BiologyQ1–100 of 170 questions
Page 1 of 2 · English
1
BiologyMediumMCQAIPMT · 2002
In the five-kingdom system,the main basis of classification is
A
Nutrition
B
Nucleus structure
C
Cell wall structure
D
Asexual reproduction
Solution
(A) The five-kingdom classification system was proposed by $R.H. Whittaker$ in $1969$. The main criteria used by him for classification include cell structure,body organization,mode of nutrition,reproduction,and phylogenetic relationships. Among these,the mode of nutrition is considered a primary basis for distinguishing organisms across different kingdoms.
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2
BiologyMediumMCQAIPMT · 2002
$A$ true species consists of a population that is:
A
Sharing the same niche
B
Interbreeding
C
Feeding over the same food
D
Reproductively isolated from other groups
Solution
(B) According to the biological species concept,a species is defined as a group of individuals that can interbreed among themselves to produce fertile offspring.
Members of the same species share a common gene pool and are reproductively isolated from other such groups.
Therefore,the most fundamental characteristic that defines a species is the ability to interbreed.
Members of the same species share a common gene pool and are reproductively isolated from other such groups.
Therefore,the most fundamental characteristic that defines a species is the ability to interbreed.
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3
BiologyEasyMCQAIPMT · 2002
$A$ mature ligule,having a prominent basal portion,is called
A
Trichocyst
B
Heterocyst
C
Rhizophore
D
Glossopodium
Solution
(D) The correct answer is $(d)$.
At the base of the ligule,there is a sheath of elongated cells known as the glossopodium,which is secretory in nature.
The ligule serves both secretory and protective functions.
It secretes water or mucilage to keep the growing point of the stem and young leaves moist.
Additionally,it protects the young leaves during their development.
At the base of the ligule,there is a sheath of elongated cells known as the glossopodium,which is secretory in nature.
The ligule serves both secretory and protective functions.
It secretes water or mucilage to keep the growing point of the stem and young leaves moist.
Additionally,it protects the young leaves during their development.
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4
BiologyEasyMCQAIPMT · 2002
Which of the following animals has a nervous system but no brain?
A
Pheretima
B
Hydra
C
Amoeba
D
Periplaneta
Solution
(B) $Hydra$ possesses a very primitive nervous system consisting of a nerve net of bipolar and multipolar nerve cells.
Unlike higher animals,it lacks a centralized nervous system or a brain.
Therefore,$Hydra$ has a nervous system but no brain.
Unlike higher animals,it lacks a centralized nervous system or a brain.
Therefore,$Hydra$ has a nervous system but no brain.
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5
BiologyMediumMCQAIPMT · 2002
Which one of the following animals possesses nerve cells but no nerves?
A
Hydra
B
Tapeworm
C
Earthworm
D
Frog's tadpole
Solution
(A) $Hydra$ belongs to the phylum $Cnidaria$ (or $Coelenterata$).
In $Hydra$,the nervous system consists of a network of nerve cells (neurons) distributed throughout the body,but these cells are not organized into distinct nerves or a central nervous system.
This is known as a diffuse or nerve net type of nervous system.
In contrast,organisms like $Tapeworms$,$Earthworms$,and $Frog's$ $tadpoles$ possess more complex nervous systems that include organized nerves and ganglia.
In $Hydra$,the nervous system consists of a network of nerve cells (neurons) distributed throughout the body,but these cells are not organized into distinct nerves or a central nervous system.
This is known as a diffuse or nerve net type of nervous system.
In contrast,organisms like $Tapeworms$,$Earthworms$,and $Frog's$ $tadpoles$ possess more complex nervous systems that include organized nerves and ganglia.
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6
BiologyEasyMCQAIPMT · 2002
Starfish belongs to which class?
A
Pisces
B
Cephalopoda
C
Asteroidea
D
Ophiuroidea
Solution
(C) Starfish belongs to the class $Asteroidea$ under the phylum $Echinodermata$.
These organisms are characterized by a star-shaped body and a water vascular system.
These organisms are characterized by a star-shaped body and a water vascular system.
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7
BiologyMediumMCQAIPMT · 2002
Which of the following is a characteristic feature of fishes?
A
Tail and venous heart
B
Venous heart and gills
C
Epidermal scales and tail
D
Epidermal scales and gills
Solution
(B) The correct answer is $B$. Fishes possess a $2$-chambered heart consisting of one auricle and one ventricle. This heart is known as a venous heart because it receives only deoxygenated (venous) blood and pumps it to the gills for oxygenation (purification).
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8
BiologyEasyMCQAIPMT · 2002
Flightless birds belong to:
A
Ratites
B
Neornithes
C
Archaeornithes
D
None of these
Solution
(A) Flightless birds are classified under the group known as $Ratites$. These birds belong to the superorder $Palaeognathae$. Unlike most birds,they lack a keel on their sternum,which is necessary for the attachment of flight muscles,thus rendering them flightless.
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9
BiologyMediumMCQAIPMT · 2002
Reptiles share which of the following characteristics with birds and mammals?
A
Amnion
B
Diaphragm
C
Homeothermy
D
All of these
Solution
(A) Reptiles,birds,and mammals are collectively known as amniotes because they possess an extra-embryonic membrane called the $Amnion$ during their development.
$Pisces$ and $Amphibia$ are anamniotic,meaning they lack this membrane.
$Diaphragm$ is a characteristic feature unique to mammals.
$Homeothermy$ (warm-bloodedness) is a characteristic feature of birds and mammals,but not reptiles.
$Pisces$ and $Amphibia$ are anamniotic,meaning they lack this membrane.
$Diaphragm$ is a characteristic feature unique to mammals.
$Homeothermy$ (warm-bloodedness) is a characteristic feature of birds and mammals,but not reptiles.
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10
BiologyMediumMCQAIPMT · 2002
Axillary bud and terminal bud are derived from the activity of
A
Parenchyma
B
Lateral meristem
C
Apical meristem
D
Intercalary meristem
Solution
(C) The $Apical \text{ meristem}$ is located at the growing tips of stems and roots.
During the formation of leaves and the elongation of the stem, some cells are left behind from the shoot apical meristem.
These cells constitute the $Axillary \text{ bud}$, which is present in the axils of leaves and is capable of forming a branch or a flower.
The $Terminal \text{ bud}$ (or shoot apex) is the primary meristematic region responsible for the longitudinal growth of the plant.
Therefore, both axillary and terminal buds are derived from the activity of the $Apical \text{ meristem}$.
During the formation of leaves and the elongation of the stem, some cells are left behind from the shoot apical meristem.
These cells constitute the $Axillary \text{ bud}$, which is present in the axils of leaves and is capable of forming a branch or a flower.
The $Terminal \text{ bud}$ (or shoot apex) is the primary meristematic region responsible for the longitudinal growth of the plant.
Therefore, both axillary and terminal buds are derived from the activity of the $Apical \text{ meristem}$.
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11
BiologyMediumMCQAIPMT · 2002
Vessels are found in
A
All pteridophytes
B
All angiosperms
C
Some gymnosperms
D
Both $(b)$ and $(c)$
Solution
(D) Vessels are a characteristic feature of the xylem in angiosperms. While most gymnosperms lack vessels,members of the order Gnetales (such as $Gnetum$,$Ephedra$,and $Welwitschia$) possess vessels in their xylem. Pteridophytes generally lack vessels. Therefore,vessels are found in all angiosperms and some gymnosperms.
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12
BiologyMediumMCQAIPMT · 2002
If four radial vascular bundles are present,then the structure will be
A
Monocot stem
B
Monocot root
C
Dicot stem
D
Dicot root
Solution
(D) In plants,radial vascular bundles are a characteristic feature of roots.
In a dicot root,the number of vascular bundles (xylem and phloem patches) is typically $2$ to $4$.
When there are $4$ radial vascular bundles,the condition is known as tetrarch.
Monocot roots typically have more than $6$ vascular bundles,which is known as polyarch.
Stems possess conjoint vascular bundles,not radial ones.
Therefore,a structure with four radial vascular bundles is a dicot root.
In a dicot root,the number of vascular bundles (xylem and phloem patches) is typically $2$ to $4$.
When there are $4$ radial vascular bundles,the condition is known as tetrarch.
Monocot roots typically have more than $6$ vascular bundles,which is known as polyarch.
Stems possess conjoint vascular bundles,not radial ones.
Therefore,a structure with four radial vascular bundles is a dicot root.
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13
BiologyMediumMCQAIPMT · 2002
Collagen is
A
Lipid
B
Carbohydrate
C
Globular protein
D
Fibrous protein
Solution
(D) $Collagen$ is the most abundant protein in the animal kingdom. It is a structural protein that forms the primary component of connective tissues. Structurally,it is classified as a fibrous protein because its polypeptide chains are arranged in long,parallel strands that form fibers,providing tensile strength to tissues. Unlike globular proteins,which are soluble in water,fibrous proteins like $Collagen$ are generally insoluble.
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14
BiologyEasyMCQAIPMT · 2002
Which of the following cartilage is present on the ends of long bones?
A
Elastic cartilage
B
Fibrous cartilage
C
Calcified cartilage
D
Hyaline cartilage
Solution
(D) . Hyaline cartilage is the most abundant type of cartilage in the body. It forms the articular surfaces at the joints of long bones,where it is specifically referred to as articular cartilage. This smooth surface reduces friction and absorbs shock during movement.
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15
BiologyMediumMCQAIPMT · 2002
Middle lamella is made up of
A
Cellulose
B
Suberin
C
Calcium and magnesium pectate
D
Lignin
Solution
(C) The middle lamella is a thin,cementing layer that acts as a glue to hold adjacent plant cells together.
