BiologyQ1–100 of 171 questions
Page 1 of 3 · English
1
BiologyMediumMCQAIPMT · 1990
The basic unit of classification / taxonomy is
A
Genus
B
Species
C
Family
D
Order
Solution
(B) The correct answer is $B$.
Species occupies a key position in biological classification.
It is considered the fundamental or basic unit of classification/taxonomy because it represents a group of individual organisms with fundamental similarities.
Species occupies a key position in biological classification.
It is considered the fundamental or basic unit of classification/taxonomy because it represents a group of individual organisms with fundamental similarities.
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2
BiologyMediumMCQAIPMT · 1990
$A$ group of plants with similar traits of any rank is known as:
A
Species
B
Genus
C
Order
D
Taxon
Solution
(D) In biological classification,a $Taxon$ (plural: $Taxa$) represents a group of organisms of any rank that share similar traits or characteristics. It is a fundamental unit of classification used in taxonomy to categorize living beings at various levels,such as species,genus,family,order,class,phylum,or kingdom.
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3
BiologyMediumMCQAIPMT · 1990
Who proposed the Binomial Nomenclature System?
A
Whittaker
B
Mendel
C
Carolus Linnaeus
D
Tippo
Solution
(C) The Binomial Nomenclature System was proposed by Carolus Linnaeus.
This system provides a distinct scientific name to each organism,consisting of two components: the generic name (genus) and the specific epithet (species).
This system provides a distinct scientific name to each organism,consisting of two components: the generic name (genus) and the specific epithet (species).
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4
BiologyMediumMCQAIPMT · 1990
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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5
BiologyEasyMCQAIPMT · 1990
$A$ free-living anaerobic bacterium capable of $N_2$ fixation in soil is
A
Rhizobium
B
Azotobacter
C
Streptococcus
D
Clostridium
Solution
(D) $Clostridium$ is a free-living,anaerobic,nitrogen-fixing bacterium.
It is a saprophytic organism that possesses the ability to fix atmospheric $N_2$ into $NH_3$ in the soil.
$Rhizobium$ is a symbiotic nitrogen fixer,while $Azotobacter$ is a free-living aerobic nitrogen fixer.
It is a saprophytic organism that possesses the ability to fix atmospheric $N_2$ into $NH_3$ in the soil.
$Rhizobium$ is a symbiotic nitrogen fixer,while $Azotobacter$ is a free-living aerobic nitrogen fixer.
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6
BiologyMediumMCQAIPMT · 1990
Plasmodium,the malarial parasite,belongs to which class?
A
Sarcodina
B
Ciliata
C
Sporozoa
D
Dinophyceae
Solution
(C) $Plasmodium$ is a well-known genus of protozoan parasites that cause malaria in humans.
In the traditional classification of protozoans,$Plasmodium$ is placed under the class $Sporozoa$ (also known as $Apicomplexa$ in modern taxonomy).
Members of this class are characterized by having an infectious spore-like stage in their life cycle and lacking locomotor organelles like cilia or flagella in their adult stages.
Therefore,the correct option is $C$.
In the traditional classification of protozoans,$Plasmodium$ is placed under the class $Sporozoa$ (also known as $Apicomplexa$ in modern taxonomy).
Members of this class are characterized by having an infectious spore-like stage in their life cycle and lacking locomotor organelles like cilia or flagella in their adult stages.
Therefore,the correct option is $C$.
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7
BiologyMediumMCQAIPMT · 1990
Protonema is found in the life cycle of
A
Spirogyra
B
Rhizopus
C
Funaria
D
Escherichia
Solution
(C) The life cycle of mosses,such as $Funaria$,includes a distinct stage known as the $Protonema$.
$Protonema$ is a creeping,green,branched,and frequently filamentous stage that develops directly from a germinating spore.
It represents the juvenile gametophyte phase in the life cycle of bryophytes,specifically mosses.
$Spirogyra$ is an alga,$Rhizopus$ is a fungus,and $Escherichia$ is a bacterium; none of these organisms possess a $Protonema$ stage.
$Protonema$ is a creeping,green,branched,and frequently filamentous stage that develops directly from a germinating spore.
It represents the juvenile gametophyte phase in the life cycle of bryophytes,specifically mosses.
$Spirogyra$ is an alga,$Rhizopus$ is a fungus,and $Escherichia$ is a bacterium; none of these organisms possess a $Protonema$ stage.
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8
BiologyEasyMCQAIPMT · 1990
The antherozoids of $Selaginella$ have:
A
Elongated body with many flagella
B
An elongated body with two flagella at one end
C
Top-shaped body with many flagella
D
Oval body with two flagella at one end
Solution
(B) $Selaginella$ is a heterosporous pteridophyte. The male gametes,known as antherozoids,are produced within the antheridium. These antherozoids are biflagellate,meaning they possess two flagella at one end,and they have an elongated,spirally coiled body to facilitate movement in water to reach the female gamete.
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9
BiologyEasyMCQAIPMT · 1990
The $Onchosphere$ larva occurs in:
A
$Ascaris$
B
$Fasciola$
C
$Taenia$
D
$Planaria$
Solution
(C) The fertilized egg of $Taenia$ $solium$ develops into an embryo that is covered by a protective shell.
These shelled embryos are known as $Onchospheres$ (or $Hexacanth$ larvae).
Therefore, the correct option is $C$.
These shelled embryos are known as $Onchospheres$ (or $Hexacanth$ larvae).
Therefore, the correct option is $C$.
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10
BiologyMediumMCQAIPMT · 1990
$Pheretima$ $posthuma$ is highly useful as:
A
Their burrows make the soil loose.
B
They make the soil porous,leave their castings,and take organic debris into the soil.
C
They are used as fish meal.
D
They kill the birds due to biomagnification of chlorinated hydrocarbons.
Solution
(B) $Pheretima$ $posthuma$ (earthworm) is known as the 'friend of farmers'.
They make the soil porous by burrowing,which helps in the penetration of air and water into the soil.
They also consume organic debris and excrete nitrogen-rich castings,which act as natural fertilizers,thereby increasing soil fertility.
They make the soil porous by burrowing,which helps in the penetration of air and water into the soil.
They also consume organic debris and excrete nitrogen-rich castings,which act as natural fertilizers,thereby increasing soil fertility.
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11
BiologyEasyMCQAIPMT · 1990
The blood of $Pheretima$ is:
A
Blue with haemocyanin in corpuscles
B
Blue with haemocyanin in plasma
C
Red with haemoglobin in corpuscles
D
Red with haemoglobin in plasma
Solution
(D) In $Pheretima$ (earthworm),the blood is red in color.
This red color is due to the presence of the respiratory pigment haemoglobin.
Unlike vertebrates,where haemoglobin is contained within red blood corpuscles,in $Pheretima$,the haemoglobin is dissolved directly in the blood plasma.
Therefore,the correct description is that the blood is red with haemoglobin in the plasma.
This red color is due to the presence of the respiratory pigment haemoglobin.
Unlike vertebrates,where haemoglobin is contained within red blood corpuscles,in $Pheretima$,the haemoglobin is dissolved directly in the blood plasma.
Therefore,the correct description is that the blood is red with haemoglobin in the plasma.
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12
BiologyEasyMCQAIPMT · 1990
$Taenia$ $saginata$ differs from $Taenia$ $solium$ in
A
Absence of scolex hooks
B
Absence of scolex hooks and uterine branching
C
Absence of scolex hooks and presence of both male and female reproductive organs
D
Presence of scolex hooks
Solution
(A) $Taenia$ $saginata$ is commonly known as the 'unarmed tapeworm' because its scolex lacks hooks.
In contrast, $Taenia$ $solium$ is known as the 'armed tapeworm' because its scolex possesses a rostellum with hooks.
Therefore, the primary morphological difference mentioned in the options is the absence of hooks on the scolex of $Taenia$ $saginata$.
In contrast, $Taenia$ $solium$ is known as the 'armed tapeworm' because its scolex possesses a rostellum with hooks.
Therefore, the primary morphological difference mentioned in the options is the absence of hooks on the scolex of $Taenia$ $saginata$.
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13
BiologyMediumMCQAIPMT · 1990
$A$ family delimited by the type of inflorescence is
A
Fabaceae
B
Asteraceae
C
Solanaceae
D
Liliaceae
Solution
(B) The family $Asteraceae$ (also known as $Compositae$) is primarily characterized and delimited by its unique type of inflorescence,which is the $capitulum$ or $head$ inflorescence.
In this type of inflorescence,numerous small,sessile flowers called $florets$ are arranged on a common receptacle,often surrounded by an involucre of bracts.
In this type of inflorescence,numerous small,sessile flowers called $florets$ are arranged on a common receptacle,often surrounded by an involucre of bracts.
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14
BiologyMediumMCQAIPMT · 1990
Vascular cambium and cork cambium are examples of
A
Lateral meristem
B
Apical meristem
C
Elements of xylem and phloem
D
Intercalary meristem
Solution
(A) Lateral meristems occur laterally in the axis,parallel to the sides of stems and roots.
They are responsible for secondary growth,which increases the girth of the plant.
The vascular cambium (including fascicular,interfascicular,and extrastelar cambium) and the cork cambium (phellogen) are primary examples of lateral meristems.
They are responsible for secondary growth,which increases the girth of the plant.
The vascular cambium (including fascicular,interfascicular,and extrastelar cambium) and the cork cambium (phellogen) are primary examples of lateral meristems.
