BiologyQ1–100 of 132 questions
Page 1 of 2 · English
1
BiologyMediumMCQAIPMT · 2003
Species are considered as
A
Real units of classification devised by taxonomists
B
Real basic units of classification
C
The lowest units of classification
D
Artificial concept of human mind which cannot be defined in absolute terms
Solution
(B) In biological classification,the species is considered the fundamental or basic unit of classification.
It represents a group of organisms that can interbreed and produce fertile offspring.
While other taxonomic categories like genus,family,etc.,are often considered artificial constructs created by taxonomists for convenience,the species is recognized as a 'real' biological entity in nature.
Therefore,species are considered the real basic units of classification.
It represents a group of organisms that can interbreed and produce fertile offspring.
While other taxonomic categories like genus,family,etc.,are often considered artificial constructs created by taxonomists for convenience,the species is recognized as a 'real' biological entity in nature.
Therefore,species are considered the real basic units of classification.
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2
BiologyMediumMCQAIPMT · 2003
Biosystematics aims at
A
Identification and arrangement of organisms on the basis of their cytological characteristics
B
The classification of organisms based on broad morphological characters
C
Delimiting various taxa of organisms and establishing their relationships
D
The classification of organisms based on their evolutionary history and establishing their phylogeny on the totality of various parameters from all fields of studies
Solution
(D) Biosystematics is a branch of taxonomy that aims to classify organisms based on their evolutionary history and phylogeny.
It considers the totality of various parameters,including morphological,cytological,genetic,ecological,and biochemical data from all fields of study to establish the relationships between organisms.
Therefore,option $D$ is the most comprehensive and accurate definition.
It considers the totality of various parameters,including morphological,cytological,genetic,ecological,and biochemical data from all fields of study to establish the relationships between organisms.
Therefore,option $D$ is the most comprehensive and accurate definition.
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3
BiologyMediumMCQAIPMT · 2003
Sycon belongs to a group of animals which are best described as
A
Unicellular or acellular
B
Multicellular without any tissue organisation
C
Multicellular with a gastrovascular cavity
D
Multicellular having tissue organisation,but no body cavity
Solution
(B) $Sycon$ belongs to the phylum $Porifera$,commonly known as sponges.
These organisms are multicellular but exhibit a cellular level of organization,meaning they lack true tissues or organs.
Their body plan is characterized by a water transport or canal system,which is distinct from a gastrovascular cavity.
Therefore,they are best described as multicellular organisms without any tissue organization.
These organisms are multicellular but exhibit a cellular level of organization,meaning they lack true tissues or organs.
Their body plan is characterized by a water transport or canal system,which is distinct from a gastrovascular cavity.
Therefore,they are best described as multicellular organisms without any tissue organization.
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4
BiologyEasyMCQAIPMT · 2003
Ommatidia serve the purpose of photoreception in
A
Sunflower
B
Cockroach
C
Frog
D
Humans
Solution
(B) . Arthropods possess compound eyes. Each compound eye is composed of numerous independent visual units called ommatidia. These structures are responsible for photoreception in insects like the cockroach.
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5
BiologyMediumMCQAIPMT · 2003
Which one of the following is a matching pair of an animal and a certain phenomenon it exhibits?
A
$Taenia$ - Polymorphism
B
$Pheretima$ - Sexual dimorphism
C
$Musca$ - Complete metamorphosis
D
$Chamaeleon$ - Parthenogenesis
Solution
(C) $Musca$ $domestica$ (housefly) exhibits complete metamorphosis,which involves four distinct stages: egg,larva,pupa,and adult. This process is known as holometabolous development. Therefore,option $C$ is the correct matching pair.
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6
BiologyMediumMCQAIPMT · 2003
During its life cycle,$Fasciola$ $hepatica$ $(Liver$ $Fluke)$ infects its intermediate host and primary host at the following larval stages respectively:
A
Redia and miracidium
B
Cercaria and redia
C
Metacercaria and cercaria
D
Miracidium and metacercaria
Solution
(D) The life cycle of $Fasciola$ $hepatica$ involves two hosts.
$1$. The intermediate host is a snail (e.g.,$Lymnaea$),which is infected by the $Miracidium$ larva.
$2$. The primary host is a sheep or human,which is infected by the $Metacercaria$ larva (encysted stage) found on aquatic vegetation.
Therefore,the intermediate host is infected by the $Miracidium$ stage,and the primary host is infected by the $Metacercaria$ stage.
$1$. The intermediate host is a snail (e.g.,$Lymnaea$),which is infected by the $Miracidium$ larva.
$2$. The primary host is a sheep or human,which is infected by the $Metacercaria$ larva (encysted stage) found on aquatic vegetation.
Therefore,the intermediate host is infected by the $Miracidium$ stage,and the primary host is infected by the $Metacercaria$ stage.
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7
BiologyMediumMCQAIPMT · 2003
Given below are four matchings of an animal and its kind of respiratory organ:
$1.$ Silver fish - trachea
$2.$ Scorpion - book lung
$3.$ Sea squirt - pharyngeal gills
$4.$ Dolphin - skin
Choose the correct option.
$1.$ Silver fish - trachea
$2.$ Scorpion - book lung
$3.$ Sea squirt - pharyngeal gills
$4.$ Dolphin - skin
Choose the correct option.
A
$3$ and $4$
B
$1$ and $4$
C
$1, 2$ and $3$
D
$2$ and $4$
Solution
(C) The correct option is $C$.
$1.$ Silver fish belongs to the class Insecta,which uses trachea for respiration.
$2.$ Scorpion belongs to the class Arachnida,which uses book lungs for respiration.
$3.$ Sea squirt (Herdmania) belongs to the subphylum Urochordata,which uses pharyngeal gills for respiration.
$4.$ Dolphin is a mammal,and it breathes through lungs,not skin. Therefore,the matching for the dolphin is incorrect.
$1.$ Silver fish belongs to the class Insecta,which uses trachea for respiration.
$2.$ Scorpion belongs to the class Arachnida,which uses book lungs for respiration.
$3.$ Sea squirt (Herdmania) belongs to the subphylum Urochordata,which uses pharyngeal gills for respiration.
$4.$ Dolphin is a mammal,and it breathes through lungs,not skin. Therefore,the matching for the dolphin is incorrect.
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8
BiologyMediumMCQAIPMT · 2003
Sycon belongs to a group of animals,which are best described as
A
Multicellular having tissue organization,but not body cavity
B
Unicellular or acellular
C
Multicellular without any tissue organization
D
Multicellular with a gastrovascular system
Solution
(C) $Sycon$ belongs to the phylum $Porifera$ (sponges).
These organisms are multicellular but do not possess true tissues or organs.
They exhibit a cellular level of organization,where cells are arranged as loose cell aggregates.
Therefore,they are best described as multicellular without any tissue organization.
These organisms are multicellular but do not possess true tissues or organs.
They exhibit a cellular level of organization,where cells are arranged as loose cell aggregates.
Therefore,they are best described as multicellular without any tissue organization.
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9
BiologyMediumMCQAIPMT · 2003
Which one of the following contains the largest quantity of extracellular material?
A
Myelinated nerve fibres
B
Striated muscle
C
Areolar tissue
D
Stratified epithelium
Solution
(C) The correct answer is $C$.
Areolar tissue is a type of loose connective tissue.
It contains a large amount of extracellular matrix (ground substance) along with various types of cells (like fibroblasts,mast cells,and macrophages) and protein fibres (collagen and elastin).
In contrast,tissues like muscle and epithelium are primarily composed of cells packed closely together with very little extracellular material.
Areolar tissue is a type of loose connective tissue.
It contains a large amount of extracellular matrix (ground substance) along with various types of cells (like fibroblasts,mast cells,and macrophages) and protein fibres (collagen and elastin).
In contrast,tissues like muscle and epithelium are primarily composed of cells packed closely together with very little extracellular material.
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10
BiologyEasyMCQAIPMT · 2003
Nissl's granules are found in the cyton of nerve cells. These have an affinity for basic dyes. The granules are made up of:
A
Mitochondria
B
Cell metabolites
C
Fat granules
D
Ribosomes
Solution
(D) Nissl's granules are characteristic features of the nerve cell body (cyton) and dendrites.
They are composed of rough endoplasmic reticulum $(RER)$ and free ribosomes.
Because they contain a high concentration of ribosomal $RNA$ $(rRNA)$,which is acidic,they show a strong affinity for basic dyes (basophilic).
Therefore,the correct answer is Ribosomes.
They are composed of rough endoplasmic reticulum $(RER)$ and free ribosomes.
Because they contain a high concentration of ribosomal $RNA$ $(rRNA)$,which is acidic,they show a strong affinity for basic dyes (basophilic).
Therefore,the correct answer is Ribosomes.
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11
BiologyMediumMCQAIPMT · 2003
Stomata of $CAM$ plants:
A
Never open
B
Are always open
C
Open during the day and close at night
D
Open during the night and close during the day
Solution
(D) $CAM$ (Crassulacean Acid Metabolism) plants exhibit a unique adaptation to arid environments.
They possess scotoactive stomata,which means their stomata open during the night to take in $CO_2$ and close during the day to minimize water loss through transpiration.
This mechanism allows them to survive in extremely dry conditions.
They possess scotoactive stomata,which means their stomata open during the night to take in $CO_2$ and close during the day to minimize water loss through transpiration.
This mechanism allows them to survive in extremely dry conditions.
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12
BiologyEasyMCQAIPMT · 2003
Which element forms part of the structure of the chlorophyll molecule?
A
$Fe$
B
$Mg$
C
$K$
D
$Mn$
Solution
(B) The chlorophyll molecule consists of a porphyrin ring (head) and a phytol tail.
At the center of the porphyrin ring,a magnesium ion $(Mg^{2+})$ is coordinated with four nitrogen atoms.
Therefore,$Mg$ is an essential structural component of the chlorophyll molecule.
At the center of the porphyrin ring,a magnesium ion $(Mg^{2+})$ is coordinated with four nitrogen atoms.
Therefore,$Mg$ is an essential structural component of the chlorophyll molecule.
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13
BiologyEasyMCQAIPMT · 2003
The major portion of the dry weight of plants comprises of
A
Carbon,hydrogen and oxygen
B
Nitrogen,Phosphorus and potassium
C
Calcium,magnesium and sulphur
D
Carbon,nitrogen and hydrogen
Solution
(A) Carbon,hydrogen,and oxygen are the primary structural components of organic molecules in plants,such as carbohydrates,proteins,lipids,and nucleic acids.
These three elements together constitute approximately $94\%$ of the total dry weight of a plant.
Therefore,the correct option is $A$.
These three elements together constitute approximately $94\%$ of the total dry weight of a plant.
Therefore,the correct option is $A$.
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14
BiologyEasyMCQAIPMT · 2003
Which of the following is a widely used metal cofactor?
A
$Ca^{2+}$
B
$Al^{3+}$
C
$Ni^{2+}$
D
$Mg^{2+}$
Solution
(C) The correct answer is $C$.