It is primarily composed of calcium and magnesium pectates.
These substances provide structural stability and adhesion between the cell walls of neighboring cells.
It is primarily composed of calcium and magnesium pectates.
These substances provide structural stability and adhesion between the cell walls of neighboring cells.
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16
BiologyMediumMCQAIPMT · 2002
The plasma membrane is composed of:
A
$A$ protein layer between two lipid layers
B
$A$ lipid layer between two protein layers
C
$A$ protein,a lipid,and a cellulose layer
D
$A$ bimolecular lipid layer surrounded by protein layers
Solution
(D) According to the Davson-Danielli model (also known as the sandwich model),the plasma membrane consists of a bimolecular lipid layer (phospholipid bilayer) that is sandwiched between two layers of globular proteins. This structure is often described as a protein-lipid-protein sandwich.
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17
BiologyMediumMCQAIPMT · 2002
Which is the latest model that is proposed to explain the structure of the plasma membrane?
A
Fluid mosaic model
B
Molecular model
C
Unit membrane model
D
None of the above
Solution
(A) The $Fluid \, mosaic \, model$ is the most widely accepted model for the structure of the plasma membrane.
It was proposed by $S.J. \, Singer$ and $G.L. \, Nicolson$ in $1972$.
According to this model, the plasma membrane is composed of a phospholipid bilayer with proteins embedded within it, which can move laterally, giving it a fluid nature.
It was proposed by $S.J. \, Singer$ and $G.L. \, Nicolson$ in $1972$.
According to this model, the plasma membrane is composed of a phospholipid bilayer with proteins embedded within it, which can move laterally, giving it a fluid nature.
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18
BiologyMediumMCQAIPMT · 2002
Lipids are insoluble in water because lipid molecules are:
A
Neutral
B
Zwitter ions
C
Hydrophobic
D
Hydrophilic
Solution
(C) Lipids are organic compounds that are non-polar in nature.
Water is a polar solvent.
According to the principle of 'like dissolves like',polar substances dissolve in polar solvents,while non-polar substances do not.
Because lipid molecules are hydrophobic (water-fearing),they do not interact with water molecules and are therefore insoluble in water.
Water is a polar solvent.
According to the principle of 'like dissolves like',polar substances dissolve in polar solvents,while non-polar substances do not.
Because lipid molecules are hydrophobic (water-fearing),they do not interact with water molecules and are therefore insoluble in water.
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19
BiologyMediumMCQAIPMT · 2002
The process of mitosis can be studied in
A
Onion root tip
B
Garlic root tip
C
Tendril tip
D
All of the above
Solution
(D) Mitosis is a type of cell division that occurs in meristematic tissues where active cell growth and division take place.
Root tips and shoot tips are classic examples of meristematic regions in plants.
Onion root tips,garlic root tips,and tendril tips all contain actively dividing meristematic cells,making them ideal specimens for observing the various stages of mitosis under a microscope.
Therefore,all the given options are correct.
Root tips and shoot tips are classic examples of meristematic regions in plants.
Onion root tips,garlic root tips,and tendril tips all contain actively dividing meristematic cells,making them ideal specimens for observing the various stages of mitosis under a microscope.
Therefore,all the given options are correct.
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20
BiologyMediumMCQAIPMT · 2002
When a potato piece is kept in a highly concentrated salt solution,then
A
Potato is plasmolysed
B
Potato is deplasmolysed
C
Potato cells get bursted
D
There is no effect due to isotonic solution
Solution
(A) When a potato piece is placed in a highly concentrated salt solution (hypertonic solution),the water potential of the surrounding solution is lower than that of the potato cells.
This causes water to move out of the potato cells into the surrounding solution through the process of exosmosis.
As water leaves the cells,the protoplast shrinks away from the cell wall,a phenomenon known as plasmolysis.
Therefore,the potato piece becomes plasmolysed.
This causes water to move out of the potato cells into the surrounding solution through the process of exosmosis.
As water leaves the cells,the protoplast shrinks away from the cell wall,a phenomenon known as plasmolysis.
Therefore,the potato piece becomes plasmolysed.
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21
BiologyMediumMCQAIPMT · 2002
Stomata in angiosperms open and close due to
A
Their genetic constitution
B
Effect of hormones
C
Changes of turgor pressure in guard cells
D
Pressure of gases inside the leaves
Solution
(C) The opening and closing of stomata in angiosperms are primarily regulated by changes in the turgor pressure of the guard cells.
When the guard cells become turgid due to the influx of water,the stomatal aperture opens.
Conversely,when the guard cells lose water and become flaccid,the stomatal aperture closes.
When the guard cells become turgid due to the influx of water,the stomatal aperture opens.
Conversely,when the guard cells lose water and become flaccid,the stomatal aperture closes.
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22
BiologyMediumMCQAIPMT · 2002
In photosynthesis,energy from the light reaction to the dark reaction is transferred in the form of:
A
$ADP$
B
$ATP$
C
$RuDP$
D
Chlorophyll
Solution
(B) During the light reaction of photosynthesis,photophosphorylation occurs,which results in the production of $ATP$ and $NADPH_2$.
These molecules act as energy carriers.
$ATP$ and $NADPH_2$ are transported from the thylakoid membranes (where light reactions occur) to the stroma (where dark reactions occur).
Together,$ATP$ and $NADPH_2$ are referred to as assimilatory power,which provides the necessary energy and reducing power to fix $CO_2$ into carbohydrates during the dark reaction (Calvin cycle).
These molecules act as energy carriers.
$ATP$ and $NADPH_2$ are transported from the thylakoid membranes (where light reactions occur) to the stroma (where dark reactions occur).
Together,$ATP$ and $NADPH_2$ are referred to as assimilatory power,which provides the necessary energy and reducing power to fix $CO_2$ into carbohydrates during the dark reaction (Calvin cycle).
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23
BiologyMediumMCQAIPMT · 2002
The role of chlorophyll in photosynthesis is
A
Absorption of $CO_2$
B
Absorption of light
C
Absorption of light and photochemical decomposition of water
D
Absorption of water
Solution
(C) Chlorophyll is the primary photosynthetic pigment that captures light energy.
This absorbed light energy is used to excite electrons,which initiates the process of photosynthesis.
The energy is further utilized for the photolysis (photochemical decomposition) of water molecules into protons,electrons,and oxygen $(2H_2O \rightarrow 4H^+ + 4e^- + O_2)$.
Therefore,chlorophyll plays a dual role in light absorption and the subsequent splitting of water.
This absorbed light energy is used to excite electrons,which initiates the process of photosynthesis.
The energy is further utilized for the photolysis (photochemical decomposition) of water molecules into protons,electrons,and oxygen $(2H_2O \rightarrow 4H^+ + 4e^- + O_2)$.
Therefore,chlorophyll plays a dual role in light absorption and the subsequent splitting of water.
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24
BiologyMediumMCQAIPMT · 2002
In germinating castor seeds,the $R.Q.$ is
A
One
B
More than one
C
Less than one
D
Zero
Solution
(C) The $R.Q.$ (Respiratory Quotient) is defined as the ratio of the volume of $CO_2$ evolved to the volume of $O_2$ consumed during respiration.
In germinating castor seeds,the primary respiratory substrate is fat (lipids).
Since fats are highly reduced compounds,they require more oxygen for complete oxidation compared to carbohydrates.
For fats,the $R.Q.$ is typically around $0.7$,which is less than one.
In germinating castor seeds,the primary respiratory substrate is fat (lipids).
Since fats are highly reduced compounds,they require more oxygen for complete oxidation compared to carbohydrates.
For fats,the $R.Q.$ is typically around $0.7$,which is less than one.
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25
BiologyEasyMCQAIPMT · 2002
The pigment involved in red-far red light interconversion is
A
Cytochrome
B
Xanthophyll
C
Lycopene
D
Phytochrome
Solution
(D) $Phytochrome$ is a proteinaceous pigment that exists in two interconvertible forms: $Pr$ (which absorbs red light at $660 \ nm$) and $Pfr$ (which absorbs far-red light at $730 \ nm$).
This interconversion between $Pr$ and $Pfr$ forms acts as a biological switch,allowing plants to sense light quality and regulate various developmental processes such as flowering and seed germination.
This interconversion between $Pr$ and $Pfr$ forms acts as a biological switch,allowing plants to sense light quality and regulate various developmental processes such as flowering and seed germination.
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26
BiologyEasyMCQAIPMT · 2002
To get sufficient carbohydrates,one should take:
A
Meat
B
Rice
C
Carrots
D
Ground nuts
Solution
(B) The correct answer is $B$.
Carbohydrates are the primary source of energy for most animals.
They are organic compounds composed of carbon,hydrogen,and oxygen.
Rich sources of carbohydrates include cereals (like rice,wheat,and maize),fruits,milk,and potatoes.
Rice is a staple food that provides a high concentration of starch,which is a complex carbohydrate.
Carbohydrates are the primary source of energy for most animals.
They are organic compounds composed of carbon,hydrogen,and oxygen.
Rich sources of carbohydrates include cereals (like rice,wheat,and maize),fruits,milk,and potatoes.
Rice is a staple food that provides a high concentration of starch,which is a complex carbohydrate.
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27
BiologyMediumMCQAIPMT · 2002
Which of the following is a reducing sugar?
