AIIMS 1993 Biology Question Paper with Answer and Solution

69 QuestionsEnglishWith Solutions

BiologyQ169 of 69 questions

Page 1 of 1 · English

1
BiologyMediumMCQAIIMS · 1993
The scientist who created the group Protista for both unicellular plants and animals is
A
Haeckel
B
Pasteur
C
Lister
D
Koch

Solution

(A) The correct answer is $A$.
In $1866$,the German biologist Ernst Haeckel proposed a third kingdom,Protista,to include all unicellular organisms,such as bacteria,protozoans,algae,and fungi,which did not fit clearly into the plant or animal kingdoms.
2
BiologyMediumMCQAIIMS · 1993
Protonema is
A
Fossil pteridophyte
B
$A$ part of the sporophyte of $Funaria$
C
The juvenile phase of the moss gametophyte
D
None of the above

Solution

(C) The correct answer is $C$.
Protonema is the juvenile phase of the moss life cycle.
It is a green,branched,frequently filamentous structure that develops directly from the germination of a haploid spore.
It represents the early stage of the gametophyte in mosses,which eventually gives rise to the leafy stage.
3
BiologyEasyMCQAIIMS · 1993
In earthworm,the ovary is situated in which segment?
A
$13$
B
$9$
C
$10$
D
$26$

Solution

(A) In the earthworm ($Pheretima$ $posthuma$),the female reproductive system consists of a pair of ovaries.
These ovaries are attached to the intersegmental septum of the $12^{th}$ and $13^{th}$ segments.
Specifically,the ovaries are located in the $13^{th}$ segment.
4
BiologyMediumMCQAIIMS · 1993
Which of the following is common among a leech,a mosquito,a bed bug,and a rat?
A
All have anticoagulants
B
All lack a nucleus
C
All lack a cellular membrane
D
All have a sexual phase

Solution

(A) The common feature among a leech,a mosquito,a bed bug,and a rat is that they all possess anticoagulants in their saliva or physiological systems to facilitate blood feeding. Leeches,mosquitoes,and bed bugs are hematophagous (blood-feeding) ectoparasites that inject anticoagulants to prevent blood clotting while feeding. While rats are not ectoparasites,they are mammals,and their physiological systems naturally involve complex blood-clotting mechanisms regulated by various factors,but in the context of biological classification questions of this type,the presence of anticoagulants is the intended commonality regarding their interaction with blood or blood-related processes.
5
BiologyMediumMCQAIIMS · 1993
The mollusc which is considered to be a living fossil and also shows characters of annelids like nephridia and internal segmentation is
A
Pinctada vulgaris
B
Nautilus
C
Neopilina galatheca
D
None of these

Solution

(C) $Neopilina$ is considered a primitive mollusc that exhibits characteristics of annelids,such as internal metameric segmentation and $5$ pairs of nephridia. It is regarded as a living fossil and serves as a connecting link between annelids and molluscs.
6
BiologyMediumMCQAIIMS · 1993
Absence of excretory organs,great power of regeneration,and exclusively marine animals belong to the phylum:
A
Mollusca
B
Echinodermata
C
Fishes
D
Arthropoda

Solution

(B) The phylum $Echinodermata$ is characterized by the following features:
$1$. They are exclusively marine animals.
$2$. They lack specialized excretory organs; excretion occurs via diffusion through the body surface or tube feet.
$3$. They possess a remarkable power of regeneration,allowing them to regrow lost body parts.
Therefore,the correct option is $B$.
7
BiologyMediumMCQAIIMS · 1993
Why do mammals lack mucus glands in their skin?
A
The skin is not slippery
B
The skin is tough
C
The epidermis has many layers of cells
D
The skin is not respiratory

Solution

(D) The correct answer is $D$. Mucus glands are primarily found in animals that use their skin for cutaneous respiration (e.g.,amphibians). Mucus keeps the skin moist,which is essential for the diffusion of gases. Mammals do not use their skin for respiration; instead,they rely on lungs,and their skin is adapted for protection and thermoregulation,thus lacking mucus glands.
8
BiologyEasyMCQAIIMS · 1993
$12$ pairs of cranial nerves are present in
A
Reptilia
B
Birds only
C
Mammals only
D
All the above

Solution

(D) In vertebrates,the number of cranial nerves varies across different classes.
$12$ pairs of cranial nerves are a characteristic feature of amniotes,which include Reptiles,Birds,and Mammals.
Therefore,all the groups mentioned in the options possess $12$ pairs of cranial nerves.
9
BiologyMediumMCQAIIMS · 1993
Whales are warm-blooded animals that live in cold seas. Which part of their body helps them maintain their body temperature?
A
Blubber
B
Pelage
C
Muscles
D
Blood vessels

Solution

(A) Whales are mammals that maintain a constant body temperature despite living in cold aquatic environments. The primary adaptation for thermoregulation in whales is a thick layer of fat called $Blubber$ located beneath the skin. This layer acts as an effective insulator,preventing heat loss from the body to the cold surrounding water,thereby helping the whale remain warm.
10
BiologyEasyMCQAIIMS · 1993
Sprain is caused due to stretching of
A
Muscle
B
Ligament
C
Tendon
D
Nerve

