NEET 2024 Zoology Question Paper with Answers and Solutions PDF T4

Step 1: Common cold is caused by Rhinoviruses. (Matching A \(\to\) III)

Step 2: Haemozoin is a malarial pigment produced by Plasmodium during its life cycle. (Matching B \(\to\) I)

Step 3: Widal test is a diagnostic test for Typhoid, caused by \textit{Salmonella typhi. (Matching C \(\to\) II)

Step 4: Allergies are triggered by allergens such as dust mites. (Matching D \(\to\) IV)


Thus, the correct answer is option (1), as the correct matches are A-III, B-I, C-II, D-IV. Quick Tip: Rhinoviruses cause the common cold, \textit{Plasmodium produces haemozoin, Widal test diagnoses typhoid, and dust mites trigger allergies.

Step 1: Convergent evolution occurs when unrelated organisms evolve similar traits due to similar environmental pressures.

Step 2: Penguins (birds) and Dolphins (mammals) belong to different groups but have flippers adapted for swimming, an example of convergent evolution.

Step 3: Divergent evolution leads to different traits in related species, adaptive radiation involves diversification from a common ancestor, and natural selection drives evolution but does not always lead to convergence.


Thus, the correct answer is option (1), as flippers in Penguins and Dolphins evolved due to convergent evolution. Quick Tip: Convergent evolution leads to similar structures in unrelated species due to similar environmental pressures, as seen in Penguins and Dolphins.

Step 1: The correct sequence of human evolution is:
- Homo habilis (earliest tool users) → \textit{Homo erectus (upright walkers) →
- \textit{Homo neanderthalensis (Neanderthals) → \textit{Homo sapiens (modern humans).

Step 2: \textit{Homo habilis was the earliest species, followed by \textit{Homo erectus, then \textit{Homo neanderthalensis, and finally \textit{Homo sapiens.


Thus, the correct answer is option (2), as A-D-C-B represents the correct evolutionary order. Quick Tip: The sequence of human evolution follows: \textit{Homo habilis → Homo erectus → Homo neanderthalensis → Homo sapiens.

Step 1: The Hardy-Weinberg equilibrium states that allele frequencies remain constant in a population under stable conditions.

Step 2: Gene migration, genetic recombination, and genetic drift cause changes in allele frequencies, disturbing equilibrium.

Step 3: A constant gene pool means no changes in allele frequencies, keeping the population in equilibrium.


Thus, the correct answer is option (2), as a constant gene pool maintains Hardy-Weinberg equilibrium. Quick Tip: Hardy-Weinberg equilibrium remains stable if the gene pool is constant, meaning no evolutionary forces act on the population.

Step 1: Oxyhaemoglobin (HbO\(_2\)) formation occurs in the alveoli, where oxygen binds to hemoglobin.

Step 2: High pO\(_2\) (partial pressure of oxygen) favors oxygen binding to hemoglobin.

Step 3: Low pCO\(_2\) and lesser H\(^+\) concentration create an alkaline environment, promoting HbO\(_2\) formation.

Step 4: High pCO\(_2\) and high H\(^+\) concentration (acidic environment) promote oxygen release (Bohr effect), making options (1) and (3) incorrect.


Thus, the correct answer is option (4), as high pO\(_2\) and low H\(^+\) concentration favor oxyhaemoglobin formation. Quick Tip: In alveoli, high pO\(_2\) and lower H\(^+\) concentration favor oxyhaemoglobin formation, while high pCO\(_2\) promotes oxygen release.

Step 1: Natural contraceptive methods involve preventing pregnancy without external devices or hormones.

Step 2: Lactational amenorrhea, coitus interruptus, and periodic abstinence are natural contraceptive methods.

Step 3: Vaults are barrier contraceptives (diaphragms), which physically prevent sperm entry, making them a non-natural method.


Thus, the correct answer is option (2), as vaults are not a natural contraceptive method. Quick Tip: Natural contraceptive methods rely on behavioral practices, while vaults (diaphragms) are physical barriers used for contraception.

Step 1: Pons acts as a bridge connecting different brain regions. (Matching A \(\to\) III)

Step 2: Hypothalamus contains neurosecretory cells, playing a key role in hormonal regulation. (Matching B \(\to\) IV)

Step 3: Medulla oblongata controls respiration and gastric secretions. (Matching C \(\to\) II)

Step 4: Cerebellum regulates posture, balance, and coordination. (Matching D \(\to\) I)


Thus, the correct answer is option (4), as the correct matches are A-III, B-IV, C-II, D-I. Quick Tip: The cerebellum controls balance, the medulla regulates involuntary functions, the hypothalamus has neurosecretory cells, and the pons connects brain regions.

