BiologyQ1–100 of 173 questions
Page 1 of 2 · English
1
BiologyMediumMCQAIPMT · 1998
In the five kingdom system of classification,which single kingdom out of the following can include blue-green algae,nitrogen-fixing bacteria,and methanogenic archaebacteria?
A
Monera
B
Fungi
C
Plantae
D
Protista
Solution
(A) The five kingdom system of classification was proposed by $R.H. Whittaker$.
According to this system,all prokaryotic organisms are placed under the kingdom $Monera$.
Blue-green algae (cyanobacteria),nitrogen-fixing bacteria,and methanogenic archaebacteria are all prokaryotic organisms.
Therefore,they are all included in the kingdom $Monera$.
According to this system,all prokaryotic organisms are placed under the kingdom $Monera$.
Blue-green algae (cyanobacteria),nitrogen-fixing bacteria,and methanogenic archaebacteria are all prokaryotic organisms.
Therefore,they are all included in the kingdom $Monera$.
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2
BiologyMediumMCQAIPMT · 1998
Floral features are commonly used for identification of angiosperms because
A
Reproductive parts are more conservative
B
Flowers can be safely pressed
C
Flowers are nice to work with
D
Flowers have various colours and scents
Solution
(A) In taxonomy,floral features are preferred for the identification of angiosperms because reproductive parts are more conservative than vegetative parts.
Vegetative parts (like leaves,stems,and roots) are highly influenced by environmental conditions and show significant variations.
In contrast,reproductive structures (flowers) are genetically more stable and show less variation due to environmental factors,making them more reliable for classification and identification.
Vegetative parts (like leaves,stems,and roots) are highly influenced by environmental conditions and show significant variations.
In contrast,reproductive structures (flowers) are genetically more stable and show less variation due to environmental factors,making them more reliable for classification and identification.
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3
BiologyEasyMCQAIPMT · 1998
The habitat of $E. coli$ is
A
Water
B
Colon
C
Soil
D
Organic food
Solution
(B) $E. coli$ (Escherichia coli) is a type of bacterium that normally lives in the intestines of people and animals.
It is a primary inhabitant of the human colon (large intestine),where it helps in the production of Vitamin $K$ and Vitamin $B$ complex.
Therefore,the correct habitat among the given options is the colon.
It is a primary inhabitant of the human colon (large intestine),where it helps in the production of Vitamin $K$ and Vitamin $B$ complex.
Therefore,the correct habitat among the given options is the colon.
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4
BiologyEasyMCQAIPMT · 1998
The $DNA$ of $E. coli$ is
A
Single stranded and linear
B
Single stranded and circular
C
Double stranded and linear
D
Double stranded and circular
Solution
(D) $E. coli$ (Escherichia coli) is a prokaryotic organism.
Prokaryotic $DNA$ is typically organized as a single,double-stranded,circular molecule located in the nucleoid region.
Therefore,the correct structure of $E. coli$ $DNA$ is double-stranded and circular.
Prokaryotic $DNA$ is typically organized as a single,double-stranded,circular molecule located in the nucleoid region.
Therefore,the correct structure of $E. coli$ $DNA$ is double-stranded and circular.
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5
BiologyMediumMCQAIPMT · 1998
Transfer of genetic information from one bacterium to another in the transduction process is through
A
Physical contact between donor and recipient strains
B
Conjugation between opposite strain bacterium
C
Bacteriophages released from the donor bacterial strain
D
Another bacterium having special organ for conjugation
Solution
(C) Transduction is a mechanism of horizontal gene transfer in bacteria.
In this process,genetic material is transferred from one bacterium to another by a virus,specifically a bacteriophage.
When a bacteriophage infects a donor bacterium,it may accidentally package host bacterial $DNA$ into its viral capsid.
Upon lysing the donor cell,these phages are released and subsequently infect a recipient bacterium,injecting the donor's genetic material into it.
Therefore,the correct mechanism is the involvement of bacteriophages released from the donor bacterial strain.
In this process,genetic material is transferred from one bacterium to another by a virus,specifically a bacteriophage.
When a bacteriophage infects a donor bacterium,it may accidentally package host bacterial $DNA$ into its viral capsid.
Upon lysing the donor cell,these phages are released and subsequently infect a recipient bacterium,injecting the donor's genetic material into it.
Therefore,the correct mechanism is the involvement of bacteriophages released from the donor bacterial strain.
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6
BiologyMediumMCQAIPMT · 1998
$A$ few organisms are known to grow and multiply at temperatures of $100-105^{\circ}C$. They belong to:
A
Thermophilic subaerial fungi
B
Marine archaebacteria
C
Thermophilic sulphur bacteria
D
Hot spring blue-green algae
Solution
(B) The correct answer is $B$. Marine archaebacteria (specifically hyperthermophiles) possess unique cell membrane structures,such as branched-chain lipids and ether-linked lipids,which provide high stability at extreme temperatures. This allows them to survive and multiply in environments like hydrothermal vents where temperatures reach $100-105^{\circ}C$.
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7
BiologyMediumMCQAIPMT · 1998
Bryophytes are dependent on water because
A
Archegonium has to remain filled with water for fertilization
B
Water is essential for fertilization for their homosporous nature
C
Water is essential for their vegetative propagation
D
The sperms can easily reach up to the egg in the archegonium
Solution
(D) Bryophytes are known as the amphibians of the plant kingdom because they require water for the completion of their life cycle. Specifically,the male gametes (antherozoids) are flagellated and motile. They require a film of water to swim from the antheridium to the archegonium to reach the egg for fertilization.
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8
BiologyMediumMCQAIPMT · 1998
Transfusion tissue,a modified vascular tissue,is present in the leaves of
A
Pinus
B
Dryopteris
C
Lycopodium
D
Dalbergia
Solution
(A) Transfusion tissue is a specialized tissue found in the leaves of gymnosperms,particularly in the genus $Pinus$.
It consists of tracheids with bordered pits and parenchyma cells,which facilitate the lateral conduction of water and nutrients from the vascular bundles to the mesophyll cells.
In $Pinus$ leaves,the vascular strand is unbranched and covered by a pericycle.
It generally contains two conjoint and collateral vascular bundles separated by $T$-shaped sclerenchyma.
It consists of tracheids with bordered pits and parenchyma cells,which facilitate the lateral conduction of water and nutrients from the vascular bundles to the mesophyll cells.
In $Pinus$ leaves,the vascular strand is unbranched and covered by a pericycle.
It generally contains two conjoint and collateral vascular bundles separated by $T$-shaped sclerenchyma.
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9
BiologyEasyMCQAIPMT · 1998
In which plant are the largest sperms found?
A
Cycas
B
Pinus
C
Mango
D
Sunflower
Solution
(A) The sperms of $Cycas$ are the largest $(300 \ \mu m)$ in nature and are visible to the naked eye.
$Cycas$ belongs to the group Gymnosperms,which are known for producing motile,multiciliated male gametes that are significantly larger than those of other seed plants.
$Cycas$ belongs to the group Gymnosperms,which are known for producing motile,multiciliated male gametes that are significantly larger than those of other seed plants.
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10
BiologyMediumMCQAIPMT · 1998
Which of the following statements is incorrect regarding $Cycas$?
A
Its xylem has vessels.
B
It has circinate vernation.
C
It does not have a well-organized female flower.
D
Its roots possess some blue-green algae.
Solution
(A) The correct answer is $A$. $Cycas$ belongs to the group $Gymnospermae$. $A$ characteristic feature of gymnosperms is the absence of vessels in the xylem (except in the order $Gnetales$). Therefore,the statement that $Cycas$ xylem has vessels is incorrect. $Cycas$ exhibits circinate vernation in its leaves,lacks a true female flower (forming megasporophylls instead),and its coralloid roots contain symbiotic blue-green algae like $Anabaena$ and $Nostoc$.
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11
BiologyEasyMCQAIPMT · 1998
Solenocytes or flame cells are the excretory structures of:
A
Echinoderms
B
Annelids
C
Platyhelminthes
D
Molluscs
Solution
(C) Flame cells,also known as protonephridia or solenocytes,are specialized excretory cells found in the simplest freshwater invertebrates,such as flatworms (Phylum $Platyhelminthes$).
These cells function primarily in osmoregulation and the excretion of nitrogenous waste by filtering fluids through a tuft of cilia that resembles a flickering flame.
These cells function primarily in osmoregulation and the excretion of nitrogenous waste by filtering fluids through a tuft of cilia that resembles a flickering flame.
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12
BiologyEasyMCQAIPMT · 1998
What is common among silver fish,scorpion,crab,and honey bee?
A
Compound eyes
B
Poison glands
C
Jointed legs
D
Metamorphosis
Solution
(C) Silver fish,scorpion,crab,and honey bee all belong to the phylum $Arthropoda$. The most characteristic feature of all members of the phylum $Arthropoda$ is the presence of jointed appendages (jointed legs).
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13
BiologyMediumMCQAIPMT · 1998
Which of the following meristems is responsible for extrastelar secondary growth in dicotyledonous stems?
A
Phellogen
B
Intrafascicular cambium
C
Interfascicular cambium
D
Intercalary meristem
Solution
(A) The extrastelar secondary growth involves the formation of tissues outside the stele,specifically in the cortex region.
This process is initiated by the formation of the cork cambium,also known as $Phellogen$.
$Phellogen$ cuts off cells on both sides: the cells produced towards the outside become $Phellem$ (cork),and the cells produced towards the inside become $Phelloderm$ (secondary cortex).
Together,$Phellem$,$Phellogen$,and $Phelloderm$ are known as the $Periderm$.
This process is initiated by the formation of the cork cambium,also known as $Phellogen$.
$Phellogen$ cuts off cells on both sides: the cells produced towards the outside become $Phellem$ (cork),and the cells produced towards the inside become $Phelloderm$ (secondary cortex).
Together,$Phellem$,$Phellogen$,and $Phelloderm$ are known as the $Periderm$.