Many enzymes require metal ions as cofactors to function efficiently.
These metal ions act as activators or structural components of the enzyme.
$Ni^{2+}$ is a well-known metal cofactor,for example,it is an essential component of the enzyme urease.
Note: While $Mg^{2+}$ is also a very common cofactor,in the context of specific textbook examples often provided for metal cofactors,$Ni^{2+}$ is frequently cited.
Many enzymes require metal ions as cofactors to function efficiently.
These metal ions act as activators or structural components of the enzyme.
$Ni^{2+}$ is a well-known metal cofactor,for example,it is an essential component of the enzyme urease.
Note: While $Mg^{2+}$ is also a very common cofactor,in the context of specific textbook examples often provided for metal cofactors,$Ni^{2+}$ is frequently cited.
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15
BiologyEasyMCQAIPMT · 2003
The important contribution of molybdenum is in:
A
Flower growth
B
Nitrogen fixation
C
Chromosome condensation
D
Carbon fixation
Solution
(B) Molybdenum is an essential micronutrient for plants.
It is a structural component of the enzyme nitrogenase,which is responsible for the biological nitrogen fixation process.
Nitrogenase catalyzes the conversion of atmospheric nitrogen $(N_2)$ into ammonia $(NH_3)$.
Therefore,molybdenum plays a critical role in nitrogen metabolism and nitrogen fixation in plants.
It is a structural component of the enzyme nitrogenase,which is responsible for the biological nitrogen fixation process.
Nitrogenase catalyzes the conversion of atmospheric nitrogen $(N_2)$ into ammonia $(NH_3)$.
Therefore,molybdenum plays a critical role in nitrogen metabolism and nitrogen fixation in plants.
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16
BiologyEasyMCQAIPMT · 2003
Gray speck disease in oats takes place by the deficiency of
A
Zinc
B
Copper
C
Potassium
D
Manganese
Solution
(D) The "Gray speck disease" in oats is caused by the deficiency of the micronutrient manganese $(Mn)$.
This deficiency leads to the development of gray-brown spots on the leaves, which can eventually cause the total failure of the crop.
This deficiency leads to the development of gray-brown spots on the leaves, which can eventually cause the total failure of the crop.
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17
BiologyEasyMCQAIPMT · 2003
Boron in green plants assists in
A
Sugar transport
B
Activation of enzymes
C
Acting as enzyme cofactor
D
Photosynthesis
Solution
(A) Boron is absorbed by plants in the form of $BO_3^{3-}$ or $B_4O_7^{2-}$ ions.
It plays a crucial role in the translocation of carbohydrates (sugars) across the plant body.
It is also involved in pollen germination,cell elongation,cell differentiation,and carbohydrate metabolism.
Therefore,the primary function listed among the options is sugar transport.
It plays a crucial role in the translocation of carbohydrates (sugars) across the plant body.
It is also involved in pollen germination,cell elongation,cell differentiation,and carbohydrate metabolism.
Therefore,the primary function listed among the options is sugar transport.
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18
BiologyEasyMCQAIPMT · 2003
Which of the following trace elements is essential for auxin synthesis in plants?
A
Molybdenum
B
Chlorine
C
Zinc
D
Boron
Solution
(C) Zinc is a crucial micronutrient required by plants for various physiological processes.
One of its most significant roles is in the biosynthesis of the plant growth hormone auxin,specifically Indole$-3-$Acetic Acid $(IAA)$.
Zinc acts as a cofactor for enzymes involved in the synthesis of tryptophan,which is the primary precursor amino acid for $IAA$ production.
One of its most significant roles is in the biosynthesis of the plant growth hormone auxin,specifically Indole$-3-$Acetic Acid $(IAA)$.
Zinc acts as a cofactor for enzymes involved in the synthesis of tryptophan,which is the primary precursor amino acid for $IAA$ production.
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19
BiologyMediumMCQAIPMT · 2003
Which one of the following is wrong in relation to photorespiration?
A
It is a characteristic of $C_3$ plants.
B
It occurs in chloroplasts.
C
It occurs in day time only.
D
It is a characteristic of $C_4$ plants.
Solution
(D) Photorespiration is a process that occurs in $C_3$ plants,not $C_4$ plants. In $C_4$ plants,photorespiration does not occur because they have a specialized leaf anatomy known as $Kranz$ anatomy,which increases the concentration of $CO_2$ around the enzyme $RuBisCO$,thereby minimizing the oxygenase activity of the enzyme. Therefore,the statement that it is a characteristic of $C_4$ plants is incorrect.
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20
BiologyMediumMCQAIPMT · 2003
In which one of the following do the two names refer to one and the same thing?
A
Tricarboxylic acid cycle and urea cycle
B
Kreb's cycle and Calvin cycle
C
Tricarboxylic acid cycle and citric acid cycle
D
Citric acid cycle and Calvin cycle
Solution
(C) The $Tricarboxylic$ $acid$ $cycle$ ($TCA$ $cycle$) is another name for the $Citric$ $acid$ $cycle$. Both terms refer to the same metabolic pathway that occurs in the mitochondrial matrix,where acetyl-$CoA$ is oxidized to produce $CO_2$,$ATP$,$NADH$,and $FADH_2$. This cycle is also commonly known as the $Krebs$ $cycle$ in honor of Sir Hans Krebs.
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21
BiologyMediumMCQAIPMT · 2003
In alcohol fermentation,
A
Oxygen is the electron acceptor
B
Triose phosphate is the electron donor while acetaldehyde is the electron acceptor
C
Triose phosphate is the electron donor while pyruvic acid is the electron acceptor
D
There is no electron donor
Solution
(B) In alcohol fermentation,glucose is broken down into $2$ molecules of pyruvic acid through glycolysis,producing $NADH + H^+$.
In the subsequent steps,pyruvic acid is converted into acetaldehyde,which then acts as the final electron acceptor.
Acetaldehyde accepts electrons from $NADH + H^+$ (which originated from the oxidation of triose phosphate) to be reduced into ethanol.
Therefore,triose phosphate acts as the electron donor (indirectly via $NADH$) and acetaldehyde acts as the electron acceptor.
In the subsequent steps,pyruvic acid is converted into acetaldehyde,which then acts as the final electron acceptor.
Acetaldehyde accepts electrons from $NADH + H^+$ (which originated from the oxidation of triose phosphate) to be reduced into ethanol.
Therefore,triose phosphate acts as the electron donor (indirectly via $NADH$) and acetaldehyde acts as the electron acceptor.
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22
BiologyMediumMCQAIPMT · 2003
Which of the following is a coconut milk factor?
A
Auxin
B
Cytokinin
C
Morphactin
D
None of the above
Solution
(B) The coconut milk factor is a substance that promotes cell division. It was identified as a cytokinin,specifically zeatin,which is present in the liquid endosperm of coconut. Therefore,the correct option is $B$.
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23
BiologyMediumMCQAIPMT · 2003
During prolonged fasting,which of the following is the correct order of utilization of energy sources in the human body?
A
The first to be used up are carbohydrates,next fat is withdrawn and proteins are metabolised at the last.
B
The first to be used up are the fats,next carbohydrates are withdrawn from stored glycogen in the liver and muscles and proteins are withdrawn at the last.
C
First lipids are used up,then proteins and finally carbohydrate.
D
None of these.
Solution
(A) During prolonged fasting,the body follows a specific hierarchy to maintain energy homeostasis.
$1$. First,the body utilizes readily available blood glucose and glycogen stores (carbohydrates) from the liver and muscles.
$2$. Once glycogen stores are depleted,the body shifts to mobilizing adipose tissue to break down stored fats (lipids) into fatty acids and glycerol for energy.
$3$. Proteins are the last resort for energy production,as they are essential structural components of the body. They are broken down into amino acids through gluconeogenesis only when other energy reserves are exhausted.
$1$. First,the body utilizes readily available blood glucose and glycogen stores (carbohydrates) from the liver and muscles.
$2$. Once glycogen stores are depleted,the body shifts to mobilizing adipose tissue to break down stored fats (lipids) into fatty acids and glycerol for energy.
$3$. Proteins are the last resort for energy production,as they are essential structural components of the body. They are broken down into amino acids through gluconeogenesis only when other energy reserves are exhausted.
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24
BiologyEasyMCQAIPMT · 2003
Which one of the following pairs is not correctly matched?
A
Vitamin $B_6$ - Beri-beri
B
Vitamin $C$ - Scurvy
C
Vitamin $B_5$ - Pellagra
D
Vitamin $B_{12}$ - Pernicious anaemia
Solution
(A) . Beri-beri is caused by the deficiency of Vitamin $B_1$ (thiamine).
Vitamin $B_6$ (pyridoxine) deficiency typically leads to skin conditions,irritability,and anaemia,not Beri-beri.
Therefore,the pair Vitamin $B_6$ - Beri-beri is incorrectly matched.
Vitamin $B_6$ (pyridoxine) deficiency typically leads to skin conditions,irritability,and anaemia,not Beri-beri.
Therefore,the pair Vitamin $B_6$ - Beri-beri is incorrectly matched.
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25
BiologyMediumMCQAIPMT · 2003
Systemic heart refers to
A
The two ventricles together in humans
B
The heart that contracts under stimulation from the nervous system
C
Left auricle and left ventricle in higher vertebrates
D
Entire heart in lower vertebrates
Solution
(C) The systemic heart refers to the left side of the heart in higher vertebrates (mammals and birds).
It consists of the left atrium (auricle) and the left ventricle.
This part of the heart receives oxygenated blood from the lungs and pumps it to the entire body through the systemic circulation.
It consists of the left atrium (auricle) and the left ventricle.
This part of the heart receives oxygenated blood from the lungs and pumps it to the entire body through the systemic circulation.
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26
BiologyMediumMCQAIPMT · 2003
Bundle of His is a network of
A
Nerve fibres found throughout the heart
B
Muscle fibres distributed throughout the heart walls
C
Muscle fibres found only in the ventricle wall
D
Nerve fibres distributed in ventricles
Solution
(C) The Bundle of His (also known as the atrioventricular bundle) is a specialized collection of modified cardiac muscle fibres.
These fibres originate from the atrioventricular node $(AVN)$ and pass through the interventricular septum.
They are responsible for conducting electrical impulses from the atria to the ventricles,ensuring coordinated contraction of the ventricular walls.
Therefore,they are muscle fibres located specifically within the ventricular region.
These fibres originate from the atrioventricular node $(AVN)$ and pass through the interventricular septum.
They are responsible for conducting electrical impulses from the atria to the ventricles,ensuring coordinated contraction of the ventricular walls.
Therefore,they are muscle fibres located specifically within the ventricular region.
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27
BiologyMediumMCQAIPMT · 2003
If Henle's loop were absent from mammalian nephron,which of the following is to be expected?
A
The urine will be more dilute.
B
There will be no urine formation.
