BiologyQ1–100 of 177 questions
Page 1 of 2 · English
1
BiologyMediumMCQAIPMT · 1996
$A$ group of plants with similar traits of any rank is known as:
A
Species
B
Genus
C
Order
D
Taxon
Solution
(D) In biological classification,a $Taxon$ (plural: $Taxa$) represents a group of organisms of any rank that share similar traits or characteristics. It is a fundamental unit of classification used in taxonomy to categorize living beings at various levels,such as species,genus,family,order,class,phylum,or kingdom.
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2
BiologyMediumMCQAIPMT · 1996
What is the meaning of 'taxon' in the classification of animals?
A
$A$ group of the same species
B
$A$ group of animals based on the number of chromosomes
C
$A$ group of any rank in the classification of organisms
D
$A$ group of similar genera
Solution
(C) In biological classification,a 'taxon' (plural: taxa) represents a taxonomic group of any rank.
It is a unit of classification that represents a group of organisms that are considered to be related and are placed together in a specific category (e.g.,species,genus,family,order,class,phylum,or kingdom).
Therefore,a taxon is a group of any rank in the classification of organisms.
It is a unit of classification that represents a group of organisms that are considered to be related and are placed together in a specific category (e.g.,species,genus,family,order,class,phylum,or kingdom).
Therefore,a taxon is a group of any rank in the classification of organisms.
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3
BiologyEasyMCQAIPMT · 1996
The influenza virus contains which of the following genetic materials?
A
$RNA$
B
$DNA$
C
Neither $RNA$ nor $DNA$
D
Both $DNA$ and $RNA$
Solution
(A) The influenza virus is an enveloped virus that contains a single-stranded $RNA$ genome. Viruses are obligate parasites and contain either $DNA$ or $RNA$ as their genetic material,but never both. Therefore,the correct option is $A$.
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4
BiologyMediumMCQAIPMT · 1996
Nucleic acids in chromosomes in bacteria are
A
Two types of $DNA$ and $RNA$
B
Linear $DNA$
C
Circular $DNA$
D
Linear $RNA$
Solution
(C) Bacteria are prokaryotic organisms. Their genetic material is not enclosed within a nuclear envelope. The bacterial chromosome consists of a single,large,circular $DNA$ molecule,which is often referred to as the nucleoid. Therefore,the nucleic acid present in the bacterial chromosome is circular $DNA$.
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5
BiologyMediumMCQAIPMT · 1996
The sex factor in bacteria is:
A
$F-$ replicon
B
Chromosomal replicon
C
$RNA$
D
Sex pilus
Solution
(A) The sex factor,also known as the $F-$factor (fertility factor),is an extra-chromosomal,circular $DNA$ molecule found in certain bacteria.
It can exist either as an independent plasmid or integrated into the bacterial chromosome.
Bacteria possessing this factor are referred to as $F^+$ cells and can act as donors during the process of conjugation.
It can exist either as an independent plasmid or integrated into the bacterial chromosome.
Bacteria possessing this factor are referred to as $F^+$ cells and can act as donors during the process of conjugation.
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6
BiologyMediumMCQAIPMT · 1996
Blue-green algae are included in
A
Eukaryotes
B
Rhodophyceae
C
Prokaryotes
D
Chlorophyceae
Solution
(C) Blue-green algae,also known as cyanobacteria,are photosynthetic organisms that lack a membrane-bound nucleus and other membrane-bound organelles.
Because they possess a primitive cellular structure,they are classified under the kingdom $Monera$.
Therefore,blue-green algae are considered $Prokaryotes$.
Because they possess a primitive cellular structure,they are classified under the kingdom $Monera$.
Therefore,blue-green algae are considered $Prokaryotes$.
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7
BiologyEasyMCQAIPMT · 1996
Which statement is wrong about lichens?
A
Some species are eaten by reindeers
B
Lichens are indicators of pollution
C
They grow rapidly about $2 \ cm$ per day
D
They have a symbiotic relationship between alga and fungus
Solution
(C) The correct answer is $C$. Lichens are extremely slow-growing organisms. They do not grow at a rate of $2 \ cm$ per day; rather,they typically grow at a very slow rate,often around $1 \ mm$ per year. The other statements are correct: some species like $Cladonia$ (reindeer moss) are consumed by reindeers,they are well-known bio-indicators of air pollution (especially $SO_2$),and they represent a symbiotic association between an algal component (phycobiont) and a fungal component (mycobiont).
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8
BiologyMediumMCQAIPMT · 1996
Seed habit originated firstly in some
A
Pteridophytes
B
Pines
C
Monocots
D
Dicots
Solution
(A) The seed habit is an evolutionary process that leads to the development of seeds.
It was first observed in certain pteridophytes,such as $Selaginella$ and $Marsilea$.
These plants exhibit heterospory (production of two types of spores) and the retention of the megaspore within the parent plant,which are the essential precursors to the development of a seed habit.
It was first observed in certain pteridophytes,such as $Selaginella$ and $Marsilea$.
These plants exhibit heterospory (production of two types of spores) and the retention of the megaspore within the parent plant,which are the essential precursors to the development of a seed habit.
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9
BiologyEasyMCQAIPMT · 1996
Which of the following is a living fossil?
A
Pinus
B
Ginkgo
C
Thuja
D
Deodar
Solution
(B) $Ginkgo$ $biloba$, a gymnosperm, is a classic example of a living fossil. It is a monotypic genus, meaning it is the only surviving member of its entire division, $Ginkgophyta$. At present, it is naturally confined to the eastern part of China and Japan.
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10
BiologyMediumMCQAIPMT · 1996
Radial symmetry is often exhibited by animals having
A
One opening of alimentary canal
B
Aquatic mode of living
C
Benthos/sedentary nature
D
Ciliary mode of feeding
Solution
(C) Radial symmetry is a body plan where the organism can be divided into identical halves by any plane passing through the central axis. This type of symmetry is most commonly observed in animals that are sessile (sedentary) or slow-moving,such as Cnidarians and adult Echinoderms. Being sedentary allows these animals to interact with their environment from all directions equally,making radial symmetry an evolutionary advantage for such a lifestyle.
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11
BiologyMediumMCQAIPMT · 1996
The coelom is the body cavity between the alimentary canal and the body wall,which is lined by:
A
Ectoderm and endoderm
B
Mesoderm and ectoderm
C
Ectoderm on both sides
D
Mesoderm on both sides
Solution
(D) The coelom is a fluid-filled body cavity that lies between the body wall and the digestive tract (alimentary canal).
It is specifically defined as a cavity that is lined on all sides by the mesoderm.
If the body cavity is lined by mesoderm on both sides,the organism is considered a true coelomate (eucoelomate).
It is specifically defined as a cavity that is lined on all sides by the mesoderm.
If the body cavity is lined by mesoderm on both sides,the organism is considered a true coelomate (eucoelomate).
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12
BiologyMediumMCQAIPMT · 1996
Which characteristic is universal for sponges (Phylum Porifera)?
A
Marine habitat
B
Calcareous spicules
C
Radial symmetry
D
High regenerative power
Solution
(D) Sponges belong to the Phylum $Porifera$.
$A$. Most sponges are marine,but some like $Spongilla$ are freshwater. Thus,it is not universal.
$B$. Spicules can be calcareous or siliceous,or they may be composed of spongin fibers. Thus,it is not universal.
$C$. Sponges are generally asymmetrical; they do not exhibit radial symmetry.
$D$. Sponges possess a high regenerative power,which is a characteristic feature of their body organization,allowing them to regrow lost parts or reproduce asexually through fragmentation. This is considered a universal trait among sponges.
$A$. Most sponges are marine,but some like $Spongilla$ are freshwater. Thus,it is not universal.
$B$. Spicules can be calcareous or siliceous,or they may be composed of spongin fibers. Thus,it is not universal.
$C$. Sponges are generally asymmetrical; they do not exhibit radial symmetry.
$D$. Sponges possess a high regenerative power,which is a characteristic feature of their body organization,allowing them to regrow lost parts or reproduce asexually through fragmentation. This is considered a universal trait among sponges.
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13
BiologyMediumMCQAIPMT · 1996
In sponges,the canal system develops due to:
A
Gastrovascular system
B
Folding of inner walls
C
Porous walls
D
Reproduction
Solution
(B) In sponges (Phylum $Porifera$),the body wall consists of two layers: an outer $pinacoderm$ and an inner $choanoderm$. The complex canal system,also known as the $aquiferous$ system,is formed due to the extensive folding and invagination of these inner body walls. This system facilitates water circulation,which is essential for nutrition,respiration,and excretion.
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14
BiologyMediumMCQAIPMT · 1996
Nephridia of Earthworm are analogous to
A
Nematoblasts of Hydra
B
Tracheae of insects
C
Flame cells of Planaria
D
Gills of Prawn
Solution
(C) Nephridia are the excretory organs of earthworms (Phylum $Annelida$).
Flame cells (protonephridia) are the excretory organs of flatworms like Planaria (Phylum $Platyhelminthes$).
Since both structures perform the same function of excretion and osmoregulation despite having different evolutionary origins,they are considered analogous organs.
Flame cells (protonephridia) are the excretory organs of flatworms like Planaria (Phylum $Platyhelminthes$).
Since both structures perform the same function of excretion and osmoregulation despite having different evolutionary origins,they are considered analogous organs.
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15
BiologyEasyMCQAIPMT · 1996
The flightless bird,cassowary,is found in:
A
Australia
B
New Zealand
C
Indonesia
D
Mauritius
Solution
(A) The cassowary is a large flightless bird belonging to the genus $Casuarius$. It is primarily native to the tropical forests of New Guinea,nearby islands,and northeastern Australia. Therefore,among the given options,Australia is the correct geographical region where they are found.
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16
BiologyMediumMCQAIPMT · 1996
Pneumatic bones are found in
A
Domestic lizard
B
Tadpole of frog
C
Flying lizard
D
Pigeon
Solution
(D) Pneumatic bones are hollow, air-filled bones that reduce body weight, which is an essential adaptation for flight in birds.
