BiologyQ1–100 of 174 questions
Page 1 of 3 · English
1
BiologyMediumMCQAIPMT · 2001
Which of the following taxonomic categories is less general in characters as compared to genus?
A
Species
B
Family
C
Class
D
Division
Solution
(A) In the taxonomic hierarchy,the categories are arranged in descending order of generality: $Kingdom > Phylum/Division > Class > Order > Family > Genus > Species$.
As we move from $Kingdom$ to $Species$,the number of common characters decreases,meaning the groups become more specific.
Conversely,as we move from $Species$ to $Kingdom$,the number of common characters decreases,making the groups more general.
Since $Species$ is lower in the hierarchy than $Genus$,it contains organisms with more specific shared characteristics and is therefore less general than $Genus$.
As we move from $Kingdom$ to $Species$,the number of common characters decreases,meaning the groups become more specific.
Conversely,as we move from $Species$ to $Kingdom$,the number of common characters decreases,making the groups more general.
Since $Species$ is lower in the hierarchy than $Genus$,it contains organisms with more specific shared characteristics and is therefore less general than $Genus$.
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2
BiologyMediumMCQAIPMT · 2001
The main difference between Gram-positive and Gram-negative bacteria lies in the composition of
A
Cilia
B
Cell wall
C
Nucleolus
D
Cytoplasm
Solution
(B) The primary difference between Gram-positive $(G+)$ and Gram-negative $(G-)$ bacteria is found in their cell wall structure.
In $G+$ bacteria,the cell wall is $200-300 \ \mathring{A}$ thick,consisting of approximately $85\%$ mucopeptides (peptidoglycan) and $1-2\%$ lipids.
In contrast,the cell wall of $G-$ bacteria is $100-200 \ \mathring{A}$ thick,containing only $10-12\%$ mucopeptides and a significantly higher lipid content of $80-90\%$.
In $G+$ bacteria,the cell wall is $200-300 \ \mathring{A}$ thick,consisting of approximately $85\%$ mucopeptides (peptidoglycan) and $1-2\%$ lipids.
In contrast,the cell wall of $G-$ bacteria is $100-200 \ \mathring{A}$ thick,containing only $10-12\%$ mucopeptides and a significantly higher lipid content of $80-90\%$.
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3
BiologyMediumMCQAIPMT · 2001
Which protozoan is unlikely to have a contractile vacuole?
A
Euglena
B
Paramecium
C
Amoeba
D
Plasmodium
Solution
(D) contractile vacuole is an organelle involved in osmoregulation,primarily found in freshwater protozoans to expel excess water.
$Euglena$,$Paramecium$,and $Amoeba$ are typically found in freshwater environments and possess contractile vacuoles for osmoregulation.
$Plasmodium$ is an endoparasite that lives within the isotonic environment of the host's blood cells or tissues. Because it lives in an isotonic environment,it does not face the problem of excess water influx and therefore lacks a contractile vacuole.
$Euglena$,$Paramecium$,and $Amoeba$ are typically found in freshwater environments and possess contractile vacuoles for osmoregulation.
$Plasmodium$ is an endoparasite that lives within the isotonic environment of the host's blood cells or tissues. Because it lives in an isotonic environment,it does not face the problem of excess water influx and therefore lacks a contractile vacuole.
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4
BiologyMediumMCQAIPMT · 2001
Fungal hyphae are able to penetrate the host with the help of
A
Mechanical pressure
B
Softening by enzymes
C
Both $(a)$ and $(b)$
D
Suckers and hooks
Solution
(C) Fungal hyphae penetrate the host tissue through a combination of two primary mechanisms:
$1$. Mechanical pressure: The hyphal tip exerts physical force against the host cell wall.
$2$. Enzymatic degradation: Fungi secrete various extracellular enzymes,such as cellulases,pectinases,and proteases,which soften and degrade the host cell wall components,facilitating entry.
$1$. Mechanical pressure: The hyphal tip exerts physical force against the host cell wall.
$2$. Enzymatic degradation: Fungi secrete various extracellular enzymes,such as cellulases,pectinases,and proteases,which soften and degrade the host cell wall components,facilitating entry.
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5
BiologyMediumMCQAIPMT · 2001
Common bread mould is
A
Yeast
B
Rhizopus
C
Bacteria
D
Virus
Solution
(B) $Rhizopus$ (commonly known as pin mould or black mould) grows on bread and is therefore referred to as bread mould.
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6
BiologyMediumMCQAIPMT · 2001
The symbiotic association of fungi and algae is called
A
Lichen
B
Mycorrhiza
C
Both $(A)$ and $(B)$
D
Mycoplasma
Solution
(A) The symbiotic association between fungi and algae is known as a $Lichen$. In this relationship,the algal component is called $phycobiont$ and the fungal component is called $mycobiont$. They live in a mutually beneficial association where the algae prepare food through photosynthesis,and the fungi provide protection,anchorage,and absorption of minerals and water for the algae. $Mycorrhiza$,on the other hand,is a symbiotic association between fungi and the roots of higher plants.
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7
BiologyEasyMCQAIPMT · 2001
In the majority of lichens,there is an association of:
A
Green algae and ascomycetes
B
Green algae and basidiomycetes
C
Blue-green algae and ascomycetes
D
Blue-green algae and basidiomycetes
Solution
(C) Lichens represent a symbiotic association between a fungal partner (mycobiont) and an algal partner (phycobiont or photobiont).
In the majority of lichens,the fungal partner belongs to the class $Ascomycetes$.
The algal partner is usually a $Cyanobacteria$ (blue-green algae) or a green alga.
However,the most common association found in the majority of lichens involves blue-green algae and $Ascomycetes$.
In the majority of lichens,the fungal partner belongs to the class $Ascomycetes$.
The algal partner is usually a $Cyanobacteria$ (blue-green algae) or a green alga.
However,the most common association found in the majority of lichens involves blue-green algae and $Ascomycetes$.
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8
BiologyEasyMCQAIPMT · 2001
Which one is a parasitic algae?
A
Vaucheria
B
Polysiphonia
C
Cephaleuros
D
Batrachospermum
Solution
(C) $Cephaleuros \text{ } virescens$ is a member of the class $Chlorophyceae$. It acts as a parasitic alga and causes the disease known as 'red rust of tea' in tea leaves $(Thea \text{ } sinensis)$.
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9
BiologyMediumMCQAIPMT · 2001
In $Funaria$,stomata are present on the
A
Leaf
B
Stem
C
Upper part of capsule
D
Lower part of capsule
Solution
(D) In $Funaria$,the sporophyte consists of a foot,seta,and capsule. The capsule is differentiated into three parts: the basal apophysis,the middle theca,and the terminal operculum. Stomata are present only in the epidermis of the apophysis region,which is the lower part of the capsule. These stomata facilitate gaseous exchange.
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10
BiologyMediumMCQAIPMT · 2001
Sporangia and spore-bearing leaves in ferns are called:
A
Ramentum
B
Sorus
C
Indusium
D
Sporophyll/Frond
Solution
(D) In ferns,the leaves that bear sporangia are known as sporophylls. These leaves are often referred to as fronds. The sporangia are typically arranged in clusters called sori on the dorsal or ventral surface of these sporophylls. Examples include $Pteris$ and $Dryopteris$.
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11
BiologyMediumMCQAIPMT · 2001
$Pinus$ shows:
A
Heterospory
B
Polyembryony
C
Homospory
D
One type of branching
Solution
(B) $Pinus$ is a gymnosperm that exhibits both heterospory and polyembryony.
$1$. Heterospory: It produces two types of spores,microspores and megaspores.
$2$. Polyembryony: It is the phenomenon of having more than one embryo in a single seed,which is a characteristic feature of many gymnosperms including $Pinus$.
Since both $A$ and $B$ are correct,in the context of typical competitive biology questions,$Pinus$ is widely recognized for exhibiting polyembryony as a distinct feature.
$1$. Heterospory: It produces two types of spores,microspores and megaspores.
$2$. Polyembryony: It is the phenomenon of having more than one embryo in a single seed,which is a characteristic feature of many gymnosperms including $Pinus$.
Since both $A$ and $B$ are correct,in the context of typical competitive biology questions,$Pinus$ is widely recognized for exhibiting polyembryony as a distinct feature.
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12
BiologyMediumMCQAIPMT · 2001
$Cycas$ has two cotyledons but is not included under angiosperms because it has:
A
Circinate ptyxis
B
Compound leaves
C
Monocot-like stem
D
Naked seeds
Solution
(D) $Cycas$ belongs to the group $Gymnospermae$. In $Gymnosperms$,the ovules are not enclosed by any ovary wall and remain exposed both before and after fertilization. Therefore,they produce naked seeds. In contrast,$Angiosperms$ are characterized by seeds that are enclosed within fruits (developed from the ovary). Although $Cycas$ may exhibit two cotyledons,the presence of naked seeds classifies it as a $Gymnosperm$ rather than an $Angiosperm$.
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13
BiologyMediumMCQAIPMT · 2001
In the archegonium of a fern,how many neck canal cells are present?
A
There are $4$ neck canal cells
B
There are $3$ neck canal cells
C
There is $1$ neck canal cell with one nucleus
D
There is $1$ neck canal cell with two nuclei
Solution
(D) In the archegonium of ferns (Pteridophytes),the neck canal cells are typically present as a single cell that is often binucleate. This cell is formed by the division of the neck canal initial. Therefore,there is $1$ neck canal cell containing two nuclei.
