NEET 2024 Zoology Question Paper with Answers and Solutions PDF S3

Step 1: Understanding the Descending Limb of the Loop of Henle


- The descending limb of the loop of Henle is actually permeable to water and impermeable to electrolytes.


- Therefore, Statement I is false.



Step 2: Understanding the Proximal Convoluted Tubule (PCT)

- The proximal convoluted tubule (PCT) is lined by simple cuboidal epithelium with a brush border, not simple columnar epithelium.


- This brush border helps in increasing the surface area for reabsorption.
- Therefore, Statement II is also false.
Quick Tip: - The descending limb of the loop of Henle is permeable to water but impermeable to salts and solutes. - The proximal convoluted tubule is lined with simple cuboidal epithelium with microvilli to maximize absorption.

Step 1: Understanding the Role of FSH
- Follicle Stimulating Hormone (FSH) acts on ovarian follicles in females and Sertoli cells in males, not Leydig cells.


- Leydig cells are stimulated by Luteinizing Hormone (LH) to produce androgens.


- Therefore, Assertion (A) is false.



Step 2: Understanding the Hormonal Secretion

- Growing ovarian follicles secrete estrogen in females.

- Interstitial cells (Leydig cells) secrete androgen in males.

- This statement is factually correct.

- Therefore, Reason (R) is true.
Quick Tip: - FSH stimulates ovarian follicle growth in females and Sertoli cells in males. - LH stimulates Leydig cells to secrete testosterone in males. - Estrogen is secreted by growing ovarian follicles in females.

Step 1: Understanding Hardy-Weinberg Equilibrium

- The Hardy-Weinberg principle states that allele and genotype frequencies in a population remain constant from generation to generation in the absence of evolutionary forces.



Step 2: Factors Affecting Hardy-Weinberg Equilibrium

- Factors that disrupt equilibrium include:


- Genetic drift: Random changes in allele frequency.


- Genetic recombination: Can lead to variation and changes in allele frequency.


- Gene migration: Introduction of new alleles into a population.



Step 3: Why Constant Gene Pool Does Not Affect Equilibrium

- A constant gene pool means that allele frequencies do not change over generations.


- This maintains the Hardy-Weinberg equilibrium, fulfilling the principle's assumptions.
Quick Tip: The Hardy-Weinberg equilibrium remains stable when: - No mutation occurs. - No natural selection acts on the population. - No gene flow (migration) occurs. - No genetic drift occurs. - Random mating happens within the population.

Step 1: Understanding Autoimmune Disorders
Autoimmune diseases occur when the body's immune system mistakenly attacks its own cells, tissues, or organs.



Step 2: Evaluating Each Condition
- Myasthenia gravis (A): An autoimmune disorder where antibodies attack neuromuscular junctions, causing muscle weakness.


- Rheumatoid arthritis (B): An autoimmune disorder leading to joint inflammation and damage.


- Gout (C): Not an autoimmune disease; it results from uric acid accumulation.


- Muscular dystrophy (D): A genetic disorder, not an autoimmune condition.


- Systemic Lupus Erythematosus (E): A severe autoimmune disease affecting multiple organs.



Step 3: Conclusion
- The correct autoimmune disorders from the list are Myasthenia gravis (A), Rheumatoid arthritis (B), and SLE (E).
- Therefore, the correct answer is (3) A, B \& E only. Quick Tip: Common autoimmune disorders include: - Type 1 Diabetes, where the immune system attacks insulin-producing cells. - Multiple Sclerosis (MS), affecting the nervous system. - Hashimoto’s Thyroiditis, where the immune system attacks the thyroid gland.

Step 1: Understanding Human Evolution
The evolution of modern humans occurred in a sequence from primitive hominins to Homo sapiens. The correct chronological order follows:

1. \textit{Homo habilis (A): Earliest member of the genus Homo, known for using primitive tools.


2. \textit{Homo erectus (D): More advanced than \textit{Homo habilis, developed better tools and control over fire.


3. \textit{Homo neanderthalensis (C): A close relative of modern humans, exhibiting advanced social behavior.


4. \textit{Homo sapiens (B): The most recent species, representing modern humans.



Step 2: Conclusion
- The correct chronological order of human evolution is A → D → C → B.
- Therefore, the correct answer is (1) A-D-C-B. Quick Tip: Key evolutionary traits: - \textit{Homo habilis: First tool users. - Homo erectus: First to use fire. - Homo neanderthalensis: Adapted to cold climates. - Homo sapiens: Modern humans with advanced cognitive abilities.

Step 1: Understanding the Hymen and Virginity

- The hymen is a thin membrane that partially covers the vaginal opening.


- Its presence or absence is not a reliable indicator of virginity, as it can be stretched or torn due to various physical activities such as sports, cycling, or medical examinations.