A
Sucrose
B
Galactose
C
Gluconic acid
D
$\beta - \text{methyl galactoside}$
Solution
(B) reducing sugar is a carbohydrate that possesses a free aldehyde or ketone group, which allows it to act as a reducing agent.
All monosaccharides, such as glucose, fructose, and galactose, are reducing sugars because they contain a free anomeric carbon.
Sucrose is a non-reducing sugar because its glycosidic bond involves the anomeric carbons of both glucose and fructose, leaving no free aldehyde or ketone group.
Gluconic acid is an oxidation product of glucose and does not function as a reducing sugar in the same manner.
$\beta - \text{methyl galactoside}$ is a glycoside where the anomeric hydroxyl group is replaced by a methoxy group, making it non-reducing.
Therefore, Galactose $(C_6H_{12}O_6)$ is a monosaccharide and acts as a reducing sugar.
All monosaccharides, such as glucose, fructose, and galactose, are reducing sugars because they contain a free anomeric carbon.
Sucrose is a non-reducing sugar because its glycosidic bond involves the anomeric carbons of both glucose and fructose, leaving no free aldehyde or ketone group.
Gluconic acid is an oxidation product of glucose and does not function as a reducing sugar in the same manner.
$\beta - \text{methyl galactoside}$ is a glycoside where the anomeric hydroxyl group is replaced by a methoxy group, making it non-reducing.
Therefore, Galactose $(C_6H_{12}O_6)$ is a monosaccharide and acts as a reducing sugar.
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28
BiologyEasyMCQAIPMT · 2002
Continuous bleeding from an injured part of the body is due to the deficiency of:
A
Vitamin $A$
B
Vitamin $B$
C
Vitamin $K$
D
Vitamin $E$
Solution
(C) Vitamin $K$ is essential for the synthesis of clotting factors in the liver,such as prothrombin and factors $VII, IX,$ and $X$.
These factors are necessary for the blood coagulation process.
Therefore,a deficiency of Vitamin $K$ prevents the blood from clotting effectively,leading to prolonged or continuous bleeding from an injured part of the body.
These factors are necessary for the blood coagulation process.
Therefore,a deficiency of Vitamin $K$ prevents the blood from clotting effectively,leading to prolonged or continuous bleeding from an injured part of the body.
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29
BiologyEasyMCQAIPMT · 2002
Organisms,which obtain energy by the oxidation of reduced inorganic compounds,are called:
A
Phototrophs
B
Saprozoic
C
Copro-heterotrophs
D
Chemo-autotrophs
Solution
(D) Organisms that obtain energy by the oxidation of reduced inorganic compounds (such as $NH_3$,$H_2S$,$Fe^{2+}$,etc.) are known as chemo-autotrophs.
These organisms use the energy released from these chemical reactions to synthesize organic compounds from $CO_2$.
Phototrophs obtain energy from light,while saprozoic organisms feed on decaying organic matter.
These organisms use the energy released from these chemical reactions to synthesize organic compounds from $CO_2$.
Phototrophs obtain energy from light,while saprozoic organisms feed on decaying organic matter.
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30
BiologyEasyMCQAIPMT · 2002
Secretin hormone is secreted by
A
Liver
B
Pancreas
C
Intestine
D
Brunner's glands
Solution
(C) Secretin is a peptide hormone produced by the $S$-cells of the duodenum,which is a part of the small intestine.
It is released into the bloodstream in response to the acidic chyme entering the duodenum from the stomach.
Its primary function is to stimulate the pancreas to secrete a bicarbonate-rich fluid,which helps neutralize the acidity of the chyme in the small intestine.
It is released into the bloodstream in response to the acidic chyme entering the duodenum from the stomach.
Its primary function is to stimulate the pancreas to secrete a bicarbonate-rich fluid,which helps neutralize the acidity of the chyme in the small intestine.
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31
BiologyEasyMCQAIPMT · 2002
Which of the following is a respiratory organ of a scorpion?
A
Gill
B
Lung
C
Ctenidia
D
Book lung
Solution
(D) Scorpions belong to the class $Arachnida$ of the phylum $Arthropoda$.
They possess specialized respiratory structures known as $Book$ $lungs$.
These structures consist of a series of thin,leaf-like folds of the body wall arranged like the pages of a book,which are located within an internal chamber and are adapted for gas exchange in terrestrial environments.
They possess specialized respiratory structures known as $Book$ $lungs$.
These structures consist of a series of thin,leaf-like folds of the body wall arranged like the pages of a book,which are located within an internal chamber and are adapted for gas exchange in terrestrial environments.
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32
BiologyEasyMCQAIPMT · 2002
The respiratory pigment haemocyanin is found in which of the following phyla?
A
Chordata
B
Annelida
C
Mollusca
D
Echinodermata
Solution
(C) Haemocyanin is a copper-containing respiratory pigment that gives blood a blue color when oxygenated.
It is commonly found in the hemolymph of many mollusks,such as cephalopods (e.g.,squids,octopuses) and some gastropods.
While some annelids use haemoglobin or chlorocruorin,haemocyanin is a characteristic feature of many members of the phylum $Mollusca$.
It is commonly found in the hemolymph of many mollusks,such as cephalopods (e.g.,squids,octopuses) and some gastropods.
While some annelids use haemoglobin or chlorocruorin,haemocyanin is a characteristic feature of many members of the phylum $Mollusca$.
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33
BiologyMediumMCQAIPMT · 2002
Heartbeat can be initiated by
A
Sino-atrial node
B
Atrio-ventricular node
C
Sodium ion
D
Purkinje's fibres
Solution
(A) The human heart is myogenic,meaning the heartbeat is initiated by specialized cardiac muscle cells. The $Sino-atrial$ $node$ $(SAN)$,also known as the pacemaker,is located in the right atrium. It generates action potentials that initiate the heartbeat,making it the primary pacemaker of the heart.
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34
BiologyMediumMCQAIPMT · 2002
Heartbeat originates from
A
Pacemaker
B
Cardiac muscles
C
Left atrium
D
Right ventricle
Solution
(A) The heartbeat originates from the pacemaker,which is the $S.A.$ node (Sinoatrial node).
It is a specialized patch of cardiac muscle tissue located in the right atrium that generates electrical impulses to initiate the contraction of the heart.
It is a specialized patch of cardiac muscle tissue located in the right atrium that generates electrical impulses to initiate the contraction of the heart.
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35
BiologyMediumMCQAIPMT · 2002
The stool of a person contains a whitish-grey colour due to the malfunction of which organ?
A
Liver
B
Spleen
C
Kidney
D
Pancreas
Solution
(A) The normal brown colour of faeces is primarily due to the presence of bile pigments,specifically stercobilin,which is derived from the breakdown of bilirubin and biliverdin produced by the liver.
If the liver malfunctions or if there is an obstruction in the bile duct,bile pigments do not reach the intestine.
Without these pigments,the stool loses its characteristic brown colour and appears pale,whitish,or clay-coloured.
Therefore,the correct answer is the liver.
If the liver malfunctions or if there is an obstruction in the bile duct,bile pigments do not reach the intestine.
Without these pigments,the stool loses its characteristic brown colour and appears pale,whitish,or clay-coloured.
Therefore,the correct answer is the liver.
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36
BiologyEasyMCQAIPMT · 2002
The number of cervical vertebrae in camels is
A
Same as that in rabbit
B
Same as that in frog
C
Less than that in giraffe
D
More than that in horse
Solution
(A) Almost all mammals,including camels,rabbits,horses,and giraffes,possess $7$ cervical vertebrae. This is a characteristic feature of the mammalian class. Since the number of cervical vertebrae is $7$ in both camels and rabbits,the correct answer is that it is the same as that in a rabbit.
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37
BiologyMediumMCQAIPMT · 2002
Acromegaly results after adolescence due to excess production of one of the following hormones:
A
Prolactin
B
Thyroxin
C
Insulin
D
$STH$
Solution
(D) Acromegaly is a condition that occurs due to the hypersecretion of $STH$ (Somatotropic Hormone or Growth Hormone) in adults after adolescence.
It is characterized by the abnormal growth of bones,particularly in the limbs,lower jaw,and facial features,often resulting in a gorilla-like appearance,kyphosis (hunchback),and a protruding bony ridge over the eyes.
It is characterized by the abnormal growth of bones,particularly in the limbs,lower jaw,and facial features,often resulting in a gorilla-like appearance,kyphosis (hunchback),and a protruding bony ridge over the eyes.
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38
BiologyMediumMCQAIPMT · 2002
If both ovaries are removed from a rat,then which hormone level will decrease in the blood?
A
Oxytocin
B
Estrogen
C
Prolactin
D
Gonadotrophic
Solution
(B) The ovaries are the primary female reproductive organs responsible for the production of steroid hormones,specifically estrogen and progesterone.
Estrogen is primarily secreted by the theca interna cells of the developing Graafian follicles in the ovary.
If both ovaries are surgically removed (ovariectomy),the source of estrogen production is eliminated,leading to a significant decrease in the concentration of estrogen in the blood.
Therefore,the correct option is $B$.
Estrogen is primarily secreted by the theca interna cells of the developing Graafian follicles in the ovary.
If both ovaries are surgically removed (ovariectomy),the source of estrogen production is eliminated,leading to a significant decrease in the concentration of estrogen in the blood.
Therefore,the correct option is $B$.
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39
BiologyMediumMCQAIPMT · 2002
If a diploid cell is treated with colchicine,then it becomes
A
Tetraploid
B
Diploid
C
Triploid
D
Monoploid
Solution
(A) Colchicine is a chemical agent that inhibits the formation of the spindle apparatus during cell division (mitosis).