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15
BiologyMediumMCQAIPMT · 1990
Collenchymatous tissue is found in
A
Climbing plants
B
Aquatic plants
C
Woody climbers
D
Herbaceous climbers
Solution
(A) $Collenchyma$ is a living mechanical tissue that provides both mechanical support and flexibility to the growing parts of plants,such as young stems and petioles of leaves.
In climbing plants,$collenchyma$ is particularly abundant because these plants require significant flexibility and tensile strength to grow and support themselves while climbing on other structures without breaking.
In climbing plants,$collenchyma$ is particularly abundant because these plants require significant flexibility and tensile strength to grow and support themselves while climbing on other structures without breaking.
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16
BiologyMediumMCQAIPMT · 1990
Multiple epidermis on the dorsal and ventral sides of the leaf is found in:
A
Zea mays
B
Ficus benghalensis
C
Mangifera indica
D
Nerium oleander
Solution
(D) $Nerium \ oleander$ is a xerophytic plant. In this plant,a multiple epidermis (multiseriate epidermis) is present on both the dorsal and ventral sides of the leaf. This structural adaptation helps in reducing the rate of transpiration and prevents excessive water loss from the leaves.
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17
BiologyMediumMCQAIPMT · 1990
In roots,the xylem is:
A
Mesarch
B
Exarch
C
Placed at different places in different plants
D
Endarch
Solution
(B) In roots,the xylem arrangement is $Exarch$.
In an $Exarch$ condition,the $protoxylem$ is directed towards the periphery (outside) and the $metaxylem$ is directed towards the center of the organ.
This is a characteristic feature of the root anatomy in vascular plants.
In an $Exarch$ condition,the $protoxylem$ is directed towards the periphery (outside) and the $metaxylem$ is directed towards the center of the organ.
This is a characteristic feature of the root anatomy in vascular plants.
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18
BiologyMediumMCQAIPMT · 1990
In a monocot leaf:
A
Bulliform cells are absent from the epidermis
B
Veins form a network
C
Mesophyll is well differentiated into these parts
D
Mesophyll is not differentiated into palisade and spongy parenchyma
Solution
(D) In a monocot leaf,the mesophyll tissue located between the upper and lower epidermis is not differentiated into palisade and spongy parenchyma. Instead,all the mesophyll cells are similar in structure and function.
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19
BiologyMediumMCQAIPMT · 1990
Vascular cambium is a meristematic layer that cuts off
A
Primary xylem and primary phloem
B
Xylem vessels and xylem tracheids
C
Primary xylem and secondary xylem
D
Secondary xylem and secondary phloem
Solution
(D) The vascular cambium is a lateral meristematic tissue responsible for secondary growth in plants.
It undergoes periclinal divisions to cut off cells towards the inner side and the outer side.
The cells cut off towards the inner side differentiate into secondary xylem.
The cells cut off towards the outer side differentiate into secondary phloem.
Therefore,the vascular cambium produces secondary xylem and secondary phloem.
It undergoes periclinal divisions to cut off cells towards the inner side and the outer side.
The cells cut off towards the inner side differentiate into secondary xylem.
The cells cut off towards the outer side differentiate into secondary phloem.
Therefore,the vascular cambium produces secondary xylem and secondary phloem.
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20
BiologyMediumMCQAIPMT · 1990
$A$ professor kept some moist seeds in an airtight jar and started lecturing. At the end of the experiment,an explosion occurred in the jar. What did the professor want to explain?
A
Osmosis
B
Diffusion
C
Anaerobic respiration
D
Imbibition
Solution
(D) The correct answer is $D$.
Imbibition is a special type of diffusion where water is absorbed by solids (colloids),causing them to increase in volume.
When moist seeds are placed in an airtight jar,they absorb water and swell significantly due to the process of imbibition.
This swelling exerts a tremendous amount of pressure,known as imbibition pressure,which can be strong enough to break the container,leading to an explosion.
Imbibition is a special type of diffusion where water is absorbed by solids (colloids),causing them to increase in volume.
When moist seeds are placed in an airtight jar,they absorb water and swell significantly due to the process of imbibition.
This swelling exerts a tremendous amount of pressure,known as imbibition pressure,which can be strong enough to break the container,leading to an explosion.
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21
BiologyMediumMCQAIPMT · 1990
The root system in a plant is well developed:
A
Due to deficiency of auxin
B
Due to deficiency of cytokinins
C
Due to deficiency of minerals
D
For increased absorption of water
Solution
(D) plant with a deep and elaborate root system can absorb more water.
Moreover,the number of root hairs will be higher in a highly branched and elaborate root system,thus providing a larger surface area in contact with water for efficient absorption.
Moreover,the number of root hairs will be higher in a highly branched and elaborate root system,thus providing a larger surface area in contact with water for efficient absorption.
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22
BiologyEasyMCQAIPMT · 1990
The principal pathways by which water is translocated in angiosperms is:
A
Xylem and phloem together
B
Sieve tubes and members of phloem
C
Sieve cells of phloem
D
Xylem vessel system
Solution
(D) In angiosperms,water and dissolved minerals are transported from the roots to the aerial parts of the plant through the xylem tissue.
Specifically,the translocation of water occurs through the lumen of xylem vessels and tracheids.
The xylem vessel system provides a continuous,low-resistance pathway for the upward movement of sap,driven primarily by transpiration pull.
Specifically,the translocation of water occurs through the lumen of xylem vessels and tracheids.
The xylem vessel system provides a continuous,low-resistance pathway for the upward movement of sap,driven primarily by transpiration pull.
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23
BiologyMediumMCQAIPMT · 1990
Which of the following is the $CO_2$ acceptor in $C_4$ plants?
A
Phosphoenol pyruvate $(PEP)$
B
Ribulose $1,5$-bisphosphate $(RuBP)$
C
Oxaloacetic acid $(OAA)$
D
Phosphoglyceric acid $(PGA)$
Solution
(A) In $C_4$ plants,the primary $CO_2$ acceptor is a $3$-carbon molecule called Phosphoenol pyruvate $(PEP)$.
This reaction occurs in the mesophyll cells and is catalyzed by the enzyme $PEP$ carboxylase.
The product formed after $CO_2$ fixation is Oxaloacetic acid $(OAA)$,which is a $4$-carbon compound.
This reaction occurs in the mesophyll cells and is catalyzed by the enzyme $PEP$ carboxylase.
The product formed after $CO_2$ fixation is Oxaloacetic acid $(OAA)$,which is a $4$-carbon compound.
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24
BiologyMediumMCQAIPMT · 1990
Kranz type of anatomy is found in
A
$C_2$ plants
B
$C_3$ plants
C
$C_4$ plants
D
$CAM$ plants
Solution
(C) The correct answer is $C_4$ plants.
$A$ fundamental characteristic of $C_4$ plants is the presence of "Kranz" (a German term meaning halo or wreath) anatomy in their leaves.
In $C_4$ leaves, the vascular bundles are surrounded by a layer of bundle sheath cells that contain a large number of chloroplasts.
The chloroplasts in $C_4$ leaves are dimorphic, meaning they exist in two morphologically distinct types: those in the mesophyll cells and those in the bundle sheath cells.
$A$ fundamental characteristic of $C_4$ plants is the presence of "Kranz" (a German term meaning halo or wreath) anatomy in their leaves.
In $C_4$ leaves, the vascular bundles are surrounded by a layer of bundle sheath cells that contain a large number of chloroplasts.
The chloroplasts in $C_4$ leaves are dimorphic, meaning they exist in two morphologically distinct types: those in the mesophyll cells and those in the bundle sheath cells.
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25
BiologyMediumMCQAIPMT · 1990
Total $ATP$ production during $EMP$ pathway is
A
$24$ $ATP$ molecules
B
$8$ $ATP$ molecules
C
$38$ $ATP$ molecules
D
$6$ $ATP$ molecules
Solution
(B) The $EMP$ pathway (Glycolysis) involves the breakdown of one molecule of glucose into two molecules of pyruvate.
During this process,$4$ $ATP$ molecules are produced by substrate-level phosphorylation,and $2$ $ATP$ molecules are consumed in the preparatory phase.
Additionally,$2$ $NADH + H^+$ molecules are produced.
In aerobic respiration,each $NADH$ molecule yields $3$ $ATP$ molecules via the electron transport system ($2 \times 3 = 6$ $ATP$).
Therefore,the total energy yield is $2$ (net $ATP$ from glycolysis) $+ 6$ ($ATP$ from $NADH$) $= 8$ $ATP$ molecules.
During this process,$4$ $ATP$ molecules are produced by substrate-level phosphorylation,and $2$ $ATP$ molecules are consumed in the preparatory phase.
Additionally,$2$ $NADH + H^+$ molecules are produced.
In aerobic respiration,each $NADH$ molecule yields $3$ $ATP$ molecules via the electron transport system ($2 \times 3 = 6$ $ATP$).
Therefore,the total energy yield is $2$ (net $ATP$ from glycolysis) $+ 6$ ($ATP$ from $NADH$) $= 8$ $ATP$ molecules.
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26
BiologyMediumMCQAIPMT · 1990
Several horticultural techniques are followed for the production of 'bonsai' plants. One of them is the drastic pruning of the root system. Which of the following physiological factors is involved in that method?
A
Inadequacy of mineral nutrients
B
Deficiency of auxins
C
Impairment of water absorption
D
Deficiency of cytokinins
Solution
(D) The production of 'bonsai' plants involves restricting the growth of the plant to keep it small.