Solution

(B) sprain is an injury that occurs when a ligament is stretched or torn beyond its normal range of motion.
Ligaments are tough,fibrous bands of connective tissue that connect bones to other bones at a joint.
When these tissues are subjected to excessive force,it results in a sprain.
11
BiologyEasyMCQAIIMS · 1993
The intervertebral disc is made up of:
A
Elastic cartilage
B
Fibrous cartilage
C
Calcified cartilage
D
Hyaline cartilage

Solution

(B) The intervertebral disc is composed of white fibrous cartilage (fibrocartilage). This type of cartilage is characterized by a dense network of collagen fibers,which provides high tensile strength and acts as a shock absorber between the vertebrae.
12
BiologyMediumMCQAIIMS · 1993
Red bone marrow of the bone produces
A
Lymphocytes
B
Eosinophils
C
Plasma
D
$RBC$

Solution

(D) In adults, red bone marrow $(myeloid \text{ tissue})$ is the primary site for hematopoiesis. It is responsible for the production of red blood cells $(RBCs)$, granular leucocytes, and platelets.
13
BiologyMediumMCQAIIMS · 1993
The Golgi apparatus contains:
A
$DNA$
B
$RNA$
C
Phospholipids,proteins,enzymes,and vitamin $C$
D
Protein-lipid-protein

Solution

(C) The Golgi apparatus is a membrane-bound organelle primarily involved in the modification,sorting,and packaging of proteins and lipids.
It contains various components such as phospholipids,proteins,enzymes (involved in glycosylation),and vitamin $C$ (ascorbic acid),which is concentrated within the Golgi cisternae for various metabolic processes.
Therefore,option $C$ is the correct answer.
14
BiologyMediumMCQAIIMS · 1993
The fluid part of the cell called cell sap is the:
A
Non-living contents of a cell
B
Living contents of a cell
C
Non-living contents of the vacuole of a cell
D
Living contents of the vacuole of a cell

Solution

(C) The cell sap is the liquid content found within the vacuole of a plant cell.
It consists of water,minerals,sugars,amino acids,and waste products.
Since the vacuole is a membrane-bound organelle and the sap itself is a solution of solutes,it is considered the non-living content of the vacuole.
15
BiologyEasyMCQAIIMS · 1993
Which of the following is not a human salivary gland?
A
Parotid
B
Submaxillary
C
Sublingual
D
Infra-orbital

Solution

(D) The human salivary glands consist of three pairs: the Parotid,Submaxillary (also known as Submandibular),and Sublingual glands.
$Infra-orbital$ glands are not present in humans; they are found in other mammals such as rabbits.
Therefore,the correct option is $D$.
16
BiologyMediumMCQAIIMS · 1993
Which of the following statements is correct?
A
Though secretin is an enzyme,it is not involved in digestion.
B
Secretin is an enzyme and so it helps digestion.
C
Secretin is a hormone but it plays a role in digestion.
D
Secretin is a hormone and hence it does not play any role in digestion.

Solution

(C) $Secretin$ is a hormone secreted by the mucosa of the duodenum.
It stimulates the secretion of pancreatic juice and increases duodenal movements.
Thus,it plays an important role in the process of digestion.
17
BiologyEasyMCQAIIMS · 1993
Which of the following can be called 'animal starch'?
A
Hemicellulose
B
Glucose
C
Glycogen
D
Chitin

Solution

(C) $Glycogen$ is a branched polymer of glucose. It is stored primarily in the muscles and liver of animals, and due to its structural similarity to starch, it is commonly referred to as 'animal starch'.
18
BiologyEasyMCQAIIMS · 1993
The intestine is different from the stomach by the presence of
A
Digestive gland
B
Villi
C
Sub-mucosa
D
Serosa

Solution

(B) The correct answer is $B$.
Intestinal villi are small,finger-like projections that significantly increase the surface area for the absorption of nutrients.
These structures are present in the small intestine but are completely absent in the stomach.
19
BiologyMediumMCQAIIMS · 1993
The vital capacity of the lung signifies the volume of air
A
Breathed in during normal inspiration
B
Breathed out with forcible expiration
C
Breathed in with forcible inspiration
D
With deep inspiration and forcible expiration

Solution

(D) Vital capacity $(VC)$ is defined as the maximum volume of air a person can breathe in after a forced expiration. This includes the Inspiratory Reserve Volume $(IRV)$,Tidal Volume $(TV)$,and Expiratory Reserve Volume $(ERV)$. Mathematically,$VC = IRV + TV + ERV$. It represents the total amount of air that can be exhaled after a deep inspiration followed by a forcible expiration.
20
BiologyEasyMCQAIIMS · 1993
In mammals,how much $CO_2$ is transported as bicarbonates of sodium and potassium in the blood?
A
$5-10\%$
B
$10-90\%$
C
$70-72\%$
D
$90-95\%$