- Thus, Statement I is true.



Step 2: Examining Statement II
- While the hymen may tear during first coitus, it is not the only reason for its rupture.


- Many women are born with a minimal hymenal membrane, while others may experience its rupture due to non-sexual activities.


- Therefore, Statement II is false.



Step 3: Conclusion
- Since Statement I is true and Statement II is false, the correct answer is (4) Statement I is true but Statement II is false. Quick Tip: The hymen can be stretched or torn due to physical activities such as sports, horse riding, or tampon use. Its presence or absence should not be considered an indicator of virginity.

Step 1: Understanding the Types of Intrauterine Devices (IUDs) and Implants


- Non-medicated IUD (A): The Lippes loop is an example of a non-medicated IUD, which prevents implantation by inducing a local inflammatory response.


- Copper releasing IUD (B): Multiload 375 is a copper-releasing IUD that enhances contraceptive effectiveness by releasing copper ions, which reduce sperm motility.


- Hormone releasing IUD (C): LNG-20 (Levonorgestrel-releasing IUD) is a hormone-based IUD that releases progestogen, preventing implantation.


- Implants (D): Progestogens are used in contraceptive implants, which release hormones steadily into the bloodstream to prevent ovulation.



Step 2: Conclusion
- Based on the correct matching:
- A → III (Lippes loop)

- B → I (Multiload 375)

- C → IV (LNG-20)

- D → II (Progestogens)



Thus, the correct answer is (1) A-III, B-I, C-IV, D-II. Quick Tip: Hormone-releasing IUDs release synthetic progestogens like Levonorgestrel, whereas copper IUDs work by releasing copper ions that reduce sperm viability and mobility.

Step 1: Understanding Coelomates, Pseudocoelomates, and Acoelomates

- Coelomates: These animals have a true coelom lined with mesoderm. Example: Annelids (Correct Statement A).


- Pseudocoelomates: These animals have a body cavity but it is not lined entirely with mesoderm. Example: Aschelminthes (Nematodes), not Poriferans. (Statement B is incorrect).


- Acoelomates: These animals lack a coelom. Example: Platyhelminthes, not Aschelminthes (Statement C is incorrect).


- Platyhelminthes are acoelomates, not pseudocoelomates (Statement D is incorrect).



Step 2: Conclusion
- The only correct statement is A (Annelids are true coelomates).


- Hence, the correct answer is (3) A only. Quick Tip: Acoelomates (e.g., Platyhelminthes) lack a body cavity, pseudocoelomates (e.g., Nematodes) have a false coelom, and true coelomates (e.g., Annelids) have a coelom lined entirely by mesoderm.

Step 1: Understanding Oxyhaemoglobin Formation

- Oxyhaemoglobin (HbO\textsubscript{2}) formation occurs when oxygen binds to haemoglobin in the alveoli.


- This process is favoured by high partial pressure of oxygen (pO\textsubscript{2) and low concentration of H\textsuperscript{+ ions (less acidity).



Step 2: Evaluating the Given Options

- Option (1): High temperature does not favour oxyhaemoglobin formation, as higher temperatures cause dissociation of HbO\textsubscript{2}.


- Option (2): High pCO\textsubscript{2 increases acidity, which promotes oxygen dissociation rather than binding.


- Option (3) (Correct): High pO\textsubscript{2 enhances oxygen binding to haemoglobin, and lower H\textsuperscript{+ concentration prevents oxygen dissociation.


- Option (4): High H\textsuperscript{+ concentration leads to oxyhaemoglobin dissociation rather than formation.



Step 3: Conclusion
- The correct answer is (3) High pO\textsubscript{2 and Lesser H\textsuperscript{+ concentration. Quick Tip: In the alveoli, oxygen binding to haemoglobin is favoured by: 1. \textbf{High pO\textsubscript{2}}, ensuring efficient loading of oxygen. 2. \textbf{Low pCO\textsubscript{2}}, reducing carbonic acid formation. 3. \textbf{Lower H\textsuperscript{+} concentration}, maintaining haemoglobin's affinity for oxygen.

Step 1: Understanding the Respiratory Capacities

1. Expiratory Capacity (A): The sum of Tidal Volume and Expiratory Reserve Volume, which corresponds to (II).


2. Functional Residual Capacity (B): The sum of Expiratory Reserve Volume and Residual Volume, which matches (IV).


3. Vital Capacity (C): The sum of Expiratory Reserve Volume, Tidal Volume, and Inspiratory Reserve Volume, which corresponds to (I).


