NEET 2024 Zoology Question Paper with Answers and Solutions PDF R5

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Sahaj Anand

Content Writer | Journalism Graduate | Updated 3+ months ago

NEET 2024 Zoology Question Paper with Solutions PDF R5 is available for download. NEET 2024 R5 Zoology Question Paper comprises 50 MCQs out of which only 45 are to be attempted. NEET 2024 question R5 Zoology is divided into 2 sections- A (35 questions) and B (15 questions). You can download NEET 2024 zoology question paper with answer key and solutions PDF for R5 using the links given below.

NEET 2024 Zoology Question Paper with Solutions PDF R5

NEET 2024 Zoology Question Paper R5 Download PDF Check Solution
NEET 2024 Zoology Question Paper with Answers and Solutions PDF R5


Question 1:

Given below are two statements: one is labelled as Assertion A and the other as Reason R:
Assertion A: FSH acts upon ovarian follicles in females and Leydig cells in males.
Reason R: Growing ovarian follicles secrete estrogen in females while interstitial cells secrete androgen in male human beings.

In the light of the above statements, choose the correct answer from the options given below:

(1) Both A and R are true but R is NOT the correct explanation of A
(2) A is true but R is false
(3) A is false but R is true
(4) Both A and R are true and R is the correct explanation of A

Correct Answer: (3) A is false but R is true
View Solution

Step 1: Understanding FSH Action


- FSH (Follicle-Stimulating Hormone) acts on ovarian follicles in females to stimulate their growth and maturation.


- However, in males, FSH stimulates the Sertoli cells (not Leydig cells) to support spermatogenesis.



Step 2: Understanding Estrogen and Androgen Secretion


- Ovarian follicles secrete estrogen, which is correct.


- Leydig cells secrete testosterone (androgen) in males, which is also correct.


Conclusion: Since A states that FSH acts on Leydig cells (which is incorrect), the correct answer is \( \mathbf{(3)} \). Quick Tip: \textbf{FSH acts on Sertoli cells, while LH (Luteinizing Hormone) acts on Leydig cells to stimulate testosterone production.}


Question 2:

Match List I with List II

Q2
(1) A-III, B-II, C-I, D-IV
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-I, C-III, D-II
(4) A-IV, B-II, C-III, D-I

Correct Answer: (2) A-II, B-IV, C-I, D-III
View Solution




Understanding the Enzyme Functions :


- Lipase hydrolyzes fats, breaking down ester bonds in lipids.


- Nuclease breaks phosphodiester bonds in nucleic acids.


- Protease hydrolyzes peptide bonds in proteins.


- Amylase breaks glycosidic bonds in carbohydrates.


Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Enzymes catalyze specific biochemical reactions by targeting specific bonds in macromolecules.}


Question 3:

Following are the stages of the pathway for conduction of an action potential through the heart:

A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node

Choose the correct sequence of the pathway from the options given below:

(1) A-E-C-B-D
(2) B-D-E-C-A
(3) E-A-D-B-C
(4) E-C-A-D-B

Correct Answer: (4) E-C-A-D-B
View Solution

Cardiac Conduction Pathway


SA node (Sinoatrial node): Acts as the pacemaker of the heart, initiating the heartbeat.
AV node (Atrioventricular node): Delays the electrical impulse before transmitting it to the ventricles.
AV bundle (Bundle of His): Conducts impulses from the atria to the ventricles.
Bundle branches: Transmit the impulse to both ventricles.
Purkinje fibers: Distribute the impulse throughout the ventricles, leading to ventricular contraction.


Conclusion:

The correct sequence is \( E \rightarrow C \rightarrow A \rightarrow D \rightarrow B \), corresponding to option \( \mathbf{(4)} \). Quick Tip: \textbf{The SA node is the natural pacemaker of the heart, setting the rhythm of cardiac contractions.}


Question 4:

Match List I with List II:

(1) A-III, B-IV, C-II, D-I
(2) A-I, B-III, C-II, D-IV
(3) A-II, B-I, C-III, D-IV
(4) A-II, B-III, C-I, D-IV

Correct Answer: (1) A-III, B-IV, C-II, D-I
View Solution

Understanding Brain Regions


- Pons connects different brain regions, facilitating communication.


- Hypothalamus contains neurosecretory cells regulating homeostasis.


- Medulla oblongata controls involuntary functions such as respiration and digestion.


- Cerebellum coordinates posture, balance, and fine motor movements.


Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{The cerebellum is essential for motor coordination, while the medulla controls vital autonomic functions.}


Question 5:

Which one of the following factors will not affect the Hardy-Weinberg equilibrium?

(1) Genetic drift
(2) Gene migration
(3) Constant gene pool
(4) Genetic recombination

Correct Answer: (3) Constant gene pool
View Solution

Understanding Hardy-Weinberg Equilibrium


- Hardy-Weinberg equilibrium states that allele frequencies in a population remain constant unless disturbed by evolutionary forces.


- Factors affecting equilibrium include genetic drift, gene migration, mutation, selection, and recombination.



Conclusion: A constant gene pool maintains equilibrium, so the correct option is \( \mathbf{(3)} \). Quick Tip: \textbf{For a population to be in Hardy-Weinberg equilibrium, it must have random mating, large population size, and no mutation, selection, or migration.}


Question 6:

In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on:

(1) 10th segment
(2) 8th and 9th segment
(3) 11th segment
(4) 5th segment

Correct Answer: (1) 10th segment
View Solution

Understanding Cockroach Anatomy


- Anal cerci are sensory appendages present on the 10th segment in both male and female cockroaches.


- They detect vibrations and help in reflexive responses to threats.