Solution

(C) The transport of $CO_2$ in the blood occurs in three forms:
$1$. About $7\%$ of $CO_2$ is transported in a dissolved state through the blood plasma.
$2$. About $20-25\%$ of $CO_2$ is transported as carbaminohaemoglobin bound to haemoglobin.
$3$. About $70\%$ of $CO_2$ is transported as bicarbonates (mainly sodium and potassium bicarbonates) in the blood plasma.
Therefore,the correct percentage range is $70-72\%$.
21
BiologyMediumMCQAIIMS · 1993
What is the quantity of $CO_2$ transported in the form of bicarbonates by blood (including both plasma and $RBC$) (in $\%$)?
A
$75$
B
$85$
C
$2$
D
$98$

Solution

(A) Carbon dioxide $(CO_2)$ is transported in the blood in three forms:
$1$. Dissolved in plasma: Approximately $7\%$ of $CO_2$ is carried in a dissolved state through the plasma.
$2$. As carbaminohemoglobin: About $20-25\%$ of $CO_2$ binds with hemoglobin to form carbaminohemoglobin.
$3$. As bicarbonates: The majority of $CO_2$ (approximately $70-75\%$) is transported as bicarbonates $(HCO_3^-)$ in the plasma and $RBCs$.
Therefore,the correct option is $75\%$.
22
BiologyEasyMCQAIIMS · 1993
The enzyme mediating the reactions of the $Krebs$ cycle during cellular respiration in animals is found in:
A
Cytoplasm
B
Ribosome
C
Lysosome
D
Mitochondria

Solution

(D) The $Krebs$ cycle,also known as the citric acid cycle or $TCA$ cycle,takes place in the mitochondrial matrix of eukaryotic cells.
All the enzymes required for the $Krebs$ cycle reactions are located within the mitochondrial matrix,except for succinate dehydrogenase,which is bound to the inner mitochondrial membrane.
Therefore,the correct location for these enzymes is the mitochondria.
23
BiologyEasyMCQAIIMS · 1993
Hamburger phenomenon is also known as
A
Hydrogen shift mechanism
B
Chloride shift mechanism
C
Carbonic acid shift mechanism
D
Sodium-potassium pump

Solution

(B) The Hamburger phenomenon is also known as the chloride shift.
To maintain the electrostatic neutrality of the plasma,chloride ions $(Cl^-)$ diffuse from the plasma into the $RBCs$ in exchange for bicarbonate ions $(HCO_3^-)$ that pass out of the $RBCs$ into the plasma.
This process occurs when oxygenated blood becomes deoxygenated,leading to an increase in the chloride content of the $RBCs$.
24
BiologyMediumMCQAIIMS · 1993
When a man inhales air containing a normal concentration of ${O_2}$ as well as $CO$,he suffers from suffocation because:
A
$CO$ reacts with ${O_2}$ reducing its percentage in air.
B
Haemoglobin combines with $CO$ instead of ${O_2}$ and the product cannot dissociate.
C
$CO$ affects the diaphragm and intercostal muscles.
D
$CO$ affects the nerves of the lungs.

Solution

(B) Carbon monoxide $(CO)$ has approximately $210$ times more affinity for haemoglobin compared to oxygen $({O_2})$.
When $CO$ is inhaled,it binds with haemoglobin to form a very stable compound called carboxyhaemoglobin.
Because this bond is much stronger than the oxyhaemoglobin bond,the haemoglobin is unable to transport oxygen to the body tissues effectively.
This leads to cellular hypoxia,which the body perceives as suffocation.
25
BiologyMediumMCQAIIMS · 1993
If an animal is made anaemic in an experiment,the production of which hormone will be increased?
A
Erythrocytin
B
Erythroblastin
C
Erythropoietin
D
Enkephalin

Solution

(C) When an animal is anaemic,the oxygen-carrying capacity of the blood decreases,leading to tissue hypoxia.
In response to low oxygen levels,the kidneys release a hormone called erythropoietin.
Erythropoietin stimulates the bone marrow to increase the production of red blood cells (erythropoiesis) to restore oxygen levels in the blood.
26
BiologyEasyMCQAIIMS · 1993
The protein which prevents coagulation of blood in the blood vessel is
A
Platelets
B
Globular protein
C
Albumin
D
Heparin

Solution

(D) Heparin is a natural anticoagulant produced by the liver and mast cells.
It prevents the clotting of blood inside the blood vessels by inhibiting the formation of thrombin from prothrombin.
Therefore,it keeps the blood in a fluid state within the circulatory system.
27
BiologyEasyMCQAIIMS · 1993
Epiphyseal discs,which are present at the ends of long bones,are responsible for:
A
Bone elongation
B
Growth of thickness of the bone
C
Remodelling the shape of bone
D
Formation of Haversian canal

Solution

(A) The epiphyseal plate,also known as the epiphyseal disc or growth plate,is a layer of hyaline cartilage located between the diaphysis (shaft) and the epiphysis (end) of long bones.
During the period of growth,the chondrocytes in this plate undergo rapid division and ossification,which leads to the lengthening of the bone.
Therefore,epiphyseal discs are primarily responsible for bone elongation.
28
BiologyMediumMCQAIIMS · 1993
The thermoregulatory centre is situated in
A
Spinal cord
B
Pituitary body
C
Cerebellum
D
Hypothalamus

