of diffusible ions inside the compartment with the protein must be greater than the sum of ions in the outer compartment. This is known as the "Donnan Excess." Therefore, the total number of osmotically active particles is higher in the ICF. While Cl- is indeed greater in ECF, option (c) describes the fundamental thermodynamic consequence of the effect regarding total osmolarity. Therefore, the correct answer is c) Total ions are more in ICF.

1. According to the Gibbs-Donnan equilibrium principle, which of the following mathematical relationships is true regarding diffusible ions K+ and Cl-?

a) $[K^+]_{in} = [K^+]_{out}$

b) $[K^+]_{in} \times [Cl^-]_{out} = [K^+]_{out} \times [Cl^-]_{in}$

c) $[K^+]_{in} \times [Cl^-]_{in} = [K^+]_{out} \times [Cl^-]_{out}$

d) $[K^+]_{in} + [Cl^-]_{in} = [K^+]_{out} + [Cl^-]_{out}$

Explanation: The Gibbs-Donnan equilibrium dictates that at equilibrium, the product of the concentrations of the diffusible cation and anion on one side of the membrane must equal the product of their concentrations on the other side. This ensures that the chemical potential gradients are balanced by the electrical potential. Therefore, the correct relationship is the product of the internal ions equals the product of the external ions. Option (d) is incorrect because the sums are unequal (Donnan excess). Therefore, the correct answer is c) $[K^+]_{in} \times [Cl^-]_{in} = [K^+]_{out} \times [Cl^-]_{out}$.

2. The presence of intracellular non-diffusible anions results in the intracellular pH being:

a) Slightly more acidic than ECF

b) Slightly more alkaline than ECF

c) Exactly equal to ECF

d) Independent of the Donnan effect

Explanation: Hydrogen ions ($H^+$) are diffusible cations. The Donnan effect predicts that diffusible cations will be attracted to the compartment containing the non-diffusible anions (the ICF). Therefore, $H^+$ tends to accumulate inside the cell, following the same distribution pattern as $K^+$. A higher concentration of $H^+$ corresponds to a lower pH. Consequently, the intracellular fluid is typically slightly more acidic (pH ~7.0-7.2) compared to the extracellular fluid (pH ~7.4), partly due to this passive distribution. Therefore, the correct answer is a) Slightly more acidic than ECF.

3. Which cellular mechanism functions as a "functional anti-Donnan pump" to prevent cell swelling and lysis?

a) Na+-H+ Exchanger

b) Cl--HCO3- Exchanger

c) Na+-K+ ATPase

d) Ca2+ ATPase

Explanation: The Donnan effect creates a situation where the total concentration of ions (osmolarity) is higher inside the cell than outside. This creates an osmotic gradient that favors water entry, which would naturally cause the cell to swell and burst. To counteract this, the cell uses the Na+-K+ ATPase. By actively pumping 3 $Na^+$ ions out for every 2 $K^+$ ions in, the pump effectively makes the membrane functionally impermeable to Sodium (keeping $Na^+$ out). This removal of osmotically active particles ("double Donnan effect") maintains cell volume. Therefore, the correct answer is c) Na+-K+ ATPase.

4. In a system where proteins are negatively charged inside the cell, what is the expected distribution of Chloride (Cl-) ions?

a) $[Cl^-]_{in} > [Cl^-]_{out}$

b) $[Cl^-]_{in} = [Cl^-]_{out}$

c) $[Cl^-]_{out} > [Cl^-]_{in}$

d) Chloride is impermeable

Explanation: Chloride is a diffusible anion. The intracellular proteins are non-diffusible anions ($Pr^-$). Since like charges repel, the negative charge of the intracellular proteins repels the Chloride ions, driving them out of the cell. Additionally, to maintain the Donnan equilibrium product, if cations ($K^+$) are high inside, anions ($Cl^-$) must be low inside. Thus, the concentration of Chloride is significantly higher in the Extracellular Fluid (ECF) than in the Intracellular Fluid (ICF). Therefore, the correct answer is c) $[Cl^-]_{out} > [Cl^-]_{in}$.

5. The Gibbs-Donnan effect contributes approximately how much to the Resting Membrane Potential (RMP) of a nerve fiber?

a) -90 mV (The entire potential)

b) -60 mV

c) -10 mV to -20 mV

d) 0 mV

Explanation: The Resting Membrane Potential (typically -70 to -90 mV) is primarily determined by the selective permeability of the membrane to Potassium (Nernst potential of K+) and the electrogenic Na+/K+ pump. The passive distribution of ions due to the Donnan effect (fixed protein anions) does contribute to the negativity, but it is not the sole generator. If the Na+/K+ pump stops, the potential eventually collapses to the Donnan potential, which is much smaller, roughly -10 mV to -20 mV. The active pump is required to reach physiological RMP. Therefore, the correct answer is c) -10 mV to -20 mV.