Step 1: The hymen can be torn due to various non-sexual activities like sports, cycling, or physical exertion. Statement I is true.

Step 2: The belief that the hymen tears only during the first coitus is incorrect, as it may already be absent due to other factors. Statement II is false.


Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: The hymen is not a definitive indicator of virginity, as it can be torn due to various physical activities unrelated to sexual intercourse.

Step 1: Axoneme is the structural core of cilia and flagella, consisting of microtubules. (Matching A \(\to\) II)

Step 2: Cartwheel pattern is a feature of centrioles, seen in their early formation stage. (Matching B \(\to\) I)

Step 3: Crista are the folds of the inner mitochondrial membrane, increasing the surface area for ATP production. (Matching C \(\to\) IV)

Step 4: Satellite DNA refers to repetitive DNA sequences found in chromosomes. (Matching D \(\to\) III)


Thus, the correct answer is option (2), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: Axoneme forms the core of cilia and flagella, cartwheel pattern is found in centrioles, cristae are folds of mitochondria, and satellite DNA is a part of chromosomes.

Step 1: Typhoid is caused by the bacterium \textit{Salmonella typhi. (Matching A \(\to\) IV)

Step 2: Leishmaniasis is caused by the protozoan parasite \textit{Leishmania donovani. (Matching B \(\to\) III)

Step 3: Ringworm is a fungal infection caused by \textit{Trichophyton, \textit{Microsporum, or \textit{Epidermophyton. (Matching C \(\to\) I)

Step 4: Filariasis is caused by nematode worms such as \textit{Wuchereria bancrofti. (Matching D \(\to\) II)


Thus, the correct answer is option (4), as the correct matches are A-IV, B-III, C-I, D-II. Quick Tip: Typhoid is caused by bacteria, Leishmaniasis by protozoa, ringworm by fungi, and filariasis by nematodes.

Step 1: The descending limb of the loop of Henle is permeable to water but impermeable to electrolytes. Thus, Statement I is false.

Step 2: The proximal convoluted tubule (PCT) is lined by simple cuboidal brush border epithelium, not columnar epithelium. Thus, Statement II is false.


Thus, the correct answer is option (4), as both Statement I and Statement II are false. Quick Tip: The descending limb of the loop of Henle is water-permeable, while the proximal convoluted tubule has simple cuboidal brush border epithelium for increased absorption.

Step 1: \(\alpha\)-1 antitrypsin is used in the treatment of emphysema. (Matching A \(\to\) III)

Step 2: Cry IAb is a Bt toxin that is effective against corn borers. (Matching B \(\to\) IV)

Step 3: Cry IAc is another Bt toxin, specifically targeting cotton bollworm. (Matching C \(\to\) I)

Step 4: Enzyme replacement therapy is used for ADA deficiency, a genetic disorder affecting the immune system. (Matching D \(\to\) II)


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: \(\alpha\)-1 antitrypsin treats emphysema, Cry IAb targets corn borers, Cry IAc affects cotton bollworms, and enzyme replacement therapy helps in ADA deficiency.

Step 1: Non-medicated IUDs include Lippes loop. (Matching A \(\to\) III)

Step 2: Copper-releasing IUDs include Multiload 375. (Matching B \(\to\) I)

Step 3: Hormone-releasing IUDs include LNG-20. (Matching C \(\to\) IV)

Step 4: Implants release progestogens, acting as long-term contraceptives. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Non-medicated IUDs include Lippes loop, Copper IUDs include Multiload 375, hormone-releasing IUDs include LNG-20, and implants release progestogens.

Step 1: Annelids (e.g., earthworms) have a true coelom, which is mesodermally derived. Statement A is correct.

Step 2: Poriferans (sponges) lack body cavities altogether, making Statement B incorrect.

Step 3: Aschelminthes (Nematodes) are pseudocoelomates, not acoelomates, making Statement C incorrect.

Step 4: Platyhelminthes (Flatworms) are acoelomates, not pseudocoelomates, making Statement D incorrect.


