25 questions in 25 minutes on the NTA computer-based test interface, all from Breathing and Exchange of Gases. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Breathing and Exchange of Gases, of which 13 are real NEET previous-year questions and 0 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Breathing and Exchange of Gases mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 13 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Breathing and Exchange of Gases?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Zoology
- Q1The chemosensitive area regulating respiration is primarily sensitive to changes in:
- Glucose levels
- O2 only
- N2 levels
- CO2 and H+ ions
Answer: (D) CO2 and H+ ions
The chemosensitive area near the rhythm centre responds strongly to CO2 and H+ ions; a rise stimulates increased breathing to expel CO2. Oxygen plays an insignificant direct role in routine regulation.
- Q2The largest fraction of carbon dioxide is transported in blood as:
- Bicarbonate ions
- Dissolved CO2 in plasma
- Carbonic acid in RBCs
- Carbamino-haemoglobin
Answer: (A) Bicarbonate ions
About 70% of CO2 is carried as bicarbonate (HCO3-), about 20-25% as carbamino-haemoglobin, and about 7% dissolved in plasma.
- Q3A person has TV = 500 mL, IRV = 3000 mL and ERV = 1100 mL. What is the person’s Inspiratory Capacity (IC)?
- 1600 mL
- 3500 mL
- 4600 mL
- 3000 mL
Answer: (B) 3500 mL
Inspiratory Capacity = Tidal Volume + Inspiratory Reserve Volume = 500 + 3000 = 3500 mL. ERV is not part of IC (it belongs to expiratory capacity), so it is not added. IC = 3500 mL.
- Q4Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A person goes to high altitude and experiences “Altitude Sickness” with symptoms like breathing difficulty and heart palpitations. Reason (R) : Due to low atmospheric pressure at high altitude, the body does not get sufficient oxygen. In the light of the above statements, choose the correct answer from the options given below :
- Both (A) and (R) are true but (R) is not the correct explanation of (A).
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- (A) is true but (R) is false
- (A) is false but (R) is true
Answer: (B) Both (A) and (R) are true and (R) is the correct explanation of (A).
C. Tests: linking altitude sickness to the partial pressure argument rather than to a change in air composition. Why C: Statement A is true, altitude sickness with breathlessness and palpitations is the standard description. Statement R is also true, and it is the mechanism. Oxygen stays about 21 percent of air at every altitude, but partial pressure equals that fraction multiplied by the total atmospheric pressure. At high altitude the total pressure falls, so inspired pO₂ falls with it, alveolar pO₂ falls, the gradient driving O₂ into pulmonary blood shrinks and haemoglobin saturation drops. The body responds by raising ventilation and heart rate, which the person feels as breathing difficulty and palpitations. R therefore explains A. Why not A: R is not false, low atmospheric pressure genuinely lowers the oxygen actually available to the blood [misconception A: “the percentage of oxygen is unchanged, so pressure is irrelevant”]. Why not B: A is not false, those are the recognised symptoms of altitude sickness [misconception B: “altitude sickness is only nausea and headache, not respiratory”]. Why not D: R is not an incidental true statement, it is the direct cause of the symptoms in A [misconception D: “two true statements in an assertion-reason pair are usually unconnected”]. Remember: at altitude the oxygen percentage is unchanged, it is the partial pressure that collapses.
- Q5The oxygen-haemoglobin dissociation curve will show a right shift in case of
- Low pCO₂
- High pO₂
- Less H⁺ concentration
- High pCO₂
Answer: (D) High pCO₂
A. Tests: converting each condition into a direction of shift on the dissociation curve. Why A: a right shift means lower affinity, that is, less saturation at the same pO₂. High pCO₂ drives the carbonic anhydrase reaction forward inside the RBC and raises H⁺; those protons stabilise the deoxy form of haemoglobin (Bohr effect), so oxygen is released more readily. This is exactly the tissue capillary condition, where unloading is wanted. Why not B: high pO₂ is the alveolar condition and it favours loading; the curve there sits left, with saturation held at about 97%. Why not C: low pCO₂ means fewer H⁺ and a more alkaline blood, which raises affinity and shifts the curve left. Why not D: less H⁺ is by definition a higher pH, and higher pH shifts the curve left, the mirror image of the Bohr effect. Remember: right shift is the tissue signature, hot, acidic and CO₂ rich, so oxygen gets released.
