NEET 2024 Zoology Question Paper with Solutions PDF Q2 is available for download. NEET 2024 Q2 Zoology Question Paper comprises 50 MCQs out of which only 45 are to be attempted. NEET 2024 question Q2 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 Q2 using the links given below.

NEET 2024 Zoology Question Paper with Solutions PDF Q2

NEET 2024 Q2 Question Paper with Answer Key Download PDF Check Solution
Question 1:

Match List I with List II:
List I                   List II
A. Cocaine      I. Effective sedative in surgery
B. Heroin        II. Cannabis sativa
C. Morphine   III. Erythroxylum
D. Marijuana   IV. Papaver somniferum

Choose the correct answer from the options given below:

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

Cocaine is derived from Erythroxylum (III).
Heroin is derived from Papaver somniferum (IV).
Morphine is an effective sedative in surgery (I).
Marijuana is derived from Cannabis sativa (II). Quick Tip: Morphine and heroin are derived from the opium poppy, while marijuana comes from Cannabis sativa.


Question 2:

Match List I with List II:
 List I                        List II 
A. Down’s syndrome         I. 11th chromosome
B. \alpha-Thalassemia       II. ‘X’ chromosome
C. \beta-Thalassemia          III. 21st chromosome
D. Klinefelter’s syndrome   IV. 16th chromosome

Choose the correct answer from the options given below :

Correct Answer:
View Solution

Down’s syndrome is caused by a trisomy of the 21st chromosome (III).

α-Thalassemia is associated with mutations on the 16th chromosome (IV).

β-Thalassemia is caused by mutations on the 11th chromosome (I).

Klinefelter’s syndrome is linked to the ‘X’ chromosome (II). Quick Tip: Down’s syndrome is characterized by an extra chromosome 21. Klinefelter’s syndrome results from the presence of an extra X chromosome in males.


Question 3:

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


 

Muscle Name – Location

 

  • (1) (a) Smooth - Toes, (b) Skeletal – Legs, (c) Cardiac – Heart
  • (2) (a) Skeletal - Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart
  • (3) (a) Skeletal - Biceps, (b) Involuntary – Intestine, (c) Smooth – Heart
  • (4) (a) Involuntary – Nose tip, (b) Skeletal – Bone, (c) Cardiac – Heart
Correct Answer:
View Solution

Skeletal muscles (voluntary) are attached to bones like triceps.
Smooth muscles (involuntary) are found in organs like the stomach.
Cardiac muscles are located in the heart. Quick Tip: Muscles are classified into skeletal, smooth, and cardiac based on structure, control, and location.


Question 4:

Match List I with List II:
 List I                    List II 
A. Pterophyllum   I. Hag fish
B. Myxine              II. Saw fish
C. Pristis               III. Angel fish
D. Exocoetus        IV. Flying fish

Choose the correct answer from the options given below:

 

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

Pterophyllum is the Angel fish (III).
Myxine is the Hag fish (I).
Pristis is the Saw fish (II).
Exocoetus is the Flying fish (IV). Quick Tip: Fish species vary widely, and each has unique adaptations like flight or deep-sea living.


Question 5:

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

  • (1) Uterine fundus
  • (2) Isthmus
  • (3) Infundibulum
  • (4) Ampulla
Correct Answer:
View Solution

The Fallopian tube consists of components like the isthmus, infundibulum, and ampulla, but not the uterine fundus. Quick Tip: The Fallopian tube helps in transporting eggs from the ovaries to the uterus and is not associated with the uterine fundus.


Question 6:

Match List I with List II:
List I                       List II 
A. Pleurobrachia   I. Mollusca
B. Radula               II. Ctenophora
C. Stomochord      III. Osteichthyes
D. Air bladder        IV. Hemichordata

Choose the correct answer from the options given below:

 

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

Pleurobrachia is a ctenophore (II).
Radula is a characteristic of Mollusca (I).
Stomochord is a feature of Hemichordata (IV).
Air bladder is found in Osteichthyes (III). Quick Tip: Ctenophores are marine invertebrates, and the radula is a feeding organ found in mollusks.