6. In the capillaries, the Donnan effect involving plasma proteins (Albumin) creates an extra osmotic pressure that is approximately what percentage of the total oncotic pressure?

a) 1%

b) 50%

c) 33-50%

d) 99%

Explanation: Plasma oncotic pressure (Colloid Osmotic Pressure) is crucial for retaining fluid in capillaries. It is roughly 28 mmHg. About 19 mmHg of this is caused by the dissolved proteins themselves (van 't Hoff component). However, because proteins are negative (at pH 7.4), they hold extra cations (mostly $Na^+$) in the plasma due to the Donnan effect. These extra trapped cations exert their own osmotic pressure, which accounts for the remaining ~9 mmHg. Thus, the Donnan effect contributes roughly 33-50% (specifically about 1/3) of the total effective oncotic pressure. Therefore, the correct answer is c) 33-50%.

7. Which ion acts as the "Non-diffusible" ion in the Donnan equilibrium established across the Red Blood Cell membrane?

a) Sodium

b) Hemoglobin

c) Bicarbonate

d) Potassium

Explanation: In the context of the Red Blood Cell (RBC), the membrane is permeable to water and small anions (Cl-, HCO3-) via the Band 3 exchanger, but it is impermeable to large proteins. The primary large, negatively charged protein trapped inside the RBC is Hemoglobin. The negative charge on hemoglobin dictates the passive distribution of diffusible ions like Chloride (the Chloride shift) according to Donnan principles. Sodium and Potassium are regulated by pumps, not just passive Donnan forces. Therefore, the correct answer is b) Hemoglobin.

8. If $[K^+]_{out} = 5$ mM and $[Cl^-]_{out} = 100$ mM, and due to the Donnan effect the $[Cl^-]_{in} = 5$ mM, what must be the concentration of $[K^+]_{in}$ at equilibrium?

a) 5 mM

b) 20 mM

c) 100 mM

d) 500 mM

Explanation: Apply the Donnan product rule: $[K^+]_{in} \times [Cl^-

1048. NEET PG

Chapter: General Physiology; Topic: Transport Across Cell Membranes; Subtopic: Primary Active Transport and ATPase Pumps

Key Definitions & Concepts

Primary Active Transport: Transport of solutes against their electrochemical gradient utilizing energy derived directly from the hydrolysis of ATP.
ATPase (Adenosine Triphosphatase): An enzyme class that catalyzes the decomposition of ATP into ADP and a free phosphate ion, releasing energy to drive transport.
Electrogenic Pump: A pump that generates a net flow of charge across the membrane (unequal movement of cations/anions), contributing directly to the membrane potential.
Na+/K+ ATPase: The ubiquitous P-type ATPase that maintains the resting membrane potential and cell volume by pumping 3 Na+ out and 2 K+ in.
P-type ATPase: A family of pumps that form a phosphorylated intermediate during the transport cycle (e.g., Na+/K+ pump, Ca2+ pump, H+/K+ pump).
SERCA (Sarcoplasmic Endoplasmic Reticulum Calcium ATPase): A pump responsible for sequestering Ca2+ back into the sarcoplasmic reticulum during muscle relaxation.
Cardiac Glycosides (Digoxin/Ouabain): Drugs that specifically inhibit the Na+/K+ ATPase by binding to the extracellular domain of the alpha subunit.
ABC Transporters (ATP-Binding Cassette): A large superfamily of primary active transporters that transport a wide variety of substrates (e.g., MDR1, CFTR).
H+/K+ ATPase (Proton Pump): Located in gastric parietal cells, responsible for acid secretion; inhibited by Omeprazole.
Secondary Active Transport: Uses the energy stored in an ion gradient (created by primary active transport) rather than direct ATP hydrolysis (e.g., SGLT).

[Image of Na-K ATPase pump mechanism]

Lead Question - 2016

ATPase is which type of pump?

a) Secondary active

b) Electrogenic

c) Symport

d) All of the above

Explanation: The question refers to the general properties of ion-motive ATPases, most notably the Na+/K+ ATPase. These pumps function via Primary Active Transport because they directly hydrolyze ATP to move ions against gradients. Therefore, option (a) is incorrect. They are generally antiports (moving ions in opposite directions), not symports, so (c) is incorrect. A key feature of the Na+/K+ ATPase is that it moves 3 positive charges (Na+) out for every 2 positive charges (K+) in. This net movement of positive charge out of the cell creates an electrical potential difference, making the pump Electrogenic. It contributes directly to the negative resting membrane potential. Therefore, the correct answer is b) Electrogenic.