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14
BiologyEasyMCQAIPMT · 2001
The famous Indian zoologist who wrote a memoir upon $Pheretima$ $posthuma$ is:
A
$J$.$C$. Bose
B
$M$.$L$. Bhatia
C
$K$.$N$. Bahl
D
Beni Prasad
Solution
(C) The correct answer is $C$. Prof. Karm Narayan Bahl of Lucknow University published a famous memoir on the Indian earthworm $Pheretima$ $posthuma$ in $1926$. He was a distinguished zoologist and was awarded the Joy Govind Law Memorial Gold Medal in $1942$ for his notable research in Asiatic Zoology.
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15
BiologyMediumMCQAIPMT · 2001
Which phylum belongs to Deuterostomia?
A
Echinodermata
B
Mollusca
C
Arthropoda
D
Annelida
Solution
(A) In animals,the development of the blastopore during gastrulation determines the classification into Protostomia and Deuterostomia.
In Protostomes,the blastopore develops into the mouth.
In Deuterostomes,the blastopore develops into the anus,and the mouth forms later.
Among the given options,$Echinodermata$ and $Chordata$ are the primary phyla that belong to the Deuterostomia group.
Therefore,the correct option is $A$.
In Protostomes,the blastopore develops into the mouth.
In Deuterostomes,the blastopore develops into the anus,and the mouth forms later.
Among the given options,$Echinodermata$ and $Chordata$ are the primary phyla that belong to the Deuterostomia group.
Therefore,the correct option is $A$.
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16
BiologyMediumMCQAIPMT · 2001
The echinoderms,hemichordates,and chordates had which of the following larvae as a common ancestral form?
A
Tornaria
B
Trochophore
C
Dipleurula
D
Bipinnaria
Solution
(C) The $Dipleurula$ larva is considered the hypothetical common ancestral larval form for the phyla $Echinodermata$,$Hemichordata$,and $Chordata$.
This theory is supported by the similarities in the early developmental stages and the basic body plan shared by these deuterostome groups.
This theory is supported by the similarities in the early developmental stages and the basic body plan shared by these deuterostome groups.
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17
BiologyMediumMCQAIPMT · 2001
Most favourable land adaptation for reptiles is:
A
Moist skin
B
Scales on body
C
Pulmonary respiration
D
None of these
Solution
(B) The correct answer is $B$.
Reptiles are primarily terrestrial animals.
The most significant adaptation for life on land is the presence of dry and cornified skin with epidermal scales or scutes.
These scales prevent excessive water loss from the body through evaporation,which is crucial for survival in dry terrestrial environments.
Reptiles are primarily terrestrial animals.
The most significant adaptation for life on land is the presence of dry and cornified skin with epidermal scales or scutes.
These scales prevent excessive water loss from the body through evaporation,which is crucial for survival in dry terrestrial environments.
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18
BiologyMediumMCQAIPMT · 2001
Post-anal tail is present in
A
Cobra
B
Earthworm
C
Scorpion
D
Lower invertebrate
Solution
(A) The post-anal tail is a characteristic feature of the phylum $Chordata$.
Among the given options,$Cobra$ belongs to the class $Reptilia$,which is a sub-group of the phylum $Chordata$.
$Earthworm$ belongs to the phylum $Annelida$,$Scorpion$ belongs to the phylum $Arthropoda$,and lower invertebrates do not possess a post-anal tail.
Therefore,the correct answer is $Cobra$.
Among the given options,$Cobra$ belongs to the class $Reptilia$,which is a sub-group of the phylum $Chordata$.
$Earthworm$ belongs to the phylum $Annelida$,$Scorpion$ belongs to the phylum $Arthropoda$,and lower invertebrates do not possess a post-anal tail.
Therefore,the correct answer is $Cobra$.
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19
BiologyEasyMCQAIPMT · 2001
Without teats,mammary glands are found in
A
Prototheria
B
Metatheria
C
Eutheria
D
Theria
Solution
(A) In $Prototheria$ (e.g.,$Ornithorhynchus$ or platypus),the mammary glands are modified sudoriferous (sweat) glands that lack nipples or teats.
Instead,the milk is secreted onto the skin or hair of the mother,from where the young ones lick it up.
Instead,the milk is secreted onto the skin or hair of the mother,from where the young ones lick it up.
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20
BiologyEasyMCQAIPMT · 2001
Which group of animals possesses a marsupium?
A
Monotremata
B
Eutheria
C
Metatheria
D
Prototheria
Solution
(C) The marsupium,also known as the marsupial pouch,is a defining characteristic of the infraclass $Metatheria$ (commonly known as marsupials). These mammals give birth to relatively undeveloped young,which then crawl into the pouch to complete their development while feeding on milk.
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21
BiologyMediumMCQAIPMT · 2001
In $Hydra$,egestion of undigested food and excretion of nitrogenous wastes occur through
A
Mouth and tentacles
B
Mouth and body wall
C
Mouth and mouth
D
Body wall and body wall
Solution
(B) $Hydra$ belongs to the phylum $Cnidaria$ (or $Coelenterata$).
It possesses an incomplete digestive system with a single opening,the mouth (hypostome),which serves for both ingestion and egestion.
Undigested food residues are egested through the mouth.
Nitrogenous wastes,such as ammonia,are excreted primarily through the general body surface (body wall) via simple diffusion.
Therefore,the correct combination is mouth and body wall.
It possesses an incomplete digestive system with a single opening,the mouth (hypostome),which serves for both ingestion and egestion.
Undigested food residues are egested through the mouth.
Nitrogenous wastes,such as ammonia,are excreted primarily through the general body surface (body wall) via simple diffusion.
Therefore,the correct combination is mouth and body wall.
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22
BiologyMediumMCQAIPMT · 2001
The eye of a potato is a/an:
A
Apical bud
B
Axillary bud
C
Accessory bud
D
Adventitious bud
Solution
(B) The potato is a modified underground stem known as a tuber.
On the surface of the potato,there are small depressions known as 'eyes'.
Each eye represents a node that contains one or more axillary buds.
These axillary buds are capable of sprouting and developing into new potato plants under favorable conditions.
On the surface of the potato,there are small depressions known as 'eyes'.
Each eye represents a node that contains one or more axillary buds.
These axillary buds are capable of sprouting and developing into new potato plants under favorable conditions.
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23
BiologyMediumMCQAIPMT · 2001
Tetradynamous condition is characteristic of
A
Liliaceae (Allium,Asphodelus)
B
Cruciferae (Mustard,Iberis)
C
Malvaceae (Althea,Hibiscus)
D
Solanaceae (Nicotiana,Petunia)
Solution
(B) The tetradynamous condition refers to the presence of six stamens in a flower,where the inner four stamens are long and the outer two stamens are short.
This arrangement is a diagnostic feature of the family $Cruciferae$ (also known as $Brassicaceae$),which includes plants like Mustard $(Brassica)$ and $Iberis$.
This arrangement is a diagnostic feature of the family $Cruciferae$ (also known as $Brassicaceae$),which includes plants like Mustard $(Brassica)$ and $Iberis$.
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24
BiologyMediumMCQAIPMT · 2001
During an injury,the nasal septum gets damaged. Which type of cartilage is present in the nasal septum for its structure and recovery?
A
Hyaline cartilage
B
Elastic cartilage
C
Calcified cartilage
D
Fibrocartilage
Solution
(A) The nasal septum is primarily composed of hyaline cartilage.
Hyaline cartilage is the most abundant type of cartilage in the body.
It provides structural support and flexibility to the nasal septum,larynx,trachea,and the ends of long bones.
Therefore,in the event of an injury to the nasal septum,hyaline cartilage is the tissue involved in its structure and recovery.
Hyaline cartilage is the most abundant type of cartilage in the body.
It provides structural support and flexibility to the nasal septum,larynx,trachea,and the ends of long bones.
Therefore,in the event of an injury to the nasal septum,hyaline cartilage is the tissue involved in its structure and recovery.
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25
BiologyMediumMCQAIPMT · 2001
The Krause membrane or $Z$-line is a structure in a myofibril that separates two adjacent:
A
Sarcomeres
B
$H$-zones
C
$I$-bands
D
$A$-bands
Solution
(A) myofibril is composed of repeating functional units called sarcomeres.
Each sarcomere is defined as the segment of a myofibril between two successive $Z$-lines (also known as Krause's membrane).
These $Z$-lines are dense,dark-colored protein discs that bisect the $I$-bands and anchor the actin filaments.
Therefore,the $Z$-line acts as the boundary that separates two adjacent sarcomeres.
Each sarcomere is defined as the segment of a myofibril between two successive $Z$-lines (also known as Krause's membrane).
These $Z$-lines are dense,dark-colored protein discs that bisect the $I$-bands and anchor the actin filaments.
Therefore,the $Z$-line acts as the boundary that separates two adjacent sarcomeres.
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26
BiologyMediumMCQAIPMT · 2001
Which of the following cell organelles has a single-layered unit membrane?
A
Centrosome
B
Lysosome
C
Mesosome
D
Nucleus
Solution
(B) The $Lysosome$ is a membrane-bound organelle that contains digestive enzymes. It is characterized by a single-layered unit membrane.
In contrast,the $Nucleus$ has a double-membrane envelope.
The $Centrosome$ is a non-membrane-bound organelle.
The $Mesosome$ is an infolding of the plasma membrane in prokaryotes,not a distinct organelle with a unit membrane.
In contrast,the $Nucleus$ has a double-membrane envelope.
The $Centrosome$ is a non-membrane-bound organelle.
The $Mesosome$ is an infolding of the plasma membrane in prokaryotes,not a distinct organelle with a unit membrane.
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27
BiologyMediumMCQAIPMT · 2001
Ribosomes,similar to those of bacteria,are found in
A
Plant nuclei
B
Pancreatic mitochondria
C
Liver endoplasmic reticulum
D
Cardiac muscle cytoplasm
Solution
(B) The ribosomes found in bacteria are of the $70S$ type.