Conclusion: The correct answer is \( \mathbf{(1)} \). Quick Tip: \textbf{Anal cerci are sensitive to vibrations and function as sensory organs in cockroaches.}


Question 7:

Match List I with List II:

Q7

Choose the correct answer from the options given below:

(1) A-III, B-II, C-IV, D-I
(2) A-II, B-I, C-IV, D-III
(3) A-I, B-III, C-II, D-IV
(4) A-II, B-IV, C-I, D-III

Correct Answer: (4) A-II, B-IV, C-I, D-III
View Solution

Step 1: Understanding Lung Capacities and Their Components

- Expiratory capacity (A-II): The total volume of air that can be exhaled after normal inspiration.
\[ Expiratory capacity = Tidal volume + Expiratory reserve volume \]

- Functional residual capacity (B-IV): The volume of air remaining in the lungs after normal expiration.
\[ Functional residual capacity = Expiratory reserve volume + Residual volume \]

- Vital capacity (C-I): The maximum volume of air a person can exhale after a deep inspiration.
\[ Vital capacity = Expiratory reserve volume + Tidal volume + Inspiratory reserve volume \]

- Inspiratory capacity (D-III): The maximum volume of air that can be inhaled after a normal expiration.
\[ Inspiratory capacity = Tidal volume + Inspiratory reserve volume \]

Step 2: Matching the correct pairs \[ \begin{aligned} A & \rightarrow II \quad (Tidal volume + Expiratory reserve volume)
B & \rightarrow IV \quad (Expiratory reserve volume + Residual volume)
C & \rightarrow I \quad (Expiratory reserve volume + Tidal volume + Inspiratory reserve volume)
D & \rightarrow III \quad (Tidal volume + Inspiratory reserve volume) \end{aligned} \]

Conclusion: The correct answer is (4) A-II, B-IV, C-I, D-III. Quick Tip: Lung capacities are combinations of different lung volumes. Expiratory capacity is the sum of tidal volume and expiratory reserve volume, while functional residual capacity includes the residual volume. Vital capacity is the total air exhaled after deep inspiration, and inspiratory capacity is the sum of tidal volume and inspiratory reserve volume.


Question 8:

The flippers of Penguins and Dolphins are an example of:

(1) Natural selection
(2) Convergent evolution
(3) Divergent evolution
(4) Adaptive radiation

Correct Answer: (2) Convergent evolution
View Solution

Understanding Evolutionary Trends


- Convergent evolution occurs when unrelated species evolve similar traits due to adaptation to similar environments.


- Flippers of Penguins (birds) and Dolphins (mammals) evolved independently for swimming, despite their different ancestral origins.



Conclusion: Since Penguins and Dolphins evolved similar features independently, the correct answer is \( \mathbf{(2)} \).
Quick Tip: \textbf{Convergent evolution results in analogous structures, while divergent evolution results in homologous structures.}


Question 9:

Match List I with List II:
Q9


Choose the correct answer from the options given below:
(1) A-III, B-I, C-II, D-IV
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-IV, C-I, D-III
(4) A-II, B-I, C-IV, D-III

Correct Answer: (2) A-III, B-IV, C-I, D-II
View Solution

Step 1: Understanding the Functions of the Given Terms


- \( \alpha \)-1 antitrypsin (A-III): It is a protein that protects lung tissue from damage and its deficiency leads to emphysema.


- Cry IAb (B-IV): This gene from Bacillus thuringiensis (Bt) produces a toxin effective against the corn borer.


- Cry IAc (C-I): This Bt toxin specifically targets cotton bollworm.


- Enzyme replacement therapy (D-II): This therapy is used to treat ADA deficiency, a disorder affecting the immune system.



Step 2: Matching the Correct Pairs \[ \begin{aligned} A & \rightarrow III \quad (Emphysema)
B & \rightarrow IV \quad (Corn borer)
C & \rightarrow I \quad (Cotton bollworm)
D & \rightarrow II \quad (ADA deficiency) \end{aligned} \]


Conclusion: The correct answer is (2) A-III, B-IV, C-I, D-II. Quick Tip: \textbf{\( \alpha \)-1 antitrypsin} prevents lung damage; \textbf{Cry IAb} targets \textbf{corn borer}, and \textbf{Cry IAc} targets \textbf{cotton bollworm}. \textbf{Enzyme replacement therapy} is a treatment for \textbf{ADA deficiency}.


Question 10:

Match List I with List II:

(1) A-II, B-III, C-IV, D-I
(2) A-I, B-IV, C-III, D-II
(3) A-I, B-II, C-IV, D-III
(4) A-IV, B-II, C-III, D-I

Correct Answer: (2) A-I, B-IV, C-III, D-II
View Solution

Understanding Chromosomal Disorders


- Down’s syndrome results from trisomy of the 21st chromosome.


- \(\alpha\)-Thalassemia is associated with deletions in the 16th chromosome, affecting hemoglobin production.


- \(\beta\)-Thalassemia is linked to mutations in the X chromosome, causing anemia.


- Klinefelter’s syndrome (XXY condition) occurs due to an extra X chromosome in males.


Conclusion: The correct match is \( \mathbf{(2)} \). Quick Tip: \textbf{Genetic disorders can be caused by trisomy (e.g., Down’s syndrome) or chromosomal mutations affecting specific gene loci.}


Question 11:

Given below are two statements:

Statement I: The presence or absence of hymen is not a reliable indicator of virginity.

Statement II: The hymen is torn during the first coitus only.

In the light of the above statements, choose the correct answer from the options given below:

(1) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true

Correct Answer: (2) Statement I is true but Statement II is false
View Solution

Understanding the Role of Hymen


- The hymen can be torn due to several activities, including sports, cycling, or medical examinations, apart from sexual intercourse.