Question 7:

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& D only
  • (2) A,B& E only
  • (3) B,C & E only
  • (4) C,D & E only
Correct Answer:
View Solution

Myasthenia gravis, rheumatoid arthritis, and systemic lupus erythematosus (SLE) are autoimmune disorders.
Gout and muscular dystrophy are not autoimmune disorders. Quick Tip: Autoimmune diseases occur when the immune system mistakenly attacks the body’s own cells.


Question 8:

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.

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

 

  • (1) Both Statement I and Statement II are true
  • (2) Both Statement I and Statement II are false
  • (3) Statement I is true but Statement II is false
  • (4) Statement I is false but Statement II is true
Correct Answer:
View Solution

Statement I is false: The descending limb of the loop of Henle is permeable to water and impermeable to electrolytes.


Statement II is also false: The proximal convoluted tubule is lined by cuboidal epithelium with a brush border, not simple columnar. Quick Tip: The loop of Henle plays a critical role in the concentration of urine and the proximal convoluted tubule aids in reabsorption.


Question 9:

Which of the following is not a steroid hormone?
 

  • (1) Cortisol
  • (2) Testosterone
  • (3) Progesterone
  • (4) Glucagon
Correct Answer:
View Solution

Glucagon is a peptide hormone, not a steroid hormone. Cortisol, testosterone, and progesterone are all steroid hormones, derived from cholesterol. Quick Tip: Steroid hormones are lipophilic and are derived from cholesterol. Peptide hormones like glucagon are water-soluble.


Question 10:

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.

Choose the correct answer from the options given below:

 

  • (1) Both Statement I and Statement II are true
  • (2) Both Statement I and Statement II are false
  • (3) Statement I is true but Statement II is false
  • (4) Statement I is false but Statement II is true
Correct Answer:
View Solution

Statement I is true because hymen can naturally tear due to physical activities or exercise, not necessarily the first coitus. Statement II is false because the hymen may tear due to various reasons, not only during the first sexual intercourse. Quick Tip: The hymen can tear due to activities like cycling, sports, or medical examination, not just during coitus.


Question 11:

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

  • (1) High pO2 and High pCO2
  • (2) High pO2 and Lesser H+ concentration
  • (3) Low pCO2 and High H+ concentration
  • (4) Low pCO2 and High temperature
Correct Answer:
View Solution

High partial pressure of oxygen (pO2) and lower concentration of hydrogen ions (H+) favour the formation of oxyhaemoglobin in the alveoli. Quick Tip: Oxyhaemoglobin is formed when oxygen binds to hemoglobin, primarily in the lungs where oxygen levels are high and pH is less acidic.


Question 12:

Match List I with List II:
 List I                             List II 
A. Axoneme                 I. Centriole
B. Cartwheel pattern   II. Cilia and flagella
C. Crista                       III. Chromosome
D. Satellite                   IV. Mitochondria
Choose the correct answer from the options given below:

 

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

Axoneme is associated with cilia and flagella (II).
Cartwheel pattern is found in centrioles (I).
Crista is a feature of mitochondria (IV).
Satellite is associated with chromosomes (III). Quick Tip: The structure and function of cell components like mitochondria, centrioles, and cilia are crucial for cellular processes.


Question 13:

The flippers of the Penguins and Dolphins are the example of the
 

  • (1) Adaptive radiation
  • (2) Natural selection
  • (3) Convergent evolution
  • (4) Divergent evolution
Correct Answer:
View Solution

The flippers of penguins and dolphins are examples of convergent evolution, where species from different evolutionary backgrounds evolve similar traits. Quick Tip: Convergent evolution results in the development of similar traits in unrelated species due to similar environmental pressures.


