AIPMT 1993 Biology Question Paper with Answer and Solution

228 QuestionsEnglishWith Solutions

BiologyQ51140 of 228 questions

Page 2 of 3 · English

51
BiologyMediumMCQAIPMT · 1993
$CO_2$ is transported in blood primarily as:
A
Carbonates
B
Bicarbonates
C
Carbaminohaemoglobin
D
Carboxyhaemoglobin

Solution

(B) Carbon dioxide $(CO_2)$ is transported in the blood in three forms:
$1$. As dissolved gas in plasma (about $7\%$).
$2$. As carbaminohaemoglobin,where $CO_2$ binds to the amino groups of haemoglobin (about $20-25\%$).
$3$. As bicarbonates $(HCO_3^-)$,which is the primary mode of transport (about $70\%$). The enzyme carbonic anhydrase in red blood cells facilitates the conversion of $CO_2$ and $H_2O$ into $HCO_3^-$ and $H^+$ ions.
52
BiologyMediumMCQAIPMT · 1993
The tricuspid valve is present at the origin of
A
Carotid arch
B
Pulmonary arch
C
Truncus arteriosus
D
Systemic arch

Solution

(C) In the heart of amphibians and some reptiles,the $Truncus \ arteriosus$ is a large vessel that arises from the ventricle. The base of this vessel contains valves that prevent the backflow of blood. Specifically,the $Truncus \ arteriosus$ is guarded by spiral valves or tricuspid valves at its origin to regulate blood flow into the systemic and pulmonary arches.
53
BiologyEasyMCQAIPMT · 1993
The atrio-ventricular valves of the heart are prevented from turning inside out by tough strands of connective tissue called:
A
Tendinous cords
B
Tricuspid
C
Pocket valve
D
Mitral valve

Solution

(A) The atrio-ventricular valves (tricuspid and bicuspid valves) are attached to the papillary muscles of the heart wall by tough,fibrous strands of connective tissue known as $Chordae$ $tendineae$ or $Tendinous$ $cords$.
These structures prevent the valves from prolapsing or turning inside out into the atria during ventricular systole (contraction).
54
BiologyMediumMCQAIPMT · 1993
The process of blood clot formation within the circulatory system is known as:
A
Thrombosis
B
Thrombocytes
C
Thrombin
D
Thrombocytopenia

Solution

(A) . Thrombosis is the process of formation of a blood clot (thrombus) inside a blood vessel,which obstructs the flow of blood through the circulatory system.
$B$. Thrombocytes are blood platelets involved in clotting,not the process itself.
$C$. Thrombin is an enzyme that converts fibrinogen to fibrin during the clotting cascade.
$D$. Thrombocytopenia is a condition characterized by an abnormally low level of platelets in the blood.
55
BiologyMediumMCQAIPMT · 1993
$A$ vein differs from an artery in having
A
Narrow lumen
B
Strong cuticular and muscular wall
C
Valves to control direction of flow
D
Dark pigmented wall

Solution

(C) Veins differ from arteries primarily because they contain valves.
These valves are essential to prevent the backflow of blood,as blood in the veins flows at a lower pressure compared to the high-pressure flow in arteries.
Arteries typically have thicker,more muscular walls and a narrower lumen,whereas veins have thinner walls and a wider lumen.
56
BiologyMediumMCQAIPMT · 1993
The trilobed valve present between the right atrium and the right ventricle in the mammalian heart is known as:
A
Triac
B
Triad
C
Tricuspid valve
D
Trigeminal

Solution

(C) In the mammalian heart,the right atrium and the right ventricle are separated by a valve consisting of three flaps or cusps.
This valve is known as the $Tricuspid$ valve.
Its primary function is to prevent the backflow of blood from the right ventricle into the right atrium during ventricular systole.
57
BiologyEasyMCQAIPMT · 1993
The special fibrous joint occurring exclusively in the skull,which is the tightest type of joint,is:
A
Suspensorium
B
Suspensory ligament
C
Suture
D
Occipital

Solution

(C) The correct answer is $C$.
Sutures are a type of fibrous joint found exclusively in the skull.
These joints are characterized by their lack of movement and are considered the tightest type of joint,as they fuse the flat bones of the cranium together to provide protection for the brain.
58
BiologyMediumMCQAIPMT · 1993
The vertebra which bears the whole weight of the skull is
A
Axis
B
Sacral
C
Cervical
D
Atlas

Solution

(D) The correct answer is $D$.
According to Greek mythology,the god Atlas supports the world on his shoulders.
Since the skull acts as the globe (or world) resting on the vertebral column,the first cervical vertebra,which supports the entire weight of the skull,is named the Atlas vertebra.
This vertebra articulates with the occipital condyles of the skull to allow for nodding movements.
59
BiologyMediumMCQAIPMT · 1993
$A$ shallow depression in the scapula which receives the head of the upper arm bone is known as the
A
Acetabulum
B
Neural arch
C
Glenoid cavity
D
None of the above

Solution

(C) The scapula is a large,triangular,flat bone situated in the dorsal part of the thorax between the $2^{nd}$ and the $7^{th}$ ribs.
It contains a shallow articular surface called the glenoid cavity.
The glenoid cavity articulates with the head of the humerus (the upper arm bone) to form the shoulder joint.
60
BiologyMediumMCQAIPMT · 1993
The cavity between the coracoid and scapula is known as:
A
Vomer
B
Acetabulum
C
Glenoid
D
Orbital

Solution

(C) The scapula is a large triangular flat bone situated in the dorsal part of the thorax between the second and the seventh ribs. The dorsal,flat,triangular body of the scapula has a slightly elevated ridge called the spine which projects as a flat,expanded process called the acromion. Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint. The coracoid process is a small hook-like structure on the lateral edge of the superior anterior portion of the scapula.
61
BiologyEasyMCQAIPMT · 1993
Metacoel is the cavity in the
A
Cerebral hemispheres
B
Diencephalon
C
Cerebellum
D
Medulla oblongata

Solution

(D) The brain contains several interconnected cavities called ventricles.
$I$. The ventricles of the cerebral hemispheres are known as paracoels or lateral ventricles.
$II$. The cavity of the diencephalon is known as the diocoel or third ventricle.
$III$. The cavity of the medulla oblongata is known as the metacoel or fourth ventricle.
Therefore,the correct answer is $D$.
62
BiologyMediumMCQAIPMT · 1993
The control of blood sugar level,osmoregulation,and thermoregulation are the functions of:
A
Medulla oblongata
B
Cerebellum
C
Hypothalamus
D
Diencephalon

Solution

(C) The $Hypothalamus$ is a vital part of the $Diencephalon$ located at the base of the brain.
It contains a number of centers which control body temperature,urge for eating and drinking (osmoregulation),and emotional reactions.
It also plays a crucial role in the regulation of blood sugar levels through the control of the autonomic nervous system and the endocrine system (via the pituitary gland).
63
BiologyEasyMCQAIPMT · 1993
The $Malleus$ is present in the:
A
Inner ear
B
Outer ear
C
Middle ear
D
Eye

Solution

(C) The middle ear contains three tiny bones called ear ossicles,which are the $Malleus$ (hammer),$Incus$ (anvil),and $Stapes$ (stirrup). Therefore,the $Malleus$ is located in the middle ear.
64
BiologyEasyMCQAIPMT · 1993
Vitreous humour is seen in
A
Ear
B
Eye
C
Brain
D
Bone marrow

