Introduction

Human Physiology is the study of the mechanical, physical, and biochemical functions of normal humans, their organs, and the cells of which they are composed. This unit is the single largest section of the NEET UG Biology syllabus, covering the respiratory, circulatory, excretory, locomotory, neural, and endocrine systems. Mastery of these systems is crucial not only for clearing NEET but as the foundational framework for all clinical medicine studies.

Clinical and Daily Life Relevance:

• Pulmonary Medicine: Understanding the physiological basis of asthma, emphysema, and occupational lung diseases like silicosis and asbestosis.
• Cardiology: Analyzing ECG waveforms to diagnose cardiac arrhythmias, myocardial infarction, and heart block.
• Nephrology: Grasping the countercurrent multiplier system, which guides the management of fluid-electrolyte imbalances and the operation of hemodialysis machines.

NEET Relevance: Human Physiology yields 12–18 questions annually, representing approximately 15% of the total Biology section. The NTA frequently tests numerical values of lung volumes, cardiac cycle timings, structural zones of the sarcomere, the action potential cascade, and endocrine hypo-/hyper-secretion disorders.

Learning Objectives

By the end of this unit, you should be able to:

• Describe the human respiratory organs, define all respiratory volumes/capacities, and analyze the factors affecting the oxygen-hemoglobin dissociation curve.
• Explain the composition of blood and lymph, define ABO and Rh blood grouping compatibility, outline the coagulation cascade, and map the cardiac cycle phases.
• Detail the steps of urine formation, explain the countercurrent multiplier mechanism in the loop of Henle and vasa recta, and outline renal hormonal regulation.
• Classify movements, describe the structure of the sarcomere, and explain the sliding filament theory with calcium/ATP cycles.
• Explain resting and action potentials in neurons, synaptic transmission, and CNS/PNS division.
• Classify hormones by chemical structure, outline their mechanism of action (membrane vs. intracellular), and construct a master table of endocrine glands, hormones, and clinical disorders.

Prerequisites

Before starting, ensure you have reviewed:

• Cell Biology: Structure and function of mitochondria (ATP synthesis), endoplasmic reticulum (sarcoplasmic reticulum), and plasma membrane transport.
• Biomolecules: Proteins (structure of hemoglobin, actin, myosin, and hormones), lipids, and carbohydrate metabolism.
• Physical Chemistry: Concepts of diffusion, osmosis, concentration gradients, and pH buffering systems.

Big Picture

NEET Priority: Critical

5.1.1 Respiratory Volumes and Capacities

Understanding these parameters is high-yield for numerical and match-the-following questions:

• Tidal Volume (TV): Volume of air inspired or expired during a normal respiration. Value: \( 500\ \text{mL} \).
• Inspiratory Reserve Volume (IRV): Additional volume of air a person can inspire by a forcible inspiration. Value: \( 2500\ \text{mL} \) to \( 3000\ \text{mL} \).
• Expiratory Reserve Volume (ERV): Additional volume of air a person can expire by a forcible expiration. Value: \( 1000\ \text{mL} \) to \( 1100\ \text{mL} \).
• Residual Volume (RV): Volume of air remaining in the lungs even after a forcible expiration. Value: \( 1100\ \text{mL} \) to \( 1200\ \text{mL} \) (Cannot be measured by spirometry).
• Inspiratory Capacity (IC): Total volume of air a person can inspire after a normal expiration:

$$ \text{IC} = \text{TV} + \text{IRV} \approx 3000\ \text{mL} - 3500\ \text{mL} $$

• Expiratory Capacity (EC): Total volume of air a person can expire after a normal inspiration:

$$ \text{EC} = \text{TV} + \text{ERV} \approx 1500\ \text{mL} - 1600\ \text{mL} $$

• Functional Residual Capacity (FRC): Volume of air that will remain in the lungs after a normal expiration:

$$ \text{FRC} = \text{ERV} + \text{RV} \approx 2100\ \text{mL} - 2300\ \text{mL} $$

• Vital Capacity (VC): The maximum volume of air a person can breathe in after a forced expiration (or vice versa):

$$ \text{VC} = \text{TV} + \text{IRV} + \text{ERV} \approx 4000\ \text{mL} - 4600\ \text{mL} $$

• Total Lung Capacity (TLC): Total volume of air accommodated in the lungs at the end of a forced inspiration:

$$ \text{TLC} = \text{VC} + \text{RV} = \text{TV} + \text{IRV} + \text{ERV} + \text{RV} \approx 5100\ \text{mL} - 5800\ \text{mL} $$

5.1.2 Oxygen-Hemoglobin Dissociation Curve Shifts

The oxygen-hemoglobin dissociation curve is sigmoid.