- Thus, it is not a reliable indicator of virginity.


Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Virginity is a social concept and cannot be medically determined by the presence or absence of the hymen.}


Question 12:

Match List I with List II:

Q12
(1) A-I, B-III, C-II, D-IV
(2) A-III, B-I, C-II, D-IV
(3) A-IV, B-II, C-III, D-I
(4) A-II, B-IV, C-III, D-I

Correct Answer: (2) A-III, B-I, C-II, D-IV
View Solution

Understanding Disease and Diagnostic Associations


- Common cold is caused by rhinoviruses.


- Haemozoin is a byproduct of Plasmodium, the parasite responsible for malaria.


- Widal test is used for diagnosing typhoid fever.


- Allergies can be triggered by dust mites and other allergens.


Conclusion: The correct match is \( \mathbf{(2)} \). Quick Tip: \textbf{Plasmodium causes malaria, and its breakdown of hemoglobin results in haemozoin, which causes fever recurrence.}


Question 13:

Which of the following is not a component of the Fallopian tube?

(1) Isthmus
(2) Infundibulum
(3) Ampulla
(4) Uterine fundus

Correct Answer: (4) Uterine fundus
View Solution

Structure of the Fallopian Tube


The Fallopian tube consists of four distinct regions:

Infundibulum: A funnel-shaped opening near the ovary.
Ampulla: The widest section where fertilization occurs.
Isthmus: A narrow segment connecting to the uterus.
Interstitial (Intramural) part: The portion passing through the uterine wall.

The Uterine Fundus is not a part of the Fallopian tube; instead, it is the topmost portion of the uterus.


Conclusion:

The correct answer is \( \mathbf{(4)} \). Quick Tip: \textbf{Fertilization occurs in the ampulla of the Fallopian tube, making it a crucial structure in reproduction.}


Question 14:

Following are the stages of cell division:

A. Gap 2 phase

B. Cytokinesis

C. Synthesis phase

D. Karyokinesis

E. Gap 1 phase
Choose the correct sequence of stages from the options given below:

(1) E-B-D-A-C
(2) B-D-E-A-C
(3) E-C-A-D-B
(4) C-E-D-A-B

Correct Answer: (3) E-C-A-D-B
View Solution

Understanding Cell Cycle Phases


- Gap 1 phase (G1): The first growth phase, preparing for DNA replication.


- Synthesis phase (S): DNA replication occurs.


- Gap 2 phase (G2): Preparation for mitosis.


- Karyokinesis: Nuclear division (mitosis).


- Cytokinesis: Division of cytoplasm leading to two daughter cells.



Conclusion: The correct sequence is \( \mathbf{E \to C \to A \to D \to B} \). Quick Tip: \textbf{The cell cycle consists of interphase (G1, S, G2) followed by mitotic phase (M), which includes karyokinesis and cytokinesis.}


Question 15:

Match List I with List II:

Q15
(1) A-IV, B-II, C-III, D-I
(2) A-II, B-IV, C-I, D-III
(3) A-II, B-I, C-IV, D-III
(4) A-IV, B-III, C-II, D-I

Correct Answer: (3) A-II, B-I, C-IV, D-III
View Solution

Understanding the Cellular Structures


- Axoneme is the structural core of cilia and flagella.


- Cartwheel pattern is seen in centrioles during cell division.


- Crista are folds in the inner membrane of mitochondria.


- Satellite regions are part of chromosomes involved in rRNA synthesis.


Conclusion: The correct match is \( \mathbf{(3)} \). Quick Tip: \textbf{Cilia and flagella are supported by axonemes, while mitochondria have crista to increase surface area for ATP synthesis.}


Question 16:

Given below are two statements:

Assertion A: Breast-feeding during the initial period of infant growth is recommended by doctors for bringing up a healthy baby.

Reason R: Colostrum contains several antibodies absolutely essential to develop resistance for the newborn baby.

In the light of the above statements, choose the most appropriate answer from the options given below:

(1) Both A and R are correct but R is NOT the correct explanation of A
(2) A is correct but R is not correct
(3) A is not correct but R is correct
(4) Both A and R are correct and R is the correct explanation of A

Correct Answer: (4) Both A and R are correct and R is the correct explanation of A
View Solution

Understanding the Role of Breastfeeding


- Colostrum, the first milk produced by the mother, is rich in immunoglobulins (IgA) that help protect the newborn.


- Breastfeeding also provides essential nutrients for growth and brain development.



Conclusion: Since colostrum helps in immunity development, it supports the assertion. Hence, \( \mathbf{(4)} \) is correct. Quick Tip: \textbf{Colostrum is crucial for newborn immunity, providing passive immunity against infections.}


Question 17:

Match List I with List II:



Choose the correct answer from the options given below:

(1) A-I, B-II, C-IV, D-III
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-III, C-II, D-I
(4) A-IV, B-II, C-III, D-I

Correct Answer: (2) A-II, B-IV, C-I, D-III
View Solution

Understanding the Sub Phases of Prophase I


- Leptotene: Chromosomes appear as thin threads.


- Zygotene: Synaptonemal complex formation occurs.


- Pachytene: Recombination nodules appear, crossing over occurs.


- Diakinesis: Terminalisation of chiasmata completes, preparing for metaphase.


Conclusion: The correct match is \( \mathbf{(2)} \). Quick Tip: \textbf{Prophase I is the longest meiotic phase, crucial for genetic variation due to recombination.}


Question 18:

Match List I with List II:

Q18
Choose the correct answer from the options given below:

(1) A-I, B-III, C-II, D-IV
(2) A-II, B-III, C-I, D-IV
(3) A-III, B-I, C-IV, D-II
(4) A-IV, B-II, C-III, D-I

Correct Answer: (3) A-III, B-I, C-IV, D-II
View Solution

Understanding Types of Joints


- Fibrous Joints: Found in the skull, with no movement.