Solution

(B) The Vitreous humour is a transparent, jelly-like substance that fills the space between the lens and the retina of the eye.
It helps in maintaining the shape of the eyeball and supports the retina.
Therefore, it is a characteristic component of the eye.
65
BiologyEasyMCQAIPMT · 1993
The central opening of the iris is called:
A
Pupil
B
Cornea
C
Lens
D
Fovea centralis

Solution

(A) The central aperture or opening of the iris is known as the pupil.
It regulates the amount of light entering the eye by adjusting its size through the contraction or relaxation of the iris muscles.
66
BiologyMediumMCQAIPMT · 1993
The intermediate lobe of the pituitary gland produces a secretion which causes a dramatic darkening of the skin of many fishes,amphibians and reptiles. It is
A
Adrenocorticotropic hormone $(ACTH)$
B
Follicle stimulating hormone $(FSH)$
C
Melanocyte stimulating hormone $(MSH)$
D
Luteinizing hormone $(LH)$

Solution

(C) The intermediate lobe of the pituitary gland,also known as the pars intermedia,secretes Melanocyte Stimulating Hormone $(MSH)$.
$MSH$ acts on melanocytes (melanin-containing cells) and regulates skin pigmentation.
In many fishes,amphibians,and reptiles,$MSH$ stimulates the dispersion of melanin granules within melanocytes,which results in the darkening of the skin.
67
BiologyEasyMCQAIPMT · 1993
The disease caused by the deficiency of iodine is:
A
Goitre
B
Myxedema
C
Cretinism
D
Tetany

Solution

(A) Simple goitre is caused by a lower intake of iodine through the diet.
Goitre is the swelling of the neck due to the enlargement of the thyroid gland.
Iodine is essential for the synthesis of thyroid hormones ($T_3$ and $T_4$).
Deficiency leads to hypothyroidism,resulting in the compensatory enlargement of the thyroid gland.
68
BiologyMediumMCQAIPMT · 1993
Cushing's syndrome and myxoedema are associated with these glands respectively:
A
Thyroid,adrenal
B
Adrenal,thyroid
C
Parathyroid,thyroid
D
Adrenal,pituitary

Solution

(B) $Cushing's$ syndrome is caused by the hypersecretion of glucocorticoids (like cortisol) from the adrenal cortex of the adrenal gland.
Myxoedema is a condition caused by the hyposecretion of thyroid hormones (thyroxine) from the thyroid gland.
Therefore,$Cushing's$ syndrome is associated with the adrenal gland and myxoedema is associated with the thyroid gland.
Thus,the correct sequence is adrenal and thyroid.
69
BiologyMediumMCQAIPMT · 1993
Glucagon hormone is secreted by
A
Pituitary
B
Adrenal
C
Beta cells of islets of Langerhans
D
Alpha cells of islets of Langerhans

Solution

(D) Glucagon is a hyperglycaemic or diabetogenic polypeptide hormone secreted by $\alpha$-cells of the islets of Langerhans in the pancreas.
Its primary function is to increase blood glucose levels by stimulating glycogenolysis and gluconeogenesis in the liver.
Its secretion is stimulated by low blood sugar levels (hypoglycaemia).
70
BiologyMediumMCQAIPMT · 1993
Which of the following hormones governs the metabolism of carbohydrates?
A
Corticoids
B
Glucagon
C
Insulin
D
Glucagon and insulin

Solution

(D) Carbohydrate metabolism in the body is primarily regulated by the hormones secreted by the pancreas,namely $Glucagon$ and $Insulin$.
$Glucagon$ is a hyperglycemic hormone that increases blood glucose levels by stimulating glycogenolysis and gluconeogenesis.
$Insulin$ is a hypoglycemic hormone that decreases blood glucose levels by facilitating the uptake of glucose by cells and promoting glycogenesis.
Together,these two hormones maintain glucose homeostasis in the blood.
71
BiologyMediumMCQAIPMT · 1993
$A$ polypeptide secreted into the blood by the cells in the stomach wall,which stimulates the production of $HCl$ by the parietal cells of the stomach,is:
A
Gastrin
B
Secretin
C
Pancreozymin
D
Renin

Solution

(A) The hormone $Gastrin$ is a polypeptide hormone secreted by the $G$-cells located in the pyloric region of the stomach wall.
It is released into the bloodstream and acts on the gastric glands.
Its primary function is to stimulate the parietal cells (also known as oxyntic cells) of the stomach to secrete $HCl$ and pepsinogen,which are essential for the digestion of food.
72
BiologyEasyMCQAIPMT · 1993
The development of the structure and shape of an organism is known as:
A
Morphology
B
Multiplication
C
Morphogenesis
D
Budding

Solution

(C) Morphogenesis is the biological process that causes an organism to develop its shape. It involves the controlled growth and differentiation of cells and tissues during embryonic development to form specific structures and organs.
73
BiologyEasyMCQAIPMT · 1993
Who was the scientist to introduce $ABO$ blood groups?
A
Wiener
B
Levine
C
Fisher
D
Landsteiner

Solution

(D) The $ABO$ blood group system was discovered by Karl Landsteiner in $1900$.
He identified the presence of antigens on the surface of red blood cells.
For this groundbreaking discovery,he was awarded the Nobel Prize in Physiology or Medicine in $1930$.
74
BiologyMediumMCQAIPMT · 1993
Disorders of amino acid metabolism result in which of the following conditions?
A
Alkaptonuria
B
Phenylketonuria
C
Albinism
D
All of the above

Solution

(D) Disorders of amino acid metabolism are genetic conditions caused by the deficiency of specific enzymes required for the breakdown of amino acids.
$1$. $Alkaptonuria$ is caused by a deficiency of the enzyme homogentisate $1,2$-dioxygenase, leading to the accumulation of homogentisic acid.
$2$. $Phenylketonuria$ $(PKU)$ is caused by a deficiency of the enzyme phenylalanine hydroxylase, which prevents the conversion of phenylalanine to tyrosine.
$3$. $Albinism$ is often caused by a defect in the enzyme tyrosinase, which is required for the synthesis of melanin from tyrosine.
Since all three conditions are metabolic disorders related to amino acid pathways, the correct answer is $All \text{ of the above}$.
75
BiologyMediumMCQAIPMT · 1993
An animal that is initially male and then changes to female is called:
A
Protandrous
B
Apomixis
C
Profixation
D
None of these

Solution

(A) In biology,organisms that exhibit sequential hermaphroditism where they function as males first and later transition into females are known as $Protandrous$.
$Protandry$ is a type of sequential hermaphroditism commonly observed in various invertebrates and some fish species.
Since the question describes an organism that starts as male and changes to female,the correct term is $Protandrous$.
76
BiologyEasyMCQAIPMT · 1993
Depending upon size and centromere position,the $46$ chromosomes have been divided into a number of groups.
A
$6$
B
$5$
C
$7$
D
$10$

Solution

(C) The human chromosomes are classified based on their size and the position of the centromere.
According to the Denver classification system,the $46$ human chromosomes are arranged in descending order of size and grouped into $7$ groups,labeled from $A$ to $G$.
These groups are categorized based on the centromere position (metacentric,submetacentric,and acrocentric) and the relative length of the chromosome arms.
77
BiologyMediumMCQAIPMT · 1993
The grouping of human chromosomes is based on
A
Secondary constrictions alone
B
Dot-like satellites alone
C
Banding patterns alone
D
All the above