- Q6Identify the region of human brain which has pneumotaxic centre that alters respiratory rate by reducing the duration of inspiration.
- Pons
- Thalamus
- Medulla
- Cerebrum
Answer: (A) Pons
C. Tests: placing each respiratory control centre in its correct part of the brain. Why C: NCERT states that a specialised centre present in the pons region of the brain, called the pneumotaxic centre, can moderate the function of the respiratory rhythm centre. Its signals shorten the duration of inspiration, which cuts each breath short and therefore raises the respiratory rate. Pons is the location asked for. Why not A: the cerebrum gives voluntary control, letting you hold or hurry a breath, but it houses no pneumotaxic centre [misconception A: “the cerebrum controls everything voluntary and involuntary”]. Why not B: the medulla holds the respiratory rhythm centre and, near it, the chemosensitive area, and the pneumotaxic centre acts on that rhythm centre from outside rather than sitting in it [misconception B: “every respiratory centre is medullary”]. Why not D: the thalamus is a relay station for sensory input to the cortex and has no respiratory centre [misconception D: “any deep brain structure can be a control centre for breathing”]. Remember: pons shortens inspiration, medulla sets the rhythm, cerebrum overrides it voluntarily.
- Q7Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
- Low pCO₂ and High temperature
- High pO₂ and Lesser H⁺ concentration
- High pO₂ and High pCO₂
- Low pCO₂ and High H⁺ concentration
Answer: (B) High pO₂ and Lesser H⁺ concentration
B. Tests: listing the four conditions that push the oxygen dissociation curve towards binding, and spotting the option where both named factors point the same way. Why B: Oxyhaemoglobin formation is favoured by high pO₂, low pCO₂, low H⁺ concentration (that is, higher pH) and low temperature. The alveolus supplies all four. Option B names high pO₂, which drives O₂ onto the haem groups, together with a lesser H⁺ concentration, which removes the Bohr effect and raises haemoglobin’s affinity for O₂. Both factors favour binding, so B is the only internally consistent pair. Why not A: high pO₂ helps, but high pCO₂ is a dissociation factor, since it lowers pH and shifts the curve to the right, so the pair contradicts itself [misconception A: “more of both gases means more binding”]. Why not C: low pCO₂ helps, but a high H⁺ concentration is the Bohr effect working against binding, so again the pair fights itself [misconception C: “H⁺ and CO₂ act on haemoglobin in opposite directions”]. Why not D: low pCO₂ helps, but a high temperature shifts the curve to the right and promotes dissociation, which is what happens in exercising muscle, not in the alveolus [misconception D: “temperature does not affect haemoglobin”]. Remember: alveolus means high pO₂, low pCO₂, low H⁺, low temperature, and every one of them loads oxygen.
- Q8Which one of the following organs in the human body is most affected due to a shortage of oxygen?
- Intestine
- Skin
- Kidney
- Brain
Answer: (D) Brain
Brain (nerve) cells are highly specialised and cannot regenerate; they also have a very high, continuous demand for oxygen and cannot respire anaerobically for long. So an oxygen shortage (hypoxia) damages and kills brain cells first. Intestine, skin and kidney are far more tolerant to brief hypoxia, so they are not the most affected.
- Q9Assertion (A): Normal expiration does not require active muscular contraction. Reason (R): During expiration the diaphragm and external intercostal muscles relax, decreasing thoracic volume.
- A is false but R is true
- A is true but R is false
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A
Answer: (C) Both A and R are true and R is the correct explanation of A
Normal expiration is passive. Relaxation of the diaphragm and external intercostals decreases thoracic volume, raising pulmonary pressure above atmospheric so air is pushed out – R correctly explains A.
- Q10Vital capacity is equal to:
- ERV + RV
- TV + IRV + ERV
- TV + IRV + ERV + RV
- TV + IRV
Answer: (B) TV + IRV + ERV
Vital Capacity = TV + IRV + ERV, i.e., the maximum air a person can exhale after a maximal inhalation. Adding RV gives Total Lung Capacity, not Vital Capacity.
- Q11Which lung parameter CANNOT be measured directly by a simple spirometer?
- Tidal volume
- Residual volume
- Vital capacity
- Inspiratory reserve volume
Answer: (B) Residual volume
Residual volume is the air remaining after a forced expiration and never leaves the lungs, so a spirometer cannot measure it (nor capacities containing it, like FRC and TLC). It is measured indirectly.