Solution

(D) The $Hypothalamus$ is a part of the forebrain that contains several centers which control body temperature,urge for eating and drinking. It acts as the body's thermostat,maintaining homeostasis by regulating heat production and heat loss.
29
BiologyMediumMCQAIIMS · 1993
The nerves leading to the central nervous system are called:
A
Efferent
B
Afferent
C
Motor
D
None

Solution

(B) The nerves that carry impulses from the receptor organs to the $CNS$ (Central Nervous System) are known as afferent or sensory nerves. These nerves consist exclusively of sensory nerve fibers that transmit sensory information toward the brain and spinal cord.
30
BiologyEasyMCQAIIMS · 1993
Which part of the mammalian brain controls muscular coordination?
A
Cerebrum
B
Medulla oblongata
C
Cerebellum
D
Corpus callosum

Solution

(C) The $Cerebellum$ is the part of the hindbrain responsible for maintaining posture, orientation, and equilibrium of the body.
It achieves this by coordinating and regulating the tone and contraction of voluntary muscles, ensuring smooth and precise motor movements.
31
BiologyMediumMCQAIIMS · 1993
The anterior choroid plexus in the brain of man covers:
A
Corpora bigemina
B
Medulla oblongata
C
Diencephalon
D
Mesencephalon

Solution

(C) The anterior choroid plexus is a vascular structure found in the roof of the $III$rd ventricle of the brain.
Since the $III$rd ventricle is located within the diencephalon,the anterior choroid plexus covers the diencephalon.
32
BiologyEasyMCQAIIMS · 1993
Meissner's corpuscles are located in
A
Pancreas and secrete trypsinogen
B
Adrenal and secrete epinephrine
C
Spleen and destroy worn out erythrocytes
D
Skin and perceive gentle pressure

Solution

(D) $Meissner$ corpuscles are specialized sensory receptors located in the dermis of the skin,specifically in the dermal papillae.
They are mechanoreceptors that are highly sensitive to light touch and low-frequency vibration.
Therefore,they are primarily responsible for perceiving gentle pressure on the skin surface.
33
BiologyEasyMCQAIIMS · 1993
The cavity of the vitreous humour in the eyes is situated:
A
Behind the lens
B
In front of the lens
C
Behind the retina
D
Between the retina and sclerotic

Solution

(A) The human eye contains two main chambers separated by the lens.
$1$. The anterior chamber (aqueous chamber) is located between the cornea and the lens,containing aqueous humour.
$2$. The posterior chamber (vitreous chamber) is the large space located behind the lens and in front of the retina,which is filled with a transparent gel-like substance called vitreous humour.
34
BiologyEasyMCQAIIMS · 1993
The space between the lens and the cornea of the human eye is known as:
A
Vitreous chamber
B
Aqueous chamber
C
Retina
D
Iris

Solution

(B) The human eye is divided into two main chambers by the lens.
$1$. The space between the cornea and the lens is called the $Aqueous \ chamber$.
$2$. This chamber is filled with a thin,watery fluid called $Aqueous \ humor$.
$3$. The space between the lens and the retina is called the $Vitreous \ chamber$,which is filled with a transparent gel called $Vitreous \ humor$.
35
BiologyMediumMCQAIIMS · 1993
Which of the following is not necessarily a property of all hormones?
A
Information carrying
B
Secreted in low amounts
C
Short half-life
D
Protein in nature

Solution

(D) Hormones are chemical messengers that carry information to target cells. They are typically secreted in very low amounts and generally have a short half-life to ensure precise regulation. However,hormones are not exclusively proteinaceous in nature. They belong to diverse chemical groups,including peptides,proteins,amines,and steroids. Therefore,being proteinaceous is not a universal property of all hormones.
36
BiologyEasyMCQAIIMS · 1993
Diabetes insipidus is caused due to the deficiency of
A
Oxytocin
B
Insulin
C
Vasopressin
D
Glucagon

Solution

(C) $Vasopressin$, also known as $Antidiuretic$ $Hormone$ $(ADH)$, is secreted by the posterior pituitary gland.
Its primary function is to stimulate the reabsorption of water and electrolytes by the distal tubules of the kidney, thereby reducing water loss through urine.
$A$ deficiency of $Vasopressin$ leads to the inability of the kidneys to conserve water, resulting in the excretion of a large volume of dilute urine. This condition is known as $Diabetes$ $insipidus$.
37
BiologyMediumMCQAIIMS · 1993
Nucleotides are building blocks of nucleic acids. Each nucleotide is a composite molecule formed by:
A
$(Base-sugar-phosphate)_n$
B
Base-sugar-$OH$
C
Base-sugar-phosphate
D
Sugar-phosphate