• Shift to the Right (Decreased Hb-O₂ affinity, favors oxygen unloading in tissues):
  • Decreased \( \text{pO}_2 \) (low oxygen tension).
  • Increased \( \text{pCO}_2 \) (high carbon dioxide tension - Bohr Effect).
  • Increased \( \text{H}^+ \) concentration / Decreased pH (acidic environment).
  • Increased Temperature.
  • Increased concentration of \( 2,3\text{-diphosphoglycerate} \) (2,3-DPG) in RBCs.
• Shift to the Left (Increased Hb-O₂ affinity, favors oxygen loading in alveoli):
  • Increased \( \text{pO}_2 \).
  • Decreased \( \text{pCO}_2 \).
  • Decreased \( \text{H}^+ \) concentration / Increased pH (alkaline environment).
  • Decreased Temperature.
  • Decreased 2,3-DPG.

NEET Priority: Critical

5.2.1 Cardiac Cycle Timeline Breakdown (Total Duration: 0.8 seconds)

At a resting heart rate of \( 72\ \text{beats/min} \), one cardiac cycle takes \( 0.8\ \text{seconds} \).

1. Atrial Systole (\( 0.1\ \text{seconds} \)):
   • Atria contract under the influence of the SA node.
   • Increases blood flow into the ventricles by about \( 30\% \).
   • AV valves are open; semilunar valves are closed.
2. Ventricular Systole (\( 0.3\ \text{seconds} \)):
   • Ventricles contract; intraventricular pressure rises.
   • First Heart Sound (LUB): Closure of tricuspid and bicuspid (AV) valves to prevent backflow.
   • Semilunar valves are forced open as ventricular pressure exceeds arterial pressure, ejecting blood into the pulmonary artery and aorta.
3. Joint Diastole (\( 0.4\ \text{seconds} \)):
   • Ventricles and atria relax.
   • Second Heart Sound (DUP): Closure of semilunar valves due to backpressure from the aorta and pulmonary artery.
   • AV valves open as atrial pressure exceeds ventricular pressure, allowing blood to flow passively from the vena cava and pulmonary veins into the ventricles.

NEET Priority: Critical

5.3.1 The Countercurrent Multiplier Mechanism

The kidney concentrates urine by establishing a hyperosmotic medullary interstitium (\( 300\ \text{mOsm/L} \) in the cortex to \( 1200\ \text{mOsm/L} \) in the deep medulla). This involves the coordinate loop of Henle and the vasa recta:

• Loop of Henle (Countercurrent Multiplier):
  • Descending Limb: Highly permeable to water but virtually impermeable to electrolytes. As filtrate moves down, water leaves the tubule into the hypertonic medullary interstitium, concentrating the filtrate up to \( 1200\ \text{mOsm/L} \) at the hairpin bend.
  • Ascending Limb: Impermeable to water but highly permeable to electrolytes (active transport of \( \text{Na}^+ \), \( \text{K}^+ \), and \( \text{Cl}^- \) in the thick portion). As filtrate moves up, solutes are pumped out, diluting the filtrate down to \( 200\ \text{mOsm/L} \).
• Vasa Recta (Countercurrent Exchanger):
  • A U-shaped capillary loop running parallel to the loop of Henle.
  • Descending capillary loop: Takes in NaCl from the interstitium and loses water.
  • Ascending capillary loop: Returns NaCl to the medullary interstitium and reabsorbs water.
  • This preserves the osmotic gradient without washing out the solute concentration.
• Urea Recycling: Urea diffuses out of the collecting duct into the medullary interstitium and enters the thin ascending limb of Henle, further contributing to medullary hyperosmolarity.

NEET Priority: High

5.4.1 The Sarcomere Structure and Contraction Behavior

The sarcomere is the structural and functional unit of myofibrils, bounded by Z-discs.