Thus, the correct answer is option (4), as only Statement A is correct. Quick Tip: Annelids have a true coelom, Poriferans lack body cavities, Aschelminthes are pseudocoelomates, and Platyhelminthes are acoelomates.

Step 1: Down’s syndrome is caused by trisomy of the 21\(^{st}\) chromosome. (Matching A \(\to\) III)

Step 2: \(\alpha\)-Thalassemia is associated with mutations in the 16\(^{th}\) chromosome. (Matching B \(\to\) IV)

Step 3: \(\beta\)-Thalassemia is caused by mutations in the 11\(^{th}\) chromosome. (Matching C \(\to\) I)

Step 4: Klinefelter’s syndrome occurs due to the presence of an extra ‘X’ chromosome (XXY). (Matching D \(\to\) II)


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Down’s syndrome is due to trisomy 21, \(\alpha\)-Thalassemia is linked to chromosome 16, \(\beta\)-Thalassemia to chromosome 11, and Klinefelter’s syndrome to the ‘X’ chromosome.

Step 1: The conduction pathway of the heart follows this sequence:
- SA node initiates the impulse. (E)
- AV node delays the impulse slightly. (C)
- AV bundle (Bundle of His) carries the signal to the ventricles. (A)
- Bundle branches transmit impulses through the ventricles. (D)
- Purkinje fibers distribute the impulse, leading to contraction. (B)

Thus, the correct answer is option (3), as the correct sequence is E-C-A-D-B. Quick Tip: The conduction pathway of the heart: SA node → AV node → AV bundle → Bundle branches → Purkinje fibers.

Step 1: Lipase breaks down lipids by hydrolyzing ester bonds. (Matching A \(\to\) II)

Step 2: Nuclease hydrolyzes nucleic acids by breaking phosphodiester bonds in DNA/RNA. (Matching B \(\to\) IV)

Step 3: Protease digests proteins by hydrolyzing peptide bonds between amino acids. (Matching C \(\to\) I)

Step 4: Amylase breaks down polysaccharides (starch) by hydrolyzing glycosidic bonds. (Matching D \(\to\) III)


Thus, the correct answer is option (1), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Lipase hydrolyzes ester bonds in fats, nucleases break phosphodiester bonds in DNA/RNA, proteases cleave peptide bonds in proteins, and amylase breaks glycosidic bonds in carbohydrates.

The vector pBR322 contains the X and Y genes.

- Gene X regulates the copy number of the plasmid DNA.

- Gene Y encodes a protein involved in the replication of the plasmid. This ensures that the plasmid can replicate inside the host cell.


Thus, the correct answer is (1) The gene ‘X’ is responsible for controlling the copy number of the linked DNA and ‘Y’ for protein involved in the replication of Plasmid. Quick Tip: In plasmid vectors like pBR322, genes involved in replication control the number of copies of the plasmid within the host cell.

Step 1: The Ti plasmid is a tumor-inducing plasmid found in Agrobacterium tumefaciens, a plant pathogen.

Step 2: It is responsible for causing crown gall disease in dicot plants.

Step 3: The T-DNA (transfer DNA) segment of the Ti plasmid integrates into the host genome, leading to uncontrolled cell division.


Thus, the correct answer is option (1), as the Ti plasmid stands for Tumor Inducing plasmid. Quick Tip: The Ti plasmid of \textit{Agrobacterium tumefaciens is widely used in plant genetic engineering due to its ability to transfer genes into plants.

Step 1: Pleurobrachia belongs to Ctenophora (comb jellies). (Matching A \(\to\) II)

Step 2: Radula is a rasping organ found in Mollusca for feeding. (Matching B \(\to\) I)

Step 3: Stomachord is a structure found in Hemichordates. (Matching C \(\to\) IV)

Step 4: Air bladder is found in Osteichthyes (bony fish) and helps in buoyancy. (Matching D \(\to\) III)


Thus, the correct answer is option (4), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: Pleurobrachia belongs to Ctenophora, radula is a feeding organ in Mollusca, stomachord is found in Hemichordata, and air bladder helps in buoyancy in bony fishes.

Step 1: Bio-reactors are used for large-scale production of biological products, not small-scale. Statement (1) is incorrect.

Step 2: They contain an agitator system, oxygen delivery system, and foam control system to maintain optimal growth conditions. Statements (2), (3), and (4) are correct.