4. Inspiratory Capacity (D): The sum of Tidal Volume and Inspiratory Reserve Volume, which matches (III).



Step 2: Evaluating the Given Options

- Option (1): A-I, B-III, C-II, D-IV (Incorrect)

- Option (2) (Correct): A-II, B-IV, C-I, D-III

- Option (3): A-III, B-II, C-IV, D-I (Incorrect)

- Option (4): A-II, B-I, C-IV, D-III (Incorrect)



Step 3: Conclusion
- The correct answer is (2) A-II, B-IV, C-I, D-III. Quick Tip: Respiratory capacities are derived from the combination of different lung volumes: 1. \textbf{Vital Capacity (VC)} = IRV + TV + ERV 2. \textbf{Inspiratory Capacity (IC)} = TV + IRV 3. \textbf{Expiratory Capacity (EC)} = TV + ERV 4. \textbf{Functional Residual Capacity (FRC)} = ERV + RV

Step 1: Understanding the Muscle Types and Locations

1. Skeletal Muscle (a): These are voluntary muscles attached to bones and responsible for movement. Example: Triceps.


2. Smooth Muscle (b): These are involuntary muscles found in internal organs like the stomach.


3. Cardiac Muscle (c): Found only in the heart, responsible for pumping blood.



Step 2: Evaluating the Given Options

- Option (1): Incorrect, as nose tip does not contain involuntary muscle.


- Option (2): Incorrect, as toes do not primarily have smooth muscle.


- Option (3) (Correct): Skeletal muscle in triceps, smooth muscle in stomach, and cardiac muscle in heart.


- Option (4): Incorrect, as intestines do not belong in this specific classification.



Step 3: Conclusion
- The correct answer is (3) Skeletal - Triceps, Smooth - Stomach, Cardiac - Heart. Quick Tip: Muscles are classified into three types: 1. Skeletal Muscle: Voluntary, striated, multinucleated, attached to bones. 2. Smooth Muscle: Involuntary, non-striated, found in organs like intestines, stomach. 3. Cardiac Muscle: Involuntary, striated, intercalated discs, found only in the heart.

N/A

N/A

N/A

Step 1: Understanding the functions of the given enzymes:


Lipase: Breaks ester bonds in lipids \(\rightarrow\) II. Ester bond.
Nuclease: Breaks phosphodiester bonds in nucleic acids \(\rightarrow\) IV. Phosphodiester bond.
Protease: Breaks peptide bonds in proteins \(\rightarrow\) I. Peptide bond.
Amylase: Breaks glycosidic bonds in polysaccharides \(\rightarrow\) III. Glycosidic bond.


Thus, the correct matching is:
\[ A-II, B-IV, C-I, D-III \] Quick Tip: Understanding enzyme specificity and the bonds they act upon is crucial for solving biochemical matching problems efficiently.

Step 1: Understanding the Ti Plasmid

The Ti plasmid (Tumor-inducing plasmid) is found in \textit{Agrobacterium tumefaciens, a soil bacterium known for causing crown gall disease in plants. The plasmid carries \textit{T-DNA, which integrates into the plant genome and leads to tumor formation.



Step 2: Analyzing the Statements


Statement 1: Incorrect. The Ti plasmid is not related to temperature independence.
Statement 2: Incorrect. The plasmid does not inhibit tumors; rather, it induces them.
Statement 3: Incorrect. It is not tumor-independent; it actively promotes tumor formation.
Statement 4: Correct. The Ti plasmid is responsible for inducing tumors in plants.


Conclusion: The correct answer is option (4). Quick Tip: The Ti plasmid is widely used in genetic engineering to transfer desirable genes into plant cells.

Step 1: Analyze the options by identifying the correct matching between the terms in List I and List II.



- Common cold is caused by Rhinoviruses, which matches with III.


- Hemozoin is produced by Plasmodium, matching with I.


- Widal test is used to diagnose Typhoid, which matches with II.


- Allergy is related to Dust mites, matching with IV.



Thus, the correct matching is:
\[ A-III, \, B-I, \, C-II, \, D-IV. \] Quick Tip: Understanding the causes of diseases and the tests used for diagnosis can help match the right terms effectively.

Step 1: Analyze the stages of cell division:

- Gap 1 phase (E) occurs first, as it is the stage before synthesis.


- Synthesis phase (C) comes after Gap 1 phase, where DNA replication occurs.


- Gap 2 phase (A) follows synthesis phase, where the cell prepares for mitosis.


- Karyokinesis (D) is the division of the nucleus during mitosis.


- Cytokinesis (B) occurs last, where the cytoplasm divides, resulting in two daughter cells.



Thus, the correct sequence is:
\[ E-C-A-D-B. \] Quick Tip: Remember the order of the stages in cell division: Gap 1 phase (E), Synthesis phase (C), Gap 2 phase (A), Karyokinesis (D), and Cytokinesis (B).

Step 1: Understand the natural/traditional contraceptive methods:



- Coitus interruptus (2), also known as withdrawal method, is a natural contraceptive method where the male partner withdraws before ejaculation.