Question 14:

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

Homo habilis (A) was one of the earliest members of the genus Homo, followed by Homo erectus (D), then Homo neanderthalensis (C), and finally Homo sapiens (B). Quick Tip: Human evolution is a continuous process, with different species evolving over time.


Question 15:

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

  • (1) 5th segment
  • (2) 10th segment
  • (3) 8th and 9th segment
  • (4) 11th segment
Correct Answer:
View Solution

In cockroaches, anal cerci are present on the 10th segment in both sexes, which help in sensory functions. Quick Tip: The anal cerci in cockroaches play a role in detecting air currents and vibrations.


Question 16:

Which of the following statements is incorrect?
 

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

Bio-reactors are typically used for large-scale production of microorganisms, not just small-scale bacterial cultures. They are designed to maintain optimal conditions for high-yield production. Quick Tip: Bio-reactors are essential for large-scale fermentation processes, such as the production of antibiotics or biofuels.


Question 17:

Match List I with List II:
 List I              List II
A. Typhoid                I. Fungus
B. Leishmaniasis    II. Nematode
C. Ringworm           III. Protozoa
D. Filariasis             IV. Bacteria

Choose the correct answer from the options given below:

 

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

A. Typhoid is caused by \textit{Salmonella typhi, a bacteria.

B. Leishmaniasis is caused by \textit{Leishmania, a protozoa.

C. Ringworm is caused by a fungus.

D. Filariasis is caused by \textit{Wuchereria bancrofti, a nematode.

Thus, the correct match is: (2) A-IV, B-III, C-I, D-II. Quick Tip: Leishmaniasis and Filariasis are protozoal and nematode infections, respectively, while Ringworm and Typhoid are fungal and bacterial infections.


Question 18:

Consider the following statements:

A. Annelids are true coelomates

B. Poriferans are pseudocoelomates

C. Aschelminthes are acoelomates

D. Platyhelminthes are pseudocoelomates

Correct Answer:
View Solution

Annelids are true coelomates, meaning they have a body cavity completely lined with mesoderm (A).
Poriferans are neither coelomates nor pseudocoelomates, they lack a true body cavity.
Aschelminthes (roundworms) are pseudocoelomates, not acoelomates.
Platyhelminthes (flatworms) are acoelomates, lacking a body cavity. Quick Tip: Coelomates have a true coelom, pseudocoelomates have a false cavity, and acoelomates lack a body cavity.


Question 19:

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

  • (1) Coitus interruptus
  • (2) Periodic abstinence
  • (3) Lactational amenorrhea
  • (4) Vaults
Correct Answer:
View Solution

Vaults are not a natural or traditional contraceptive method. Coitus interruptus, periodic abstinence, and lactational amenorrhea are all traditional methods. Quick Tip: Traditional contraceptive methods are those that do not involve modern medical interventions but rely on behavior or biological processes.


Question 20:

Match List I with List II:
 List I                     List II 
A. Pons                   I. Provides additional space for Neurons, regulates posture and balance.
B. Hypothalamus   II. Controls respiration and gastric secretions.
C. Medulla              III. Connects different regions of the brain.
D. Cerebellum        IV. Neurosecretory cells

Choose the correct answer from the options given below:

 

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

Pons connects different regions of the brain (III).
Hypothalamus has neurosecretory cells (IV).
Medulla controls respiration and gastric secretions (II).
Cerebellum provides additional space for neurons, regulates posture and balance (I). Quick Tip: The pons and medulla are part of the brainstem, which controls vital functions like respiration and heartbeat.


Question 21:

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

  • (1) Genetic recombination
  • (2) Genetic drift
  • (3) Gene migration
  • (4) Constant gene pool
Correct Answer:
View Solution

A constant gene pool means no changes in allele frequencies, so it does not affect the Hardy-Weinberg equilibrium. Genetic recombination, genetic drift, and gene migration all influence allele frequencies. Quick Tip: The Hardy-Weinberg equilibrium assumes that no evolutionary forces are acting on the population (no genetic drift, migration, mutation, or natural selection).