Thus, the correct answer is option (1), as bio-reactors are used for large-scale production, not small-scale. Quick Tip: Bio-reactors are used for large-scale production of enzymes, antibiotics, and recombinant proteins, not small-scale bacterial cultures.

Step 1: DNA-dependent RNA polymerase synthesizes mRNA complementary to the DNA template.

Step 2: The given template strand (3’→5’ direction) results in an mRNA strand in the 5’→3’ direction.


Thus, the correct answer is option (3), as the correct mRNA sequence is 5’ AUGUACCGUUUUAUAGGAAGU 3’. Quick Tip: DNA-dependent RNA polymerase synthesizes mRNA complementary to the template strand, replacing thymine (T) with uracil (U).

Step 1: Cocaine is derived from Erythroxylum coca. (Matching A \(\to\) III)

Step 2: Heroin is synthesized from morphine, which is obtained from \textit{Papaver somniferum (opium poppy). (Matching B \(\to\) IV)

Step 3: Morphine is an effective sedative used in surgery. (Matching C \(\to\) I)

Step 4: Marijuana is derived from \textit{Cannabis sativa. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Cocaine is derived from \textit{Erythroxylum coca, heroin from Papaver somniferum, morphine is a sedative, and marijuana comes from Cannabis sativa.

Step 1: Diakinesis is the final stage of prophase I, where chiasmata terminalization is completed. (Matching A \(\to\) II)

Step 2: Pachytene is the stage where recombination nodules appear. (Matching B \(\to\) IV)

Step 3: Zygotene is characterized by synaptonemal complex formation. (Matching C \(\to\) I)

Step 4: Leptotene is the first stage where chromosomes appear as thin threads. (Matching D \(\to\) III)


Thus, the correct answer is option (1), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Prophase I stages: Leptotene (thin chromosomes), Zygotene (synaptonemal complex), Pachytene (crossing over), Diakinesis (chiasmata terminalization).

Step 1: Cockroaches possess anal cerci, which are jointed filamentous structures that function as sensory organs.

Step 2: These cerci are present in both male and female cockroaches on the 10\(^{th}\) abdominal segment.

Step 3: They help in detecting vibrations and movements in the surrounding environment.


Thus, the correct answer is option (4), as anal cerci are present on the 10\(^{th}\) segment of the cockroach. Quick Tip: Anal cerci in cockroaches are located on the 10\(^{th}\) abdominal segment and serve as sensory organs for detecting movement and vibrations.

Step 1: Breastfeeding is recommended as it provides essential nutrients and immunity to the infant. Assertion A is correct.

Step 2: Colostrum, the first milk produced after birth, is rich in maternal antibodies (IgA, IgG, IgM), which help develop immunity in the newborn. Reason R is correct.

Step 3: Since colostrum provides antibodies essential for infant immunity, it directly supports the reason given in Assertion A.


Thus, the correct answer is option (3), as both A and R are correct, and R is the correct explanation of A. Quick Tip: Colostrum is rich in maternal antibodies and helps develop passive immunity in newborns, making breastfeeding highly beneficial.

Step 1: Pterophyllum is commonly known as the Angel fish. (Matching A \(\to\) III)

Step 2: \textit{Myxine is commonly known as the Hag fish, which is a jawless fish. (Matching B \(\to\) I)

Step 3: \textit{Pristis is known as the Saw fish, characterized by its elongated snout with teeth. (Matching C \(\to\) II)

Step 4: \textit{Exocoetus is called the Flying fish, as it can glide over water surfaces. (Matching D \(\to\) IV)


Thus, the correct answer is option (4), as the correct matches are A-III, B-I, C-II, D-IV. Quick Tip: \textit{Pterophyllum (Angel fish), Myxine (Hag fish - jawless), Pristis (Saw fish), and Exocoetus (Flying fish) are well-known aquatic species.

Step 1: The three major types of muscles in the human body are:
- Skeletal muscle (voluntary and striated, found in biceps, triceps, etc.).
- Smooth muscle (involuntary, non-striated, found in internal organs like the stomach and intestines).
- Cardiac muscle (striated, involuntary, found in the heart).

Step 2:
- Image (a) represents skeletal muscle, found in triceps.
- Image (b) represents smooth muscle, found in stomach.
- Image (c) represents cardiac muscle, found in heart.

Thus, the correct answer is option (4), as (a) Skeletal – Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart. Quick Tip: Skeletal muscles are voluntary and striated, smooth muscles are involuntary and non-striated, and cardiac muscles are striated but involuntary.