- Periodic abstinence (3) involves avoiding sexual intercourse during the fertile period of the woman's menstrual cycle.


- Lactational amenorrhea (4) is the temporary natural infertility that occurs after childbirth when a woman is breastfeeding, which can act as a contraceptive.



Vaults (1), however, is not a traditional contraceptive method; it is not recognized as a natural or traditional way to prevent pregnancy.



Thus, the correct answer is:
\[ Vaults (1). \] Quick Tip: Traditional contraceptive methods often rely on natural cycles or physical withdrawal, whereas Vaults are not a recognized method.

Step 1: Match the items from List I with the corresponding items in List II:

- Pleurobrachia (A) is a member of the phylum Ctenophora, so it matches with II.


- Radula (B) is an organ found in Mollusca, so it matches with I.


- Stomochord (C) is a characteristic of Hemichordata, so it matches with IV.


- Air bladder (D) is found in Osteichthyes (bony fish), so it matches with III.



Thus, the correct matching is:
\[ A-II, \, B-I, \, C-IV, \, D-III. \] Quick Tip: Understand the classification and key characteristics of various phyla to help match terms correctly.

Step 1: The DNA template strand is given, and we need to transcribe it to RNA. During transcription, RNA polymerase synthesizes the RNA strand in the 5' to 3' direction, using the DNA template strand in the 3' to 5' direction.



The RNA sequence is complementary to the template strand, where:



- A (Adenine) pairs with U (Uracil) in RNA.

- T (Thymine) pairs with A (Adenine) in RNA.

- C (Cytosine) pairs with G (Guanine) in RNA.

- G (Guanine) pairs with C (Cytosine) in RNA.



By applying these rules, we transcribe the given DNA template strand (3' TACATGGCAAATATCCATTCA 5') to the RNA sequence:
\[ 5' \, AUGACCGUUUAGGUG3' \]



Thus, the correct RNA sequence is:
\[ 5' \, AUGACCGUUUAGGUG3'. \] Quick Tip: Remember the base pairing rules during transcription: - A (DNA) pairs with U (RNA), - T (DNA) pairs with A (RNA), - C (DNA) pairs with G (RNA), - G (DNA) pairs with C (RNA).

Step 1: Analyze each term in List I and match it with the correct term from List II:



- α-I antitrypsin (A) is related to Emphysema (III). α-I antitrypsin deficiency leads to emphysema, a lung condition.


- Cry IAb (B) is related to Cotton bollworm (I). Cry proteins are used in genetically modified crops like cotton to control pests like the bollworm.


- Cry IAc (C) is related to Corn borer (IV). Cry proteins also provide protection against pests like the corn borer.


- Enzyme replacement therapy (D) is related to ADA deficiency (II). Enzyme replacement therapy is used to treat ADA (Adenosine deaminase) deficiency, a genetic disorder.



Thus, the correct matching is:
\[ A-III, \, B-IV, \, C-I, \, D-II. \]
Quick Tip: Understanding the relationship between genetic disorders, enzyme deficiencies, and their treatments, along with the use of Cry proteins in pest control, is essential for matching the terms correctly.

Step 1: Match the terms from List I with the correct terms from List II:



- Pterophyllum (A) is the scientific name for the Angel fish (III).


- Myxine (B) is the scientific name for the Hag fish (I).


- Pristis (C) is the scientific name for the Saw fish (II).


- Exocoetus (D) is the scientific name for the Flying fish (IV).



Thus, the correct matching is:
\[ A-III, \, B-I, \, C-II, \, D-IV. \] Quick Tip: Remember the unique characteristics of different fish species, such as the flying ability of Exocoetus and the distinct appearance of Pristis, to help identify the correct matches.

Step 1: The Fallopian tube consists of the following parts:


- Ampulla: The part of the Fallopian tube where fertilization usually occurs.


- Isthmus: The narrow part of the Fallopian tube that connects to the uterus.


- Infundibulum: The funnel-shaped opening of the Fallopian tube near the ovary, which catches the egg.



Uterine fundus, however, is a part of the uterus, not the Fallopian tube. Therefore, it is not a component of the Fallopian tube.



Thus, the correct answer is:
\[ Uterine fundus (2). \] Quick Tip: The Fallopian tube includes the ampulla, isthmus, and infundibulum, but the uterine fundus is part of the uterus.

Step 1: In the pBR322 cloning vector:

- Gene X: It is responsible for controlling the copy number of the linked plasmid DNA. It ensures that the plasmid is replicated at the proper rate to maintain a suitable number of plasmids within the host cell.


- Gene Y: This gene codes for a protein that is involved in the replication of the plasmid. It plays a critical role in the DNA replication machinery, ensuring the plasmid DNA is copied.