1. The Na+/K+ ATPase pump consists of alpha and beta subunits. The specific binding site for ATP is located on the:

a) Extracellular side of the alpha subunit

b) Intracellular side of the alpha subunit

c) Extracellular side of the beta subunit

d) Intracellular side of the beta subunit

Explanation: The Na+/K+ ATPase is a heterodimer. The large Alpha subunit is the catalytic unit responsible for the transport activity. It spans the membrane and has binding sites for Na+ and ATP on the Intracellular side and binding sites for K+ and cardiac glycosides (like ouabain) on the extracellular side. The Beta subunit is a glycoprotein essential for the proper folding and trafficking of the pump to the plasma membrane but does not bind ATP. Phosphorylation of the alpha subunit is the key step in the transport cycle. Therefore, the correct answer is b) Intracellular side of the alpha subunit.

2. Which of the following is a classic P-type ATPase found in the Sarcoplasmic Reticulum of muscle cells, essential for relaxation?

a) Na+/H+ Exchanger

b) H+/K+ ATPase

c) SERCA pump

d) MDR1 protein

Explanation: Muscle relaxation requires the rapid removal of Calcium (Ca2+) from the cytosol. This is achieved primarily by pumping Ca2+ back into the Sarcoplasmic Reticulum (SR) against a massive concentration gradient. The pump responsible for this is the SERCA (Sarcoplasmic Endoplasmic Reticulum Calcium ATPase) pump. It is a P-type ATPase that consumes ATP to transport 2 Ca2+ ions per cycle from the cytosol into the SR lumen. Dysfunction of this pump can lead to delayed relaxation (Brody's disease) or altered contractility in heart failure. Therefore, the correct answer is c) SERCA pump.

3. Digoxin increases cardiac contractility (positive inotropy) by inhibiting the Na+/K+ ATPase. The immediate consequence of this inhibition that leads to increased contractility is:

a) Increased intracellular K+

b) Decreased intracellular Ca2+

c) Increased intracellular Na+

d) Membrane hyperpolarization

Explanation: Digoxin inhibits the Na+/K+ ATPase. Normally, this pump keeps intracellular Sodium low. When inhibited, Intracellular Na+ accumulates. This rise in [Na+]in reduces the gradient for the Na+/Ca2+ Exchanger (NCX), which normally pumps Ca2+ out by bringing Na+ in. With a reduced Na+ gradient driving it, the NCX slows down, leading to an accumulation of Intracellular Calcium. This extra Ca2+ is stored in the SR and released during subsequent beats, increasing the force of contraction (inotropy). Thus, the primary effect is increased Na+. Therefore, the correct answer is c) Increased intracellular Na+.

4. The gastric H+/K+ ATPase (Proton Pump) is responsible for acid secretion. It is an electroneutral pump because it exchanges:

a) 1 H+ for 1 K+

b) 2 H+ for 1 K+

c) 1 H+ for 2 K+

d) 3 H+ for 2 K+

Explanation: The H+/K+ ATPase located in the parietal cells of the stomach is responsible for the final step of acid secretion. It pumps Hydrogen ions (H+) into the gastric lumen and Potassium ions (K+) into the cell, using ATP hydrolysis. The stoichiometry of this exchange is 1 H+ for 1 K+ (some sources say 2:2, but the ratio is 1:1). Because equal amounts of positive charge are moved in opposite directions, there is no net movement of charge across the membrane. Thus, unlike the Na+/K+ pump, the proton pump is Electroneutral. Therefore, the correct answer is a) 1 H+ for 1 K+.

5. Multidrug Resistance Protein 1 (MDR1 or P-glycoprotein) pumps chemotherapeutic agents out of cancer cells. It belongs to which family of primary active transporters?

a) P-type ATPases

b) V-type ATPases

c) F-type ATPases

d) ABC Transporters

Explanation: Transporters are classified by structure and mechanism. The P-type ATPases (Na/K, Ca, H/K) involve phosphorylation. The ABC (ATP-Binding Cassette) Transporters utilize the energy of ATP binding and hydrolysis to transport a wide variety of substrates (lipids, drugs, ions) across membranes without being phosphorylated themselves. MDR1 (P-glycoprotein) and CFTR (Cystic Fibrosis Transmembrane Conductance Regulator) are prominent members of the ABC Transporter superfamily. Overexpression of MDR1 causes resistance to chemotherapy by actively pumping the drugs out of the tumor cells. Therefore, the correct answer is d) ABC Transporters.

6. In addition to maintaining membrane potential, the Na+/K+ ATPase is critical for regulating:

a) Cell division

b) Cell volume

c) DNA replication

d) Protein synthesis

Explanation: Cells contain high concentrations of imperatnt organic molecules (proteins, phosphates) that are negatively charged and osmotically active. This creates a "Donnan effect" that tends to pull water into the cell, threatening to burst it. The Na+/K+ ATPase acts as a functional "pump-leak" mechanism. By actively pumping 3 Na+ out (and keeping the membrane effectively impermeable to Na+), it removes osmotically active particles from the cytoplasm. This counteracts the osmotic pull of intracellular proteins, preventing cellular swelling and maintaining Cell Volume. Inhibition of the pump leads to cell swelling and lysis. Therefore, the correct answer is b) Cell volume.