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14
BiologyMediumMCQAIPMT · 1998
The periderm includes:
A
Cork
B
Cambium
C
Secondary phloem
D
All of these
Solution
(A) The periderm is a collective term for the protective tissues formed during secondary growth in plants. It consists of three layers:
$1$. Phellogen (cork cambium),
$2$. Phellem (cork),and
$3$. Phelloderm (secondary cortex).
Since the phellogen is a type of cambium,and the cork is included,the most appropriate answer among the given choices is $A$ (Cork),as it is a primary component of the periderm. However,if the question implies components of the periderm,it specifically includes phellogen,phellem,and phelloderm.
$1$. Phellogen (cork cambium),
$2$. Phellem (cork),and
$3$. Phelloderm (secondary cortex).
Since the phellogen is a type of cambium,and the cork is included,the most appropriate answer among the given choices is $A$ (Cork),as it is a primary component of the periderm. However,if the question implies components of the periderm,it specifically includes phellogen,phellem,and phelloderm.
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15
BiologyEasyMCQAIPMT · 1998
In mammals,histamine is secreted by
A
Fibroblasts
B
Histiocytes
C
Lymphocytes
D
Mast cells
Solution
(D) $Mast$ cells are specialized cells found in the ground substance of areolar connective tissue.
They play a crucial role in the immune response by secreting chemical mediators such as heparin (an anticoagulant) and histamine (a vasodilator involved in inflammatory responses).
They play a crucial role in the immune response by secreting chemical mediators such as heparin (an anticoagulant) and histamine (a vasodilator involved in inflammatory responses).
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16
BiologyMediumMCQAIPMT · 1998
Which of the following is not the main function of lymph glands?
A
Forming $WBC$
B
Forming antibodies
C
Forming $RBC$
D
Destroying bacteria
Solution
(C) The correct answer is $(C)$.
Lymph glands,also known as lymph nodes,are essential components of the lymphatic system.
They are composed of lymphatic tissue and are responsible for the production of lymphocytes (a type of $WBC$) and antibodies,which are crucial for the immune response.
They act as a defense barrier by filtering the lymph and destroying bacteria and other foreign particles.
However,lymph glands are not involved in the formation of $RBC$ (red blood cells); $RBC$ production occurs primarily in the bone marrow.
Lymph glands,also known as lymph nodes,are essential components of the lymphatic system.
They are composed of lymphatic tissue and are responsible for the production of lymphocytes (a type of $WBC$) and antibodies,which are crucial for the immune response.
They act as a defense barrier by filtering the lymph and destroying bacteria and other foreign particles.
However,lymph glands are not involved in the formation of $RBC$ (red blood cells); $RBC$ production occurs primarily in the bone marrow.
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17
BiologyMediumMCQAIPMT · 1998
The structural and functional unit of the striated muscle fibre is called
A
Sarcolemma
B
Sarcomere
C
Sarcoplasm
D
Myofibril
Solution
(B) The portion of the myofibril between two successive $Z$-lines is considered as the functional unit of contraction and is called a sarcomere.
Each sarcomere consists of one complete $A$-band and two halves of $I$-bands on either side,represented as $(I/2 + A + I/2)$.
Each sarcomere consists of one complete $A$-band and two halves of $I$-bands on either side,represented as $(I/2 + A + I/2)$.
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18
BiologyMediumMCQAIPMT · 1998
Which of the following is the contractile protein of a muscle?
A
Tubulin
B
Myosin
C
Tropomyosin
D
All of these
Solution
(B) Muscle contraction is primarily driven by the interaction of two key proteins: $Actin$ and $Myosin$.
$Myosin$ is the primary contractile protein that forms the thick filaments in muscle fibers.
$Tropomyosin$ is a regulatory protein that covers the binding sites on $Actin$ filaments,and $Tubulin$ is a structural protein involved in the cytoskeleton and microtubule formation,not muscle contraction.
Therefore,$Myosin$ is the correct contractile protein among the given options.
$Myosin$ is the primary contractile protein that forms the thick filaments in muscle fibers.
$Tropomyosin$ is a regulatory protein that covers the binding sites on $Actin$ filaments,and $Tubulin$ is a structural protein involved in the cytoskeleton and microtubule formation,not muscle contraction.
Therefore,$Myosin$ is the correct contractile protein among the given options.
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19
BiologyMediumMCQAIPMT · 1998
Microtubules are involved in which of the following processes?
A
Cell division
B
$DNA$ recognition
C
Muscle contraction
D
Membrane architecture
Solution
(A) Microtubules are essential components of the cytoskeleton. During cell division,they form the spindle fibers that attach to the kinetochores of chromosomes to facilitate their movement and segregation into daughter cells. Therefore,they play a critical role in cell division.
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20
BiologyMediumMCQAIPMT · 1998
Which one of the following statements about cytochrome $P_{450}$ is wrong?
A
It has an important role in metabolism
B
It contains iron
C
It is a coloured cell
D
It is an enzyme involved in oxidation reactions
Solution
(C) Cytochrome $P_{450}$ is a large and diverse superfamily of hemoproteins (containing iron) that function as monooxygenases.
They play a crucial role in the metabolism of various endogenous and exogenous substances,including drugs and toxins,primarily through oxidation reactions.
Cytochrome $P_{450}$ is an enzyme,not a cell.
Therefore,the statement that it is a 'coloured cell' is incorrect.
They play a crucial role in the metabolism of various endogenous and exogenous substances,including drugs and toxins,primarily through oxidation reactions.
Cytochrome $P_{450}$ is an enzyme,not a cell.
Therefore,the statement that it is a 'coloured cell' is incorrect.
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21
BiologyMediumMCQAIPMT · 1998
Which one among the following chemicals is used for causing defoliation of forest trees?
A
Amo $-1618$
B
Phosphon $-D$
C
Maleic hydrazide
D
$2, 4-D$
Solution
(D) $2, 4-D$ ($2$,$4$-Dichlorophenoxyacetic acid) is a synthetic auxin that acts as a herbicide. It is widely used to cause defoliation in forest trees and to control broad-leaved weeds.
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22
BiologyMediumMCQAIPMT · 1998
$A$ plant hormone used for inducing morphogenesis in plant tissue culture is
A
Abscisic acid
B
Gibberellins
C
Cytokinins
D
Ethylene
Solution
(C) Cytokinins are plant hormones that promote cell division and are essential for organogenesis (morphogenesis) in plant tissue culture.
In tissue culture,the ratio of cytokinin to auxin determines the development of shoots and roots.
$A$ higher concentration of cytokinin relative to auxin induces shoot formation,while a higher concentration of auxin relative to cytokinin induces root formation.
In tissue culture,the ratio of cytokinin to auxin determines the development of shoots and roots.
$A$ higher concentration of cytokinin relative to auxin induces shoot formation,while a higher concentration of auxin relative to cytokinin induces root formation.
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23
BiologyEasyMCQAIPMT · 1998
Which combination of gases is suitable for fruit ripening?
A
$80\% \ C_2H_4$ and $20\% \ CO_2$
B
$80\% \ CO_2$ and $20\% \ CH_2$
C
$80\% \ CH_4$ and $20\% \ CO_2$
D
$80\% \ CO_2$ and $20\% \ O_2$
Solution
(A) Ethylene $(C_2H_4)$ is a gaseous plant hormone that promotes fruit ripening.
In commercial applications,a mixture containing $80\% \ C_2H_4$ and $20\% \ CO_2$ is often used to accelerate the ripening process of fruits.
Therefore,the correct combination is $80\% \ C_2H_4$ and $20\% \ CO_2$.
In commercial applications,a mixture containing $80\% \ C_2H_4$ and $20\% \ CO_2$ is often used to accelerate the ripening process of fruits.
Therefore,the correct combination is $80\% \ C_2H_4$ and $20\% \ CO_2$.
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24
BiologyEasyMCQAIPMT · 1998
The response of different organisms to the environmental rhythms of light and darkness is called
A
Phototaxis
B
Phototropism
C
Vernalization
D
Photoperiodism
Solution
(D) The correct answer is $D$.
Photoperiodism is the physiological reaction of organisms to the length of night or a dark period.
It is a biological response to the changes in the proportion of light and darkness in a $24$ hour cycle,which influences various developmental processes such as flowering.
Photoperiodism is the physiological reaction of organisms to the length of night or a dark period.
It is a biological response to the changes in the proportion of light and darkness in a $24$ hour cycle,which influences various developmental processes such as flowering.
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25
BiologyEasyMCQAIPMT · 1998
Phytochrome becomes active in
A
Green light
B
Blue light
C
Red light
D
None of these
Solution
(C) Phytochrome is a photoreceptor pigment that exists in two interconvertible forms: $Pr$ (inactive form) and $Pfr$ (active form).
$Pr$ absorbs red light (wavelength $660 \ nm$) and gets converted into the active $Pfr$ form.
$Pfr$ absorbs far-red light (wavelength $730 \ nm$) and gets converted back into the inactive $Pr$ form.
Therefore,phytochrome becomes active in the presence of red light.
$Pr$ absorbs red light (wavelength $660 \ nm$) and gets converted into the active $Pfr$ form.
$Pfr$ absorbs far-red light (wavelength $730 \ nm$) and gets converted back into the inactive $Pr$ form.
Therefore,phytochrome becomes active in the presence of red light.
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26
BiologyEasyMCQAIPMT · 1998
In vertebrates,lacteals are found in
A
Ileum
B
Ischium
C
Oesophagus
D
Ear
Solution
(A) Lacteals are specialized lymphatic capillaries located in the villi of the small intestine,specifically in the $Ileum$.
They are responsible for the absorption of dietary fats and fat-soluble vitamins from the digested food.
Therefore,the correct option is $A$.
They are responsible for the absorption of dietary fats and fat-soluble vitamins from the digested food.
Therefore,the correct option is $A$.