Solution

(D) The classification and grouping of human chromosomes (karyotyping) are primarily based on their size,the position of the centromere,and the specific banding patterns observed after staining (such as $G$-banding). While secondary constrictions and satellites are features used to identify specific chromosomes,the systematic grouping relies on a combination of these morphological characteristics. Therefore,the grouping is based on all the mentioned features.
78
BiologyMediumMCQAIPMT · 1993
The genophore,bacterial genome,or nucleoid is made up of ...........
A
Histones and non-histones
B
$RNA$ and histones
C
$A$ single double-stranded $DNA$
D
$A$ single-stranded $DNA$

Solution

(C) The bacterial genome,also known as the genophore or nucleoid,consists of a single,circular,double-stranded $DNA$ molecule. Unlike eukaryotic cells,prokaryotic $DNA$ is not enclosed within a nuclear membrane and lacks histone proteins,which are characteristic of eukaryotic chromatin. Therefore,the correct description is a single double-stranded $DNA$.
79
BiologyMediumMCQAIPMT · 1993
The protonema is formed during the life cycle of $..........$.
A
Riccia
B
Funaria
C
Chlamydomonas
D
Spirogyra

Solution

(B) The life cycle of mosses (Bryophyta) includes a juvenile stage known as the protonema.
$1$. The protonema is a creeping,green,branched,and frequently filamentous stage.
$2$. It develops directly from a spore.
$3$. Among the given options,$Funaria$ is a moss,whereas $Riccia$ is a liverwort,and $Chlamydomonas$ and $Spirogyra$ are algae.
$4$. Therefore,the protonema is characteristic of the life cycle of $Funaria$.
80
BiologyEasyMCQAIPMT · 1993
Pyrenoids are centers for the formation of . . . . . . .
A
Porphyra
B
Enzymes
C
Fats
D
Starch

Solution

(D) Pyrenoids are specialized proteinaceous structures found in the chloroplasts of many algae,such as $Chlamydomonas$ and $Spirogyra$.
They act as centers for the storage and formation of starch.
Therefore,the correct option is $D$.
81
BiologyMediumMCQAIPMT · 1993
From an evolutionary perspective,which of the following represents the most advanced form?
A
Selaginella
B
Funaria
C
Chlamydomonas
D
Pinus

Solution

(D) The evolutionary sequence of plants from simple to complex is: Algae $(Chlamydomonas)$ $\rightarrow$ Bryophytes $(Funaria)$ $\rightarrow$ Pteridophytes $(Selaginella)$ $\rightarrow$ Gymnosperms $(Pinus)$.
$Chlamydomonas$ is a unicellular green alga (simplest).
$Funaria$ is a moss (bryophyte),which is more complex than algae.
$Selaginella$ is a pteridophyte,which has vascular tissues.
$Pinus$ is a gymnosperm,which produces seeds and is the most advanced among the given options.
82
BiologyMediumMCQAIPMT · 1993
Pteridophytes differ from algae/bryophytes in possessing .............
A
Independent gametophyte
B
Well-developed vascular system
C
Archegonia
D
Flagellated spermatozoa

Solution

(B) Pteridophytes are the first terrestrial plants to possess vascular tissues ($xylem$ and $phloem$).
While algae and bryophytes lack a true vascular system,pteridophytes have a well-developed vascular system for the conduction of water and nutrients.
Therefore,the presence of a well-developed vascular system is a key feature that distinguishes pteridophytes from algae and bryophytes.
83
BiologyMediumMCQAIPMT · 1993
Pinus differs from Mangifera (Mango) due to the presence of:
A
Arborescent habit
B
Green leaves
C
Ovules not enclosed in an ovary
D
Wood

Solution

(C) $Pinus$ belongs to the group Gymnosperms,while $Mangifera$ (Mango) belongs to the group Angiosperms.
In Gymnosperms,the ovules are naked,meaning they are not enclosed within an ovary wall,and therefore,they do not form fruits.
In contrast,Angiosperms have ovules that are enclosed within an ovary,which develops into a fruit after fertilization.
Therefore,$Pinus$ differs from $Mangifera$ because its ovules are not enclosed in an ovary.
84
BiologyMediumMCQAIPMT · 1993
In $Ulothrix$ / $Spirogyra$,the reduction division (meiosis) occurs at the time of:
A
Gamete formation
B
Zoospore formation
C
Zygospore germination
D
Vegetative reproduction

Solution

(C) In $Ulothrix$ and $Spirogyra$,the plant body is haploid $(n)$.
Sexual reproduction involves the fusion of two gametes to form a diploid $(2n)$ zygote.
Since the organism is haploid,the zygote must undergo meiosis (reduction division) to restore the haploid state.
This process occurs during the germination of the zygospore $(2n)$,which results in the formation of haploid zoospores or new filaments.
85
BiologyEasyMCQAIPMT · 1993
The chloroplast of $Chlamydomonas$ is . . . . . . .
A
Stellate
B
Cup-shaped
C
Collar-shaped
D
Spiral

Solution

(B) $Chlamydomonas$ is a unicellular green alga belonging to the class $Chlorophyceae$.
It is characterized by the presence of a single,large,cup-shaped chloroplast that occupies a significant portion of the cell.
This chloroplast contains a pyrenoid,which is a center for starch storage.
86
BiologyEasyMCQAIPMT · 1993
In humans,the number of cervical vertebrae is . . . . . . .
A
the same as in whales.
B
more than in rabbits.
C
double that of horses.
D
less than in giraffes.

Solution

(A) In mammals,the number of cervical vertebrae is almost constant,which is $7$. This includes humans,whales,rabbits,horses,and giraffes. Therefore,the number of cervical vertebrae in humans is the same as in whales $(7)$.
87
BiologyEasyMCQAIPMT · 1993
Gorillas,chimpanzees,monkeys,and humans belong to the same ..........
A
Species
B
Genus
C
Family
D
Order

Solution

(D) Gorillas,chimpanzees,monkeys,and humans are all classified under the same order,which is $Primates$. While they belong to different families (e.g.,$Hominidae$ for humans,gorillas,and chimpanzees; $Cercopithecidae$ for Old World monkeys),they share the same taxonomic order $Primates$ due to shared evolutionary characteristics such as grasping hands and forward-facing eyes.
88
BiologyMediumMCQAIPMT · 1993
What is common among whales,bats,and rats?
A
Absence of neck
B
Muscular diaphragm between thorax and abdomen
C
Extra-abdominal testes to protect from high body temperature
D
Presence of external ears

Solution

(B) Whales,bats,and rats are all members of the class $Mammalia$.
One of the defining characteristics of mammals is the presence of a muscular diaphragm that separates the thoracic cavity from the abdominal cavity.
This structure plays a crucial role in respiration.
While some mammals have external ears (pinnae),not all do (e.g.,whales have internal ear openings).
Extra-abdominal testes are not a universal feature of all mammals (e.g.,in whales,testes are abdominal).
Therefore,the presence of a muscular diaphragm is the most accurate common feature among these organisms.
89
BiologyMediumMCQAIPMT · 1993
Which of the following helps in locomotion?
A
Trichocysts in Paramecium
B
Pedicellariae in Starfish
C
Clitellum in Pheretima
D
Posterior sucker in Hirudinaria