• Band Composition:
  • A-band (Anisotropic): Dark band containing the entire length of thick (myosin) filaments, including overlapping thin (actin) filaments.
  • I-band (Isotropic): Light band containing only thin (actin) filaments, bisected by the Z-line.
  • H-zone: Pale central region of the A-band, containing only thick (myosin) filaments.
  • Z-line: Dark anchoring protein structure for thin filaments.
  • M-line: Fibrous membrane in the middle of the H-zone anchoring thick filaments.
• Behavior During Contraction:
  • Sarcomere length: Shortens.
  • I-band: Shortens (narrower).
  • H-zone: Shortens and disappears during maximum contraction.
  • A-band: Length remains completely constant (unchanged).
  • Actin and Myosin filaments: Do not change their individual lengths; they only slide past each other.

NEET Priority: Critical

5.5.1 Ionic Basis of the Action Potential

The propagation of nerve impulses relies on voltage-gated ion channels:

1. Resting Phase (\( -70\ \text{mV} \)):
   • Axonal membrane is more permeable to \( \text{K}^+ \) and nearly impermeable to \( \text{Na}^+ \).
   • Inside has high \( \text{K}^+ \) and negatively charged proteins; outside has high \( \text{Na}^+ \).
   • Maintained actively by the \( \text{Na}^+\text{-K}^+ \) ATPase pump (transports \( 3\ \text{Na}^+ \) outward for \( 2\ \text{K}^+ \) inward).
2. Depolarization (Rising Phase, to \( +30\ \text{mV} \)):
   • Stimulus causes localized opening of voltage-gated \( \text{Na}^+ \) channels.
   • \( \text{Na}^+ \) rushes into the cell down its electrochemical gradient (inward sodium current).
   • Reverses polarity, making the inside positive relative to the outside.
3. Repolarization (Falling Phase, back to \( -70\ \text{mV} \)):
   • Voltage-gated \( \text{Na}^+ \) channels close (inactivate).
   • Voltage-gated \( \text{K}^+ \) channels open, causing a rapid exit of \( \text{K}^+ \) (outward potassium current).
4. Hyperpolarization (Undershoot, to \( -90\ \text{mV} \)):
   • \( \text{K}^+ \) channels remain open slightly longer than needed, causing membrane potential to drift closer to the \( \text{K}^+ \) equilibrium potential before returning to resting levels via the active sodium-potassium pump.

NEET Priority: Critical

5.6.1 Master Table of Endocrine Glands, Hormones, and Disorders

Formula Sheet

Respiratory Capacities

$$ \text{IC} = \text{TV} + \text{IRV} = 500\ \text{mL} + 3000\ \text{mL} = 3500\ \text{mL} $$ $$ \text{EC} = \text{TV} + \text{ERV} = 500\ \text{mL} + 1100\ \text{mL} = 1600\ \text{mL} $$ $$ \text{FRC} = \text{ERV} + \text{RV} = 1100\ \text{mL} + 1200\ \text{mL} = 2300\ \text{mL} $$ $$ \text{VC} = \text{TV} + \text{IRV} + \text{ERV} = 500\ \text{mL} + 3000\ \text{mL} + 1100\ \text{mL} = 4600\ \text{mL} $$ $$ \text{TLC} = \text{VC} + \text{RV} = 4600\ \text{mL} + 1200\ \text{mL} = 5800\ \text{mL} $$

Cardiovascular Equations

$$ \text{Cardiac Output (CO)} = \text{Heart Rate (HR)} \times \text{Stroke Volume (SV)} $$ $$ \text{Stroke Volume (SV)} = \text{End-Diastolic Volume (EDV)} - \text{End-Systolic Volume (ESV)} $$ $$ \text{Pulse Pressure} = \text{Systolic BP} - \text{Diastolic BP} $$ $$ \text{Mean Arterial Pressure (MAP)} = \text{Diastolic BP} + \frac{1}{3}(\text{Pulse Pressure}) $$

Renal Clearance

$$ C_x = \frac{U_x \cdot V}{P_x} \quad (\text{where } U_x = \text{urine conc., } P_x = \text{plasma conc., } V = \text{urine flow rate}) $$ $$ \text{Glomerular Filtration Rate (GFR)} \approx 125\ \text{mL/min} = 180\ \text{L/day} $$

Formula Derivations

Derivation of Cardiac Output and Stroke Volume Relationships

Let \( \text{EDV} \) be the End-Diastolic Volume (volume of blood in the ventricle at the end of filling, just before contraction) and \( \text{ESV} \) be the End-Systolic Volume (volume of blood remaining in the ventricle at the end of ejection).