Step 1: Steroid hormones are derived from cholesterol and include sex hormones and adrenal cortex hormones.

Step 2:
- Progesterone (option 1) is a steroid hormone involved in pregnancy.
- Cortisol (option 3) is a steroid hormone secreted by the adrenal cortex.
- Testosterone (option 4) is a steroid hormone responsible for male secondary sexual characteristics.
- Glucagon (option 2) is a peptide hormone, not a steroid. It is secreted by the pancreas and regulates blood glucose levels.

Thus, the correct answer is option (2), as Glucagon is a peptide hormone, not a steroid hormone. Quick Tip: Steroid hormones are derived from cholesterol (e.g., testosterone, cortisol, progesterone), while peptide hormones (e.g., glucagon, insulin) are made of amino acids.

Step 1: Fibrous joints (A) are immovable and found in the skull (Matching A \(\to\) III).

Step 2: Cartilaginous joints (B) provide limited movement and are found in the vertebral column (Matching B \(\to\) I).

Step 3: Hinge joints (C) allow movement in one plane and are found in the knee (Matching C \(\to\) IV).

Step 4: Ball and socket joints (D) allow rotational movement and are found in the humerus and pectoral girdle (Matching D \(\to\) II).


Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Fibrous joints (skull - no movement), Cartilaginous joints (vertebral column - limited movement), Hinge joints (knee - unidirectional), Ball and socket joints (shoulder - rotational).

Step 1: FSH (Follicle Stimulating Hormone) acts on Sertoli cells, not Leydig cells, in males. It regulates spermatogenesis. Thus, Assertion A is false.

Step 2: Leydig cells in males produce androgens (testosterone). In females, growing ovarian follicles secrete estrogen. Thus, Reason R is correct.


Thus, the correct answer is option (2), as Assertion A is incorrect, but Reason R is correct. Quick Tip: FSH acts on Sertoli cells (not Leydig cells), while growing ovarian follicles secrete estrogen, and Leydig cells secrete testosterone.

Step 1: Expiratory capacity (A) is the sum of tidal volume + expiratory reserve volume. (Matching A \(\to\) II)

Step 2: Functional residual capacity (B) is the sum of expiratory reserve volume + residual volume. (Matching B \(\to\) IV)

Step 3: Vital capacity (C) is the sum of tidal volume + inspiratory reserve volume + expiratory reserve volume. (Matching C \(\to\) I)

Step 4: Inspiratory capacity (D) is the sum of tidal volume + inspiratory reserve volume. (Matching D \(\to\) III)


Thus, the correct answer is option (3), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Expiratory capacity = TV + ERV, Functional residual capacity = ERV + RV, Vital capacity = TV + IRV + ERV, Inspiratory capacity = TV + IRV.

Step 1: The correct sequence of cell cycle stages is:
- G1 phase (E): Cell growth occurs.
- S phase (C): DNA replication happens.
- G2 phase (A): Preparation for mitosis.
- Karyokinesis (D): Nuclear division occurs.
- Cytokinesis (B): Cytoplasm divides, forming two daughter cells.

Thus, the correct answer is option (2), as the correct order is E-C-A-D-B. Quick Tip: The cell cycle follows the order: G1 → S → G2 → M (Karyokinesis + Cytokinesis).

Step 1: Autoimmune disorders occur when the immune system attacks its own body tissues.

Step 2:
- Myasthenia gravis (A): Autoimmune attack on neuromuscular junctions.
- Rheumatoid arthritis (B): Autoimmune inflammation of joints.
- SLE (E): A systemic autoimmune disease affecting multiple organs.
- Gout (C) is due to uric acid crystal accumulation (not autoimmune).
- Muscular dystrophy (D) is a genetic disorder, not autoimmune.

Thus, the correct answer is option (4), as A, B, and E are autoimmune disorders. Quick Tip: Autoimmune diseases include Myasthenia gravis, Rheumatoid arthritis, and Systemic Lupus Erythematosus (SLE), but not gout or muscular dystrophy.

Step 1: The Fallopian tube consists of four parts:
- Infundibulum (1): Funnel-shaped structure with fimbriae.
- Ampulla (2): The widest and longest part, where fertilization occurs.
- Isthmus (4): A narrow region connecting to the uterus.

Step 2: The Uterine fundus (3) is the uppermost part of the uterus, not the Fallopian tube.