Solution

(D) Locomotion refers to the movement of an organism from one place to another.
$A$. Trichocysts in Paramecium are defensive organelles,not involved in locomotion.
$B$. Pedicellariae in Starfish are small wrench-like structures used for cleaning the body surface and protection,not for locomotion.
$C$. Clitellum in Pheretima (earthworm) is a glandular structure involved in cocoon formation,not locomotion.
$D$. Hirudinaria (leech) uses its anterior and posterior suckers for attachment and locomotion (looping movement). Therefore,the posterior sucker is involved in its locomotion.
90
BiologyEasyMCQAIPMT · 1993
What is common among Ostrich,Kiwi,and Penguin?
A
Running birds
B
Migratory birds
C
Flightless birds
D
Four-toed birds

Solution

(C) Ostrich $(Struthio)$,Kiwi $(Apteryx)$,and Penguin $(Aptenodytes)$ are all examples of flightless birds. These birds have evolved to lose the ability to fly due to various evolutionary adaptations,such as reduced wing size and the absence of a keel on the sternum,which is necessary for the attachment of large flight muscles. Therefore,the common characteristic among them is that they are flightless birds.
91
BiologyMediumMCQAIPMT · 1993
Which of the following animals possesses neurons but lacks a nervous system (nerve net/ganglia/brain)?
A
Hydra
B
Tapeworm
C
Earthworm
D
Frog tadpole

Solution

(A) Hydra belongs to the phylum $Cnidaria$.
It possesses a primitive nervous system consisting of a network of neurons (nerve net) spread throughout the body.
However, it lacks a centralized nervous system, such as a brain or distinct nerve cords, which are characteristic of higher animals like flatworms $(Platyhelminthes)$, annelids $(Earthworm)$, and chordates $(Frog$ tadpole).
Therefore, among the given options, $Hydra$ is the most primitive organism with a diffuse nerve net rather than a complex nervous system.
92
BiologyMediumMCQAIPMT · 1993
Which of the following is true for $Taenia$ $saginata$?
A
Pig is the intermediate host in its life cycle.
B
It has two large suckers on the scolex.
C
Rostellar hooks are absent.
D
There is a double row of hooks on the rostellum.

Solution

(C) $Taenia$ $saginata$,commonly known as the beef tapeworm,is a parasite that infects humans.
Its primary intermediate host is cattle ($Bos$ $taurus$),not pigs.
The scolex of $Taenia$ $saginata$ possesses four suckers but lacks a rostellum and rostellar hooks.
Therefore,the statement that rostellar hooks are absent is correct.
93
BiologyMediumMCQAIPMT · 1993
All vertebrates possess ......... .
A
Renal portal system
B
Dorsal,hollow,central nervous system
C
Four-chambered ventral heart
D
Pharyngeal gill slits

Solution

(B) All vertebrates belong to the phylum $Chordata$.
Key characteristics of chordates include the presence of a notochord,a dorsal hollow nerve cord,and pharyngeal gill slits at some stage of their life cycle.
While not all vertebrates have a renal portal system or a four-chambered heart (e.g.,fish have a two-chambered heart),the presence of a dorsal,hollow,central nervous system is a fundamental diagnostic feature of all vertebrates.
94
BiologyMediumMCQAIPMT · 1993
In which of the following groups do all members exhibit a larval stage in their life cycle?
A
Frog,Lizard,Cockroach
B
Ascaris,Housefly,Frog
C
Housefly,Earthworm,Mosquito
D
Butterfly,Frog,Mosquito

Solution

(D) larval stage is a distinct juvenile form that many animals undergo before metamorphosis into adults.
$1$. Butterfly: Undergoes complete metamorphosis (Egg $\rightarrow$ Larva/Caterpillar $\rightarrow$ Pupa $\rightarrow$ Adult).
$2$. Frog: The larval stage is known as a tadpole,which is aquatic and undergoes metamorphosis to become an adult frog.
$3$. Mosquito: Undergoes complete metamorphosis (Egg $\rightarrow$ Larva/Wriggler $\rightarrow$ Pupa $\rightarrow$ Adult).
Therefore,all members of the group (Butterfly,Frog,and Mosquito) exhibit a larval stage in their life cycle.
95
BiologyMediumMCQAIPMT · 1993
The trachea in mammals and the trachea in cockroaches are similar in that they both have:
A
paired structures
B
non-collapsible walls
C
ciliated inner lining
D
origin from the head

Solution

(B) In mammals,the trachea is supported by cartilaginous rings that prevent it from collapsing,ensuring a continuous airway. Similarly,the trachea in cockroaches is supported by spiral thickenings of the cuticle called intima (taenidia),which also prevent the tracheal tubes from collapsing. Therefore,both structures share the characteristic of having non-collapsible walls.
96
BiologyEasyMCQAIPMT · 1993
In frogs,mucus helps to keep the skin ..........
A
thick
B
dry
C
soft
D
moist

Solution

(D) In frogs,the skin is smooth and slippery due to the presence of mucus glands. The mucus secreted by these glands helps to keep the skin moist,which is essential for cutaneous respiration (breathing through the skin).
97
BiologyMediumMCQAIPMT · 1993
Which type of wood is more susceptible to decay?
A
Sapwood
B
Softwood
C
Fibrous wood
D
Heartwood

Solution

(A) . Sapwood (also known as alburnum) is the peripheral,living part of the secondary xylem.
It contains living cells and conducts water and minerals.
Because it is less dense and contains stored food materials like starch and sugars,it is more susceptible to attack by microorganisms,fungi,and insects,making it decay faster than heartwood.
Heartwood is the central,dead part of the secondary xylem,which is impregnated with tannins,resins,and oils,making it highly resistant to decay.
98
BiologyEasyMCQAIPMT · 1993
Periderm is produced from ..........
A
Vascular cambium
B
Fascicular cambium
C
Phellogen
D
Intrafascicular cambium

Solution

(C) The periderm is a collective term for the secondary protective tissues formed in the stem and root of woody plants.
It consists of three layers: the phellogen (cork cambium),the phellem (cork),and the phelloderm (secondary cortex).
The phellogen is a meristematic tissue that gives rise to the periderm by producing phellem towards the outside and phelloderm towards the inside.
Therefore,the periderm originates from the phellogen.
99
BiologyMediumMCQAIPMT · 1993
The narrow layer of thin-walled cells found in dicots (between the phloem and xylem) is known as ..........
A
Phellogen
B
Vascular cambium
C
Endodermis
D
Pericycle

Solution

(B) In dicot stems, the meristematic layer that exists between the primary xylem and primary phloem is called the $Vascular \text{ } cambium$.
This layer is responsible for secondary growth by producing secondary xylem towards the inside and secondary phloem towards the outside.
Therefore, the correct option is $B$.
100
BiologyEasyMCQAIPMT · 1993
Bordered pits are found in . . . . . . .
A
Sieve cells
B
Vessel elements
C
Companion cells
D
Sieve tube elements