7. Which class of ATPases is primarily responsible for acidifying intracellular organelles like lysosomes and endosomes?

a) P-type ATPases

b) V-type ATPases

c) F-type ATPases

d) ABC Transporters

Explanation: The acidification of intracellular compartments (lysosomes, endosomes, secretory vesicles) is achieved by proton pumps. These pumps are structurally distinct from the P-type pumps of the plasma membrane. They are classified as V-type (Vacuolar) ATPases. V-type ATPases pump protons (H+) from the cytoplasm into the organelle lumen, consuming ATP. They do not undergo phosphorylation during the cycle. F-type ATPases are the mitochondrial ATP synthases (running in reverse to synthesize ATP). P-type are the ion pumps like Na/K. Therefore, the correct answer is b) V-type ATPases.

8. The Cystic Fibrosis Transmembrane Conductance Regulator (CFTR) is unique among ABC transporters because it functions as:

a) An active pump for Chloride

b) An ion channel for Chloride

c) An active pump for Sodium

d) An exchanger for Bicarbonate

Explanation: Most ABC transporters (like MDR1) use ATP to actively pump substrates against a gradient. CFTR is a notable exception. Although it possesses the structural Nucleotide Binding Domains (NBDs) characteristic of the ABC family and binds/hydrolyzes ATP, it does not function as an active pump. Instead, ATP binding/hydrolysis regulates the opening and closing (gating) of the protein, allowing it to function as a passive Ion Channel for Chloride. Cl- flows down its electrochemical gradient. This unique "broken pump" mechanism is vital for epithelial fluid secretion. Therefore, the correct answer is b) An ion channel for Chloride.

9. A significant portion of the body's Basal Metabolic Rate (BMR) is accounted for by the energy consumption of the:

a) Ca2+ ATPase

b) H+ ATPase

c) Na+/K+ ATPase

d) Myosin ATPase

Explanation: The maintenance of ion gradients is a metabolically expensive process. The Na+/K+ ATPase is constitutively active in almost all cells of the body. In neurons and kidney tubules, its activity is particularly high. It is estimated that the Na+/K+ pump consumes approximately 20-40% of the total ATP produced in a resting individual. Consequently, it is a major contributor to the Basal Metabolic Rate (BMR) and basal thermogenesis. Thyroid hormones increase BMR in part by upregulating the expression of Na+/K+ pump units. Therefore, the correct answer is c) Na+/K+ ATPase.

10. Omeprazole and Pantoprazole are clinically used to treat GERD. They function by forming a covalent disulfide bond with, and irreversibly inhibiting, which pump?

a) Gastric Na+/K+ ATPase

b) Gastric H+/K+ ATPase

c) Esophageal Ca2+ ATPase

d) Intestinal Na+/Glucose Symport

Explanation: Proton Pump Inhibitors (PPIs) like Omeprazole are prodrugs that are activated in the acidic environment of the parietal cell canaliculus. Once activated, they bind irreversibly to cysteine residues on the alpha subunit of the Gastric H+/K+ ATPase (Proton Pump). This inhibition blocks the final common pathway of acid secretion, regardless of the stimulus (histamine, gastrin, or acetylcholine). Because the inhibition is irreversible, acid secretion is suppressed until new pump molecules are synthesized, allowing for once-daily dosing. Therefore, the correct answer is b) Gastric H+/K+ ATPase.

1049. NEET PG

Chapter: Respiratory Physiology; Topic: Gas Exchange; Subtopic: Diffusion of Oxygen and Carbon Dioxide

Key Definitions & Concepts

Simple Diffusion: The random thermal movement of molecules from an area of higher concentration (or partial pressure) to an area of lower concentration directly through the membrane lipid bilayer or channel.
Respiratory Membrane: The thin barrier (0.6 micrometers) separating alveolar air from pulmonary capillary blood, consisting of alveolar epithelium, basement membranes, and capillary endothelium.
Partial Pressure Gradient: The driving force for gas diffusion; Oxygen moves from alveoli (PO2 ~104 mmHg) to venous blood (PO2 ~40 mmHg).
Solubility Coefficient: A measure of how easily a gas dissolves in a fluid. CO2 is about 20 times more soluble in plasma than O2.
Diffusion Capacity (DLCO): A clinical test using Carbon Monoxide to measure the lung's ability to transfer gas; dependent on surface area, thickness, and perfusion.
Perfusion-Limited Exchange: Gas exchange limited by blood flow (e.g., N2O, O2 in normal conditions); equilibrium is reached rapidly.
Diffusion-Limited Exchange: Gas exchange limited by the properties of the membrane (e.g., CO, or O2 in severe fibrosis/exercise); equilibrium is not reached.
Fick’s Law of Diffusion: Rate of diffusion is proportional to Surface Area x Concentration Gradient x Solubility / (Thickness x √Molecular Weight).
Graham's Law: Rate of diffusion is inversely proportional to the square root of molecular weight.
Transit Time: The time a red blood cell spends in the pulmonary capillary (~0.75 seconds); equilibration of O2 normally takes only ~0.25 seconds.