C
There will be hardly any change in the quality and quantity of urine formed.
D
The urine will be more concentrated.
Solution
(A) The primary function of the loop of Henle is the reabsorption of water and electrolytes,which helps in the concentration of urine.
In the absence of the loop of Henle,the counter-current mechanism cannot function effectively.
Consequently,water cannot be reabsorbed from the filtrate in the collecting duct,leading to the excretion of highly dilute urine.
Therefore,the correct expectation is that the urine will be more dilute.
In the absence of the loop of Henle,the counter-current mechanism cannot function effectively.
Consequently,water cannot be reabsorbed from the filtrate in the collecting duct,leading to the excretion of highly dilute urine.
Therefore,the correct expectation is that the urine will be more dilute.
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28
BiologyMediumMCQAIPMT · 2003
Which of the following plants are used as green manure in crop fields and in sandy soils?
A
Dicanthium annulatum and Azolla pinnata
B
Crotalaria juncea and Alhagi camelorum
C
Calotropis procera and Phyllanthus niruri
D
Saccharum munja and Lantana camara
Solution
(B) Green manure consists of plants that are grown and then incorporated into the soil to improve its fertility and structure.
$Crotalaria juncea$ (Sunn hemp) is a well-known leguminous plant widely used as green manure due to its nitrogen-fixing ability.
$Alhagi camelorum$ (Camel thorn) is also used as a green manure,particularly in sandy soils,as it helps in improving soil organic matter and moisture retention.
Therefore,the correct pair is $Crotalaria juncea$ and $Alhagi camelorum$.
$Crotalaria juncea$ (Sunn hemp) is a well-known leguminous plant widely used as green manure due to its nitrogen-fixing ability.
$Alhagi camelorum$ (Camel thorn) is also used as a green manure,particularly in sandy soils,as it helps in improving soil organic matter and moisture retention.
Therefore,the correct pair is $Crotalaria juncea$ and $Alhagi camelorum$.
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29
BiologyMediumMCQAIPMT · 2003
Species are considered as
A
Real units of classification devised by taxonomists
B
Real basic units of classification
C
The lowest units of classification
D
Artificial concept of human mind which cannot be defined in absolute terms
Solution
(B) . Species is the basic unit of classification.
Only the species has a real existence in nature,while other units of classification (like genus,family,order,etc.) are man-made artificial groups created for the convenience of study.
Only the species has a real existence in nature,while other units of classification (like genus,family,order,etc.) are man-made artificial groups created for the convenience of study.
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30
BiologyEasyMCQAIPMT · 2003
Which one of the following is not a micronutrient?
A
Molybdenum
B
Magnesium
C
Zinc
D
Boron
Solution
(B) The correct answer is $B$.
Macronutrients are essential elements present in plant tissues in large amounts,typically $1-10 \ mg$ per gram of dry weight.
These include carbon,hydrogen,oxygen,nitrogen,phosphorus,sulfur,potassium,calcium,and magnesium.
Micronutrients,or trace elements,are required in very small amounts,typically equal to or less than $0.1 \ mg$ per gram of dry matter.
These include iron,manganese,copper,molybdenum,zinc,boron,chlorine,and nickel.
Since magnesium is a macronutrient,it is not a micronutrient.
Macronutrients are essential elements present in plant tissues in large amounts,typically $1-10 \ mg$ per gram of dry weight.
These include carbon,hydrogen,oxygen,nitrogen,phosphorus,sulfur,potassium,calcium,and magnesium.
Micronutrients,or trace elements,are required in very small amounts,typically equal to or less than $0.1 \ mg$ per gram of dry matter.
These include iron,manganese,copper,molybdenum,zinc,boron,chlorine,and nickel.
Since magnesium is a macronutrient,it is not a micronutrient.
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31
BiologyEasyMCQAIPMT · 2003
The Tobacco Mosaic Virus $(TMV)$ is a tubular filament of size:
A
$700 \times 30 \, nm$
B
$300 \times 10 \, nm$
C
$300 \times 5 \, nm$
D
$300 \times 18 \, nm$
Solution
(D) The Tobacco Mosaic Virus $(TMV)$ is a well-studied plant virus.
It has a rod-shaped or tubular structure.
The dimensions of the $TMV$ particle are approximately $300 \, nm$ in length and $18 \, nm$ in diameter.
Therefore,the correct size is $300 \times 18 \, nm$.
It has a rod-shaped or tubular structure.
The dimensions of the $TMV$ particle are approximately $300 \, nm$ in length and $18 \, nm$ in diameter.
Therefore,the correct size is $300 \times 18 \, nm$.
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32
BiologyMediumMCQAIPMT · 2003
The number of chromosomes in bacterial cells is $1-3$ and they are ..........
A
can be circular or linear in the same cell.
B
always circular.
C
always linear.
D
can be circular or linear but not both in the same cell.
Solution
(D) Bacterial cells typically contain a single,circular,double-stranded $DNA$ molecule,which is often referred to as the bacterial chromosome or nucleoid. While most bacteria possess a single circular chromosome,some species have been found to contain linear chromosomes or multiple chromosomes. However,within a single bacterial cell,the chromosomal structure is consistent,meaning they are either circular or linear,but not both simultaneously.
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33
BiologyMediumMCQAIPMT · 2003
Which of the following statements is true for viruses?
A
The nucleic acid of a virus is called a capsid.
B
Viruses possess their own metabolic system.
C
All viruses contain both $RNA$ and $DNA$.
D
Viruses are obligate parasites.
Solution
(D) Viruses are non-cellular organisms characterized by having an inert crystalline structure outside the living cell.
They are obligate parasites,meaning they require a living host cell to replicate as they lack their own metabolic machinery.
The protein coat of a virus is called a capsid,not the nucleic acid.
Viruses contain either $RNA$ or $DNA$,never both.
They are obligate parasites,meaning they require a living host cell to replicate as they lack their own metabolic machinery.
The protein coat of a virus is called a capsid,not the nucleic acid.
Viruses contain either $RNA$ or $DNA$,never both.
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34
BiologyMediumMCQAIPMT · 2003
Viruses are no more alive than isolated chromosomes because .........
A
Both require a cellular environment for replication.
B
They both require $DNA$ and $RNA$.
C
They both require food molecules.
D
They both require oxygen for respiration.
Solution
(A) Viruses are considered to be at the border of living and non-living organisms. They are inert outside a host cell because they lack the metabolic machinery required for self-replication. Similarly,isolated chromosomes are just genetic material that cannot replicate or perform metabolic functions on their own. Therefore,both viruses and isolated chromosomes require a living cellular environment to replicate their genetic material.
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35
BiologyEasyMCQAIPMT · 2003
Mycorrhiza represents a/an ...... relationship.
A
Symbiosis
B
Endemism
C
Antibiosis
D
Parasitism
Solution
(A) Mycorrhiza is a symbiotic association between a fungus and the roots of higher plants.
In this relationship,the fungus helps the plant in the absorption of essential nutrients like phosphorus from the soil,while the plant provides carbohydrates (food) to the fungus.
Since both organisms benefit from this association,it is classified as a symbiotic relationship (mutualism).
In this relationship,the fungus helps the plant in the absorption of essential nutrients like phosphorus from the soil,while the plant provides carbohydrates (food) to the fungus.
Since both organisms benefit from this association,it is classified as a symbiotic relationship (mutualism).
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36
BiologyMediumMCQAIPMT · 2003
Sexual reproduction in $Spirogyra$ is considered an advanced feature because it shows:
A
Physiologically differentiated reproductive organs
B
Motile gametes of different sizes
C
Motile gametes of similar sizes
D
Morphologically distinct sex organs
Solution
(A) In $Spirogyra$,sexual reproduction occurs through conjugation. Although the gametes are morphologically similar (isogamous),they are physiologically differentiated into male $(+)$ and female $(-)$ strains. This physiological differentiation is considered an evolutionary advancement over simple vegetative or asexual reproduction,as it introduces genetic recombination while maintaining the simplicity of the thalloid body structure.
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37
BiologyMediumMCQAIPMT · 2003
According to a system of plant classification,which of the following pairs represents the group $Spermatophyta$ correctly?
A
$Rhizopus, Triticum$
B
$Ginkgo, Pisum$
C
$Acacia, Saccharum$
D
$Pinus, Cycas$
Solution
(B) The group $Spermatophyta$ (seed-bearing plants) includes both $Gymnosperms$ and $Angiosperms$.
$Ginkgo$ is a $Gymnosperm$ and $Pisum$ (pea) is an $Angiosperm$.
Both are seed-bearing plants,thus they belong to the $Spermatophyta$ group.
$Rhizopus$ is a fungus,$Pinus$ and $Cycas$ are $Gymnosperms$,and $Acacia$ and $Saccharum$ are $Angiosperms$. However,the pair $Ginkgo$ and $Pisum$ represents the diversity within the $Spermatophyta$ group effectively.
$Ginkgo$ is a $Gymnosperm$ and $Pisum$ (pea) is an $Angiosperm$.
Both are seed-bearing plants,thus they belong to the $Spermatophyta$ group.
$Rhizopus$ is a fungus,$Pinus$ and $Cycas$ are $Gymnosperms$,and $Acacia$ and $Saccharum$ are $Angiosperms$. However,the pair $Ginkgo$ and $Pisum$ represents the diversity within the $Spermatophyta$ group effectively.
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38
BiologyMediumMCQAIPMT · 2003
Which of the following pairs of plants does not show seed formation?
A
Ficus and Chlamydomonas
B
Punica and Pinus
C
Dryopteris and Funaria
D
Funaria and Ficus
Solution
(C) Seed formation is a characteristic feature of Spermatophytes (Gymnosperms and Angiosperms).
$1$. $Chlamydomonas$ is an Alga,$Funaria$ is a Bryophyte,and $Dryopteris$ (Fern) is a Pteridophyte. None of these produce seeds.
$2$. $Ficus$ (Angiosperm),$Punica$ (Angiosperm),and $Pinus$ (Gymnosperm) are seed-bearing plants.
$3$. In option $C$,$Dryopteris$ (Pteridophyte) and $Funaria$ (Bryophyte) both reproduce via spores and do not form seeds.
Therefore,the correct pair is $Dryopteris$ and $Funaria$.
$1$. $Chlamydomonas$ is an Alga,$Funaria$ is a Bryophyte,and $Dryopteris$ (Fern) is a Pteridophyte. None of these produce seeds.
$2$. $Ficus$ (Angiosperm),$Punica$ (Angiosperm),and $Pinus$ (Gymnosperm) are seed-bearing plants.
$3$. In option $C$,$Dryopteris$ (Pteridophyte) and $Funaria$ (Bryophyte) both reproduce via spores and do not form seeds.
Therefore,the correct pair is $Dryopteris$ and $Funaria$.
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39
BiologyEasyMCQAIPMT · 2003
Which of the following is a living fossil?