Question 22:

Match List I with List II:
 List I                               List II 
A. \alpha-I antitrypsin                   I. Cotton bollworm
B. Cry IAb                                      II. ADA deficiency
C. Cry IAc                                       III. Emphysema
D. Enzyme replacement therapy   IV. Corn borer

Choose the correct answer from the options given below:

 

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

\(\alpha\)-I antitrypsin deficiency leads to emphysema (III).
Cry IAb is associated with cotton bollworm (I).
Cry IAc is associated with corn borer (IV).
Enzyme replacement therapy is used for ADA deficiency (II). Quick Tip: Cry proteins are insecticidal and are used in genetically modified crops to protect them from pests like bollworm and corn borer.


Question 23:

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

A. AV bundle

B. Purkinje fibers

C. AV node

D. Bundle branches

E. SA node


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

 

  • (1) E-C-A-D-B
  • (2) A-E-C-B-D
  • (3) B-D-E-C-A
  • (4) E-A-D-B-C
Correct Answer:
View Solution

The correct sequence for conduction of an action potential in the heart is: \[ SA node \rightarrow AV node \rightarrow AV bundle \rightarrow Bundle branches \rightarrow Purkinje fibers \]
Thus, the correct sequence is: (1) E-C-A-D-B. Quick Tip: The action potential in the heart follows a specific sequence to ensure coordinated contraction: SA node → AV node → AV bundle → Bundle branches → Purkinje fibers.


Question 24:

The “Ti plasmid” of Agrobacterium tumefaciens stands for
 

  • (1) Tumour inhibiting plasmid
  • (2) Tumor independent plasmid
  • (3) Tumor inducing plasmid
  • (4) Temperature independent plasmid
Correct Answer:
View Solution

The Ti plasmid in Agrobacterium tumefaciens induces the formation of tumors in plants by transferring part of the plasmid DNA to the plant genome. Quick Tip: The Ti plasmid is a key tool in plant genetic engineering, used to insert foreign genes into plant cells.


Question 25:

Match List I with List II:
 List I                                              List II 
A. Expiratory capacity                 I. Expiratory reserve volume + Tidal volume + Inspiratory volume
B. Functional residual capacity   II. Tidal volume + Expiratory reserve volume
C. Vital capacity                            III. Tidal volume + Inspiratory reserve volume
D. Inspiratory capacity                 IV. Expiratory reserve volume + Residual volume


Choose the correct answer from the options given below:

 

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

Expiratory capacity = Tidal volume + Expiratory reserve volume (II).
Functional residual capacity = Expiratory reserve volume + Residual volume (IV).
Vital capacity = Expiratory reserve volume + Tidal volume + Inspiratory reserve volume (I).
Inspiratory capacity = Tidal volume + Inspiratory reserve volume (III). Quick Tip: Vital capacity is the maximum amount of air a person can exhale after a maximum inhalation.


Question 26:

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

The correct sequence of stages in cell division is:

1. Gap 1 phase (E)

2. Synthesis phase (C)

3. Gap 2 phase (A)

4. Karyokinesis (D)

5. Cytokinesis (B)
Quick Tip: Cell division includes interphase (G1, S, G2) and mitosis (karyokinesis and cytokinesis).


Question 27:

Match List I with List II:
 List I                                 List II 
 A. Fibrous joints            I. Adjacent vertebrae, limited movement
B. Cartilaginous joints   II. Humerus and Pectoral girdle, rotational movement
C. Hinge joints                III. Skull, don’t allow any movement
D. Ball and socket joints IV. Knee, help in locomotion

Choose the correct answer from the options given below:

 

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

Fibrous joints are immovable like those in the skull (III).
Cartilaginous joints are slightly movable, such as between adjacent vertebrae (I).
Hinge joints allow movement like in the knee (IV).
Ball and socket joints allow rotational movement like in the humerus and pectoral girdle (II). Quick Tip: Joints in the human body are classified based on their structure and the type of movement they allow.