Thus, the correct answer is option (3), as the uterine fundus is not part of the Fallopian tube. Quick Tip: The Fallopian tube consists of Infundibulum, Ampulla, Isthmus, and the Uterine part, but not the Uterine fundus.

Step 1: Blood type inheritance follows the Mendelian principles of codominance. The possible genotypes are:
- Blood group A: I\(^A\)I\(^A\) or I\(^A\)i
- Blood group B: I\(^B\)I\(^B\) or I\(^B\)i
- Blood group O: ii

Step 2: Since the child has O\(^-\) blood, they must have received the ii genotype (one i from each parent).

Step 3:
- The father’s genotype must be I\(^B\)i or I\(^B\)I\(^B\).
- The mother’s genotype must be I\(^A\)i or I\(^A\)I\(^A\).
- The only correct parental genotype pair that produces an O\(^-\) child is A (I\(^B\)i/I\(^B\)i) & D (I\(^A\)i/I\(^B\)i).

Thus, the correct answer is option (3), as A & D represent the correct possible genotypes. Quick Tip: For a child with O blood type, both parents must contribute an "i" allele (i.e., they must be heterozygous A or B).

Step 1: Exophthalmic goiter is due to hyperthyroidism, leading to protruding eyeballs. (Matching A \(\to\) III)

Step 2: Acromegaly occurs due to excess secretion of growth hormone. (Matching B \(\to\) IV)

Step 3: Cushing’s syndrome is due to excess cortisol secretion, leading to moon face and hyperglycemia. (Matching C \(\to\) I)

Step 4: Cretinism results from thyroid hormone deficiency, leading to stunted growth. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Exophthalmic goiter (hyperthyroidism), Acromegaly (excess GH), Cushing’s syndrome (excess cortisol), and Cretinism (thyroid hormone deficiency).

Step 1: Mesozoic Era is known as the "Age of Reptiles and Birds". (Matching A \(\to\) III)

Step 2: Proterozoic Era had the first lower invertebrates. (Matching B \(\to\) I)

Step 3: Cenozoic Era is known as the "Age of Mammals". (Matching C \(\to\) IV)

Step 4: Paleozoic Era was the period of early Fish and Amphibia. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Mesozoic Era (Reptiles \& Birds), Proterozoic Era (Lower invertebrates), Cenozoic Era (Mammals), Paleozoic Era (Fish \& Amphibians).

Step 1: Non-chordates lack notochord (B), post-anal tail (E), and have a dorsal heart (D).

Step 2: Chordates have pharyngeal gill slits (A) and a dorsal CNS (C), so these statements do not apply to non-chordates.

Step 3: Since non-chordates lack notochord, post-anal tail, and have a dorsal heart, the correct answer is B, D, and E only.

Thus, the correct answer is option (1), as B, D, and E are true for non-chordates. Quick Tip: Non-chordates lack a notochord and post-anal tail, while their heart (if present) is dorsal. Chordates have pharyngeal gill slits and a dorsal CNS.

Step 1: The corpus callosum is a bundle of nerve fibers that connects the left and right cerebral hemispheres, making Statement I correct.

Step 2: The brainstem consists of the medulla oblongata, pons, and midbrain (not cerebrum), making Statement II incorrect.


Thus, the correct answer is option (1), as Statement I is correct, but Statement II is incorrect. Quick Tip: The brainstem includes the medulla oblongata, pons, and midbrain but not the cerebrum. The corpus callosum connects the cerebral hemispheres.

Step 1: Unicellular glandular epithelium (A) consists of goblet cells in the alimentary canal. (Matching A \(\to\) III)

Step 2: Compound epithelium (B) lines moist surfaces like the buccal cavity. (Matching B \(\to\) IV)

Step 3: Multicellular glandular epithelium (C) includes salivary glands. (Matching C \(\to\) I)

Step 4: Endocrine glandular epithelium (D) includes the pancreas. (Matching D \(\to\) II)


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Unicellular epithelium (goblet cells), compound epithelium (moist surfaces), multicellular epithelium (salivary glands), endocrine epithelium (pancreas).

Step 1: P wave represents depolarisation of atria. (Matching A \(\to\) III)

Step 2: QRS complex represents depolarisation of ventricles. (Matching B \(\to\) II)

Step 3: T wave represents repolarisation of ventricles. (Matching C \(\to\) IV)

Step 4: T-P gap indicates electrical silence in heart muscles. (Matching D \(\to\) I)


Thus, the correct answer is option (4), as the correct matches are A-III, B-II, C-IV, D-I. Quick Tip: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation), T-P gap (electrical silence).