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27
BiologyEasyMCQAIPMT · 1998
The contraction of the gall bladder is due to:
A
Gastrin
B
Secretin
C
Cholecystokinin
D
Enterogastrone
Solution
(C) The hormone $Cholecystokinin$ $(CCK)$ is secreted by the duodenal mucosa. Its primary function is to stimulate the contraction of the gall bladder and the relaxation of the sphincter of $Oddi$,which facilitates the release of bile into the duodenum to aid in the digestion of fats.
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28
BiologyEasyMCQAIPMT · 1998
Lactose is composed of
A
Glucose + fructose
B
Glucose + glucose
C
Glucose + galactose
D
Fructose + galactose
Solution
(C) Lactose $({C_{12}}{H_{22}}{O_{11}})$ is known as milk sugar.
It is a disaccharide formed by the glycosidic linkage between one molecule of glucose and one molecule of galactose.
It is a disaccharide formed by the glycosidic linkage between one molecule of glucose and one molecule of galactose.
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29
BiologyEasyMCQAIPMT · 1998
To which of the following families do folic acid and pantothenic acid belong?
A
Vitamin $K$
B
Vitamin $A$
C
Vitamin $C$
D
Vitamin $B$ complex
Solution
(D) Folic acid is known as Vitamin $B_9$ and pantothenic acid is known as Vitamin $B_5$.
Both of these vitamins are water-soluble and are essential components of the Vitamin $B$ complex group.
Therefore,they belong to the Vitamin $B$ complex family.
Both of these vitamins are water-soluble and are essential components of the Vitamin $B$ complex group.
Therefore,they belong to the Vitamin $B$ complex family.
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30
BiologyEasyMCQAIPMT · 1998
One of the factors required for the maturation of erythrocytes is
A
Vitamin $D$
B
Vitamin $A$
C
Vitamin $B_{12}$
D
Vitamin $C$
Solution
(C) The correct answer is $C$.
Vitamin $B_{12}$,also known as cyanocobalamin,is an essential water-soluble vitamin.
It plays a critical role in $DNA$ synthesis and the maturation of erythrocytes (red blood cells) in the bone marrow.
$A$ deficiency of Vitamin $B_{12}$ leads to megaloblastic anemia,where immature,large,and dysfunctional red blood cells are produced.
Vitamin $B_{12}$,also known as cyanocobalamin,is an essential water-soluble vitamin.
It plays a critical role in $DNA$ synthesis and the maturation of erythrocytes (red blood cells) in the bone marrow.
$A$ deficiency of Vitamin $B_{12}$ leads to megaloblastic anemia,where immature,large,and dysfunctional red blood cells are produced.
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31
BiologyMediumMCQAIPMT · 1998
The exchange of gases in the alveoli of the lungs takes place by
A
Osmosis
B
Simple diffusion
C
Passive transport
D
Active transport
Solution
(B) The exchange of gases in the alveoli occurs via simple diffusion.
Oxygen moves from the alveolar air into the blood,while $CO_2$ moves from the blood into the alveolar air.
This process is driven by partial pressure gradients of the respective gases across the respiratory membrane.
Oxygen moves from the alveolar air into the blood,while $CO_2$ moves from the blood into the alveolar air.
This process is driven by partial pressure gradients of the respective gases across the respiratory membrane.
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32
BiologyMediumMCQAIPMT · 1998
Total number of bones in the hind limb of a man is
A
$14$
B
$21$
C
$24$
D
$30$
Solution
(D) The correct answer is $(d)$.
Each hind limb of a human consists of $30$ bones.
The distribution of these bones is as follows:
$1$ Femur (thigh bone),
$1$ Patella (knee cap),
$1$ Tibia (shin bone),
$1$ Fibula,
$7$ Tarsals (ankle bones),
$5$ Metatarsals (sole bones),
$14$ Phalanges (toe bones).
Total = $1 + 1 + 1 + 1 + 7 + 5 + 14 = 30$ bones.
Each hind limb of a human consists of $30$ bones.
The distribution of these bones is as follows:
$1$ Femur (thigh bone),
$1$ Patella (knee cap),
$1$ Tibia (shin bone),
$1$ Fibula,
$7$ Tarsals (ankle bones),
$5$ Metatarsals (sole bones),
$14$ Phalanges (toe bones).
Total = $1 + 1 + 1 + 1 + 7 + 5 + 14 = 30$ bones.
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33
BiologyEasyMCQAIPMT · 1998
The embryo in sunflower has
A
No cotyledon
B
One cotyledon
C
Two cotyledons
D
Many cotyledons
Solution
(C) Sunflower $(Helianthus \ annuus)$ belongs to the family Asteraceae,which is a group of dicotyledonous plants.
In dicotyledonous plants,the embryo typically possesses two cotyledons.
Therefore,the embryo in a sunflower seed has two cotyledons.
In dicotyledonous plants,the embryo typically possesses two cotyledons.
Therefore,the embryo in a sunflower seed has two cotyledons.
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34
BiologyMediumMCQAIPMT · 1998
How many genome types are present in a typical green plant cell?
A
Two
B
Three
C
More than five
D
More than ten
Solution
(B) In a typical green plant cell,there are three distinct types of genomes present:
$1$. Nuclear genome: Located within the nucleus.
$2$. Chloroplast genome: Located within the chloroplasts.
$3$. Mitochondrial genome: Located within the mitochondria.
Therefore,the correct answer is $B$ (Three).
$1$. Nuclear genome: Located within the nucleus.
$2$. Chloroplast genome: Located within the chloroplasts.
$3$. Mitochondrial genome: Located within the mitochondria.
Therefore,the correct answer is $B$ (Three).
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35
BiologyMediumMCQAIPMT · 1998
Which of the following is a free-living,aerobic,and non-photosynthetic nitrogen-fixing bacterium?
A
Rhizobium
B
Nostoc
C
Azospirillum
D
Azotobacter
Solution
(D) $Azotobacter$ is a free-living,aerobic,and non-photosynthetic nitrogen-fixing bacterium found in the soil.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.
$Nostoc$ is a photosynthetic cyanobacterium that can fix nitrogen.
$Azospirillum$ is an associative symbiotic nitrogen-fixing bacterium.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.
$Nostoc$ is a photosynthetic cyanobacterium that can fix nitrogen.
$Azospirillum$ is an associative symbiotic nitrogen-fixing bacterium.
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36
BiologyMediumMCQAIPMT · 1998
How does carbon monoxide,emitted by automobiles,prevent the transport of oxygen in the body tissues?
A
By changing oxygen into carbon dioxide
B
By destroying the haemoglobin
C
By forming a stable compound with haemoglobin
D
By obstructing the reaction of oxygen with haemoglobin
Solution
(C) Carbon monoxide $(CO)$ has a much higher affinity for haemoglobin than oxygen. When inhaled,it binds with haemoglobin to form a very stable compound called carboxyhaemoglobin. This process prevents haemoglobin from binding with oxygen,thereby blocking oxygen transport to the body tissues. Prolonged inhalation of $CO$ can lead to severe hypoxia and even death.
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37
BiologyMediumMCQAIPMT · 1998
Some microorganisms grow and multiply at temperatures of $100-105^{\circ}C$. They are included in:
A
Marine archaebacteria
B
Thermophilic sulfur bacteria
C
Hot spring blue-green algae (cyanobacteria)
D
Thermophilic,suberian fungi
Solution
(A) The microorganisms that can survive and multiply at extremely high temperatures $(100-105^{\circ}C)$ are known as hyperthermophiles.
These organisms belong to the domain $Archaea$.
Specifically,certain species of archaebacteria,such as those found in deep-sea hydrothermal vents,are adapted to thrive in these extreme conditions due to the unique structure of their cell walls and membranes.
Therefore,the correct category is marine archaebacteria.
These organisms belong to the domain $Archaea$.
Specifically,certain species of archaebacteria,such as those found in deep-sea hydrothermal vents,are adapted to thrive in these extreme conditions due to the unique structure of their cell walls and membranes.
Therefore,the correct category is marine archaebacteria.
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38
BiologyMediumMCQAIPMT · 1998
$Puccinia$ produces which of the following on wheat leaves?
A
Uredia and Aecia on wheat leaves
B
Uredia and Telia on wheat leaves
C
Uredia and Aecia on barberry leaves
D
Uredia and Pycnia on barberry leaves
Solution
(B) $Puccinia$ $graminis$ $tritici$ is a heteroecious fungus that completes its life cycle on two hosts: wheat and barberry.
On wheat leaves,it produces two types of spores: urediospores (in uredia) and teliospores (in telia).
On barberry leaves,it produces pycniospores (in pycnia) and aeciospores (in aecia).
Therefore,the structures produced on wheat leaves are uredia and telia.
On wheat leaves,it produces two types of spores: urediospores (in uredia) and teliospores (in telia).
On barberry leaves,it produces pycniospores (in pycnia) and aeciospores (in aecia).
Therefore,the structures produced on wheat leaves are uredia and telia.
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39
BiologyMediumMCQAIPMT · 1998
The name "Walking Fern" is given because of:
A
its propagation by motile animal agencies.
B
it shows vegetative growth through the tips of its leaves.
C
it knows how to walk by itself.
D
its spores are motile (can walk).
Solution
(B) The "Walking Fern" (scientifically known as $Adiantum \, caudatum$) is named so because it reproduces vegetatively through the tips of its leaves. When the leaf tip touches the soil, it develops adventitious roots and a new plantlet, allowing the fern to "walk" or spread across the ground over time.
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40
BiologyMediumMCQAIPMT · 1998
$Ulothrix$ can be described as:
A
$A$ non-motile colonial alga lacking zoospores.
B
$A$ filamentous alga lacking a flagellated reproductive stage.
C
$A$ membranous alga producing zoospores.
D
$A$ filamentous alga with a flagellated reproductive stage.
Solution
(D) $Ulothrix$ is a genus of non-branching filamentous green algae.
It is found in fresh and marine water.
The life cycle of $Ulothrix$ includes a motile reproductive stage where it produces flagellated zoospores and gametes.
Since it possesses flagellated reproductive cells,it is classified as a filamentous alga with a flagellated reproductive stage.