Solution

(C) Nucleotides are the fundamental building blocks or monomeric units of nucleic acids ($DNA$ and $RNA$).
Each nucleotide consists of three distinct components:
$1$. $A$ nitrogenous base (Purines or Pyrimidines).
$2$. $A$ pentose sugar (Ribose or Deoxyribose).
$3$. $A$ phosphate group (derived from phosphoric acid).
Therefore,the correct composition is Base-sugar-phosphate.
38
BiologyEasyMCQAIIMS · 1993
Antibody formation and immunity production is done by a protein called globulin present in the
A
Stroma of $RBC$
B
Haemoglobin of $RBC$
C
Plasma
D
Blood platelets

Solution

(C) Globulins are a group of proteins present in the blood plasma.
They are primarily synthesized by the liver and the immune system.
Globulins are essential for liver function,blood clotting,and fighting infection.
Specifically,immunoglobulins (also known as antibodies) are a type of globulin that plays a crucial role in the immune response by identifying and neutralizing foreign pathogens like bacteria and viruses.
39
BiologyMediumMCQAIIMS · 1993
Newborn babies generally do not shiver despite low temperatures because of:
A
The skin is not covered by hair
B
Brown fat,which has $20$ times greater heat value than white fat
C
White fat,which has $20$ times greater heat value than brown fat
D
The skeleton is not very strong

Solution

(B) Newborn babies do not shiver in response to cold because they possess a specialized tissue called 'brown fat' (or brown adipose tissue).
Brown fat is rich in mitochondria and has a dense supply of blood vessels and sympathetic adrenergic nerves.
When infants are exposed to cold,the metabolism of brown fat is activated,which generates heat through a process called non-shivering thermogenesis.
This tissue provides significantly more heat energy compared to white fat,helping the newborn maintain body temperature without the need for shivering.
40
BiologyMediumMCQAIIMS · 1993
Which of the following was used by Hershey and Chase to prove that $DNA$ is the chemical basis of heredity?
A
$TMV$
B
Cauliflower mosaic virus
C
Dahlia mosaic virus
D
$T_2$ bacteriophage

Solution

(D) Alfred Hershey and Martha Chase conducted experiments in $1952$ to determine whether $DNA$ or protein was the genetic material.
They used the $T_2$ bacteriophage,a virus that infects bacteria $(E. coli)$.
They grew some viruses in a medium containing radioactive phosphorus $(^{32}P)$ to label $DNA$ and others in a medium containing radioactive sulfur $(^{35}S)$ to label proteins.
After allowing the bacteriophages to infect the bacteria,they found that only radioactive phosphorus $(^{32}P)$ was transferred into the bacterial cells,proving that $DNA$ is the genetic material.
41
BiologyEasyMCQAIIMS · 1993
The plant whose seeds are known to have the longest viability period is
A
Nelumbo nucifera (lotus)
B
Triticum vulgare (wheat)
C
Zizyphus jujuba (ber)
D
Carica papaya (papaya)

Solution

(A) Seed viability refers to the ability of a seed to germinate after a certain period of time.
Among the given options,the seeds of $Nelumbo\, nucifera$ (lotus) exhibit the longest viability,which ranges from $400$ to $2000$ years.
Therefore,the correct option is $A$.
42
BiologyMediumMCQAIIMS · 1993
The chemical which causes damage to $WBC$, bone marrow, spleen, lymph nodes, and lungs is
A
Iodine-$131$
B
Calcium
C
Strontium-$90$
D
Iodine-$127$

Solution

(C) $Strontium-90$ is a radioactive isotope that behaves chemically similar to calcium.
Due to this similarity, it is absorbed by the body and deposited in the bones instead of calcium.
Once deposited, it continuously emits ionizing radiation, which damages the bone marrow, $WBC$ production, spleen, lymph nodes, and lungs.
This prolonged exposure can lead to severe health issues, including tumors and bone cancer.
43
BiologyEasyMCQAIIMS · 1993
Infertility is believed to be due to the lack of which vitamin?
A
$A$
B
$B$
C
$C$
D
$E$

Solution

(D) Vitamin $E$ is essential for the normal functioning of reproductive organs,which is why it is often referred to as the fertility vitamin.
Deficiency of vitamin $E$ can lead to sterility (infertility) and muscular atrophy.
44
BiologyMediumMCQAIIMS · 1993
After culturing the anther of a plant,a few diploid plants were found along with haploid plants. The diploid plants could have arisen from
A
Generative cell of pollen
B
Cells of anther wall
C
Vegetative cell of pollen
D
Exine of pollen wall

Solution

(B) In anther culture,haploid plants are typically produced from the microspores (pollen grains). However,if diploid plants are observed,they usually arise from the somatic cells of the anther wall (such as the tapetum,middle layers,or endothecium). These cells are diploid $(2n)$ in nature and can undergo callus formation and subsequent organogenesis to produce diploid plants,unlike the haploid microspores which produce haploid plants.
45
BiologyMediumMCQAIIMS · 1993
How many meiotic divisions are necessary to produce $100$ pollen grains?
A
$100$
B
$25$
C
$50$
D
$20$