Solution

(B) Bordered pits are specialized structures found in the cell walls of xylem elements,particularly in tracheids and vessel elements. These pits facilitate the lateral transport of water and minerals between adjacent cells. Sieve cells,companion cells,and sieve tube elements are components of the phloem and do not possess bordered pits.
101
BiologyMediumMCQAIPMT · 1993
The cells of $Leydig$ produce and secrete .......... .
A
Ovary and Estrogens
B
Liver and Cholesterol
C
Pancreas and Glucagon
D
Testis and Testosterone

Solution

(D) The $Leydig$ cells,also known as interstitial cells,are located in the connective tissue spaces between the seminiferous tubules of the testis.
These cells are responsible for the synthesis and secretion of a group of hormones called androgens,primarily $Testosterone$.
$Testosterone$ is the main male sex hormone responsible for the development of secondary sexual characteristics in males.
102
BiologyEasyMCQAIPMT · 1993
The study of the formation,growth,and development of a new organism from an egg is known as:
A
Apomixis
B
Embryology
C
Embryogenesis
D
Cytology

Solution

(B) Embryology is the branch of biology that deals with the study of the formation,growth,and development of a new organism from an egg (zygote).
$1$. Apomixis is a form of asexual reproduction that mimics sexual reproduction.
$2$. Embryogenesis refers specifically to the process of development of an embryo from the zygote.
$3$. Cytology is the study of the structure and function of cells.
Therefore,the correct answer is Embryology.
103
BiologyMediumMCQAIPMT · 1993
How many meiotic divisions are required to produce $200$ and $400$ seeds in pea plants,respectively?
A
$200$ and $400$
B
$250$ and $500$
C
$300$ and $600$
D
$400$ and $800$

Solution

(B) In angiosperms,one meiotic division produces $4$ microspores (pollen grains) and $1$ functional megaspore (embryo sac).
Each seed is formed by the fusion of one male gamete and one female gamete.
To produce $n$ seeds,$n$ meiotic divisions are required for the male gametes (pollen grains) and $n$ meiotic divisions are required for the female gametes (embryo sacs).
Therefore,the total number of meiotic divisions required is $n + n/4$ (since $4$ pollen grains are produced per meiosis,but only $1$ embryo sac is produced per meiosis).
However,in standard textbook problems of this type,the calculation is often simplified as $n + n/4$ for the total divisions.
For $200$ seeds: $200$ (for megaspores) + $200/4$ (for microspores) = $200 + 50 = 250$ divisions.
For $400$ seeds: $400$ (for megaspores) + $400/4$ (for microspores) = $400 + 100 = 500$ divisions.
Thus,the correct answer is $250$ and $500$.
104
BiologyMediumMCQAIPMT · 1993
What is an ovule called when the funicle,hilum,micropyle,and chalaza lie in a single straight line?
A
Orthotropous
B
Anatropous
C
Campylotropous
D
Amphitropous

Solution

(A) In an $Orthotropous$ (or $Atropous$) ovule,the body of the ovule is straight,and the funicle,hilum,micropyle,and chalaza lie in a single straight vertical line. This is considered the most primitive type of ovule and is found in plants like $Polygonum$ and $Piper$.
105
BiologyEasyMCQAIPMT · 1993
Who discovered double fertilization and triple fusion?
A
Hofmeister
B
Nawaschin and Guignard
C
Leeuwenhoek
D
Strasburger

Solution

(B) Double fertilization and triple fusion were discovered by $S.G. \text{Nawaschin}$ and $L. \text{Guignard}$ in $1898$ in the plants $Lilium$ and $Fritillaria$.
Double fertilization involves the fusion of one male gamete with the egg cell to form a diploid zygote,and the fusion of the second male gamete with two polar nuclei to form the primary endosperm nucleus (triple fusion).
106
BiologyMediumMCQAIPMT · 1993
The gastrula stage is characterized by which of the following?
A
Invagination of the blastoderm
B
Invagination of the archenteron
C
Closure of the blastopore
D
Closure of the neural tube

Solution

(B) The gastrula stage is a critical phase in embryonic development where the single-layered blastula is reorganized into a multi-layered structure called the gastrula. This process is known as gastrulation. $A$ key characteristic of this stage is the formation of the primitive gut,known as the $archenteron$. This occurs through the process of invagination,where cells move inward to form the cavity. Therefore,the gastrula stage is characterized by the invagination of the $archenteron$.
107
BiologyMediumMCQAIPMT · 1993
When is the second polar body extruded from the oocyte nucleus?
A
After the entry of the sperm but before the completion of fertilization
B
After fertilization
C
Before the entry of the sperm
D
Without any relation to the entry of the sperm

Solution

(A) The secondary oocyte remains arrested in the metaphase-$II$ stage of meiosis-$II$. The entry of the sperm into the secondary oocyte triggers the completion of meiosis-$II$. This process results in the formation of a large haploid ovum (ootid) and a small second polar body. Therefore,the extrusion of the second polar body occurs after the entry of the sperm but before the fusion of the male and female pronuclei (completion of fertilization).
108
BiologyMediumMCQAIPMT · 1993
Where is the yolk found in a telolecithal egg?
A
Distributed throughout the egg
B
Concentrated at one pole
C
Concentrated at both poles
D
Concentrated in the center

Solution

(B) In a $telolecithal$ egg,the yolk is very abundant and is concentrated at one pole,known as the $vegetal$ pole,while the active cytoplasm is restricted to the opposite pole,known as the $animal$ pole. This type of egg is commonly found in birds and reptiles.
109
BiologyMediumMCQAIPMT · 1993
The process of the acrosome in a sperm can be associated with which of the following?
A
Capacitation
B
Release of lysins
C
Entry of $Na^+$
D
Release of fertilizin

Solution

(B) The acrosome is a cap-like structure present on the anterior portion of the sperm head.
It is derived from the Golgi apparatus and contains hydrolytic enzymes,collectively known as sperm lysins (e.g.,hyaluronidase and acrosin).
During fertilization,the acrosome undergoes the acrosome reaction,which involves the release of these lysins.
These enzymes help the sperm to penetrate the protective layers of the ovum (zona pellucida and corona radiata),thereby facilitating fertilization.
110
BiologyMediumMCQAIPMT · 1993
Polygenic inheritance in humans is represented by ....
A
Skin color
B
Phenylketonuria
C
Color blindness
D
Sickle cell anemia

Solution

(A) Polygenic inheritance is a type of inheritance in which a single trait is controlled by two or more genes.
In humans,skin color is a classic example of polygenic inheritance.
It is governed by three pairs of genes $(A, B, C)$.
The intensity of skin color depends on the number of dominant alleles present; more dominant alleles result in darker skin,while fewer result in lighter skin.
Other options like Phenylketonuria,Color blindness,and Sickle cell anemia are examples of Mendelian disorders caused by mutations in single genes.
111
BiologyDifficultMCQAIPMT · 1993
In a normal couple,half of the sons are hemophilic and half of the daughters are carriers. Where is the gene located?
A
On the father's $X$ chromosome
B
On the father's $Y$ chromosome
C
On one of the mother's $X$ chromosomes
D
On both of the mother's $X$ chromosomes