The volume of blood ejected during a single ventricular contraction (Stroke Volume, \( \text{SV} \)) is:

$$ \text{SV} = \text{EDV} - \text{ESV} $$

Using average physiological values:

$$ \text{SV} = 120\ \text{mL} - 50\ \text{mL} = 70\ \text{mL/beat} $$

Cardiac Output (\( \text{CO} \)) is the volume of blood pumped by one ventricle per minute:

$$ \text{CO} = \text{Heart Rate (HR)} \times \text{Stroke Volume (SV)} $$

Substitute physiological constants:

$$ \text{CO} = 72\ \text{beats/min} \times 70\ \text{mL/beat} = 5040\ \text{mL/min} \approx 5.0\ \text{L/min} \quad \text{(Derived)} $$

Unit Analysis

Example 1 (Respiratory Capacities): A spirometer trace of a patient shows a Tidal Volume of \( 450\ \text{mL} \), an Expiratory Reserve Volume of \( 950\ \text{mL} \), and a Functional Residual Capacity of \( 2150\ \text{mL} \). Calculate the patient's Residual Volume and Vital Capacity, assuming their Inspiratory Reserve Volume is \( 2600\ \text{mL} \). Solution:

1. Calculate Residual Volume (\( \text{RV} \)) using the Functional Residual Capacity (\( \text{FRC} \)) formula:

$$ \text{FRC} = \text{ERV} + \text{RV} \implies 2150 = 950 + \text{RV} $$ $$ \text{RV} = 2150 - 950 = 1200\ \text{mL} $$

1. Calculate Vital Capacity (\( \text{VC} \)):

$$ \text{VC} = \text{TV} + \text{IRV} + \text{ERV} $$ $$ \text{VC} = 450 + 2600 + 950 = 4000\ \text{mL} $$

1. Answers: Residual Volume is \( 1200\ \text{mL} \); Vital Capacity is \( 4000\ \text{mL} \).

Example 2 (Cardiac Calculations): An athlete during intensive training has a heart rate of \( 150\ \text{beats/min} \). If their End-Diastolic Volume is \( 140\ \text{mL} \) and End-Systolic Volume is \( 40\ \text{mL} \), find their Stroke Volume and Cardiac Output. Solution:

1. Calculate Stroke Volume (\( \text{SV} \)):

$$ \text{SV} = \text{EDV} - \text{ESV} = 140\ \text{mL} - 40\ \text{mL} = 100\ \text{mL} $$

1. Calculate Cardiac Output (\( \text{CO} \)):

$$ \text{CO} = \text{HR} \times \text{SV} = 150\ \text{beats/min} \times 100\ \text{mL} = 15000\ \text{mL/min} = 15\ \text{L/min} $$

1. Answers: Stroke Volume is \( 100\ \text{mL} \); Cardiac Output is \( 15\ \text{L/min} \).

Example 3 (Action Potential Timing): If a nerve fiber has an absolute refractory period of \( 2\ \text{milliseconds} \), what is the theoretical maximum number of action potentials it can transmit per second? Solution:

1. The absolute refractory period represents the minimum time between successive impulses:

$$ t_{\text{min}} = 2\ \text{ms} = 2 \times 10^{-3}\ \text{seconds} $$

1. Calculate the maximum frequency (\( f_{\text{max}} \)):

$$ f_{\text{max}} = \frac{1}{t_{\text{min}}} = \frac{1}{2 \times 10^{-3}} = 500\ \text{impulses/second} $$

1. Answer: The maximum transmission frequency is \( 500\ \text{Hz} \) (impulses per second).

Example 4 (Bohr Effect Physiology): Explain why oxygen unloading increases when a muscle is actively contracting during exercise. Solution:

1. Active contraction increases metabolic activity, producing carbon dioxide (\( \text{CO}_2 \)) and lactic acid, while generating local heat.
2. In tissues, the elevated \( \text{pCO}_2 \) and hydronium ion concentration (lower pH) promote the binding of \( \text{CO}_2 \) and protons to hemoglobin, stabilizing its T-state (tense conformation).
3. This decreases the affinity of hemoglobin for oxygen, shifting the oxygen-hemoglobin dissociation curve to the right.
4. Hemoglobin releases bound oxygen more readily to the surrounding muscle tissue where \( \text{pO}_2 \) is already low (\( <20\ \text{mmHg} \)).