Among the given options, the pigeon $(Columba \text{ livia})$ belongs to the class $Aves$.
Birds possess pneumatic bones to facilitate flight.
Therefore, the correct option is $D$.
Among the given options, the pigeon $(Columba \text{ livia})$ belongs to the class $Aves$.
Birds possess pneumatic bones to facilitate flight.
Therefore, the correct option is $D$.
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17
BiologyMediumMCQAIPMT · 1996
The organelles whose major function is the storage of hydrolytic enzymes are:
A
Centrioles
B
Chromoplasts
C
Lysosomes
D
Chloroplasts
Solution
(C) Lysosomes are membrane-bound vesicular structures formed by the process of packaging in the Golgi apparatus.
These organelles are rich in hydrolytic enzymes (hydrolases like lipases,proteases,and carbohydrases) that are optimally active at an acidic $pH$.
These enzymes are capable of digesting carbohydrates,proteins,lipids,and nucleic acids,which is why lysosomes are often referred to as the 'suicide bags' of the cell.
These organelles are rich in hydrolytic enzymes (hydrolases like lipases,proteases,and carbohydrases) that are optimally active at an acidic $pH$.
These enzymes are capable of digesting carbohydrates,proteins,lipids,and nucleic acids,which is why lysosomes are often referred to as the 'suicide bags' of the cell.
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18
BiologyMediumMCQAIPMT · 1996
The largest physical and chemical molecules in biological systems are:
A
Carbohydrates
B
Lipids
C
Proteins
D
Nucleic acids
Solution
(D) Nucleic acids,specifically $DNA$ and $RNA$,are the largest macromolecules found in living organisms. $DNA$ molecules can reach extremely high molecular weights,often consisting of millions of base pairs,making them significantly larger than typical proteins,carbohydrates,or lipids.
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19
BiologyMediumMCQAIPMT · 1996
During glycolysis,the enzyme hexokinase converts glucose to glucose-$6$-phosphate. The activity of hexokinase is inhibited by glucose-$6$-phosphate through which of the following mechanisms?
A
Feedback inhibition
B
Positive feedback
C
Competitive inhibition
D
Non-competitive inhibition
Solution
(A) In glycolysis,hexokinase catalyzes the phosphorylation of glucose to glucose-$6$-phosphate.
Glucose-$6$-phosphate acts as an allosteric inhibitor of the enzyme hexokinase.
This is a classic example of feedback inhibition (or end-product inhibition),where the product of a metabolic pathway inhibits an enzyme that acts early in the pathway to regulate the flux of the reaction and prevent the over-accumulation of the product.
Glucose-$6$-phosphate acts as an allosteric inhibitor of the enzyme hexokinase.
This is a classic example of feedback inhibition (or end-product inhibition),where the product of a metabolic pathway inhibits an enzyme that acts early in the pathway to regulate the flux of the reaction and prevent the over-accumulation of the product.
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20
BiologyEasyMCQAIPMT · 1996
$DNA$ replication takes place in,or the $DNA$ molecule of each chromosome becomes double in:
A
$G_1$ phase
B
$G_2$ phase
C
$S$ phase
D
Mitotic phase
Solution
(C) During the cell cycle,the $S$ phase (Synthesis phase) is the period of interphase where $DNA$ replication occurs.
In this phase,the amount of $DNA$ per cell doubles,although the number of chromosomes remains the same.
Therefore,each chromosome consists of two sister chromatids after the $S$ phase.
In this phase,the amount of $DNA$ per cell doubles,although the number of chromosomes remains the same.
Therefore,each chromosome consists of two sister chromatids after the $S$ phase.
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21
BiologyEasyMCQAIPMT · 1996
Which of the following elements is not essential for plants?
A
Iron
B
Zinc
C
Potassium
D
Iodine
Solution
(D) The essential elements for plants are classified into macronutrients and micronutrients based on their quantitative requirements.
Iron $(Fe)$,Zinc $(Zn)$,and Potassium $(K)$ are well-documented essential nutrients for plant growth and development.
Iodine $(I)$ is not considered an essential element for plants,although it is a vital trace element for animals,particularly for the synthesis of thyroid hormones.
Iron $(Fe)$,Zinc $(Zn)$,and Potassium $(K)$ are well-documented essential nutrients for plant growth and development.
Iodine $(I)$ is not considered an essential element for plants,although it is a vital trace element for animals,particularly for the synthesis of thyroid hormones.
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22
BiologyEasyMCQAIPMT · 1996
Which of the following is a micro-element in plants?
A
Manganese
B
Nitrogen
C
Magnesium
D
Calcium
Solution
(A) Essential elements are classified into two categories based on their quantitative requirements: macronutrients and micronutrients.
$1$. Macronutrients are generally present in plant tissues in large amounts (in excess of $10 \text{ mmol kg}^{-1}$ of dry matter).
$2$. Micronutrients or trace elements are needed in very small amounts (less than $10 \text{ mmol kg}^{-1}$ of dry matter).
$3$. Among the given options, $Nitrogen$, $Magnesium$, and $Calcium$ are macronutrients.
$4$. $Manganese$ $(Mn)$ is a micronutrient required by plants for various metabolic processes, including the photolysis of water during photosynthesis.
$1$. Macronutrients are generally present in plant tissues in large amounts (in excess of $10 \text{ mmol kg}^{-1}$ of dry matter).
$2$. Micronutrients or trace elements are needed in very small amounts (less than $10 \text{ mmol kg}^{-1}$ of dry matter).
$3$. Among the given options, $Nitrogen$, $Magnesium$, and $Calcium$ are macronutrients.
$4$. $Manganese$ $(Mn)$ is a micronutrient required by plants for various metabolic processes, including the photolysis of water during photosynthesis.
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23
BiologyEasyMCQAIPMT · 1996
Entry of mineral ions in plant root cells by diffusion is
A
Passive absorption
B
Active absorption
C
Osmosis
D
Endocytosis
Solution
(A) The movement of mineral ions into plant root cells via diffusion occurs along the concentration gradient without the expenditure of metabolic energy $(ATP)$.
This process is known as passive absorption.
Passive absorption is driven by purely physical forces such as diffusion and mass flow,rather than active transport mechanisms.
This process is known as passive absorption.
Passive absorption is driven by purely physical forces such as diffusion and mass flow,rather than active transport mechanisms.
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24
BiologyEasyMCQAIPMT · 1996
The law of limiting factors for photosynthesis was enunciated by
A
Blackman
B
Hill
C
Ruben
D
Kalmen
Solution
(A) The law of limiting factors for photosynthesis was proposed by $F.F. Blackman$ in $1905$.
According to this law,if a chemical process is affected by more than one factor,then its rate will be determined by the factor which is nearest to its minimal value,as it is the factor which directly affects the process if its quantity is changed.
$Blackman$ also proposed the existence of a light-independent (dark) phase in photosynthesis.
According to this law,if a chemical process is affected by more than one factor,then its rate will be determined by the factor which is nearest to its minimal value,as it is the factor which directly affects the process if its quantity is changed.
$Blackman$ also proposed the existence of a light-independent (dark) phase in photosynthesis.
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25
BiologyEasyMCQAIPMT · 1996
Photosynthetically active radiation $(PAR)$ represents the following range of wavelength:
A
$340-450 \,nm$
B
$400-700 \,nm$
C
$500-600 \,nm$
D
$450-950 \,nm$
Solution
(B) Photosynthetically active radiation $(PAR)$ is the spectral range of solar radiation from $400$ to $700 \,nm$ that photosynthetic organisms are able to use in the process of photosynthesis.
This range corresponds to the visible light spectrum,which is essential for the excitation of chlorophyll molecules to initiate the light-dependent reactions.
This range corresponds to the visible light spectrum,which is essential for the excitation of chlorophyll molecules to initiate the light-dependent reactions.
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26
BiologyMediumMCQAIPMT · 1996
How many turns of the Calvin cycle are required to form one hexose molecule?
A
$2$
B
$6$
C
$4$
D
$8$
Solution
(B) The Calvin cycle fixes one molecule of $CO_2$ per turn.
Since a hexose molecule (glucose,$C_6H_{12}O_6$) contains $6$ carbon atoms,the cycle must complete $6$ turns to incorporate $6$ molecules of $CO_2$ to produce one molecule of glucose.
Since a hexose molecule (glucose,$C_6H_{12}O_6$) contains $6$ carbon atoms,the cycle must complete $6$ turns to incorporate $6$ molecules of $CO_2$ to produce one molecule of glucose.
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27
BiologyEasyMCQAIPMT · 1996
The $CO_2$ acceptor in $C_3$ plants is:
A
Xylulose-$5$-phosphate
B
$3$-phosphoglyceric acid
C
Ribulose $1,5$-bisphosphate
D
Phosphoenolpyruvic acid
Solution
(C) In $C_3$ plants,the primary $CO_2$ acceptor is a $5$-carbon ketose sugar called Ribulose $1,5$-bisphosphate $(RuBP)$.
This reaction is catalyzed by the enzyme RuBisCO,which leads to the formation of two molecules of $3$-phosphoglyceric acid ($3$-$PGA$),which is the first stable product of the Calvin cycle.
This reaction is catalyzed by the enzyme RuBisCO,which leads to the formation of two molecules of $3$-phosphoglyceric acid ($3$-$PGA$),which is the first stable product of the Calvin cycle.
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28
BiologyMediumMCQAIPMT · 1996
Which of the following is present in the Calvin cycle?
A
Photophosphorylation
B
Oxidative carboxylation
C
Reductive carboxylation
D
Oxidative phosphorylation
Solution
(C) The Calvin cycle involves the process of reductive carboxylation.
In the carboxylation phase,$CO_2$ is fixed to a $5$-carbon sugar,Ribulose $1,5$-bisphosphate $(RuBP)$,to form $3$-phosphoglycerate ($3$-$PGA$).
This reaction is catalyzed by the enzyme $RuBisCO$.