- Cartilaginous Joints: Found between vertebrae, allowing limited movement.


- Hinge Joints: Present in knees, allowing movement in one plane.


- Ball and Socket Joints: Shoulder and hip joints allow multi-directional movement.


Conclusion: The correct match is \( \mathbf{(3)} \). Quick Tip: \textbf{Joints allow mobility while maintaining structural integrity; different types provide different movement ranges.}


Question 19:

Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?

(1) High pO2 and Lesser H+ concentration
(2) Low pCO2 and High H+ concentration
(3) Low pCO2 and High temperature
(4) High pO2 and High pCO2

Correct Answer: (1) High pO2 and Lesser H+ concentration
View Solution

Conditions for Oxyhaemoglobin Formation


- High pO2 promotes oxygen binding with haemoglobin.


- Low H+ concentration prevents haemoglobin from releasing oxygen.


- Low pCO2 prevents the Bohr effect, which enhances oxygen loading.


Conclusion: \( \mathbf{(1)} \) is correct. Quick Tip: \textbf{Oxygen binds to haemoglobin in high pO2 environments (lungs) and is released in low pO2 tissues.}


Question 20:

Which of the following is not a natural/traditional contraceptive method?

(1) Periodic abstinence
(2) Lactational amenorrhea
(3) Vaults
(4) Coitus interruptus

Correct Answer: (3) Vaults
View Solution

Understanding Contraceptive Methods


- Periodic abstinence: Avoiding intercourse during the fertile window.


- Lactational amenorrhea: Temporary infertility during breastfeeding.


- Coitus interruptus: Withdrawal before ejaculation.


- Vaults: Barrier method, not natural.


Conclusion: \( \mathbf{(3)} \) is correct. Quick Tip: \textbf{Natural contraceptive methods rely on physiological changes and awareness, while barrier methods like vaults are artificial.}


Question 21:

Which of the following are Autoimmune disorders?

A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)

Choose the most appropriate answer from the options given below:

(1) A, B E only
(2) B, C E only
(3) C, D E only
(4) A, B D only

Correct Answer: (1) A, B & E only
View Solution

Understanding Autoimmune Disorders


- Myasthenia gravis: An autoimmune disorder affecting neuromuscular function.


- Rheumatoid arthritis: An autoimmune disease causing joint inflammation.


- Systemic Lupus Erythematosus (SLE): A systemic autoimmune disease.


- Gout: A metabolic disorder, not autoimmune.


- Muscular dystrophy: A genetic disorder, not autoimmune.



Conclusion: The correct answer is \( \mathbf{(1)} \). Quick Tip: \textbf{Autoimmune disorders occur when the immune system mistakenly attacks the body's own cells.}


Question 22:

The “Ti plasmid” of Agrobacterium tumefaciens stands for

(1) Tumor independent plasmid
(2) Tumor inducing plasmid
(3) Temperature independent plasmid
(4) Tumour inhibiting plasmid

Correct Answer: (2) Tumor inducing plasmid
View Solution

Understanding Ti Plasmid


- Agrobacterium tumefaciens is a bacterium that causes crown gall disease in plants.


- The Ti plasmid (Tumor-inducing plasmid) carries genes responsible for transforming plant cells.


- It is widely used in genetic engineering to transfer foreign genes into plants.



Conclusion: The correct answer is \( \mathbf{(2)} \). Quick Tip: \textbf{Ti plasmid is an essential tool in plant biotechnology, enabling genetic modifications in crops.}


Question 23:

Which one is the correct product of DNA-dependent RNA polymerase to the given template?
3'-TACATGGCAAATATCCATTCA-5'

(1) 5’-AUGUAAAGUUUAUAGGUAAGU-3’
(2) 5’-AUGUACCGUUUAUAGGGAAGU-3’
(3) 5’-ATGTACCGTTTATAGGTAAGT-3’
(4) 5’-AUGUACCGUUUAUAGGUAAGU-3’

Correct Answer: (4) 5'-AUGUACCGUUUAUAGGUAAGU-3'
View Solution

Step 1: Understanding Transcription


- The given DNA template is 3' to 5', and RNA polymerase synthesizes mRNA in the 5' to 3' direction.


- Base pairing rules for transcription:

- A (Adenine) → U (Uracil)

- T (Thymine) → A (Adenine)

- C (Cytosine) → G (Guanine)

- G (Guanine) → C (Cytosine)



Step 2: Complementary mRNA Formation \[ DNA template: 3'-TACATGGCAAATATCCATTCA-5' \] \[ mRNA strand: 5'-AUGUACCGUUUAUAGGUAAGU-3' \]



Conclusion: The correct answer is \( \mathbf{(4)} \). Quick Tip: \textbf{Transcription produces mRNA, which is complementary to the DNA template and replaces thymine (T) with uracil (U).}


Question 24:

Which of the following statements is incorrect?

(1) Most commonly used bio-reactors are of stirring type
(2) Bio-reactors are used to produce small scale bacterial cultures
(3) Bio-reactors have an agitator system, an oxygen delivery system and foam control system
(4) A bio-reactor provides optimal growth conditions for achieving the desired product

Correct Answer: (2) Bio-reactors are used to produce small scale bacterial cultures
View Solution

Understanding Bio-reactors


- Bio-reactors are used for large-scale production of microbial cultures.


- Small-scale cultures are grown in flasks or test tubes rather than in bio-reactors.


- Bio-reactors provide controlled temperature, pH, aeration, and mixing to maximize product yield.