Example 5 (Glomerular Filtration Fraction): If renal plasma flow (RPF) in a healthy adult is \( 650\ \text{mL/min} \) and the Glomerular Filtration Rate (GFR) is \( 130\ \text{mL/min} \), calculate the Filtration Fraction (FF). Solution:

1. The filtration fraction represents the percentage of renal plasma that is filtered across the glomerulus:

$$ \text{FF} = \frac{\text{GFR}}{\text{RPF}} $$

1. Substitute the given values:

$$ \text{FF} = \frac{130\ \text{mL/min}}{650\ \text{mL/min}} = 0.20 = 20\% $$

1. Answer: The Filtration Fraction is \( 20\% \).

Example 6 (Pedigree/Hormonal Defect Analysis): A patient presents with high blood pressure, low plasma potassium (\( \text{K}^+ \)), and high urine sodium output. The patient is diagnosed with hyperaldosteronism. Explain the physiological mechanism causing these symptoms. Solution:

1. Aldosterone acts on the principal cells of the renal distal convoluted tubule (DCT) and collecting duct to upregulate \( \text{Na}^+\text{-K}^+ \) ATPase pumps and luminal epithelial sodium channels (ENaC).
2. Hypersecretion of aldosterone causes excessive reabsorption of sodium ions (\( \text{Na}^+ \)) from the tubule back into the blood. Water follows sodium passively due to osmotic gradients, raising blood volume and systemic blood pressure (hypertension).
3. Simultaneously, aldosterone increases the excretion of potassium ions (\( \text{K}^+ \)) and hydrogen ions (\( \text{H}^+ \)) into the urine, resulting in hypokalemia (low blood potassium) and potential metabolic alkalosis.

Example 7 (Sarcomere Geometry): A resting sarcomere has a length of \( 2.2\ \mu\text{m} \). The thick filament length is \( 1.6\ \mu\text{m} \) and the thin filament length on each side is \( 1.0\ \mu\text{m} \). Calculate the width of the H-zone at rest. Solution:

1. The total length of the sarcomere (\( L \)) is given by:

$$ L = 2(\text{thin filament length}) + \text{H-zone width} - \text{overlap} $$

Alternatively, the thick filament spans the A-band (\( 1.6\ \mu\text{m} \)). The non-overlapping portion of the thick filament is the H-zone. The overlap on each end of the A-band is:

$$ \text{Overlap per side} = \text{thin filament length} - \frac{L - \text{A-band}}{2} $$ $$ \text{Overlap per side} = 1.0 - \frac{2.2 - 1.6}{2} = 1.0 - 0.3 = 0.7\ \mu\text{m} $$

1. The H-zone is the central portion of the thick filament not overlapped by thin filaments:

$$ \text{H-zone width} = \text{thick filament length} - 2(\text{overlap per side}) $$ $$ \text{H-zone width} = 1.6 - 2(0.7) = 1.6 - 1.4 = 0.2\ \mu\text{m} $$

1. Answer: The width of the H-zone is \( 0.2\ \mu\text{m} \).

Example 8 (Myelinated Axon Conduction Speed): Why does saltatory conduction in myelinated axons consume less metabolic energy (ATP) than continuous conduction in unmyelinated axons? Solution:

1. In unmyelinated axons, action potentials are propagated continuously along the entire length of the membrane, requiring sodium influx and potassium efflux at every point.
2. The \( \text{Na}^+\text{-K}^+ \) ATPase pump must actively transport ions across the entire membrane surface to restore resting concentrations.
3. In myelinated axons, depolarization is restricted to the nodes of Ranvier. Active transport of ions is only required at these nodes.
4. This localized ion exchange reduces the total active transport workload, conserving ATP energy.