It is found in fresh and marine water.
The life cycle of $Ulothrix$ includes a motile reproductive stage where it produces flagellated zoospores and gametes.
Since it possesses flagellated reproductive cells,it is classified as a filamentous alga with a flagellated reproductive stage.
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41
BiologyMediumMCQAIPMT · 1998
Which of the following statements is incorrect for $Cycas$?
A
It lacks an organized female flower.
B
It has circinate vernation.
C
Its xylem contains only xylem tracheids.
D
Its roots contain some blue-green algae.
Solution
(C) In $Cycas$,the female reproductive structures are not organized into a compact cone or flower; instead,they are megasporophylls arranged loosely. Thus,statement $A$ is correct. $Cycas$ leaves exhibit circinate vernation (young leaves are coiled),so statement $B$ is correct. The roots of $Cycas$ form a symbiotic association with nitrogen-fixing cyanobacteria (blue-green algae like $Anabaena$ or $Nostoc$) in coralloid roots,so statement $D$ is correct. Regarding statement $C$,the xylem of $Cycas$ consists of tracheids,but it also contains xylem parenchyma. The statement that it contains 'only' xylem tracheids is incorrect because it ignores the presence of parenchyma,making $C$ the incorrect statement.
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42
BiologyMediumMCQAIPMT · 1998
Bryophytes are dependent on water because...
A
Water is essential for their homosporous nature.
B
Water is required for vegetative growth.
C
Male gametes can easily reach the egg cells within the archegonium.
D
The archegonium must be filled with water for fertilization to occur.
Solution
(C) Bryophytes are often called the 'amphibians of the plant kingdom' because they require water for fertilization.
In bryophytes,the male gametes (antherozoids) are flagellated and motile.
These male gametes require a film of water to swim from the antheridium to the archegonium,where the egg is located.
Without water,the male gametes cannot reach the egg,and fertilization cannot take place.
In bryophytes,the male gametes (antherozoids) are flagellated and motile.
These male gametes require a film of water to swim from the antheridium to the archegonium,where the egg is located.
Without water,the male gametes cannot reach the egg,and fertilization cannot take place.
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43
BiologyEasyMCQAIPMT · 1998
The largest ovule in the plant kingdom is found in $......$.
A
$Pinus$
B
$Banyan$
C
$Cycas$
D
$Thuja$
Solution
(C) The plant kingdom exhibits a wide range of variation in the size of reproductive structures. Among gymnosperms,the genus $Cycas$ is known for producing the largest ovules. Specifically,$Cycas$ $revoluta$ is noted for having very large ovules compared to other seed-bearing plants.
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44
BiologyMediumMCQAIPMT · 1998
The most appropriate term to describe the life cycle of $Obelia$ is .....
A
Neoteny
B
Metagenesis
C
Metamorphosis
D
All of the above
Solution
(B) $Obelia$ exhibits an alternation of generations between the asexual polyp form and the sexual medusa form. This phenomenon,where the asexual phase alternates with the sexual phase in the life cycle of certain cnidarians,is specifically termed as Metagenesis.
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45
BiologyMediumMCQAIPMT · 1998
Long bones are hollow and connected by air passages; these are the characteristics of ......... .
A
Aves
B
Mammals
C
Reptiles
D
Terrestrial vertebrates
Solution
(A) The presence of hollow long bones that are connected by air passages is known as pneumatic bones.
These pneumatic bones are a characteristic feature of birds (Class $Aves$).
This adaptation helps in reducing body weight,which facilitates flight.
These pneumatic bones are a characteristic feature of birds (Class $Aves$).
This adaptation helps in reducing body weight,which facilitates flight.
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46
BiologyMediumMCQAIPMT · 1998
The embryo of a sunflower contains ..........
A
one cotyledon
B
two cotyledons
C
many cotyledons
D
no cotyledons
Solution
(B) Sunflower $(Helianthus \text{ } annuus)$ belongs to the family $Asteraceae$ (or $Compositae$).
It is a dicotyledonous plant.
In dicotyledonous plants, the embryo typically consists of an embryonal axis and two cotyledons.
Therefore, the embryo of a sunflower contains two cotyledons.
It is a dicotyledonous plant.
In dicotyledonous plants, the embryo typically consists of an embryonal axis and two cotyledons.
Therefore, the embryo of a sunflower contains two cotyledons.
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47
BiologyMediumMCQAIPMT · 1998
Floral characters are used for identification of angiosperms because ...........
A
Flowers are of different colors.
B
Flowers can be easily pressed.
C
Reproductive parts are more stable and conservative than vegetative parts.
D
Flowers can be worked with beautifully.
Solution
(C) Floral characters are preferred for the identification and classification of angiosperms because vegetative characters (such as leaves,stems,and roots) are highly plastic and can change rapidly due to environmental factors. In contrast,floral parts (reproductive organs) are more stable and conservative,meaning they show less variation over evolutionary time. Therefore,they provide more reliable and consistent taxonomic markers for classification.
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48
BiologyMediumMCQAIPMT · 1998
The leaf primordium develops into a mature leaf by the activity of ................
A
Apical meristem
B
Lateral meristem
C
Marginal meristem
D
Initially apical meristem and later mostly marginal meristem
Solution
(D) The development of a leaf primordium into a mature leaf involves specific meristematic activities.
Initially,the leaf primordium arises from the shoot apical meristem.
However,the subsequent growth and expansion of the leaf blade (lamina) are primarily driven by the activity of the marginal meristem,which is located along the edges of the developing leaf primordium.
Therefore,the correct sequence is that it starts with the apical meristem and is followed by the activity of the marginal meristem.
Initially,the leaf primordium arises from the shoot apical meristem.
However,the subsequent growth and expansion of the leaf blade (lamina) are primarily driven by the activity of the marginal meristem,which is located along the edges of the developing leaf primordium.
Therefore,the correct sequence is that it starts with the apical meristem and is followed by the activity of the marginal meristem.
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49
BiologyMediumMCQAIPMT · 1998
Which of the following meristems is responsible for the formation of the extra-stelar secondary growth in a dicot stem?
A
Interfascicular cambium
B
Intrafascicular cambium
C
Intercalary meristem
D
Phellogen
Solution
(D) In a dicot stem,secondary growth occurs in two regions: the stelar region and the extra-stelar region.
$1$. The stelar secondary growth is caused by the vascular cambium (formed by intrafascicular and interfascicular cambium).
$2$. The extra-stelar secondary growth (also known as cork formation or periderm formation) is responsible for replacing the epidermis and cortex as the stem increases in girth.
$3$. This process is initiated by the $Phellogen$ (cork cambium),which is a meristematic tissue that develops in the cortex region.
$4$. Therefore,$Phellogen$ is responsible for the extra-stelar secondary growth.
$1$. The stelar secondary growth is caused by the vascular cambium (formed by intrafascicular and interfascicular cambium).
$2$. The extra-stelar secondary growth (also known as cork formation or periderm formation) is responsible for replacing the epidermis and cortex as the stem increases in girth.
$3$. This process is initiated by the $Phellogen$ (cork cambium),which is a meristematic tissue that develops in the cortex region.
$4$. Therefore,$Phellogen$ is responsible for the extra-stelar secondary growth.
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50
BiologyMediumMCQAIPMT · 1998
The functional unit of the contractile system for a striated muscle is:
A
Myofibril
B
Sarcomere
C
$Z$-lines
D
Cross bridges
Solution
(B) The functional unit of contraction in striated muscle is the $Sarcomere$.
It is defined as the segment of a myofibril between two successive $Z$-lines.
Each $Sarcomere$ contains the contractile proteins,actin and myosin,arranged in a specific pattern that allows for muscle shortening during contraction.
It is defined as the segment of a myofibril between two successive $Z$-lines.
Each $Sarcomere$ contains the contractile proteins,actin and myosin,arranged in a specific pattern that allows for muscle shortening during contraction.
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51
BiologyEasyMCQAIPMT · 1998
Species living in different geographical areas are known as:
A
Sibling species
B
Morphospecies
C
Sympatric species
D
Allopatric species
Solution
(D) The correct answer is $D$.
Species that inhabit different geographical areas and are separated by physical barriers are known as allopatric species.
These species evolve independently due to geographical isolation,which prevents gene flow between populations.
Species that inhabit different geographical areas and are separated by physical barriers are known as allopatric species.
These species evolve independently due to geographical isolation,which prevents gene flow between populations.
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52
BiologyMediumMCQAIPMT · 1998
The main role of bacteria in the carbon cycle involves:
A
Photosynthesis
B
Assimilation of nitrogenous compounds
C
Chemosynthesis
D
Digestion or breakdown of organic compounds
Solution
(D) The main role of bacteria in the carbon cycle is that of a decomposer.
Bacteria break down complex organic compounds into simpler substances through the process of decomposition.
These simpler substances are then released back into the environment and reused in the carbon cycle,thereby maintaining the balance of carbon in the ecosystem.
Bacteria break down complex organic compounds into simpler substances through the process of decomposition.
These simpler substances are then released back into the environment and reused in the carbon cycle,thereby maintaining the balance of carbon in the ecosystem.
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53
BiologyMediumMCQAIPMT · 1998
Which of the following is a non-symbiotic biofertilizer?
A
$VAM$
B
$Azotobacter$
C
$Anabaena$
D
$Rhizobium$
Solution
(B) Non-symbiotic biofertilizers are free-living microorganisms that fix atmospheric nitrogen independently without forming a symbiotic association with plants.
$Azotobacter$ is a well-known free-living (non-symbiotic) nitrogen-fixing bacterium.
$VAM$ (Vesicular-Arbuscular Mycorrhiza) is a symbiotic association between fungi and plant roots.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium found in the root nodules of legumes.
$Anabaena$ can be free-living but is most commonly associated with the water fern $Azolla$ in a symbiotic relationship.
Therefore,$Azotobacter$ is the correct answer.
$Azotobacter$ is a well-known free-living (non-symbiotic) nitrogen-fixing bacterium.