Thus, the correct answer is:
\[ Gene 'X' controls copy number of linked DNA and gene 'Y' is involved in replication of plasmid. \] Quick Tip: In plasmid vectors like pBR322, the replication control gene (X) and replication initiation proteins (Y) work together to maintain plasmid stability and replication within the host.

Step 1: Match the terms from List I with the correct terms from List II:

- Fibrous joints (A): These joints are immovable and found between adjacent vertebrae, allowing limited movement. Therefore, they match with I.


- Cartilaginous joints (B): These joints are semi-movable and are found in areas like the humerus and pectoral girdle, allowing rotational movement. Thus, they match with II.


- Hinge joints (C): These joints, like the knee, help in movement and are located in places where bending occurs. Hence, they match with IV.


- Ball and socket joints (D): These joints, like the skull, allow no movement and provide stability. Therefore, they match with III.



Thus, the correct matching is:
\[ A-I, \, B-II, \, C-IV, \, D-III. \] Quick Tip: Fibrous joints allow limited movement, cartilaginous joints allow rotational movement, hinge joints enable bending, and ball-and-socket joints offer a wide range of motion.

Step 1: Understand the classification of hormones:

- Steroid hormones are derived from cholesterol and include hormones like Cortisol, Testosterone, and Progesterone. These hormones are lipid-soluble and can pass through cell membranes to bind with intracellular receptors.



- Glucagon (1) is a peptide hormone, not a steroid hormone. It is produced by the pancreas and plays a key role in regulating blood sugar levels.



Thus, the correct answer is:
\[ Glucagon (1). \] Quick Tip: Steroid hormones are derived from cholesterol, while peptide hormones like glucagon are made of amino acids and are water-soluble.

Step 1: Match the terms from List I with the correct terms from List II:

- Axoneme (A) is the central structure of cilia and flagella, so it matches with II (Cilia and flagella).


- Cartwheel pattern (B) is seen in centrioles, so it matches with I (Centrioles).


- Crista (C) is the inner membrane structure found in mitochondria, so it matches with IV (Mitochondria).


- Satellite (D) is related to chromosomes, so it matches with III (Chromosome).

Thus, the correct matching is:
\[ A-II, \, B-I, \, C-IV, \, D-III. \] Quick Tip: The structure and patterns in cellular organelles are unique. For instance, the axoneme is crucial for cilia and flagella, while cartwheel patterns are characteristic of centrioles.

Step 1: Match the terms from List I with the correct terms from List II:



- Cocaine (A) is derived from Erythroxylum (III), a plant that produces the stimulant.


- Heroin (B) is derived from Papaver somniferum (IV), also known as the opium poppy.


- Morphine (C) is derived from Papaver somniferum (IV), and it is often used as an effective sedative in surgery.


- Marijuana (D) is derived from Cannabis sativa (II), a plant that produces the drug.



Thus, the correct matching is:
\[ A-III, \, B-IV, \, C-I, \, D-II. \] Quick Tip: Understanding the sources of various drugs can help match them with their respective plants or derivatives, such as cocaine from Erythroxylum and marijuana from Cannabis sativa.

Step 1: The anal cerci are sensory structures found in cockroaches. These paired jointed filamentous structures are present on the 10^th segment of both male and female cockroaches.



Thus, the correct answer is:
\[ 10^th segment (3). \] Quick Tip: In cockroaches, anal cerci play an important role in detecting air currents and other environmental signals. They are located on the 10^th segment.

Step 1: The correct matching between the sub-phases of prophase I and their specific characters is as follows:

- A. Diakinesis corresponds to II. Completion of terminalisation of chiasmata. In this stage, the chromosomes are fully condensed, and the chiasmata (points where homologous chromosomes cross over) are completed.


- B. Pachytene corresponds to IV. Appearance of recombination nodules. During pachytene, the homologous chromosomes are fully paired, and recombination nodules (sites of crossover) appear.


- C. Zygotene corresponds to I. Synaptonemal complex formation. In this stage, the synaptonemal complex is formed between homologous chromosomes, facilitating their pairing.


- D. Leptotene corresponds to III. Chromosomes look like thin threads. During leptotene, chromosomes begin to condense but still appear as thin threads under a microscope.



Thus, the correct matching is:
\[ A-II, B-IV, C-I, D-III. \] Quick Tip: In meiosis, the prophase I stages are characterized by the pairing and recombination of homologous chromosomes. Each stage has specific morphological characteristics that help in identifying the process.

Step 1: Assertion A is correct. Breastfeeding is recommended by doctors during the initial period of infant growth because it provides essential nutrients and immunity to the baby.



Step 2: Reason R is also correct. Colostrum, the first milk produced by the mother after childbirth, is rich in antibodies that are crucial for the newborn's immune system to resist infections.



