BiologyQ1–100 of 171 questions
Page 1 of 2 · English
1
BiologyMediumMCQAIPMT · 1997
$A$ group of plants with similar traits of any rank is known as:
A
Species
B
Genus
C
Order
D
Taxon
Solution
(D) In biological classification,a $Taxon$ (plural: $Taxa$) represents a group of organisms of any rank that share similar traits or characteristics. It is a fundamental unit of classification used in taxonomy to categorize living beings at various levels,such as species,genus,family,order,class,phylum,or kingdom.
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2
BiologyMediumMCQAIPMT · 1997
The genetic material in viruses is
A
Only $RNA$
B
Only $DNA$
C
$RNA$ and $DNA$ both
D
$RNA$ or $DNA$ i.e. one nucleic acid in a virus
Solution
(D) Viruses are nucleoprotein entities where the genetic material is infectious.
In viruses,the genetic material is either $RNA$ or $DNA$,but never both.
Therefore,a virus contains only one type of nucleic acid as its genetic material.
In viruses,the genetic material is either $RNA$ or $DNA$,but never both.
Therefore,a virus contains only one type of nucleic acid as its genetic material.
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3
BiologyEasyMCQAIPMT · 1997
The site of respiration in bacteria is
A
Episome
B
Mesosome (cytoplasmic membrane)
C
Ribosome
D
Microsome
Solution
(B) In bacteria,the plasma membrane forms specialized invaginations known as $Mesosomes$.
These structures are rich in respiratory enzymes and are the primary sites for aerobic respiration,as bacteria lack mitochondria.
These structures are rich in respiratory enzymes and are the primary sites for aerobic respiration,as bacteria lack mitochondria.
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4
BiologyEasyMCQAIPMT · 1997
The $DNA$ of $E. coli$ is
A
Single stranded and linear
B
Single stranded and circular
C
Double stranded and linear
D
Double stranded and circular
Solution
(D) $E. coli$ (Escherichia coli) is a prokaryotic organism.
Prokaryotic $DNA$ is typically organized as a single,double-stranded,circular molecule located in the nucleoid region.
Therefore,the correct structure of $E. coli$ $DNA$ is double-stranded and circular.
Prokaryotic $DNA$ is typically organized as a single,double-stranded,circular molecule located in the nucleoid region.
Therefore,the correct structure of $E. coli$ $DNA$ is double-stranded and circular.
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5
BiologyEasyMCQAIPMT · 1997
Which of the following is a free-living,aerobic,non-photosynthetic,nitrogen-fixing bacterium?
A
Rhizobium
B
Azotobacter
C
Nostoc
D
Azospirillum
Solution
(B) $Azotobacter$ is a free-living,aerobic,and non-photosynthetic bacterium that fixes atmospheric nitrogen.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.
$Nostoc$ is a photosynthetic cyanobacterium that can fix nitrogen.
$Azospirillum$ is an associative symbiotic nitrogen-fixing bacterium.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.
$Nostoc$ is a photosynthetic cyanobacterium that can fix nitrogen.
$Azospirillum$ is an associative symbiotic nitrogen-fixing bacterium.
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6
BiologyMediumMCQAIPMT · 1997
Algae are in the same major group of plants as are the
A
Mosses
B
Liverworts
C
Fungi
D
Ferns
Solution
(C) Algae and fungi were historically classified together under the group $Thallophyta$ in the older classification systems because both possess a simple,undifferentiated plant body known as a thallus. While modern taxonomy places fungi in a separate kingdom $(Kingdom \ Fungi)$,in the context of traditional botanical classification questions,they are often grouped together based on their simple thalloid structure.
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7
BiologyMediumMCQAIPMT · 1997
Brown algae is characterised by the presence of
A
Phycocyanin
B
Phycoerythrin
C
Fucoxanthin
D
Haematochrome
Solution
(C) Brown algae (Phaeophyceae) are characterised by the presence of chlorophyll $a$,$c$,carotenoids,and xanthophylls.
Fucoxanthin is the dominant pigment that gives them their characteristic brown color.
Examples include $Fucus$ and $Sargassum$.
Fucoxanthin is the dominant pigment that gives them their characteristic brown color.
Examples include $Fucus$ and $Sargassum$.
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8
BiologyMediumMCQAIPMT · 1997
In $Ulothrix$,sexual reproduction is by
A
Isogamy
B
Anisogamy
C
Oogamy
D
Conjugation
Solution
(A) In $Ulothrix$,sexual reproduction occurs through the fusion of morphologically similar gametes,a process known as $Isogamy$.
These gametes are flagellated and motile.
Therefore,the correct mode of sexual reproduction in $Ulothrix$ is $Isogamy$.
These gametes are flagellated and motile.
Therefore,the correct mode of sexual reproduction in $Ulothrix$ is $Isogamy$.
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9
BiologyMediumMCQAIPMT · 1997
Bryophytes can be separated from algae,because they
A
Are thalloid forms
B
Have no conducting tissue
C
Possess archegonia
D
Contain chloroplast
Solution
(C) . The female sex organ in bryophytes is the archegonium. It is a flask-shaped structure that is distinguishable into a long neck and a globular swollen venter. Algae lack such specialized multicellular sex organs,whereas bryophytes possess them,which is a key evolutionary advancement.
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10
BiologyEasyMCQAIPMT · 1997
Which of the following is a living fossil?