$VAM$ (Vesicular-Arbuscular Mycorrhiza) is a symbiotic association between fungi and plant roots.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium found in the root nodules of legumes.
$Anabaena$ can be free-living but is most commonly associated with the water fern $Azolla$ in a symbiotic relationship.
Therefore,$Azotobacter$ is the correct answer.
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54
BiologyMediumMCQAIPMT · 1998
Typhoid fever is caused by
A
Giardia
B
Salmonella
C
Shigella
D
Escherichia
Solution
(B) Typhoid fever is a bacterial infection caused by the bacterium $Salmonella$ $\text{typhi}$.
It is transmitted through the ingestion of contaminated food and water.
The pathogen enters the small intestine and migrates to other organs through the blood.
Common symptoms include high fever, weakness, stomach pain, constipation, headache, and loss of appetite.
It is transmitted through the ingestion of contaminated food and water.
The pathogen enters the small intestine and migrates to other organs through the blood.
Common symptoms include high fever, weakness, stomach pain, constipation, headache, and loss of appetite.
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55
BiologyMediumMCQAIPMT · 1998
Farmers have reported over $50\%$ higher yields of rice by using which of the following biofertilizer?
A
Mycorrhiza
B
Azolla pinnata
C
Cyanobacteria
D
Legume-Rhizobium symbiosis
Solution
(B) $Azolla pinnata$ is a small aquatic fern that has a symbiotic relationship with the nitrogen-fixing cyanobacterium $Anabaena azollae$.
When used as a biofertilizer in rice fields,it provides a significant amount of nitrogen to the soil.
Farmers have reported that the application of $Azolla$ can increase rice yields by over $50\%$ due to its nitrogen-fixing capabilities and its role as a green manure.
When used as a biofertilizer in rice fields,it provides a significant amount of nitrogen to the soil.
Farmers have reported that the application of $Azolla$ can increase rice yields by over $50\%$ due to its nitrogen-fixing capabilities and its role as a green manure.
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56
BiologyMediumMCQAIPMT · 1998
$A$ bacterium divides every $35$ minutes. If a culture containing $10^5$ cells/ml is grown for $175$ minutes,what will be the cell concentration/ml after $175$ minutes?
A
$175 \times 10^5$ cells/ml
B
$85 \times 10^5$ cells/ml
C
$35 \times 10^5$ cells/ml
D
$32 \times 10^5$ cells/ml
Solution
(D) The bacterium divides every $35$ minutes.
Given total time = $175$ minutes.
Number of divisions $(n)$ = $\frac{175}{35} = 5$.
Since one bacterium produces two cells per division (binary fission),the population grows exponentially according to the formula $N_t = N_0 \times 2^n$,where $N_0$ is the initial concentration and $n$ is the number of divisions.
Initial concentration $(N_0)$ = $10^5$ cells/ml.
Final concentration $(N_t)$ = $10^5 \times 2^5 = 10^5 \times 32 = 32 \times 10^5$ cells/ml.
Given total time = $175$ minutes.
Number of divisions $(n)$ = $\frac{175}{35} = 5$.
Since one bacterium produces two cells per division (binary fission),the population grows exponentially according to the formula $N_t = N_0 \times 2^n$,where $N_0$ is the initial concentration and $n$ is the number of divisions.
Initial concentration $(N_0)$ = $10^5$ cells/ml.
Final concentration $(N_t)$ = $10^5 \times 2^5 = 10^5 \times 32 = 32 \times 10^5$ cells/ml.
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57
BiologyEasyMCQAIPMT · 1998
Animals or organisms floating on the surface of water are known as:
A
Plankton
B
Pelagic
C
Benthon
D
Neritic
Solution
(A) Organisms that float on the surface of water due to their inability to swim against water currents are called $Plankton$.
$Pelagic$ organisms inhabit the open water column.
$Benthon$ (or $Benthos$) refers to organisms living on the bottom of a water body.
$Neritic$ refers to the relatively shallow part of the ocean above the drop-off of the continental shelf.
$Pelagic$ organisms inhabit the open water column.
$Benthon$ (or $Benthos$) refers to organisms living on the bottom of a water body.
$Neritic$ refers to the relatively shallow part of the ocean above the drop-off of the continental shelf.
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58
BiologyEasyMCQAIPMT · 1998
Which one of the following cells,found in the testes of a rabbit,secretes male hormones?
A
Leydig's cells
B
Sertoli cells
C
Epithelial cells
D
Spermatocytes
Solution
(A) The endocrine part of the testes is composed of groups of cells known as interstitial cells or Leydig's cells.
These Leydig's cells are located in the interstitial spaces between the seminiferous tubules.
They are stimulated by Luteinizing Hormone $(LH)$ to secrete male sex hormones,primarily androgens (such as testosterone).
These Leydig's cells are located in the interstitial spaces between the seminiferous tubules.
They are stimulated by Luteinizing Hormone $(LH)$ to secrete male sex hormones,primarily androgens (such as testosterone).
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59
BiologyMediumMCQAIPMT · 1998
The role of double fertilization in angiosperms is to produce
A
Endosperm
B
Integuments
C
Cotyledons
D
Endocarp
Solution
(A) Double fertilization is a characteristic feature of angiosperms. It involves two fusion events:
$1$. Syngamy: One male gamete fuses with the egg cell to form a diploid zygote $(2n)$,which develops into the embryo.
$2$. Triple Fusion: The second male gamete fuses with the two polar nuclei (or the secondary nucleus) to form the primary endosperm nucleus $(PEN)$,which is triploid $(3n)$.
This $PEN$ subsequently develops into the endosperm,which provides nourishment to the developing embryo. Therefore,the primary role of double fertilization is the production of the endosperm.
$1$. Syngamy: One male gamete fuses with the egg cell to form a diploid zygote $(2n)$,which develops into the embryo.
$2$. Triple Fusion: The second male gamete fuses with the two polar nuclei (or the secondary nucleus) to form the primary endosperm nucleus $(PEN)$,which is triploid $(3n)$.
This $PEN$ subsequently develops into the endosperm,which provides nourishment to the developing embryo. Therefore,the primary role of double fertilization is the production of the endosperm.
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60
BiologyAdvancedMCQAIPMT · 1998
If Mendel had studied the seven traits using a plant with $12$ chromosomes instead of $14$,in what way would his interpretation have been different?
A
He could have mapped the chromosome
B
He would have discovered blending or incomplete dominance
C
He would not have discovered the law of independent assortment
D
He would have discovered sex linkage
Solution
(C) Mendel studied $7$ traits in pea plants,which have $14$ chromosomes $(2n = 14)$.
These $7$ traits were located on the $7$ pairs of chromosomes such that each trait was either on a different chromosome or far apart on the same chromosome,allowing them to assort independently.
If the plant had only $12$ chromosomes $(2n = 12)$,there would be only $6$ pairs of chromosomes.
Since there are $7$ traits and only $6$ pairs of chromosomes,at least two traits would have to be located on the same chromosome pair.
If two traits are located on the same chromosome,they are linked and do not show independent assortment.
Therefore,Mendel would not have been able to formulate the Law of Independent Assortment.
These $7$ traits were located on the $7$ pairs of chromosomes such that each trait was either on a different chromosome or far apart on the same chromosome,allowing them to assort independently.
If the plant had only $12$ chromosomes $(2n = 12)$,there would be only $6$ pairs of chromosomes.
Since there are $7$ traits and only $6$ pairs of chromosomes,at least two traits would have to be located on the same chromosome pair.
If two traits are located on the same chromosome,they are linked and do not show independent assortment.
Therefore,Mendel would not have been able to formulate the Law of Independent Assortment.
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61
BiologyMediumMCQAIPMT · 1998
How many different types of genetically different gametes will be produced by a heterozygous plant having the genotype $AABbCc$?
A
Two
B
Four
C
Six
D
Nine
Solution
(B) To determine the number of different types of gametes produced by an organism,we use the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
In the genotype $AABbCc$:
- The gene pair $AA$ is homozygous.
- The gene pair $Bb$ is heterozygous.
- The gene pair $Cc$ is heterozygous.
Thus,the number of heterozygous pairs $(n)$ is $2$ ($Bb$ and $Cc$).
Using the formula $2^n = 2^2 = 4$.
The four types of gametes produced are $ABC$,$ABc$,$AbC$,and $Abc$.
In the genotype $AABbCc$:
- The gene pair $AA$ is homozygous.
- The gene pair $Bb$ is heterozygous.
- The gene pair $Cc$ is heterozygous.
Thus,the number of heterozygous pairs $(n)$ is $2$ ($Bb$ and $Cc$).
Using the formula $2^n = 2^2 = 4$.
The four types of gametes produced are $ABC$,$ABc$,$AbC$,and $Abc$.
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62
BiologyMediumMCQAIPMT · 1998
Crossing over in a diploid organism is responsible for
A
Dominance of genes
B
Linkage between genes
C
Segregation of alleles
D
Recombination of linked alleles
Solution
(D) Crossing over is a biological process that occurs during the pachytene stage of prophase-$I$ of meiosis.
It involves the exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process leads to the formation of new combinations of alleles on the chromosomes,which is known as genetic recombination.
Therefore,crossing over is responsible for the recombination of linked alleles,increasing genetic diversity in the offspring.
It involves the exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process leads to the formation of new combinations of alleles on the chromosomes,which is known as genetic recombination.
Therefore,crossing over is responsible for the recombination of linked alleles,increasing genetic diversity in the offspring.
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63
BiologyDifficultMCQAIPMT · 1998
The formation of multivalents at meiosis in a diploid organism is due to
A
Monosomy
B
Inversion
C
Deletion
D
Reciprocal translocation
Solution
(D) The correct answer is $D$.
Multivalents are associations of more than two homologous chromosomes during meiosis.
They typically occur in organisms that are heterozygous for reciprocal translocations.
In these individuals,segments of non-homologous chromosomes are exchanged,leading to a cross-shaped configuration (quadrivalent) during the pachytene stage of prophase-$I$ of meiosis,as the homologous regions attempt to pair.