Conclusion: The correct answer is \( \mathbf{(2)} \). Quick Tip: \textbf{Bio-reactors are essential in industrial biotechnology for large-scale microbial culture and fermentation processes.}


Question 25:

Match List I with List II
Q25

(1) A-I, B-III, C-II, D-IV
(2) A-II, B-I, C-III, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-IV, B-III, C-I, D-II

Correct Answer: (3) A-III, B-IV, C-I, D-II
View Solution

- Cocaine is derived from Erythroxylum plant.


- Heroin is synthesized from morphine, which is obtained from Papaver somniferum.


- Morphine is an effective sedative used in surgery.


- Marijuana comes from Cannabis sativa.



Conclusion: The correct answer is \( \mathbf{(3)} \). Quick Tip: \textbf{Drugs derived from plants have significant medicinal and recreational impacts.}


Question 26:

Which of the following is not a steroid hormone?

(1) Testosterone
(2) Progesterone
(3) Glucagon
(4) Cortisol

Correct Answer: (3) Glucagon
View Solution

- Steroid hormones include testosterone, progesterone, and cortisol.


- Glucagon is a peptide hormone, not a steroid hormone.


- It is produced by pancreatic alpha cells and regulates blood glucose.



Conclusion: The correct answer is \( \mathbf{(3)} \). Quick Tip: \textbf{Steroid hormones are derived from cholesterol and regulate metabolism, inflammation, and reproductive functions.}


Question 27:

Match List I with List II
Q26

(1) A-I, B-III, C-IV, D-II
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-I, C-IV, D-II
(4) A-III, B-I, C-II, D-IV

Correct Answer: (3) A-III, B-I, C-IV, D-II
View Solution

- Non-medicated IUD: Lippes loop.


- Copper releasing IUD: Multiload 375.


- Hormone releasing IUD: LNG-20.


- Implants: Progestogens.



Conclusion: The correct answer is \( \mathbf{(3)} \). Quick Tip: \textbf{IUDs (Intrauterine Devices) are effective birth control methods with varying mechanisms.}


Question 28:

Given below are two statements:
Statement I: In the nephron, the descending limb of the loop of Henle is impermeable to water and permeable to electrolytes.

Statement II: The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.

(1) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true

Correct Answer: (1) Both Statement I and Statement II are false
View Solution

Understanding Nephron Structure and Function


The descending limb of the loop of Henle is permeable to water but partially permeable to electrolytes, making the given statement incorrect.
The proximal convoluted tubule (PCT) is lined with simple cuboidal brush border epithelium, not columnar epithelium, which makes the given statement incorrect.


Conclusion:

The correct answer is \( \mathbf{(1)} \). Quick Tip: \textbf{Nephrons regulate water and electrolyte balance through selective permeability in different segments.}


Question 29:

Given below are some stages of human evolution. Arrange them in correct sequence (Past to Recent).

A. Homo habilis
B. Homo sapiens
C. Homo neanderthalensis
D. Homo erectus

Choose the correct sequence of human evolution from the options given below:

(1) B-A-D-C
(2) C-B-D-A
(3) A-D-C-B
(4) D-A-C-B

Correct Answer: (3) A-D-C-B
View Solution

- Homo habilis (Earliest tool-user, appeared ~2.1 million years ago).


- Homo erectus (More advanced, first to use fire).


- Homo neanderthalensis (Coexisted with early Homo sapiens, adapted to cold).


- Homo sapiens (Modern humans, dominant species today).


Conclusion: The correct sequence is A-D-C-B. Quick Tip: \textbf{Human evolution involved gradual brain development, tool usage, and social complexity.}


Question 30:

Three types of muscles are given as a, b, and c. Identify the correct matching pair along with their location in the human body:


(1) (a) Skeletal - Triceps,
(b) Smooth - Stomach,
(c) Cardiac - Heart
(2) (a) Skeletal - Biceps,
(b) Involuntary - Intestine,
(c) Smooth - Heart
(3) (a) Involuntary - Nose tip,
(b) Skeletal - Bone,
(c) Cardiac - Heart
(4) (a) Smooth - Toes,
(b) Skeletal - Legs,
(c) Cardiac - Heart

Correct Answer: (1) Skeletal - Triceps, Smooth - Stomach, Cardiac - Heart
View Solution

- Skeletal muscles: Voluntary, attached to bones (e.g., triceps, biceps).


- Smooth muscles: Involuntary, found in digestive organs (e.g., stomach, intestine).


- Cardiac muscles: Specialized involuntary muscles in the heart.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{Muscle types differ in function: Skeletal (voluntary), Smooth (involuntary), and Cardiac (automatic rhythmic contractions).}


Question 31:

Match List I with List II



Choose the correct answer from the options given below :

(1) A-II, B-I, C-IV, D-III
(2) A-II, B-IV, C-I, D-III
(3) A-IV, B-III, C-II, D-I
(4) A-IV, B-II, C-III, D-I

Correct Answer: (1) A-II, B-I, C-IV, D-III
View Solution



- Pleurobrachia: Ctenophora, known as comb jellies.


- Radula: Mollusca, rasping organ used for feeding.


- Stomochord: Hemichordata, structure similar to notochord.


- Air bladder: Osteichthyes (bony fish), used for buoyancy.


Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{Animal classification is based on body structure, function, and evolutionary lineage.}


Question 32:

Match List I with List II
Q32

Choose the correct answer from the options given below :

(1) A-III, B-I, C-II, D-IV
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-II, C-I, D-IV
(4) A-II, B-I, C-III, D-IV

Correct Answer: (1) A-III, B-I, C-II, D-IV
View Solution

- Pterophyllum (Angel fish) is a popular aquarium fish.