Drastic pruning of the root system is a key technique used for this purpose.
Roots are the primary site for the synthesis of cytokinins,which are plant hormones that promote cell division and shoot growth.
By pruning the roots,the plant's ability to synthesize and transport adequate amounts of cytokinins to the aerial parts is significantly reduced.
This deficiency of cytokinins leads to stunted growth and the characteristic small size of 'bonsai' plants.
Drastic pruning of the root system is a key technique used for this purpose.
Roots are the primary site for the synthesis of cytokinins,which are plant hormones that promote cell division and shoot growth.
By pruning the roots,the plant's ability to synthesize and transport adequate amounts of cytokinins to the aerial parts is significantly reduced.
This deficiency of cytokinins leads to stunted growth and the characteristic small size of 'bonsai' plants.
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27
BiologyMediumMCQAIPMT · 1990
Phytohormones control:
A
Growth
B
Physiological functions
C
Rooting
D
All of the above
Solution
(D) Phytohormones (plant growth regulators) are chemical substances that regulate various aspects of plant life.
They control growth,development,physiological functions,rooting,flowering,and responses to environmental stimuli.
Therefore,they control all the listed processes.
They control growth,development,physiological functions,rooting,flowering,and responses to environmental stimuli.
Therefore,they control all the listed processes.
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28
BiologyMediumMCQAIPMT · 1990
Which of the following movements is not related to changes in auxin levels?
A
Nyctinastic leaf movement
B
Movement of roots towards soil
C
Movement of sunflower tracking the direction of the sun
D
Movement of shoot towards light
Solution
(A) Auxin is a plant hormone primarily responsible for growth-related movements such as phototropism (movement of shoot towards light) and geotropism (movement of roots towards soil).
Nyctinastic movements (sleep movements) are primarily caused by changes in turgor pressure within the pulvinus cells,not by auxin-mediated growth.
Therefore,nyctinastic leaf movement is not related to changes in auxin levels.
Nyctinastic movements (sleep movements) are primarily caused by changes in turgor pressure within the pulvinus cells,not by auxin-mediated growth.
Therefore,nyctinastic leaf movement is not related to changes in auxin levels.
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29
BiologyMediumMCQAIPMT · 1990
Highest concentration of auxins exists in
A
At the base of various plant organs
B
Growing tip of plants
C
In leaves
D
In xylem and phloem cells only
Solution
(B) Auxins are plant hormones that are primarily synthesized in the shoot apical meristems and young leaves.
Because they are responsible for cell elongation and apical dominance,their concentration is highest in the growing tips (apical buds) of plants.
Therefore,the correct option is $B$.
Because they are responsible for cell elongation and apical dominance,their concentration is highest in the growing tips (apical buds) of plants.
Therefore,the correct option is $B$.
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30
BiologyMediumMCQAIPMT · 1990
Auxins are abundantly produced in
A
Root
B
Meristematic region of the root
C
Shoot
D
Meristematic region of the shoot
Solution
(D) Auxins are plant growth regulators that are primarily synthesized in the growing apices of stems and roots. However,they are produced in the highest concentrations in the apical meristems of the shoot,which are responsible for primary growth and apical dominance. Therefore,the meristematic region of the shoot is the primary site of abundant auxin production.
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31
BiologyMediumMCQAIPMT · 1990
Abscisic acid controls
A
Shoot elongation
B
Cell elongation and cell wall formation
C
Cell division
D
Leaf fall and dormancy
Solution
(D) Abscisic acid $(ABA)$ is a plant growth inhibitor. It plays a crucial role in the abscission of leaves,flowers,and fruits. Spraying $ABA$ induces rapid leaf fall (abscission). Additionally,it promotes dormancy in buds and seeds,helping plants survive unfavorable environmental conditions.
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32
BiologyMediumMCQAIPMT · 1990
Phototropic and geotropic movements are linked to
A
Gibberellins
B
Enzymes
C
Auxin
D
Cytokinins
Solution
(C) Phototropic and geotropic movements in plants are primarily regulated by the hormone $Auxin$.
$Auxin$ is synthesized at the shoot and root tips and is transported to the site of action.
In response to light (phototropism),$Auxin$ accumulates on the shaded side of the stem,causing cells to elongate more on that side,which results in the bending of the stem towards the light.
Similarly,in response to gravity (geotropism),$Auxin$ distribution influences the differential growth of roots and shoots,directing the growth of the plant.
$Auxin$ is synthesized at the shoot and root tips and is transported to the site of action.
In response to light (phototropism),$Auxin$ accumulates on the shaded side of the stem,causing cells to elongate more on that side,which results in the bending of the stem towards the light.
Similarly,in response to gravity (geotropism),$Auxin$ distribution influences the differential growth of roots and shoots,directing the growth of the plant.
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33
BiologyMediumMCQAIPMT · 1990
Phytohormones are:
A
Hormones regulating growth from seed to adulthood
B
Growth regulators synthesised by plants and influencing physiological processes
C
Hormones regulating flowering
D
Hormones regulating secondary growth
Solution
(B) Phytohormones,also known as plant growth regulators (PGRs),are chemical substances produced naturally by plants. They are synthesized in small quantities in various parts of the plant and are transported to other regions where they influence physiological processes such as growth,development,differentiation,and responses to environmental stimuli. Therefore,option $(b)$ is the most comprehensive and accurate definition.
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34
BiologyMediumMCQAIPMT · 1990
In the pancreas,pancreatic juice and hormones are secreted by
A
Same cells
B
Different cells
C
Same cells at different times
D
None of these
Solution
(B) The pancreas is a mixed gland (heterocrine gland).
Its exocrine part,consisting of acinar cells,secretes pancreatic juice containing enzymes.
Its endocrine part,consisting of the Islets of Langerhans,secretes hormones like insulin and glucagon.
Therefore,pancreatic juice and hormones are secreted by different cells.
Its exocrine part,consisting of acinar cells,secretes pancreatic juice containing enzymes.
Its endocrine part,consisting of the Islets of Langerhans,secretes hormones like insulin and glucagon.
Therefore,pancreatic juice and hormones are secreted by different cells.
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35
BiologyEasyMCQAIPMT · 1990
Emulsification of fats is brought about by
A
Bile pigments
B
Bile salts
C
Pancreatic juice
D
$HCl$
Solution
(B) The emulsification of fats is the process of breaking down large fat droplets into smaller,fine droplets to increase the surface area for the action of enzymes like lipase.
This process is carried out by bile salts (such as sodium glycocholate and sodium taurocholate) present in the bile juice secreted by the liver.
Bile pigments (bilirubin and biliverdin) do not participate in the emulsification of fats; they are merely excretory products.
This process is carried out by bile salts (such as sodium glycocholate and sodium taurocholate) present in the bile juice secreted by the liver.
Bile pigments (bilirubin and biliverdin) do not participate in the emulsification of fats; they are merely excretory products.
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36
BiologyEasyMCQAIPMT · 1990
Zymogen cells and chief cells secrete
A
Hydrochloric acid
B
Mucus
C
Pepsinogen
D
Trypsin
Solution
(C) Gastric glands contain three main types of secretory cells: zymogen (also known as chief or peptic) cells,parietal (oxyntic) cells,and mucous cells.
Zymogen or chief cells are responsible for secreting inactive digestive proenzymes,specifically pepsinogen and prorennin.
$HCl$ (Hydrochloric acid),secreted by parietal cells,converts these inactive proenzymes into their active forms,pepsin and rennin,respectively.
Zymogen or chief cells are responsible for secreting inactive digestive proenzymes,specifically pepsinogen and prorennin.
$HCl$ (Hydrochloric acid),secreted by parietal cells,converts these inactive proenzymes into their active forms,pepsin and rennin,respectively.
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37
BiologyEasyMCQAIPMT · 1990
The hormone 'secretin' stimulates the secretion of:
A
Pancreatic juice
B
Bile juice
C
Salivary juice
D
Gastric juice
Solution
(A) The hormone 'secretin' is produced by the duodenal mucosa in response to acidic chyme entering the small intestine.
It primarily acts on the exocrine pancreas to stimulate the secretion of water and bicarbonate ions,which form the bulk of pancreatic juice.
Therefore,the correct option is $A$.
It primarily acts on the exocrine pancreas to stimulate the secretion of water and bicarbonate ions,which form the bulk of pancreatic juice.
Therefore,the correct option is $A$.
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38
BiologyMediumMCQAIPMT · 1990
Chloride shift is essential for the transport of
A
$CO_2$ and $O_2$
B
$N_2$
C
$CO_2$
D
$O_2$
Solution
(C) The chloride shift,also known as the $Hamburger$ phenomenon,is a process that occurs in the blood to facilitate the transport of $CO_2$.
When $CO_2$ enters the red blood cells,it reacts with water to form carbonic acid $(H_2CO_3)$,which dissociates into bicarbonate $(HCO_3^-)$ and hydrogen ions $(H^+)$.
To maintain electrical neutrality as bicarbonate ions diffuse out of the red blood cells into the plasma,chloride ions $(Cl^-)$ move from the plasma into the red blood cells.
This exchange is essential for the efficient transport of $CO_2$ from the tissues to the lungs in the form of bicarbonate ions.
When $CO_2$ enters the red blood cells,it reacts with water to form carbonic acid $(H_2CO_3)$,which dissociates into bicarbonate $(HCO_3^-)$ and hydrogen ions $(H^+)$.