Multivalents are associations of more than two homologous chromosomes during meiosis.
They typically occur in organisms that are heterozygous for reciprocal translocations.
In these individuals,segments of non-homologous chromosomes are exchanged,leading to a cross-shaped configuration (quadrivalent) during the pachytene stage of prophase-$I$ of meiosis,as the homologous regions attempt to pair.
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64
BiologyMediumMCQAIPMT · 1998
Mental retardation in humans,associated with sex chromosomal abnormality,is usually due to:
A
Reduction in $X$ complement
B
Increase in $X$ complement
C
Moderate increase in $Y$ complement
D
Large increase in $Y$ complement
Solution
(B) Mental retardation associated with sex chromosomal abnormalities is most commonly observed in conditions like Klinefelter syndrome $(47, XXY)$ or Triple $X$ syndrome $(47, XXX)$.
In these conditions,there is an increase in the number of $X$ chromosomes (aneuploidy).
Specifically,the presence of extra $X$ chromosomes interferes with normal development,often leading to cognitive impairment or mental retardation.
Therefore,an increase in the $X$ complement is the primary cause associated with these chromosomal disorders.
In these conditions,there is an increase in the number of $X$ chromosomes (aneuploidy).
Specifically,the presence of extra $X$ chromosomes interferes with normal development,often leading to cognitive impairment or mental retardation.
Therefore,an increase in the $X$ complement is the primary cause associated with these chromosomal disorders.
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65
BiologyMediumMCQAIPMT · 1998
Albinism is known to be due to an autosomal recessive mutation. The first child of a couple with normal skin pigmentation was an albino. What is the probability that their second child will also be an albino (in $\%$)?
A
$100$
B
$25$
C
$50$
D
$75$
Solution
(B) Albinism is an autosomal recessive trait. Let the normal allele be $A$ and the albino allele be $a$.
Since the parents have normal skin pigmentation but produced an albino child $(aa)$,both parents must be carriers (heterozygous,$Aa$).
The cross is $Aa \times Aa$.
The Punnett square results in the following genotypes: $AA$ (normal),$Aa$ (normal carrier),$Aa$ (normal carrier),and $aa$ (albino).
The probability of having an albino child $(aa)$ in any given pregnancy is $1/4$ or $25\%$.
Therefore,the probability that their second child will be an albino is $25\%$.
Since the parents have normal skin pigmentation but produced an albino child $(aa)$,both parents must be carriers (heterozygous,$Aa$).
The cross is $Aa \times Aa$.
The Punnett square results in the following genotypes: $AA$ (normal),$Aa$ (normal carrier),$Aa$ (normal carrier),and $aa$ (albino).
The probability of having an albino child $(aa)$ in any given pregnancy is $1/4$ or $25\%$.
Therefore,the probability that their second child will be an albino is $25\%$.
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66
BiologyMediumMCQAIPMT · 1998
Loss of a $X$ chromosome in a particular cell during its development results in:
A
Diploid individual
B
Triploid individual
C
Gynandromorphs
D
Both $(a)$ and $(b)$
Solution
(C) $Gynandromorph$ is an organism that contains both male and female tissues. In $Drosophila$,sex is determined by the ratio of $X$ chromosomes to autosomes. If a $XX$ (female) zygote loses one $X$ chromosome in one of its cells during early embryonic development,that cell becomes $XO$ (male). As the organism develops,the descendants of the $XO$ cell will exhibit male characteristics,while the descendants of the original $XX$ cells will exhibit female characteristics. This mosaic of male and female tissues in a single individual is known as a $Gynandromorph$.
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67
BiologyMediumMCQAIPMT · 1998
$A$ woman carrying two genes for haemophilia and one gene for colour blindness on one of her $X$ chromosomes marries a normal man. What will be the phenotype of the progeny?
A
All sons and daughters are haemophilic and colourblind
B
Haemophilic and colourblind daughters
C
$50\%$ haemophilic colourblind sons and $50\%$ normal sons
D
$50\%$ haemophilic daughters and $50\%$ colourblind daughters
Solution
(C) Let the $X$ chromosome carrying the genes for haemophilia and colour blindness be denoted as $X^{hc}$. The normal $X$ chromosome is $X$. The woman's genotype is $X^{hc}X$. The normal man's genotype is $XY$.
Cross: $X^{hc}X \times XY$
Gametes: $(X^{hc}, X)$ and $(X, Y)$
Progeny genotypes: $X^{hc}X$ (carrier daughter),$X^{hc}Y$ (haemophilic and colourblind son),$XX$ (normal daughter),$XY$ (normal son).
Thus,$50\%$ of the sons will be haemophilic and colourblind,and $50\%$ of the sons will be normal.
Cross: $X^{hc}X \times XY$
Gametes: $(X^{hc}, X)$ and $(X, Y)$
Progeny genotypes: $X^{hc}X$ (carrier daughter),$X^{hc}Y$ (haemophilic and colourblind son),$XX$ (normal daughter),$XY$ (normal son).
Thus,$50\%$ of the sons will be haemophilic and colourblind,and $50\%$ of the sons will be normal.
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68
BiologyEasyMCQAIPMT · 1998
Recombinant $DNA$ is achieved by cleaving the pro-$DNA$ by
A
Primase
B
Exonucleases
C
Ligase
D
Restriction endonuclease
Solution
(D) Restriction endonucleases are enzymes that are specialized to cut $DNA$ at specific sites in regions having palindromic sequences. These enzymes act as molecular scissors to create recombinant $DNA$ molecules.
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69
BiologyEasyMCQAIPMT · 1998
The chemical knives of $DNA$ are:
A
Ligases
B
Polymerases
C
Endonucleases
D
Transcriptases
Solution
(C) The correct answer is $(c)$.
Endonucleases are enzymes that cut $DNA$ at specific,desired locations.
Because they act as molecular scissors to cleave the $DNA$ backbone,they are commonly referred to as the chemical knives of $DNA$.
Endonucleases are enzymes that cut $DNA$ at specific,desired locations.
Because they act as molecular scissors to cleave the $DNA$ backbone,they are commonly referred to as the chemical knives of $DNA$.
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70
BiologyMediumMCQAIPMT · 1998
Restriction endonucleases are:
A
Synthesized by bacteria
B
Present in mammalian cells for degradation of $DNA$
C
Used in genetic engineering
D
Used for in vitro $DNA$ synthesis
Solution
(C) Restriction endonucleases are enzymes that cut $DNA$ at specific recognition sequences.
They are naturally produced by bacteria as a defense mechanism against bacteriophages.
In the field of biotechnology,they are essential tools used in genetic engineering to create recombinant $DNA$ molecules by cutting $DNA$ at precise locations.
They are naturally produced by bacteria as a defense mechanism against bacteriophages.
In the field of biotechnology,they are essential tools used in genetic engineering to create recombinant $DNA$ molecules by cutting $DNA$ at precise locations.
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71
BiologyEasyMCQAIPMT · 1998
$DNA$ elements, which can switch their position, are called:
A
Exons
B
Introns
C
Cistrons
D
Transposons/Jumping genes
Solution
(D) . Transposons are genetic elements that can change their position within the genome. They were originally discovered in maize plants by $B. McClintock$. These elements are often referred to as "jumping genes" and can influence gene expression by turning it on or off or by causing mutations when they insert into a gene.
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72
BiologyMediumMCQAIPMT · 1998
Which one of the following statements is correct?
A
$Homo$ $erectus$ is the ancestor of man.
B
Cro-magnon man's fossil has been found in Ethiopia.
C
$Australopithecus$ is the real ancestor of modern man.
D
Neanderthal man is the direct ancestor of $Homo$ $sapiens$.
Solution
(A) The correct statement is that $Homo$ $erectus$ is considered an ancestor of modern man.
$Australopithecus$ is an early hominid but not the direct ancestor of modern humans.
Cro-magnon fossils were primarily found in France,not Ethiopia.
Neanderthal man ($Homo$ $neanderthalensis$) is considered a separate species or subspecies that coexisted with early $Homo$ $sapiens$ rather than being their direct ancestor.
$Australopithecus$ is an early hominid but not the direct ancestor of modern humans.
Cro-magnon fossils were primarily found in France,not Ethiopia.
Neanderthal man ($Homo$ $neanderthalensis$) is considered a separate species or subspecies that coexisted with early $Homo$ $sapiens$ rather than being their direct ancestor.
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73
BiologyEasyMCQAIPMT · 1998
The age of the fossil of $Dryopithecus$ on the geological time scale is approximately:
A
$75 \times 10^6$ years ago
B
$25 \times 10^6$ years ago
C
$2.5 \times 10^6$ years ago
D
$50 \times 10^6$ years ago
Solution
(B) $Dryopithecus$ is considered one of the earliest ancestors of humans and apes.
According to the geological time scale,$Dryopithecus$ lived during the Miocene epoch.
The fossils of $Dryopithecus$ are estimated to be approximately $20-25$ million years old.
Therefore,the correct option is $25 \times 10^6$ years ago.
According to the geological time scale,$Dryopithecus$ lived during the Miocene epoch.
The fossils of $Dryopithecus$ are estimated to be approximately $20-25$ million years old.
Therefore,the correct option is $25 \times 10^6$ years ago.
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74
BiologyMediumMCQAIPMT · 1998
In the developmental history of the mammalian heart,it is observed that it passes through a two-chambered fish-like heart,a three-chambered frog-like heart,and finally a four-chambered stage. To which hypothesis can this above-cited statement be approximated?
A
Biogenetic law
B
Hardy-Weinberg law
C
Lamarck's principle
D
Mendelian principle
Solution
(A) The observation that the embryonic development of an organism reflects its evolutionary history is known as the Biogenetic law or the theory of recapitulation,proposed by Ernst Haeckel.
This law states that 'ontogeny repeats phylogeny'.