According to the endosymbiotic theory,mitochondria and chloroplasts are believed to have originated from prokaryotic organisms that were engulfed by ancestral eukaryotic cells.
Therefore,these organelles contain $70S$ ribosomes,which are structurally and functionally similar to bacterial ribosomes,unlike the $80S$ ribosomes found in the cytoplasm of eukaryotic cells.
Thus,pancreatic mitochondria contain ribosomes similar to those of bacteria.
According to the endosymbiotic theory,mitochondria and chloroplasts are believed to have originated from prokaryotic organisms that were engulfed by ancestral eukaryotic cells.
Therefore,these organelles contain $70S$ ribosomes,which are structurally and functionally similar to bacterial ribosomes,unlike the $80S$ ribosomes found in the cytoplasm of eukaryotic cells.
Thus,pancreatic mitochondria contain ribosomes similar to those of bacteria.
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28
BiologyEasyMCQAIPMT · 2001
The enzyme responsible for atmospheric nitrogen fixation is
A
Nitrogenase
B
Hydrogenase
C
Oxygenase
D
Carboxylase
Solution
(A) The enzyme $Nitrogenase$ is responsible for the biological fixation of atmospheric nitrogen into ammonia. This enzyme is found exclusively in certain prokaryotes,such as $Rhizobium$ and $Azotobacter$. It catalyzes the reduction of $N_2$ to $NH_3$ using $ATP$ and electrons.
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29
BiologyMediumMCQAIPMT · 2001
Synaptonemal complex is formed during
A
Meiosis
B
Amitosis
C
Mitosis
D
Cytokinesis
Solution
(A) The synaptonemal complex is a protein structure that forms between homologous chromosomes during the $Zygotene$ stage of $Prophase-I$ of $Meiosis$.
It facilitates the process of synapsis,which is the pairing of homologous chromosomes.
Therefore,the correct answer is $Meiosis$.
It facilitates the process of synapsis,which is the pairing of homologous chromosomes.
Therefore,the correct answer is $Meiosis$.
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30
BiologyEasyMCQAIPMT · 2001
$DNA$ 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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31
BiologyMediumMCQAIPMT · 2001
When water moves through a semipermeable membrane,which of the following is created?
A
$OP$
B
$SP$
C
$TP$
D
$WP$
Solution
(A) Osmotic pressure $(OP)$ is defined as the pressure that must be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.
When water moves through a semipermeable membrane due to a concentration gradient,the pressure required to stop this movement is known as the osmotic pressure $(OP)$.
When water moves through a semipermeable membrane due to a concentration gradient,the pressure required to stop this movement is known as the osmotic pressure $(OP)$.
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32
BiologyEasyMCQAIPMT · 2001
The most widely accepted explanation for the ascent of sap in trees is:
A
Capillarity
B
Role of atmospheric pressure
C
Pulsating action of living cells
D
Transpiration cohesion theory of Dixon
Solution
(D) The most widely accepted theory for the ascent of sap is the $Transpiration$ $Cohesion-Tension$ theory,proposed by $Dixon$ and $Joly$ in $1894$.
According to this theory,the ascent of sap is driven by the transpiration pull generated at the leaf surface due to the evaporation of water.
This pull is transmitted down the water column in the $Xylem$ due to the cohesive and adhesive properties of water molecules,which prevent the column from breaking under tension.
According to this theory,the ascent of sap is driven by the transpiration pull generated at the leaf surface due to the evaporation of water.
This pull is transmitted down the water column in the $Xylem$ due to the cohesive and adhesive properties of water molecules,which prevent the column from breaking under tension.
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33
BiologyEasyMCQAIPMT · 2001
The term water potential was proposed by
A
Bose
B
Dixon
C
Godlewski
D
Slatyer and Taylor
Solution
(D) The term water potential was proposed by $Slatyer$ and $Taylor$ in $1960$.
Water potential is defined as the chemical potential of water in a system,compared to pure water at atmospheric pressure and the same temperature.
It is denoted by the Greek symbol $Psi$ $(\Psi)$.
It is equivalent to the $DPD$ (Diffusion Pressure Deficit) but with a negative sign,i.e.,$\Psi_w = -DPD$.
Water potential is defined as the chemical potential of water in a system,compared to pure water at atmospheric pressure and the same temperature.
It is denoted by the Greek symbol $Psi$ $(\Psi)$.
It is equivalent to the $DPD$ (Diffusion Pressure Deficit) but with a negative sign,i.e.,$\Psi_w = -DPD$.
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34
BiologyMediumMCQAIPMT · 2001
The net gain of energy,from one molecule of sucrose in aerobic respiration,is (in $ATP$)
A
$18$
B
$38$
C
$76$
D
$80$
Solution
(C) Sucrose is a disaccharide composed of one molecule of glucose and one molecule of fructose.
In aerobic respiration,both glucose and fructose are broken down through glycolysis,the Krebs cycle,and the electron transport system.
One molecule of glucose yields $38$ $ATP$ (assuming the malate-aspartate shuttle).
Since sucrose consists of two hexose sugars,the total energy yield is $38 + 38 = 76$ $ATP$.
Therefore,the net gain of energy from one molecule of sucrose is $76$ $ATP$.
In aerobic respiration,both glucose and fructose are broken down through glycolysis,the Krebs cycle,and the electron transport system.
One molecule of glucose yields $38$ $ATP$ (assuming the malate-aspartate shuttle).
Since sucrose consists of two hexose sugars,the total energy yield is $38 + 38 = 76$ $ATP$.
Therefore,the net gain of energy from one molecule of sucrose is $76$ $ATP$.
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35
BiologyMediumMCQAIPMT · 2001
Cytochromes are
A
Lipid
B
Glycoprotein
C
Metalloporphyrins
D
$Fe^{++}$ containing with porphyrin pigment
Solution
(D) Cytochromes are iron-containing hemoproteins that act as electron carriers in the electron transport chain.
They contain a heme group,which is a $Fe$-containing porphyrin pigment.
During the electron transport process,the iron atom in the cytochrome undergoes reversible oxidation and reduction between the $Fe^{3+}$ and $Fe^{2+}$ states $(Fe^{3+} + e^- \to Fe^{2+})$.
They contain a heme group,which is a $Fe$-containing porphyrin pigment.
During the electron transport process,the iron atom in the cytochrome undergoes reversible oxidation and reduction between the $Fe^{3+}$ and $Fe^{2+}$ states $(Fe^{3+} + e^- \to Fe^{2+})$.
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36
BiologyMediumMCQAIPMT · 2001
Which of the following plant hormones prevents the premature falling of fruits?
A
$GA_3$
B
$NAA$
C
Ethylene
D
Zeatin
Solution
(B) Auxins,such as $NAA$ (Naphthalene Acetic Acid),are synthetic plant hormones that are widely used in agriculture to prevent the premature abscission (falling) of fruits and leaves.
$NAA$ helps in retaining fruits on the plant by inhibiting the formation of the abscission layer at the pedicel.
In contrast,Ethylene promotes abscission,while $GA_3$ (Gibberellic acid) and Zeatin (a type of cytokinin) have different primary roles in growth and cell division.
$NAA$ helps in retaining fruits on the plant by inhibiting the formation of the abscission layer at the pedicel.
In contrast,Ethylene promotes abscission,while $GA_3$ (Gibberellic acid) and Zeatin (a type of cytokinin) have different primary roles in growth and cell division.
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37
BiologyEasyMCQAIPMT · 2001
Which of the following hormones is responsible for senescence?
A
$GA$
B
$ABA$
C
Auxin
D
Cytokinin
Solution
(B) $ABA$ (Abscisic acid) stimulates senescence of leaves by causing the destruction of chlorophyll and the inhibition of protein and $RNA$ synthesis.
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38
BiologyMediumMCQAIPMT · 2001
The proteinaceous pigment that acts as the center of activities concerned with light is:
A
Phytochrome
B
Chlorophyll
C
Anthocyanin
D
Carotenoids
Solution
(A) Phytochrome is a proteinaceous pigment that acts as a photoreceptor in plants. It exists in two interconvertible forms: $Pr$ (which absorbs red light) and $Pfr$ (which absorbs far-red light). It plays a crucial role in various light-mediated developmental processes such as photoperiodism,seed germination,and stem elongation,making it the central pigment for light-related activities.
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39
BiologyEasyMCQAIPMT · 2001
Which of the following is a long day plant?
A
Mirabilis
B
Glycine max
C
Mirabilis jalapa
D
Spinacia oleracea
Solution
(D) Long day plants $(LDP)$ are those that require a light period longer than a critical duration to flower.
Among the given options, $Spinacia \text{ } oleracea$ (Spinach) is a classic example of a long day plant.
$Glycine \text{ } max$ (Soybean) is a short day plant $(SDP)$, and $Mirabilis \text{ } jalapa$ is a day-neutral plant.
Among the given options, $Spinacia \text{ } oleracea$ (Spinach) is a classic example of a long day plant.
$Glycine \text{ } max$ (Soybean) is a short day plant $(SDP)$, and $Mirabilis \text{ } jalapa$ is a day-neutral plant.
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40
BiologyEasyMCQAIPMT · 2001
Rickets in children and osteomalacia in adults is caused by the deficiency of
A
Vitamin $A$
B
Vitamin $B$
C
Vitamin $C$
D
Vitamin $D$
Solution
(D) Rickets is a condition that affects bone development in children,causing them to become soft and weak,which leads to skeletal deformities.
Osteomalacia is the adult equivalent of rickets,characterized by the softening of bones due to defective bone mineralization.