[Image of Diffusion of gases across respiratory membrane]

Lead Question - 2016

Diffusion related to O2 transport across respiratory membrane is an example of?

a) Simple diffusion

b) Facilitated diffusion

c) Active diffusion

d) Osmotic diffusion

Explanation: The movement of respiratory gases (Oxygen and Carbon Dioxide) across the respiratory membrane occurs solely by Simple Diffusion. This process is passive, driven entirely by the partial pressure gradient of the gases between the alveolar air and the pulmonary capillary blood. Oxygen molecules are small, non-polar, and lipid-soluble, allowing them to pass directly through the lipid bilayers of the alveolar and endothelial cells without the need for carrier proteins (facilitated diffusion) or energy expenditure (active transport). Therefore, the correct answer is a) Simple diffusion.

1. According to Fick's Law, the rate of gas diffusion across the respiratory membrane is inversely proportional to the:

a) Surface area of the membrane

b) Partial pressure gradient

c) Thickness of the membrane

d) Solubility of the gas

Explanation: Fick's Law governs gas exchange. The rate of diffusion (V_gas) is directly proportional to the Surface Area (A), the Diffusion Coefficient (D), and the Pressure Gradient (P1-P2). However, it is Inversely proportional to the Thickness (T) of the membrane. This means that any condition increasing the thickness of the blood-gas barrier, such as pulmonary edema (fluid in interstitial space) or pulmonary fibrosis (scarring), will significantly decrease the rate of gas exchange, leading to hypoxemia. Therefore, the correct answer is c) Thickness of the membrane.

2. Carbon Dioxide (CO2) diffuses across the respiratory membrane approximately 20 times faster than Oxygen (O2). This is primarily because CO2 has a much higher:

a) Partial pressure gradient

b) Molecular weight

c) Solubility coefficient

d) Active transport rate

Explanation: The diffusion coefficient of a gas depends on its solubility in the fluid of the membrane and its molecular weight (D ∝ Solubility / √MW). Although CO2 is heavier than O2 (which would slightly slow it down), its Solubility coefficient in plasma and tissue fluids is drastically higher (about 24 times that of O2). This high solubility allows CO2 to traverse the aqueous layers of the membrane extremely rapidly, even with a much smaller partial pressure gradient (5-6 mmHg) compared to Oxygen (60 mmHg). Therefore, the correct answer is c) Solubility coefficient.

3. Under normal resting conditions, the transfer of Oxygen from the alveoli to the blood is considered to be:

a) Diffusion-limited

b) Perfusion-limited

c) Solubility-limited

d) Ventilation-limited

Explanation: A gas exchange process is "Perfusion-limited" if the partial pressure of the gas in the blood equilibrates with the alveolar pressure before the blood exits the capillary. For Oxygen, equilibration occurs very early (within the first 0.25 seconds of the 0.75-second transit time). Once equilibrium is reached, no more net diffusion can occur unless fresh blood enters (perfusion increases). Therefore, under normal healthy conditions, O2 transport is Perfusion-limited. It only becomes diffusion-limited in disease states (fibrosis) or extreme exercise. Therefore, the correct answer is b) Perfusion-limited.

4. Carbon Monoxide (CO) is used clinically to measure the Diffusion Capacity of the Lung (DLCO) because its uptake is:

a) Perfusion-limited

b) Active transport dependent

c) Diffusion-limited

d) Flow-limited

Explanation: Carbon Monoxide binds extremely avidly to hemoglobin (200x affinity of O2). Consequently, as CO moves across the membrane, it is immediately bound by Hb, keeping the partial pressure of free dissolved CO in the plasma near zero. Because a back-pressure never builds up, the gradient across the membrane remains maximal throughout the entire transit time. The only factor limiting the rate of uptake is the properties of the membrane itself (area/thickness). Thus, CO is the classic example of a Diffusion-limited gas. Therefore, the correct answer is c) Diffusion-limited.