Step 3: Reason R provides the correct explanation for Assertion A because the antibodies in colostrum contribute to the newborn's immunity, supporting the recommendation for breastfeeding during the initial period.



Thus, both Assertion A and Reason R are correct, and Reason R correctly explains Assertion A. Quick Tip: Breastfeeding in the initial stages not only provides nutrients but also helps in building the infant's immune system through the antibodies present in colostrum.

Step 1: Down's syndrome is caused by an extra copy of the 21st chromosome. Hence, A. Down's syndrome corresponds to III. 21st chromosome.



Step 2: \(\alpha\)-Thalassemia is associated with a deletion or mutation on chromosome 16. Hence, B. \(\alpha\)-Thalassemia corresponds to IV. 16th chromosome.



Step 3: \(\beta\)-Thalassemia is associated with mutations on chromosome 11. Hence, C. \(\beta\)-Thalassemia corresponds to I. 11th chromosome.



Step 4: Klinefelter's syndrome is caused by the presence of an extra X chromosome, making males XXY. Hence, D. Klinefelter's syndrome corresponds to II. 'X' chromosome.



Thus, the correct matching is:
\[ A-III, B-IV, C-I, D-II. \] Quick Tip: Genetic disorders are often linked to mutations or anomalies in specific chromosomes. Understanding the chromosomal basis of these disorders helps in their diagnosis and study.

Step 1: A. Pons connects different regions of the brain, particularly involved in communication between the cerebellum and cerebrum. Hence, A corresponds to III. Connects different regions of the brain.



Step 2: B. Hypothalamus controls important functions such as respiration and gastric secretions. Hence, B corresponds to II. Controls respiration and gastric secretions.



Step 3: C. Medulla contains centers that regulate vital autonomic functions like heartbeat, blood pressure, and respiration. Hence, C corresponds to I. Provides additional space for neurons, regulates posture and balance.



Step 4: D. Cerebellum plays a key role in motor control, balance, and posture regulation. Hence, D corresponds to IV. Neurosecretory cells.



Thus, the correct matching is:
\[ A-III, B-IV, C-II, D-I. \] Quick Tip: The brain is divided into various regions, each responsible for specific functions like coordination, balance, respiration, and communication between other parts of the brain.

Step 1: The father has blood group B, which means his genotype could either be \(|B^i|/|B^i|\) or \(|B^i|/|A^i|\). Since the child is of blood group O, the father must have a genotype \(|B^i|/|A^i|\), with the possibility of carrying the O allele.



Step 2: The mother has blood group A, which means her genotype could either be \(|A^i|/|A^i|\) or \(|A^i|/|B^i|\). Since the child has blood group O, the mother must also carry an O allele, so her genotype must be \(|A^i|/ii\).



Step 3: The child has blood group O, which means the child must inherit an O allele from both parents. This matches the genotypes given in option A, \(|B^i|/|A^i|\) for the father and \(|A^i|/ii\) for the mother.



Thus, the correct answer is A only. Quick Tip: In ABO blood grouping, a person with blood group O must have two O alleles (ii), whereas a person with blood group A or B may carry one O allele and one A or B allele.

Step 1: A. Unicellular glandular epithelium consists of a single cell type that secretes mucus, and it is present in the III. Goblet cells of the alimentary canal.



Step 2: B. Compound epithelium consists of multiple layers and is found in areas requiring protection and covering, such as IV. Moist surface of the buccal cavity.



Step 3: C. Multicellular glandular epithelium is composed of multiple cells and includes structures like I. Salivary glands.



Step 4: D. Endocrine glandular epithelium releases hormones directly into the bloodstream and is present in organs such as the II. Pancreas.



Thus, the correct matching is:
\[ A-III, B-IV, C-I, D-II. \] Quick Tip: Glandular epithelium can be unicellular (e.g., goblet cells), multicellular (e.g., salivary glands), or endocrine (e.g., pancreas), each serving distinct functions such as secretion and hormone release.

Step 1: The question describes the hormonal regulation of spermatogenesis, beginning with GnRH (Gonadotropin-releasing hormone). GnRH stimulates the release of LH (Luteinizing hormone), which acts on Leydig cells to stimulate the production of androgens (testosterone).



Step 2: Androgens then promote the function of Sertoli cells, which are responsible for the nourishment and development of sperm cells.



Step 3: FSH (Follicle-stimulating hormone) also plays a role in regulating Sertoli cells and aiding in the process of spermiogenesis, which is the final stage of sperm cell maturation, forming mature sperm from spermatids.