- Q12Mark the true statement among the following with reference to normal breathing
- Inspiration and expiration are passive processes
- Inspiration is a passive process where as expiration is active
- Inspiration and expiration are active processes
- Inspiration is a active process where as expiration is passive
Answer: (D) Inspiration is a active process where as expiration is passive
B. Tests: deciding which phase of a quiet breath needs muscle contraction and which runs on elastic recoil. Why B: Inspiration requires the diaphragm and the external intercostals to contract. The diaphragm flattens and increases the vertical space of the thorax while the ribs and sternum are lifted, enlarging thoracic volume, so intrapulmonary pressure falls below atmospheric and air is pulled in. That contraction costs energy, so inspiration is active. In quiet expiration those same muscles simply relax, and the stretched lungs and thoracic wall recoil to their resting size, raising intrapulmonary pressure above atmospheric and pushing air out with no contraction at all, so normal expiration is passive. Why not A: this is the exact reversal, and it would require air to enter the lungs without any pressure gradient being created for it [misconception A: “breathing out feels like the effortful part, so it must be the active one”]. Why not C: quiet expiration recruits no muscle contraction; only forced expiration brings in the internal intercostals and the abdominal muscles, and the question specifies normal breathing [misconception C: “any body movement must be actively powered”]. Why not D: inspiration cannot be passive, since without an actively created sub-atmospheric intrapulmonary pressure there is nothing to draw air in [misconception D: “air flows into the lungs on its own”]. Remember: quiet breathing spends energy going in and gets the way out free from elastic recoil.
- Q13The exchange of gases in the alveoli of the lungs takes place by:
- simple diffusion
- active transport
- osmosis
- passive transport involving carriers
Answer: (A) simple diffusion
Gas exchange in the alveoli occurs purely by SIMPLE DIFFUSION down partial-pressure gradients of O₂ and CO₂ – no energy and no carriers. Osmosis applies to water movement; active transport needs ATP; carrier-mediated transport is not involved. Hence simple diffusion.
- Q14People living at sea level have around 5 million RBC per cubic millimetre of their blood whereas those living at an altitude of 5400 metres have around 8 million. This is because at high altitude
- people eat more nutritive food, therefore more RBCs are formed
- people get pollution-free air to breath and more oxygen is available
- atmospheric O₂ level is less and hence more RBCs are needed to absorb the required amount of O₂ to survive
- there is more UV radiation which enhances RBC production
Answer: (C) atmospheric O₂ level is less and hence more RBCs are needed to absorb the required amount of O₂ to survive
C. Tests: explaining a measured rise in RBC count from the drop in oxygen partial pressure with altitude. Why C: air is still 21% oxygen at 5400 m, but barometric pressure is roughly half the sea level value, so the partial pressure of oxygen in inspired air is roughly halved and haemoglobin loads less oxygen per pass. The resulting tissue hypoxia makes the kidney secrete erythropoietin, bone marrow steps up erythropoiesis, and RBC number rises from about 5 million to about 8 million per cubic millimetre so that total oxygen carrying capacity is restored. Why not A: diet does not set RBC count at these numbers; high altitude diets are generally poorer, and adequate iron only permits erythropoiesis, it does not drive it. Why not B: if more oxygen were available the hypoxic stimulus would vanish and no extra RBCs would be needed; cleanliness of the air is irrelevant to erythropoietin. Why not D: UV exposure is indeed higher at altitude, but UV has no role in erythropoiesis; the signal that raises RBC count is hypoxia sensed by the kidney. Remember: altitude lowers pO₂, the kidney releases erythropoietin, and RBC count climbs.
- Q15Although much CO₂ is carried in blood, yet blood does not become acidic, because:
- it combines with water to form H₂CO₃ which is neutralised by Na₂CO₃
- it is continuously diffused through tissues and is not allowed to accumulate
- blood buffers play an important role in CO₂ transport
- it is absorbed by the leucocytes
Answer: (C) blood buffers play an important role in CO₂ transport
Blood buffers (chiefly the bicarbonate buffer, plus haemoglobin and plasma proteins) neutralise the H⁺ produced when CO₂ forms carbonic acid, keeping blood pH nearly constant despite large CO₂ loads. Leucocytes do not absorb CO₂; option C invents a wrong neutralising reaction; and continuous diffusion (D) does not itself prevent acidity. So buffering is the reason.