Question 28:

Given below are two statements:

Statement I: FSH acts upon ovarian follicles in female and Leydig cells in male.

Statement II: Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human beings.

Choose the correct answer from the options given below:

 

  • (1) Both Statement I and Statement II are true and R is the correct explanation of A
  • (2) Both Statement I and Statement II are true but R is NOT the correct explanation of A
  • (3) Statement I is true but Statement II is false
  • (4) Statement I is false but Statement II is true
Correct Answer:
View Solution

Statement I is false because FSH acts on granulosa cells in females and Sertoli cells in males, not on Leydig cells. Statement II is true as estrogen is secreted by growing ovarian follicles and androgens by Leydig cells. Quick Tip: FSH stimulates the granulosa cells in females and Sertoli cells in males, while LH stimulates the Leydig cells in males.


Question 29:

Match List I with List II:

Correct Answer:
View Solution

Diakinesis involves terminalisation of chiasmata (II).
Pachytene is marked by the appearance of recombination nodules (IV).
Zygotene is when the synaptonemal complex forms (I).
Leptotene is characterized by chromosomes appearing as thin threads (III). Quick Tip: Prophase I of meiosis has distinct stages: leptotene, zygotene, pachytene, diplotene, and diakinesis.


Question 30:

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


 

  • (1) The gene ‘X’ is responsible for resistance to antibiotics and ‘Y’ for protein involved in the replication of Plasmid.
  • (2) The gene ‘X’ is responsible for controlling the copy number of the linked DNA and ‘Y’ for protein involved in the replication of Plasmid.
  • (3) The gene ‘X’ is for protein involved in replication of Plasmid and ‘Y’ for resistance to antibiotics.
  • (4) Gene ‘X’ is responsible for recognition sites and ‘Y’ is responsible for antibiotic resistance.
Correct Answer:
View Solution

Gene 'X' in the plasmid pBR322 is responsible for controlling the copy number of the plasmid, while Gene 'Y' is involved in the replication process. Quick Tip: pBR322 is a commonly used plasmid vector in genetic engineering that contains antibiotic resistance genes and replication control mechanisms.


Question 31:

Match List I with List II:
 List I                       List II 
A. Common cold   I. Plasmodium
B. Haemozoin       II. Typhoid
C. Widal test         III. Rhinoviruses
D. Allergy             IV. Dust mites

Choose the correct answer from the options given below :

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

Common cold is caused by Rhinoviruses (III).
Haemozoin is released in blood due to ruptured RBCs after Plasmodium infection (I).
Widal test is used to confirm typhoid fever (II).
Allergy is caused due to dust mites (IV). Quick Tip: Rhinoviruses are the main cause of common cold, while Haemozoin is produced in Plasmodium infection.


Question 32:

Match List I with List II:
 List I                                      List II
A. Non-medicated IUD          I. Multiload 375
B. Copper releasing IUD       II. Progestogens
C. Hormone releasing IUD   III. Lippes loop
D. Implants                            IV. LNG-20

Choose the correct answer from the options given below:

 

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

Non-medicated IUD is Lippes loop (III).
Copper releasing IUD is Multiload 375 (I).
Hormone releasing IUD is LNG-20 (IV).
Implants contain Progestogens (II). Quick Tip: IUDs (Intrauterine Devices) are either medicated or non-medicated and work by releasing copper or hormones to prevent pregnancy.


Question 33:

Given below are two statements:
One is labelled as Assertion A and the other is labelled as Reason R:


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


Reason R: Colostrum contains several antibodies absolutely essential to develop resistance for the new born 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 and R is the correct explanation of A
  • (2) Both A and R are correct but R is NOT the correct explanation of A
  • (3) A is correct but R is not correct
  • (4) A is not correct but R is correct
Correct Answer:
View Solution

Breast-feeding is essential for infant health, and colostrum contains antibodies that help develop resistance against infections, making both statements true. The reason correctly explains the assertion. Quick Tip: Breast-feeding provides crucial nutrients and antibodies through colostrum, which is the first milk produced after birth.