A
Cycas
B
Algae
C
Saccharomyces
D
Spirogyra
Solution
(A) living fossil is an extant taxon that closely resembles related species known only from the fossil record.
Among the given options,$Cycas$ is considered a living fossil because it has remained morphologically unchanged over millions of years and represents an ancient group of gymnosperms.
$Algae$ is a broad group of photosynthetic organisms,$Saccharomyces$ is a yeast (fungus),and $Spirogyra$ is a green alga; none of these are classified as living fossils.
Among the given options,$Cycas$ is considered a living fossil because it has remained morphologically unchanged over millions of years and represents an ancient group of gymnosperms.
$Algae$ is a broad group of photosynthetic organisms,$Saccharomyces$ is a yeast (fungus),and $Spirogyra$ is a green alga; none of these are classified as living fossils.
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40
BiologyMediumMCQAIPMT · 2003
Which of the following pairs correctly represent an animal and its respiratory organ?
$(a)$ Silverfish $-$ Trachea
$(b)$ Scorpion $-$ Book lungs
$(c)$ Sea squirt $-$ Pharyngeal gills
$(d)$ Dolphin $-$ Skin
$(a)$ Silverfish $-$ Trachea
$(b)$ Scorpion $-$ Book lungs
$(c)$ Sea squirt $-$ Pharyngeal gills
$(d)$ Dolphin $-$ Skin
A
$a$ and $b$
B
$c$ and $d$
C
$a$ and $d$
D
$a, b$ and $c$
Solution
(D) Let us analyze the respiratory organs of the given animals:
$1$. Silverfish (an insect) respires through a tracheal system (Trachea). This is correct.
$2$. Scorpion (an arachnid) respires through book lungs. This is correct.
$3$. Sea squirt (a Urochordate/Tunicate) respires through pharyngeal gill slits. This is correct.
$4$. Dolphin (a mammal) respires through lungs,not skin. This is incorrect.
Therefore,the correct pairs are $(a)$,$(b)$,and $(c)$.
$1$. Silverfish (an insect) respires through a tracheal system (Trachea). This is correct.
$2$. Scorpion (an arachnid) respires through book lungs. This is correct.
$3$. Sea squirt (a Urochordate/Tunicate) respires through pharyngeal gill slits. This is correct.
$4$. Dolphin (a mammal) respires through lungs,not skin. This is incorrect.
Therefore,the correct pairs are $(a)$,$(b)$,and $(c)$.
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41
BiologyMediumMCQAIPMT · 2003
Which of the following is a correct match of an animal and its specific characteristic?
A
Taenia $-$ Metamerism
B
Pheretima $-$ Sexual dimorphism
C
Musca $-$ Complete metamorphosis
D
Chameleon $-$ Mimicry
Solution
(C) The correct match is $C$.
$Musca$ (housefly) belongs to the class $Insecta$ and undergoes complete metamorphosis,which includes four stages: egg,larva,pupa,and adult.
$Taenia$ (tapeworm) is an acoelomate and does not show metamerism.
$Pheretima$ (earthworm) is hermaphrodite (bisexual),not showing sexual dimorphism.
$Chameleon$ is known for color change (chromatophores) rather than mimicry as its primary defining characteristic in this context.
$Musca$ (housefly) belongs to the class $Insecta$ and undergoes complete metamorphosis,which includes four stages: egg,larva,pupa,and adult.
$Taenia$ (tapeworm) is an acoelomate and does not show metamerism.
$Pheretima$ (earthworm) is hermaphrodite (bisexual),not showing sexual dimorphism.
$Chameleon$ is known for color change (chromatophores) rather than mimicry as its primary defining characteristic in this context.
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42
BiologyMediumMCQAIPMT · 2003
To which group does the animal $Sycon$ belong?
A
Multicellular,with tissue-level organization but without a body cavity.
B
Unicellular or acellular.
C
Multicellular,without tissue-level organization.
D
Multicellular and with a coelenteron cavity.
Solution
(C) $Sycon$ belongs to the phylum $Porifera$.
Members of the phylum $Porifera$ are commonly known as sponges.
They are multicellular organisms but exhibit a cellular level of organization,meaning they lack true tissues or organs.
Therefore,they are characterized as multicellular without tissue-level organization.
Members of the phylum $Porifera$ are commonly known as sponges.
They are multicellular organisms but exhibit a cellular level of organization,meaning they lack true tissues or organs.
Therefore,they are characterized as multicellular without tissue-level organization.
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43
BiologyMediumMCQAIPMT · 2003
During its life cycle,the liver fluke ($Fasciola$ $hepatica$) infects its intermediate host and primary host,respectively,through which of the following larval stages?
A
Miracidium and Metacercaria
B
Redia and Miracidium
C
Cercaria and Redia
D
Metacercaria and Cercaria
Solution
(A) The life cycle of the liver fluke ($Fasciola$ $hepatica$) involves two hosts.
$1$. The primary host is a sheep or human (vertebrate),and the intermediate host is a snail (mollusc).
$2$. The $Miracidium$ larva infects the intermediate host (snail) to continue its development.
$3$. After passing through various stages ($Sporocyst$,$Redia$,$Cercaria$) within the snail,the $Cercaria$ larvae emerge and encyst on aquatic vegetation to form the $Metacercaria$ stage.
$4$. The $Metacercaria$ is the infective stage for the primary host (sheep/human) when ingested with vegetation.
Therefore,the liver fluke infects the intermediate host as a $Miracidium$ and the primary host as a $Metacercaria$.
$1$. The primary host is a sheep or human (vertebrate),and the intermediate host is a snail (mollusc).
$2$. The $Miracidium$ larva infects the intermediate host (snail) to continue its development.
$3$. After passing through various stages ($Sporocyst$,$Redia$,$Cercaria$) within the snail,the $Cercaria$ larvae emerge and encyst on aquatic vegetation to form the $Metacercaria$ stage.
$4$. The $Metacercaria$ is the infective stage for the primary host (sheep/human) when ingested with vegetation.
Therefore,the liver fluke infects the intermediate host as a $Miracidium$ and the primary host as a $Metacercaria$.
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44
BiologyMediumMCQAIPMT · 2003
The juicy hair-like structures found in lemon fruits develop from the .......... .
A
Mesocarp and endocarp
B
Epicarp
C
Mesocarp
D
Endocarp
Solution
(D) In citrus fruits like lemon,the edible part consists of juicy,hair-like structures known as juice sacs. These juice sacs are unicellular outgrowths that develop from the inner epidermis of the carpel,which corresponds to the endocarp of the fruit. Therefore,the correct answer is the endocarp.
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45
BiologyMediumMCQAIPMT · 2003
What is the characteristic of the cells in the quiescent center?
A
They divide regularly to increase the tunica.
B
They possess dense cytoplasm and a prominent nucleus.
C
They possess thin cytoplasm and a small nucleus.
D
They divide continuously to increase the growth of the body.
Solution
(C) The quiescent center is a region found in the root apical meristem.
It consists of cells that have a very low rate of cell division.
These cells are characterized by having a thin cytoplasm and a small nucleus,which indicates their low metabolic activity compared to the surrounding actively dividing meristematic cells.
Therefore,option $C$ is the correct description.
It consists of cells that have a very low rate of cell division.
These cells are characterized by having a thin cytoplasm and a small nucleus,which indicates their low metabolic activity compared to the surrounding actively dividing meristematic cells.
Therefore,option $C$ is the correct description.
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46
BiologyMediumMCQAIPMT · 2003
The root apical meristem is present in ............ .
A
All roots
B
Only embryonic roots
C
Only taproots
D
Only adventitious roots
Solution
(A) The root apical meristem is a region of actively dividing cells located at the tip of the root. It is responsible for the primary growth of the root system. Since all types of roots (taproots,fibrous roots,and adventitious roots) grow in length from their tips,the root apical meristem is present in all roots.
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47
BiologyMediumMCQAIPMT · 2003
In which of the following does chlorenchyma tissue develop?
A
Pollen tube of Pinus
B
Cytoplasm of Chlorella
C
Green fungi like Aspergillus
D
Capsule of moss
Solution
(D) Chlorenchyma is a type of parenchyma tissue that contains chloroplasts and is specialized for photosynthesis. In the life cycle of mosses (Bryophytes),the sporophyte consists of a foot,seta,and capsule. The capsule is the photosynthetic part of the sporophyte,which contains chlorenchyma tissue to produce food for the developing spores. Therefore,the correct answer is the capsule of moss.
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48
BiologyMediumMCQAIPMT · 2003
Which of the following contains an abundant amount of extracellular matrix?
A
Striated muscle
B
Areolar connective tissue
C
Epithelial tissue
D
Nervous tissue
Solution
(B) Connective tissues are characterized by the presence of a significant amount of extracellular matrix (ground substance and fibers) in which the cells are embedded.
Among the given options,areolar connective tissue is a type of loose connective tissue that contains a large amount of intercellular ground substance and fibers,making it rich in extracellular matrix.
Striated muscle,epithelial tissue,and nervous tissue are primarily composed of cells with very little extracellular matrix.
Among the given options,areolar connective tissue is a type of loose connective tissue that contains a large amount of intercellular ground substance and fibers,making it rich in extracellular matrix.
Striated muscle,epithelial tissue,and nervous tissue are primarily composed of cells with very little extracellular matrix.
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49
BiologyMediumMCQAIPMT · 2003
If the Loop of Henle were absent from the mammalian nephron,which of the following is to be expected?
A
The urine will be more dilute.
B
Urine formation will not take place.
C
There will be little change in the quality and quantity of urine produced.
D
The urine will be more concentrated.
Solution
(A) The Loop of Henle plays a critical role in the counter-current mechanism,which is responsible for the concentration of urine in mammals.
It creates a hyperosmotic medullary interstitium by reabsorbing solutes (mainly $NaCl$) from the filtrate.
If the Loop of Henle were absent,the kidney would lose its ability to create this osmotic gradient.
Consequently,the water reabsorption from the collecting duct would be significantly reduced,leading to the production of highly dilute urine.
It creates a hyperosmotic medullary interstitium by reabsorbing solutes (mainly $NaCl$) from the filtrate.
If the Loop of Henle were absent,the kidney would lose its ability to create this osmotic gradient.
Consequently,the water reabsorption from the collecting duct would be significantly reduced,leading to the production of highly dilute urine.
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50
BiologyEasyMCQAIPMT · 2003
In a neuron,the components known as Nissl's granules are now identified as:
A
Ribosomes
B
Mitochondria
C
Cell metabolites
D
Fat granules
Solution
(A) Nissl's granules are irregular masses of rough endoplasmic reticulum $(RER)$ with numerous ribosomes attached to them,found in the cell body $(cyton)$ and dendrites of neurons. These structures are primarily sites of protein synthesis. Modern cell biology identifies these granules as clusters of ribosomes and rough endoplasmic reticulum.
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51
BiologyMediumMCQAIPMT · 2003
Pollen tube discharges its gametes in
A
Synergids
B
Antipodals
C
Central cell
D
None of these
Solution
(A) The pollen tube enters the ovule,typically through the micropyle,and reaches one of the two synergids.