- Myxine (Hag fish) is a jawless fish known for its slime production.


- Pristis (Saw fish) has a long, saw-like rostrum.


- Exocoetus (Flying fish) can glide above water using wing-like fins.


Conclusion: The correct option is (1). Quick Tip: \textbf{Different fish species have unique adaptations for survival, such as gliding, slime production, and specialized feeding structures.}


Question 33:

Match List I with List II



Choose the correct answer from the options given below:

(1) A-IV, B-III, C-I, D-II
(2) A-III, B-I, C-IV, D-II
(3) A-II, B-IV, C-III, D-I
(4) A-I, B-III, C-II, D-IV

Correct Answer: (1) A-IV, B-III, C-I, D-II
View Solution



- Typhoid is caused by the bacterium Salmonella typhi.


- Leishmaniasis is caused by the protozoan \textit{Leishmania.


- Ringworm is a fungal infection affecting the skin.


- Filariasis is caused by a nematode (Wuchereria bancrofti).


Conclusion: The correct option is (1). Quick Tip: \textbf{Different types of pathogens (bacteria, protozoa, fungi, and nematodes) cause various infectious diseases in humans.


Question 34:

The following diagram shows restriction sites in E. coli cloning vector pBR322. Find the role of ‘X’ and ‘Y’ genes:


(1) The gene ‘X’ controls the copy number of the linked DNA, and ‘Y’ codes for a protein
26
involved in plasmid replication.
(2) The gene ‘X’ codes for a protein involved in plasmid replication, and ‘Y’ provides
antibiotic resistance.
(3) Gene ‘X’ is responsible for recognition sites, and ‘Y’ is responsible for antibiotic
resistance.
(4) The gene ‘X’ provides antibiotic resistance, and ‘Y’ codes for a protein involved in
plasmid replication.

Correct Answer: (1) The gene ‘X’ controls the copy number of the linked DNA, and ‘Y’ codes for a protein involved in plasmid replication.
View Solution

- Gene ‘X’: Regulates plasmid copy number, crucial for cloning efficiency.


- Gene ‘Y’: Produces a protein essential for plasmid replication.



Conclusion: The correct option is (1). Quick Tip: \textbf{Plasmid vectors such as pBR322 play a critical role in genetic engineering by allowing gene insertion and controlled replication.}


Question 35:

Consider the following statements:

A. Annelids are true coelomates

B. Poriferans are pseudocoelomates

C. Aschelminthes are acoelomates

D. Platyhelminthes are pseudocoelomates

Choose the correct answer from the options given below :

(1) A only
(2) C only
(3) D only
(4) B only

Correct Answer: (1) A only
View Solution

- Annelids (e.g., earthworms) have a true coelom (coelomates).


- Poriferans (sponges) lack body cavities and are acoelomates.


- Aschelminthes (roundworms) have a pseudocoelom, not an acoelom.


- Platyhelminthes (flatworms) are also acoelomates, not pseudocoelomates.



Conclusion: The correct statement is (A: Annelids are true coelomates). Quick Tip: \textbf{Coelom classification: Coelomates (Annelids), Pseudocoelomates (Aschelminthes), Acoelomates (Platyhelminthes, Porifera).}


Question 36:

Choose the correct statement given below regarding juxta medullary nephron.

(1) Renal corpuscle of juxta medullary nephron lies in the outer portion of the renal medulla.
(2) Loop of Henle of juxta medullary nephron runs deep into medulla.
(3) Juxta medullary nephrons outnumber the cortical nephrons.
(4) Juxta medullary nephrons are located in the columns of Bertini.

Correct Answer: (2) Loop of Henle of juxta medullary nephron runs deep into medulla.
View Solution

Juxta Medullary and Cortical Nephrons: Key Differences


Juxta medullary nephrons have their renal corpuscle positioned deep within the renal cortex.
Their Loop of Henle extends significantly into the renal medulla, enabling enhanced water reabsorption.
Cortical nephrons are more abundant than juxta medullary nephrons.
The Columns of Bertini do not house nephrons; they primarily serve as support structures for the renal pyramids.


Conclusion:

The correct answer is \( \mathbf{(2)} \). Quick Tip: \textbf{Juxta medullary nephrons play a crucial role in urine concentration by maintaining the osmotic gradient in the medulla.}


Question 37:

Given below are two statements:
Statement I: Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II: Both bone marrow and thymus provide microenvironments for the development and maturation of T-lymphocytes.

In the light of above statements, choose the most appropriate answer from the options given below :

(1) Both Statement I and Statement II are incorrect.
(2) Statement I is correct but Statement II is incorrect.
(3) Statement I is incorrect but Statement II is correct.
(4) Both Statement I and Statement II are correct.

Correct Answer: (4) Both Statement I and Statement II are correct.
View Solution




- Bone marrow is the primary lymphoid organ responsible for blood cell formation, including lymphocytes.


- T-lymphocytes mature in the thymus, which provides the necessary microenvironment for their differentiation.


- Both bone marrow and thymus play key roles in immune system development.


Conclusion: The correct option is (4). Quick Tip: \textbf{Bone marrow is the primary site of hematopoiesis, while the thymus is essential for T-cell maturation.}


Question 38:

Match List I with List II related to the digestive system of cockroach.


(1) A-I, B-II, C-III, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-II, C-IV, D-I
(4) A-IV, B-II, C-III, D-I
Choose the correct answer from the options given below:

Correct Answer: (4) A-IV, B-II, C-III, D-I
View Solution




- Crop (A-IV): Stores food before digestion.


- Gastric Caeca (B-II): Secretes digestive enzymes.


- Malpighian Tubules (C-III): Excretory structures involved in osmoregulation.