Following this,the reduction phase uses $ATP$ and $NADPH$ (produced in the light reaction) to convert $3$-$PGA$ into triose phosphate,which is a reductive process.
Therefore,the overall pathway is characterized as reductive carboxylation.
In the carboxylation phase,$CO_2$ is fixed to a $5$-carbon sugar,Ribulose $1,5$-bisphosphate $(RuBP)$,to form $3$-phosphoglycerate ($3$-$PGA$).
This reaction is catalyzed by the enzyme $RuBisCO$.
Following this,the reduction phase uses $ATP$ and $NADPH$ (produced in the light reaction) to convert $3$-$PGA$ into triose phosphate,which is a reductive process.
Therefore,the overall pathway is characterized as reductive carboxylation.
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29
BiologyMediumMCQAIPMT · 1996
The chlorophyll $a$ molecule has which of the following at its carbon atom $3$ of the pyrrole ring $II$?
A
Aldehyde group
B
Methyl group
C
Carboxylic group
D
Magnesium
Solution
(B) Chlorophyll $a$ is a pigment molecule consisting of a porphyrin ring with a central magnesium atom.
In the structure of chlorophyll $a$,the pyrrole ring $II$ contains a methyl group $(-CH_3)$ attached to its carbon atom $3$.
In contrast,chlorophyll $b$ has an aldehyde group $(-CHO)$ at the same position.
In the structure of chlorophyll $a$,the pyrrole ring $II$ contains a methyl group $(-CH_3)$ attached to its carbon atom $3$.
In contrast,chlorophyll $b$ has an aldehyde group $(-CHO)$ at the same position.
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30
BiologyEasyMCQAIPMT · 1996
The products of anaerobic fermentation are:
A
Alcohol and lipoprotein
B
Ether and nucleic acid
C
Protein and nucleic acid
D
Alcohol,lactic acid,and similar compounds
Solution
(D) Anaerobic respiration or fermentation is a metabolic process that converts sugar to acids,gases,or alcohol in the absence of oxygen.
In yeast,fermentation results in the production of ethanol $(C_2H_5OH)$ and carbon dioxide $(CO_2)$.
In certain bacteria and animal muscle cells,fermentation results in the production of lactic acid $(C_3H_6O_3)$.
Therefore,the products of fermentation include alcohol,lactic acid,and similar organic compounds.
In yeast,fermentation results in the production of ethanol $(C_2H_5OH)$ and carbon dioxide $(CO_2)$.
In certain bacteria and animal muscle cells,fermentation results in the production of lactic acid $(C_3H_6O_3)$.
Therefore,the products of fermentation include alcohol,lactic acid,and similar organic compounds.
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31
BiologyMediumMCQAIPMT · 1996
In oxidative phosphorylation,oxidation and phosphorylation take place simultaneously and form
A
$NADP$
B
$DPN$
C
$Pyruvic\ acid$
D
$ATP$
Solution
(D) Oxidative phosphorylation is the process in which $ATP$ is synthesized during the electron transport system $(ETS)$.
In this process,the energy released during the oxidation of reduced coenzymes ($NADH$ and $FADH_2$) is used to phosphorylate $ADP$ to form $ATP$.
Therefore,the final product formed through this coupled reaction is $ATP$.
In this process,the energy released during the oxidation of reduced coenzymes ($NADH$ and $FADH_2$) is used to phosphorylate $ADP$ to form $ATP$.
Therefore,the final product formed through this coupled reaction is $ATP$.
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32
BiologyMediumMCQAIPMT · 1996
The reactions of the $Krebs$ cycle take place in:
A
In cytoplasm
B
In endoplasmic reticulum
C
In the matrix of mitochondria
D
On the surface of the mitochondrion
Solution
(C) The $Krebs$ cycle,also known as the citric acid cycle or $TCA$ cycle,is a series of chemical reactions used by all aerobic organisms to generate energy.
In eukaryotic cells,the enzymes required for the $Krebs$ cycle are located within the mitochondrial matrix.
Therefore,the entire process of the $Krebs$ cycle occurs in the matrix of the mitochondria.
In eukaryotic cells,the enzymes required for the $Krebs$ cycle are located within the mitochondrial matrix.
Therefore,the entire process of the $Krebs$ cycle occurs in the matrix of the mitochondria.
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33
BiologyMediumMCQAIPMT · 1996
Which one of the following is the correct match of digestive enzyme and its substrate?
A
Lactose-Renin
B
Starch-Maltose
C
Fat-Steapsin
D
Casein-Trypsin
Solution
(C) is the correct answer.
Pancreatic lipase,historically known as $Steapsin$,is the primary enzyme responsible for the digestion of fats.
It hydrolyzes fats (triglycerides) into glycerol and fatty acids.
In other options: $Lactose$ is digested by $Lactase$,$Starch$ is digested by $Amylase$ (into $Maltose$),and $Casein$ is digested by $Renin$ (in infants) or $Pepsin$.
Pancreatic lipase,historically known as $Steapsin$,is the primary enzyme responsible for the digestion of fats.
It hydrolyzes fats (triglycerides) into glycerol and fatty acids.
In other options: $Lactose$ is digested by $Lactase$,$Starch$ is digested by $Amylase$ (into $Maltose$),and $Casein$ is digested by $Renin$ (in infants) or $Pepsin$.
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34
BiologyEasyMCQAIPMT · 1996
Which one of the following sets is a polysaccharide group?
A
Glucose,fructose,lactose
B
Starch,glycogen,cellulose
C
Sucrose,maltose,glucose
D
Galactose,starch,sucrose
Solution
(B) The correct answer is $B$.
Carbohydrates are classified based on the number of sugar units:
$1$. Monosaccharides: These are simple sugars that cannot be hydrolyzed further (e.g.,glucose,fructose,galactose).
$2$. Disaccharides: These consist of two monosaccharide units linked together (e.g.,sucrose,maltose,lactose).
$3$. Polysaccharides: These are complex carbohydrates consisting of long chains of monosaccharide units (e.g.,starch,glycogen,cellulose,dextrins).
Therefore,starch,glycogen,and cellulose are all polysaccharides.
Carbohydrates are classified based on the number of sugar units:
$1$. Monosaccharides: These are simple sugars that cannot be hydrolyzed further (e.g.,glucose,fructose,galactose).
$2$. Disaccharides: These consist of two monosaccharide units linked together (e.g.,sucrose,maltose,lactose).
$3$. Polysaccharides: These are complex carbohydrates consisting of long chains of monosaccharide units (e.g.,starch,glycogen,cellulose,dextrins).
Therefore,starch,glycogen,and cellulose are all polysaccharides.
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35
BiologyEasyMCQAIPMT · 1996
Pellagra is caused due to the deficiency of
A
Thiamine
B
Ascorbic acid
C
Niacin $(B_3)$
D
Calciferol
Solution
(C) Pellagra is a disease caused by the deficiency of Niacin,which is also known as Vitamin $B_3$ or Nicotinic acid.
It is often referred to as the $PP$ factor (Pellagra-Preventing factor).
The symptoms of pellagra are commonly described by the 'four Ds': dermatitis,diarrhea,dementia,and death.
It is often referred to as the $PP$ factor (Pellagra-Preventing factor).
The symptoms of pellagra are commonly described by the 'four Ds': dermatitis,diarrhea,dementia,and death.
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36
BiologyMediumMCQAIPMT · 1996
What does the doctor advise to patients suffering from high blood cholesterol?
A
Red mutton with fat layer
B
Vegetable and margarine
C
Vegetable oil such as ground-nut oil
D
Pure deshi ghee or butter
Solution
(C) Patients with high blood cholesterol are advised to consume vegetable oils because they contain unsaturated fatty acids. These oils have low molecular weight lipids and are healthier compared to saturated fats found in animal products like red meat,ghee,or butter,which increase the risk of cardiovascular diseases.
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37
BiologyMediumMCQAIPMT · 1996
$500\,ml$ respiratory volume in a normal adult human is related to:
A
Residual volume
B
Total lung capacity
C
Inspiratory reserve volume
D
Tidal volume
Solution
(D) The volume of air inspired or expired during a normal respiration is called Tidal Volume $(TV)$.
For a healthy adult human,the $TV$ is approximately $500\,ml$ per breath.
Therefore,$500\,ml$ respiratory volume corresponds to the Tidal volume.
For a healthy adult human,the $TV$ is approximately $500\,ml$ per breath.
Therefore,$500\,ml$ respiratory volume corresponds to the Tidal volume.
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38
BiologyMediumMCQAIPMT · 1996
Which of the following performs the same function as nephridia in earthworms?
A
Flame cells in liver fluke
B
Myotomes in fish
C
Statocysts in prawn
D
Parotid gland in toad
Solution
(A) Nephridia are the excretory organs in earthworms,responsible for osmoregulation and the removal of nitrogenous wastes.
Flame cells (protonephridia) perform the same excretory and osmoregulatory functions in Platyhelminthes (e.g.,liver fluke).
Myotomes are involved in locomotion in fish.
Statocysts are balancing organs in prawns.
Parotid glands are associated with poison secretion in toads.
Flame cells (protonephridia) perform the same excretory and osmoregulatory functions in Platyhelminthes (e.g.,liver fluke).
Myotomes are involved in locomotion in fish.
Statocysts are balancing organs in prawns.
Parotid glands are associated with poison secretion in toads.
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39
BiologyMediumMCQAIPMT · 1996
Which one of the following pairs of waste substances is removed from the blood in the ornithine cycle?
A
$CO_2$ and urea
B
Ammonia and urea
C
$CO_2$ and ammonia
D
Urea and sodium salt
Solution
(C) The ornithine cycle,also known as the urea cycle,occurs in the liver.
It is a metabolic pathway that converts toxic ammonia $(NH_3)$ and carbon dioxide $(CO_2)$ into urea.
Therefore,these two waste substances are removed from the blood during this process.
It is a metabolic pathway that converts toxic ammonia $(NH_3)$ and carbon dioxide $(CO_2)$ into urea.