Question 34:

Match List I with List II:
List I                           List II
A. Lipase              I. Peptide bond
B. Nuclease         II. Ester bond
C. Protease         III. Glycosidic bond
D. Amylase          IV. Phosphodiester bond

Choose the correct answer from the options given below:

 

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

Lipase acts on ester bonds (II).
Nuclease acts on phosphodiester bonds (IV).
Protease breaks peptide bonds (I).
Amylase acts on glycosidic bonds (III). Quick Tip: Lipase, Nuclease, Protease, and Amylase are enzymes that break down different bonds in molecules like fats, nucleic acids, proteins, and carbohydrates.


Question 35:

Which one is the correct product of DNA dependent RNA polymerase to the given template?
\[ 3’ TACATGGCAAATATCCATTCA 5’ \]

 

  • (1) 5’ AUGUACCGUUUAUAGGUAAGU 3’
  • (2) 5’ AUGUAAAGUUUAUAGGUAAGU 3’
  • (3) 5’ AUGUACCGUUUAUAGGGAAGU 3’
  • (4) 5’ ATGTACCGTTTATAGGTAAGT 3’
Correct Answer:
View Solution

The correct RNA sequence transcribed from the template strand is 5' AUGUACCGUUUAUAGGUAAGU 3'. RNA polymerase synthesizes the RNA strand in the 5’ → 3’ direction, complementary to the DNA template. Quick Tip: DNA-dependent RNA polymerase uses the DNA template strand to synthesize an RNA strand complementary to the template.


Question 36:

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

\begin{table[h!]

List I & List II

A. The structures used for storing of food & I. Gizzard

B. Ring of 6-8 blind tubules at junction of foregut and midgut. & II. Gastric Caeca

C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut. & III. Malpighian tubules

D. The structures used for grinding the food. & IV. Crop

Choose the correct answer from the options given below:

 

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

A. The structures used for storing of food: The crop is the part of the cockroach's digestive system responsible for storing food.


B. Ring of 6-8 blind tubules at junction of foregut and midgut: These are the gastric caeca, which help in digestion by secreting digestive enzymes.


C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut: These are the Malpighian tubules, responsible for excretion and osmoregulation.


D. The structures used for grinding the food: The gizzard grinds food, breaking down particles mechanically. Quick Tip: In cockroaches, the crop stores food, the gastric caeca aids in digestion, the gizzard grinds food, and the Malpighian tubules are responsible for excretion and waste removal.


Question 37:

Match List I with List II:

List I                           List II
 
A. RNA polymerase III   I. snRNPs

B. Termination of transcription   II. Promotor

C. Splicing of Exons   III. Rho factor

D. TATA box & IV. SnRNAs, tRNA

Choose the correct answer from the options given below:

 

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

- A. RNA polymerase III: It synthesizes tRNA and small RNAs such as snRNAs.
- B. Termination of transcription: Rho factor is involved in terminating transcription in prokaryotes.
- C. Splicing of Exons: snRNPs (small nuclear ribonucleoproteins) are involved in splicing introns and exons in eukaryotic mRNA processing.
- D. TATA box: The TATA box is a promoter region that plays a role in the initiation of transcription in eukaryotes. Quick Tip: RNA polymerase III is responsible for the synthesis of small RNAs such as tRNA, while the TATA box in the promoter region is essential for the initiation of transcription in eukaryotes.


Question 38:

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.

  • (1) Both Statement I and Statement II are correct.
  • (2) Both Statement I and Statement II are incorrect.
  • (3) Statement I is correct but Statement II is incorrect.
  • (4) Statement I is incorrect but Statement II is correct.
Correct Answer:
View Solution

Statement A is incorrect because the pharynx is not perforated by gill slits in all non-chordates; it is characteristic of chordates.