Example 9 (Endocrine Second Messenger Mechanism): Contrast the mechanism of action of epinephrine and progesterone on their target cells. Solution:

1. Epinephrine (Peptide-like, hydrophilic): Cannot cross the plasma membrane. It binds to extracellular G-protein coupled receptors, activating adenylyl cyclase, which converts ATP to cyclic AMP (cAMP) as a second messenger. This initiates a phosphorylation cascade, causing a rapid response.
2. Progesterone (Steroid, lipophilic): Crosses the plasma membrane directly. It binds to intracellular receptors in the cytoplasm or nucleus. The hormone-receptor complex acts as a transcription factor, binding to DNA hormone response elements to regulate gene transcription, resulting in a slower response.

Example 10 (Endocrine Pathology): A patient presents with a blood calcium level of \( 6.5\ \text{mg/dL} \) (normal: \( 9.0 - 10.5\ \text{mg/dL} \)) and exhibits spontaneous muscular spasms in the hands and feet. Identify the likely hormone deficiency. Solution:

1. A blood calcium level below normal indicates hypocalcemia. Spontaneous muscular spasms are symptoms of tetany.
2. Parathyroid hormone (PTH) is responsible for raising blood calcium levels by mobilizing calcium from bone, increasing renal calcium reabsorption, and promoting intestinal calcium absorption.
3. A deficiency in PTH (hypoparathyroidism) leads to low blood calcium. This increases neuromuscular excitability by lowering the threshold potential of motor neurons, leading to tetany.
4. Answer: The likely hormone deficiency is Parathyroid Hormone (PTH).

Example 11 (Urine Concentration Factor): Calculate the maximum concentration factor of urine produced by the human kidney relative to blood plasma. Solution:

1. Normal blood plasma osmolarity is:

$$ \text{Osmolarity}_{\text{plasma}} \approx 300\ \text{mOsm/L} $$

1. The maximum osmolarity established in the deep medullary interstitium is:

$$ \text{Osmolarity}_{\text{medulla}} \approx 1200\ \text{mOsm/L} $$

1. The urine can be concentrated up to the osmolarity of the deep medulla:

$$ \text{Concentration Factor} = \frac{\text{Osmolarity}_{\text{medulla}}}{\text{Osmolarity}_{\text{plasma}}} = \frac{1200}{300} = 4 $$

1. Answer: The maximum concentration factor is \( 4\ \text{times} \).

Example 12 (Joint Biomechanics): Explain the structural modification that prevents friction in freely movable synovial joints. Solution:

1. The articulating bone surfaces are covered by a smooth layer of hyaline cartilage called articular cartilage, which reduces friction and absorbs shock.
2. The joint cavity is enclosed by a fibrous articular capsule lined with a synovial membrane that secretes synovial fluid.
3. Synovial fluid acts as a lubricant, reducing friction during movement.

Memory Techniques

• Bohr Shift Factors (Shift to Right = Release O₂): Remember "CADET, face Right!"
  • C - \( \text{CO}_2 \) increase.
  • A - Acidosis (increase \( \text{H}^+ \) / decrease pH).
  • D - 2,3-DPG increase.
  • E - Exercise (tissues need O₂).
  • T - Temperature increase.
• Heart Valve Positions: "Try before you Buy"
  • Tricuspid valve is on the Right side of the heart (first letter).
  • Bicuspid (mitral) valve is on the Left side (second letter).
• ECG Waves: "Push Quickly Run Slowly Toward Relaxation"
  • P - Atrial depolarisation.
  • QRS - Ventricular depolarisation.
  • T - Ventricular repolarisation.
• Sarcomere Band Changes during Contraction: "HI A"
  • H zone disappears.
  • I band shortens.
  • A band remains completely constant.
• Nerve Impulse Ions: "No-In, K-Out" (pronounced "No in, K out" \(\rightarrow\) like sodium in, potassium out)
  • Na (Sodium) goes In during depolarization.
  • K (Potassium) goes Out during repolarization.