In the case of the mammalian heart,the transition from a two-chambered structure (similar to fish) to a three-chambered structure (similar to amphibians) and finally to a four-chambered structure (mammals) serves as an example of this evolutionary recapitulation.
This law states that 'ontogeny repeats phylogeny'.
In the case of the mammalian heart,the transition from a two-chambered structure (similar to fish) to a three-chambered structure (similar to amphibians) and finally to a four-chambered structure (mammals) serves as an example of this evolutionary recapitulation.
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75
BiologyMediumMCQAIPMT · 1998
The reason for using vegetatively reproducing crop plants to maintain hybrid vigour is that:
A
They can be easily propagated
B
They have a longer life span
C
They are more resistant to diseases
D
Once a desirable hybrid is produced,there is no risk of losing it
Solution
(D) Hybrid vigour (heterosis) is the phenomenon where the progeny of a cross between two genetically diverse parents shows superior traits compared to the parents. In sexually reproducing plants,hybrid vigour is lost in subsequent generations due to segregation and recombination during meiosis. Vegetative propagation is an asexual method of reproduction where the offspring are genetically identical clones of the parent plant. Therefore,if a desirable hybrid is produced,vegetative propagation ensures that the hybrid vigour is maintained indefinitely without any genetic change or loss of traits in the progeny.
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76
BiologyEasyMCQAIPMT · 1998
Two bacteria found to be very useful in genetic engineering experiments are
A
Nitrosomonas and Klebsiella
B
Escherichia and Agrobacterium
C
Nitrobacter and Azotobacter
D
Rhizobium and Diplococcus
Solution
(B) The correct answer is $B$.
$Escherichia coli$ $(E. coli)$ is widely used in genetic engineering as a host organism for cloning and the production of recombinant proteins like insulin (humulin) and growth hormones.
$Agrobacterium tumefaciens$ is a soil bacterium that contains the $Ti$ (Tumor-inducing) plasmid,which is extensively used as a vector for gene transfer in plants.
$Escherichia coli$ $(E. coli)$ is widely used in genetic engineering as a host organism for cloning and the production of recombinant proteins like insulin (humulin) and growth hormones.
$Agrobacterium tumefaciens$ is a soil bacterium that contains the $Ti$ (Tumor-inducing) plasmid,which is extensively used as a vector for gene transfer in plants.
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77
BiologyMediumMCQAIPMT · 1998
Genetic engineering is possible,because
A
The phenomenon of transduction in bacteria is well understood.
B
We can see $DNA$ by electron microscope.
C
We can cut $DNA$ at specific sites by endonucleases like $DNAase-I$.
D
Restriction endonucleases purified from bacteria can be used in vitro.
Solution
(D) Genetic engineering involves the manipulation of the genetic material of an organism using specific enzymes.
Restriction endonucleases are enzymes that cut $DNA$ at specific recognition sequences.
These enzymes were discovered in bacteria,where they serve as a defense mechanism against viral infection.
By purifying these enzymes,scientists can use them in vitro to cut and join $DNA$ fragments,which is the foundation of recombinant $DNA$ technology.
$Nathans$ and $Smith$ $(1970)$ isolated the first restriction endonuclease.
$Jackson$,$Symons$,and $Paul$ $Berg$ $(1972)$ successfully generated recombinant $DNA$ molecules in vitro.
Restriction endonucleases are enzymes that cut $DNA$ at specific recognition sequences.
These enzymes were discovered in bacteria,where they serve as a defense mechanism against viral infection.
By purifying these enzymes,scientists can use them in vitro to cut and join $DNA$ fragments,which is the foundation of recombinant $DNA$ technology.
$Nathans$ and $Smith$ $(1970)$ isolated the first restriction endonuclease.
$Jackson$,$Symons$,and $Paul$ $Berg$ $(1972)$ successfully generated recombinant $DNA$ molecules in vitro.
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78
BiologyMediumMCQAIPMT · 1998
Which of the following pesticides is an acetylcholinesterase inhibitor?
A
Aldrin
B
$ \gamma-BHC $
C
Endosulfan
D
Malathion
Solution
(D) Malathion is an organophosphate pesticide that has a strong effect on the nervous system.
It functions by inhibiting the activity of the enzyme acetylcholinesterase.
This inhibition leads to the accumulation of the neurotransmitter acetylcholine at the synaptic cleft, causing continuous nerve stimulation and eventually paralysis or death of the pest.
It functions by inhibiting the activity of the enzyme acetylcholinesterase.
This inhibition leads to the accumulation of the neurotransmitter acetylcholine at the synaptic cleft, causing continuous nerve stimulation and eventually paralysis or death of the pest.
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79
BiologyMediumMCQAIPMT · 1998
Farmers have reported $50\%$ higher yield of rice by using which biofertilizer?
A
Azolla pinnata
B
Legume-Rhizobium symbiosis
C
Cyanobacteria
D
Mycorrhiza
Solution
(A) $Azolla$ $pinnata$ is an aquatic fern that has a symbiotic association with the nitrogen-fixing cyanobacterium $Anabaena$ $azollae$.
This association helps in fixing atmospheric nitrogen,which significantly enriches the soil with nutrients.
When used as a biofertilizer in rice fields,it provides a continuous supply of nitrogen to the crop,leading to a substantial increase in yield,often reported up to $50\%$ higher compared to fields without such biofertilizers.
This association helps in fixing atmospheric nitrogen,which significantly enriches the soil with nutrients.
When used as a biofertilizer in rice fields,it provides a continuous supply of nitrogen to the crop,leading to a substantial increase in yield,often reported up to $50\%$ higher compared to fields without such biofertilizers.
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80
BiologyEasyMCQAIPMT · 1998
Third generation pesticides are:
A
Insect repellents
B
Pheromones
C
Pathogens
D
Insect hormone analogues
Solution
(D) Third generation pesticides refer to insect hormone analogues,such as juvenile hormone mimics.
These substances interfere with the normal development and metamorphosis of insects,preventing them from reaching the adult stage.
Unlike first-generation (inorganic) and second-generation (synthetic organic) pesticides,these are highly specific and generally considered more environmentally friendly.
These substances interfere with the normal development and metamorphosis of insects,preventing them from reaching the adult stage.
Unlike first-generation (inorganic) and second-generation (synthetic organic) pesticides,these are highly specific and generally considered more environmentally friendly.
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81
BiologyMediumMCQAIPMT · 1998
Yeast is used in the production of
A
Ethyl alcohol
B
Acetic acid
C
Cheese
D
Curd
Solution
(A) Yeast $(Saccharomyces \text{ } cerevisiae)$ is used in the industrial production of ethanol $(C_2H_5OH)$.
When yeast is added to a sugar solution, it performs fermentation, which converts sugars into alcohol and carbon dioxide $(CO_2)$.
When yeast is added to a sugar solution, it performs fermentation, which converts sugars into alcohol and carbon dioxide $(CO_2)$.
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82
BiologyMediumMCQAIPMT · 1998
Penicillin was discovered by
A
Waksman
B
Dubois
C
Robert Koch
D
$A$. Fleming
Solution
(D) $A. Fleming$ $(1929)$ discovered the antibiotic penicillin while working on $Staphylococcus$ bacteria. He observed that the mold $Penicillium$ $notatum$ inhibited the growth of these bacteria,marking the discovery of the first antibiotic.
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83
BiologyMediumMCQAIPMT · 1998
Citric acid is produced by
A
Aspergillus niger
B
Streptococcus lactis
C
Acetobacter suboxydans
D
Candida utilis
Solution
(A) Citric acid is commercially produced by the aerobic fermentation of sucrose present in beet molasses using the fungus $Aspergillus$ $niger$.
$Aspergillus$ $niger$ is a filamentous fungus widely used in the biotechnology industry for the production of organic acids.
$Aspergillus$ $niger$ is a filamentous fungus widely used in the biotechnology industry for the production of organic acids.
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84
BiologyMediumMCQAIPMT · 1998
If there was no $CO_2$ in the earth's atmosphere,the temperature of earth's surface would be
A
Same as present
B
Less than the present
C
Higher than the present
D
Dependent on the amount of oxygen in the atmosphere
Solution
(B) $CO_2$ acts as a greenhouse gas in the earth's atmosphere.
It traps the infrared radiation emitted by the earth's surface,which is known as the greenhouse effect.
This process acts as an insulator,preventing heat from escaping into space and keeping the earth's surface warm.
If there were no $CO_2$,the heat would escape into space,leading to a significant decrease in the earth's surface temperature compared to the present.
It traps the infrared radiation emitted by the earth's surface,which is known as the greenhouse effect.
This process acts as an insulator,preventing heat from escaping into space and keeping the earth's surface warm.
If there were no $CO_2$,the heat would escape into space,leading to a significant decrease in the earth's surface temperature compared to the present.
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85
BiologyMediumMCQAIPMT · 1998
In a terrestrial ecosystem such as a forest, which trophic level contains the maximum energy?
A
$T_1$
B
$T_2$
C
$T_3$
D
$T_4$
Solution
(A) In any ecosystem, energy flow follows the $10\%$ law, where only $10\%$ of the energy is transferred from one trophic level to the next.
$T_1$ represents the producers (green plants), which capture solar energy and convert it into chemical energy.
Since energy is lost as heat at each successive trophic level, the primary producers $(T_1)$ possess the maximum amount of energy available in the ecosystem.
Therefore, the correct trophic level is $T_1$.
$T_1$ represents the producers (green plants), which capture solar energy and convert it into chemical energy.
Since energy is lost as heat at each successive trophic level, the primary producers $(T_1)$ possess the maximum amount of energy available in the ecosystem.
Therefore, the correct trophic level is $T_1$.
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86
BiologyMediumMCQAIPMT · 1998
The rate at which light energy is converted into chemical energy of organic molecules is the ecosystem's
A
Net primary productivity
B
Gross secondary productivity
C
Net secondary productivity
D
Gross primary productivity
Solution
(D) $Gross \text{ primary productivity}$ is the total rate of photosynthesis.