Because the spindle fibers do not form,the sister chromatids fail to separate during anaphase.
However,the replication of $DNA$ and the duplication of chromosomes continue normally.
As a result,the cell enters the next interphase with double the number of chromosomes,turning a $2n$ (diploid) cell into a $4n$ (tetraploid) cell.
Because the spindle fibers do not form,the sister chromatids fail to separate during anaphase.
However,the replication of $DNA$ and the duplication of chromosomes continue normally.
As a result,the cell enters the next interphase with double the number of chromosomes,turning a $2n$ (diploid) cell into a $4n$ (tetraploid) cell.
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40
BiologyMediumMCQAIPMT · 2002
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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41
BiologyEasyMCQAIPMT · 2002
Organisms which obtain energy by the oxidation of reduced inorganic compounds are called
A
Saprozoic
B
Chemoautotrophs
C
Photoautotrophs
D
Coproheterotroph
Solution
(B) Organisms that derive energy from the oxidation of reduced inorganic compounds (such as ammonia,nitrites,or hydrogen sulfide) are known as $Chemoautotrophs$.
These organisms use this chemical energy to fix carbon dioxide into organic compounds,as they do not rely on sunlight for energy production.
These organisms use this chemical energy to fix carbon dioxide into organic compounds,as they do not rely on sunlight for energy production.
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42
BiologyEasyMCQAIPMT · 2002
In bacteria,plasmids are ............... .
A
Extra-chromosomal material
B
Main $DNA$
C
Non-functional $DNA$
D
Repetitive genes
Solution
(A) Plasmids are small,circular,double-stranded $DNA$ molecules that are distinct from a cell's chromosomal $DNA$.
They are found in bacteria and some other microorganisms.
These are considered extra-chromosomal genetic elements because they exist and replicate independently of the main bacterial chromosome.
Plasmids often carry genes that provide advantages to the bacteria,such as antibiotic resistance.
They are found in bacteria and some other microorganisms.
These are considered extra-chromosomal genetic elements because they exist and replicate independently of the main bacterial chromosome.
Plasmids often carry genes that provide advantages to the bacteria,such as antibiotic resistance.
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43
BiologyMediumMCQAIPMT · 2002
Organisms that obtain energy by the oxidation of reduced inorganic compounds are called .............
A
Photosynthetic
B
Chemosynthetic
C
Saprophytic
D
Heterotrophic
Solution
(B) Organisms that derive energy from the oxidation of reduced inorganic substances (such as ammonia, nitrites, or sulfides) are known as $Chemosynthetic$ autotrophs. These organisms play a vital role in recycling nutrients like nitrogen, phosphorus, iron, and sulfur in the ecosystem. Unlike photosynthetic organisms, they do not require sunlight for energy production.
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44
BiologyEasyMCQAIPMT · 2002
In $Amoeba$ and $Paramecium$,osmoregulation occurs through . . . . . . .
A
Pseudopodia
B
Nucleus
C
Contractile vacuole
D
General body surface
Solution
(C) Osmoregulation is the process of maintaining the water and salt balance within an organism. In freshwater protozoans like $Amoeba$ and $Paramecium$,the contractile vacuole is the specialized organelle responsible for osmoregulation. It collects excess water from the cytoplasm and expels it out of the cell to prevent the organism from bursting due to the influx of water by osmosis.
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45
BiologyEasyMCQAIPMT · 2002
The presence of macronucleus and micronucleus is a characteristic feature of $......$.
A
Paramecium and Vorticella
B
Opalina and Nyctotherus
C
Hydra and Balantidium
D
Vorticella and Nyctotherus
Solution
(A) The presence of two types of nuclei,a large macronucleus and a small micronucleus,is a characteristic feature of the phylum $Ciliophora$ (ciliates).
$Paramecium$,$Vorticella$,$Nyctotherus$,and $Balantidium$ are all examples of ciliates.
However,among the given options,$Paramecium$ and $Vorticella$ are the most classic examples of organisms exhibiting nuclear dimorphism (macronucleus and micronucleus).
Therefore,the correct option is $A$.
$Paramecium$,$Vorticella$,$Nyctotherus$,and $Balantidium$ are all examples of ciliates.
However,among the given options,$Paramecium$ and $Vorticella$ are the most classic examples of organisms exhibiting nuclear dimorphism (macronucleus and micronucleus).
Therefore,the correct option is $A$.
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46
BiologyEasyMCQAIPMT · 2002
Which of the following plants produce seeds but are flowerless?
A
Maize
B
Mint
C
Peepal
D
Pinus
Solution
(D) Plants that produce seeds but do not bear flowers are known as Gymnosperms.
In Gymnosperms,the ovules are not enclosed by any ovary wall and remain exposed,both before and after fertilization.
Among the given options,$A$ (Maize),$B$ (Mint),and $C$ (Peepal) are Angiosperms,which are flowering plants.
$D$ (Pinus) is a Gymnosperm,which produces seeds but is flowerless.
In Gymnosperms,the ovules are not enclosed by any ovary wall and remain exposed,both before and after fertilization.
Among the given options,$A$ (Maize),$B$ (Mint),and $C$ (Peepal) are Angiosperms,which are flowering plants.
$D$ (Pinus) is a Gymnosperm,which produces seeds but is flowerless.
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47
BiologyEasyMCQAIPMT · 2002
In which of the following is the notochord present during the embryonic stage?
A
All chordates
B
Some chordates
C
Vertebrates
D
Non-chordates
Solution
(A) The phylum $Chordata$ is characterized by the presence of a notochord,a dorsal hollow nerve cord,and paired pharyngeal gill slits at some stage of their life cycle.
In all chordates,the notochord is present during the embryonic stage. It is a mesodermally derived rod-like structure formed on the dorsal side during embryonic development.
In all chordates,the notochord is present during the embryonic stage. It is a mesodermally derived rod-like structure formed on the dorsal side during embryonic development.
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48
BiologyEasyMCQAIPMT · 2002
In which organism is a dimorphic nucleus found?
A
Amoeba
B
Trypanosoma gambiense
C
Plasmodium vivax
D
Paramecium caudatum
Solution
(D) Dimorphic nucleus refers to the presence of two types of nuclei within a single cell, specifically a large macronucleus and a small micronucleus.
This condition is a characteristic feature of the phylum Ciliophora.
Among the given options, $Paramecium \text{ } caudatum$ belongs to the class Ciliata (Ciliophora) and exhibits nuclear dimorphism.
The macronucleus controls vegetative functions, while the micronucleus is involved in reproductive processes.
This condition is a characteristic feature of the phylum Ciliophora.
Among the given options, $Paramecium \text{ } caudatum$ belongs to the class Ciliata (Ciliophora) and exhibits nuclear dimorphism.
The macronucleus controls vegetative functions, while the micronucleus is involved in reproductive processes.
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49
BiologyEasyMCQAIPMT · 2002
Geocarpic fruit is .......
A
Potato
B
Groundnut
C
Onion
D
Garlic
Solution
(B) geocarpic fruit is a fruit that develops underground.
In the case of groundnut ($Arachis$ $hypogaea$),after fertilization,the flower stalk (gynophore) elongates and grows downwards into the soil,pushing the developing ovary into the ground where the fruit matures.
Therefore,groundnut is known as a geocarpic fruit.
In the case of groundnut ($Arachis$ $hypogaea$),after fertilization,the flower stalk (gynophore) elongates and grows downwards into the soil,pushing the developing ovary into the ground where the fruit matures.
Therefore,groundnut is known as a geocarpic fruit.
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50
BiologyMediumMCQAIPMT · 2002
The edible part of the mango is . . . . . . .
A
Mesocarp
B
Epicarp
C
Endocarp
D
Epidermis
Solution
(A) In mango $(Mangifera \text{ } indica)$, the fruit is a drupe.
It develops from a monocarpellary superior ovary and is one-seeded.
The fruit wall is differentiated into three layers:
$1$. The outer thin epicarp (skin).
$2$. The middle fleshy, edible mesocarp.
$3$. The inner stony hard endocarp.
Therefore, the edible part of the mango is the mesocarp.
It develops from a monocarpellary superior ovary and is one-seeded.
The fruit wall is differentiated into three layers:
$1$. The outer thin epicarp (skin).
$2$. The middle fleshy, edible mesocarp.
$3$. The inner stony hard endocarp.
Therefore, the edible part of the mango is the mesocarp.
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51
BiologyEasyMCQAIPMT · 2002
The transformation experiment was performed on which of the following bacteria?
A
$E$. coli
B
Salmonella
C
Pasturella pestis
D
Diplococcus pneumoniae
Solution
(D) The transformation experiment was conducted by Frederick Griffith in $1928$. He used the bacterium $Diplococcus$ $pneumoniae$ (also known as $Streptococcus$ $pneumoniae$) to demonstrate the phenomenon of bacterial transformation. In this experiment,he observed that a non-virulent strain of bacteria could be transformed into a virulent strain by taking up genetic material from heat-killed virulent bacteria.
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52
BiologyEasyMCQAIPMT · 2002
Which of the following secretes toxins during storage conditions of crop plants?
A
Fusarium
B
Penicillium
C
Aspergillus
D
Colletotrichum
Solution
(C) $Aspergillus$ $flavus$ grows on stored cereal grains,groundnut,and areca nut,and produces a carcinogenic toxin called aflatoxin. Therefore,$Aspergillus$ is the correct answer.