Both conditions are primarily caused by a deficiency of Vitamin $D$,which is essential for the absorption of calcium and phosphorus from the diet.
Therefore,the correct option is $D$.
Osteomalacia is the adult equivalent of rickets,characterized by the softening of bones due to defective bone mineralization.
Both conditions are primarily caused by a deficiency of Vitamin $D$,which is essential for the absorption of calcium and phosphorus from the diet.
Therefore,the correct option is $D$.
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41
BiologyEasyMCQAIPMT · 2001
Which of the following is the correct match?
A
Vitamin $A$ - calciferol
B
Vitamin $E$ - tocopherol
C
Vitamin $D$ - thiamine
D
Vitamin $K$ - ascorbic acid
Solution
(B) The correct match is Vitamin $E$ - tocopherol.
Vitamin $A$ is retinol.
Vitamin $D$ is calciferol.
Vitamin $K$ is phylloquinone.
Vitamin $E$ is known as tocopherol and is often referred to as the anti-sterility vitamin.
Vitamin $A$ is retinol.
Vitamin $D$ is calciferol.
Vitamin $K$ is phylloquinone.
Vitamin $E$ is known as tocopherol and is often referred to as the anti-sterility vitamin.
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42
BiologyEasyMCQAIPMT · 2001
Which of the following sugars is absorbed from the small intestine by facilitated diffusion?
A
Fructose
B
Glucose
C
Sucrose
D
Lactose
Solution
(A) The absorption of monosaccharides occurs in the small intestine.
$Glucose$ and $galactose$ are absorbed into the blood via active transport, which requires energy.
$Fructose$ is absorbed through the process of facilitated diffusion, where it moves across the membrane with the help of carrier proteins like $GLUT5$ without the expenditure of energy.
$Glucose$ and $galactose$ are absorbed into the blood via active transport, which requires energy.
$Fructose$ is absorbed through the process of facilitated diffusion, where it moves across the membrane with the help of carrier proteins like $GLUT5$ without the expenditure of energy.
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43
BiologyMediumMCQAIPMT · 2001
During forced expiration,which muscles actively contract?
A
Diaphragm
B
External intercostals
C
Abdominal muscles
D
Diaphragm and intestinal muscles
Solution
(C) During normal breathing,expiration is a passive process. However,during forced expiration,the process becomes active. This involves the contraction of the internal intercostal muscles and the abdominal muscles. These muscles work to increase the pressure within the thoracic cavity by reducing its volume,thereby pushing air out of the lungs more forcefully. Therefore,the correct option is $C$.
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44
BiologyMediumMCQAIPMT · 2001
Which of the following nitrogenous substances is highly toxic?
A
Urea
B
Uric acid
C
Amino acid
D
Ammonia
Solution
(D) The correct answer is $D$. Ammonia $(NH_3)$ is highly toxic because it is a strong base and can significantly alter the $pH$ of body fluids.
To prevent toxicity,organisms must either convert ammonia into less toxic forms like urea or uric acid,or expel it immediately from the body.
Due to its high toxicity,it requires a large amount of water for its elimination,which is why it is primarily excreted by aquatic organisms.
To prevent toxicity,organisms must either convert ammonia into less toxic forms like urea or uric acid,or expel it immediately from the body.
Due to its high toxicity,it requires a large amount of water for its elimination,which is why it is primarily excreted by aquatic organisms.
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45
BiologyEasyMCQAIPMT · 2001
In birds,some of the vertebrae are fused to form:
A
Keel
B
Synsacrum
C
Syncytium
D
Furcula
Solution
(B) The correct answer is $B$.
In birds,the posterior thoracic,lumbar,sacral,and anterior caudal vertebrae are fused together to form a single structure known as the $Synsacrum$.
This structure provides rigidity to the skeleton,which is essential for flight and for supporting the weight of the body on the hind limbs.
In birds,the posterior thoracic,lumbar,sacral,and anterior caudal vertebrae are fused together to form a single structure known as the $Synsacrum$.
This structure provides rigidity to the skeleton,which is essential for flight and for supporting the weight of the body on the hind limbs.
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46
BiologyMediumMCQAIPMT · 2001
The joint between the femur and the pelvic girdle is known as:
A
Pivotal
B
Ball and socket
C
Hinge
D
Saddle
Solution
(B) The joint between the femur (thigh bone) and the pelvic girdle (acetabulum of the hip bone) is a $Ball$ and $socket$ joint.
In this type of joint,the rounded head of one bone (femur) fits into the cup-like cavity (acetabulum) of another bone (pelvic girdle).
This configuration allows for multi-axial movement,including rotation,flexion,extension,abduction,and adduction.
In this type of joint,the rounded head of one bone (femur) fits into the cup-like cavity (acetabulum) of another bone (pelvic girdle).
This configuration allows for multi-axial movement,including rotation,flexion,extension,abduction,and adduction.
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47
BiologyEasyMCQAIPMT · 2001
Parkinsonism is related to which part of the nervous system?
A
Brain
B
Spinal cord
C
Cranial nerves
D
Spinal nerves
Solution
(A) Parkinsonism,also known as Parkinson's disease or paralysis agitans,is a progressive neurodegenerative disorder.
It is caused by the degeneration of dopamine-producing neurons in the substantia nigra,a region located in the midbrain.
Therefore,it is a disorder primarily related to the brain.
It is caused by the degeneration of dopamine-producing neurons in the substantia nigra,a region located in the midbrain.
Therefore,it is a disorder primarily related to the brain.
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48
BiologyMediumMCQAIPMT · 2001
The enzyme required for the conduction of nerve impulses across the synapse is:
A
Peroxidase
B
Choline acetylase
C
Ascorbic acid oxidase
D
Succinic dehydrogenase
Solution
(B) The conduction of nerve impulses across a chemical synapse involves the release of neurotransmitters like acetylcholine $(ACh)$.
Choline acetylase (also known as choline acetyltransferase) is the enzyme responsible for the synthesis of acetylcholine from choline and acetyl-CoA in the presynaptic terminal.
This neurotransmitter is then released into the synaptic cleft to propagate the nerve impulse to the postsynaptic neuron.
Therefore,choline acetylase is essential for the chemical transmission of nerve impulses.
Choline acetylase (also known as choline acetyltransferase) is the enzyme responsible for the synthesis of acetylcholine from choline and acetyl-CoA in the presynaptic terminal.
This neurotransmitter is then released into the synaptic cleft to propagate the nerve impulse to the postsynaptic neuron.
Therefore,choline acetylase is essential for the chemical transmission of nerve impulses.
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49
BiologyEasyMCQAIPMT · 2001
Proprioceptors are those,which give the sense of
A
Chemicals
B
Temperature
C
Taste
D
Changes in the internal environment of the body
Solution
(D) . Proprioceptors are specialized sensory receptors located within muscles,tendons,and joints. They provide information about the position,movement,and orientation of the body parts,effectively detecting changes in the internal environment of the body regarding posture and mechanical state.
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50
BiologyMediumMCQAIPMT · 2001
Insulin increases glucose uptake in all the following structures except:
A
Cardiac muscle
B
Skeletal muscle
C
Adipose tissue
D
Intestinal mucosa
Solution
(D) Insulin is a peptide hormone that regulates blood glucose levels by facilitating the uptake of glucose into cells.
It primarily acts on insulin-sensitive tissues such as skeletal muscle,cardiac muscle,and adipose tissue by promoting the translocation of $GLUT4$ glucose transporters to the plasma membrane.
However,glucose uptake in the intestinal mucosa,brain,and red blood cells is insulin-independent,as these tissues utilize different glucose transporters (such as $GLUT2$ or $GLUT1$) that do not require insulin for their activity.
Therefore,the correct answer is intestinal mucosa.
It primarily acts on insulin-sensitive tissues such as skeletal muscle,cardiac muscle,and adipose tissue by promoting the translocation of $GLUT4$ glucose transporters to the plasma membrane.
However,glucose uptake in the intestinal mucosa,brain,and red blood cells is insulin-independent,as these tissues utilize different glucose transporters (such as $GLUT2$ or $GLUT1$) that do not require insulin for their activity.
Therefore,the correct answer is intestinal mucosa.
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51
BiologyMediumMCQAIPMT · 2001
The first organisms to evolve on the earth were
A
Saprotrophs
B
Chemoheterotrophs
C
Photoautotrophs
D
Chemoautotrophs
Solution
(B) According to the Oparin-Haldane hypothesis of chemical evolution,the primitive atmosphere of the Earth was reducing and lacked free oxygen. The first life forms were simple,anaerobic,and heterotrophic in nature. They obtained energy by consuming the organic molecules present in the 'primordial soup' of the oceans. These organisms are classified as chemoheterotrophs because they relied on pre-existing organic compounds for their carbon and energy needs.
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52
BiologyEasyMCQAIPMT · 2001
Marine algae flourished well during which period?
A
Triassic
B
Devonian
C
Permian
D
Ordovician
Solution
(D) The $Ordovician$ period is often referred to as the 'Age of Invertebrates'.
During this period,there was a significant diversification of marine life.
Marine algae,particularly green algae,flourished extensively in the shallow seas during the $Ordovician$ period,contributing to the primary productivity of the marine ecosystems.
During this period,there was a significant diversification of marine life.
Marine algae,particularly green algae,flourished extensively in the shallow seas during the $Ordovician$ period,contributing to the primary productivity of the marine ecosystems.
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53
BiologyEasyMCQAIPMT · 2001
The aquatic fern which is an excellent biofertilizer is
A
Azolla
B
Salvinia
C
Marsilia
D
Pteridium
Solution
(A) $Azolla$ is an aquatic fern that lives in a symbiotic relationship with the nitrogen-fixing cyanobacterium $Anabaena$ $azollae$.