Solution

(C) Hemophilia is an $X$-linked recessive disorder.
If the mother is a carrier $(X^HX^h)$ and the father is normal $(X^HY)$,the cross results in:
$1$. $X^HX^H$ (Normal daughter)
$2$. $X^HX^h$ (Carrier daughter)
$3$. $X^HY$ (Normal son)
$4$. $X^hY$ (Hemophilic son)
In this scenario,$50\%$ of sons are hemophilic and $50\%$ of daughters are carriers. This confirms that the mother carries the recessive gene on one of her $X$ chromosomes.
112
BiologyMediumMCQAIPMT · 1993
When does sex determination occur in humans?
A
At the time of egg formation
B
At the time of fertilization
C
$40$ days after fertilization
D
At the 7th to 8th week when the differentiation of the embryo's reproductive organs occurs

Solution

(B) In humans,sex determination is genetic and occurs at the time of fertilization.
When a sperm carrying either an $X$ or a $Y$ chromosome fertilizes an egg (which always carries an $X$ chromosome),the sex of the zygote is decided.
If an $X$-bearing sperm fertilizes the egg,the zygote becomes $XX$ (female).
If a $Y$-bearing sperm fertilizes the egg,the zygote becomes $XY$ (male).
Therefore,the sex is determined at the moment of fusion of the male and female gametes.
113
BiologyMediumMCQAIPMT · 1993
In Mr. Kapoor,there is a pair of autosomal genes $Bb$ and a sex-linked factor $d$. What will be the proportion of $Bd$ in his sperm?
A
$0$
B
$1/2$
C
$1/4$
D
$1/8$

Solution

(C) Mr. Kapoor is a male,so his genotype for the sex chromosomes is $XY$. The sex-linked factor $d$ is located on the $X$ chromosome. Thus,his genotype is $BbXdY$.
During meiosis,the autosomal genes $B$ and $b$ segregate independently.
The possible combinations of autosomal genes and sex chromosomes in the sperm are:
$1$. $B$ with $X^d$
$2$. $B$ with $Y$
$3$. $b$ with $X^d$
$4$. $b$ with $Y$
Each of these four combinations has an equal probability of $1/4$.
Therefore,the proportion of sperm carrying $Bd$ (where $d$ is the sex-linked factor) is $1/4$.
114
BiologyMediumMCQAIPMT · 1993
What is the probability of having a color-blind male child from two normal parents?
A
Not at all.
B
If all four grandparents have normal vision.
C
Only possible if the father's mother is color-blind.
D
Possible if the mother's father is color-blind.

Solution

(D) Color blindness is an $X$-linked recessive disorder.
Let the normal vision allele be $X^C$ and the color-blind allele be $X^c$.
$A$ normal male is $X^CY$ and a normal female is $X^CX^C$ or $X^CX^c$ (carrier).
If the mother is a carrier $(X^CX^c)$ and the father is normal $(X^CY)$,the cross is $X^CX^c \times X^CY$.
The offspring genotypes are $X^CX^C$ (normal female),$X^CX^c$ (carrier female),$X^CY$ (normal male),and $X^cY$ (color-blind male).
Thus,a color-blind male child is possible if the mother is a carrier. $A$ woman is a carrier if her father was color-blind (since she receives one $X$ chromosome from her father).
115
BiologyMediumMCQAIPMT · 1993
Mendel studied seven pairs of traits in pea plants,which allow for $21$ possible combinations. If you are told that one of these combinations does not show independent assortment in a subsequent study,what would be your reaction?
A
The Law of Independent Assortment is incorrect.
B
Mendel did not study all the combinations.
C
It is possible.
D
The subsequent study might be incorrect.

Solution

(C) मेंडल के स्वतंत्र अपव्यूहन के नियम के अनुसार,जीन एक-दूसरे से स्वतंत्र रूप से वंशागत होते हैं। हालाँकि,यह नियम केवल तभी लागू होता है जब जीन अलग-अलग गुणसूत्रों पर स्थित हों या एक ही गुणसूत्र पर बहुत दूर स्थित हों। यदि दो जीन एक ही गुणसूत्र पर बहुत पास स्थित होते हैं,तो वे 'सहलग्नता' (Linkage) प्रदर्शित करते हैं और स्वतंत्र अपव्यूहन नहीं दिखाते हैं। मेंडल द्वारा चुने गए सात लक्षण संयोग से अलग-अलग गुणसूत्रों पर थे या बहुत दूर थे,इसलिए उन्होंने स्वतंत्र अपव्यूहन देखा। यदि किसी अन्य अध्ययन में स्वतंत्र अपव्यूहन नहीं देखा जाता है,तो इसका अर्थ है कि उन जीनों के बीच 'सहलग्नता' (Linkage) है,जो जैविक रूप से संभव है।
116
BiologyMediumMCQAIPMT · 1993
Two dominant non-allelic genes are $50$ map units apart. The linkage present is:
A
$cis$ type
B
$trans$ type
C
Complete
D
Absent/Incomplete

Solution

(D) The distance between two genes is measured in map units or centimorgans $(cM)$.
$1$ map unit corresponds to $1\%$ recombination frequency.
If two genes are $50$ map units apart,the recombination frequency between them is $50\%$.
According to the principle of linkage,if the recombination frequency is $50\%$,the genes behave as if they are assorting independently.
Therefore,linkage is considered to be absent or effectively incomplete to the point of independent assortment.
117
BiologyMediumMCQAIPMT · 1993
Which of the following is suitable for the study of linkage?
A
$aaBB \times aaBB$
B
$AABB \times aabb$
C
$AaBb \times AaBb$
D
$AAbb \times AaBB$

Solution

(B) To study linkage,a test cross is performed between a dihybrid individual $(AaBb)$ and a homozygous recessive individual $(aabb)$.
However,in the given options,we look for a cross that produces a dihybrid $(AaBb)$ in the $F_1$ generation,which is then used for linkage analysis.
The cross $AABB \times aabb$ produces $AaBb$ in the $F_1$ generation.
This $F_1$ dihybrid is then test-crossed with a double recessive parent $(aabb)$ to observe the linkage between genes.
Therefore,the cross $AABB \times aabb$ is the standard starting point for creating the dihybrid required to study linkage.
118
BiologyMediumMCQAIPMT · 1993
What was the transforming principle identified by Avery,MacLeod,and McCarty in Pneumococcus?
A
$m.RNA$
B
$DNA$
C
Protein
D
Polysaccharide

Solution

(B) Oswald Avery,Colin MacLeod,and Maclyn McCarty worked to determine the biochemical nature of the 'transforming principle' in Griffith's experiment.
They purified biochemicals (proteins,$DNA$,$RNA$,etc.) from the heat-killed $S$ strain of Streptococcus pneumoniae.
They discovered that $DNA$ alone from the heat-killed $S$ strain caused the transformation of non-virulent $R$ strain bacteria into virulent $S$ strain bacteria.
They also observed that protein-digesting enzymes (proteases) and $RNA$-digesting enzymes (RNases) did not affect transformation,whereas $DNA$-digesting enzymes (DNases) inhibited the process.
Thus,they concluded that the genetic material is $DNA$.
119
BiologyEasyMCQAIPMT · 1993
The arrangement of nucleotides in $DNA$ can be seen by which of the following techniques?
A
$X$-ray crystallography
B
Electron microscope
C
Ultracentrifuge
D
Light microscope