Therefore,these two waste substances are removed from the blood during this process.
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40
BiologyMediumMCQAIPMT · 1996
Neural stimulation in visceral organs in human beings is done by:
A
Sympathetic and parasympathetic nerves and is under involuntary action
B
Sympathetic nerves and is under voluntary action
C
Sympathetic and parasympathetic nerves and is under voluntary action
D
Parasympathetic nerves and is under voluntary action
Solution
(A) The human neural system is divided into the Central Neural System $(CNS)$ and the Peripheral Neural System $(PNS)$.
The $PNS$ is further divided into the Somatic Neural System and the Autonomic Neural System $(ANS)$.
The $ANS$ transmits impulses from the $CNS$ to the involuntary organs and smooth muscles of the body.
The $ANS$ is classified into the Sympathetic and Parasympathetic neural systems.
These systems regulate the activities of visceral organs,which are not under the conscious control of the individual,meaning they are under involuntary action.
The $PNS$ is further divided into the Somatic Neural System and the Autonomic Neural System $(ANS)$.
The $ANS$ transmits impulses from the $CNS$ to the involuntary organs and smooth muscles of the body.
The $ANS$ is classified into the Sympathetic and Parasympathetic neural systems.
These systems regulate the activities of visceral organs,which are not under the conscious control of the individual,meaning they are under involuntary action.
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41
BiologyMediumMCQAIPMT · 1996
Corneal grafting is especially successful because:
A
Its technique is very easy.
B
Preservation of the cornea is easy.
C
The cornea is not connected to the circulatory system and the immune system.
D
The cornea is easily available.
Solution
(C) The cornea is an avascular tissue,meaning it does not have a direct blood supply. Because it lacks blood vessels,it is not exposed to the immune system's surveillance cells or circulating antibodies. This lack of vascularization makes the cornea an 'immune-privileged' site,significantly reducing the risk of graft rejection during corneal transplantation.
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42
BiologyMediumMCQAIPMT · 1996
According to one of the theories of ageing, the decline and disappearance of which gland by late middle age is the primary cause of ageing?
A
Parathyroid
B
Thyroid
C
Thymus
D
Posterior lobe of pituitary
Solution
(C) The correct answer is $(C)$. According to the immunity theory of ageing, there is a direct link between the ageing process and the gradual involution (disappearance) of the thymus gland by late middle age in humans.
The thymus gland plays a crucial role in the development of the immune system by maturing $T$-lymphocytes.
As the thymus gland degenerates with age, the body's natural defense mechanism against foreign pathogens weakens, and the production of abnormal $(defective)$ cells increases, which contributes to the physiological decline associated with ageing.
The thymus gland plays a crucial role in the development of the immune system by maturing $T$-lymphocytes.
As the thymus gland degenerates with age, the body's natural defense mechanism against foreign pathogens weakens, and the production of abnormal $(defective)$ cells increases, which contributes to the physiological decline associated with ageing.
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43
BiologyEasyMCQAIPMT · 1996
$A$ true coelom is a body cavity that is located between the body wall and the alimentary canal. It is lined by:
A
Mesoderm on one side and ectoderm on the other side
B
Endoderm on one side and ectoderm on the other side
C
Mesoderm on both the sides
D
Ectoderm on both the sides
Solution
(C) true coelom,or eucoelom,is a fluid-filled body cavity that is completely lined by mesoderm on both sides. This mesodermal lining allows for the separation of the body wall from the gut,facilitating independent movement and the development of complex organ systems. In contrast,a pseudocoelom is only partially lined by mesoderm.
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44
BiologyMediumMCQAIPMT · 1996
Exchange of genetic material between chromatids of homologous chromosomes during meiosis is called
A
Synapsis
B
Chiasmata
C
Transformation
D
Crossing over
Solution
(D) The exchange of genetic material between non-sister chromatids of homologous chromosomes during meiosis is known as $Crossing \ over$.
It occurs during the $Pachytene$ sub-stage of $Prophase-I$ of meiosis.
Although the process occurs in $Pachytene$, the $X$-shaped structures formed as a result, known as $Chiasmata$, become visible during the $Diplotene$ sub-stage of $Prophase-I$.
It occurs during the $Pachytene$ sub-stage of $Prophase-I$ of meiosis.
Although the process occurs in $Pachytene$, the $X$-shaped structures formed as a result, known as $Chiasmata$, become visible during the $Diplotene$ sub-stage of $Prophase-I$.
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45
BiologyMediumMCQAIPMT · 1996
Lampbrush chromosomes are visible in:
A
Diplotene of meiosis
B
Prophase of meiosis
C
Interphase
D
Metaphase of meiosis
Solution
(A) Lampbrush chromosomes are a special type of chromosome found in the oocytes of most vertebrates (except mammals) and some invertebrates. They are highly extended and are most clearly visible during the $diplotene$ stage of $prophase-I$ of meiosis. During this stage,the chromosomes exhibit a characteristic 'lampbrush' appearance due to the presence of lateral loops that are sites of active transcription.
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46
BiologyMediumMCQAIPMT · 1996
$A$ cholera patient is administered a 'saline drip' because
A
$Na^+$ ions are essential for the transport of substances across the membrane
B
$Na^+$ ions are helpful in conserving water in the body
C
$Cl^-$ ions are helpful in the formation of $HCl$ for digestion
D
$Cl^-$ ions are a significant component of blood plasma
Solution
(A) Cholera is caused by the bacterium $Vibrio \text{ cholerae}$, which produces an enterotoxin that leads to severe diarrhea and vomiting.
This results in massive loss of water and electrolytes, particularly sodium $(Na^+)$ and chloride $(Cl^-)$ ions, from the body.
Dehydration and electrolyte imbalance can lead to circulatory collapse.
$A$ saline drip (isotonic saline) is administered to replenish the lost water and electrolytes.
$Na^+$ ions are particularly crucial because they play a vital role in the active transport of various substances (like glucose and amino acids) across the intestinal membrane, which helps in the reabsorption of water, thereby preventing further dehydration.
This results in massive loss of water and electrolytes, particularly sodium $(Na^+)$ and chloride $(Cl^-)$ ions, from the body.
Dehydration and electrolyte imbalance can lead to circulatory collapse.
$A$ saline drip (isotonic saline) is administered to replenish the lost water and electrolytes.
$Na^+$ ions are particularly crucial because they play a vital role in the active transport of various substances (like glucose and amino acids) across the intestinal membrane, which helps in the reabsorption of water, thereby preventing further dehydration.
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47
BiologyMediumMCQAIPMT · 1996
$2, 4-D$ is an effective:
A
Insecticide
B
Herbicide
C
Fungicide
D
Rodenticide
Solution
(B) $2, 4-D$ stands for $2, 4$-dichlorophenoxyacetic acid.
It is a synthetic auxin that acts as a systemic herbicide.
It is widely used in agriculture to control broad-leaved weeds without affecting monocot crops like wheat,maize,and rice.
It is a synthetic auxin that acts as a systemic herbicide.
It is widely used in agriculture to control broad-leaved weeds without affecting monocot crops like wheat,maize,and rice.
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48
BiologyMediumMCQAIPMT · 1996
$Azotobacter$ and $Bacillus \text{ } polymyxa$ are:
A
Decomposers
B
Nonsymbiotic nitrogen fixers
C
Symbiotic nitrogen fixers
D
Pathogenic bacteria
Solution
(B) $Azotobacter$ and $Bacillus \text{ } polymyxa$ are free-living or non-symbiotic nitrogen-fixing bacteria found in the soil.
These bacteria convert atmospheric nitrogen into ammonia, which is then utilized by plants.
By enriching the soil with nitrogen, they increase soil fertility and consequently improve the yield of crop plants.
These bacteria convert atmospheric nitrogen into ammonia, which is then utilized by plants.
By enriching the soil with nitrogen, they increase soil fertility and consequently improve the yield of crop plants.
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49
BiologyMediumMCQAIPMT · 1996
The cerebrum,which wraps around the thalamus,is a major coordinating center for:
A
Sensory signaling.
B
Motor signaling.
C
Motor and sensory signaling.
D
Not a neural part of the brain.
Solution
(C) The cerebrum is the largest part of the human brain. It wraps around the thalamus. The thalamus is a major coordinating center for sensory and motor signaling. Therefore,the cerebrum,which surrounds the thalamus,acts as a major coordinating center for both motor and sensory signaling.
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50
BiologyMediumMCQAIPMT · 1996
$A$ patient suffering from cholera is given a saline drip because:
A
$Cl^-$ ions are an important component of blood plasma.
B
$Na^+$ ions help to retain water in the body.
C
$Na^+$ ions are important in the transport of substances across membranes.
D
$Cl^-$ ions help in the formation of $HCl$ in the stomach for digestion.
Solution
(B) Cholera is a severe diarrheal disease caused by the bacterium $Vibrio$ $cholerae$.
During cholera,the patient loses a significant amount of water and electrolytes (especially $Na^+$ and $Cl^-$) through watery stools.
$Na^+$ ions are the primary extracellular cations that play a crucial role in maintaining osmotic pressure and fluid balance in the body.
By administering a saline drip (an isotonic solution of $NaCl$),the body's fluid volume is restored,and the $Na^+$ ions help in retaining water within the extracellular fluid,preventing severe dehydration and circulatory collapse.
During cholera,the patient loses a significant amount of water and electrolytes (especially $Na^+$ and $Cl^-$) through watery stools.
$Na^+$ ions are the primary extracellular cations that play a crucial role in maintaining osmotic pressure and fluid balance in the body.
By administering a saline drip (an isotonic solution of $NaCl$),the body's fluid volume is restored,and the $Na^+$ ions help in retaining water within the extracellular fluid,preventing severe dehydration and circulatory collapse.
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51
BiologyEasyMCQAIPMT · 1996
Two or more species occupying identical or overlapping areas are known as
A
Sympatric species
B
Allopatric species
C
Sibling species
D
Subspecies
Solution
(A) Two or more species that inhabit the same geographical area and are reproductively isolated from each other are known as $Sympatric$ species.