This association allows $Azolla$ to fix atmospheric nitrogen,making it an excellent biofertilizer,particularly in rice paddies.
This association allows $Azolla$ to fix atmospheric nitrogen,making it an excellent biofertilizer,particularly in rice paddies.
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54
BiologyMediumMCQAIPMT · 2001
Pinus shows
A
Simple polyembryony
B
Cleavage polyembryony
C
Both simple and cleavage polyembryony
D
None of the above
Solution
(C) Pinus exhibits both simple and cleavage polyembryony.
In simple polyembryony,multiple embryos develop due to the fertilization of different archegonia within the same ovule.
In cleavage polyembryony,the zygote undergoes cleavage to form multiple embryos from a single fertilized egg.
In simple polyembryony,multiple embryos develop due to the fertilization of different archegonia within the same ovule.
In cleavage polyembryony,the zygote undergoes cleavage to form multiple embryos from a single fertilized egg.
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55
BiologyMediumMCQAIPMT · 2001
The spleen is a:
A
Haemopoietic organ
B
Lymphoid organ
C
Reproductive organ
D
Celluloid organ
Solution
(B) The spleen is a large,bean-shaped organ which primarily acts as a filter for the blood.
It contains lymphocytes and phagocytes,which help in destroying pathogens.
Therefore,the spleen is classified as a secondary lymphoid organ in the human body.
It contains lymphocytes and phagocytes,which help in destroying pathogens.
Therefore,the spleen is classified as a secondary lymphoid organ in the human body.
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56
BiologyMediumMCQAIPMT · 2001
To obtain a haploid callus,one can culture:
A
Embryo
B
Leaf tissue
C
Stigma
D
Pollen grain
Solution
(D) haploid callus is derived from cells that contain only one set of chromosomes $(n)$.
In plants,the only cells that are naturally haploid are the gametophytes,such as pollen grains (microspores) or embryo sacs (megaspores).
By culturing pollen grains (microspore culture) on a suitable nutrient medium,the cells undergo division to form a mass of undifferentiated haploid cells known as a haploid callus.
Embryos,leaf tissues,and stigmas are diploid $(2n)$ structures and would result in a diploid callus.
In plants,the only cells that are naturally haploid are the gametophytes,such as pollen grains (microspores) or embryo sacs (megaspores).
By culturing pollen grains (microspore culture) on a suitable nutrient medium,the cells undergo division to form a mass of undifferentiated haploid cells known as a haploid callus.
Embryos,leaf tissues,and stigmas are diploid $(2n)$ structures and would result in a diploid callus.
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57
BiologyMediumMCQAIPMT · 2001
Anemophilous pollination is mainly observed in
A
Gramineae
B
Annonaceae
C
Papilionaceae
D
Euphorbiaceae
Solution
(A) Anemophilous pollination refers to pollination by wind.
In the family $Gramineae$ (also known as $Poaceae$),flowers are small,inconspicuous,and often lack showy petals or nectar,which are adaptations for wind pollination.
The pollen grains in these plants are produced in large quantities,and they are small,light,smooth,and dry,allowing them to be easily carried by wind currents.
In the family $Gramineae$ (also known as $Poaceae$),flowers are small,inconspicuous,and often lack showy petals or nectar,which are adaptations for wind pollination.
The pollen grains in these plants are produced in large quantities,and they are small,light,smooth,and dry,allowing them to be easily carried by wind currents.
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58
BiologyMediumMCQAIPMT · 2001
Which cells in the testis secrete testosterone?
A
Interstitial cells or cells of Leydig
B
Cells of the germinal epithelium
C
Sertoli cells
D
Secondary spermatocytes
Solution
(A) The $Leydig$ cells,also known as interstitial cells,are located in the interstitial spaces of the testis.
These cells are stimulated to secrete the male sex hormone,testosterone,by the Interstitial Cell Stimulating Hormone $(ICSH)$,which is released from the anterior lobe of the pituitary gland.
These cells are stimulated to secrete the male sex hormone,testosterone,by the Interstitial Cell Stimulating Hormone $(ICSH)$,which is released from the anterior lobe of the pituitary gland.
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59
BiologyMediumMCQAIPMT · 2001
The $9 : 3 : 3 : 1$ ratio is modified to $9 : 7$ ratio due to
A
Complementary gene
B
Epistatic gene
C
Hypostatic gene
D
Supplementary gene
Solution
(A) The $9 : 7$ ratio is a classic example of complementary gene interaction.
In this interaction,two pairs of non-allelic genes interact to produce a single phenotypic trait.
If either gene is present alone,it cannot express the trait; both dominant alleles must be present together to produce the phenotype.
This results in a modification of the dihybrid $9 : 3 : 3 : 1$ ratio to $9 : 7$ in the $F_2$ generation.
In this interaction,two pairs of non-allelic genes interact to produce a single phenotypic trait.
If either gene is present alone,it cannot express the trait; both dominant alleles must be present together to produce the phenotype.
This results in a modification of the dihybrid $9 : 3 : 3 : 1$ ratio to $9 : 7$ in the $F_2$ generation.
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60
BiologyMediumMCQAIPMT · 2001
When dominant and recessive alleles express themselves together,it is called
A
Dominance
B
Co-dominance
C
Amphidominance
D
Pseudodominance
Solution
(B) Codominance is a condition in a heterozygote where both members of an allelic pair contribute to the phenotype.
This results in a mixture of the phenotypic traits produced in either homozygous condition.
For example,in cattle,the cross of red $\times$ white produces roan offspring,whose coat consists of both red and white hair.
This results in a mixture of the phenotypic traits produced in either homozygous condition.
For example,in cattle,the cross of red $\times$ white produces roan offspring,whose coat consists of both red and white hair.
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61
BiologyEasyMCQAIPMT · 2001
The Mongoloid condition is related to,or in mongolism,a patient shows:
A
Monosomy
B
Trisomy
C
Nullisomy
D
None of the above
Solution
(B) Mongolism is also known as $Down$ syndrome.
It is a chromosomal disorder caused by the presence of an extra copy of chromosome number $21$.
This condition is known as trisomy of chromosome $21$,where the individual has $47$ chromosomes instead of the normal $46$.
Therefore,the correct answer is trisomy.
It is a chromosomal disorder caused by the presence of an extra copy of chromosome number $21$.
This condition is known as trisomy of chromosome $21$,where the individual has $47$ chromosomes instead of the normal $46$.
Therefore,the correct answer is trisomy.
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62
BiologyMediumMCQAIPMT · 2001
The enzyme which can cut molecules of $DNA$ into segments is known as
A
$DNA$ polymerase
B
$DNA$ ligase
C
Restriction enzyme
D
$DNA$ gyrase
Solution
(C) Restriction enzymes are a class of enzymes that act as molecular scissors.
They recognize specific nucleotide sequences in a $DNA$ molecule,known as recognition sites,and cleave the phosphodiester backbone at these specific locations to produce $DNA$ segments.
$DNA$ polymerase is involved in $DNA$ replication,$DNA$ ligase acts as a molecular glue to join $DNA$ fragments,and $DNA$ gyrase is involved in relieving supercoiling during $DNA$ replication.
They recognize specific nucleotide sequences in a $DNA$ molecule,known as recognition sites,and cleave the phosphodiester backbone at these specific locations to produce $DNA$ segments.
$DNA$ polymerase is involved in $DNA$ replication,$DNA$ ligase acts as a molecular glue to join $DNA$ fragments,and $DNA$ gyrase is involved in relieving supercoiling during $DNA$ replication.
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63
BiologyMediumMCQAIPMT · 2001
Which of the following is the correct chronological order of the evolutionary history of man?
A
Peking man,Heidelberg man,Neanderthal,Cro-Magnon
B
Peking man,Homo sapiens,Cro-Magnon,Neanderthal
C
Peking man,Neanderthal,Homo sapiens,Heidelberg
D
Peking man,Cro-Magnon,Homo sapiens,Neanderthal
Solution
(A) The evolutionary sequence of human ancestors is generally accepted as follows:
$1$. Peking man ($Homo$ $erectus$ $pekinensis$) lived approximately $750,000$ to $200,000$ years ago.
$2$. Heidelberg man ($Homo$ $heidelbergensis$) lived approximately $600,000$ to $200,000$ years ago.
$3$. Neanderthal man ($Homo$ $neanderthalensis$) lived approximately $400,000$ to $40,000$ years ago.
$4$. Cro-Magnon man ($Homo$ $sapiens$ $fossilis$) appeared approximately $40,000$ to $10,000$ years ago.
Therefore, the correct chronological order is Peking man $\rightarrow$ Heidelberg man $\rightarrow$ Neanderthal $\rightarrow$ Cro-Magnon.
$1$. Peking man ($Homo$ $erectus$ $pekinensis$) lived approximately $750,000$ to $200,000$ years ago.
$2$. Heidelberg man ($Homo$ $heidelbergensis$) lived approximately $600,000$ to $200,000$ years ago.
$3$. Neanderthal man ($Homo$ $neanderthalensis$) lived approximately $400,000$ to $40,000$ years ago.
$4$. Cro-Magnon man ($Homo$ $sapiens$ $fossilis$) appeared approximately $40,000$ to $10,000$ years ago.
Therefore, the correct chronological order is Peking man $\rightarrow$ Heidelberg man $\rightarrow$ Neanderthal $\rightarrow$ Cro-Magnon.
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64
BiologyEasyMCQAIPMT · 2001
Agriculture originated in the Mesolithic age about:
A
$2000$ years ago
B
$3000-10000$ years ago
C
$7000-13000$ years ago
D
$20000-25000$ years ago
Solution
(C) The origin of agriculture is estimated to have occurred approximately $7000-13000$ years ago.