Upon reaching the synergid,the tip of the pollen tube ruptures,and it releases its two male gametes into the cytoplasm of the synergid cell.
From there,one male gamete moves towards the egg cell for syngamy,and the other moves towards the central cell for triple fusion.
Upon reaching the synergid,the tip of the pollen tube ruptures,and it releases its two male gametes into the cytoplasm of the synergid cell.
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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52
BiologyEasyMCQAIPMT · 2003
Bartholin's glands are situated
A
On either side of vas deferens in humans
B
On the sides of the head of frog
C
At the reduced tail end of birds
D
On either side of vagina in humans
Solution
(D) $Bartholin's$ glands are a pair of small glands located on either side of the vaginal opening in human females. They are homologous to the bulbourethral glands in males and are responsible for secreting mucus to provide lubrication during sexual intercourse.
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53
BiologyMediumMCQAIPMT · 2003
Down's syndrome is caused by an extra copy of chromosome number $21$. What percentage of offspring produced by an affected mother and a normal father would be affected by this disorder (in $\%$)?
A
$25$
B
$100$
C
$75$
D
$50$
Solution
(D) Down's syndrome is a chromosomal disorder caused by trisomy of chromosome $21$.
An affected mother has $47$ chromosomes (including an extra $21$st chromosome),while a normal father has $46$ chromosomes.
During meiosis in the mother,the gametes produced will have either $23$ or $24$ chromosomes. Specifically,$50\%$ of the eggs will carry the extra $21$st chromosome.
When these eggs are fertilized by normal sperm ($23$ chromosomes) from the father,$50\%$ of the resulting zygotes will have $47$ chromosomes (trisomy $21$),leading to Down's syndrome.
Therefore,$50\%$ of the offspring are expected to be affected.
An affected mother has $47$ chromosomes (including an extra $21$st chromosome),while a normal father has $46$ chromosomes.
During meiosis in the mother,the gametes produced will have either $23$ or $24$ chromosomes. Specifically,$50\%$ of the eggs will carry the extra $21$st chromosome.
When these eggs are fertilized by normal sperm ($23$ chromosomes) from the father,$50\%$ of the resulting zygotes will have $47$ chromosomes (trisomy $21$),leading to Down's syndrome.
Therefore,$50\%$ of the offspring are expected to be affected.
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54
BiologyMediumMCQAIPMT · 2003
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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55
BiologyMediumMCQAIPMT · 2003
In $Drosophila$,the sex is determined by
A
Whether the egg is fertilized or develops parthenogenetically
B
The ratio of number of $X$-chromosomes to the sets of autosomes
C
$X$ and $Y$ chromosomes
D
The ratio of pairs of $X$-chromosomes to the pairs of autosomes
Solution
(B) In $Drosophila$,sex determination is based on the genic balance theory proposed by $C.B. Bridges$.
According to this theory,the sex of an individual is determined by the ratio of the number of $X$-chromosomes to the number of sets of autosomes $(A)$.
This ratio is known as the sex index ratio $(X/A)$.
For example,in a superfemale,the number of $X$-chromosomes is $3$ and the number of sets of autosomes is $2$,resulting in a ratio of $3/2 = 1.5$.
According to this theory,the sex of an individual is determined by the ratio of the number of $X$-chromosomes to the number of sets of autosomes $(A)$.
This ratio is known as the sex index ratio $(X/A)$.
For example,in a superfemale,the number of $X$-chromosomes is $3$ and the number of sets of autosomes is $2$,resulting in a ratio of $3/2 = 1.5$.
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56
BiologyMediumMCQAIPMT · 2003
Pattern baldness,moustaches and beard in human males are examples of
A
Sex-determining traits
B
Sex-linked traits
C
Sex-limited traits
D
Sex-differentiating traits
Solution
(C) $Sex-limited$ genes express their effects in only one sex,and their action is clearly related to the sex hormones.
For example,the development of beard and moustaches in human males is a $sex-limited$ character because these traits are phenotypically expressed only in males,even though the genes for these traits may be present in both sexes.
For example,the development of beard and moustaches in human males is a $sex-limited$ character because these traits are phenotypically expressed only in males,even though the genes for these traits may be present in both sexes.
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57
BiologyMediumMCQAIPMT · 2003
When a cluster of genes shows linkage behaviour,they:
A
Induce cell division
B
Do not show a chromosome map
C
Show recombination during meiosis
D
Do not show independent assortment
Solution
(D) When a cluster of genes shows linkage behavior,they do not show independent assortment.
This is because linked genes are located very close to each other on the same chromosome and tend to be inherited together as a single unit,thereby violating Mendel's Law of Independent Assortment.
This is because linked genes are located very close to each other on the same chromosome and tend to be inherited together as a single unit,thereby violating Mendel's Law of Independent Assortment.
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58
BiologyMediumMCQAIPMT · 2003
The linkage map of $X$-chromosome of fruit fly has $66$ units with yellow body gene $(y)$ at one end and bobbed hair $(b)$ gene at the other end. The recombination frequency between these two genes ($y$ and $b$) should be
A
$100\%$
B
$66\%$
C
$> 50\%$
D
$5.50\%$
Solution
(B) The distance between two genes on a chromosome is measured in map units (centimorgans),where $1$ map unit corresponds to $1\%$ recombination frequency.
Since the linkage map distance between the yellow body gene $(y)$ and the bobbed hair gene $(b)$ is $66$ units,the theoretical recombination frequency is $66\%$.
However,in practice,recombination frequency cannot exceed $50\%$ because multiple crossovers between distant genes lead to an observed frequency that approaches $50\%$.
Given the options provided and the definition of map units,the correct representation of the map distance is $66\%$.
Since the linkage map distance between the yellow body gene $(y)$ and the bobbed hair gene $(b)$ is $66$ units,the theoretical recombination frequency is $66\%$.
However,in practice,recombination frequency cannot exceed $50\%$ because multiple crossovers between distant genes lead to an observed frequency that approaches $50\%$.
Given the options provided and the definition of map units,the correct representation of the map distance is $66\%$.
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59
BiologyMediumMCQAIPMT · 2003
Which one of the following sequences was proposed by Darwin and Wallace for organic evolution?
A
Variations, natural selection, overproduction, constancy of population size.
B
Overproduction, variations, constancy of population size, natural selection.
C
Variations, constancy of population size, overproduction, natural selection.
D
Overproduction, constancy of population size, variations, natural selection.
Solution
(D) Darwin and Wallace's theory of natural selection follows a logical sequence:
$1$. $Overproduction$: Organisms tend to produce more offspring than the environment can support.
$2$. $Constancy \text{ of population size}$: Despite high reproductive rates, population sizes remain relatively stable due to limited resources.
$3$. $Variations$: Individuals within a population exhibit variations in their traits.
$4$. $Natural \text{ selection}$: Individuals with advantageous variations are more likely to survive and reproduce, leading to evolutionary change over time.
Therefore, the correct sequence is $Overproduction, \text{ constancy of population size, variations, natural selection}$.
$1$. $Overproduction$: Organisms tend to produce more offspring than the environment can support.
$2$. $Constancy \text{ of population size}$: Despite high reproductive rates, population sizes remain relatively stable due to limited resources.
$3$. $Variations$: Individuals within a population exhibit variations in their traits.
$4$. $Natural \text{ selection}$: Individuals with advantageous variations are more likely to survive and reproduce, leading to evolutionary change over time.
Therefore, the correct sequence is $Overproduction, \text{ constancy of population size, variations, natural selection}$.
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60
BiologyMediumMCQAIPMT · 2003
In recent years,$DNA$ sequences (nucleotide sequence) of $mt-DNA$ and $Y$-chromosomes were considered for the study of human evolution,because
A
They can be studied from the samples of fossil remains
B
They are small,and therefore,easy to study
C
They are uniparental in origin and do not take part in recombination
D
Their structure is known in greater detail
Solution
(C) $mt-DNA$ (mitochondrial $DNA$) is inherited maternally,and the $Y$-chromosome is inherited paternally. Because they are uniparental in origin,they do not undergo recombination during sexual reproduction. This lack of recombination allows scientists to trace lineage and evolutionary history accurately over many generations without the genetic shuffling that occurs in autosomal $DNA$.
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61
BiologyEasyMCQAIPMT · 2003
The term "antibiotic" was coined by
A
Alexander Fleming
B
Edward Jenner
C
Louis Pasteur
D
Selman Waksman
Solution
(D) The term "antibiotic" was coined by the American microbiologist $Selman Waksman$ in $1942$.
An antibiotic is a substance produced by a microorganism that, in low concentrations, inhibits or kills other microorganisms.
While $Alexander Fleming$ discovered the first antibiotic, $Penicillin$, it was $Selman Waksman$ who introduced the term to describe these compounds.
An antibiotic is a substance produced by a microorganism that, in low concentrations, inhibits or kills other microorganisms.
While $Alexander Fleming$ discovered the first antibiotic, $Penicillin$, it was $Selman Waksman$ who introduced the term to describe these compounds.
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62
BiologyEasyMCQAIPMT · 2003
The technique involving the insertion of a desired gene into the $DNA$ of a plasmid vector is known as:
A
Dressing
B
Cloning
C
Splicing
D
Drafting
Solution
(B) The technique of inserting a desired gene into the $DNA$ of a plasmid vector is called gene cloning or recombinant $DNA$ technology.
In this process,the target gene is inserted into a vector (like a plasmid) to create a recombinant $DNA$ molecule,which is then introduced into a host organism for replication and expression.
Therefore,the correct term among the given options is cloning.
In this process,the target gene is inserted into a vector (like a plasmid) to create a recombinant $DNA$ molecule,which is then introduced into a host organism for replication and expression.
Therefore,the correct term among the given options is cloning.
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63
BiologyEasyMCQAIPMT · 2003
Which group of vertebrates comprises the highest number of endangered species?
A
Birds
B
Mammals
C
Fishes
D
Reptiles
Solution
(B) According to the $IUCN$ Red List data,mammals represent the highest number of endangered species among vertebrates.
Specifically,the number of endangered species of mammals is approximately $62$,while for reptiles and birds,the numbers are $6$ and $11$ respectively.
Therefore,the correct option is $B$.
Specifically,the number of endangered species of mammals is approximately $62$,while for reptiles and birds,the numbers are $6$ and $11$ respectively.
Therefore,the correct option is $B$.
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64
BiologyEasyMCQAIPMT · 2003
Which endangered animal is the source of the world's finest, lightest, warmest, and most expensive wool, the shahtoosh?
A
Chiru
B
Nilgai
C
Cheetal
D
Kashmiri goat
Solution
(A) The $Chiru$ $(Pantholops hodgsonii)$, also known as the Tibetan antelope, is the source of $shahtoosh$ wool.
$Shahtoosh$ is considered the finest, lightest, and warmest wool in the world.