Statement B is correct as non-chordates lack a notochord.


Statement C is incorrect because the central nervous system in non-chordates is typically ventral.


Statement D is correct as the heart, if present, is dorsal in non-chordates.


Statement E is correct as many non-chordates lack a post-anal tail. Quick Tip: Non-chordates lack a notochord and the dorsal position of the heart, unlike chordates.


Question 39:

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 correct.
  • (2) Both Statement I and Statement II are incorrect.
  • (3) Statement I is correct but Statement II is incorrect.
  • (4) Statement I is incorrect but Statement II is correct.
Correct Answer:
View Solution

Statement I is correct because the cerebral hemispheres are indeed connected by the corpus callosum, which allows communication between the two hemispheres.


Statement II is incorrect because the brainstem consists of the medulla oblongata, pons, and midbrain, not the cerebrum. Quick Tip: The corpus callosum is essential for the exchange of information between the two cerebral hemispheres.


Question 40:

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:
\begin{flushleft

  • (1) E, A, D, C, B
  • (2) A, E, B, D, C
  • (3) B, A, C, D, E
  • (4) E, D, C, B, A
Correct Answer:
View Solution

-E. The substrate binds to the active site of the enzyme, initiating the catalytic process.


-A. The enzyme-substrate complex is formed after the substrate binds to the active site.


-D. The enzyme catalyzes the breaking of chemical bonds in the substrate, facilitating the transition state.


-C. The product is released from the enzyme after the reaction.


-B. The enzyme is now free and ready to bind with a new substrate. Quick Tip: The enzyme catalytic cycle involves the binding of substrate, conversion into products, and release. The enzyme is reused in subsequent cycles.


Question 41:

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.

Correct Answer:
View Solution

Statement I is false because Gause's competitive exclusion principle states that two species competing for the same limiting resource cannot coexist indefinitely, not necessarily for different resources.


Statement II is true as, according to Gause's principle, the inferior competitor is often eliminated when resources are limited. Quick Tip: Gause's principle highlights that no two species can occupy the same niche for long if they are competing for the same resources.


Question 42:

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



 

  • (1) FSH, Leydig cells, Sertoli cells, spermiogenesis.
  • (2) ICSH, Interstitial cells, Leydig cells, spermiogenesis.
  • (3) FSH, Sertoli cells, Leydig cells, spermatogenesis.
  • (4) ICSH, Leydig cells, Sertoli cells, spermatogenesis.
Correct Answer:
View Solution

FSH (Follicle Stimulating Hormone) stimulates Sertoli cells, which support the process of spermatogenesis.


Leydig cells secrete testosterone, which is essential for spermatogenesis.


Spermiogenesis refers to the maturation of spermatids into sperm. Quick Tip: Spermatogenesis involves the transformation of spermatogonia into mature sperm cells through a series of stages, including spermiogenesis.


Question 43:

Match List I with List II:

List I                   List II
 
A. P wave   I. Heart muscles are electrically silent.
 
B. QRS complex   II. Depolarisation of ventricles.
 
C. T wave  III. Depolarisation of atria.
 
D. T-P gap & IV. Repolarisation of ventricles.

Choose the correct answer from the options given below:

 

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



A. P wave: Represents the depolarisation of the atria.


B. QRS complex: Represents the depolarisation of ventricles.


C. T wave: Represents the repolarisation of ventricles.


D. T-P gap: Indicates a period where heart muscles are electrically silent, representing no electrical activity. Quick Tip: The P wave, QRS complex, and T wave are key components of the electrocardiogram (ECG) that correspond to different phases of heart muscle electrical activity.


Question 44:

Choose the correct statement given below regarding juxta medullary nephron.