To maintain electrical neutrality as bicarbonate ions diffuse out of the red blood cells into the plasma,chloride ions $(Cl^-)$ move from the plasma into the red blood cells.
This exchange is essential for the efficient transport of $CO_2$ from the tissues to the lungs in the form of bicarbonate ions.
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39
BiologyMediumMCQAIPMT · 1990
Kidney crystals (stones) are solid clusters of:
A
Calcium nitrate and uric acid
B
Calcium oxalate and uric acid
C
Calcium carbonate and uric acid
D
Calcium metabisulphite and uric acid
Solution
(B) Kidney stones, also known as renal calculi, are hard deposits made of minerals and salts that form inside the kidneys.
The most common composition of these crystals includes calcium salts, such as calcium oxalate or calcium phosphate, often combined with uric acid.
Among the given options, calcium oxalate is the most clinically significant component found in kidney stones.
Therefore, the correct answer is $B$.
The most common composition of these crystals includes calcium salts, such as calcium oxalate or calcium phosphate, often combined with uric acid.
Among the given options, calcium oxalate is the most clinically significant component found in kidney stones.
Therefore, the correct answer is $B$.
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40
BiologyMediumMCQAIPMT · 1990
The proximal convoluted tubule has a brush border which is due to
A
Microvilli
B
Minute hairs
C
Endothelium
D
Folded tubes
Solution
(A) The epithelial cells of the $PCT$ (Proximal Convoluted Tubule) possess a brush border.
This brush border is formed by the presence of numerous microvilli on the apical surface of the cells.
The primary function of these microvilli is to increase the surface area for the reabsorption of water,electrolytes,and nutrients from the filtrate.
This brush border is formed by the presence of numerous microvilli on the apical surface of the cells.
The primary function of these microvilli is to increase the surface area for the reabsorption of water,electrolytes,and nutrients from the filtrate.
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41
BiologyMediumMCQAIPMT · 1990
Proximal and distal convoluted tubules are parts of a
A
Nephron
B
Oviduct
C
Vas deferens
D
Caecum
Solution
(A) nephron is the structural and functional unit of the kidney.
It consists of two main parts: the Malpighian corpuscle and the renal tubule.
The renal tubule begins with a double-walled cup-like structure called Bowman's capsule,which further continues into a highly coiled network called the Proximal Convoluted Tubule $(PCT)$.
This is followed by the Henle's loop and then the Distal Convoluted Tubule $(DCT)$,which finally opens into the collecting duct.
Therefore,both $PCT$ and $DCT$ are integral parts of a nephron.
It consists of two main parts: the Malpighian corpuscle and the renal tubule.
The renal tubule begins with a double-walled cup-like structure called Bowman's capsule,which further continues into a highly coiled network called the Proximal Convoluted Tubule $(PCT)$.
This is followed by the Henle's loop and then the Distal Convoluted Tubule $(DCT)$,which finally opens into the collecting duct.
Therefore,both $PCT$ and $DCT$ are integral parts of a nephron.
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42
BiologyMediumMCQAIPMT · 1990
Urine is acidic in nature as it contains:
A
$HCl$
B
$H_2SO_4$
C
$NaH_2PO_4$
D
$HNO_3$
Solution
(C) The human urine is typically acidic in nature,with a normal $pH$ range of approximately $4.5$ to $8.0$,often averaging around $6.0$.
This acidity is primarily due to the presence of acidic salts,specifically sodium acid phosphate $(NaH_2PO_4)$,which is excreted by the kidneys to help maintain the body's acid-base balance.
Therefore,the correct option is $C$.
This acidity is primarily due to the presence of acidic salts,specifically sodium acid phosphate $(NaH_2PO_4)$,which is excreted by the kidneys to help maintain the body's acid-base balance.
Therefore,the correct option is $C$.
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43
BiologyEasyMCQAIPMT · 1990
Bone marrow is largely composed of
A
Periosteum and osteoblast
B
Adipose tissue and blood vessels
C
Yellow and elastic tissue
D
Cartilage and elastic tissue
Solution
(B) Bone marrow is the soft,spongy tissue found in the hollow cavities of bones.
It is primarily composed of adipose tissue (fat cells),blood vessels,and hematopoietic stem cells.
In adults,red bone marrow is found in flat bones and the ends of long bones,while yellow bone marrow,which is largely composed of adipose tissue,is found in the shafts of long bones.
It is primarily composed of adipose tissue (fat cells),blood vessels,and hematopoietic stem cells.
In adults,red bone marrow is found in flat bones and the ends of long bones,while yellow bone marrow,which is largely composed of adipose tissue,is found in the shafts of long bones.
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44
BiologyEasyMCQAIPMT · 1990
Typically,all mammals have seven cervical vertebrae,except in:
A
Elephant
B
Man
C
Kangaroo
D
Sea cow
Solution
(D) The correct answer is $D$.
Typically,all mammals possess seven cervical vertebrae.
However,in certain aquatic mammals like the sea cow (manatee),the cervical vertebrae are fused into a solid bony mass due to the reduction of the neck.
In some species of sea cows,there are only six cervical vertebrae,and the neural arches are often incomplete.
Typically,all mammals possess seven cervical vertebrae.
However,in certain aquatic mammals like the sea cow (manatee),the cervical vertebrae are fused into a solid bony mass due to the reduction of the neck.
In some species of sea cows,there are only six cervical vertebrae,and the neural arches are often incomplete.
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45
BiologyMediumMCQAIPMT · 1990
Which one of the following is essential for the formation of the myelin sheath?
A
Zinc
B
Sodium
C
Iron
D
Phosphorus
Solution
(D) The myelin sheath is a lipid-rich insulating layer that surrounds the axons of neurons.
Phosphorus is a critical component of phospholipids,which are the primary structural lipids forming the myelin sheath.
Therefore,phosphorus is essential for the synthesis and maintenance of the myelin sheath.
Phosphorus is a critical component of phospholipids,which are the primary structural lipids forming the myelin sheath.
Therefore,phosphorus is essential for the synthesis and maintenance of the myelin sheath.
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46
BiologyEasyMCQAIPMT · 1990
The third ventricle of the rabbit's brain is called:
A
Rhinocoel
B
Rhombocoel
C
Diocoel
D
None of these
Solution
(C) The third ventricle of the brain is known as the $Diocoel$.
It is a small,narrow,slit-like cavity located within the $Diencephalon$ of the forebrain.
$Rhinocoel$ refers to the cavities in the olfactory lobes,while $Rhombocoel$ (or the fourth ventricle) is the cavity of the medulla oblongata.
It is a small,narrow,slit-like cavity located within the $Diencephalon$ of the forebrain.
$Rhinocoel$ refers to the cavities in the olfactory lobes,while $Rhombocoel$ (or the fourth ventricle) is the cavity of the medulla oblongata.
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47
BiologyMediumMCQAIPMT · 1990
Which of the following is a function of the parasympathetic nervous system in mammals?
A
Acceleration of heart beat
B
Constriction of pupil
C
Stimulation of sweat glands
D
Contraction of arrector pili muscles
Solution
(B) The parasympathetic nervous system is responsible for 'rest and digest' activities.
$1$. It promotes the constriction of the pupil (miosis).
$2$. It decreases the heart rate,whereas the sympathetic nervous system increases it.
$3$. Stimulation of sweat glands and contraction of arrector pili muscles are functions of the sympathetic nervous system.
Therefore,the correct function of the parasympathetic nervous system among the given options is the constriction of the pupil.
$1$. It promotes the constriction of the pupil (miosis).
$2$. It decreases the heart rate,whereas the sympathetic nervous system increases it.
$3$. Stimulation of sweat glands and contraction of arrector pili muscles are functions of the sympathetic nervous system.
Therefore,the correct function of the parasympathetic nervous system among the given options is the constriction of the pupil.
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48
BiologyMediumMCQAIPMT · 1990
Cholecystokinin and secretin are:
A
Hormones liberated by the mucosa of the duodenum and stimulate the gall bladder and pancreas respectively.
B
Hormones stimulating the liver.
C
Hormones stimulating the pancreas.
D
Enzymes.
Solution
(A) The gastrointestinal tract secretes several hormones that regulate digestive processes.
$1$. Cholecystokinin $(CCK)$ is secreted by the duodenal mucosa and acts on the gall bladder to stimulate the contraction and release of bile.
$2$. Secretin is also secreted by the duodenal mucosa and acts on the exocrine pancreas to stimulate the secretion of water and bicarbonate ions.
Therefore,these are hormones liberated by the mucosa of the duodenum that stimulate the gall bladder and pancreas respectively.
$1$. Cholecystokinin $(CCK)$ is secreted by the duodenal mucosa and acts on the gall bladder to stimulate the contraction and release of bile.
$2$. Secretin is also secreted by the duodenal mucosa and acts on the exocrine pancreas to stimulate the secretion of water and bicarbonate ions.
Therefore,these are hormones liberated by the mucosa of the duodenum that stimulate the gall bladder and pancreas respectively.
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49
BiologyEasyMCQAIPMT · 1990
True coelom develops as a split in
A
Mesoderm
B
Endoderm
C
Ectoderm
D
Between ectoderm and endoderm
Solution
(A) true coelom is a body cavity that is completely lined by mesoderm. In schizocoelous development,the coelom is formed by the splitting of the mesodermal layer. Therefore,the true coelom develops as a split in the mesoderm.