These species occupy the same or overlapping ranges but do not interbreed due to various ecological,behavioral,or genetic barriers.
In contrast,$Allopatric$ species are those that inhabit different geographical areas.
Therefore,the correct option is $A$.
These species occupy the same or overlapping ranges but do not interbreed due to various ecological,behavioral,or genetic barriers.
In contrast,$Allopatric$ species are those that inhabit different geographical areas.
Therefore,the correct option is $A$.
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52
BiologyEasyMCQAIPMT · 1996
Which one of the following is an exotic fish species introduced in India?
A
Clarias
B
Labeo
C
Cyprinus
D
Daphnia
Solution
(C) An exotic species is a species that is not native to a particular region. Among the given options,$Cyprinus$ (specifically $Cyprinus$ $carpio$,the common carp) is an exotic fish species that was introduced into Indian waters for aquaculture purposes. $Clarias$ (catfish) and $Labeo$ (rohu/labio) are indigenous to India. $Daphnia$ is a genus of small planktonic crustaceans,not a fish.
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53
BiologyMediumMCQAIPMT · 1996
Which hormone stimulates the secretion of milk during the sucking of milk by a baby?
A
Oxytocin
B
Relaxin
C
Prolactin
D
Progesterone
Solution
(A) $Oxytocin$ is the hormone responsible for the milk ejection reflex. When a baby sucks at the breast,sensory impulses are sent to the hypothalamus,which triggers the posterior pituitary to release $Oxytocin$. This hormone causes the contraction of myoepithelial cells surrounding the alveoli of the mammary glands,leading to the ejection (let-down) of milk. Note that $Prolactin$ is responsible for the production (synthesis) of milk,while $Oxytocin$ is responsible for its secretion (ejection).
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54
BiologyMediumMCQAIPMT · 1996
How many pollen grains are formed from $10$ microspore mother cells by meiosis?
A
$80$
B
$40$
C
$20$
D
$10$
Solution
(B) During microsporogenesis,each microspore mother cell $(MMC)$ undergoes meiosis to produce a cluster of $4$ haploid microspores,known as a microspore tetrad.
Each microspore develops into a pollen grain.
Therefore,$1$ microspore mother cell produces $4$ pollen grains.
For $10$ microspore mother cells,the total number of pollen grains formed is $10 \times 4 = 40$.
Each microspore develops into a pollen grain.
Therefore,$1$ microspore mother cell produces $4$ pollen grains.
For $10$ microspore mother cells,the total number of pollen grains formed is $10 \times 4 = 40$.
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55
BiologyMediumMCQAIPMT · 1996
In angiosperms,triple fusion is necessary for the formation of:
A
Seed coat
B
Fruit wall
C
Embryo
D
Endosperm
Solution
(D) In angiosperms,double fertilization involves two processes: syngamy and triple fusion.
Syngamy involves the fusion of one male gamete with the egg cell to form a diploid zygote,which develops into the embryo.
Triple fusion involves the fusion of the second male gamete with two polar nuclei (or a secondary nucleus) to form a primary endosperm nucleus $(PEN)$.
The $PEN$ develops into the endosperm,which provides nutrition to the developing embryo.
Therefore,triple fusion is necessary for the formation of the endosperm.
Syngamy involves the fusion of one male gamete with the egg cell to form a diploid zygote,which develops into the embryo.
Triple fusion involves the fusion of the second male gamete with two polar nuclei (or a secondary nucleus) to form a primary endosperm nucleus $(PEN)$.
The $PEN$ develops into the endosperm,which provides nutrition to the developing embryo.
Therefore,triple fusion is necessary for the formation of the endosperm.
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56
BiologyMediumMCQAIPMT · 1996
$A$ person who shows the secondary sexual characters of both male and female is called:
A
Intersex
B
Hermaphrodite
C
Bisexual
D
Gynandromorph
Solution
(D) $Gynandromorph$ is an organism that contains both male and female tissues. In the context of humans,this term is often used to describe individuals who exhibit secondary sexual characteristics of both sexes. While 'Intersex' is a broader term for variations in sex characteristics,'Gynandromorph' specifically refers to the presence of both male and female phenotypic traits.
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57
BiologyMediumMCQAIPMT · 1996
Alleles which show independent effect are called
A
Supplementary alleles
B
Codominant alleles
C
Epistatic alleles
D
Complementary alleles
Solution
(B) In codominance,both the alleles of an allelomorphic pair express themselves equally in $F_1$ hybrids. This results in a $1:2:1$ ratio both genotypically and phenotypically in the $F_2$ generation,where each allele maintains its independent effect.
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58
BiologyMediumMCQAIPMT · 1996
$A$ person who has $47$ chromosomes due to an extra $X$ chromosome is affected by
A
Turner's syndrome
B
Klinefelter's syndrome
C
Super female
D
Down's syndrome
Solution
(B) The correct answer is $B$.
Individuals with Klinefelter's syndrome possess $47$ chromosomes due to the presence of an additional $X$ chromosome,resulting in an $XXY$ genotype.
This condition occurs in males.
It is generally observed that as the number of $X$ chromosomes increases,the severity of mental defects and physical abnormalities also tends to increase.
Individuals with Klinefelter's syndrome possess $47$ chromosomes due to the presence of an additional $X$ chromosome,resulting in an $XXY$ genotype.
This condition occurs in males.
It is generally observed that as the number of $X$ chromosomes increases,the severity of mental defects and physical abnormalities also tends to increase.
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59
BiologyMediumMCQAIPMT · 1996
In human females,Barr bodies are formed by:
A
Inactivation of mother's $X$ chromosome
B
Inactivation of father's $X$ chromosome
C
Inactivation of both mother's and father's $X$ chromosomes
D
Inactivation of either mother's or father's $X$ chromosome
Solution
(D) According to the Lyon hypothesis,in human females,one of the two $X$ chromosomes in each somatic cell is randomly inactivated during early embryonic development to ensure dosage compensation.
This inactivated $X$ chromosome becomes highly condensed and is visible as a dark-staining structure known as a Barr body.
Since the inactivation is random,it can involve either the $X$ chromosome inherited from the mother or the $X$ chromosome inherited from the father in different cells.
This inactivated $X$ chromosome becomes highly condensed and is visible as a dark-staining structure known as a Barr body.
Since the inactivation is random,it can involve either the $X$ chromosome inherited from the mother or the $X$ chromosome inherited from the father in different cells.
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60
BiologyDifficultMCQAIPMT · 1996
If all the sons are haemophilic and all the daughters are normal when a haemophilic father mates with a normal mother,this character is:
A
$X$-linked recessive
B
$Y$-linked
C
$X$-linked dominant
D
Autosomal recessive
Solution
(B) Haemophilia is typically an $X$-linked recessive disorder. However,the question describes a scenario where a haemophilic father $(X^hY)$ mates with a normal mother $(XX)$.
In this case,the father passes his $Y$ chromosome to all his sons and his $X^h$ chromosome to all his daughters.
If the trait were $X$-linked recessive,the daughters would be carriers $(X^hX)$ and the sons would be normal $(XY)$.
Since the question states all sons are haemophilic and daughters are normal,this indicates the trait is passed from father to son via the $Y$ chromosome.
Therefore,this character is $Y$-linked (holandric).
In this case,the father passes his $Y$ chromosome to all his sons and his $X^h$ chromosome to all his daughters.
If the trait were $X$-linked recessive,the daughters would be carriers $(X^hX)$ and the sons would be normal $(XY)$.
Since the question states all sons are haemophilic and daughters are normal,this indicates the trait is passed from father to son via the $Y$ chromosome.
Therefore,this character is $Y$-linked (holandric).
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61
BiologyMediumMCQAIPMT · 1996
Which triplet codon acts as a stop signal for the process of translation?
A
$UCG$
B
$UAA$
C
$UAC$
D
$UGG$
Solution
(B) The process of translation is terminated when the ribosome encounters a stop codon (also known as a termination codon) on the $mRNA$ strand.
There are three stop codons in the genetic code: $UAA$,$UAG$,and $UGA$.
These codons do not code for any amino acid and signal the release of the polypeptide chain from the ribosome.
Among the given options,$UAA$ is a stop codon.
There are three stop codons in the genetic code: $UAA$,$UAG$,and $UGA$.
These codons do not code for any amino acid and signal the release of the polypeptide chain from the ribosome.
Among the given options,$UAA$ is a stop codon.
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62
BiologyMediumMCQAIPMT · 1996
Which enzyme is responsible for linking the fragments of $DNA$?
A
$DNA$ polymerase $III$
B
Endonuclease
C
$DNA$ polymerase $I$
D
$DNA$ ligase
Solution
(D) $DNA$ ligase is the enzyme responsible for joining $DNA$ fragments, specifically the Okazaki fragments, by catalyzing the formation of a phosphodiester bond between the $3'-OH$ end of one nucleotide and the $5'-phosphate$ end of another. This process is essential during $DNA$ replication to seal the nicks in the sugar-phosphate backbone.
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63
BiologyMediumMCQAIPMT · 1996
Purines of $DNA$ are represented by
A
Uracil and thymine
B
Guanine and adenine
C
Uracil and cytosine
D
Thymine and cytosine
Solution
(B) Nitrogenous bases in $DNA$ are classified into two types: Purines and Pyrimidines.
Purines are double-ring structures,which include Adenine $(A)$ and Guanine $(G)$.
Pyrimidines are single-ring structures,which include Cytosine $(C)$ and Thymine $(T)$ in $DNA$.
Therefore,the purines of $DNA$ are Adenine and Guanine.
Purines are double-ring structures,which include Adenine $(A)$ and Guanine $(G)$.
Pyrimidines are single-ring structures,which include Cytosine $(C)$ and Thymine $(T)$ in $DNA$.
Therefore,the purines of $DNA$ are Adenine and Guanine.