Thus, the correct sequence is:
\[ FSH, Leydig cells, Sertoli cells, spermiogenesis. \] Quick Tip: Spermatogenesis involves the process where spermatogonia develop into mature sperm cells. Spermiogenesis is the final maturation of spermatids into spermatozoa, which involves the action of hormones such as FSH and LH.

Step 1: Statement I is incorrect. Gause's competitive exclusion principle states that two closely related species that compete for the same resources cannot coexist indefinitely in the same ecological niche. It does not specifically state that they cannot compete for different resources.



Step 2: Statement II is correct. According to Gause's principle, in competitive situations, one species may be more efficient at utilizing the resources, and the inferior species may be eliminated if the resources are limited. This is consistent with the idea that competition can lead to the elimination of one species when resources are scarce.



Thus, the correct answer is Statement I is false but Statement II is true. Quick Tip: The competitive exclusion principle suggests that when two species compete for identical resources in an ecological niche, one species will outcompete the other. This highlights the importance of niche differentiation in maintaining biodiversity.

Step 1: A. The structures used for storing food refers to the IV. Crop. The crop stores food temporarily before digestion.



Step 2: B. Ring of 6-8 blind tubules at junction of foregut and midgut refers to II. Gastric Caeca. These structures are involved in digestion and absorption in cockroaches.



Step 3: C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut refers to the III. Malpighian tubules. These tubules are involved in excretion and osmoregulation.



Step 4: D. The structures used for grinding the food refers to the I. Gizzard, which grinds food in the digestive system of the cockroach.



Thus, the correct matching is:
\[ A-IV, B-II, C-III, D-I. \] Quick Tip: The digestive system of the cockroach is efficient in storing, digesting, and excreting food through specialized structures like the crop, gastric caeca, malpighian tubules, and gizzard.

Step 1: The Mesozoic Era is known for the rise of birds and reptiles, making it correspond to III. Birds \& Reptiles.



Step 2: The Proterozoic Era is characterized by the development of lower invertebrates, hence it corresponds to I. Lower invertebrates.



Step 3: The Cenozoic Era saw the rise of mammals, so it corresponds to IV. Mammals.



Step 4: The Paleozoic Era is known for the development of fish and amphibians, thus it corresponds to II. Fish \& Amphibia.



Thus, the correct matching is:
\[ A-III, B-I, C-IV, D-II. \] Quick Tip: The geological time scale is divided into different eras based on the predominant life forms and evolutionary stages, with each era contributing to the rise of distinct groups of organisms.

Step 1: The P wave represents the depolarisation of atria, so it corresponds to III. Depolarisation of atria.



Step 2: The QRS complex represents the depolarisation of ventricles, so it corresponds to II. Depolarisation of ventricles.



Step 3: The T wave represents the repolarisation of ventricles, so it corresponds to IV. Repolarisation of ventricles.



Step 4: The T-P gap represents the period when heart muscles are electrically silent, so it corresponds to I. Heart muscles are electrically silent.



Thus, the correct matching is:
\[ A-III, B-I, C-IV, D-I. \] Quick Tip: In an electrocardiogram (ECG), the P wave, QRS complex, and T wave represent different phases of electrical activity in the heart: depolarisation of the atria, depolarisation of the ventricles, and repolarisation of the ventricles, respectively.

Step 1: Statement A: Pharynx is perforated by gill slits. This is true for non-chordates, such as invertebrates like amphioxus.



Step 2: Statement B: Notochord is absent in non-chordates, as they do not possess this structure, which is a characteristic of chordates.



Step 3: Statement C: Central nervous system is dorsal. This is characteristic of chordates, but non-chordates do not have a dorsal central nervous system.



Step 4: Statement D: The heart, if present, is dorsal. This is a feature of non-chordates, where the heart is located on the dorsal side in certain species.



Step 5: Statement E: Post anal tail is absent in non-chordates, as this feature is typically found in chordates.



Thus, the correct matching is:
\[ B, D \& E only. \] Quick Tip: Non-chordates are a diverse group that lack a notochord, and their anatomy can differ greatly from chordates, such as having a ventral heart and absence of post-anal tails.

Step 1: The correct sequence for the catalytic cycle of an enzyme is as follows:



- E. Substrate binding to the active site.


- A. Substrate enzyme complex formation.


- D. Chemical bonds of the substrate broken.


- C. Release of products.


- B. Free enzyme ready to bind with another substrate.



Thus, the correct sequential order is:
\[ E, A, D, C, B. \] Quick Tip: The catalytic cycle of an enzyme follows a series of steps: first, the substrate binds to the enzyme's active site, then the enzyme catalyzes the reaction, breaking bonds, releasing products, and returning to its free form to bind to another substrate.

Step 1: Statement I: Bone marrow is indeed the main lymphoid organ where all blood cells, including lymphocytes, are produced. This statement is correct.