Solution

(A) The structure of $DNA$ and the arrangement of its nucleotides were famously determined by Rosalind Franklin and Maurice Wilkins using $X$-ray crystallography. This technique involves directing $X$-rays at a crystallized sample of $DNA$,which then diffracts the rays to create a pattern on a photographic plate. By analyzing this diffraction pattern,scientists were able to deduce the helical structure and the spacing of the nucleotides within the $DNA$ molecule.
120
BiologyMediumMCQAIPMT · 1993
Since most amino acids are represented by more than one codon,the genetic code is said to be . . . . . . .
A
Overlapping
B
Wobbling
C
Degenerate
D
Productive

Solution

(C) The genetic code is described as $degenerate$ because some amino acids are encoded by more than one codon. For example,there are $64$ possible codons,but only $20$ amino acids. This redundancy allows multiple codons to specify the same amino acid,which provides a buffer against mutations.
121
BiologyEasyMCQAIPMT · 1993
Who proved that $DNA$ is the genetic material?
A
Griffith
B
Watson
C
Boveri and Sutton
D
Hershey and Chase

Solution

(D) Alfred Hershey and Martha Chase $(1952)$ conducted experiments using bacteriophages. They grew some phages in a medium containing radioactive phosphorus $(^{32}P)$ to label $DNA$ and others in a medium containing radioactive sulfur $(^{35}S)$ to label proteins. They allowed these phages to infect $E. coli$ bacteria. After agitation and centrifugation,they observed that radioactive phosphorus $(^{32}P)$ was found inside the bacterial cells,while radioactive sulfur $(^{35}S)$ remained in the supernatant. This proved that $DNA$,not protein,entered the bacteria and acted as the genetic material.
122
BiologyEasyMCQAIPMT · 1993
During $DNA$ replication,the strands are separated by which enzyme?
A
$DNA$ polymerase
B
Topoisomerase
C
Helicase
D
Gyrase

Solution

(C) During $DNA$ replication,the enzyme $Helicase$ is responsible for unwinding the $DNA$ double helix by breaking the hydrogen bonds between the complementary nitrogenous bases.
This separation creates a replication fork,allowing each strand to serve as a template for the synthesis of a new complementary strand.
$DNA$ polymerase is involved in the synthesis of new strands,while Topoisomerase and Gyrase help in relieving the supercoiling tension ahead of the replication fork.
123
BiologyMediumMCQAIPMT · 1993
The process of translation is .....
A
Ribosome synthesis
B
Protein synthesis
C
$DNA$ synthesis
D
$RNA$ synthesis

Solution

(B) Translation is the process in which the genetic information present in $mRNA$ is used to synthesize a polypeptide chain (protein).
This process occurs in the cytoplasm on ribosomes,where $tRNA$ molecules bring specific amino acids based on the codon sequence of the $mRNA$.
Therefore,translation is essentially the process of protein synthesis.
124
BiologyMediumMCQAIPMT · 1993
In $DNA$,the unequal nitrogenous bases will primarily be ...
A
Single-stranded
B
Double-stranded
C
Triple-stranded
D
Quadruple-stranded

Solution

(A) According to $Chargaff's$ rule,in a double-stranded $DNA$ molecule,the amount of adenine $(A)$ is equal to thymine $(T)$,and the amount of guanine $(G)$ is equal to cytosine $(C)$. Therefore,the ratio $(A+T)/(G+C)$ is constant for a species. If the nitrogenous bases are unequal (i.e.,$A+T \neq G+C$),it indicates that the $DNA$ molecule is single-stranded,as the base-pairing rules of $Watson$ and $Crick$ do not apply to single-stranded structures.
125
BiologyEasyMCQAIPMT · 1993
The nucleosome core is made up of ............ .
A
$H_1, H_2A, H_2B$ and $H_3$
B
$H_1, H_2A, H_2B$ and $H_4$
C
$H_1, H_2A, H_2B, H_3$ and $H_4$
D
$H_2A, H_2B, H_3$ and $H_4$

Solution

(D) nucleosome is the basic structural unit of $DNA$ packaging in eukaryotes.
It consists of a segment of $DNA$ wound around a core of histone proteins.
This core particle is an octamer,which is composed of two molecules each of four types of histone proteins: $H_2A, H_2B, H_3$,and $H_4$.
The $H_1$ histone protein is not part of the core octamer; instead,it binds to the $DNA$ where it enters and leaves the nucleosome,helping to stabilize the structure.
Therefore,the nucleosome core is made up of $H_2A, H_2B, H_3$,and $H_4$.
126
BiologyMediumMCQAIPMT · 1993
Which is the initiation codon for protein synthesis in eukaryotes?
A
$GUA$
B
$GCA$
C
$CCA$
D
$AUG$

Solution

(D) In protein synthesis,the process of translation begins with an initiation codon.
For both prokaryotes and eukaryotes,the codon $AUG$ serves as the initiation codon.
$AUG$ codes for the amino acid Methionine.
Therefore,the correct option is $D$.
127
BiologyDifficultMCQAIPMT · 1993
Which of the following captures the maximum amount of solar energy?
A
Planted trees
B
Cultivated crops
C
Algae grown in a tank
D
Grown grass

Solution

(C) The efficiency of solar energy conversion depends on the photosynthetic surface area and the rate of primary productivity.
Algae grown in a tank (often in controlled environments like bioreactors) can be maintained at high densities with optimal light exposure,nutrient availability,and $CO_2$ concentration.
Compared to terrestrial plants like trees,crops,or grass,which are limited by seasonal changes,soil nutrients,and structural biomass (non-photosynthetic parts like wood),algae in a tank represent a highly efficient system for capturing solar energy per unit area.
128
BiologyMediumMCQAIPMT · 1993
What is genetic drift a change in?
A
Gene frequency in the same generation
B
Appearance of recessive genes
C
Gene frequency from one generation to the next
D
None of the above

Solution

(C) Genetic drift is defined as the change in the frequency of an existing gene variant (allele) in a population due to random sampling of organisms.
It occurs when chance events cause allele frequencies to fluctuate unpredictably from one generation to the next.
Unlike natural selection,which is adaptive,genetic drift is a stochastic (random) process that can lead to the loss of alleles or the fixation of others,especially in small populations.
129
BiologyMediumMCQAIPMT · 1993
Weismann cut off the tails of mice for several generations. However,the tail did not disappear or become shorter,which shows that ...........
A
Darwin was correct.
B
The tail is an essential organ.
C
The mutation theory is incorrect.
D
Lamarckism - the inheritance of acquired characters was incorrect.