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50
BiologyMediumMCQAIPMT · 1990
Biological marriage of one of the following should be avoided:
A
$Rh^+$ male and $Rh^-$ female
B
$Rh^+$ male and $Rh^+$ female
C
$Rh^-$ male and $Rh^+$ female
D
$Rh^-$ male and $Rh^-$ female
Solution
(A) The marriage between an $Rh^+$ male and an $Rh^-$ female should be avoided because it can lead to Erythroblastosis fetalis in the offspring.
If the mother is $Rh^-$ and the fetus is $Rh^+$,the mother's immune system may develop antibodies against the $Rh$ antigen during the first pregnancy.
In subsequent pregnancies,these maternal antibodies can cross the placenta and destroy the fetal red blood cells,causing severe anemia and jaundice in the newborn.
If the mother is $Rh^-$ and the fetus is $Rh^+$,the mother's immune system may develop antibodies against the $Rh$ antigen during the first pregnancy.
In subsequent pregnancies,these maternal antibodies can cross the placenta and destroy the fetal red blood cells,causing severe anemia and jaundice in the newborn.
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51
BiologyEasyMCQAIPMT · 1990
$Archaeopteryx$,called a connecting link,carried the characters of:
A
Reptile and bird
B
Reptile and mammal
C
Fish and amphibian
D
Amphibian and reptile
Solution
(A) $Archaeopteryx$ is considered a connecting link between reptiles and birds.
It exhibits reptilian characteristics such as teeth in jaws,a long tail,and claws on digits.
It also exhibits avian characteristics such as the presence of feathers and wings.
Therefore,the presence of both reptilian and avian traits indicates that birds have evolved from reptiles.
It exhibits reptilian characteristics such as teeth in jaws,a long tail,and claws on digits.
It also exhibits avian characteristics such as the presence of feathers and wings.
Therefore,the presence of both reptilian and avian traits indicates that birds have evolved from reptiles.
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52
BiologyEasyMCQAIPMT · 1990
Penguins are found in:
A
Africa
B
Australia
C
America
D
Antarctica
Solution
(D) Penguins are flightless birds that are primarily found in the Southern Hemisphere,particularly in the Antarctic region. While some species inhabit islands near South Africa,South America,and Australia,they are most characteristically associated with the cold climate of Antarctica.
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53
BiologyMediumMCQAIPMT · 1990
New banana plants develop from
A
Rhizome
B
Stolon
C
Seed
D
All of these
Solution
(A) New banana plants develop from a specialized underground stem modification known as a rhizome.
Specifically,the banana plant reproduces vegetatively through a type of rhizome that grows vertically,often referred to as a rootstock or sucker.
Therefore,the correct option is $A$.
Specifically,the banana plant reproduces vegetatively through a type of rhizome that grows vertically,often referred to as a rootstock or sucker.
Therefore,the correct option is $A$.
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54
BiologyEasyMCQAIPMT · 1990
Tegmen develops from
A
Outer integument
B
Inner integument
C
Chalaza
D
Funicle
Solution
(B) The seed coat is formed by the integuments of the ovule.
After fertilization,the ovule matures into a seed.
The outer integument hardens to form the $Testa$,which is the outer seed coat.
The inner integument becomes thin and papery to form the $Tegmen$,which is the inner seed coat.
Therefore,the correct option is $B$.
After fertilization,the ovule matures into a seed.
The outer integument hardens to form the $Testa$,which is the outer seed coat.
The inner integument becomes thin and papery to form the $Tegmen$,which is the inner seed coat.
Therefore,the correct option is $B$.
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55
BiologyMediumMCQAIPMT · 1990
For a successful graft,the adhesion between stock and scion is a must. Which one of the following is the earliest event towards a good graft?
A
Production of plasmodesmata in the cells at the interface of stock and scion
B
Coordinated differentiation of vascular tissue between the stock and scion
C
Regeneration of cortex and epidermis over the union of stock and scion
D
Production of callus tissue between the cells of stock and scion
Solution
(D) The process of grafting involves joining two plant parts,the stock and the scion. The first step in the union of these two parts is the proliferation of parenchyma cells from both the stock and the scion at the cut surface. This mass of undifferentiated,proliferating cells is known as callus. The formation of callus is the earliest event that bridges the gap between the stock and the scion,eventually leading to vascular connection and successful grafting. Therefore,option $D$ is correct.
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56
BiologyMediumMCQAIPMT · 1990
In anther culture,the androgenic haploid plants are obtained from
A
Young pollen grain
B
Connective tissue
C
Anther tapetum
D
Anther wall
Solution
(A) Anther culture is a technique used to produce haploid plants.
In this process,the immature or young pollen grains (microspores) are cultured on a nutrient medium.
These microspores undergo embryogenesis to develop into haploid plants,which are referred to as androgenic haploid plants because they originate from the male gametophyte (pollen).
In this process,the immature or young pollen grains (microspores) are cultured on a nutrient medium.
These microspores undergo embryogenesis to develop into haploid plants,which are referred to as androgenic haploid plants because they originate from the male gametophyte (pollen).
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57
BiologyMediumMCQAIPMT · 1990
In monocots,the male gametophyte is represented by:
A
Microspore
B
Megaspore
C
Tetrad
D
Nucellus
Solution
(A) In flowering plants,including monocots,the pollen grain represents the male gametophyte. The pollen grain develops from the microspore. Therefore,the microspore is the initial stage of the male gametophyte.
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58
BiologyMediumMCQAIPMT · 1990
The embryo sac is:
A
Megasporangium
B
Megaspore
C
Female gametophyte
D
Female gamete
Solution
(C) The embryo sac is the female gametophyte in angiosperms.
It develops from a functional megaspore through the process of megagametogenesis.
Therefore,the correct option is $C$.
It develops from a functional megaspore through the process of megagametogenesis.
Therefore,the correct option is $C$.
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59
BiologyMediumMCQAIPMT · 1990
Tegmen develops from
A
Outer integument
B
Inner integument
C
Chalaza
D
Funicle
Solution
(B) In the development of a seed,the ovule transforms into the seed. The integuments of the ovule form the seed coat. The outer integument develops into the $Testa$ (outer seed coat),while the inner integument develops into the $Tegmen$ (inner seed coat).
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60
BiologyMediumMCQAIPMT · 1990
The nuclei of the sperm and egg fuse as a result of
A
Base pair relation of $DNA$ and $RNA$
B
Formation of hydrogen bonds
C
Mutual attraction caused by differences in electrical charges
D
Attraction of protoplasts of egg and sperm
Solution
(D) During fertilization,the male gamete (sperm) and the female gamete (egg) come into contact.
The fusion of their nuclei is facilitated by the attraction between the protoplasts of the egg and the sperm.
This process allows the two haploid nuclei to merge,forming a diploid zygote.
The fusion of their nuclei is facilitated by the attraction between the protoplasts of the egg and the sperm.
This process allows the two haploid nuclei to merge,forming a diploid zygote.
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61
BiologyEasyMCQAIPMT · 1990
Cryptorchidism is the condition in man when
A
There are two testis in each scrotum
B
Testis do not descend into the scrotum
C
Testis enlarge in the scrotum
D
Testis degenerate in the scrotum
Solution
(B) $Cryptorchidism$ is a condition in which one or both of the testes fail to descend from the abdomen into the scrotum.
In mammals,the testes are located in the extra-abdominal scrotal sac to maintain a temperature $2-2.5^{\circ}C$ lower than the internal body temperature,which is necessary for spermatogenesis.
If the testes do not descend,the higher body temperature prevents normal sperm production,often leading to sterility.
In mammals,the testes are located in the extra-abdominal scrotal sac to maintain a temperature $2-2.5^{\circ}C$ lower than the internal body temperature,which is necessary for spermatogenesis.
If the testes do not descend,the higher body temperature prevents normal sperm production,often leading to sterility.
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62
BiologyEasyMCQAIPMT · 1990
The mammalian follicle was first described by
A
Von Baer
B
de Graaf
C
Robert Brown
D
Spallanzani
Solution
(B) The correct answer is $B$.
Regner de Graaf $(1641-1673)$ first described the ovarian follicles in mammals in $1671$.
Although he initially mistook them for eggs,these structures were later named Graafian follicles in his honor.
Regner de Graaf $(1641-1673)$ first described the ovarian follicles in mammals in $1671$.
Although he initially mistook them for eggs,these structures were later named Graafian follicles in his honor.
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63
BiologyEasyMCQAIPMT · 1990
Gonads are derived from which embryonic germ layer?
A
Mesoderm
B
Endoderm
C
Ectoderm
D
Mesoderm and endoderm
Solution
(A) The gonads (testes and ovaries) are derived from the embryonic mesoderm. Specifically,they develop from the genital ridges,which are thickenings of the intermediate mesoderm located along the dorsal body wall of the embryo.
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64
BiologyEasyMCQAIPMT · 1990
The concept that the organiser is essential for embryonic development was given by,or for the 'Theory of organiser',the Nobel Prize was given to:
A
$J$. Axelrod
B
$C$. Landsteiner
C
$H$. Spemann
D
$I$.$P$. Pavlov
Solution
(C) The concept of the 'organiser' in embryonic development was proposed by Hans Spemann and Hilde Mangold. Hans Spemann of Germany was awarded the Nobel Prize in Physiology or Medicine in $1935$ for his discovery of the 'organiser effect' in embryonic development,which explains how specific regions of the embryo direct the differentiation of surrounding tissues.