5. A patient with Emphysema has a reduced Diffusion Capacity (DLCO). The primary pathophysiological mechanism for this reduction is:

a) Increased membrane thickness

b) Decreased surface area for diffusion

c) Reduced hemoglobin concentration

d) Increased alveolar dead space

Explanation: Emphysema is characterized by the destruction of alveolar septa (walls). This destruction results in the coalescence of small alveoli into larger air sacs (bullae). While lung volume may increase, the critical Surface Area available for gas exchange is drastically reduced. According to Fick's Law, diffusion is directly proportional to surface area. The loss of alveolar capillary bed also contributes. In contrast, Pulmonary Fibrosis reduces DLCO primarily by increasing membrane thickness. Therefore, the correct answer is b) Decreased surface area for diffusion.

6. At high altitude, the partial pressure of atmospheric oxygen decreases. How does this affect the diffusion of oxygen across the respiratory membrane?

a) Rate decreases due to decreased solubility

b) Rate decreases due to reduced pressure gradient

c) Rate increases due to compensatory hyperventilation

d) Rate remains unchanged

Explanation: Diffusion is driven by the pressure gradient (P1 - P2), where P1 is Alveolar PO2 and P2 is Venous PO2. At high altitude, the barometric pressure drops, lowering the inspired PO2 and consequently the Alveolar PO2. This narrows the difference between alveolar and venous oxygen tension. Since the driving force (gradient) is reduced, the Rate of diffusion decreases. This can lead to hypoxemia, especially during exertion when transit time is shortened. Therefore, the correct answer is b) Rate decreases due to reduced pressure gradient.

7. The movement of Oxygen from the plasma into the Red Blood Cell (RBC) involves crossing the RBC membrane. This step is:

a) Active transport via pumps

b) Simple diffusion

c) Facilitated diffusion via GLUT1

d) Endocytosis

Explanation: The path of oxygen involves: Alveolar epithelium -> Interstitium -> Capillary endothelium -> Plasma -> RBC membrane -> Binding to Hemoglobin. Throughout this entire pathway, including entry into the erythrocyte, Oxygen moves by Simple Diffusion. The RBC membrane is a lipid bilayer permeable to non-polar gases. There are no specific O2 transporters. The reaction rate of O2 with Hemoglobin can influence the overall uptake rate (theta), but the transport mechanism remains diffusion. Therefore, the correct answer is b) Simple diffusion.

8. During heavy exercise, the blood moves faster through the pulmonary capillaries (reduced transit time). Despite this, a healthy person maintains full oxygen saturation because:

a) Diffusion capacity increases (recruitment/distension)

b) The membrane becomes thinner

c) Active transport of O2 is induced

d) Hemoglobin affinity decreases

Explanation: During exercise, cardiac output increases, reducing capillary transit time from 0.75s to as low as 0.25s. Normally, O2 equilibration is complete within 0.25s ("diffusion reserve"). Additionally, the body compensates by recruiting previously closed apical capillaries and distending open ones. This effectively increases the Surface Area for diffusion, thereby increasing the overall Diffusion Capacity (DLCO). This ensures that even with less time available, sufficient oxygen enters the blood to saturate hemoglobin. Therefore, the correct answer is a) Diffusion capacity increases (recruitment/distension).

9. Which component of the respiratory membrane presents the greatest barrier (path length) to the diffusion of gases?

a) Surfactant layer

b) Alveolar epithelium

c) Capillary endothelium

d) Plasma layer and Intracellular path to Hb

Explanation: The respiratory membrane itself (tissue barrier) is extremely thin (~0.3-0.6 microns). However, once the gas crosses the tissue, it must diffuse through the plasma and into the RBC to bind Hemoglobin. The diffusion distance through the Plasma layer and the RBC interior actually constitutes a significant portion of the resistance to diffusion, roughly equivalent to the tissue resistance. In anemia (fewer RBCs), this resistance increases because the effective path length between available heme binding sites increases. Therefore, the correct answer is d) Plasma layer and Intracellular path to Hb.

10. Nitrous Oxide (N2O) is used to measure pulmonary blood flow because its uptake is purely:

a) Diffusion-limited

b) Perfusion-limited

c) Reaction-limited

d) Transport-limited

Explanation: Nitrous Oxide (N2O) is a gas that does not bind to hemoglobin. It is purely dissolved in plasma. Because it doesn't bind, the partial pressure in the blood rises very rapidly, equilibrating with the alveolar pressure almost instantly (within 0.1s). Once equilibrium is reached, no more N2O can enter the blood unless new blood flows in. Thus, the total amount of N2O taken up is entirely dependent on the volume of blood flowing past the alveoli (Cardiac Output). It is the classic example of a Perfusion-limited gas. Therefore, the correct answer is b) Perfusion-limited.