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53
BiologyMediumMCQAIPMT · 2002
In the manufacture of bread, it becomes porous due to the release of $CO_2$ by the action of:
A
Virus
B
Yeast
C
Bacteria
D
Protozoans
Solution
(B) The process of bread making involves the use of yeast, specifically $Saccharomyces$ $cerevisiae$, which is commonly known as baker's yeast.
During the fermentation process, yeast breaks down the sugars present in the dough into ethanol and $CO_2$.
The release of $CO_2$ gas creates bubbles within the dough, which causes the bread to rise and become porous and soft.
During the fermentation process, yeast breaks down the sugars present in the dough into ethanol and $CO_2$.
The release of $CO_2$ gas creates bubbles within the dough, which causes the bread to rise and become porous and soft.
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54
BiologyMediumMCQAIPMT · 2002
Which of the following fungal diseases spreads by seeds and flowers?
A
Corn stunt
B
Covered smut
C
Potato root
D
Loose smut of wheat
Solution
(D) The correct answer is $D$. Loose smut of wheat is caused by the fungus $Ustilago$ $tritici$. This disease is internally seed-borne, where the pathogen infects the developing ovary of the flower, remains dormant in the seed, and manifests when the infected seed is planted in the next season.
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55
BiologyMediumMCQAIPMT · 2002
The reason for the formation of an embryoid from a pollen grain in a tissue culture medium is:
A
Organogenesis
B
Double fertilization
C
Test tube culture
D
Cellular totipotency
Solution
(D) The formation of an embryoid from a pollen grain (microspore) in a tissue culture medium is a classic example of cellular totipotency.
Cellular totipotency is the ability of a single plant cell to divide and differentiate into all the specialized cells of a mature plant.
In this process,the haploid pollen grain is induced to undergo embryogenesis,forming a haploid embryo (embryoid) instead of developing into a male gametophyte.
This phenomenon demonstrates that even specialized cells like pollen grains retain the genetic potential to regenerate an entire organism under appropriate in vitro conditions.
Cellular totipotency is the ability of a single plant cell to divide and differentiate into all the specialized cells of a mature plant.
In this process,the haploid pollen grain is induced to undergo embryogenesis,forming a haploid embryo (embryoid) instead of developing into a male gametophyte.
This phenomenon demonstrates that even specialized cells like pollen grains retain the genetic potential to regenerate an entire organism under appropriate in vitro conditions.
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56
BiologyEasyMCQAIPMT · 2002
In angiosperms,all $4$ microspores of a tetrad are covered by a layer which is formed by:
A
Callose
B
Cellulose
C
Sporopollenin
D
Pectocellulose
Solution
(A) During microsporogenesis,the microspore mother cell undergoes meiosis to form a group of $4$ haploid cells known as a microspore tetrad.
These $4$ microspores are held together by a special layer of callose.
As the anther matures and dehydrates,the enzyme callase (secreted by the tapetum) breaks down this callose layer,allowing the microspores to dissociate and develop into pollen grains.
These $4$ microspores are held together by a special layer of callose.
As the anther matures and dehydrates,the enzyme callase (secreted by the tapetum) breaks down this callose layer,allowing the microspores to dissociate and develop into pollen grains.
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57
BiologyMediumMCQAIPMT · 2002
What is the direction of the micropyle in an anatropous ovule?
A
Left
B
Right
C
Upward
D
Downward
Solution
(D) In an anatropous ovule,the body of the ovule becomes completely inverted due to unilateral growth of the funicle.
As a result of this $180^{\circ}$ rotation,the micropyle comes to lie close to the hilum,pointing downwards towards the placenta.
Therefore,the micropyle is directed downwards.
As a result of this $180^{\circ}$ rotation,the micropyle comes to lie close to the hilum,pointing downwards towards the placenta.
Therefore,the micropyle is directed downwards.
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58
BiologyEasyMCQAIPMT · 2002
Pollination by slugs and snails is called:
A
Ornithophilous
B
Malacophilous
C
Anemophilous
D
Chiropterophilous
Solution
(B) The correct answer is $B$.
Pollination by snails and slugs is known as $Malacophily$.
$Ornithophilous$ refers to pollination by birds.
$Anemophilous$ refers to pollination by wind.
$Chiropterophilous$ refers to pollination by bats.
An example of a plant pollinated by snails is $Arisaema$ (cobra plant).
Pollination by snails and slugs is known as $Malacophily$.
$Ornithophilous$ refers to pollination by birds.
$Anemophilous$ refers to pollination by wind.
$Chiropterophilous$ refers to pollination by bats.
An example of a plant pollinated by snails is $Arisaema$ (cobra plant).
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59
BiologyMediumMCQAIPMT · 2002
Which of the following is without exception in angiosperms?
A
Secondary growth
B
Presence of vessels
C
Double fertilization
D
Autotrophic nutrition
Solution
(C) $Double$ $fertilization$ is a unique and defining characteristic found exclusively in angiosperms, with no exceptions.
During this process, one male gamete fuses with the egg cell to form a diploid zygote $(2n)$, while the second male gamete fuses with the two polar nuclei (or the secondary nucleus) to form a triploid primary endosperm nucleus $(3n)$.
The triploid nucleus develops into the endosperm, which provides nourishment to the developing embryo.
During this process, one male gamete fuses with the egg cell to form a diploid zygote $(2n)$, while the second male gamete fuses with the two polar nuclei (or the secondary nucleus) to form a triploid primary endosperm nucleus $(3n)$.
The triploid nucleus develops into the endosperm, which provides nourishment to the developing embryo.
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60
BiologyMediumMCQAIPMT · 2002
If in a garden pea plant,a cross is made between red flowered and white flowered plants,what will be the phenotypic ratio in the $F_2$ generation?
A
$1:2:1$
B
$9:3:3:1$
C
$3:1$
D
$1:3$
Solution
(C) In garden pea plants,red flower color is dominant over white flower color. Let $R$ represent the dominant allele for red flowers and $r$ represent the recessive allele for white flowers.
$P$ generation: $RR$ (Red) $\times$ $rr$ (White)
$F_1$ generation: All offspring are $Rr$ (Red).
When $F_1$ plants $(Rr)$ are self-pollinated $(Rr \times Rr)$:
Using a Punnett square:
| | $R$ | $r$ |
|---|---|---|
| $R$ | $RR$ | $Rr$ |
| $r$ | $Rr$ | $rr$ |
Genotypes: $1 RR : 2 Rr : 1 rr$
Phenotypes: $3$ Red $(RR, Rr, Rr)$ : $1$ White $(rr)$.
Therefore,the phenotypic ratio in the $F_2$ generation is $3:1$.
$P$ generation: $RR$ (Red) $\times$ $rr$ (White)
$F_1$ generation: All offspring are $Rr$ (Red).
When $F_1$ plants $(Rr)$ are self-pollinated $(Rr \times Rr)$:
Using a Punnett square:
| | $R$ | $r$ |
|---|---|---|
| $R$ | $RR$ | $Rr$ |
| $r$ | $Rr$ | $rr$ |
Genotypes: $1 RR : 2 Rr : 1 rr$
Phenotypes: $3$ Red $(RR, Rr, Rr)$ : $1$ White $(rr)$.
Therefore,the phenotypic ratio in the $F_2$ generation is $3:1$.
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61
BiologyEasyMCQAIPMT · 2002
Which pair of chromosomes is concerned with Down's syndrome?
A
$7^{th}$
B
$18^{th}$
C
$21^{st}$
D
$23^{rd}$
Solution
(C) Down's syndrome is a genetic disorder caused by the presence of an extra copy of chromosome number $21$.
This condition is known as trisomy of the $21^{st}$ chromosome.
Therefore,the correct pair of chromosomes associated with Down's syndrome is the $21^{st}$ pair.
This condition is known as trisomy of the $21^{st}$ chromosome.
Therefore,the correct pair of chromosomes associated with Down's syndrome is the $21^{st}$ pair.
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62
BiologyMediumMCQAIPMT · 2002
Random genetic drift in a population probably results from
A
Large population size
B
Highly genetically variable individuals
C
Interbreeding within small isolated population
D
Constant low mutation rate
Solution
(C) Random genetic drift refers to the change in allele frequencies in a population due to chance events.
It is most pronounced in small,isolated populations where chance fluctuations have a significant impact on the gene pool.
Therefore,interbreeding within a small,isolated population is the primary factor that leads to random genetic drift.
It is most pronounced in small,isolated populations where chance fluctuations have a significant impact on the gene pool.
Therefore,interbreeding within a small,isolated population is the primary factor that leads to random genetic drift.
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63
BiologyMediumMCQAIPMT · 2002
$A$ diseased man marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is
A
Autosomal dominant
B
Sex linked recessive
C
Sex limited character
D
Sex linked dominant
Solution
(D) In this scenario,the father is diseased and the mother is normal.
All daughters inherit the $X$ chromosome from their father,while all sons inherit the $Y$ chromosome from their father.
Since all daughters are diseased and all sons are normal,the disease-causing gene must be located on the $X$ chromosome inherited from the father.
Because the trait is expressed in all daughters (who receive the $X$ chromosome from the father),it indicates that the gene is $X$-linked dominant.
If it were $X$-linked recessive,the daughters would be carriers but not necessarily diseased unless the mother was also a carrier or affected.
Therefore,the pattern of inheritance is $X$-linked dominant.
All daughters inherit the $X$ chromosome from their father,while all sons inherit the $Y$ chromosome from their father.
Since all daughters are diseased and all sons are normal,the disease-causing gene must be located on the $X$ chromosome inherited from the father.
Because the trait is expressed in all daughters (who receive the $X$ chromosome from the father),it indicates that the gene is $X$-linked dominant.