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64
BiologyEasyMCQAIPMT · 1996
Okazaki segments are formed during
A
Replication
B
Transduction
C
Transcription
D
Translation
Solution
(A) Okazaki segments are formed during $DNA$ replication.
During the process of $DNA$ replication,one strand of $DNA$ is synthesized continuously (leading strand),while the other strand is synthesized in short,discontinuous segments known as Okazaki fragments (lagging strand).
These fragments are later joined together by the enzyme $DNA$ ligase to form a continuous strand.
During the process of $DNA$ replication,one strand of $DNA$ is synthesized continuously (leading strand),while the other strand is synthesized in short,discontinuous segments known as Okazaki fragments (lagging strand).
These fragments are later joined together by the enzyme $DNA$ ligase to form a continuous strand.
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65
BiologyMediumMCQAIPMT · 1996
Which one of the following is the direct ancestor of modern man?
A
Australopithecus
B
Ramapithecus
C
Homo erectus
D
Homo habilis
Solution
(C) The evolutionary lineage of modern humans ($Homo$ $sapiens$) is complex. Among the given options,$Homo$ $erectus$ is widely considered a direct ancestor in the human evolutionary tree. $Homo$ $erectus$ evolved from earlier hominids like $Homo$ $habilis$ and is believed to be the ancestor from which $Homo$ $sapiens$ eventually emerged. While $Australopithecus$ and $Ramapithecus$ are early hominids,they represent earlier branches in the evolutionary timeline.
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66
BiologyAdvancedMCQAIPMT · 1996
During the evolution of man, many changes have taken place in his ancestral characters. Which one of the following is an insignificant change?
A
Change of diet from hard, tough fruits and roots to soft food
B
Qualitative improvement in the structure of hands and skills for making tools
C
Disappearance of tail
D
Improvement in speech for communication and social behaviour
Solution
(NONE) In the context of human evolution, all the listed options (dietary changes, tool-making skills, loss of tail, and development of speech) represent significant evolutionary milestones that contributed to the survival and advancement of the human species. However, if this question is evaluated based on standard biological curriculum, it is often considered a flawed or ambiguous question because none of these changes are 'insignificant'. Each played a crucial role in the transition from ancestral primates to modern humans $(Homo \text{ } sapiens)$. Therefore, there is no correct option among the choices provided as all are highly significant.
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67
BiologyEasyMCQAIPMT · 1996
Animals living in colder regions have shorter tails and ears compared to animals living in warmer regions. This phenomenon is called:
A
Bergmann's law
B
Gloger's law
C
Allen's law
D
Jordan's law
Solution
(C) The correct answer is $C$. According to $Allen's$ $Law$,mammals living in colder climates generally have shorter ears,tails,and limbs compared to those living in warmer climates. This adaptation helps in minimizing heat loss from the body surface,thereby conserving body heat in cold environments.
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68
BiologyMediumMCQAIPMT · 1996
Sensitivity to any allergen is related to
A
Deviation from the process of immunity
B
Age of the person
C
Eating habit
D
Rise in environmental temperature
Solution
(A) Allergy is an exaggerated response of the immune system to certain antigens present in the environment,which are called allergens.
This response is a result of the immune system reacting abnormally to substances that are generally harmless to most people.
Therefore,sensitivity to an allergen is a deviation from the normal process of immunity,where the body's defense mechanism overreacts.
This response is a result of the immune system reacting abnormally to substances that are generally harmless to most people.
Therefore,sensitivity to an allergen is a deviation from the normal process of immunity,where the body's defense mechanism overreacts.
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69
BiologyEasyMCQAIPMT · 1996
Passive immunity was discovered by
A
Robert Koch
B
$L$. Pasteur
C
Edward Jenner
D
Emil Von Behring
Solution
(D) Passive immunity is a type of immunity where pre-formed antibodies are directly introduced into the body to provide immediate protection.
It was discovered by Emil Von Behring and Kitasato Shibasaburo in $1890$.
They demonstrated that serum from animals immunized against diphtheria could protect other animals from the disease,a concept that led to the development of antitoxin therapy.
It was discovered by Emil Von Behring and Kitasato Shibasaburo in $1890$.
They demonstrated that serum from animals immunized against diphtheria could protect other animals from the disease,a concept that led to the development of antitoxin therapy.
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70
BiologyMediumMCQAIPMT · 1996
Which one of the following is a pair of viral diseases?
A
Tetanus and typhoid
B
Syphilis and $AIDS$
C
Whooping cough and sleeping sickness
D
Measles and rabies
Solution
(D) viral disease is caused by a virus.
$1$. Tetanus and typhoid are caused by bacteria ($Clostridium tetani$ and $Salmonella typhi$ respectively).
$2$. Syphilis is caused by a bacterium ($Treponema pallidum$), while $AIDS$ is caused by a virus ($HIV$).
$3$. Whooping cough is caused by a bacterium ($Bordetella pertussis$), and sleeping sickness is caused by a protozoan ($Trypanosoma brucei$).
$4$. Measles is caused by the Measles virus, and rabies is caused by the Rabies virus ($Lyssavirus$).
Therefore, both diseases in option $D$ are viral.
$1$. Tetanus and typhoid are caused by bacteria ($Clostridium tetani$ and $Salmonella typhi$ respectively).
$2$. Syphilis is caused by a bacterium ($Treponema pallidum$), while $AIDS$ is caused by a virus ($HIV$).
$3$. Whooping cough is caused by a bacterium ($Bordetella pertussis$), and sleeping sickness is caused by a protozoan ($Trypanosoma brucei$).
$4$. Measles is caused by the Measles virus, and rabies is caused by the Rabies virus ($Lyssavirus$).
Therefore, both diseases in option $D$ are viral.
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71
BiologyMediumMCQAIPMT · 1996
In human beings,retroviruses are considered a cause of cancer because:
A
Their genome contains an oncogene.
B
Their hereditary material is made up of single-stranded $RNA$.
C
They have a gene for reverse transcriptase.
D
Their genome may contain a cellular proto-oncogene.
Solution
(A) Retroviruses are known to cause cancer because they possess viral oncogenes $(v-onc)$.
These viral oncogenes are derived from cellular proto-oncogenes $(c-onc)$ present in the host genome.
When these viruses infect a host cell,they can integrate their genetic material into the host genome,potentially activating these proto-oncogenes or introducing viral oncogenes,which leads to uncontrolled cell division and cancer.
These viral oncogenes are derived from cellular proto-oncogenes $(c-onc)$ present in the host genome.
When these viruses infect a host cell,they can integrate their genetic material into the host genome,potentially activating these proto-oncogenes or introducing viral oncogenes,which leads to uncontrolled cell division and cancer.
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72
BiologyMediumMCQAIPMT · 1996
When a natural predator is used to control pathogenic organisms,this process is called:
A
Biological control
B
Genetic engineering
C
Confusion technique
D
Artificial control
Solution
(A) Biological control refers to the use of biological methods for controlling plant diseases and pests. This method relies on natural predation,parasitism,or other natural mechanisms rather than the introduction of chemicals. Examples include using ladybird beetles to control aphids or Bacillus thuringiensis to control butterfly caterpillars.
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73
BiologyEasyMCQAIPMT · 1996
Cochineal insects have proved very useful for the control of
A
Eichhornia
B
Cactus
C
Weeds
D
Parthenium
Solution
(B) Cochineal insects,specifically $Dactylopius$ $ceylonicus$,are biological control agents used to manage the growth of invasive $Opuntia$ (prickly pear) $Cactus$ species. These insects feed on the $Cactus$ tissues,effectively reducing their population in areas where they have become invasive weeds.
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74
BiologyEasyMCQAIPMT · 1996
The cochineal insect has been used in checking the wild growth of:
A
Opuntia
B
Eichhornia
C
Aphids
D
Screw worm
Solution
(A) The cochineal insect,specifically $Cactoblastis cactorum$,is a biological control agent.
In countries like India and Australia,the rapid and invasive spread of the prickly pear cactus,known as $Opuntia$,caused significant ecological damage.
To control this,the cochineal insect was introduced as a biological control method,which successfully fed on the $Opuntia$ plants and checked their wild growth.
In countries like India and Australia,the rapid and invasive spread of the prickly pear cactus,known as $Opuntia$,caused significant ecological damage.
To control this,the cochineal insect was introduced as a biological control method,which successfully fed on the $Opuntia$ plants and checked their wild growth.
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75
BiologyMediumMCQAIPMT · 1996
The phenomenon of using a predator for controlling a pest is known as:
A
Biological control
B
Genetic engineering
C
Artificial control
D
Confusion technique
Solution
(A) The use of natural predators or parasites to control pest populations is known as $Biological \ control$.
This method relies on natural predation rather than chemical pesticides.
For example, the $Ladybird$ beetle is used to control aphids, and the $Dragonfly$ is used to control mosquitoes.
These organisms act as biocontrol agents to maintain ecological balance.
This method relies on natural predation rather than chemical pesticides.
For example, the $Ladybird$ beetle is used to control aphids, and the $Dragonfly$ is used to control mosquitoes.
These organisms act as biocontrol agents to maintain ecological balance.
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76
BiologyMediumMCQAIPMT · 1996
Interferons are:
A
Antiviral proteins
B
Complex proteins
C
Anti-bacterial proteins
D
Anti-cancer proteins
Solution
(A) Interferons are a group of signaling proteins made and released by host cells in response to the presence of several viruses.
They are classified as $Antiviral \text{ proteins}$ because they inhibit viral replication within the host cells.
When a cell is infected by a virus, it secretes interferons, which signal neighboring cells to heighten their antiviral defenses.
They are classified as $Antiviral \text{ proteins}$ because they inhibit viral replication within the host cells.
When a cell is infected by a virus, it secretes interferons, which signal neighboring cells to heighten their antiviral defenses.
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77
BiologyMediumMCQAIPMT · 1996
The cessation of reproduction in an organism by creating hurdles in its biology or physiology,or its destruction by the use of another organism,is known as:
A
Predation
B
Competition
C
Biological control
D
Physiological control
Solution
(C) is the correct answer.