1050. NEET PG

Chapter: General Physiology; Topic: Body Fluids; Subtopic: Measurement of Body Fluid Compartments

Key Definitions & Concepts

Indicator Dilution Principle: The method used to measure the volume of a fluid compartment. Formula: Volume = (Amount of indicator injected - Amount excreted) / Concentration of indicator.
Total Body Water (TBW): Measured using substances that distribute freely in all body fluids, such as Deuterium Oxide (D2O), Tritium Oxide, or Antipyrine.
Extracellular Fluid (ECF): Measured using substances that cross capillaries but not cell membranes, such as Inulin, Mannitol, or Sucrose.
Intracellular Fluid (ICF): Cannot be measured directly because no substance distributes exclusively in the ICF. It is calculated as: ICF = Total Body Water - Extracellular Fluid.
Plasma Volume: Measured using substances that remain in the vessels, such as Evans Blue dye (T-1824) or Radio-labeled Albumin (131I-Albumin).
Blood Volume: Calculated from Plasma Volume and Hematocrit: Blood Volume = Plasma Volume / (1 - Hematocrit). Alternatively measured with Cr-51 labeled RBCs.
Interstitial Fluid (ISF): Cannot be measured directly. Calculated as: ISF = Extracellular Fluid - Plasma Volume.
Inulin: A polymer of fructose used as the gold standard for measuring ECF volume and GFR because it is neither metabolized nor secreted.
Evans Blue (T-1824): A dye that binds avidly to plasma albumin; used to measure plasma volume.
Deuterium Oxide (Heavy Water): An isotope of water used to measure Total Body Water.

Lead Question - 2016

Measurement of intracellular fluid in a 50 years old male is done by?

a) Dilution method

b) Evans blue

c) D20

d) Indirectly

Explanation: The measurement of body fluid volumes relies on the Indicator Dilution Principle. However, this method requires a substance that distributes *exclusively* into the specific compartment being measured. For the Intracellular Fluid (ICF), there is no known substance that can be injected into the blood and will distribute *only* into cells without remaining in the extracellular space. Therefore, the ICF volume cannot be measured directly. Instead, it is calculated Indirectly. We measure Total Body Water (using D2O/Tritium) and Extracellular Fluid (using Inulin/Mannitol), and then subtract the ECF from the TBW (ICF = TBW - ECF). Therefore, the correct answer is d) Indirectly.

1. Which substance is considered the "Gold Standard" for measuring Extracellular Fluid (ECF) volume?

a) Sodium thiosulfate

b) Inulin

c) Deuterium Oxide

d) Evans Blue

Explanation: To measure the ECF, a substance must pass freely through capillary walls (to leave plasma) but must not cross cell membranes (to avoid entering the ICF). It should also not be metabolized rapidly. Inulin, a large polysaccharide (polymer of fructose), fits these criteria perfectly. It distributes into the plasma and interstitial fluid but does not enter cells. While other substances like Mannitol, Sucrose, and radioactive Sodium are used, Inulin is widely regarded as the physiological standard. D2O measures TBW. Evans Blue measures plasma. Therefore, the correct answer is b) Inulin.

2. To measure Plasma Volume specifically, the indicator substance must possess which property?

a) It must be a small, non-polar molecule

b) It must bind tightly to plasma proteins

c) It must cross the capillary membrane freely

d) It must be actively secreted by the renal tubules

Explanation: Plasma volume is the intravascular part of the ECF. To measure it, the indicator must stay within the blood vessels and not leak out into the interstitial space. Capillaries are generally permeable to small molecules but impermeable to large proteins (albumin). Therefore, dyes like Evans Blue (which binds to albumin) or Radio-iodinated Albumin (I-131 Albumin) are used. By binding to proteins, these indicators are effectively trapped in the vascular compartment, allowing for the calculation of plasma volume via dilution. Small non-polar molecules would measure TBW. Therefore, the correct answer is b) It must bind tightly to plasma proteins.

3. Total Body Water (TBW) constitutes approximately what percentage of body weight in a healthy young adult male?

a) 40%

b) 50%

c) 60%

d) 80%

Explanation: The standard reference value for Total Body Water in a young adult male is 60% of body weight. This percentage varies with age, gender, and adiposity. Females generally have a lower percentage (~50%) due to higher adipose tissue content (fat is hydrophobic, so higher fat = lower water percentage). Infants have a much higher water content (~75-80%). In the elderly, TBW decreases to about 50-55% in males. The 60-40-20 rule helps remember the distribution: 60% TBW, 40% ICF, 20% ECF. Therefore, the correct answer is c) 60%.

4. A patient has a Hematocrit of 40% and a Plasma Volume measured as 3 Liters. What is the calculated Total Blood Volume?

a) 4 Liters

b) 5 Liters

c) 6 Liters

d) 7.5 Liters

Explanation: Blood is composed of Plasma and Cells (RBCs). Hematocrit (Hct) represents the fraction of blood volume occupied by cells. Therefore, the fraction occupied by plasma is (1 - Hct). Formula: Total Blood Volume = Plasma Volume / (1 - Hematocrit). Given: Plasma Volume = 3 L, Hematocrit = 0.40 (40%). Calculation: Blood Volume = 3 / (1 - 0.40) = 3 / 0.6 = 5 Liters. This calculation is crucial clinically when only plasma volume can be measured directly using dye dilution. Therefore, the correct answer is b) 5 Liters.