Solution

(D) August Weismann performed an experiment where he cut off the tails of mice for $22$ generations.
He observed that the offspring were still born with tails of normal length.
This experiment provided evidence against the theory of 'Inheritance of Acquired Characters' proposed by Jean-Baptiste Lamarck.
Lamarck believed that traits acquired during an organism's lifetime could be passed on to the next generation.
Weismann's experiment demonstrated that somatic changes (changes in body cells) are not inherited,as only changes in germ cells (gametes) are passed to the offspring.
130
BiologyMediumMCQAIPMT · 1993
The theory of natural selection is based on which of the following?
A
Contribution of the environment to evolution
B
Natural selection acts on favorable variations
C
Genetic variations result from changes in the gene pool
D
None of the above

Solution

(B) The theory of natural selection,proposed by $Charles \ Darwin$,is primarily based on the observation that individuals within a population exhibit variations.
Some of these variations are heritable and provide a survival or reproductive advantage in a given environment.
Natural selection acts on these favorable variations,allowing individuals possessing them to survive and reproduce more successfully than others.
Over generations,this leads to the accumulation of these advantageous traits in the population,which is the core mechanism of evolution.
131
BiologyMediumMCQAIPMT · 1993
Evolutionary convergence is the development of $......$.
A
$A$ common set of traits in a group of different ancestors
B
Different traits in a group of closely related organisms
C
Common traits in a group of distantly related organisms
D
Irregular reproduction

Solution

(C) Evolutionary convergence,also known as convergent evolution,is the process whereby organisms not closely related (not monophyletic) independently evolve similar traits as a result of having to adapt to similar environments or ecological niches.
For example,the wings of butterflies and birds are analogous structures that evolved independently to serve the function of flight.
Therefore,it represents the development of common traits in a group of distantly related organisms.
132
BiologyMediumMCQAIPMT · 1993
Which cells are involved in the immune response?
A
Red blood cells
B
Lymphocytes
C
Acidophils
D
Platelets

Solution

(B) The immune system is primarily composed of specialized cells that recognize and eliminate pathogens.
Lymphocytes are the key cells of the immune system.
They are a type of white blood cell $(WBC)$ that includes $B$-lymphocytes and $T$-lymphocytes.
$B$-lymphocytes produce antibodies to fight pathogens,while $T$-lymphocytes help activate other immune cells and directly destroy infected cells.
Therefore,the correct option is $B$.
133
BiologyMediumMCQAIPMT · 1993
Which of the following is an opiate narcotic?
A
Cannabis
B
Charas
C
Heroin
D
Nicotine

Solution

(C) Opiates are drugs that are derived from the opium poppy $(Papaver \text{ somniferum})$.
Heroin, also known as diacetylmorphine, is a semi-synthetic opioid derived from morphine, which is extracted from the opium poppy.
Cannabis and Charas are cannabinoids, while Nicotine is a stimulant found in tobacco.
Therefore, Heroin is the correct answer.
134
BiologyMediumMCQAIPMT · 1993
Haploid plants are preferred over diploid plants for mutation studies because:
A
Recessive mutations are expressed immediately.
B
It is easier to induce mutations in them.
C
They are easier to grow.
D
Dominant mutations are expressed immediately.

Solution

(A) In diploid organisms,a recessive mutation is masked by the presence of a dominant allele on the homologous chromosome.
However,in haploid plants,there is only one set of chromosomes.
Therefore,any mutation,whether dominant or recessive,will be expressed in the phenotype immediately because there is no corresponding allele to mask the effect of the recessive mutation.
This makes haploid plants ideal for screening and studying induced mutations.
135
BiologyMediumMCQAIPMT · 1993
What is the total population of organisms of the same type called?
A
Colony
B
Species
C
Population
D
Community

Solution

(C) In ecology, a group of individuals of the same species living in a specific geographical area at a specific time is referred to as a $Population$.
$Species$ refers to a group of organisms that can interbreed to produce fertile offspring.
$Community$ refers to an assemblage of populations of different species living in the same area.
Therefore, the correct term for a group of organisms of the same type is $Population$.
136
BiologyMediumMCQAIPMT · 1993
What is the study of the interrelationships between living organisms and their environment called?
A
Ecology
B
Ecosystem
C
Phytogeography
D
Bioecology

Solution

(A) The study of the interrelationships between living organisms and their environment is known as $Ecology$.
$Ecology$ is derived from the Greek words $Oikos$ (meaning house or dwelling) and $Logos$ (meaning study).
An $Ecosystem$ refers to the functional unit of nature where living organisms interact among themselves and with the surrounding physical environment.
$Phytogeography$ is the study of the geographic distribution of plants.
Therefore,the correct term for the study of these interrelationships is $Ecology$.
137
BiologyMediumMCQAIPMT · 1993
In a pond ecosystem,the pyramid of numbers is ............ .
A
Irregular
B
Inverted
C
Upright
D
Spindle-shaped

Solution

(C) In a pond ecosystem,the pyramid of numbers is upright.
$1$. At the base,there are a large number of producers (phytoplankton).
$2$. These are followed by a smaller number of primary consumers (zooplankton).
$3$. Then,there are even fewer secondary consumers (small fish).
$4$. Finally,there are very few tertiary consumers (large fish) at the top.
Thus,the number of organisms decreases as we move to higher trophic levels,resulting in an upright pyramid.
138
BiologyMediumMCQAIPMT · 1993
The pyramid of numbers deals with the number of what?
A
Species in an area
B
Individuals in a community
C
Individuals at a trophic level
D
Subspecies in a community

Solution

(C) The pyramid of numbers is a graphical representation that shows the relationship between the number of producers,herbivores,and carnivores at successive trophic levels in an ecosystem.
It specifically counts the number of individual organisms present at each trophic level.
Therefore,the pyramid of numbers deals with the number of individuals at a trophic level.
139
BiologyMediumMCQAIPMT · 1993
Which aquatic plant, introduced from America, has caused significant ecological problems in India?
A
Cyperus rotundus
B
Eichhornia crassipes
C
Trapa latifolia
D
Trapa bispinosa

Solution

(B) The aquatic plant $Eichhornia$ $crassipes$, commonly known as water hyacinth, was introduced into India from America for its beautiful flowers and leaf shape.
It is an invasive species that grows rapidly in water bodies, causing clogging of waterways and depletion of dissolved oxygen, which leads to the death of fishes and other aquatic organisms.
It is often referred to as the 'Terror of Bengal' due to its uncontrollable growth in the water bodies of West Bengal.
Therefore, the correct option is $B$.
140
BiologyEasyMCQAIPMT · 1993
Ultraviolet radiation from sunlight acts on $O_2$ to produce . . . . . . .
A
$O_3$
B
$S_2$
C
$CO$
D
$CH_4$

Solution

(A) In the stratosphere, ultraviolet $(UV)$ radiation acts on molecular oxygen $(O_2)$ to split it into free oxygen atoms $(O)$.
These free oxygen atoms then react with molecular oxygen $(O_2)$ to form ozone $(O_3)$.
The chemical reaction is as follows:
$O_2 + \text{UV} \rightarrow O + O$
$O + O_2 \rightarrow O_3$
Therefore, the correct product is ozone $(O_3)$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real AIPMT style covering Biology with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D Biology papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Run live AIPMT mock exams with unlimited students, 360° analytics & white-label branding.

See Demo

Frequently Asked Questions

How many Biology questions are in AIPMT 1993?

There are 228 Biology questions from the AIPMT 1993 paper on Vedclass, each with a detailed step-by-step solution in English.

Are AIPMT 1993 Biology solutions available in English?

Yes. All solutions on this page are in English. You can also switch to English or Hindi using the language buttons above the questions.

Can I practice AIPMT 1993 Biology as a timed test?

Yes. Use the Vedclass Test Series to attempt a full AIPMT mock test covering Biology with time limits and instant score analysis.

Can teachers create Biology papers from AIPMT previous year questions?

Yes. The Vedclass Exam Paper Generator lets teachers mix AIPMT Biology questions and generate Set A/B/C/D papers in minutes.

For Teachers & Institutes

Build a Custom Biology Paper

Pick AIPMT 1993 Biology questions, set difficulty, and generate Set A/B/C/D in 2 minutes.