Biological control refers to the method of controlling pests or harmful organisms by using their natural predators,parasites,or pathogens instead of chemical pesticides.
This process involves manipulating the biology or physiology of the target organism to stop its reproduction or directly destroying it using another living organism.
Biological control refers to the method of controlling pests or harmful organisms by using their natural predators,parasites,or pathogens instead of chemical pesticides.
This process involves manipulating the biology or physiology of the target organism to stop its reproduction or directly destroying it using another living organism.
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78
BiologyMediumMCQAIPMT · 1996
The transfer of energy from one trophic level to another follows the second law of thermodynamics. The efficiency of energy transfer from herbivores to carnivores is: (in $\%$)
A
$25$
B
$50$
C
$10$
D
$5$
Solution
(C) According to the $10\%$ law of energy transfer in an ecosystem,proposed by Lindeman,only about $10\%$ of the energy available at one trophic level is transferred to the next trophic level.
This occurs because a significant portion of energy is lost as heat during metabolic processes,in accordance with the second law of thermodynamics.
Therefore,the efficiency of energy transfer from herbivores (primary consumers) to carnivores (secondary consumers) is $10\%$.
This occurs because a significant portion of energy is lost as heat during metabolic processes,in accordance with the second law of thermodynamics.
Therefore,the efficiency of energy transfer from herbivores (primary consumers) to carnivores (secondary consumers) is $10\%$.
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79
BiologyMediumMCQAIPMT · 1996
In a food chain, which of the following is produced in the largest amount?
A
Producers
B
Decomposers
C
Tertiary consumers
D
Primary consumers
Solution
(A) In a typical ecosystem, energy flows from producers to various levels of consumers. According to the $10\%$ law of energy transfer, only about $10\%$ of the energy available at one trophic level is transferred to the next. Therefore, the biomass and energy are highest at the base of the food chain. Producers (autotrophs) occupy the first trophic level and constitute the largest amount of biomass and energy in the ecosystem to support all higher trophic levels.
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80
BiologyMediumMCQAIPMT · 1996
If the forest area is reduced to half,which one of the following will be a long-term effect?
A
The natives (tribals) of that area will die on account of hunger.
B
Cattle of that area will die due to scarcity of fodder.
C
The diversity in germplasm will affect crop breeding.
D
It will be converted into a large desert.
Solution
(C) The correct answer is $C$. If the forest area is reduced to half,the habitats of a large variety of organisms would be destroyed. This leads to a significant loss of biodiversity and the disruption of food chains. Consequently,the loss of wild relatives of crops (germplasm) will negatively impact crop breeding programs,as these wild varieties are essential sources of disease resistance and other desirable traits.
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81
BiologyMediumMCQAIPMT · 1996
If we remove half of the forest cover of the Earth,the crisis that will occur is:
A
Many species would become extinct
B
Population,pollution,and ecological imbalance will rise
C
Energy crisis will commence
D
The remaining forest will correct the imbalance
Solution
(B) If half of the Earth's forest cover is removed,it would lead to the destruction of habitats for a vast variety of organisms. This loss of biodiversity disrupts food chains and food webs,which directly results in a significant rise in pollution,loss of species,and a severe ecological imbalance. Therefore,option $(B)$ is the most comprehensive consequence.
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82
BiologyEasyMCQAIPMT · 1996
Where do flamingos reproduce in India?
A
Chilka Lake
B
Sambhar Lake
C
Kutch
D
Mansarovar
Solution
(C) The Greater Flamingo $(Phoenicopterus \text{ roseus})$ is known to breed in the saline mudflats of the Rann of Kutch in Gujarat, India. This area, specifically the 'Flamingo City', provides the necessary nesting grounds for these birds during the breeding season.
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83
BiologyEasyMCQAIPMT · 1996
Which one of the following pairs is correctly matched regarding the wildlife and the national park/sanctuary?
A
Rann of Kutch - Wild ass
B
Gir forest - Tiger
C
Manas - Elephant
D
Corbett park - Asiatic lion
Solution
(A) The correct match is $A$. The Indian Wild Ass Sanctuary,also known as the Little Rann of Kutch Sanctuary,is located in the Rann of Kutch,Gujarat.
$B$ is incorrect because Gir Forest is famous for the Asiatic lion,not the tiger.
$C$ is incorrect because Manas National Park is primarily known for the one-horned rhinoceros and wild water buffalo,though elephants are present,it is not its primary defining feature compared to the specific match in $A$.
$D$ is incorrect because Corbett National Park is famous for the Bengal tiger,not the Asiatic lion.
$B$ is incorrect because Gir Forest is famous for the Asiatic lion,not the tiger.
$C$ is incorrect because Manas National Park is primarily known for the one-horned rhinoceros and wild water buffalo,though elephants are present,it is not its primary defining feature compared to the specific match in $A$.
$D$ is incorrect because Corbett National Park is famous for the Bengal tiger,not the Asiatic lion.
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84
BiologyEasyMCQAIPMT · 1996
In the last few decades,the most serious nuclear reactor accident and the $MIC$ gas tragedy occurred in:
A
Russia $(1990)$ and Bhopal $(1996)$
B
Ukraine $(1986)$ and Bhopal $(1984)$
C
Bhopal $(1994)$ and Russia $(1990)$
D
Ukraine and $USA$ $(1984)$
Solution
(B) The correct answer is $B$.
$1$. The Bhopal gas tragedy occurred on $3^{rd}$ December $1984$ due to the leakage of Methyl Isocyanate $(MIC)$ gas,which caused severe air pollution and the death of approximately $2500$ people.
$2$. The Chernobyl nuclear disaster occurred on April $26, 1986$,due to an explosion at the Chernobyl nuclear power station in Ukraine,which released a massive amount of radioactive material into the atmosphere.
$1$. The Bhopal gas tragedy occurred on $3^{rd}$ December $1984$ due to the leakage of Methyl Isocyanate $(MIC)$ gas,which caused severe air pollution and the death of approximately $2500$ people.
$2$. The Chernobyl nuclear disaster occurred on April $26, 1986$,due to an explosion at the Chernobyl nuclear power station in Ukraine,which released a massive amount of radioactive material into the atmosphere.
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85
BiologyEasyMCQAIPMT · 1996
Which of the following countries is responsible for releasing the largest amount of greenhouse gases?
A
Russia
B
Germany
C
Brazil
D
America $(USA)$
Solution
(D) The correct answer is $(d)$. The $USA$ is historically and currently one of the largest emitters of greenhouse gases due to its high industrial activity,extensive transportation networks,and massive consumption of fossil fuel energy.
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86
BiologyMediumMCQAIPMT · 1996
Fishes die due to sewage pollution because:
A
Of its bad smell
B
It replaces the food material of fishes
C
It increases the competition for oxygen among fishes
D
$CO_2$ is mixed in large amounts in water
Solution
(C) Sewage contains high amounts of organic matter. When this organic matter is discharged into water bodies,microorganisms (bacteria) decompose it. This decomposition process consumes a large amount of dissolved oxygen $(DO)$ from the water. As a result,the level of dissolved oxygen decreases significantly,leading to the death of fishes due to oxygen depletion.
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87
BiologyEasyMCQAIPMT · 1996
$H.J. Muller$ received the Nobel Prize for which of the following?
A
Inducing mutations using $X$-rays
B
Discovering the linkage of genes
C
Proving that $DNA$ is the genetic material
D
His studies on Drosophila for genetic research
Solution
(A) $H.J. Muller$ was awarded the Nobel Prize in Physiology or Medicine in $1946$ for his discovery that mutations can be induced by $X$-rays.
He demonstrated that $X$-rays have a mutagenic effect on the genetic material of organisms, specifically using $Drosophila$ $melanogaster$ as his model organism.
This discovery provided a powerful tool for genetic research and significantly advanced our understanding of the nature of mutations.
He demonstrated that $X$-rays have a mutagenic effect on the genetic material of organisms, specifically using $Drosophila$ $melanogaster$ as his model organism.
This discovery provided a powerful tool for genetic research and significantly advanced our understanding of the nature of mutations.
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88
BiologyEasyMCQAIPMT · 1996
The $10\%$ law of energy transfer in a food chain was given by
A
Tansley
B
Stanley
C
Lindeman
D
Weismann
Solution
(C) The $10\%$ law of energy transfer was proposed by Raymond Lindeman in $1942$.
This law states that during the transfer of energy from one trophic level to the next,only about $10\%$ of the energy is stored as biomass,while the remaining $90\%$ is lost as heat or used for metabolic activities.
This law states that during the transfer of energy from one trophic level to the next,only about $10\%$ of the energy is stored as biomass,while the remaining $90\%$ is lost as heat or used for metabolic activities.
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89
BiologyEasyMCQAIPMT · 1996
'The law of limiting factor' was proposed by
A
Blackman
B
Liebig
C
Hatch and Slack
D
Arnold
Solution
(A) The 'Law of Limiting Factors' was proposed by $F.F. Blackman$ in $1905$.
According to this law,if a physiological process is controlled by several factors,the rate of the process is limited by the factor that is present in the least amount relative to its optimum requirement.
This is a fundamental concept in the study of photosynthesis.
According to this law,if a physiological process is controlled by several factors,the rate of the process is limited by the factor that is present in the least amount relative to its optimum requirement.
This is a fundamental concept in the study of photosynthesis.
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90
BiologyEasyMCQAIPMT · 1996
Which of the following is an Indian medicinal plant?
A
Oryza sativa
B
Solanum melongena
C
Rauwolfia serpentina
D
Saccharum officinarum
Solution
(C) $Rauwolfia \text{ } serpentina$, commonly known as Sarpagandha, is a well-known medicinal plant native to the Indian subcontinent.
It is traditionally used in Ayurveda to treat high blood pressure, insomnia, and various mental disorders due to the presence of the alkaloid reserpine.