5. Which of the following fluid compartments cannot be measured directly by any exogenous marker?

a) Plasma Volume

b) Extracellular Fluid

c) Interstitial Fluid

d) Total Body Water

Explanation: There are two major compartments that are measured indirectly because no substance distributes exclusively into them. 1. Intracellular Fluid (ICF): No substance goes only into cells. (ICF = TBW - ECF). 2. Interstitial Fluid (ISF): No substance stays only in the interstitium without entering plasma or cells. Indicators for ECF (like Inulin) enter both plasma and ISF. Indicators for plasma (Evans Blue) stay in plasma. Thus, Interstitial Fluid is calculated as the difference: ISF = ECF - Plasma Volume. Therefore, the correct answer is c) Interstitial Fluid.

6. In the indicator dilution method, if the substance injected is metabolized or excreted by the kidney before equilibrium is reached, the calculated volume will be:

a) Correct

b) Falsely Low

c) Falsely High

d) Zero

Explanation: The formula is Volume = Amount Injected / Final Concentration. If the indicator is metabolized or excreted, the Final Concentration in the blood will be lower than it should be (numerator stays same, denominator decreases). Mathematically, dividing by a smaller number yields a larger result. Therefore, the calculated volume will be Falsely High (overestimated). To correct for this, serial measurements are taken and extrapolated back to time zero, or substances that are not metabolized (like radioactive isotopes or inulin) are preferred. Therefore, the correct answer is c) Falsely High.

7. Which compartment contains the smallest volume of fluid in the body?

a) Intracellular Fluid

b) Plasma

c) Interstitial Fluid

d) Transcellular Fluid

Explanation: The body fluids are divided as follows (for a 70kg man): TBW = 42 L. ICF = 28 L (2/3 of TBW). ECF = 14 L (1/3 of TBW). ECF is subdivided into Interstitial Fluid (~10.5 L or 3/4 of ECF) and Plasma (~3.5 L or 1/4 of ECF). However, there is a specialized "third space" called Transcellular Fluid (CSF, synovial, pleural, pericardial, intraocular fluids). This compartment is the smallest, normally constituting only about 1-2 Liters (~1.5% of body weight). Therefore, the correct answer is d) Transcellular Fluid.

8. The "20-40-60 Rule" is a useful mnemonic for body fluid distribution. What does the "40" represent?

a) Percentage of body weight that is Extracellular Fluid

b) Percentage of Total Body Water that is Intracellular

c) Percentage of body weight that is Intracellular Fluid

d) Percentage of body weight that is solid tissue

Explanation: The "60-40-20 Rule" helps estimate fluid volumes as a percentage of total body weight: 60% = Total Body Water (TBW). 40% = Intracellular Fluid (ICF). 20% = Extracellular Fluid (ECF). This rule applies to a standard "reference man." It highlights that the majority of our body weight is water, and the majority of that water (2/3) is located inside our cells. Therefore, the "40" stands for the percentage of body weight that is Intracellular Fluid. Therefore, the correct answer is c) Percentage of body weight that is Intracellular Fluid.

9. Tritiated Water (3H2O) is used to measure which body fluid compartment?

a) Plasma Volume

b) Total Body Water

c) Extracellular Fluid

d) Blood Volume

Explanation: To measure Total Body Water (TBW), the indicator must be a substance that behaves exactly like water: it must cross capillary walls AND cell membranes freely to distribute evenly throughout the entire aqueous space of the body. Isotopes of water are the ideal candidates. Tritiated Water (3H2O) (radioactive) and Deuterium Oxide (D2O) (heavy water, non-radioactive) are chemically equivalent to water and distribute in the total water pool. Antipyrine is another lipid-soluble substance used for this purpose. Therefore, the correct answer is b) Total Body Water.

10. Which condition leads to a decrease in the percentage of Total Body Water relative to body weight?

a) Increased muscle mass

b) Infancy

c) Obesity

d) Male gender

Explanation: The percentage of Total Body Water is inversely proportional to the amount of body fat. Adipose tissue contains very little water compared to lean muscle tissue. Therefore, as body fat increases, the relative fraction of the body composed of water decreases. Obesity significantly lowers the TBW percentage (sometimes to as low as 45%). Infants have very little fat and high water content (75%). Males typically have more lean muscle than females, so they have a higher water percentage. Therefore, the correct answer is c) Obesity.

BACK TO NEET PG HOME PAGE

Page updated

Google Sites

Report abuse