This development took place during the Mesolithic period in fertile,well-watered regions such as the Indus,Tigris,Nile,and Euphrates river valleys.
This development took place during the Mesolithic period in fertile,well-watered regions such as the Indus,Tigris,Nile,and Euphrates river valleys.
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65
BiologyEasyMCQAIPMT · 2001
Emasculation is a part of
A
Clonal selection
B
Mass selection
C
Hybridization
D
Pure line selection
Solution
(C) The removal of stamens from a bisexual flower before anthesis is known as emasculation.
Emasculation is a crucial step in the process of hybridization.
It is performed to prevent self-pollination in the female parent,ensuring that the desired cross-pollination occurs.
Emasculation is a crucial step in the process of hybridization.
It is performed to prevent self-pollination in the female parent,ensuring that the desired cross-pollination occurs.
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66
BiologyEasyMCQAIPMT · 2001
The process of removing stamens from the flower bud during hybridization is called
A
Crossing
B
Selfing
C
Emasculation
D
Bagging
Solution
(C) In plant breeding,specifically during hybridization,it is essential to prevent self-pollination. The process of removing stamens (anthers) from a bisexual flower bud before they dehisce is known as $Emasculation$. This ensures that the female reproductive parts are available for cross-pollination with the desired pollen grains.
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67
BiologyEasyMCQAIPMT · 2001
Triticale is the hybrid between wheat and
A
Maize
B
Barley
C
Rye
D
Bean
Solution
(C) Triticale is the first man-made cereal crop.
It is produced by crossing wheat ($Triticum$ $aestivum$) and rye ($Secale$ $cereale$).
This intergeneric hybridization combines the high yield and grain quality of wheat with the hardiness and disease resistance of rye.
It is produced by crossing wheat ($Triticum$ $aestivum$) and rye ($Secale$ $cereale$).
This intergeneric hybridization combines the high yield and grain quality of wheat with the hardiness and disease resistance of rye.
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68
BiologyEasyMCQAIPMT · 2001
The order of the lac insect is:
A
Hymenoptera
B
Lepidoptera
C
Hemiptera
D
Diptera
Solution
(C) The lac insect,scientifically known as $Laccifer$ $lacca$ (or $Kerria$ $lacca$),belongs to the order $Hemiptera$ and the family $Kerriidae$. These insects are sap-sucking insects that secrete a resinous substance known as lac.
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69
BiologyMediumMCQAIPMT · 2001
In the poultry industry,the production of hatching eggs is more expensive than the production of market eggs mainly because:
A
The cost of males and their depreciation value is high.
B
Mortality among females is usually lower when they are mated with males.
C
The number of eggs produced by the hatchery flock is to be sold only as market eggs.
D
Some of the eggs produced by hatchery flocks are not acceptable for incubation.
Solution
(A) In the poultry industry,hatching eggs are produced by a breeding flock that includes both males and females.
$1$. The maintenance cost of male birds is significant,and they do not contribute to egg production,which increases the overall cost per egg.
$2$. The depreciation value of these breeding males is high.
$3$. Furthermore,the management requirements for a breeding flock are more rigorous than for a commercial layer flock,leading to higher operational expenses.
Therefore,the primary reason for the higher cost is the investment in and maintenance of the male birds.
$1$. The maintenance cost of male birds is significant,and they do not contribute to egg production,which increases the overall cost per egg.
$2$. The depreciation value of these breeding males is high.
$3$. Furthermore,the management requirements for a breeding flock are more rigorous than for a commercial layer flock,leading to higher operational expenses.
Therefore,the primary reason for the higher cost is the investment in and maintenance of the male birds.
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70
BiologyMediumMCQAIPMT · 2001
Biofertilizers include:
A
Cow dung manure and farmyard waste
B
$A$ quick-growing crop ploughed back into the soil (Green manure)
C
$BGA$ (Blue-Green Algae),Anabaena,and Azolla
D
All of the above
Solution
(D) Biofertilizers are organisms that enrich the nutrient quality of the soil.
$1$. Cow dung manure and farmyard waste are organic fertilizers that provide essential nutrients to the soil.
$2$. Green manure involves growing quick-growing crops and ploughing them back into the soil to increase fertility.
$3$. $BGA$ (Blue-Green Algae) like Anabaena and the fern Azolla are well-known biofertilizers that fix atmospheric nitrogen,thereby enhancing soil fertility.
Since all these options contribute to improving soil nutrient status,the correct answer is $D$.
$1$. Cow dung manure and farmyard waste are organic fertilizers that provide essential nutrients to the soil.
$2$. Green manure involves growing quick-growing crops and ploughing them back into the soil to increase fertility.
$3$. $BGA$ (Blue-Green Algae) like Anabaena and the fern Azolla are well-known biofertilizers that fix atmospheric nitrogen,thereby enhancing soil fertility.
Since all these options contribute to improving soil nutrient status,the correct answer is $D$.
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71
BiologyMediumMCQAIPMT · 2001
Which aquatic fern is an excellent biofertilizer?
A
Salvinia
B
Azolla
C
Marsilea
D
Pteridium
Solution
(B) $Azolla$ $pinnata$ (an aquatic fern) contains symbiotic nitrogen-fixing cyanobacteria ($Anabaena$ $azollae$) in its leaf cavities.
It is widely used as a biofertilizer in rice fields,particularly in south-eastern Asia.
It helps in nitrogen fixation and has been found to increase crop yield by up to $50\%$.
It is widely used as a biofertilizer in rice fields,particularly in south-eastern Asia.
It helps in nitrogen fixation and has been found to increase crop yield by up to $50\%$.
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72
BiologyMediumMCQAIPMT · 2001
$A$ community is defined as
A
$A$ group of birds
B
$A$ collection of species
C
Interacting populations
D
An interactive ecosystem
Solution
(C) community is an assemblage of populations of plants,animals,bacteria,and fungi that live in a specific area and interact with each other. Therefore,it is defined as interacting populations.
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73
BiologyMediumMCQAIPMT · 2001
In an aqueous environment,the microscopic animals are collectively called
A
Herbivores
B
Carnivores
C
Planktons
D
Fauna and flora
Solution
(C) The microscopic organisms that drift or float in an aqueous environment are collectively known as $Planktons$.
These are further classified into $Phytoplanktons$ (microscopic plants) and $Zooplanktons$ (microscopic animals).
Since the question asks for the collective term for microscopic animals in an aquatic environment,$Planktons$ is the correct category,specifically referring to the $Zooplankton$ component.
These are further classified into $Phytoplanktons$ (microscopic plants) and $Zooplanktons$ (microscopic animals).
Since the question asks for the collective term for microscopic animals in an aquatic environment,$Planktons$ is the correct category,specifically referring to the $Zooplankton$ component.
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74
BiologyMediumMCQAIPMT · 2001
The tip of an ecological pyramid is occupied by
A
Herbivores
B
Carnivores
C
Producers
D
Decomposers
Solution
(B) In an ecological pyramid,the base is occupied by producers (autotrophs),followed by primary consumers (herbivores),secondary consumers,and finally,the top carnivores occupy the tip or apex of the pyramid. Therefore,the tip of the ecological pyramid is occupied by carnivores.
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75
BiologyMediumMCQAIPMT · 2001
When the number of organisms at successive levels are plotted,they assume the shape of a pyramid. This is called the pyramid of
A
Energy
B
Number
C
Biomass
D
Both $(a)$ and $(c)$
Solution
(B) The correct answer is $(b)$.
In an ecological pyramid,when the number of individual organisms at each successive trophic level is plotted,it forms the pyramid of number.
This pyramid can be upright,inverted,or spindle-shaped depending on the ecosystem.
In an ecological pyramid,when the number of individual organisms at each successive trophic level is plotted,it forms the pyramid of number.
This pyramid can be upright,inverted,or spindle-shaped depending on the ecosystem.
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76
BiologyMediumMCQAIPMT · 2001
Energy enters into the ecosystem through
A
Herbivores
B
Carnivores
C
Producers
D
Decomposers
Solution
(C) The correct answer is $C$.
Producers,primarily photosynthetic plants,are the entry point for energy in an ecosystem.
They capture solar energy and convert it into chemical energy through the process of photosynthesis.
This stored chemical energy is then transferred to other trophic levels,such as herbivores and carnivores,as they consume the producers.
Producers,primarily photosynthetic plants,are the entry point for energy in an ecosystem.
They capture solar energy and convert it into chemical energy through the process of photosynthesis.
This stored chemical energy is then transferred to other trophic levels,such as herbivores and carnivores,as they consume the producers.
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77
BiologyEasyMCQAIPMT · 2001
What is the intensity of sound in normal conversation?
A
$10-20 dB$
B
$40-60 dB$
C
$90-120 dB$
D
$120-150 dB$
Solution
(B) The intensity of sound is measured in decibels $(dB)$.
Normal conversation typically occurs at an intensity level ranging between $40$ and $60 dB$.
Sounds above $80 dB$ are considered to be noise pollution and can cause hearing damage over prolonged exposure.
Normal conversation typically occurs at an intensity level ranging between $40$ and $60 dB$.
Sounds above $80 dB$ are considered to be noise pollution and can cause hearing damage over prolonged exposure.
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78
BiologyEasyMCQAIPMT · 2001
Which of the following is most poisonous?
A
$CO$
B
$CO_2$
C
$C$
D
$SO_2$
Solution
(A) $CO$ (carbon monoxide) is the most poisonous among the given options.
In human beings,carbon monoxide causes headaches,dizziness,and difficulty in breathing.