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65
BiologyMediumMCQAIPMT · 1990
Which of the following is a correct statement?
A
In blastulation,major presumptive and organ-forming areas are segregated into definite points of the blastoderm.
B
Blastulation establishes the three germinal layers.
C
Blastulation of frog is known as discoblastula.
D
The fluid-filled space in a blastula is known as the archenteron.
Solution
(A) The correct statement is $A$.
During blastulation,the cells of the blastoderm undergo rearrangement,and the major presumptive and organ-forming areas are segregated into definite regions.
Option $B$ is incorrect because the three germinal layers are established during gastrulation,not blastulation.
Option $C$ is incorrect because the blastula of a frog is called a coeloblastula,while discoblastula is found in organisms with discoidal cleavage (e.g.,birds).
Option $D$ is incorrect because the fluid-filled space in a blastula is called the blastocoel,whereas the archenteron is the primitive gut cavity formed during gastrulation.
During blastulation,the cells of the blastoderm undergo rearrangement,and the major presumptive and organ-forming areas are segregated into definite regions.
Option $B$ is incorrect because the three germinal layers are established during gastrulation,not blastulation.
Option $C$ is incorrect because the blastula of a frog is called a coeloblastula,while discoblastula is found in organisms with discoidal cleavage (e.g.,birds).
Option $D$ is incorrect because the fluid-filled space in a blastula is called the blastocoel,whereas the archenteron is the primitive gut cavity formed during gastrulation.
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66
BiologyEasyMCQAIPMT · 1990
$A$ phenomenon is termed parthenogenesis when
A
Artificial fertilization occurs
B
Egg is fertilized by a sperm
C
Egg undergoes cleavage without fertilization
D
Sperm dies before fertilization
Solution
(C) Parthenogenesis is a form of asexual reproduction in which an embryo develops from an unfertilized egg cell.
In this process,the egg undergoes cleavage and develops into a new organism without the involvement of fertilization by a sperm.
Therefore,the correct description is that the egg undergoes cleavage without fertilization.
In this process,the egg undergoes cleavage and develops into a new organism without the involvement of fertilization by a sperm.
Therefore,the correct description is that the egg undergoes cleavage without fertilization.
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67
BiologyMediumMCQAIPMT · 1990
If in a dihybrid cross, Mendel had used two such characters which were linked, he would have faced difficulty in explaining the results on the basis of his
A
Law of segregation
B
Law of multiple factor hypothesis
C
Law of independent assortment
D
Law of dominance
Solution
(C) The $Law$ of $Independent$ $Assortment$ states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.
Linkage is the phenomenon where genes located on the same chromosome tend to be inherited together, which violates the principle of independent assortment.
Therefore, if Mendel had used linked genes, he would have observed that the traits do not assort independently, making it difficult to explain his results using the $Law$ of $Independent$ $Assortment$.
Linkage is the phenomenon where genes located on the same chromosome tend to be inherited together, which violates the principle of independent assortment.
Therefore, if Mendel had used linked genes, he would have observed that the traits do not assort independently, making it difficult to explain his results using the $Law$ of $Independent$ $Assortment$.
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68
BiologyMediumMCQAIPMT · 1990
From a cross $Aa BB \times aa BB$,what genotypic ratio will be obtained in the $F_1$ generation?
A
$1 Aa BB : 1 aa BB$
B
$1 Aa BB : 3 aa BB$
C
$3 Aa BB : 1 aa BB$
D
All $Aa BB$ : No $aa BB$
Solution
(A) To determine the genotypic ratio of the cross $Aa BB \times aa BB$,we perform a Punnett square analysis:
$1$. Identify the gametes produced by each parent:
- Parent $1$ $(Aa BB)$ produces gametes: $AB$ and $aB$.
- Parent $2$ $(aa BB)$ produces gametes: $aB$ and $aB$.
$2$. Perform the cross:
- $(AB) \times (aB) = Aa BB$
- $(aB) \times (aB) = aa BB$
$3$. Resulting genotypes in the $F_1$ generation are $Aa BB$ and $aa BB$ in a ratio of $1:1$.
Therefore,the correct genotypic ratio is $1 Aa BB : 1 aa BB$.
$1$. Identify the gametes produced by each parent:
- Parent $1$ $(Aa BB)$ produces gametes: $AB$ and $aB$.
- Parent $2$ $(aa BB)$ produces gametes: $aB$ and $aB$.
$2$. Perform the cross:
- $(AB) \times (aB) = Aa BB$
- $(aB) \times (aB) = aa BB$
$3$. Resulting genotypes in the $F_1$ generation are $Aa BB$ and $aa BB$ in a ratio of $1:1$.
Therefore,the correct genotypic ratio is $1 Aa BB : 1 aa BB$.
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69
BiologyMediumMCQAIPMT · 1990
Inheritance of $ABO$ blood group system is an example of
A
Multiple allelism
B
Partial dominance
C
Epistasis
D
Dominance
Solution
(A) The $ABO$ blood group system in humans is controlled by the gene $I$. This gene has three alleles: $I^A$,$I^B$,and $i$.
Since there are more than two alleles present for a single gene locus within a population,this phenomenon is known as multiple allelism.
Different combinations of these three alleles result in four distinct phenotypes: blood types $A, B, AB$,and $O$.
Since there are more than two alleles present for a single gene locus within a population,this phenomenon is known as multiple allelism.
Different combinations of these three alleles result in four distinct phenotypes: blood types $A, B, AB$,and $O$.
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70
BiologyMediumMCQAIPMT · 1990
Which of the following is a genetic disease?
A
Phenylketonuria
B
Blindness
C
Cataract
D
Leprosy
Solution
(A) Phenylketonuria is a genetic disorder caused by a mutation in the gene that codes for the enzyme phenylalanine hydroxylase. It is inherited as an autosomal recessive trait. Blindness,cataract,and leprosy are generally not classified as genetic diseases in the context of Mendelian disorders.
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71
BiologyMediumMCQAIPMT · 1990
Sickle cell anaemia is due to
A
Hormones
B
Viruses
C
Genes
D
Bacteria
Solution
(C) Sickle-cell anaemia is a genetic disorder caused by a point mutation in the $HbA$ gene located on chromosome $11$.
This mutation leads to the substitution of glutamic acid with valine at the sixth position of the $\beta$-globin chain of haemoglobin.
Since it is caused by a change in the $DNA$ sequence,it is classified as a genetic disease.
This mutation leads to the substitution of glutamic acid with valine at the sixth position of the $\beta$-globin chain of haemoglobin.
Since it is caused by a change in the $DNA$ sequence,it is classified as a genetic disease.
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72
BiologyEasyMCQAIPMT · 1990
'Eugenics' pertains to
A
Improvement of mankind by improving his heredity
B
Preserving human sperms for artificial insemination
C
Study of human genetics
D
Controlling size of a human family
Solution
(A) Eugenics is a branch of science that focuses on the improvement of the human race by improving its genetic quality or heredity. It involves the application of laws of inheritance to improve the human population.
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73
BiologyEasyMCQAIPMT · 1990
The study of the improvement of the human race by providing ideal environmental conditions is known as:
A
Eugenics
B
Euphenics
C
Euthenics
D
None of these
Solution
(C) is the correct answer.
Euthenics is the study of the improvement of the human race by optimizing environmental conditions.
This includes providing better nutrition,unpolluted ecological conditions,and improved educational opportunities to enhance the quality of life and human potential.
Euthenics is the study of the improvement of the human race by optimizing environmental conditions.
This includes providing better nutrition,unpolluted ecological conditions,and improved educational opportunities to enhance the quality of life and human potential.
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74
BiologyMediumMCQAIPMT · 1990
$A$ husband and wife have normal vision,but the fathers of both of them were colour blind. What is the probability of their first daughter being colour blind (in $\%$)?
A
$25$
B
$50$
C
$75$
D
$0$
Solution
(D) Colour blindness is an $X$-linked recessive disorder.
Since the husband and wife have normal vision,their genotypes are:
Husband: $X^CY$ (where $C$ is normal vision,$c$ is colour blind).
Wife: $X^CX^c$ (she is a carrier because her father was colour blind,$X^cY$).
The cross is: $X^CY \times X^CX^c$.
The possible genotypes of their children are: $X^CX^C$ (normal daughter),$X^CX^c$ (carrier daughter),$X^CY$ (normal son),$X^cY$ (colour blind son).
For a daughter to be colour blind,she must inherit the $X^c$ allele from both parents $(X^cX^c)$.
Since the father has normal vision $(X^CY)$,he cannot pass the $X^c$ allele to his daughter.
Therefore,the probability of their daughter being colour blind is $0\%$.
Since the husband and wife have normal vision,their genotypes are:
Husband: $X^CY$ (where $C$ is normal vision,$c$ is colour blind).
Wife: $X^CX^c$ (she is a carrier because her father was colour blind,$X^cY$).
The cross is: $X^CY \times X^CX^c$.
The possible genotypes of their children are: $X^CX^C$ (normal daughter),$X^CX^c$ (carrier daughter),$X^CY$ (normal son),$X^cY$ (colour blind son).
For a daughter to be colour blind,she must inherit the $X^c$ allele from both parents $(X^cX^c)$.
Since the father has normal vision $(X^CY)$,he cannot pass the $X^c$ allele to his daughter.
Therefore,the probability of their daughter being colour blind is $0\%$.
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75
BiologyMediumMCQAIPMT · 1990
There are $64$ types of codons in the genetic code dictionary because:
A
There are $64$ types of $tRNA$ found in the cell.