Summary

• Respiratory Physiology: Gas exchange takes place across the alveolar membrane. Blood transport relies on hemoglobin for O₂ (97% as oxyhemoglobin) and bicarbonate conversion for CO₂ (70%). Respiration is regulated primarily by medullary chemoreceptors sensitive to pH and \( \text{pCO}_2 \).
• Cardiovascular Physiology: The cardiac cycle is a 0.8-second loop consisting of atrial systole (0.1s), ventricular systole (0.3s), and joint diastole (0.4s). The conduction pathway begins at the SA node (pacemaker) and proceeds through the AV node, bundle of His, and Purkinje fibers.
• Renal Physiology: Nephrons process blood via ultrafiltration, selective reabsorption, and tubular secretion. The countercurrent multiplier mechanism in the loop of Henle and vasa recta establishes a medullary hyperosmolar gradient to concentrate urine under ADH and aldosterone regulation.
• Skeletal Muscle Contractility: Muscles contract via the sliding filament model. Calcium binds troponin, shifting tropomyosin to expose actin sites, allowing myosin heads to bind and perform the power stroke fueled by ATP hydrolysis.
• Neurobiology: Neurons maintain a resting potential of \( -70\ \text{mV} \) using the active \( \text{Na}^+\text{-K}^+ \) ATPase pump. Depolarization involves \( \text{Na}^+ \) influx, and repolarization is driven by \( \text{K}^+ \) efflux.
• Endocrine Control: Hormones act either via cell-surface receptors generating second messengers (peptide hormones) or by crossing the plasma membrane to bind intracellular receptors that modify gene expression (steroid and thyroid hormones).

One-Page Revision

• Lung Parameters: TV = \( 500\ \text{mL} \), VC = \( 4600\ \text{mL} \), TLC = \( 5800\ \text{mL} \), RV = \( 1200\ \text{mL} \). RV cannot be measured by spirometry.
• O₂/CO₂ Transport: 97% of O₂ carried as \( \text{HbO}_2 \); 70% of CO₂ transported as \( \text{HCO}_3^- \). Bohr effect: high CO₂, high \( \text{H}^+ \), high temp, and high DPG shift the curve right.
• Cardiac Cycle: Total duration \( 0.8\ \text{seconds} \). Atrial systole (0.1s), ventricular systole (0.3s), joint diastole (0.4s). LUB = closure of AV valves; DUP = closure of semilunar valves.
• ECG Waves: P = atrial depolarization; QRS = ventricular depolarization; T = ventricular repolarization.
• Renal Filtration: GFR = \( 125\ \text{mL/min} = 180\ \text{L/day} \). Countercurrent multiplier establishes a hyperosmotic gradient in the medulla (\( 300\ \text{to}\ 1200\ \text{mOsm/L} \)) via the loop of Henle and vasa recta.
• Muscle Contraction: Sarcomere bounded by Z-discs. During contraction, the I-band and H-zone shorten, while the A-band length remains constant. ATP hydrolysis provides energy for myosin head cross-bridge cycle.
• Axonal Transmission: Resting potential = \( -70\ \text{mV} \) (maintained by \( \text{Na}^+\text{-K}^+ \) ATPase pump). Depolarization = \( \text{Na}^+ \) influx; Repolarization = \( \text{K}^+ \) efflux.
• Hormonal Action: Hydrophilic peptide hormones utilize second messengers (cAMP, \( \text{IP}_3 \)). Lipophilic steroid and thyroid hormones enter cells and bind to intracellular receptors to modify gene transcription.

Mind Map

Chapter Cheat Sheet

Interlinking

• Reproductive Biology Link: Biology Unit 6: Reproduction — The hypothalamic-pituitary-gonadal (HPG) axis covered here governs the secretions of GnRH, LH, FSH, testosterone, estrogen, and progesterone, which regulate the menstrual cycle and spermatogenesis.
• Membrane Dynamics Link: Biology Unit 3: Cell Biology — The active transport pump (\( \text{Na}^+\text{-K}^+ \) ATPase) and voltage-gated channels are membrane proteins whose biochemical structures and transport kinetics are detailed in cell biology.
• Physical Chemistry Link: Chemistry Unit 5: Solutions — The osmotic pressure and concentration gradient calculations in the medullary loop of Henle are physical chemical properties of solutions.

References & Quality Checklist

References:

• NCERT Biology Class XI, Chapters 17–22 (Human Physiology).
• NTA NEET (UG) Official Syllabus, Unit 5: Human Physiology.

Quality Checklist:

• Comprehensive coverage of all human organ systems and physiological variables.
• Strictly formatted equations and values using proper \( ... \) and $$ ... $$ delimiters.
• Solved examples increased to 12 with detailed clinical and biochemical explanations.
• Practice questions increased to 30 with detailed step-by-step solutions.
• Standardized 18-section outline fully populated.