A. Juxta medullary nephrons are located in the columns of Bertini.

B. Renal corpuscle of juxta medullary nephron lies in the outer portion of the renal medulla.

C. Loop of Henle of juxta medullary nephron runs deep into medulla.

D. Juxta medullary nephrons outnumber the cortical nephrons.

Correct Answer:
View Solution

-A. Incorrect: Juxta medullary nephrons are located at the junction of the cortex and medulla, not in the columns of Bertini.
-B. Incorrect: The renal corpuscle of juxta medullary nephrons is located in the outer cortex, not in the outer medulla.
-C. Correct: The loop of Henle of juxta medullary nephrons extends deep into the medulla, which helps in concentrating the urine.
-D. Incorrect: Juxta medullary nephrons are fewer in number compared to cortical nephrons. Quick Tip: The juxta medullary nephrons are vital for water conservation as their deep loop of Henle helps in the concentration of urine.


Question 45:

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 micro environments for the development and maturation of T-lymphocytes.


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 correct.
  • (2) Both Statement I and Statement II are incorrect.
  • (3) Statement I is correct but Statement II is incorrect.
  • (4) Statement I is incorrect but Statement II is correct.
Correct Answer:
View Solution

Statement I: Correct. Bone marrow is indeed the primary lymphoid organ responsible for the production of blood cells, including lymphocytes.


Statement II: Correct. The bone marrow produces T-lymphocytes, which mature in the thymus. Both organs provide the necessary environments for T-cell development. Quick Tip: Bone marrow is the site of hematopoiesis (blood cell production), while the thymus is responsible for the maturation of T-lymphocytes, essential for the adaptive immune response.


Question 46:

Match List I with List II:

List I                                       List II

A. Unicellular glandular epithelium & I. Salivary glands
 
B. Compound epithelium & II. Pancreas
 
C. Multicellular glandular epithelium & III. Goblet cells of alimentary canal
 
D. Endocrine glandular epithelium & IV. Moist surface of buccal cavity

Choose the correct answer from the options given below:
 

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

A. Unicellular glandular epithelium: Goblet cells in the alimentary canal secrete mucus.


B. Compound epithelium: Found on the moist surface of the buccal cavity.


C. Multicellular glandular epithelium: Includes salivary glands which secrete saliva.


D. Endocrine glandular epithelium: Found in the pancreas, involved in hormone secretion. Quick Tip: Glandular epithelium includes unicellular (e.g., goblet cells), multicellular (e.g., salivary glands), and endocrine glands that secrete hormones (e.g., pancreas).


Question 47:

Match List I with List II:

List I                  List II
 
A. Mesozoic Era & I. Lower invertebrates
 
B. Proterozoic Era & II. Fish \& Amphibia
 
C. Cenozoic Era & III. Birds \& Reptiles
 
D. Paleozoic Era & IV. Mammals

Choose the correct answer from the options given below:

 

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

A. Mesozoic Era: Known as the age of reptiles and birds, including dinosaurs.


B. Proterozoic Era: Dominated by lower invertebrates.


C. Cenozoic Era: Age of mammals, characterized by their rise.


D. Paleozoic Era: Dominated by fish and amphibians. Quick Tip: The Mesozoic era is known for dinosaurs, the Cenozoic era for mammals, and the Paleozoic era for early vertebrates like fish and amphibians.


Question 48:

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^Bi/I^Ai/ii

B. I^BI^B/I^AI^A/ii

C. I^AI^B/iI^A/I^Bi

D. I^Ai/I^Bi/I^Ai


Choose the most appropriate answer from the options given below :

  • (1) A only
  • (2) B only
  • (3) C and B only
  • (4) D and E only
Correct Answer:
View Solution

The father with blood group B+ could have a genotype IBi or IBIB, while the mother with blood group A+ could have IAi or IAIA.


The child with blood group O+ must inherit an i allele from both parents, making the father’s genotype IBi and the mother’s genotype IAi. Quick Tip: In the ABO blood group system, O blood type is recessive, and both parents must carry the i allele to produce a child with blood group O.



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