Due to the illegal hunting of $Chiru$ for this wool, it is classified as an endangered species.
Therefore, the correct option is $A$.
$Shahtoosh$ is considered the finest, lightest, and warmest wool in the world.
Due to the illegal hunting of $Chiru$ for this wool, it is classified as an endangered species.
Therefore, the correct option is $A$.
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65
BiologyMediumMCQAIPMT · 2003
Two opposite forces operate in the growth and development of every population. One of them relates to the ability to reproduce at a given rate. The force opposite to it is called:
A
Biotic control
B
Mortality
C
Fecundity
D
Environmental resistance
Solution
(D) Every population has an innate ability to grow exponentially,which is known as biotic potential. This is the ability to reproduce at a maximum rate under ideal conditions.
However,in nature,resources are limited,and various factors such as food scarcity,predation,disease,and competition act against this growth.
These limiting factors collectively are known as $Environmental \ resistance$.
Therefore,the force opposite to the biotic potential is $Environmental \ resistance$.
However,in nature,resources are limited,and various factors such as food scarcity,predation,disease,and competition act against this growth.
These limiting factors collectively are known as $Environmental \ resistance$.
Therefore,the force opposite to the biotic potential is $Environmental \ resistance$.
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66
BiologyMediumMCQAIPMT · 2003
Test tube baby means a baby born when
A
It develops from a non-fertilized egg
B
It developed in a test tube
C
It is developed through tissue culture method
D
The ovum is fertilised externally and thereafter implanted in the uterus
Solution
(D) . Test tube baby refers to a baby born through the process of In Vitro Fertilization $(IVF)$. In this procedure,the ovum is fertilized by sperm outside the mother's body in a laboratory setting. Once fertilization occurs and the embryo reaches the appropriate stage,it is implanted into the mother's uterus for further growth and development until birth.
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67
BiologyMediumMCQAIPMT · 2003
The main advantage of encystation in $Amoeba$ is .....
A
Disposal of accumulated waste products
B
Ability to survive during unfavourable physical conditions
C
Ability to live for some time without food intake
D
Protection from parasites and predators
Solution
(B) Encystation is a process in $Amoeba$ where the organism withdraws its pseudopodia and secretes a three-layered,hard,protective covering or cyst around itself. This process occurs during unfavourable environmental conditions. The primary purpose of this cyst is to protect the $Amoeba$ from harsh external factors,allowing it to survive until favourable conditions return. Once conditions become favourable,the cyst wall breaks,and the $Amoeba$ undergoes multiple fission to release many small amoebae (sporulation).
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68
BiologyMediumMCQAIPMT · 2003
In flowering plants,what does the archesporial cell produce?
A
Only the tapetum and sporogenous cells
B
Only the microsporangium wall
C
The wall and sporogenous cells
D
The wall and the tapetum
Solution
(C) In the development of the microsporangium,the archesporial cells undergo periclinal division.
This division results in the formation of two types of cells:
$1$. Primary parietal cells,which divide to form the wall layers of the microsporangium (epidermis,endothecium,middle layers,and tapetum).
$2$. Primary sporogenous cells,which directly or after a few divisions form the microspore mother cells (sporogenous tissue).
Therefore,the archesporial cell gives rise to both the wall layers and the sporogenous cells.
This division results in the formation of two types of cells:
$1$. Primary parietal cells,which divide to form the wall layers of the microsporangium (epidermis,endothecium,middle layers,and tapetum).
$2$. Primary sporogenous cells,which directly or after a few divisions form the microspore mother cells (sporogenous tissue).
Therefore,the archesporial cell gives rise to both the wall layers and the sporogenous cells.
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69
BiologyMediumMCQAIPMT · 2003
During embryonic development,the establishment of polarity such as anterior/posterior,dorsal/ventral,and medial/lateral axes is called:
A
Anamorphosis
B
Pattern formation
C
Organizer phenomenon
D
Axis formation
Solution
(B) During embryonic development,the process by which cells in a developing embryo acquire positional information and organize themselves into specific structures along defined axes is known as $Pattern \ formation$. This process ensures that the embryo develops the correct body plan,establishing the $anterior/posterior$,$dorsal/ventral$,and $medial/lateral$ axes.
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70
BiologyEasyMCQAIPMT · 2003
Bartholin's glands are situated in ..........
A
On either side of the vagina in humans
B
On either side of the vas deferens in humans
C
On the sides of the head in some amphibians
D
At the reduced tail end of birds
Solution
(A) Bartholin's glands,also known as greater vestibular glands,are two pea-sized compound alveolar glands located slightly posterior and to the left and right of the opening of the vagina in human females. They secrete mucus to provide lubrication for the vulva.
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71
BiologyMediumMCQAIPMT · 2003
$A$ test-tube baby is born when:
A
The ovum is fertilized externally and then implanted in the uterus.
B
It develops from an unfertilized ovum.
C
It develops in a test tube.
D
It is developed through tissue culture methods.
Solution
(A) test-tube baby is produced through the process of In-Vitro Fertilization $(IVF)$.
In this procedure,the ovum from the female and sperm from the male are collected and induced to form a zygote under simulated conditions in the laboratory.
The zygote or early embryo (up to $8$ blastomeres) is then transferred into the fallopian tube ($ZIFT$ - Zygote Intra Fallopian Transfer) or into the uterus ($IUT$ - Intra Uterine Transfer) for further development.
Therefore,the baby is not born in a test tube,but the initial fertilization occurs outside the body.
In this procedure,the ovum from the female and sperm from the male are collected and induced to form a zygote under simulated conditions in the laboratory.
The zygote or early embryo (up to $8$ blastomeres) is then transferred into the fallopian tube ($ZIFT$ - Zygote Intra Fallopian Transfer) or into the uterus ($IUT$ - Intra Uterine Transfer) for further development.
Therefore,the baby is not born in a test tube,but the initial fertilization occurs outside the body.
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72
BiologyDifficultMCQAIPMT · 2003
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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73
BiologyEasyMCQAIPMT · 2003
What is a genetic map?
A
Shows the stages during cell division.
B
Shows the distribution of various species in an area.
C
Establishes the position of genes on a chromosome.
D
Establishes the various stages of gene evolution.
Solution
(C) genetic map (also known as a linkage map) is a representation of the relative positions of genes or genetic markers on a chromosome. It is based on the frequency of recombination between markers during crossover in meiosis. The distance between genes on a genetic map is measured in map units or centimorgans $(cM)$. Therefore,it establishes the physical or relative location of genes on a chromosome.
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74
BiologyMediumMCQAIPMT · 2003
Currently,the $DNA$ sequence of $mtDNA$ and the $Y$-chromosome are considered for the study of human evolution because:
A
Their structure is known in greater detail.
B
They can be studied through fossil samples.
C
They are small and therefore easy to study.
D
They are inherited from a single parent and do not undergo recombination.
Solution
(D) The $mtDNA$ (mitochondrial $DNA$) is inherited maternally,and the $Y$-chromosome is inherited paternally. Because they are inherited from a single parent,they do not undergo the process of genetic recombination (crossing over) during meiosis. This allows scientists to trace lineages back through generations without the 'shuffling' of genetic material that occurs with autosomal chromosomes,making them ideal markers for studying human evolutionary history and migration patterns.
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75
BiologyMediumMCQAIPMT · 2003
Where are the cytoplasmic genes for male sterility typically located?
A
Nuclear genome
B
Cytosol
C
Chloroplast genome
D
Mitochondrial genome
Solution
(D) Cytoplasmic male sterility $(CMS)$ is a condition in plants where the plant is unable to produce functional pollen.
This trait is inherited through the cytoplasm,specifically through the extranuclear $DNA$.
In most cases,the genes responsible for $CMS$ are located in the mitochondrial genome rather than the nuclear or chloroplast genome.
Therefore,the correct option is $D$.
This trait is inherited through the cytoplasm,specifically through the extranuclear $DNA$.
In most cases,the genes responsible for $CMS$ are located in the mitochondrial genome rather than the nuclear or chloroplast genome.
Therefore,the correct option is $D$.
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76
BiologyMediumMCQAIPMT · 2003
In $Drosophila$,how is sex determination decided?
A
By the ratio of pairs of $X$ chromosomes to pairs of autosomes.
B
By whether the egg is fertilized or develops via parthenogenesis.
C
By the ratio of the number of $X$ chromosomes to the number of sets of autosomes.
D
By the presence of $X$ and $Y$ chromosomes.
Solution
(C) In $Drosophila$ (fruit fly),sex determination is based on the genic balance theory proposed by $C.B. Bridges$.
According to this theory,the sex of the individual is determined by the ratio of the number of $X$ chromosomes $(X)$ to the number of sets of autosomes $(A)$.
This ratio is denoted as $X/A$ ratio.
If the $X/A$ ratio is $1.0$,the individual is a female.
If the $X/A$ ratio is $0.5$,the individual is a male.
If the ratio is between $0.5$ and $1.0$,it results in an intersex individual.
Therefore,the correct answer is the ratio of the number of $X$ chromosomes to the number of sets of autosomes.
According to this theory,the sex of the individual is determined by the ratio of the number of $X$ chromosomes $(X)$ to the number of sets of autosomes $(A)$.
This ratio is denoted as $X/A$ ratio.
If the $X/A$ ratio is $1.0$,the individual is a female.
If the $X/A$ ratio is $0.5$,the individual is a male.
If the ratio is between $0.5$ and $1.0$,it results in an intersex individual.
Therefore,the correct answer is the ratio of the number of $X$ chromosomes to the number of sets of autosomes.
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77
BiologyMediumMCQAIPMT · 2003
When a group of genes exhibits linkage,they:
A
Do not show independent assortment.
B
Induce cell division.
C
Do not show chromosomal mapping.
D
Show recombination during meiosis.
Solution
(A) Linkage is the phenomenon where genes located on the same chromosome are inherited together because they are physically close to each other.
According to Mendel's Law of Independent Assortment,genes on different chromosomes assort independently.
However,linked genes do not follow this law because they tend to stay together during gamete formation,thus failing to show independent assortment.
According to Mendel's Law of Independent Assortment,genes on different chromosomes assort independently.
However,linked genes do not follow this law because they tend to stay together during gamete formation,thus failing to show independent assortment.
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78
BiologyMediumMCQAIPMT · 2003
Which of the following traits of pea plants studied by Mendel was recessive?
A
Green pod color
B
Round seed shape
C
Axial flower position
D
Green seed color
Solution
(D) Mendel studied $7$ pairs of contrasting traits in pea plants.
Among these,the recessive traits are:
$1$. Wrinkled seed shape
$2$. Green seed color
$3$. White flower color
$4$. Constricted pod shape
$5$. Yellow pod color
$6$. Terminal flower position
$7$. Dwarf plant height.
Therefore,among the given options,'Green seed color' is a recessive trait.