Solution

(B) In flowering plants,one pollen mother cell $(PMC)$ undergoes meiosis to produce four haploid microspores (pollen grains).
Therefore,to produce $100$ pollen grains,the number of meiotic divisions required is calculated as:
Number of meiotic divisions = $\frac{\text{Total number of pollen grains}}{4} = \frac{100}{4} = 25$.
Thus,$25$ meiotic divisions are necessary.
46
BiologyMediumMCQAIIMS · 1993
Which part of the reproductive structure produces both enzymes and hormones?
A
Archegonium
B
Middle layer
C
Tapetum
D
Endothecium

Solution

(C) The correct answer is $C$.
The $Tapetum$ is the innermost layer of the microsporangium wall.
Its primary functions include the production of enzymes,hormones like $IAA$ (Indole$-3-$acetic acid),and the secretion of nutrients required for the development of pollen grains.
47
BiologyMediumMCQAIIMS · 1993
Nucellar embryo is
A
Apomictic embryo
B
Amphimictic haploid
C
Adventitive embryony
D
Amphimictic diploid

Solution

(C) Nucellar embryony is a type of apomixis where the embryo develops directly from the diploid nucellar cells surrounding the embryo sac.
Since the nucellar cells are diploid $(2n)$ and they develop into an embryo without fertilization,this process is known as adventitive embryony.
Therefore,nucellar embryo is an example of adventitive embryony.
48
BiologyMediumMCQAIIMS · 1993
Triploid plants can be obtained from the culture of:
A
Pollen
B
Endosperm
C
Ovule
D
Megaspore

Solution

(B) In angiosperms,the endosperm is formed as a result of double fertilization,specifically through the fusion of one male gamete with two polar nuclei (triple fusion).
This process results in a triploid $(3n)$ tissue.
Since the endosperm cells are naturally triploid,culturing these cells through tissue culture techniques can lead to the development of triploid plants.
49
BiologyMediumMCQAIIMS · 1993
The sequence of development of the embryo sac is:
A
Archesporium $\rightarrow$ megaspore mother cell $\rightarrow$ megaspore $\rightarrow$ embryo sac
B
Archesporium $\rightarrow$ megaspore $\rightarrow$ megaspore mother cell $\rightarrow$ embryo sac
C
Archesporium $\rightarrow$ megaspore $\rightarrow$ megasporophyte $\rightarrow$ embryo sac
D
None of the above

Solution

(A) The development of the embryo sac (female gametophyte) follows a specific sequence:
$1$. The archesporial cell differentiates in the nucellus.
$2$. It functions as the megaspore mother cell $(MMC)$.
$3$. The $MMC$ undergoes meiosis to form four haploid megaspores.
$4$. Three megaspores degenerate,and the remaining functional megaspore develops into the embryo sac through mitosis.
50
BiologyMediumMCQAIIMS · 1993
When the pollen grains are not transferred from anthers to the stigma in a flower due to a physical barrier or fence,or when some natural barrier exists between the androecium and gynoecium to check self-pollination,it is known as:
A
Heterostyly
B
Herkogamy
C
Dichogamy
D
Cleistogamy

Solution

(B) Herkogamy is a mechanical or physical barrier that prevents self-pollination in bisexual flowers.
In this mechanism,the spatial arrangement of the anthers and the stigma is such that the pollen grains from the anthers cannot easily reach the stigma of the same flower.
This physical separation or barrier ensures cross-pollination by preventing self-pollination.
Therefore,the correct option is $B$.
51
BiologyMediumMCQAIIMS · 1993
Fertilization in humans,rabbits,and other placental mammals occurs in:
A
Ovary
B
Uterus
C
Fallopian tubes
D
Vagina

Solution

(C) In mammals,including humans and rabbits,fertilization of the ovum takes place in the fallopian tube (also known as the oviduct or uterine tube). The sperm meets the secondary oocyte in the ampullary region of the fallopian tube,where the fusion of male and female gametes occurs to form a zygote.
52
BiologyEasyMCQAIIMS · 1993
The fluid filled in the blastocoel cavity of blastula is
A
Acidic
B
Albuminous
C
Saline
D
Pure water

Solution

(B) The blastocoel is the primary cavity formed during the blastula stage of embryonic development.
This cavity is filled with an albuminous fluid.
This fluid is secreted by the surrounding blastomeres (cells of the blastula) to provide a suitable environment for further development.
53
BiologyMediumMCQAIIMS · 1993
During the fourth and sixth cleavage of the zygote in sea urchin development,what is the fate of the resulting cells?
A
Mesomeres give rise to endoderm
B
Micromeres give rise to ectoderm
C
Mesomeres give rise to ectoderm
D
Macromeres give rise to ectoderm

Solution

(C) In sea urchin development,the cleavage pattern is radial.
During the fourth cleavage,the animal hemisphere cells divide meridionally to form eight cells called mesomeres.
The vegetal hemisphere cells divide equatorially to form four macromeres and four micromeres.
Mesomeres are destined to form the ectoderm of the embryo.
Macromeres contribute to the endoderm and mesoderm,while micromeres give rise to the primary mesenchyme (mesoderm).
54
BiologyMediumMCQAIIMS · 1993
Pure homozygous offsprings in a dihybrid cross in the $F_2$ generation will be
A
$1/2$
B
$1/4$
C
$1/8$
D
$1/16$