- Q16Given TV = 500 mL, IRV = 2500 mL, ERV = 1000 mL and RV = 1200 mL, the Vital Capacity (VC) of the person is:
- 4000 mL
- 5200 mL
- 2200 mL
- 3500 mL
Answer: (A) 4000 mL
Vital Capacity = TV + IRV + ERV = 500 + 2500 + 1000 = 4000 mL. Residual volume (1200 mL) is NOT included in vital capacity (adding it would give the total lung capacity, 5200 mL). So VC = 4000 mL.
- Q17The partial pressures (in mm Hg) of oxygen (O₂) and carbon dioxide (CO₂) at the alveoli (the site of diffusion) are:
- pO₂ = 104 and pCO₂ = 40
- pO₂ = 40 and pCO₂ = 45
- pO₂ = 95 and pCO₂ = 40
- pO₂ = 159 and pCO₂ = 0.3
Answer: (A) pO₂ = 104 and pCO₂ = 40
At the alveoli pO₂ ≈ 104 mm Hg and pCO₂ ≈ 40 mm Hg. Values 40/45 are for deoxygenated blood/tissues, 95/40 for oxygenated blood, and 159/0.3 for atmospheric air. So the alveolar diffusion-site values are pO₂ = 104, pCO₂ = 40.
- Q18Listed below are four respiratory capacities (i to iv) and four jumbled respiratory volumes of a normal human adult. Respiratory capacities: (i) Residual volume, (ii) Vital capacity, (iii) Inspiratory reserve volume, (iv) Inspiratory capacity. Respiratory volumes (jumbled): 2500 mL, 3500 mL, 1200 mL, 4500 mL. Which one of the following is the correct matching of two capacities and volumes?
- (ii) 2500 mL, (iii) 4500 mL
- (i) 4500 mL, (ii) 3500 mL
- (iii) 1200 mL, (iv) 2500 mL
- (iv) 3500 mL, (i) 1200 mL
Answer: (D) (iv) 3500 mL, (i) 1200 mL
C. Tests: recalling standard adult lung values and checking a pairing against the additive definitions. Why C: place each number by what the label can physically be. 1200 mL is far too small for any capacity and is the standard residual volume of 1100 to 1200 mL, so (i) = 1200 mL. 4500 mL is the largest and is vital capacity, VC = TV + IRV + ERV = 500 + 3000 + 1000 = 4500 mL, so (ii) = 4500 mL. That leaves 2500 and 3500 for IRV and IC, and since IC = TV + IRV the inspiratory capacity must be the bigger of the pair, so (iv) IC = 3500 mL and (iii) IRV = 2500 mL. The pairing option C states, (iv) 3500 mL with (i) 1200 mL, is therefore the correct one. The book quotes IRV as a range of 2500 to 3000 mL, which is why a rounded 2500 and a rounded 3500 can sit in the same list. Why not A: it sets VC = 2500 mL and IRV = 4500 mL. Since VC = IRV + TV + ERV, that IRV alone would force VC to be at least 4500 + 500 + 1000 = 6000 mL, contradicting the 2500 mL it assigns to VC. Why not B: it sets IRV = 1200 mL and IC = 2500 mL, which forces TV = IC – IRV = 2500 – 1200 = 1300 mL, but tidal volume is about 500 mL. The 1200 mL figure is the residual volume, not IRV. Why not D: it sets RV = 4500 mL and VC = 3500 mL, so residual volume would exceed vital capacity, and total lung capacity would come to 3500 + 4500 = 8000 mL against the real adult figure of about 5700 to 6000 mL. Remember: a capacity is always a sum of volumes, so IC must exceed IRV and VC must exceed IC; the 1100 to 1200 mL slice is always residual volume.
- Q19Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively?
- Inflammation of bronchioles; Decreased respiratory surface
- Increased number of bronchioles; Increased respiratory surface
- Increased respiratory surface; Inflammation of bronchioles
- Decreased respiratory surface; Inflammation of bronchioles
Answer: (A) Inflammation of bronchioles; Decreased respiratory surface
Asthma is inflammation of the bronchi and bronchioles (with wheezing and breathing difficulty). Emphysema involves destruction of alveolar walls, so the respiratory (gas-exchange) surface is decreased. Therefore asthma → inflammation of bronchioles, emphysema → decreased respiratory surface, which is option C.