Among these,the recessive traits are:
$1$. Wrinkled seed shape
$2$. Green seed color
$3$. White flower color
$4$. Constricted pod shape
$5$. Yellow pod color
$6$. Terminal flower position
$7$. Dwarf plant height.
Therefore,among the given options,'Green seed color' is a recessive trait.
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79
BiologyEasyMCQAIPMT · 2003
Mendel studied seven traits of pea plants controlled by genes. On how many different chromosomes are these genes located?
A
Five
B
Four
C
Seven
D
Eight
Solution
(B) Gregor Mendel studied $7$ pairs of contrasting traits in pea plants $(Pisum \text{ } sativum)$.
These $7$ traits are controlled by genes located on $4$ different chromosomes.
Specifically, these genes are located on chromosome numbers $1, 4, 5,$ and $7$ of the pea plant.
Therefore, the correct answer is $4$.
These $7$ traits are controlled by genes located on $4$ different chromosomes.
Specifically, these genes are located on chromosome numbers $1, 4, 5,$ and $7$ of the pea plant.
Therefore, the correct answer is $4$.
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80
BiologyMediumMCQAIPMT · 2003
Down syndrome is caused by an extra chromosome in the $21$st pair. What percentage of offspring from an affected mother and a normal father will show the effect of this disorder (in $\%$)?
A
$50$
B
$25$
C
$100$
D
$75$
Solution
(A) Down syndrome is a chromosomal disorder caused by trisomy of chromosome $21$ ($47, XX, +21$ or $47, XY, +21$).
In an affected mother (trisomic,$2n+1$),the gametes produced will be $n$ and $n+1$ in equal proportions ($50\%$ each).
When these gametes fuse with normal gametes $(n)$ from a normal father,the resulting offspring will be $2n$ (normal) and $2n+1$ (affected) in a $1:1$ ratio.
Therefore,$50\%$ of the offspring are expected to inherit the extra chromosome and exhibit the disorder.
In an affected mother (trisomic,$2n+1$),the gametes produced will be $n$ and $n+1$ in equal proportions ($50\%$ each).
When these gametes fuse with normal gametes $(n)$ from a normal father,the resulting offspring will be $2n$ (normal) and $2n+1$ (affected) in a $1:1$ ratio.
Therefore,$50\%$ of the offspring are expected to inherit the extra chromosome and exhibit the disorder.
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81
BiologyAdvancedMCQAIPMT · 2003
In a fruit fly,the $X$-chromosome linkage map has a total length of $66$ units,with the yellow body gene $(y)$ at one end and the bobbed hair gene $(b)$ at the other end. What will be the recombination frequency between these two genes $(y$ and $b)$?
A
$\leq 50\%$
B
$66\%$
C
$100\%$
D
$> 50\%$
Solution
(A) The distance between two genes on a chromosome is measured in map units or centimorgans $(cM)$.
One map unit corresponds to $1\%$ recombination frequency.
However,the maximum observable recombination frequency between any two linked genes is capped at $50\%$,even if the map distance is greater than $50$ map units.
This is because multiple crossovers (double or higher-order crossovers) between distant genes can result in the parental combination of alleles,effectively masking the recombination events.
Since the map distance is $66$ units (which is $> 50$ units),the observed recombination frequency will be $50\%$ or less (approaching $50\%$ as the distance increases).
Therefore,the correct answer is $\leq 50\%$.
One map unit corresponds to $1\%$ recombination frequency.
However,the maximum observable recombination frequency between any two linked genes is capped at $50\%$,even if the map distance is greater than $50$ map units.
This is because multiple crossovers (double or higher-order crossovers) between distant genes can result in the parental combination of alleles,effectively masking the recombination events.
Since the map distance is $66$ units (which is $> 50$ units),the observed recombination frequency will be $50\%$ or less (approaching $50\%$ as the distance increases).
Therefore,the correct answer is $\leq 50\%$.
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82
BiologyMediumMCQAIPMT · 2003
What is a cross between two genotypes or phenotypes where the source of gametes is reversed in one cross called?
A
Reciprocal cross
B
Back cross
C
Test cross
D
Dihybrid cross
Solution
(A) $Reciprocal$ $cross$ is a breeding experiment designed to test the role of parental sex on a given inheritance pattern. In this cross, two sets of crosses are made:
$1$. In the first cross, a male of genotype $A$ is crossed with a female of genotype $B$.
$2$. In the second cross, a male of genotype $B$ is crossed with a female of genotype $A$.
Since the source of gametes (sex of the parent) is reversed, it is termed a $Reciprocal$ $cross$.
$1$. In the first cross, a male of genotype $A$ is crossed with a female of genotype $B$.
$2$. In the second cross, a male of genotype $B$ is crossed with a female of genotype $A$.
Since the source of gametes (sex of the parent) is reversed, it is termed a $Reciprocal$ $cross$.
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83
BiologyMediumMCQAIPMT · 2003
Mustache and beard in males are examples of which of the following?
A
Sex-differentiating traits
B
Sex-determining traits
C
Sex-linked traits
D
Sex-limited traits
Solution
(D) Sex-limited traits are autosomal traits that are expressed in only one sex.
Even though both males and females possess the genes for these traits,they are only expressed in the presence of sex-specific hormones.
In humans,the development of a beard and mustache is a secondary sexual characteristic in males,which is controlled by male sex hormones (androgens).
Therefore,these are classified as sex-limited traits.
Even though both males and females possess the genes for these traits,they are only expressed in the presence of sex-specific hormones.
In humans,the development of a beard and mustache is a secondary sexual characteristic in males,which is controlled by male sex hormones (androgens).
Therefore,these are classified as sex-limited traits.
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84
BiologyMediumMCQAIPMT · 2003
Which of the following triplet codons correctly represents a start or stop signal for protein synthesis?
A
$UGU$ Leucine
B
$UAC$ Tyrosine
C
$AUG$ Start
D
$UUU$ Stop
Solution
(C) In the genetic code,$AUG$ is the universal start codon,which codes for the amino acid Methionine.
Stop codons,also known as termination codons,are $UAA$,$UAG$,and $UGA$.
Among the given options,$AUG$ is correctly identified as the start codon.
$UGU$ codes for Cysteine,$UAC$ codes for Tyrosine,and $UUU$ codes for Phenylalanine.
Stop codons,also known as termination codons,are $UAA$,$UAG$,and $UGA$.
Among the given options,$AUG$ is correctly identified as the start codon.
$UGU$ codes for Cysteine,$UAC$ codes for Tyrosine,and $UUU$ codes for Phenylalanine.
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85
BiologyMediumMCQAIPMT · 2003
During the process of translation in prokaryotes,a $GTP$ molecule is required for which of the following steps?
A
Binding of $m-RNA$ with $30S$ ribosomal subunit along with formyl-methionyl $t-RNA$
B
Association of $50S$ ribosomal subunit with the initiation complex
C
Formation of formyl-methionyl $t-RNA$
D
Binding of $30S$ ribosomal subunit with $m-RNA$
Solution
(B) In prokaryotic translation initiation,the formation of the initiation complex involves several steps.
$1$. The $30S$ ribosomal subunit binds to the $m-RNA$.
$2$. The initiator $t-RNA$ (formyl-methionyl $t-RNA$) binds to the start codon.
$3$. The association of the $50S$ ribosomal subunit with the $30S$ initiation complex requires energy,which is provided by the hydrolysis of a $GTP$ molecule.
Therefore,the $GTP$ molecule is specifically required for the association of the $50S$ ribosomal subunit with the initiation complex.
$1$. The $30S$ ribosomal subunit binds to the $m-RNA$.
$2$. The initiator $t-RNA$ (formyl-methionyl $t-RNA$) binds to the start codon.
$3$. The association of the $50S$ ribosomal subunit with the $30S$ initiation complex requires energy,which is provided by the hydrolysis of a $GTP$ molecule.
Therefore,the $GTP$ molecule is specifically required for the association of the $50S$ ribosomal subunit with the initiation complex.
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86
BiologyMediumMCQAIPMT · 2003
The degenerate nature of the genetic code is primarily associated with which position of the codon?
A
First position
B
Second position
C
Third position
D
All positions equally
Solution
(C) The genetic code is described as 'degenerate' because more than one codon can code for the same amino acid.
This phenomenon is primarily due to the 'wobble hypothesis' proposed by Francis Crick.
According to this hypothesis,the base pairing between the $3'$ end of the codon and the $5'$ end of the anticodon is less strict.
Therefore,the third base (the $3'$ position) of the codon is the most flexible and is responsible for the degeneracy of the genetic code.
This phenomenon is primarily due to the 'wobble hypothesis' proposed by Francis Crick.
According to this hypothesis,the base pairing between the $3'$ end of the codon and the $5'$ end of the anticodon is less strict.
Therefore,the third base (the $3'$ position) of the codon is the most flexible and is responsible for the degeneracy of the genetic code.
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87
BiologyMediumMCQAIPMT · 2003
If the $25^{th}$ codon $UAU$ in a polypeptide chain of $50$ amino acids undergoes a mutation to $UAA$,what will be the result?
A
$A$ polypeptide of $49$ amino acids will be formed.
B
$A$ polypeptide of $25$ amino acids will be formed.
C
$A$ polypeptide of $24$ amino acids will be formed.
D
Two polypeptides,one of $24$ amino acids and another of $25$ amino acids,will be formed.
Solution
(C) The codon $UAU$ codes for the amino acid Tyrosine.
$UAA$ is a stop codon (nonsense codon) that does not code for any amino acid and signals the termination of protein synthesis.
Since the mutation occurs at the $25^{th}$ position,the translation process will stop after the $24^{th}$ amino acid is incorporated.
Therefore,a polypeptide chain consisting of only $24$ amino acids will be formed.
$UAA$ is a stop codon (nonsense codon) that does not code for any amino acid and signals the termination of protein synthesis.
Since the mutation occurs at the $25^{th}$ position,the translation process will stop after the $24^{th}$ amino acid is incorporated.
Therefore,a polypeptide chain consisting of only $24$ amino acids will be formed.
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88
BiologyMediumMCQAIPMT · 2003
In the genetic code dictionary,how many codons are used to code for the $20$ amino acids?
A
$61$
B
$60$
C
$20$
D
$64$
Solution
(A) The genetic code consists of $64$ total codons.
Out of these $64$ codons,$61$ codons code for amino acids.
The remaining $3$ codons $(UAA, UAG, UGA)$ are stop codons (also known as termination codons) and do not code for any amino acid.
Therefore,$61$ codons are used to code for the $20$ amino acids.
Out of these $64$ codons,$61$ codons code for amino acids.
The remaining $3$ codons $(UAA, UAG, UGA)$ are stop codons (also known as termination codons) and do not code for any amino acid.
Therefore,$61$ codons are used to code for the $20$ amino acids.
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89
BiologyMediumMCQAIPMT · 2003
During transcription,the site on $DNA$ where $RNA$ polymerase binds is called .......