Light energy is converted into chemical energy, including the organic matter used up in respiration during the measurement period.
Light energy is converted into chemical energy, including the organic matter used up in respiration during the measurement period.
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87
BiologyMediumMCQAIPMT · 1998
Which of the following communities is more vulnerable to invasion by outside animals and plants?
A
Mangroves
B
Tropical evergreen forests
C
Temperate forests
D
Oceanic island communities
Solution
(D) Oceanic island communities are highly vulnerable to invasion by alien species because they often have low species diversity and lack natural predators or competitors that would otherwise keep invasive populations in check. Due to their isolation,these ecosystems have evolved in the absence of many mainland species,making them ecologically fragile when new species are introduced.
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88
BiologyMediumMCQAIPMT · 1998
In desert grasslands,which type of animals are relatively more abundant?
A
Diurnal
B
Arboreal
C
Aquatic
D
Fossorial
Solution
(D) . Fossorial animals are relatively more abundant in desert grasslands. These animals are adapted to live underground inside burrows to escape the extreme heat and maintain a lower body temperature during the day.
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89
BiologyEasyMCQAIPMT · 1998
Plants such as $Prosopis$,$Acacia$,and $Capparis$ represent examples of tropical
A
Grassland
B
Thorny deserts
C
Deciduous forests
D
Evergreen forests
Solution
(B) The correct answer is $B$.
$Prosopis$ (e.g.,$Prosopis$ $cineraria$),$Acacia$ (e.g.,$Acacia$ $senegal$),and $Capparis$ (e.g.,$Capparis$ $decidua$) are characteristic plant species found in tropical thorn forests or arid desert regions.
These plants have adaptations such as deep root systems,reduced leaves,and thorns to survive in water-scarce environments.
$Prosopis$ (e.g.,$Prosopis$ $cineraria$),$Acacia$ (e.g.,$Acacia$ $senegal$),and $Capparis$ (e.g.,$Capparis$ $decidua$) are characteristic plant species found in tropical thorn forests or arid desert regions.
These plants have adaptations such as deep root systems,reduced leaves,and thorns to survive in water-scarce environments.
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90
BiologyMediumMCQAIPMT · 1998
The role of bacteria in the carbon cycle is:
A
Photosynthesis
B
Chemosynthesis
C
Breakdown of organic compounds
D
Assimilation of nitrogen compounds
Solution
(C) In the carbon cycle,bacteria and fungi act as decomposers. They break down complex organic compounds present in dead plant and animal matter into simpler inorganic substances like $CO_2$. This process releases carbon back into the atmosphere,allowing it to be reused by producers,thus completing the carbon cycle.
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91
BiologyMediumMCQAIPMT · 1998
If there was no $CO_2$ in the earth's atmosphere,the temperature of earth's surface would be
A
As such
B
Less than the present level
C
Increase from present level
D
Dependent upon oxygen amount of the environment
Solution
(B) $CO_2$ acts as a greenhouse gas in the earth's atmosphere. It traps infrared radiation reflected from the earth's surface,acting as an insulator that prevents heat from escaping into space. Therefore,in the absence of $CO_2$,the greenhouse effect would be significantly reduced,causing the earth's surface temperature to be lower than the present level.
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92
BiologyMediumMCQAIPMT · 1998
What is the major cause of the diminishing number of wildlife?
A
Cannibalism
B
Habitat destruction
C
Falling of trees
D
Paucity of drinking water
Solution
(B) The major cause of the decline in wildlife populations is habitat destruction. Habitat destruction refers to the process in which natural habitats are rendered unable to support the species present. This leads to the displacement or death of wildlife. While 'falling of trees' (deforestation) is a specific form of habitat destruction,'Habitat destruction' is the broader and more accurate scientific term that encompasses all activities leading to the loss of natural environments.
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93
BiologyEasyMCQAIPMT · 1998
Which one of the following organisms is used as an indicator of water quality?
A
Beggiatoa
B
Chlorella
C
Azospirillum
D
Escherichia
Solution
(D) $Escherichia$ $coli$ $(E. coli)$ is a bacterium commonly found in the lower intestine of warm-blooded organisms.
Its presence in water samples is used as a standard indicator of faecal contamination,which suggests that the water may contain other harmful pathogens.
Its presence in water samples is used as a standard indicator of faecal contamination,which suggests that the water may contain other harmful pathogens.
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94
BiologyMediumMCQAIPMT · 1998
Carbon monoxide is a pollutant because
A
It reacts with $O_2$
B
It inhibits glycolysis
C
Reacts with haemoglobin
D
Makes nervous system inactive
Solution
(C) $Carbon$ $monoxide$ $(CO)$ is a major air pollutant. It has a significantly higher affinity for haemoglobin than oxygen. When inhaled,it binds with haemoglobin to form carboxyhaemoglobin,which is much more stable than oxyhaemoglobin. This process reduces the oxygen-carrying capacity of the blood,leading to hypoxia and eventually asphyxiation.
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95
BiologyMediumMCQAIPMT · 1998
Two opposite forces operate in the growth and development of every population. One of them relates to the ability to reproduce at a given rate. The force opposite to it is called:
A
Biotic control
B
Mortality
C
Fecundity
D
Environmental resistance
Solution
(D) Every population has an innate ability to grow exponentially,which is known as biotic potential. This is the ability to reproduce at a maximum rate under ideal conditions.
However,in nature,resources are limited,and various factors such as food scarcity,predation,disease,and competition act against this growth.
These limiting factors collectively are known as $Environmental \ resistance$.
Therefore,the force opposite to the biotic potential is $Environmental \ resistance$.
However,in nature,resources are limited,and various factors such as food scarcity,predation,disease,and competition act against this growth.
These limiting factors collectively are known as $Environmental \ resistance$.
Therefore,the force opposite to the biotic potential is $Environmental \ resistance$.
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96
BiologyMediumMCQAIPMT · 1998
The most important component of the oral contraceptive pills is
A
Progesterone
B
Growth hormone
C
Thyroxin
D
Luteinizing hormone
Solution
(A) The most important component of oral contraceptive pills is $Progesterone$ or a combination of $Progesterone$ and $Estrogen$.
These hormones work by inhibiting the release of $Gonadotropin-releasing$ $\text{hormone}$ $(GnRH)$ from the hypothalamus, which in turn prevents the secretion of $Follicle-stimulating$ $\text{hormone}$ $(FSH)$ and $Luteinizing$ $\text{hormone}$ $(LH)$ from the anterior pituitary gland.
By suppressing the surge of $LH$, these pills effectively prevent ovulation, thereby preventing pregnancy.
These hormones work by inhibiting the release of $Gonadotropin-releasing$ $\text{hormone}$ $(GnRH)$ from the hypothalamus, which in turn prevents the secretion of $Follicle-stimulating$ $\text{hormone}$ $(FSH)$ and $Luteinizing$ $\text{hormone}$ $(LH)$ from the anterior pituitary gland.
By suppressing the surge of $LH$, these pills effectively prevent ovulation, thereby preventing pregnancy.
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97
BiologyEasyMCQAIPMT · 1998
The genetic material in the bacterium $Escherichia$ $coli$ $(E. coli)$ is:
A
Single-stranded $DNA$
B
Deoxyribose sugar
C
Double-stranded $DNA$
D
Single-stranded $RNA$
Solution
(C) $E. coli$ is a prokaryotic organism.
Prokaryotes possess a circular,double-stranded $DNA$ molecule as their primary genetic material,which is located in the nucleoid region.
Therefore,the genetic material in $E. coli$ is double-stranded $DNA$.
Prokaryotes possess a circular,double-stranded $DNA$ molecule as their primary genetic material,which is located in the nucleoid region.
Therefore,the genetic material in $E. coli$ is double-stranded $DNA$.
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98
BiologyMediumMCQAIPMT · 1998
The major contribution of bacteria in the carbon cycle includes ..........
A
Photosynthesis
B
Chemosynthesis
C
Digestion or decomposition of organic compounds
D
Assimilation of nitrogenous compounds
Solution
(C) Bacteria play a crucial role in the carbon cycle primarily through the process of decomposition.
They break down complex organic matter (dead plants and animals) into simpler inorganic substances like $CO_2$.
This process releases carbon back into the atmosphere or soil,making it available for reuse by producers.
Therefore,the digestion or decomposition of organic compounds is their most significant contribution to the carbon cycle.
They break down complex organic matter (dead plants and animals) into simpler inorganic substances like $CO_2$.
This process releases carbon back into the atmosphere or soil,making it available for reuse by producers.
Therefore,the digestion or decomposition of organic compounds is their most significant contribution to the carbon cycle.
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99
BiologyEasyMCQAIPMT · 1998
Which of the following microorganisms is used in the industry for the production of citric acid?
A
Penicillium citrinum
B
Aspergillus niger
C
Rhizopus nigricans
D
Lactobacillus bulgaricus
Solution
(B) Citric acid is a commercially important organic acid produced on a large scale using microorganisms.
Specifically,the fungus $Aspergillus$ $niger$ is widely used in industrial fermentation processes for the production of citric acid due to its high yield and efficiency.
Therefore,the correct option is $B$.
Specifically,the fungus $Aspergillus$ $niger$ is widely used in industrial fermentation processes for the production of citric acid due to its high yield and efficiency.
Therefore,the correct option is $B$.
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100
BiologyMediumMCQAIPMT · 1998
The yeast ($Saccharomyces$ $cerevisiae$) is used for the industrial production of ..........
A
Citric acid
B
Tetracycline
C
Ethanol
D
Butanol
Solution
(C) The yeast $Saccharomyces$ $cerevisiae$,commonly known as brewer's yeast,is widely used in the fermentation industry.
It performs anaerobic respiration (fermentation) to convert sugars into ethanol and carbon dioxide.
This process is the fundamental basis for the production of alcoholic beverages and industrial ethanol.
It performs anaerobic respiration (fermentation) to convert sugars into ethanol and carbon dioxide.
This process is the fundamental basis for the production of alcoholic beverages and industrial ethanol.
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