Solution

(C) In a dihybrid cross,the $F_2$ generation produces $16$ possible combinations of genotypes.
Out of these $16$ combinations,only $2$ are completely homozygous: the double dominant $(AABB)$ and the double recessive $(aabb)$.
Therefore,the proportion of pure homozygous offspring is $2/16$,which simplifies to $1/8$.
Thus,the correct option is $1/8$.
55
BiologyMediumMCQAIIMS · 1993
$A$ farmer crossed a walnut-combed chicken with a single-combed one and obtained all walnut-combed chickens in $F_1$. The genotype of the parents was
A
$Rr\ Pp \times rr\ pp$
B
$RR\ PP \times rr\ pp$
C
$RR\ pp \times rr\ pp$
D
$RR\ Pp \times rr\ pp$

Solution

(B) In chickens,comb shape is determined by two gene pairs,$R$ and $P$. The interaction is as follows: Walnut comb $(R-P-)$,Rose comb $(R-pp)$,Pea comb $(rrP-)$,and Single comb $(rrpp)$.
Since the $F_1$ generation consists entirely of walnut-combed chickens $(R-P-)$,the walnut-combed parent must be homozygous dominant for both traits $(RR\ PP)$ to ensure that all offspring receive at least one dominant allele for both genes ($R$ and $P$).
The single-combed parent is recessive for both traits,having the genotype $rr\ pp$.
Therefore,the cross is $RR\ PP \times rr\ pp$.
56
BiologyMediumMCQAIIMS · 1993
Normal maize has starchy seeds which remain smooth when dry. $A$ mutant form has sugary seeds which go crinkled when dry. When a mutant was crossed with a normal plant,an $F_1$ was produced which had smooth seeds. What would be the relative ratios of the different seed types,if the $F_1$ was allowed to self-pollinate?
A
$1$ smooth : $3$ sugary
B
$3$ smooth : $1$ sugary
C
$1$ smooth : $1$ sugary
D
All sugary

Solution

(B) $1$. The trait for smooth seeds is dominant over the trait for crinkled (sugary) seeds.
$2$. Let the dominant allele for smooth seeds be $S$ and the recessive allele for crinkled seeds be $s$.
$3$. The cross between a mutant (homozygous recessive,$ss$) and a normal plant (homozygous dominant,$SS$) produces an $F_1$ generation with the genotype $Ss$,which expresses the dominant smooth phenotype.
$4$. When the $F_1$ $(Ss)$ is allowed to self-pollinate $(Ss \times Ss)$,the offspring follow the Mendelian monohybrid ratio.
$5$. The Punnett square for $Ss \times Ss$ results in genotypes: $1 SS : 2 Ss : 1 ss$.
$6$. Phenotypically,this corresponds to $3$ smooth seeds ($SS$ and $Ss$) and $1$ crinkled (sugary) seed $(ss)$.
$7$. Therefore,the ratio is $3$ smooth : $1$ sugary.
57
BiologyMediumMCQAIIMS · 1993
During a serological test in which anti-human serum is mixed with the blood of another animal,the blood of which animal gives the thickest precipitate?
A
Gibbon
B
Chimpanzee
C
Dog
D
Mule

Solution

(B) The serological test,specifically the precipitin test,measures the degree of evolutionary relationship between species based on the similarity of their blood proteins.
Anti-human serum contains antibodies produced by injecting human blood proteins into another animal.
When this serum is mixed with the blood of other animals,the intensity of the precipitate formed depends on the degree of similarity between the animal's blood proteins and human blood proteins.
Among the given options,the $Chimpanzee$ is the closest evolutionary relative to humans.
Therefore,the blood of a $Chimpanzee$ shows the highest degree of protein similarity to human blood,resulting in the thickest precipitate.
58
BiologyMediumMCQAIIMS · 1993
The genotype of an individual with Down's syndrome is:
A
$45 + XX$
B
$44 + XY$
C
$44 + XXY$
D
$22 + XY$

Solution

(A) Down's syndrome is a chromosomal disorder caused by the presence of an additional copy of chromosome number $21$ (trisomy of $21$).
This condition results in a total of $47$ chromosomes instead of the normal $46$.
For a female with Down's syndrome,the genotype is $45 + XX$,and for a male,it is $45 + XY$.
Among the given options,$45 + XX$ represents the genotype of a female with Down's syndrome.
59
BiologyMediumMCQAIIMS · 1993
The sex determination pattern in honeybee is called
A
Female haploidy
B
Haplodiploidy
C
Gametic diploidy
D
Gametogony

Solution

(B) The sex determination system in honeybees is known as $Haplodiploidy$.
In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives.
Females ($queens$ and $workers$) are $diploid$ $(2n = 32)$, developing from fertilized eggs.
Males ($drones$) are $haploid$ $(n = 16)$, developing from unfertilized eggs through a process called $parthenogenesis$.
60
BiologyMediumMCQAIIMS · 1993
The expression of genes for the production of milk in only females is a
A
$X$-linked character
B
$Y$-linked character
C
Sex-limited genes
D
Sex-influenced genes