A
Receptor
B
Enhancer
C
Promoter
D
Regulator
Solution
(C) During the process of transcription in prokaryotes and eukaryotes,the $RNA$ polymerase enzyme recognizes and binds to a specific sequence on the $DNA$ template strand known as the promoter. The promoter is located towards the $5'$ end (upstream) of the structural gene and provides the binding site for $RNA$ polymerase to initiate transcription.
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90
BiologyMediumMCQAIPMT · 2003
In the $lac$ operon, what does the term "$lac$" refer to?
A
Lac insect
B
$1,00,000$ (a number)
C
Lactose
D
Laccase
Solution
(C) The $lac$ operon is a classic example of an inducible operon in $E. coli$ that regulates the metabolism of lactose.
In the term "$lac$ operon", "$lac$" stands for lactose.
The operon consists of genes that code for enzymes required to transport and break down lactose into glucose and galactose when lactose is present in the medium.
In the term "$lac$ operon", "$lac$" stands for lactose.
The operon consists of genes that code for enzymes required to transport and break down lactose into glucose and galactose when lactose is present in the medium.
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91
BiologyEasyMCQAIPMT · 2003
Coconut milk is used in tissue culture because it contains:
A
Gibberellin
B
Cytokinin
C
Auxin
D
Ethylene
Solution
(B) Coconut milk is widely used in plant tissue culture media because it is a rich natural source of $Cytokinin$ (specifically $Zeatin$).
$Cytokinins$ are plant hormones that promote cell division (cytokinesis) and help in the growth and differentiation of plant tissues in vitro.
Therefore,the presence of $Cytokinin$ makes coconut milk an essential supplement for successful tissue culture.
$Cytokinins$ are plant hormones that promote cell division (cytokinesis) and help in the growth and differentiation of plant tissues in vitro.
Therefore,the presence of $Cytokinin$ makes coconut milk an essential supplement for successful tissue culture.
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92
BiologyMediumMCQAIPMT · 2003
Which of the following correctly describes the structure of homologous organs?
A
Organs that appear only during embryonic development and disappear in the adult.
B
Organs that have similar internal structural anatomy but perform different functions.
C
Organs that have different internal structural anatomy but perform the same function.
D
Organs that currently perform no function but were important in their ancestors.
Solution
(B) Homologous organs are defined as those organs that share a common evolutionary origin and possess a similar basic anatomical structure,even though they may have evolved to perform different functions due to adaptation to different environments. This phenomenon is known as divergent evolution. For example,the forelimbs of whales,bats,cheetahs,and humans share the same structural plan of bones (humerus,radius,ulna,carpals,metacarpals,and phalanges) but are used for different purposes like swimming,flying,running,and grasping,respectively.
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93
BiologyMediumMCQAIPMT · 2003
Convergent evolution is typically represented by which of the following?
A
Dogfish and Whale
B
Rat and Dog
C
Bacteria and Protozoa
D
Starfish and Cuttlefish
Solution
(A) Convergent evolution occurs when unrelated species evolve similar traits as a result of having to adapt to similar environments or ecological niches.
Dogfish (a cartilaginous fish) and Whale (a marine mammal) both possess streamlined bodies adapted for swimming in an aquatic environment,despite belonging to completely different evolutionary lineages.
This similarity in body shape is an example of analogous organs,which are a result of convergent evolution.
Dogfish (a cartilaginous fish) and Whale (a marine mammal) both possess streamlined bodies adapted for swimming in an aquatic environment,despite belonging to completely different evolutionary lineages.
This similarity in body shape is an example of analogous organs,which are a result of convergent evolution.
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94
BiologyMediumMCQAIPMT · 2003
Which of the following sequences was proposed by Darwin and Wallace for biological evolution?
A
Variations,Natural selection,Overproduction of offspring,Constancy of population size
B
Overproduction of offspring,Variations,Constancy of population size,Natural selection
C
Variations,Constancy of population size,Overproduction of offspring,Natural selection
D
Overproduction of offspring,Constancy of population size,Variations,Natural selection
Solution
(D) The theory of natural selection proposed by Darwin and Wallace follows a logical sequence:
$1$. Overproduction of offspring: All organisms have a high reproductive potential.
$2$. Constancy of population size: Despite high reproductive rates,the population size of a species remains relatively constant due to limited resources.
$3$. Variations: Individuals within a population exhibit variations.
$4$. Natural selection: Individuals with favorable variations survive and reproduce,leading to evolution.
Therefore,the correct sequence is: Overproduction of offspring $\rightarrow$ Constancy of population size $\rightarrow$ Variations $\rightarrow$ Natural selection.
$1$. Overproduction of offspring: All organisms have a high reproductive potential.
$2$. Constancy of population size: Despite high reproductive rates,the population size of a species remains relatively constant due to limited resources.
$3$. Variations: Individuals within a population exhibit variations.
$4$. Natural selection: Individuals with favorable variations survive and reproduce,leading to evolution.
Therefore,the correct sequence is: Overproduction of offspring $\rightarrow$ Constancy of population size $\rightarrow$ Variations $\rightarrow$ Natural selection.
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95
BiologyMediumMCQAIPMT · 2003
In a population,random genetic drift is typically the result of which of the following?
A
Large population size
B
Organisms with high genetic variation
C
Inbreeding in a population
D
Constant mutation rate
Solution
(C) Genetic drift refers to the change in the frequency of an existing gene variant (allele) in a population due to random sampling of organisms.
It is most pronounced in small populations where chance events can significantly alter allele frequencies.
Inbreeding,which is the mating of closely related individuals,often occurs in small,isolated populations and exacerbates the effects of genetic drift by reducing genetic diversity and increasing homozygosity.
Therefore,inbreeding is a key factor associated with the occurrence and impact of genetic drift in a population.
It is most pronounced in small populations where chance events can significantly alter allele frequencies.
Inbreeding,which is the mating of closely related individuals,often occurs in small,isolated populations and exacerbates the effects of genetic drift by reducing genetic diversity and increasing homozygosity.
Therefore,inbreeding is a key factor associated with the occurrence and impact of genetic drift in a population.
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96
BiologyMediumMCQAIPMT · 2003
In a population that is not evolving, which of the following can cause a change in allele frequencies in a specific direction?
A
Migration
B
Mutation
C
Genetic drift
D
Natural selection
Solution
(D) According to the $Hardy-Weinberg$ principle, allele frequencies in a population remain constant in the absence of evolutionary forces.
$Natural selection$ is the only evolutionary force that causes a change in allele frequencies in a specific, adaptive direction.
$Migration$ (gene flow) and $Mutation$ can change allele frequencies but are not inherently directional in terms of adaptation.
$Genetic drift$ causes random, non-directional changes in allele frequencies, especially in small populations.
$Natural selection$ is the only evolutionary force that causes a change in allele frequencies in a specific, adaptive direction.
$Migration$ (gene flow) and $Mutation$ can change allele frequencies but are not inherently directional in terms of adaptation.
$Genetic drift$ causes random, non-directional changes in allele frequencies, especially in small populations.
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97
BiologyMediumMCQAIPMT · 2003
Industrial melanism is an example of what?
A
Protective adaptation of skin against ultraviolet radiation
B
Drug resistance
C
Darkening of skin due to industrial smoke
D
Protective resemblance with the surroundings
Solution
(D) Industrial melanism is a classic example of natural selection in action.
During the Industrial Revolution in England,the tree trunks became dark due to soot and smoke from factories.
The light-colored moths (Biston betularia) were easily spotted by predators on these dark trunks,while the dark-colored (melanic) moths were camouflaged.
This provided the dark-colored moths with a survival advantage,allowing them to reproduce more successfully.
Therefore,industrial melanism is an example of protective resemblance with the surroundings,which helps the organism survive in its environment.
During the Industrial Revolution in England,the tree trunks became dark due to soot and smoke from factories.
The light-colored moths (Biston betularia) were easily spotted by predators on these dark trunks,while the dark-colored (melanic) moths were camouflaged.
This provided the dark-colored moths with a survival advantage,allowing them to reproduce more successfully.
Therefore,industrial melanism is an example of protective resemblance with the surroundings,which helps the organism survive in its environment.
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98
BiologyMediumMCQAIPMT · 2003
Currently,the mitochondrial $DNA$ $(mt-DNA)$ sequence and the $Y$-chromosome are considered for the study of human evolution because ........
A
They can also be studied from fossil samples.
B
They are small,making them easy to study.
C
They are uniparentally inherited and do not undergo recombination.
D
Their structure is extensively known.
Solution
(C) Mitochondrial $DNA$ $(mt-DNA)$ is inherited maternally (from the mother),and the $Y$-chromosome is inherited paternally (from the father).
Because they are inherited from a single parent (uniparental inheritance),they do not undergo genetic recombination during fertilization.
This lack of recombination allows scientists to trace lineages back through generations without the 'shuffling' of genetic material that occurs with autosomal chromosomes.
Therefore,they serve as excellent molecular markers for tracking human migration and evolutionary history.
Because they are inherited from a single parent (uniparental inheritance),they do not undergo genetic recombination during fertilization.
This lack of recombination allows scientists to trace lineages back through generations without the 'shuffling' of genetic material that occurs with autosomal chromosomes.
Therefore,they serve as excellent molecular markers for tracking human migration and evolutionary history.
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99
BiologyMediumMCQAIPMT · 2003
Which of the following was $NOT$ considered a part of Darwin's theory of natural selection in organic evolution?
A
Discontinuous variations
B
Parasites and predators as natural enemies
C
Survival of the fittest
D
Struggle for existence
Solution
(A) Darwin's theory of natural selection is primarily based on the concepts of 'Struggle for existence','Survival of the fittest',and 'Variations'.
Darwin believed that variations are small,directional,and continuous.
He did not consider 'Discontinuous variations' (mutations) as a part of his theory,as the concept of mutation was later introduced by Hugo de Vries.
Darwin believed that variations are small,directional,and continuous.
He did not consider 'Discontinuous variations' (mutations) as a part of his theory,as the concept of mutation was later introduced by Hugo de Vries.
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100
BiologyMediumMCQAIPMT · 2003
$ELISA$ is used to detect viruses where the key reagent is . . .
A
$DNA$ probe
B
$RNA$ase
C
Alkaline phosphatase
D
Catalase
Solution
(C) $ELISA$ (Enzyme-Linked Immunosorbent Assay) is a diagnostic technique used to detect the presence of antigens or antibodies in a sample.
In the context of viral detection,$ELISA$ utilizes an enzyme-linked antibody to identify viral antigens.
The enzyme commonly used as a reporter molecule in this process is alkaline phosphatase or horseradish peroxidase.
These enzymes catalyze a reaction that produces a detectable color change,confirming the presence of the virus.
In the context of viral detection,$ELISA$ utilizes an enzyme-linked antibody to identify viral antigens.
The enzyme commonly used as a reporter molecule in this process is alkaline phosphatase or horseradish peroxidase.
These enzymes catalyze a reaction that produces a detectable color change,confirming the presence of the virus.
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