- Gizzard (D-I): Helps in mechanical grinding of food.


Conclusion: The correct option is (4). Quick Tip: \textbf{Cockroach digestive system includes the foregut (crop and gizzard), midgut (gastric caeca), and hindgut (Malpighian tubules for excretion).}


Question 39:

Match List I with List II:


Choose the correct answer from the options given below :

(1) A-III, B-II, C-IV, D-I
(2) A-II, B-III, C-I, D-IV
(3) A-IV, B-II, C-I, D-III
(4) A-I, B-III, C-IV, D-II

Correct Answer: (1) A-III, B-II, C-IV, D-I
View Solution



- P wave represents atrial depolarization, indicating contraction of the atria.


- QRS complex represents ventricular depolarization, indicating contraction of the ventricles.


- T wave represents ventricular repolarization, meaning ventricles relax.


- T-P gap is the period when the heart is electrically silent before the next cycle.


Conclusion: The correct option is (1). Quick Tip: \textbf{ECG is an important tool in cardiology, where each wave corresponds to specific electrical events in the heart.}


Question 40:

As per ABO blood grouping system, the blood group of father is B+, mother is A+, and child is O+. Their respective genotype can be:

(A) \( I^B i / I^A i \)

(B) \( I^B I^B / I^A i \)

(C) \( I^A B / I^A I^B \)

(D) \( I^A i / I^B I^A \)

(E) \( ii / I^A I^B \)

 

Choose the most appropriate answer from the options given below :
(1) B only
(2) C & B only
(3) D & E only
(4) A only

Correct Answer: (4) A only
View Solution



- Since the child has O+ blood type, both parents must have one recessive 'i' allele.


- Possible parental genotypes are:


- Father (B+): \( I^B i \)


- Mother (A+): \( I^A i \)


- This allows the child to inherit 'i' from both parents, resulting in O+ blood type.



Conclusion: The correct option is (4). Quick Tip: \textbf{The ABO blood group system follows Mendelian inheritance, where 'A' and 'B' are dominant over 'O'.}


Question 41:

Match List I with List II:

Q41
Choose the correct answer from the options given below:

(1) A-III, B-II, C-IV, D-I
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-IV, C-I, D-III

Correct Answer: (3) A-IV, B-III, C-I, D-II
View Solution




Step 1: Understanding the Functions of the Given Terms


- RNA polymerase III (A-IV): Responsible for the transcription of small nuclear RNAs (snRNAs) and transfer RNAs (tRNA).


- Termination of transcription (B-III): The Rho factor plays a role in the termination of transcription in prokaryotic cells.


- Splicing of Exons (C-I): Small nuclear ribonucleoproteins (snRNPs) participate in RNA splicing, removing introns and joining exons.


- TATA box (D-II): A conserved promoter sequence that helps in the initiation of transcription.


Step 2: Matching the Correct Pairs \[ \begin{aligned} A & \rightarrow IV \quad (SnRNAs, tRNA)
B & \rightarrow III \quad (Rho factor)
C & \rightarrow I \quad (snRNPs)
D & \rightarrow II \quad (Promoter) \end{aligned} \]


Conclusion: The correct answer is (3) A-IV, B-III, C-I, D-II. Quick Tip: RNA polymerase III transcribes snRNAs and tRNA, the Rho factor terminates transcription, snRNPs assist in splicing, and the TATA box is a crucial promoter sequence for gene expression.


Question 42:

Given below are two statements:
Statement I:Mitochondria and chloroplasts are both double-membrane bound organelles.
Statement II: Inner membrane of mitochondria is relatively less permeable compared to chloroplast.

In the light of the above statements, choose the mis appropriate answer from the options given below:

(1) Both Statement I and Statement II are incorrect.
(2) Statement I is correct but Statement II is incorrect.
(3) Statement I is incorrect but Statement II is correct.
(4) Both Statement I and Statement II are correct

Correct Answer: (2) Statement I is correct but Statement II is incorrect.
View Solution




- Mitochondria and chloroplasts are double-membrane bound organelles, confirming Statement I is correct.


- However, the inner membrane of chloroplasts is less permeable than mitochondria, making Statement II incorrect.


- The mitochondrial inner membrane is involved in ATP synthesis and has more selective transporters than chloroplasts.


Conclusion: The correct option is (2). Quick Tip: \textbf{Mitochondria are the powerhouse of the cell, while chloroplasts are responsible for photosynthesis. Both have their own DNA.}


Question 43:

Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.


Correct Answer: (4) FSH, Leydig cells, Sertoli cells, spermiogenesis.
View Solution

- FSH (Follicle-Stimulating Hormone) stimulates Sertoli cells to support spermatogenesis.


- Leydig cells, stimulated by ICSH (Interstitial Cell-Stimulating Hormone), produce testosterone for sperm development.


- Spermiogenesis is the final transformation of spermatids into mature spermatozoa.



Conclusion: The correct option is (4). Quick Tip: \textbf{FSH plays a crucial role in sperm development, while Leydig cells support testosterone production.}


Question 44:

The following are the statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post-anal tail is absent.

Choose the most appropriate answer from the options given below:

(1) A, B & D only
(2) B, D & E only
(3) B, C & D only
(4) A & C only

Correct Answer: (2) B, D \& E only
View Solution

- Non-chordates lack a notochord and post-anal tail (Statement B, E).


- If a heart is present, it is dorsal (Statement D).


- Gill slits and dorsal CNS are characteristics of chordates, so A and C are incorrect.



Conclusion: The correct option is (2). Quick Tip: \textbf{Non-chordates lack a notochord, post-anal tail, and have a ventral nerve cord instead of a dorsal one.}


Question 45:

Given below are two statements:
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below :

(1) Both Statement I and Statement II are false.
(2) Statement I is true but Statement II is false.
(3) Statement I is false but Statement II is true.
(4) Both Statement I and Statement II are true.