- Q20Carbon dioxide is transported from tissues to the respiratory surface by only:
- erythrocytes and leucocytes
- plasma
- plasma and erythrocytes
- erythrocytes
Answer: (C) plasma and erythrocytes
CO₂ travels from tissues to the lungs in the plasma (as dissolved gas and as bicarbonate) and in the erythrocytes (as carbaminohaemoglobin and where bicarbonate is generated by carbonic anhydrase). Leucocytes have no role, so D is wrong, and plasma or erythrocytes alone (B, C) is incomplete. Both plasma and erythrocytes carry it.
- Q21Which two of the following changes (1-4) usually tend to occur in plain dwellers when they move to high altitudes (3,500 m or more)? 1. Increase in red blood cell size 2. Increase in red blood cell production 3. Increased breathing rate 4. Increase in thrombocyte count
- 2 and 3
- 3 and 4
- 1 and 2
- 1 and 4
Answer: (C) 1 and 2
At high altitude the low pO₂ (hypoxia) stimulates the kidney to release erythropoietin, increasing RBC production (2); initially the new RBCs are larger, so RBC size also increases (1). The book keys 1 and 2. (Increased breathing rate also occurs physiologically, but among the paired options the keyed answer is 1 and 2.) Thrombocyte count is not characteristically raised.
- Q22Select the correct statement about the mechanism of breathing.
- Expiration occurs due to contraction of external intercostal muscles
- Expiration is initiated due to contraction of the diaphragm
- Inspiration occurs when atmospheric pressure is less than intrapulmonary pressure
- Intrapulmonary pressure is lower than the atmospheric pressure during inspiration
Answer: (D) Intrapulmonary pressure is lower than the atmospheric pressure during inspiration
During inspiration the thorax enlarges, so intra-pulmonary (intrapulmonary) pressure falls BELOW atmospheric, and air flows in – B is correct. Expiration is caused by RELAXATION (not contraction) of the diaphragm and external intercostals, so A and D are wrong. Inspiration occurs when atmospheric pressure is GREATER than intrapulmonary pressure, making C the reverse of the truth.
- Q23A person breathes in some volume of air by forced inspiration after having a forced expiration. This quantity of air taken in is
- Tidal volume
- Total lung capacity
- Vital capacity
- Inspiratory capacity
Answer: (C) Vital capacity
C. Tests: reading a described manoeuvre and naming the capacity it measures, by tracking the start point and the end point. Why C: The manoeuvre begins after a forced expiration, so the lungs are already emptied down to the residual volume. It then ends at a forced inspiration, so the lungs are filled to total lung capacity. The air moved in is therefore TLC minus RV, and that difference is vital capacity, equal to IRV + TV + ERV. Why not A: total lung capacity includes the residual volume, which was still in the lungs at the start of the manoeuvre and so was never breathed in [misconception A: “filling the lungs completely means you inhaled the total lung capacity”]. Why not B: tidal volume is the roughly 500 ml of a single quiet breath, whereas both ends of this manoeuvre are forced [misconception B: “any single breath is a tidal volume”]. Why not D: inspiratory capacity is measured from the end of a normal expiration, that is from FRC, and equals TV + IRV, so it starts higher than this manoeuvre does and is smaller than the answer by exactly ERV [misconception D: “any maximal inspiration measures inspiratory capacity”]. Remember: name the start and the end, RV to TLC is vital capacity, FRC to TLC is inspiratory capacity.
- Q24Which structure in the human respiratory system is commonly called the ‘sound box’?
- Trachea
- Pharynx
- Epiglottis
- Larynx
Answer: (D) Larynx
The larynx houses the vocal cords and is therefore called the sound box; air passing through it during exhalation produces sound. The epiglottis is a cartilaginous flap that prevents food from entering the larynx.
- Q25Functional Residual Capacity (FRC) is the volume of air remaining in the lungs after a normal expiration. It equals:
- TV + ERV
- TV + IRV
- ERV + RV
- IRV + RV
Answer: (C) ERV + RV
FRC = ERV + RV. After a normal (tidal) expiration, the air still left in the lungs is the expiratory reserve volume plus the residual volume.