Step 2: Statement II: Both bone marrow and thymus play crucial roles in the development and maturation of T-lymphocytes. Bone marrow is the site of origin for T-lymphocytes, and the thymus provides the necessary environment for their maturation. This statement is also correct.



Thus, both statements are correct. Quick Tip: The bone marrow is responsible for the production of all blood cells, including lymphocytes, while the thymus is involved in the maturation of T-lymphocytes. These two organs are key components of the immune system.

Step 1: A. RNA polymerase III is responsible for the transcription of small RNA molecules, including tRNA and snRNAs, so it corresponds to IV. SnRNAs, tRNA.



Step 2: B. Termination of transcription often involves the Rho factor, which helps terminate transcription in prokaryotes, so it corresponds to III. Rho factor.



Step 3: C. Splicing of Exons involves snRNPs (small nuclear ribonucleoproteins), which are crucial for splicing, so it corresponds to I. snRNPs.



Step 4: D. TATA box is a promoter element found in eukaryotic genes that is involved in initiating transcription, so it corresponds to II. Promotor.



Thus, the correct matching is:
\[ A-IV, B-III, C-I, D-II. \] Quick Tip: RNA polymerase III transcribes tRNA and small RNA molecules, while splicing is carried out by snRNPs. The Rho factor plays a key role in termination of transcription in prokaryotes, and the TATA box is a promoter element for gene expression.

Step 1: A. Exophthalmic goiter is a condition caused by hyper secretion of thyroid hormone, leading to protruding eye balls, so it corresponds to III. Hyper secretion of thyroid hormone \& protruding eye balls.



Step 2: B. Acromegaly is caused by the excessive secretion of growth hormone, leading to abnormal growth, so it corresponds to IV. Excessive secretion of growth hormone.



Step 3: C. Cushing's syndrome is caused by excess secretion of cortisol, resulting in symptoms such as moon face and hyperglycemia, so it corresponds to I. Excess secretion of cortisol, moon face \& hyperglycemia.



Step 4: D. Cretinism is caused by the hypo-secretion of thyroid hormone, leading to stunted growth, so it corresponds to II. Hypo-secretion of thyroid hormone and stunted growth.



Thus, the correct matching is:
\[ A-III, B-IV, C-I, D-II. \] Quick Tip: The conditions related to hormone secretion often have distinct symptoms based on whether there is hyper-secretion or hypo-secretion, such as acromegaly (excessive growth hormone) and cretinism (lack of thyroid hormone).

Step 1: Statement 1: Juxta medullary nephrons do not outnumber cortical nephrons. In fact, cortical nephrons are more numerous than juxta medullary nephrons. Hence, this statement is incorrect.



Step 2: Statement 2: Juxta medullary nephrons are not located in the columns of Bertini. They are located at the boundary of the renal cortex and medulla. Therefore, this statement is also incorrect.



Step 3: Statement 3: The renal corpuscle of juxta medullary nephron lies in the outer cortex, not in the outer portion of the renal medulla. Hence, this statement is incorrect.



Step 4: Statement 4: The Loop of Henle of juxta medullary nephrons indeed runs deep into the renal medulla. This is the characteristic feature of juxta medullary nephrons that help in the concentration of urine.



Thus, the correct statement is:
\[ (4) Loop of Henle of juxta medullary nephron runs deep into medulla. \] Quick Tip: Juxta medullary nephrons are responsible for concentrating urine and have long loops of Henle that extend deep into the medulla, enabling the kidney to conserve water.

Step 1: Statement I: The cerebral hemispheres are indeed connected by the corpus callosum, which is a large bundle of nerve fibers. Therefore, Statement I is correct.



Step 2: Statement II: The brain stem does not include the cerebrum. The brain stem consists of the medulla oblongata, pons, and midbrain, not the cerebrum. Therefore, Statement II is incorrect.



Thus, the correct answer is:
\[ (4) Statement I is correct but Statement II is incorrect. \] Quick Tip: The brain stem consists of the medulla oblongata, pons, and midbrain. The corpus callosum connects the two cerebral hemispheres, facilitating communication between them.

Step 1: Statement I: Both mitochondria and chloroplasts are indeed double membrane-bound organelles. The outer membrane is permeable to small molecules, and the inner membrane is selective. Hence, Statement I is correct.



Step 2: Statement II: The inner membrane of mitochondria is relatively more impermeable than that of chloroplasts. However, it is more permeable compared to the outer membrane. The chloroplast inner membrane, in contrast, is selectively permeable but allows more movement of metabolites. Hence, Statement II is incorrect.



Thus, the correct answer is:
\[ (4) Statement I is correct but Statement II is incorrect. \] Quick Tip: Both mitochondria and chloroplasts play vital roles in energy conversion in cells, and their structure includes double membranes with specific permeability properties to regulate transport.