Step 1: Both mitochondria and chloroplasts have double membranes. (Statement I is correct)

Step 2: The inner membrane of mitochondria is highly folded but more permeable than the chloroplast. (Statement II is incorrect)


Thus, the correct answer is option (1). Quick Tip: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation), T-P gap (electrical silence).

Step 1: Crop (A) is the structure used for storing food in cockroaches. (Matching A \(\to\) IV)

Step 2: Gastric Caeca (B) are the 6-8 blind tubules at the junction of foregut and midgut. (Matching B \(\to\) II)

Step 3: Malpighian tubules (C) are the yellow-colored filaments found at the junction of midgut and hindgut. (Matching C \(\to\) III)

Step 4: Gizzard (D) is used for grinding food. (Matching D \(\to\) I)


Thus, the correct answer is option (3), as the correct matches are A-IV, B-II, C-III, D-I. Quick Tip: The cockroach digestive system has the crop (storage), gastric caeca (digestion), malpighian tubules (excretion), and gizzard (grinding food).

Step 1: RNA polymerase III (A) synthesizes SnRNAs and tRNA. (Matching A \(\to\) IV)

Step 2: Termination of transcription (B) is regulated by the Rho factor. (Matching B \(\to\) II)

Step 3: Splicing of Exons (C) is associated with snRNPs. (Matching C \(\to\) I)

Step 4: The TATA box (D) is a promotor region that facilitates transcription. (Matching D \(\to\) III)


Thus, the correct answer is option (2), as the correct matches are A-IV, B-II, C-I, D-III. Quick Tip: RNA polymerase III produces tRNA and snRNAs. The TATA box is a core promoter for transcription initiation.

Step 1: FSH (Follicle-stimulating hormone) stimulates Sertoli cells, which support spermatogenesis.

Step 2: LH (Luteinizing hormone) stimulates Leydig cells, which secrete androgens (testosterone) necessary for spermiogenesis.


Thus, the correct answer is option (3), as FSH stimulates Sertoli cells and Leydig cells support spermiogenesis. Quick Tip: FSH acts on Sertoli cells (spermatogenesis), and LH acts on Leydig cells (spermiogenesis).

Step 1: Juxta medullary nephrons have a longer loop of Henle that extends deep into the renal medulla, aiding in concentration of urine. (Matching Option 1)

Step 2: Cortical nephrons are more numerous than juxta medullary nephrons.


Thus, the correct answer is option (1), as the loop of Henle in juxta medullary nephron extends deep into medulla. Quick Tip: Juxta medullary nephrons play a critical role in water reabsorption and urine concentration due to their long loop of Henle.

Step 1: Gause's competitive exclusion principle does not state that species with different resources cannot coexist indefinitely. It specifically states that closely related species competing for the same resources cannot coexist indefinitely. (Statement I is false)

Step 2: Statement II is true because, in cases of competition with limited resources, the inferior species is eliminated.


Thus, the correct answer is option (2), as Statement I is false, but Statement II is true. Quick Tip: Competitive exclusion occurs when two species competing for the same resources cannot coexist, and the inferior species is eliminated.

Step 1: Bone marrow is indeed the primary site for the production of all blood cells, including lymphocytes. (Statement I is correct)

Step 2: Both bone marrow and thymus play crucial roles in the development and maturation of T-lymphocytes. (Statement II is correct)


Thus, the correct answer is option (3), as both Statement I and Statement II are correct. Quick Tip: Bone marrow produces all blood cells, including lymphocytes. The thymus plays a key role in the maturation of T-lymphocytes.

Step 1: The enzyme action begins with substrate binding to the active site (E).

Step 2: This leads to substrate-enzyme complex formation (A).

Step 3: The enzyme is then ready to bind with another substrate (B).

Step 4: The chemical bonds of the substrate are broken (D) as the reaction progresses.

Step 5: Finally, the products are released (C).


Thus, the correct answer is option (3), as the correct sequence is E, A, B, C, D. Quick Tip: The enzyme catalytic cycle follows: Substrate binding → Substrate-enzyme complex → Enzyme ready for next substrate → Substrate broken → Product released.