$Oryza \text{ } sativa$ is rice, $Solanum \text{ } melongena$ is brinjal, and $Saccharum \text{ } officinarum$ is sugarcane, which are primarily food or cash crops.
It is traditionally used in Ayurveda to treat high blood pressure, insomnia, and various mental disorders due to the presence of the alkaloid reserpine.
$Oryza \text{ } sativa$ is rice, $Solanum \text{ } melongena$ is brinjal, and $Saccharum \text{ } officinarum$ is sugarcane, which are primarily food or cash crops.
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91
BiologyEasyMCQAIPMT · 1996
The earliest animal domesticated by primitive man was:
A
Goat
B
Dog
C
Horse
D
Cat
Solution
(B) The correct answer is $B$. The dog $(Canis \ lupus \ familiaris)$ was the first species to be domesticated by humans. Archaeological evidence suggests that dogs were domesticated from wolves by hunter-gatherers as early as $15,000$ to $30,000$ years ago,primarily for protection and assistance in hunting.
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92
BiologyEasyMCQAIPMT · 1996
$A$ living fossil is:
A
Gnetum
B
Ginkgo
C
Riccia
D
Pinus
Solution
(B) living fossil is an organism that has remained unchanged over a long period of geological time and has no close living relatives.
$Ginkgo$ $biloba$ is a classic example of a living fossil because it is the only surviving species of the division $Ginkgophyta$,with fossils dating back over $200$ million years.
$Gnetum$,$Riccia$,and $Pinus$ do not fit the definition of a living fossil in the same context as $Ginkgo$.
$Ginkgo$ $biloba$ is a classic example of a living fossil because it is the only surviving species of the division $Ginkgophyta$,with fossils dating back over $200$ million years.
$Gnetum$,$Riccia$,and $Pinus$ do not fit the definition of a living fossil in the same context as $Ginkgo$.
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93
BiologyMediumMCQAIPMT · 1996
In India,the population growth:
A
Like some animal species,human population is likely to reach zero population growth level
B
Can be reduced by natural calamities and birth control methods
C
Like many animal species,it is also going to take sigmoid curve
D
Can be controlled by adopting national family welfare programmes
Solution
(D) The human population in India is influenced by various socio-economic factors and government initiatives.
Option $D$ is the most accurate statement because the government of India has implemented various national family welfare programmes to manage and control population growth through education,awareness,and the provision of contraceptive services.
While natural calamities can impact population size,they are not a sustainable or ethical method for population control.
Human population growth does not strictly follow a simple sigmoid curve due to complex cultural and technological interventions.
Option $D$ is the most accurate statement because the government of India has implemented various national family welfare programmes to manage and control population growth through education,awareness,and the provision of contraceptive services.
While natural calamities can impact population size,they are not a sustainable or ethical method for population control.
Human population growth does not strictly follow a simple sigmoid curve due to complex cultural and technological interventions.
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94
BiologyEasyMCQAIPMT · 1996
Interferons are ...........
A
Antiviral proteins
B
Antibacterial proteins
C
Anticancer proteins
D
Complex proteins
Solution
(A) Interferons $(IFNs)$ are a group of signaling proteins made and released by host cells in response to the presence of several viruses.
They are a type of cytokine that acts as an innate immune response.
When a cell is infected by a virus,it secretes interferons,which signal neighboring cells to heighten their antiviral defenses,thereby preventing the spread of the viral infection.
Therefore,they are classified as antiviral proteins.
They are a type of cytokine that acts as an innate immune response.
When a cell is infected by a virus,it secretes interferons,which signal neighboring cells to heighten their antiviral defenses,thereby preventing the spread of the viral infection.
Therefore,they are classified as antiviral proteins.
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95
BiologyEasyMCQAIPMT · 1996
Which of the following is an exotic Indian fish?
A
Catla catla
B
Heteropneustes fossilis
C
Cyprinus carpio
D
Labeo rohita
Solution
(C) An exotic species is a species that is introduced to a region outside its native range.
$Cyprinus$ $carpio$ (Common carp) is an exotic fish species introduced to India from Europe and Asia for aquaculture purposes.
In contrast,$Catla$ $catla$,$Heteropneustes$ $fossilis$,and $Labeo$ $rohita$ are indigenous (native) fish species found in Indian water bodies.
$Cyprinus$ $carpio$ (Common carp) is an exotic fish species introduced to India from Europe and Asia for aquaculture purposes.
In contrast,$Catla$ $catla$,$Heteropneustes$ $fossilis$,and $Labeo$ $rohita$ are indigenous (native) fish species found in Indian water bodies.
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96
BiologyMediumMCQAIPMT · 1996
Which of the following is a true fruit?
A
Apple
B
Pear
C
Cashew
D
Coconut
Solution
(D) true fruit is defined as a fruit that develops exclusively from the ovary of a flower after fertilization.
In contrast,false fruits (pseudocarps) develop from other floral parts such as the thalamus or receptacle in addition to the ovary.
$1$. Apple and Pear are false fruits because their fleshy part develops from the thalamus.
$2$. Cashew is also considered a false fruit as the peduncle (stalk) becomes fleshy.
$3$. Coconut is a true fruit (a drupe) that develops solely from the ovary.
In contrast,false fruits (pseudocarps) develop from other floral parts such as the thalamus or receptacle in addition to the ovary.
$1$. Apple and Pear are false fruits because their fleshy part develops from the thalamus.
$2$. Cashew is also considered a false fruit as the peduncle (stalk) becomes fleshy.
$3$. Coconut is a true fruit (a drupe) that develops solely from the ovary.
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97
BiologyMediumMCQAIPMT · 1996
In angiosperms,triple fusion is required for the formation of which of the following?
A
Embryo
B
Endosperm
C
Seed coat
D
Fruit wall
Solution
(B) In angiosperms,the process of double fertilization involves two events: syngamy and triple fusion.
Syngamy involves the fusion of one male gamete with the egg cell to form a diploid zygote $(2n)$,which develops into the embryo.
Triple fusion involves the fusion of the second male gamete $(n)$ with two polar nuclei $(n+n)$ or the secondary nucleus $(2n)$ to form the primary endosperm nucleus $(PEN)$ which is triploid $(3n)$.
The $PEN$ subsequently develops into the endosperm,which provides nourishment to the developing embryo.
Therefore,triple fusion is essential for the formation of the endosperm.
Syngamy involves the fusion of one male gamete with the egg cell to form a diploid zygote $(2n)$,which develops into the embryo.
Triple fusion involves the fusion of the second male gamete $(n)$ with two polar nuclei $(n+n)$ or the secondary nucleus $(2n)$ to form the primary endosperm nucleus $(PEN)$ which is triploid $(3n)$.
The $PEN$ subsequently develops into the endosperm,which provides nourishment to the developing embryo.
Therefore,triple fusion is essential for the formation of the endosperm.
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98
BiologyMediumMCQAIPMT · 1996
How many pollen grains will be produced through the meiotic division of $10$ microspore mother cells?
A
$10$
B
$20$
C
$40$
D
$80$
Solution
(C) In flowering plants,each microspore mother cell $(MMC)$ undergoes meiosis to produce $4$ haploid microspores.
These $4$ microspores develop into $4$ pollen grains.
Therefore,the number of pollen grains produced from $10$ microspore mother cells is calculated as:
$10 \times 4 = 40$ pollen grains.
Thus,the correct option is $C$.
These $4$ microspores develop into $4$ pollen grains.
Therefore,the number of pollen grains produced from $10$ microspore mother cells is calculated as:
$10 \times 4 = 40$ pollen grains.
Thus,the correct option is $C$.
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99
BiologyMediumMCQAIPMT · 1996
Human population growth in India is:
A
Following a sigmoid curve as in many other animal species.
B
Showing a zero growth rate as in some animal species.
C
Controlled by natural disasters and birth control measures.
D
Controlled by the adoption of national family planning programs.
Solution
(C) Human population growth in India is influenced by a combination of natural factors and deliberate human interventions.
Unlike many animal species that follow a natural $S$-shaped (sigmoid) growth curve,human population dynamics are significantly altered by technological,medical,and social advancements.
Natural disasters act as density-independent limiting factors,while birth control measures and national family planning programs are deliberate strategies to regulate population growth.
Therefore,option $C$ is the most accurate description of the factors controlling population growth in the Indian context.
Unlike many animal species that follow a natural $S$-shaped (sigmoid) growth curve,human population dynamics are significantly altered by technological,medical,and social advancements.
Natural disasters act as density-independent limiting factors,while birth control measures and national family planning programs are deliberate strategies to regulate population growth.
Therefore,option $C$ is the most accurate description of the factors controlling population growth in the Indian context.
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100
BiologyMediumMCQAIPMT · 1996
$A$ test-tube baby is one that is produced by:
A
Birth through artificial insemination
B
Development inside a test tube
C
Fertilization in a laboratory setting
D
Development without fertilization
Solution
(C) test-tube baby is produced through the process of In Vitro Fertilization $(IVF)$.
In this technique,the ovum from the wife or a donor and the sperm from the husband or a donor are collected and induced to form a zygote under simulated conditions in the laboratory.
The zygote or early embryos (up to $8$ blastomeres) are then transferred into the fallopian tube ($ZIFT$ - Zygote Intra Fallopian Transfer) or embryos with more than $8$ blastomeres are transferred into the uterus ($IUT$ - Intra Uterine Transfer) to complete further development.
Therefore,the term 'test-tube baby' refers to the fertilization occurring outside the body in a laboratory.
In this technique,the ovum from the wife or a donor and the sperm from the husband or a donor are collected and induced to form a zygote under simulated conditions in the laboratory.
The zygote or early embryos (up to $8$ blastomeres) are then transferred into the fallopian tube ($ZIFT$ - Zygote Intra Fallopian Transfer) or embryos with more than $8$ blastomeres are transferred into the uterus ($IUT$ - Intra Uterine Transfer) to complete further development.
Therefore,the term 'test-tube baby' refers to the fertilization occurring outside the body in a laboratory.
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