If it were $X$-linked recessive,the daughters would be carriers but not necessarily diseased unless the mother was also a carrier or affected.
Therefore,the pattern of inheritance is $X$-linked dominant.
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64
BiologyMediumMCQAIPMT · 2002
Which one of the following is called the polynucleotide joining enzyme?
A
Polymerase $I$
B
Polymerase $II$
C
Ligase
D
Ribonuclease
Solution
(C) $DNA$ ligase is known as the polynucleotide joining enzyme.
It catalyzes the formation of phosphodiester bonds between adjacent nucleotides.
This enzyme is essential for joining $DNA$ fragments,such as Okazaki fragments during replication,and for sealing nicks in the $DNA$ backbone during repair processes.
It catalyzes the formation of phosphodiester bonds between adjacent nucleotides.
This enzyme is essential for joining $DNA$ fragments,such as Okazaki fragments during replication,and for sealing nicks in the $DNA$ backbone during repair processes.
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65
BiologyMediumMCQAIPMT · 2002
Histone occupies the major grooves of $DNA$ at an angle of
A
$15^o$
B
$90^o$
C
$45^o$ to the helix axis
D
$30^o$ to the helix axis
Solution
(C) In the structure of nucleosomes,the histone octamer is wrapped by $DNA$. The $DNA$ makes approximately $1.65$ turns around the histone core. Studies on the interaction between histone proteins and $DNA$ show that the $DNA$ helix interacts with the histone surface,and the orientation of the $DNA$ relative to the histone core involves specific angular positioning. Specifically,the $DNA$ path around the histone octamer is oriented at an angle of $45^o$ to the helix axis.
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66
BiologyMediumMCQAIPMT · 2002
The exon part of $m-RNA$ contains the code for:
A
Protein
B
Lipid
C
Phospholipid
D
Carbohydrate
Solution
(A) Exons are the coding sequences in $m-RNA$ that are expressed during protein synthesis.
Introns are the non-coding sequences that are removed during $RNA$ splicing.
Therefore,the exon part of $m-RNA$ contains the genetic code required for the synthesis of specific proteins.
Introns are the non-coding sequences that are removed during $RNA$ splicing.
Therefore,the exon part of $m-RNA$ contains the genetic code required for the synthesis of specific proteins.
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67
BiologyMediumMCQAIPMT · 2002
Out of $64$ codons,$61$ codons code for $20$ types of amino acids. This phenomenon is called:
A
Wobbling of codon
B
Overlapping of gene
C
Universality of codons
D
Degeneracy of genetic code
Solution
(D) The genetic code is degenerate,which means that more than one codon can code for the same amino acid.
Since there are $64$ possible codons and only $20$ amino acids to be coded,some amino acids are specified by more than one codon.
This redundancy or degeneracy ensures that the genetic code is robust against mutations.
Therefore,the correct answer is $D$.
Since there are $64$ possible codons and only $20$ amino acids to be coded,some amino acids are specified by more than one codon.
This redundancy or degeneracy ensures that the genetic code is robust against mutations.
Therefore,the correct answer is $D$.
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68
BiologyMediumMCQAIPMT · 2002
Jacob and Monod studied lactose metabolism in $E. coli$ and proposed the operon concept,which is applicable for:
A
Prokaryotes
B
Eukaryotes
C
Protozoans
D
All of these
Solution
(A) The operon model was proposed by Jacob and Monod in $1961$ to explain the regulation of gene expression and protein synthesis specifically in prokaryotes,such as $E. coli$.
In this model,a cluster of genes (operon) is regulated by a single promoter and operator,allowing the cell to respond efficiently to environmental changes.
While some operon-like structures exist in eukaryotes,the classic operon concept as defined by Jacob and Monod is a fundamental characteristic of prokaryotic gene regulation.
In this model,a cluster of genes (operon) is regulated by a single promoter and operator,allowing the cell to respond efficiently to environmental changes.
While some operon-like structures exist in eukaryotes,the classic operon concept as defined by Jacob and Monod is a fundamental characteristic of prokaryotic gene regulation.
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69
BiologyEasyMCQAIPMT · 2002
$A$ gene is said to be dominant,if
A
It is never expressed in any condition
B
It is expressed only in heterozygous condition
C
It expresses its effect only in homozygous stage
D
It is expressed both in homozygous and heterozygous conditions
Solution
(D) gene is defined as dominant if it expresses its phenotypic effect in both the homozygous $(AA)$ and heterozygous $(Aa)$ conditions. In a heterozygous state,the dominant allele masks the expression of the recessive allele.
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70
BiologyMediumMCQAIPMT · 2002
The evolution of man took place in
A
Central Africa
B
Central Asia
C
Australia
D
America
Solution
(A) Scientific evidence,including fossil records and molecular biology ($DNA$ analysis),indicates that the evolution of modern humans ($Homo$ $sapiens$) occurred in Africa. Specifically,the earliest ancestors of humans emerged in the East African Rift Valley,which is often referred to as the 'cradle of humankind'. From there,early humans migrated to other parts of the world.
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71
BiologyMediumMCQAIPMT · 2002
Which of the following statements is true regarding individuals of the same species?
A
They are interbreeding.
B
They live in the same niche.
C
They live in different niches.
D
They live in different habitats.
Solution
(A) species is defined as a group of individuals that are capable of interbreeding among themselves and producing fertile offspring. Therefore,the ability to interbreed is the fundamental characteristic that defines members of the same species.
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72
BiologyMediumMCQAIPMT · 2002
Which of the following types of isolation is important for speciation?
A
Seasonal
B
Tropical
C
Behavioural
D
Reproductive
Solution
(D) The correct answer is $D$. Reproductive isolation is the most critical factor for speciation. It refers to the mechanisms that prevent two different populations from interbreeding and producing fertile offspring. The interruption of gene flow between populations due to reproductive isolation is a fundamental pre-requisite for the formation of new species.
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73
BiologyMediumMCQAIPMT · 2002
$A$ plant being eaten by a herbivore,which in turn is eaten by a carnivore,constitutes a:
A
Food chain
B
Food web
C
Omnivore
D
Interdependent system
Solution
(A) food chain is a linear sequence of organisms through which nutrients and energy pass as one organism eats another.
In this scenario,the plant (producer) is consumed by a herbivore (primary consumer),which is then consumed by a carnivore (secondary consumer).
This linear flow of energy from the producer to the primary consumer and then to the secondary consumer is defined as a food chain.
In this scenario,the plant (producer) is consumed by a herbivore (primary consumer),which is then consumed by a carnivore (secondary consumer).
This linear flow of energy from the producer to the primary consumer and then to the secondary consumer is defined as a food chain.
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74
BiologyMediumMCQAIPMT · 2002
Which of the following food chains may not be directly dependent upon solar energy?
A
Grazing food chain
B
Detritus food chain
C
Soaking food chain
D
Depleting food chain
Solution
(B) The $Grazing$ food chain begins with producers (green plants) that capture solar energy through photosynthesis.
In contrast, the $Detritus$ food chain begins with dead organic matter (detritus) and decomposers.
Since the $Detritus$ food chain relies on the energy stored in dead organic matter rather than direct sunlight, it is not directly dependent on solar energy for its primary energy source.
In contrast, the $Detritus$ food chain begins with dead organic matter (detritus) and decomposers.
Since the $Detritus$ food chain relies on the energy stored in dead organic matter rather than direct sunlight, it is not directly dependent on solar energy for its primary energy source.
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75
BiologyMediumMCQAIPMT · 2002
If a bamboo plant is growing in a forest,what will be its trophic level?
A
First
B
Second
C
Third
D
Fourth
Solution
(A) Bamboo is a producer (plant) that synthesizes its own food through photosynthesis. In any food chain,producers occupy the first trophic level.
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76
BiologyEasyMCQAIPMT · 2002
Which of the following countries is responsible for releasing the largest amount of greenhouse gases?
A
Russia
B
Germany
C
Brazil
D
America $(USA)$
Solution
(D) The correct answer is $(d)$. The $USA$ is historically and currently one of the largest emitters of greenhouse gases due to its high industrial activity,extensive transportation networks,and massive consumption of fossil fuel energy.
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77
BiologyMediumMCQAIPMT · 2002
Positive pollution of soil is due to
A
Excessive use of fertilizers
B
Addition of wastes on soil
C
Reduction in soil productivity
D
All of these
Solution
(D) Positive soil pollution refers to the addition of undesirable substances to the soil that degrade its quality.
$A$. Excessive use of fertilizers adds chemicals like nitrates and phosphates,which alter soil chemistry.
$B$. Addition of industrial,agricultural,or domestic wastes introduces pollutants into the soil.
$C$. Reduction in soil productivity is a consequence of these pollutants.
Since all these factors contribute to the contamination of soil,the correct answer is $D$.
$A$. Excessive use of fertilizers adds chemicals like nitrates and phosphates,which alter soil chemistry.
$B$. Addition of industrial,agricultural,or domestic wastes introduces pollutants into the soil.
$C$. Reduction in soil productivity is a consequence of these pollutants.
Since all these factors contribute to the contamination of soil,the correct answer is $D$.
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78
BiologyMediumMCQAIPMT · 2002
The cause of mimicry is:
A
Isolation
B
Attack
C
Protection
D
Both $(b)$ and $(c)$
Solution
(D) Mimicry is an evolved resemblance between an organism and another object,often an organism of another species.
$1$. Protective mimicry: Animals imitate other animals,plants,or natural objects to protect themselves from predators.