B
There are $44$ meaningless and $20$ codons for amino acids.
C
There are $64$ amino acids for coding.
D
The genetic code is a triplet.
Solution
(D) The genetic code is a triplet,meaning each codon consists of a sequence of three nitrogenous bases. Since there are $4$ types of nitrogenous bases $(A, U, G, C)$ in $RNA$,the total number of possible combinations is $4^3 = 4 \times 4 \times 4 = 64$. Thus,there are $64$ possible codons.
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76
BiologyEasyMCQAIPMT · 1990
Which chemical compound caused the death of thousands of people in the Bhopal gas tragedy?
A
Carbon tetrachloride
B
Nitrous acid
C
Mustard gas
D
Methyl isocyanate
Solution
(D) The Bhopal gas tragedy occurred on the night of $2nd-3rd$ December $1984$ at the Union Carbide India Limited $(UCIL)$ pesticide plant in Bhopal,Madhya Pradesh.
This disaster was caused by the accidental release of $40$ tonnes of the toxic gas Methyl isocyanate $(CH_3NCO)$.
It is considered one of the world's worst industrial disasters,resulting in thousands of immediate deaths and long-term health complications for the survivors.
This disaster was caused by the accidental release of $40$ tonnes of the toxic gas Methyl isocyanate $(CH_3NCO)$.
It is considered one of the world's worst industrial disasters,resulting in thousands of immediate deaths and long-term health complications for the survivors.
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77
BiologyMediumMCQAIPMT · 1990
Ultraviolet radiation from sunlight causes the reaction that produces:
A
Carbon monoxide
B
Sulphur dioxide
C
Ozone
D
Fluorides
Solution
(C) Ultraviolet $(UV)$ radiation from sunlight plays a crucial role in the formation of ozone $(O_3)$ in the stratosphere.
When high-energy $UV$ radiation strikes oxygen molecules $(O_2)$,it causes them to split into two individual oxygen atoms $(O + O)$.
These highly reactive oxygen atoms then collide with other oxygen molecules $(O_2)$ to form ozone $(O_3)$ molecules $(O + O_2 \rightarrow O_3)$.
This process is essential for creating the ozone layer,which protects the Earth from harmful $UV$ radiation.
When high-energy $UV$ radiation strikes oxygen molecules $(O_2)$,it causes them to split into two individual oxygen atoms $(O + O)$.
These highly reactive oxygen atoms then collide with other oxygen molecules $(O_2)$ to form ozone $(O_3)$ molecules $(O + O_2 \rightarrow O_3)$.
This process is essential for creating the ozone layer,which protects the Earth from harmful $UV$ radiation.
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78
BiologyMediumMCQAIPMT · 1990
Deforestation may reduce the chances of
A
Rainfall
B
Frequent cyclones
C
Erosion of surface soil
D
Frequent landslides
Solution
(A) Deforestation refers to the large-scale removal of forest cover. Trees play a crucial role in the water cycle through the process of transpiration,where they release water vapor into the atmosphere. This water vapor contributes to cloud formation and subsequent precipitation. Therefore,the removal of forests disrupts the water cycle,leading to a significant reduction in rainfall.
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79
BiologyMediumMCQAIPMT · 1990
Polluted water can be purified by using
A
Micro-organisms
B
Algae
C
Pesticides
D
Fishes
Solution
(A) Micro-organisms,specifically aerobic bacteria,are used in the process of sewage treatment and water purification.
In the presence of oxygen,these micro-organisms oxidize complex organic pollutants present in the water into simpler inorganic forms,thereby reducing the biological oxygen demand $(BOD)$ and purifying the water.
In the presence of oxygen,these micro-organisms oxidize complex organic pollutants present in the water into simpler inorganic forms,thereby reducing the biological oxygen demand $(BOD)$ and purifying the water.
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80
BiologyEasyMCQAIPMT · 1990
Sewage water can be purified for recycling with the action of
A
Aquatic plants
B
Penicillin
C
Micro-organisms
D
Fishes
Solution
(C) Sewage water contains a high amount of organic matter.
Micro-organisms,such as bacteria and fungi,play a crucial role in the treatment of sewage.
These organisms decompose complex organic compounds present in the sewage into simpler substances through biological processes,thereby purifying the water for recycling.
Micro-organisms,such as bacteria and fungi,play a crucial role in the treatment of sewage.
These organisms decompose complex organic compounds present in the sewage into simpler substances through biological processes,thereby purifying the water for recycling.
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81
BiologyEasyMCQAIPMT · 1990
Binomial nomenclature was introduced by
A
Linnaeus
B
Darwin
C
Aristotle
D
De Candolle
Solution
(A) The system of binomial nomenclature was introduced by $Carl$ $Linnaeus$.
He proposed this system in his book $Species$ $Plantarum$ $(1753)$.
According to this system,each organism is given a scientific name consisting of two components: the genus and the specific epithet.
He proposed this system in his book $Species$ $Plantarum$ $(1753)$.
According to this system,each organism is given a scientific name consisting of two components: the genus and the specific epithet.
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82
BiologyEasyMCQAIPMT · 1990
The husk fibre,known as coir of commerce,is obtained from which part of the coconut $(Cocos\, nucifera)$?
A
Epicarp
B
Mesocarp
C
Endocarp
D
Seed coat
Solution
(B) The coconut fruit is a fibrous drupe.
In a drupe,the pericarp is differentiated into three layers: the outer epicarp,the middle mesocarp,and the inner endocarp.
The husk of the coconut,which provides the coir fibre,is derived from the fibrous mesocarp.
Therefore,the correct answer is the mesocarp.
In a drupe,the pericarp is differentiated into three layers: the outer epicarp,the middle mesocarp,and the inner endocarp.
The husk of the coconut,which provides the coir fibre,is derived from the fibrous mesocarp.
Therefore,the correct answer is the mesocarp.
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83
BiologyEasyMCQAIPMT · 1990
Cotton fibres are derived from
A
Pericarp
B
Phloem
C
Pericycle
D
Testa
Solution
(D) Cotton fibres are epidermal outgrowths of the seed coat,specifically known as the $Testa$. These are unicellular,elongated hairs that arise from the surface of the seed. Therefore,cotton fibres are derived from the $Testa$.
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84
BiologyEasyMCQAIPMT · 1990
In groundnut $(Arachis \, hypogaea)$,oil is stored in:
A
Endosperm
B
Cotyledons
C
Embryo
D
Tuber
Solution
(B) Groundnut $(Arachis \, hypogaea)$ is a dicotyledonous plant.
In dicot seeds,the endosperm is often consumed during the development of the embryo,and the food reserves are stored in the cotyledons.
Therefore,in groundnut seeds,the oil and other nutrients are stored in the cotyledons.
In dicot seeds,the endosperm is often consumed during the development of the embryo,and the food reserves are stored in the cotyledons.
Therefore,in groundnut seeds,the oil and other nutrients are stored in the cotyledons.
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85
BiologyMediumMCQAIPMT · 1990
In the case of an artificial pacemaker,the electrode is inserted into:
A
Right ventricle
B
Right auricle
C
Left ventricle
D
Left auricle
Solution
(A) An artificial pacemaker is a medical device used to regulate the heart's rhythm.
During the implantation procedure,the lead (electrode) is typically inserted through a vein and guided into the $Right$ $ventricle$ of the heart.
This position allows the electrode to directly stimulate the heart muscle to maintain a proper heartbeat,especially in patients with conditions like heart block or bradycardia.
During the implantation procedure,the lead (electrode) is typically inserted through a vein and guided into the $Right$ $ventricle$ of the heart.
This position allows the electrode to directly stimulate the heart muscle to maintain a proper heartbeat,especially in patients with conditions like heart block or bradycardia.
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86
BiologyEasyMCQAIPMT · 1990
What type of fuel are coal, petrol, and natural gas?
A
Biofuels
B
Electrical fuels
C
Fossil fuels
D
Liquid fuels
Solution
(C) Coal, petrol, and natural gas are formed from the remains of ancient plants and animals that were buried under the Earth's surface millions of years ago.
Due to high pressure and temperature over geological time, these organic materials were converted into energy-rich compounds.
Therefore, they are classified as $Fossil fuels$.
Due to high pressure and temperature over geological time, these organic materials were converted into energy-rich compounds.
Therefore, they are classified as $Fossil fuels$.
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87
BiologyEasyMCQAIPMT · 1990
The current consumption of domestic firewood in India is about:
A
$18.6$ million tonnes
B
$146.5$ million tonnes
C
$1246$ million tonnes
D
$21870$ million tonnes
Solution
(B) According to various environmental reports and forestry data regarding biomass energy in India,the annual domestic consumption of firewood is estimated to be approximately $146.5$ million tonnes. This reflects the significant reliance on traditional biomass for cooking and heating in rural and semi-urban households across the country.
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88
BiologyEasyMCQAIPMT · 1990
The application of the principles of genetics for the improvement of the human race is called ..............
A
Euthenics
B
Eugenics
C
Euphenics
D
Ethnology
Solution
(B) The application of the principles of genetics to improve the hereditary qualities of the human race is known as $Eugenics$.
$Euthenics$ refers to the improvement of human well-being by improving environmental conditions.
$Euphenics$ refers to the improvement of the human condition by modifying the expression of genes (phenotype) through medical or environmental interventions.
$Ethnology$ is the branch of anthropology that compares and analyzes the characteristics of different peoples and the relationships between them.