Haemoglobin has a significantly greater affinity for $CO$ than for oxygen.
It binds with haemoglobin to form carboxyhaemoglobin $(Hb.CO)$,which is much more stable than oxyhaemoglobin.
This reduces the oxygen-carrying capacity of the blood,leading to hypoxia and potentially death.
In human beings,carbon monoxide causes headaches,dizziness,and difficulty in breathing.
Haemoglobin has a significantly greater affinity for $CO$ than for oxygen.
It binds with haemoglobin to form carboxyhaemoglobin $(Hb.CO)$,which is much more stable than oxyhaemoglobin.
This reduces the oxygen-carrying capacity of the blood,leading to hypoxia and potentially death.
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79
BiologyEasyMCQAIPMT · 2001
Ganja and $LSD$ are classified in
A
Narcotics
B
Hallucinogens
C
Stimulants
D
Depressants
Solution
(B) Hallucinogens are drugs that alter the thoughts, feelings, and perceptions of individuals who consume them.
Ganja and $LSD$ are both classified as hallucinogens.
Ganja is obtained from the hemp plant $(Cannabis sativa)$, and $LSD$ (Lysergic acid diethylamide) is derived from the fungus $Claviceps purpurea$.
Ganja and $LSD$ are both classified as hallucinogens.
Ganja is obtained from the hemp plant $(Cannabis sativa)$, and $LSD$ (Lysergic acid diethylamide) is derived from the fungus $Claviceps purpurea$.
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80
BiologyEasyMCQAIPMT · 2001
The lightest wood is obtained from which of the following plants?
A
Hardwickia binata
B
Ochroma lagopus (Balsa)
C
Cereus giganteus
D
Cycas
Solution
(B) The lightest wood known is obtained from the tree $Ochroma$ $lagopus$,commonly known as the $Balsa$ tree.
Its wood has a very low density,making it extremely lightweight yet strong,which is why it is widely used in model building,insulation,and life-saving equipment.
Its wood has a very low density,making it extremely lightweight yet strong,which is why it is widely used in model building,insulation,and life-saving equipment.
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81
BiologyEasyMCQAIPMT · 2001
Which vegetables were absent in India before the arrival of European invaders?
A
Potato and Tomato
B
Bitter Gourd
C
Capsicum and Brinjal
D
Maize and Snake Gourd
Solution
(A) Potato,tomato,and chili were introduced to India by the Portuguese from the New World (South America). Before their arrival,these vegetables were not cultivated in India. Therefore,the correct answer is $A$.
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82
BiologyEasyMCQAIPMT · 2001
Which of the following is the closest relative of humans?
A
Chimpanzee
B
Gorilla
C
Orangutan
D
Gibbon
Solution
(A) The $Chimpanzee$ ($Pan$ $troglodytes$) is considered the closest living relative to humans ($Homo$ $sapiens$).
Genetic studies have shown that humans and chimpanzees share approximately $98.7\%$ to $99\%$ of their $DNA$ sequences.
This high degree of genetic similarity indicates a very recent common ancestor in evolutionary history,approximately $6$ to $7$ million years ago.
Genetic studies have shown that humans and chimpanzees share approximately $98.7\%$ to $99\%$ of their $DNA$ sequences.
This high degree of genetic similarity indicates a very recent common ancestor in evolutionary history,approximately $6$ to $7$ million years ago.
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83
BiologyMediumMCQAIPMT · 2001
The first life on Earth was ...
A
Cyanobacteria
B
Chemoheterotrophs
C
Autotrophs
D
Photoautotrophs
Solution
(B) According to the Oparin-Haldane hypothesis and current biological consensus,the first life forms on Earth were anaerobic chemoheterotrophs.
These organisms obtained energy by consuming organic molecules present in the 'primordial soup' of the early ocean,as they lacked the complex machinery for photosynthesis or chemosynthesis at that stage.
These organisms obtained energy by consuming organic molecules present in the 'primordial soup' of the early ocean,as they lacked the complex machinery for photosynthesis or chemosynthesis at that stage.
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84
BiologyMediumMCQAIPMT · 2001
What is the cause of the diversity observed in living organisms?
A
Mutation
B
Gradual change
C
Long-term evolutionary changes
D
Short-term evolutionary changes
Solution
(C) The diversity observed in living organisms is primarily the result of long-term evolutionary changes. Over millions of years,processes such as natural selection,genetic drift,and mutation lead to the accumulation of variations,resulting in the vast biodiversity we see today.
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85
BiologyMediumMCQAIPMT · 2001
Interferons are synthesized in response to the infection of which of the following?
A
Mycoplasma
B
Bacteria
C
Virus
D
Fungi
Solution
(C) Interferons $(IFNs)$ are a group of signaling proteins made and released by host cells in response to the presence of several pathogens,such as viruses,bacteria,parasites,or tumor cells.
However,they are most characteristically known for being synthesized in response to a $Virus$ infection.
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.
However,they are most characteristically known for being synthesized in response to a $Virus$ infection.
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.
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86
BiologyMediumMCQAIPMT · 2001
What is a plasmid?
A
$A$ piece of $DNA$ that acts as a vector
B
$A$ piece connecting two genes
C
An $mRNA$ that acts as a carrier
D
$A$ photosynthetic fragment
Solution
(A) plasmid is a small,circular,double-stranded $DNA$ molecule that is distinct from a cell's chromosomal $DNA$.
Plasmids are naturally found in bacterial cells and some eukaryotes.
In biotechnology,plasmids are extensively used as vectors to carry foreign genetic material into another cell,where it can be replicated and/or expressed.
Plasmids are naturally found in bacterial cells and some eukaryotes.
In biotechnology,plasmids are extensively used as vectors to carry foreign genetic material into another cell,where it can be replicated and/or expressed.
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87
BiologyMediumMCQAIPMT · 2001
Extranuclear inheritance is observed in .......... .
A
Killer strains of Paramecium
B
Color blindness
C
Phenylketonuria
D
Tay-Sachs disease
Solution
(A) Extranuclear inheritance,also known as cytoplasmic inheritance,refers to the transmission of genes located outside the nucleus,typically in mitochondria or chloroplasts,or through cytoplasmic particles.
In $Paramecium$,the presence of 'kappa particles' in the cytoplasm is a classic example of extranuclear inheritance.
These particles are responsible for the 'killer' trait,where individuals containing them can kill sensitive strains.
Color blindness,Phenylketonuria,and Tay-Sachs disease are all examples of nuclear (Mendelian) genetic disorders.
In $Paramecium$,the presence of 'kappa particles' in the cytoplasm is a classic example of extranuclear inheritance.
These particles are responsible for the 'killer' trait,where individuals containing them can kill sensitive strains.
Color blindness,Phenylketonuria,and Tay-Sachs disease are all examples of nuclear (Mendelian) genetic disorders.
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88
BiologyEasyMCQAIPMT · 2001
The fish used for the control of mosquitoes,known as the larvicidal fish,is ..............
A
Eel
B
Carp
C
Catfish
D
Gambusia
Solution
(D) The fish $Gambusia$ is widely known as the larvicidal fish.
It is used as a biological control agent to manage mosquito populations.
$Gambusia$ feeds on the larvae of mosquitoes,thereby preventing them from reaching the adult stage and reducing the spread of diseases like malaria and dengue.
It is used as a biological control agent to manage mosquito populations.
$Gambusia$ feeds on the larvae of mosquitoes,thereby preventing them from reaching the adult stage and reducing the spread of diseases like malaria and dengue.
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89
BiologyEasyMCQAIPMT · 2001
What is the intensity of sound in normal conversation?
A
$10-20\, dB$
B
$35-60\, dB$
C
$70-90\, dB$
D
$120-150\, dB$
Solution
(B) The intensity of sound is measured in decibels $(dB)$.
Normal conversation typically occurs at an intensity level ranging from $35$ to $60\, dB$.
Sounds below $30\, dB$ are considered very quiet (like a whisper),while sounds above $80-90\, dB$ can be harmful to the human ear over prolonged exposure.
Normal conversation typically occurs at an intensity level ranging from $35$ to $60\, dB$.
Sounds below $30\, dB$ are considered very quiet (like a whisper),while sounds above $80-90\, dB$ can be harmful to the human ear over prolonged exposure.
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90
BiologyMediumMCQAIPMT · 2001
In which of the following is pollination by wind observed?
A
Salvia
B
Bottlebrush
C
Vallisneria
D
Coconut
Solution
(D) Pollination by wind is known as $Anemophily$.
$Salvia$ is pollinated by insects $(Entomophily)$.
$Bottlebrush$ is pollinated by birds $(Ornithophily)$.
$Vallisneria$ is pollinated by water $(Hydrophily)$.
$Coconut$ is a wind-pollinated plant where flowers are small and pollen grains are light and non-sticky to be easily carried by wind currents.
$Salvia$ is pollinated by insects $(Entomophily)$.
$Bottlebrush$ is pollinated by birds $(Ornithophily)$.
$Vallisneria$ is pollinated by water $(Hydrophily)$.
$Coconut$ is a wind-pollinated plant where flowers are small and pollen grains are light and non-sticky to be easily carried by wind currents.
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91
BiologyMediumMCQAIPMT · 2001
In citrus,adventive embryony is caused by which of the following?
A
Nucellus
B
Integument
C
Zygotic embryo
D
Fertilized egg
Solution
(A) Adventive embryony is a type of polyembryony where embryos develop directly from the diploid cells of the nucellus or integuments.
In citrus,the cells of the nucellus surrounding the embryo sac start dividing,protrude into the embryo sac,and develop into embryos.
Since these cells are diploid and originate from the maternal tissue (nucellus),the resulting embryos are genetically identical to the parent plant.