$2$. Aggressive mimicry: Predatory animals mimic natural objects or other organisms so that their prey becomes confused,allowing the predator to attack or catch them easily.
Since mimicry serves both the purpose of protection from predators and facilitating the attack on prey,the correct answer is both $(b)$ and $(c)$.
$1$. Protective mimicry: Animals imitate other animals,plants,or natural objects to protect themselves from predators.
$2$. Aggressive mimicry: Predatory animals mimic natural objects or other organisms so that their prey becomes confused,allowing the predator to attack or catch them easily.
Since mimicry serves both the purpose of protection from predators and facilitating the attack on prey,the correct answer is both $(b)$ and $(c)$.
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79
BiologyEasyMCQAIPMT · 2002
Melanin protects from
A
$UV$ rays
B
$X-$ rays
C
Visible rays
D
Infrared rays
Solution
(A) Melanin is a pigment found in the skin,hair,and eyes of humans and other animals.
Its primary biological function is to act as a photoprotective agent.
Melanin absorbs harmful $UV$ (ultraviolet) radiation from the sun,which helps prevent $DNA$ damage in skin cells and protects against sunburns and skin cancer.
Its primary biological function is to act as a photoprotective agent.
Melanin absorbs harmful $UV$ (ultraviolet) radiation from the sun,which helps prevent $DNA$ damage in skin cells and protects against sunburns and skin cancer.
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80
BiologyEasyMCQAIPMT · 2002
Which of the following crops have been brought to India from the New World?
A
Coffee
B
Mango,tea
C
Tea,rubber,mango
D
Cashew nut,potato,rubber
Solution
(D) The 'New World' refers to the Americas. Many crops were introduced to India from the Americas by European explorers and traders. Among the given options,cashew nut,potato,and rubber are well-known crops that originated in the New World and were subsequently introduced to India.
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81
BiologyEasyMCQAIPMT · 2002
The physiological capacity of a population to produce offspring is known as:
A
Environmental resistance
B
Carrying capacity
C
Biotic potential
D
None
Solution
(C) Biotic potential refers to the maximum reproductive capacity of an organism under optimal environmental conditions. It represents the inherent ability of a population to increase in number when resources are unlimited and environmental resistance is absent.
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82
BiologyEasyMCQAIPMT · 2002
Life does not exist on the moon due to the absence of .......
A
$O_2$
B
Water
C
Light
D
Temperature
Solution
(B) Life as we know it requires specific conditions to exist,primarily the presence of liquid water,an atmosphere containing oxygen,and a suitable temperature range.
On the moon,there is no atmosphere to hold gases or regulate temperature,and there is no liquid water available on the surface.
Among the given options,the absence of water is the most fundamental reason why life cannot be sustained on the moon,as water is the universal solvent and essential for all biological processes.
On the moon,there is no atmosphere to hold gases or regulate temperature,and there is no liquid water available on the surface.
Among the given options,the absence of water is the most fundamental reason why life cannot be sustained on the moon,as water is the universal solvent and essential for all biological processes.
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83
BiologyMediumMCQAIPMT · 2002
Which fungal disease is transmitted through seeds and flowers?
A
Loose smut of wheat
B
Maize stunt
C
Covered smut of barley
D
Soft rot of potato
Solution
(A) The loose smut of wheat,caused by the fungus $Ustilago$ $tritici$,is a seed-borne disease. The fungus infects the flowers of the wheat plant during the flowering stage. The spores are carried by the wind to the stigma of healthy flowers,where they germinate and infect the developing ovary. The fungus then remains dormant within the seed until it is planted,at which point it grows systemically with the plant.
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84
BiologyMediumMCQAIPMT · 2002
During the production of bread,it becomes porous due to the release of $CO_2$ by the action of:
A
Yeast
B
Bacteria
C
Virus
D
Protozoans
Solution
(A) Bread making involves the process of fermentation.
$Yeast$ (Saccharomyces cerevisiae) is added to the dough.
During fermentation,yeast breaks down the sugars present in the flour into ethanol and $CO_2$.
The release of $CO_2$ gas creates bubbles in the dough,which makes the bread porous and soft.
$Yeast$ (Saccharomyces cerevisiae) is added to the dough.
During fermentation,yeast breaks down the sugars present in the flour into ethanol and $CO_2$.
The release of $CO_2$ gas creates bubbles in the dough,which makes the bread porous and soft.
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85
BiologyMediumMCQAIPMT · 2002
Which of the following is present in all angiosperms without exception?
A
Presence of vessels
B
Double fertilization
C
Secondary growth
D
Autotrophic nutrition
Solution
(B) Double fertilization is a unique and defining characteristic of angiosperms (flowering plants). It involves the fusion of one male gamete with the egg cell to form a zygote (syngamy) and the fusion of another male gamete with the two polar nuclei to form the primary endosperm nucleus (triple fusion). This process is universal in all angiosperms. Vessels are absent in some primitive angiosperms,secondary growth is absent in monocots,and while most angiosperms are autotrophic,some are parasitic (e.g.,Cuscuta),making double fertilization the only exceptionless feature among the choices.
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86
BiologyMediumMCQAIPMT · 2002
Which steroid is used for transformation?
A
Cortisol
B
Cholesterol
C
Testosterone
D
Progesterone
Solution
(B) In the context of industrial steroid production and microbial transformation,$Cholesterol$ is a fundamental precursor. Many steroid hormones are synthesized from $Cholesterol$ through various enzymatic pathways. Therefore,$Cholesterol$ is the primary steroid used as a starting material for the transformation into other steroid hormones.
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87
BiologyMediumMCQAIPMT · 2002
Which types of hormones regulate the menstrual cycle in humans?
A
$FSH$
B
$LH$
C
$FSH, LH,$ and estrogen
D
Progesterone
Solution
(C) The menstrual cycle in human females is regulated by a complex interaction of hormones.
$1$. The hypothalamus releases Gonadotropin-Releasing Hormone $(GnRH)$,which stimulates the anterior pituitary gland.
$2$. The anterior pituitary gland secretes Follicle-Stimulating Hormone $(FSH)$ and Luteinizing Hormone $(LH)$.
$3$. $FSH$ stimulates the growth of ovarian follicles,which in turn secrete estrogen.
$4$. $LH$ triggers ovulation and the formation of the corpus luteum,which secretes progesterone.
$5$. Therefore,the menstrual cycle is regulated by the combined action of $FSH, LH,$ estrogen,and progesterone. Among the given options,$FSH, LH,$ and estrogen are the primary regulators mentioned.
$1$. The hypothalamus releases Gonadotropin-Releasing Hormone $(GnRH)$,which stimulates the anterior pituitary gland.
$2$. The anterior pituitary gland secretes Follicle-Stimulating Hormone $(FSH)$ and Luteinizing Hormone $(LH)$.
$3$. $FSH$ stimulates the growth of ovarian follicles,which in turn secrete estrogen.
$4$. $LH$ triggers ovulation and the formation of the corpus luteum,which secretes progesterone.
$5$. Therefore,the menstrual cycle is regulated by the combined action of $FSH, LH,$ estrogen,and progesterone. Among the given options,$FSH, LH,$ and estrogen are the primary regulators mentioned.
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88
BiologyEasyMCQAIPMT · 2002
In angiosperms,all four microspores of a microspore tetrad are covered by a layer which is made up of ........
A
Octocellulose
B
Callose
C
Cellulose
D
Sporopollenin
Solution
(B) In angiosperms,the microspore mother cell $(MMC)$ undergoes meiosis to form a cluster of four haploid microspores,known as a microspore tetrad.
These four microspores are held together by a special layer of polysaccharide called $Callose$.
As the anther matures and dehydrates,the enzyme $Callase$ is secreted by the tapetum,which degrades the $Callose$ layer,allowing the microspores to dissociate and develop into pollen grains.
These four microspores are held together by a special layer of polysaccharide called $Callose$.
As the anther matures and dehydrates,the enzyme $Callase$ is secreted by the tapetum,which degrades the $Callose$ layer,allowing the microspores to dissociate and develop into pollen grains.
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89
BiologyEasyMCQAIPMT · 2002
In an anatropous ovule,what is the orientation of the micropyle?
A
Upward
B
Downward
C
Right side
D
Left side
Solution
(B) An anatropous ovule is the most common type of ovule in angiosperms. In this type,the body of the ovule becomes completely inverted during development,such that the micropyle comes to lie very close to the funicle (stalk). Due to this inversion,the micropyle is directed downwards towards the placenta.
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90
BiologyEasyMCQAIPMT · 2002
In angiosperms,where does the pollen tube release its male gametes?
A
Central cell
B
Antipodal cell
C
Egg cell
D
Synergid cell
Solution
(D) In angiosperms,the pollen tube enters the ovule,usually through the micropyle,and then enters one of the two synergids through the filiform apparatus.
Once inside the synergid,the pollen tube releases the two male gametes into the cytoplasm of the synergid.
From there,one male gamete moves towards the egg cell for syngamy,and the other moves towards the central cell for triple fusion.
Once inside the synergid,the pollen tube releases the two male gametes into the cytoplasm of the synergid.
From there,one male gamete moves towards the egg cell for syngamy,and the other moves towards the central cell for triple fusion.
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91
BiologyMediumMCQAIPMT · 2002
What type of relationship is observed between insect-pollinated flowers and their pollinators?
A
Mutualism
B
Commensalism
C
Cooperation
D
Co-evolution
Solution
(D) The relationship between insect-pollinated flowers and their pollinators is a classic example of $Mutualism$ (a type of positive interaction where both species benefit).