Solution

(C) The correct answer is $C$.
Milk production is a trait controlled by sex-limited genes.
Sex-limited genes are autosomal genes whose expression is restricted to only one sex due to the physiological or hormonal environment of that specific sex.
Although these genes are present in both males and females,they are only expressed in females because the hormonal regulation required for milk production is absent in males.
61
BiologyEasyMCQAIIMS · 1993
Which disease is caused by a nematode?
A
Leprosy
B
Poliomyelitis
C
Amoebiasis
D
Filariasis

Solution

(D) Filariasis, also known as $Elephantiasis$, is caused by filarial worms, which are nematodes such as $Wuchereria$ $bancrofti$ and $Wuchereria$ $malayi$.
Leprosy is caused by the bacterium $Mycobacterium$ $leprae$.
Poliomyelitis is a viral disease caused by the $Polio$ $virus$.
Amoebiasis is caused by the protozoan parasite $Entamoeba$ $histolytica$.
62
BiologyEasyMCQAIIMS · 1993
$A$ non-infectious,unnatural,and unusual reaction of a person to any substance or condition for which they are hypersensitive is termed as:
A
Infection
B
Immunity
C
Allergy
D
Toxin

Solution

(C) An allergy is an exaggerated response of the immune system to certain antigens present in the environment.
These antigens are called allergens.
The immune system produces $IgE$ antibodies in response to these allergens,leading to the release of chemicals like histamine and serotonin from mast cells,which cause allergic symptoms.
Therefore,the correct term for this hypersensitive reaction is Allergy.
63
BiologyMediumMCQAIIMS · 1993
Which of the following organisms is known to form abscesses in human liver,lungs,brain,etc.?
A
Entamoeba histolytica
B
Monocystis
C
Plasmodium
D
Fasciola hepatica

Solution

(A) $Entamoeba \ histolytica$ is a protozoan parasite in the large intestine of humans which causes amoebiasis (amoebic dysentery).
When the parasite invades the intestinal mucosa,it can enter the bloodstream and travel to other organs like the liver,lungs,and brain.
In these organs,it causes the formation of abscesses (tissue destruction and pus formation),which is a serious complication of extra-intestinal amoebiasis.
64
BiologyEasyMCQAIIMS · 1993
The insect that is not found in the wild state is
A
Lac insect
B
Cochineal insect
C
Honey bee
D
Silk moth

Solution

(D) The $Bombyx mori$, commonly known as the silk moth, is a completely domesticated insect. It has been selectively bred for thousands of years for the production of silk. Consequently, it has lost its ability to fly and survive in the wild, making it entirely dependent on humans for its life cycle and reproduction.
65
BiologyEasyMCQAIIMS · 1993
Which of the following animals has become almost extinct in India?
A
Wolf
B
Rhinoceros
C
Hippopotamus
D
Cheetah

Solution

(D) The $Cheetah$ $(Acinonyx jubatus)$ was declared extinct in India in $1952$ due to hunting and habitat loss. While recent reintroduction programs have brought some individuals to Kuno National Park, it is historically and ecologically recognized as the species that became extinct in the wild in India.
66
BiologyEasyMCQAIIMS · 1993
The living and dividing layer of epidermis is
A
Stratum granulosum
B
Stratum corneum
C
Stratum germinativum
D
Stratum spongiosum

Solution

(C) The $Stratum$ $germinativum$, also known as the $Stratum$ $basale$, is the deepest layer of the epidermis.
This layer consists of a single row of cuboidal or columnar cells that are actively undergoing mitosis.
These cells continuously divide to produce new cells that are pushed outward to replace the cells shed from the surface of the skin.
67
BiologyEasyMCQAIIMS · 1993
Stratum Malpighii is a part of
A
Epidermis
B
Stratum spongiosum
C
Stratum compactum
D
Renal tubule

Solution

(A) The $Stratum Malpighii$, also known as the $Stratum basale$ or basal layer, is the deepest layer of the $Epidermis$ in the skin.
It consists of a single layer of cuboidal or columnar cells that are actively dividing to produce new skin cells.
68
BiologyEasyMCQAIIMS · 1993
The finger nails develop from the
A
Dermis
B
Epidermis
C
Bone
D
Cartilage

Solution

(B) Finger nails are derivatives of the skin. They are formed by the keratinization of the epidermal cells. Specifically,the nail plate is derived from the stratum germinativum of the epidermis. Therefore,the correct origin of finger nails is the epidermis.
69
BiologyEasyMCQAIIMS · 1993
Modified sweat glands on the eyelid of a rabbit are known as:
A
Meibomian gland
B
Retinal gland
C
Lachrymal gland
D
Pituitary gland

Solution

(C) The correct answer is $C$.
Lachrymal glands are associated with the eyes and are considered modified sweat glands.
They secrete tears,which help in keeping the eyes and eyelids clean,lubricated,and moist.

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