Therefore,the correct answer is $A$ (Nucellus).
In citrus,the cells of the nucellus surrounding the embryo sac start dividing,protrude into the embryo sac,and develop into embryos.
Since these cells are diploid and originate from the maternal tissue (nucellus),the resulting embryos are genetically identical to the parent plant.
Therefore,the correct answer is $A$ (Nucellus).
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92
BiologyMediumMCQAIPMT · 2001
Which of the following groups represents a pair of similar structures or functions?
A
Corpus luteum - Graafian follicle
B
Sebum - Sweat
C
Bundle of His - Pacemaker
D
Vitamin $B_7$ - Niacin
Solution
(A) The correct answer is $A$.
$1$. The $Corpus luteum$ is a temporary endocrine structure in the ovary that develops from the ruptured $Graafian follicle$ after ovulation. Both are stages in the ovarian cycle related to the same follicle.
$2$. $Sebum$ is an oily secretion from sebaceous glands, while $Sweat$ is a watery secretion from sweat glands; they are distinct.
$3$. The $Bundle of His$ is a part of the cardiac conduction system, whereas the $Pacemaker$ $(SA node)$ is the initiator of the heartbeat; they are different.
$4$. Vitamin $B_7$ is $Biotin$, while $Niacin$ is Vitamin $B_3$; they are different vitamins.
$1$. The $Corpus luteum$ is a temporary endocrine structure in the ovary that develops from the ruptured $Graafian follicle$ after ovulation. Both are stages in the ovarian cycle related to the same follicle.
$2$. $Sebum$ is an oily secretion from sebaceous glands, while $Sweat$ is a watery secretion from sweat glands; they are distinct.
$3$. The $Bundle of His$ is a part of the cardiac conduction system, whereas the $Pacemaker$ $(SA node)$ is the initiator of the heartbeat; they are different.
$4$. Vitamin $B_7$ is $Biotin$, while $Niacin$ is Vitamin $B_3$; they are different vitamins.
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93
BiologyMediumMCQAIPMT · 2001
The phenotypic ratio of complementary genes is ................
A
$9 : 7$
B
$12 : 3 : 1$
C
$9 : 3 : 4$
D
$9 : 3 : 3 : 1$
Solution
(A) Complementary genes are two independent pairs of genes that interact to produce a single trait,where each gene is required for the expression of the other.
In the $F_2$ generation of a dihybrid cross involving complementary genes,the phenotypic ratio is modified from the standard $9 : 3 : 3 : 1$ to $9 : 7$.
This occurs because the presence of at least one dominant allele at both loci $(A-B-)$ results in one phenotype (ratio $9$),while any other combination (where one or both genes are homozygous recessive) results in the alternative phenotype (ratio $3+3+1 = 7$).
In the $F_2$ generation of a dihybrid cross involving complementary genes,the phenotypic ratio is modified from the standard $9 : 3 : 3 : 1$ to $9 : 7$.
This occurs because the presence of at least one dominant allele at both loci $(A-B-)$ results in one phenotype (ratio $9$),while any other combination (where one or both genes are homozygous recessive) results in the alternative phenotype (ratio $3+3+1 = 7$).
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94
BiologyMediumMCQAIPMT · 2001
Genes $A$ and $B$ are linked. What will be the genotype of the offspring produced by a cross between $AB/ab$ and $ab/ab$?
A
$AB/ab$ and $ab/ab$
B
$AaBb$ and $aabb$
C
$AA BB$ and $aa bb$
D
None of the above
Solution
(A) The cross is between a dihybrid individual with linked genes $(AB/ab)$ and a homozygous recessive individual $(ab/ab)$.
This is a test cross.
The gametes produced by the $AB/ab$ parent are $AB$ and $ab$ (assuming no crossing over).
The gametes produced by the $ab/ab$ parent are only $ab$.
Therefore,the offspring genotypes will be $AB/ab$ and $ab/ab$.
This is a test cross.
The gametes produced by the $AB/ab$ parent are $AB$ and $ab$ (assuming no crossing over).
The gametes produced by the $ab/ab$ parent are only $ab$.
Therefore,the offspring genotypes will be $AB/ab$ and $ab/ab$.
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95
BiologyMediumMCQAIPMT · 2001
When two non-allelic genes produce a new phenotype when present together, but fail to do so independently, what is this phenomenon called?
A
Epistasis
B
Polygenic inheritance
C
Non-complementary genes
D
Complementary genes
Solution
(D) Complementary genes are two non-allelic genes that interact to produce a specific phenotype, which neither gene can produce on its own. For example, in the sweet pea $(Lathyrus \, odoratus)$, the production of purple flower color requires the presence of two dominant genes ($C$ and $P$). If either gene is absent (e.g., $ccPP$ or $Ccpp$), the flower remains white. Thus, these genes complement each other to express the trait.
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96
BiologyMediumMCQAIPMT · 2001
Sometimes $XX$ males and $XY$ females are produced due to:
A
Deletion
B
Hormonal imbalance
C
Aneuploidy
D
Translocation of a segment of $X$ and $Y$ chromosomes
Solution
(D) In humans,sex determination is primarily governed by the $XY$ system,where $XY$ individuals are typically male and $XX$ individuals are female.
However,the $SRY$ gene (Sex-determining Region $Y$) located on the $Y$ chromosome is responsible for male development.
If a translocation occurs where the $SRY$ gene segment is transferred from the $Y$ chromosome to the $X$ chromosome during meiosis,an $X$ chromosome carrying the $SRY$ gene can be inherited.
Consequently,an $XX$ individual inheriting this $SRY$-carrying $X$ chromosome will develop as a male.
Conversely,an $XY$ individual who loses the $SRY$ gene segment through translocation will develop as a female.
Therefore,the correct answer is the translocation of a segment of $X$ and $Y$ chromosomes.
However,the $SRY$ gene (Sex-determining Region $Y$) located on the $Y$ chromosome is responsible for male development.
If a translocation occurs where the $SRY$ gene segment is transferred from the $Y$ chromosome to the $X$ chromosome during meiosis,an $X$ chromosome carrying the $SRY$ gene can be inherited.
Consequently,an $XX$ individual inheriting this $SRY$-carrying $X$ chromosome will develop as a male.
Conversely,an $XY$ individual who loses the $SRY$ gene segment through translocation will develop as a female.
Therefore,the correct answer is the translocation of a segment of $X$ and $Y$ chromosomes.
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97
BiologyMediumMCQAIPMT · 2001
The number of Barr bodies in a $XXXX$ female is:
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(C) The number of Barr bodies in a cell is calculated using the formula: $N - 1$,where $N$ is the total number of $X$ chromosomes.
In a $XXXX$ female,there are $4$ $X$ chromosomes.
Therefore,the number of Barr bodies = $4 - 1 = 3$.
Thus,the correct option is $C$.
In a $XXXX$ female,there are $4$ $X$ chromosomes.
Therefore,the number of Barr bodies = $4 - 1 = 3$.
Thus,the correct option is $C$.
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98
BiologyMediumMCQAIPMT · 2001
Extranuclear inheritance occurs in ...... .
A
Killer Paramecium
B
Killer Amoeba
C
Euglena
D
Hydra
Solution
$(A)$ Extranuclear inheritance, also known as cytoplasmic inheritance, refers to the transmission of genes present in cytoplasmic organelles like mitochondria or chloroplasts, or other cytoplasmic factors.
In $Paramecium$, the 'killer' trait is determined by cytoplasmic particles called $\kappa$ particles.
These $\kappa$ particles contain $DNA$ and are capable of self-replication, which is a classic example of extranuclear or cytoplasmic inheritance.
Therefore, the correct option is $A$.
In $Paramecium$, the 'killer' trait is determined by cytoplasmic particles called $\kappa$ particles.
These $\kappa$ particles contain $DNA$ and are capable of self-replication, which is a classic example of extranuclear or cytoplasmic inheritance.
Therefore, the correct option is $A$.
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99
BiologyMediumMCQAIPMT · 2001
Which of the following does not follow the law of independent assortment?
A
Non-homologous chromosomes and lack of linkage
B
Genes on homologous chromosomes
C
Linked genes on the same chromosome
D
Unlinked genes on the same chromosome
Solution
(C) The law of independent assortment states that the alleles of two (or more) different genes get sorted into gametes independently of one another.
This law holds true only when the genes are located on different chromosomes or are very far apart on the same chromosome.
When genes are located close together on the same chromosome,they are said to be 'linked'.
Linked genes do not assort independently because they tend to be inherited together as a single unit during meiosis.
Therefore,linked genes on the same chromosome violate the law of independent assortment.
This law holds true only when the genes are located on different chromosomes or are very far apart on the same chromosome.
When genes are located close together on the same chromosome,they are said to be 'linked'.
Linked genes do not assort independently because they tend to be inherited together as a single unit during meiosis.
Therefore,linked genes on the same chromosome violate the law of independent assortment.
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100
BiologyMediumMCQAIPMT · 2001
In his experiments,Mendel observed wrinkled peas. These wrinkles are caused by the accumulation of sugar instead of starch. This is due to the deficiency of which enzyme?
A
Amylase
B
Invertase
C
Diastase
D
Starch-branching enzyme
Solution
(D) In Mendel's experiments on pea plants,the round seed shape is dominant over the wrinkled seed shape. The wrinkled phenotype is caused by the mutation in the gene encoding the starch-branching enzyme $(SBE)$. Due to the deficiency or lack of this enzyme,the conversion of sugar into starch is inefficient,leading to the accumulation of sugar. This high sugar concentration causes the seeds to absorb more water during development and subsequently lose it during maturation,resulting in the wrinkled appearance of the seeds.
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