However,in the context of evolutionary biology,this relationship often leads to $Co-evolution$.
$Co-evolution$ refers to the process where two or more species reciprocally affect each other's evolution.
In the case of flowers and pollinators,the flowers evolve specific traits (like shape,color,and scent) to attract specific pollinators,while the pollinators evolve specific morphological traits (like proboscis length) to access the nectar,ensuring a tight evolutionary link.
However,in the context of evolutionary biology,this relationship often leads to $Co-evolution$.
$Co-evolution$ refers to the process where two or more species reciprocally affect each other's evolution.
In the case of flowers and pollinators,the flowers evolve specific traits (like shape,color,and scent) to attract specific pollinators,while the pollinators evolve specific morphological traits (like proboscis length) to access the nectar,ensuring a tight evolutionary link.
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92
BiologyMediumMCQAIPMT · 2002
Which of the following statements is true regarding cleavage?
A
The size of the embryo increases.
B
The size of the cells decreases.
C
The size of the cells increases.
D
The size of the embryo decreases.
Solution
(B) Cleavage is a series of rapid mitotic cell divisions of the zygote following fertilization.
During cleavage,the total volume of the embryo remains constant because there is no growth phase ($G_1$ or $G_2$ phases) between the divisions.
As the number of cells (blastomeres) increases,the size of each individual cell decreases significantly with each successive division.
Therefore,the correct statement is that the size of the cells decreases.
During cleavage,the total volume of the embryo remains constant because there is no growth phase ($G_1$ or $G_2$ phases) between the divisions.
As the number of cells (blastomeres) increases,the size of each individual cell decreases significantly with each successive division.
Therefore,the correct statement is that the size of the cells decreases.
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93
BiologyEasyMCQAIPMT · 2002
The unchecked reproductive capacity in a population is called ..........
A
Biotic potential
B
Fecundity
C
Carrying capacity
D
Natality
Solution
(A) Biotic potential refers to the maximum reproductive capacity of an organism under optimal environmental conditions,where resources are unlimited and there are no environmental resistances. It represents the inherent ability of a population to increase in size when conditions are ideal.
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94
BiologyMediumMCQAIPMT · 2002
$A$ factor (allele) is dominant if it is expressed in ...............
A
both homozygous and heterozygous conditions
B
the second generation
C
heterozygous condition only
D
homozygous condition only
Solution
(A) According to the law of dominance,a dominant allele is one that expresses its phenotypic effect in both the homozygous $(AA)$ and heterozygous $(Aa)$ conditions.
In the heterozygous condition,the dominant allele masks the expression of the recessive allele.
Therefore,the correct option is $A$.
In the heterozygous condition,the dominant allele masks the expression of the recessive allele.
Therefore,the correct option is $A$.
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95
BiologyDifficultMCQAIPMT · 2002
What is the cause of $Down's$ syndrome?
A
Crossing over
B
Linkage
C
Sex-linked inheritance
D
Non-disjunction of chromosomes
Solution
(D) $Down's$ syndrome is a chromosomal disorder caused by the presence of an additional copy of chromosome number $21$ (trisomy of $21$).
This condition arises due to the failure of homologous chromosomes or sister chromatids to separate properly during meiosis,a phenomenon known as non-disjunction.
As a result,one gamete receives an extra chromosome,and upon fertilization,the zygote develops with $47$ chromosomes instead of the normal $46$.
This condition arises due to the failure of homologous chromosomes or sister chromatids to separate properly during meiosis,a phenomenon known as non-disjunction.
As a result,one gamete receives an extra chromosome,and upon fertilization,the zygote develops with $47$ chromosomes instead of the normal $46$.
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96
BiologyMediumMCQAIPMT · 2002
$A$ man affected by a certain disease marries a normal woman. They have $8$ children ($3$ daughters and $5$ sons). All the daughters suffer from their father's disease,but none of the sons are affected. What type of inheritance can you suggest for this disease?
A
Sex-linked recessive
B
Sex-linked dominant
C
Autosomal dominant
D
Sex-limited recessive
Solution
(B) In this scenario,the father is affected and passes the trait to all his daughters but none of his sons.
Since the father passes his $X$ chromosome to all his daughters and his $Y$ chromosome to all his sons,the trait must be located on the $X$ chromosome.
Because all daughters are affected,the trait must be dominant,as a single copy of the dominant allele on the $X$ chromosome inherited from the father is sufficient to express the disease in the daughters.
If it were recessive,the daughters would be carriers (assuming the mother is normal),but they would not necessarily show the disease unless the mother was also a carrier. Thus,the pattern of inheritance is $X$-linked dominant.
Since the father passes his $X$ chromosome to all his daughters and his $Y$ chromosome to all his sons,the trait must be located on the $X$ chromosome.
Because all daughters are affected,the trait must be dominant,as a single copy of the dominant allele on the $X$ chromosome inherited from the father is sufficient to express the disease in the daughters.
If it were recessive,the daughters would be carriers (assuming the mother is normal),but they would not necessarily show the disease unless the mother was also a carrier. Thus,the pattern of inheritance is $X$-linked dominant.
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97
BiologyMediumMCQAIPMT · 2002
In a plant,the genotype of the $F_1$ generation is $AABbCC$. Upon self-pollination of this plant,the phenotypic ratio in the $F_2$ generation will be:
A
$3 : 1$
B
$1 : 1$
C
$9 : 3 : 3 : 1$
D
$27 : 9 : 9 : 9 : 3 : 3 : 3 : 1$
Solution
(A) The given genotype is $AABbCC$.
Self-pollination means the cross is $AABbCC \times AABbCC$.
We can analyze the inheritance of each gene pair independently:
$1$. For gene $A$: $AA \times AA$ results in all $AA$ offspring (monomorphic).
$2$. For gene $B$: $Bb \times Bb$ results in a phenotypic ratio of $3:1$ (dominant:recessive).
$3$. For gene $C$: $CC \times CC$ results in all $CC$ offspring (monomorphic).
Since genes $A$ and $C$ are homozygous,they do not contribute to phenotypic variation.
Therefore,the overall phenotypic ratio is determined solely by the $Bb$ gene pair,which is $3:1$.
Self-pollination means the cross is $AABbCC \times AABbCC$.
We can analyze the inheritance of each gene pair independently:
$1$. For gene $A$: $AA \times AA$ results in all $AA$ offspring (monomorphic).
$2$. For gene $B$: $Bb \times Bb$ results in a phenotypic ratio of $3:1$ (dominant:recessive).
$3$. For gene $C$: $CC \times CC$ results in all $CC$ offspring (monomorphic).
Since genes $A$ and $C$ are homozygous,they do not contribute to phenotypic variation.
Therefore,the overall phenotypic ratio is determined solely by the $Bb$ gene pair,which is $3:1$.
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98
BiologyMediumMCQAIPMT · 2002
What is a change in the nucleotide sequence of $DNA$ called?
A
Mutagen
B
Mutation
C
Recombination
D
Translation
Solution
(B) change in the nucleotide sequence of $DNA$ is known as $Mutation$.
$Mutation$ is a phenomenon which results in alteration of $DNA$ sequences and consequently results in changes in the genotype and the phenotype of an organism.
$Mutagens$ are the agents that induce mutations.
$Recombination$ is the process of forming new combinations of alleles.
$Translation$ is the process of protein synthesis from $mRNA$.
$Mutation$ is a phenomenon which results in alteration of $DNA$ sequences and consequently results in changes in the genotype and the phenotype of an organism.
$Mutagens$ are the agents that induce mutations.
$Recombination$ is the process of forming new combinations of alleles.
$Translation$ is the process of protein synthesis from $mRNA$.
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99
BiologyMediumMCQAIPMT · 2002
$A$ pleiotropic gene is responsible for which of the following conditions?
A
Hemophilia
B
Thalassemia
C
Sickle cell anemia
D
Color blindness
Solution
(C) Pleiotropy occurs when a single gene influences two or more seemingly unrelated phenotypic traits. In the case of $Sickle \ cell \ anemia$, the mutated gene $(Hb^S)$ affects the structure of hemoglobin, which leads to multiple physiological effects, including the sickling of red blood cells, anemia, and resistance to malaria. Therefore, it is a classic example of a pleiotropic gene.
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100
BiologyMediumMCQAIPMT · 2002
Which of the following is not an example of a sexually transmitted disease $(STD)$?
A
$AIDS$
B
Color blindness
C
Syphilis
D
Gonorrhea
Solution
(B) Sexually transmitted diseases $(STDs)$ are infections that are spread primarily through sexual contact.
$AIDS$ (Acquired Immuno Deficiency Syndrome) is caused by $HIV$ and is a well-known sexually transmitted disease.
Syphilis is a bacterial infection caused by $Treponema pallidum$ and is transmitted sexually.
Gonorrhea is a bacterial infection caused by $Neisseria gonorrhoeae$ and is transmitted sexually.
Color blindness is a genetic disorder (specifically an $X$-linked recessive trait) and is not a sexually transmitted disease.
Therefore, the correct answer is Color blindness.
$AIDS$ (Acquired Immuno Deficiency Syndrome) is caused by $HIV$ and is a well-known sexually transmitted disease.
Syphilis is a bacterial infection caused by $Treponema pallidum$ and is transmitted sexually.
Gonorrhea is a bacterial infection caused by $Neisseria gonorrhoeae$ and is transmitted sexually.
Color blindness is a genetic disorder (specifically an $X$-linked recessive trait) and is not a sexually transmitted disease.
Therefore, the correct answer is Color blindness.
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