Correct Answer: (3) Statement I is false but Statement II is true.
View Solution

- Gause's principle states that two species competing for the same resources cannot coexist indefinitely, but if they utilize different resources, coexistence is possible.


- Statement I is false, as competition occurs when species compete for same resources.


- Statement II is true, as the inferior competitor is eliminated when resources are limited.



Conclusion: The correct option is (3). Quick Tip: \textbf{Competitive exclusion principle states that no two species can occupy the same ecological niche for long.}


Question 46:

Regarding catalytic cycle of an enzyme action, select the correct sequential steps:
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below :

(1) A, E, B, D, C
(2) B, A, C, D, E
(3) E, D, C, B, A
(4) E, A, D, C, B

Correct Answer: (4) E, A, D, C, B
View Solution

- Step 1 (E): The substrate binds to the active site of the enzyme.


- Step 2 (A): This forms a substrate-enzyme complex.


- Step 3 (D): The enzyme catalyzes the breakage of chemical bonds in the substrate.


- Step 4 (C): The products are released from the enzyme.


- Step 5 (B): The enzyme is free to bind another substrate.


Conclusion: The correct sequence is E → A → D → C → B, option (4). Quick Tip: \textbf{Enzymes work by lowering the activation energy of a reaction and remain unchanged after the process.}


Question 47:

Given below are two statements:
Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.

In the light of the above statements, choose the most appropriate answer from the options given below:

(1) Both Statement I and Statement II are incorrect.
(2) Statement I is correct but Statement II is incorrect.
(3) Statement I is incorrect but Statement II is correct.
(4) Both Statement I and Statement II are correct.

Correct Answer: (2) Statement I is correct but Statement II is incorrect.
View Solution

- The corpus callosum is a large nerve fiber bundle that connects the two cerebral hemispheres, facilitating communication between them. Statement I is correct.


- The brain stem consists of the midbrain, pons, and medulla oblongata, but not the cerebrum. Statement II is incorrect.



Conclusion: The correct option is (2). Quick Tip: \textbf{The brainstem controls vital functions such as breathing, heart rate, and reflexes, while the cerebrum handles higher cognitive functions.}


Question 48:

Match List I with List II:
Q48

Choose the correct answer from the options given below:

(1) A-IV, B-II, C-I, D-III
(2) A-III, B-IV, C-II, D-I
(3) A-III, B-IV, C-I, D-II
(4) A-I, B-III, C-II, D-IV

Correct Answer: (3) A-III, B-IV, C-I, D-II
View Solution

- Exophthalmic goiter (Graves’ disease) results from hypersecretion of thyroid hormones, leading to protruding eyeballs.


- Acromegaly occurs due to excess growth hormone secretion in adulthood, causing enlarged extremities.


- Cushing’s syndrome is caused by excess cortisol secretion, leading to moon face and hyperglycemia.


- Cretinism is due to thyroid hormone deficiency in childhood, causing stunted growth and mental retardation.



Conclusion: The correct option is (3). Quick Tip: \textbf{Hormonal imbalances can lead to distinct disorders; thyroid hormones affect metabolism, while cortisol and growth hormone regulate stress and growth.}


Question 49:

Match List I with List II:




Choose the correct answer from the options given below:

(1) A-IV, B-III, C-I, D-II
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-I, C-IV, D-III
(4) A-II, B-I, C-III, D-IV

Correct Answer: (2) A-III, B-IV, C-I, D-II
View Solution

- Unicellular glandular epithelium consists of goblet cells found in the alimentary canal.


- Compound epithelium forms moist surfaces like the buccal cavity and provides protection.


- Multicellular glandular epithelium is found in salivary glands for secretion.


- Endocrine glandular epithelium includes the pancreas, which secretes hormones.


Conclusion: The correct option is (2). Quick Tip: \textbf{Glandular epithelium plays a major role in secretion; unicellular glands release mucus, whereas multicellular glands produce enzymes and hormones.}


Question 50:

Match List I with List II:




Choose the correct answer from the options given below:

(1) A-III, B-I, C-II, D-IV
(2) A-I, B-II, C-IV, D-III
(3) A-III, B-I, C-IV, D-II
(4) A-II, B-I, C-III, D-IV

Correct Answer: (3) A-III, B-I, C-IV, D-II
View Solution

Step 1: Understanding Geological Eras and Their Characteristic Fauna


- Mesozoic Era (A-III): Known as the "Age of Reptiles," it saw the dominance of birds and reptiles.


- Proterozoic Era (B-I): Marked by the emergence of lower invertebrates such as early multicellular life forms.


- Cenozoic Era (C-IV): Often referred to as the "Age of Mammals," characterized by the evolution and diversification of mammals.


- Paleozoic Era (D-II): Notable for the evolution of fish and amphibians, which thrived in aquatic environments.



Step 2: Matching the Correct Pairs \[ \begin{aligned} A & \rightarrow III \quad (Birds \& Reptiles)
B & \rightarrow I \quad (Lower invertebrates)
C & \rightarrow IV \quad (Mammals)
D & \rightarrow II \quad (Fish \& Amphibia) \end{aligned} \]



Conclusion: The correct answer is (3) A-III, B-I, C-IV, D-II. Quick Tip: The geological time scale classifies Earth's history into different eras. The \textbf{Proterozoic Era} saw early multicellular life, the \textbf{Paleozoic Era} was dominated by fish and amphibians, the \textbf{Mesozoic Era} was the age of reptiles and birds, and the \textbf{Cenozoic Era} saw the rise of mammals.


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