Therefore,the correct option is $B$.
$Euthenics$ refers to the improvement of human well-being by improving environmental conditions.
$Euphenics$ refers to the improvement of the human condition by modifying the expression of genes (phenotype) through medical or environmental interventions.
$Ethnology$ is the branch of anthropology that compares and analyzes the characteristics of different peoples and the relationships between them.
Therefore,the correct option is $B$.
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89
BiologyMediumMCQAIPMT · 1990
Amoebiasis can be prevented by .............
A
Eating a balanced diet
B
Eating plenty of fruits
C
Drinking boiled water
D
Using mosquito nets
Solution
(C) Amoebiasis (Amoebic dysentery) is caused by the protozoan parasite $Entamoeba \text{ } histolytica$.
This parasite is transmitted through the fecal-oral route, primarily by consuming contaminated food and water.
Therefore, the most effective way to prevent the spread of this disease is to maintain personal and public hygiene, such as drinking boiled water and ensuring food is properly covered and washed.
Using mosquito nets helps prevent diseases like malaria or dengue, not amoebiasis.
This parasite is transmitted through the fecal-oral route, primarily by consuming contaminated food and water.
Therefore, the most effective way to prevent the spread of this disease is to maintain personal and public hygiene, such as drinking boiled water and ensuring food is properly covered and washed.
Using mosquito nets helps prevent diseases like malaria or dengue, not amoebiasis.
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90
BiologyEasyMCQAIPMT · 1990
Kala-azar and Oriental sore are transmitted by ............ .
A
Housefly
B
Bedbug
C
Sandfly
D
Fruit fly
Solution
(C) Kala-azar (visceral leishmaniasis) and Oriental sore (cutaneous leishmaniasis) are caused by protozoan parasites of the genus $Leishmania$.
These diseases are transmitted to humans through the bite of an infected female phlebotomine sandfly,belonging to the genus $Phlebotomus$.
These diseases are transmitted to humans through the bite of an infected female phlebotomine sandfly,belonging to the genus $Phlebotomus$.
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91
BiologyEasyMCQAIPMT · 1990
Penguins are found in .............. .
A
Australia
B
Antarctica
C
Africa
D
America
Solution
(B) Penguins are a group of aquatic flightless birds. They live almost exclusively in the Southern Hemisphere,with the largest populations found in Antarctica. They are highly adapted for life in the water,possessing flippers for swimming and a thick layer of blubber for insulation against the extreme cold of the Antarctic environment.
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92
BiologyMediumMCQAIPMT · 1990
In hot summers and cold winters,malaria cases and $Anopheles$ mosquitoes are seen less frequently. They reappear in hot-humid conditions due to:
A
Survival of malaria parasites in human carriers
B
Presence of sporozoites in surviving mosquitoes
C
Monkeys
D
Mosquito larvae in stagnant water
Solution
(A) Malaria is caused by the protozoan parasite $Plasmodium$. During unfavorable conditions like extreme heat or cold,the adult mosquitoes may die or become inactive. However,the $Plasmodium$ parasite survives within the human host (the reservoir) during these periods. When hot-humid conditions return,the mosquito population increases,and they pick up the parasite from infected humans,leading to a resurgence of malaria cases. Thus,the survival of the parasite in human carriers is the primary reason for the reappearance of the disease.
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93
BiologyMediumMCQAIPMT · 1990
In groundnut, the source of oil is found in:
A
Embryo
B
Cotyledons
C
Endosperm
D
Nodule
Solution
(B) Groundnut $(Arachis \, hypogaea)$ is a dicotyledonous plant.
In dicot seeds, the food reserves are primarily stored in the cotyledons.
Since the endosperm is consumed during the development of the embryo, the mature seeds are non-endospermic (exalbuminous).
Therefore, the oil, which serves as a food reserve, is stored in the cotyledons of the groundnut seed.
In dicot seeds, the food reserves are primarily stored in the cotyledons.
Since the endosperm is consumed during the development of the embryo, the mature seeds are non-endospermic (exalbuminous).
Therefore, the oil, which serves as a food reserve, is stored in the cotyledons of the groundnut seed.
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94
BiologyMediumMCQAIPMT · 1990
$A$ new banana plant develops from $..........$.
A
Rhizome
B
Sucker
C
Stolon
D
Seed
Solution
(B) In banana plants,the main stem is underground and is known as a rhizome.
New banana plants develop from the lateral branches that arise from the underground portion of the main stem.
These lateral branches are technically known as suckers.
Therefore,the vegetative propagation in banana occurs through suckers.
New banana plants develop from the lateral branches that arise from the underground portion of the main stem.
These lateral branches are technically known as suckers.
Therefore,the vegetative propagation in banana occurs through suckers.
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95
BiologyMediumMCQAIPMT · 1990
In grafting,which of the following is formed first at the junction of the stock and scion?
A
Formation of callus
B
Formation of plasmodesmata
C
Differentiation of new vascular tissues
D
Regeneration of epidermis and cortex
Solution
(A) In the process of grafting,the first step after joining the stock and scion is the proliferation of parenchyma cells from both cut surfaces. This mass of undifferentiated cells is known as $Callus$. The formation of $Callus$ is essential to bridge the gap between the stock and the scion. Once the $Callus$ is formed,it facilitates the development of new vascular tissues and the reconnection of the cambium,eventually leading to a successful graft union.
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96
BiologyEasyMCQAIPMT · 1990
From which of the following does a new banana plant develop?
A
Stolon
B
Rhizome
C
Sucker
D
Seed
Solution
(C) new banana plant develops from a $Sucker$.
$Suckers$ are lateral branches that develop from the underground base of the main stem.
These grow horizontally beneath the soil and then emerge above the ground to form a new independent plant.
This is a common method of vegetative propagation in banana plants.
$Suckers$ are lateral branches that develop from the underground base of the main stem.
These grow horizontally beneath the soil and then emerge above the ground to form a new independent plant.
This is a common method of vegetative propagation in banana plants.
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97
BiologyMediumMCQAIPMT · 1990
The male gametophyte of an angiosperm is represented by the . . . . . . .
A
Microsporangium
B
Nucellus
C
Microspore
D
Stamen
Solution
(C) In angiosperms,the male gametophyte is highly reduced and is represented by the pollen grain.
$1$. The microspore (pollen grain) undergoes mitotic divisions to develop into the male gametophyte.
$2$. The microsporangium is the structure where microspores are produced.
$3$. The nucellus is the central part of the ovule.
$4$. The stamen is the male reproductive organ of the flower.
Therefore,the microspore (pollen grain) represents the male gametophyte.
$1$. The microspore (pollen grain) undergoes mitotic divisions to develop into the male gametophyte.
$2$. The microsporangium is the structure where microspores are produced.
$3$. The nucellus is the central part of the ovule.
$4$. The stamen is the male reproductive organ of the flower.
Therefore,the microspore (pollen grain) represents the male gametophyte.
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98
BiologyEasyMCQAIPMT · 1990
The entry of the pollen tube through the micropyle is known as:
A
Chalazogamy
B
Mesogamy
C
Porogamy
D
Pseudogamy
Solution
(C) When the pollen tube enters the ovule through the micropyle,it is called $Porogamy$.
$Chalazogamy$ refers to the entry of the pollen tube through the chalaza.
$Mesogamy$ refers to the entry of the pollen tube through the integuments.
$Pseudogamy$ is a type of apomixis where pollination is required for fruit development but fertilization does not occur.
$Chalazogamy$ refers to the entry of the pollen tube through the chalaza.
$Mesogamy$ refers to the entry of the pollen tube through the integuments.
$Pseudogamy$ is a type of apomixis where pollination is required for fruit development but fertilization does not occur.
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99
BiologyMediumMCQAIPMT · 1990
In angiosperms, the female gametophyte is known as the .........
A
Ovule
B
Megaspore mother cell
C
Embryo sac
D
Nucellus
Solution
(C) In angiosperms, the female gametophyte is represented by the $Embryo \text{ } sac$.
$1$. The $Megaspore \text{ } mother \text{ } cell$ undergoes meiosis to produce four megaspores.
$2$. Out of these, three degenerate and one functional megaspore develops into the $Embryo \text{ } sac$ through mitosis.
$3$. The $Embryo \text{ } sac$ typically consists of $8$ nuclei and $7$ cells, representing the mature female gametophyte.
$1$. The $Megaspore \text{ } mother \text{ } cell$ undergoes meiosis to produce four megaspores.
$2$. Out of these, three degenerate and one functional megaspore develops into the $Embryo \text{ } sac$ through mitosis.
$3$. The $Embryo \text{ } sac$ typically consists of $8$ nuclei and $7$ cells, representing the mature female gametophyte.
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100
BiologyMediumMCQAIPMT · 1990
How many spermatozoa are produced from a single secondary spermatocyte?
A
$4$
B
$8$
C
$2$
D
$1$
Solution
(C) During spermatogenesis,the primary spermatocyte undergoes meiosis-$I$ to form two haploid secondary spermatocytes.
Each secondary spermatocyte then undergoes meiosis-$II$ to produce two spermatids.
These spermatids eventually differentiate into spermatozoa.
Therefore,one secondary spermatocyte produces $2$ spermatozoa.
Each secondary spermatocyte then undergoes meiosis-$II$ to produce two spermatids.
These spermatids eventually differentiate into spermatozoa.
Therefore,one secondary spermatocyte produces $2$ spermatozoa.
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