A
Pinus
B
Ginkgo
C
Thuja
D
Deodar
Solution
(B) $Ginkgo$ $biloba$, a gymnosperm, is a classic example of a living fossil. It is a monotypic genus, meaning it is the only surviving member of its entire division, $Ginkgophyta$. At present, it is naturally confined to the eastern part of China and Japan.
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11
BiologyMediumMCQAIPMT · 1997
Radial symmetry is often exhibited by animals having
A
One opening of alimentary canal
B
Aquatic mode of living
C
Benthos/sedentary nature
D
Ciliary mode of feeding
Solution
(C) Radial symmetry is a body plan where the organism can be divided into identical halves by any plane passing through the central axis. This type of symmetry is most commonly observed in animals that are sessile (sedentary) or slow-moving,such as Cnidarians and adult Echinoderms. Being sedentary allows these animals to interact with their environment from all directions equally,making radial symmetry an evolutionary advantage for such a lifestyle.
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12
BiologyEasyMCQAIPMT · 1997
What is common among silver fish,scorpion,crab,and honey bee?
A
Compound eyes
B
Poison glands
C
Jointed legs
D
Metamorphosis
Solution
(C) Silver fish,scorpion,crab,and honey bee all belong to the phylum $Arthropoda$. The most characteristic feature of all members of the phylum $Arthropoda$ is the presence of jointed appendages (jointed legs).
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13
BiologyMediumMCQAIPMT · 1997
If juvenile hormone is absent when a silkworm moults,it will:
A
Moult into another larval stage
B
Moult into a pupa
C
Moult into an adult
D
Die
Solution
(B) In insects,the metamorphosis process is regulated by hormones. Juvenile hormone $(JH)$ maintains the larval characteristics during each moult. When the level of $JH$ is high,the larva moults into another larval stage. As the larva grows,the concentration of $JH$ decreases. When $JH$ is absent or at a very low level during a moult,the larva undergoes metamorphosis and transforms into a pupa.
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14
BiologyMediumMCQAIPMT · 1997
Which two families dominate in having the maximum number of useful plants?
A
Fabaceae and Poaceae
B
Liliaceae and Solanaceae
C
Malvaceae and Brassicaceae
D
Liliaceae and Poaceae
Solution
(A) $Fabaceae$,commonly called the legume family,is the $4^{th}$ largest and $2^{nd}$ most valuable family,containing approximately $600$ genera.
$Poaceae$,commonly called the cereal or grass family,is the $3^{rd}$ largest family globally,but in the context of Indian flora,it constitutes the largest family with $620$ genera.
Together,these two families provide the majority of food crops,pulses,and fodder,making them the most useful families for human consumption.
$Poaceae$,commonly called the cereal or grass family,is the $3^{rd}$ largest family globally,but in the context of Indian flora,it constitutes the largest family with $620$ genera.
Together,these two families provide the majority of food crops,pulses,and fodder,making them the most useful families for human consumption.
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15
BiologyMediumMCQAIPMT · 1997
At maturity,which of the following is non-nucleated?
A
Sieve cell
B
Companion cell
C
Palisade cell
D
Cortical cell
Solution
(A) In mature phloem tissue,the sieve tube elements lose their nucleus to facilitate the efficient transport of food materials.
While sieve cells (in gymnosperms) and sieve tube elements (in angiosperms) are non-nucleated at maturity,companion cells remain nucleated to regulate the metabolic activities of the sieve elements.
Palisade cells and cortical cells are parenchyma cells that retain their nucleus throughout their life.
While sieve cells (in gymnosperms) and sieve tube elements (in angiosperms) are non-nucleated at maturity,companion cells remain nucleated to regulate the metabolic activities of the sieve elements.
Palisade cells and cortical cells are parenchyma cells that retain their nucleus throughout their life.
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16
BiologyEasyMCQAIPMT · 1997
The protein present in cartilage is:
A
Chondrin
B
Osein
C
Cartilagin
D
Ossein
Solution
(A) Cartilage is a specialized connective tissue. The matrix of cartilage is composed of a protein called $Chondrin$.
$Ossein$ is the protein found in the matrix of bones.
Therefore,the correct option is $A$.
$Ossein$ is the protein found in the matrix of bones.
Therefore,the correct option is $A$.
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17
BiologyEasyMCQAIPMT · 1997
Which of the following is an agranulocyte?
A
Lymphocyte
B
Eosinophil
C
Basophil
D
Neutrophil
Solution
(A) Agranulocytes are leukocytes that lack granules in the cytoplasm.
Lymphocytes and monocytes are the two main types of agranulocytes.
Since lymphocytes do not have granules in their cytoplasm,they are classified as agranulocytes.
In contrast,eosinophils,basophils,and neutrophils are granulocytes because they contain granules in their cytoplasm.
Lymphocytes and monocytes are the two main types of agranulocytes.
Since lymphocytes do not have granules in their cytoplasm,they are classified as agranulocytes.
In contrast,eosinophils,basophils,and neutrophils are granulocytes because they contain granules in their cytoplasm.
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18
BiologyMediumMCQAIPMT · 1997
The life span of human $WBC$ is approximately
A
Less than $10$ days
B
Between $20$ to $30$ days
C
Between $2$ to $3$ months
D
More than $4$ months
Solution
(A) The life span of human $WBC$s (leukocytes) varies significantly depending on the type of cell.
Granulocytes (neutrophils,eosinophils,and basophils) typically have a short life span,circulating in the blood for $4-8$ hours and residing in tissues for $4-5$ days.
Monocytes have a life span of $10-20$ hours in the blood before migrating into tissues to become macrophages.
Lymphocytes can live for days,months,or even years depending on the immunological memory requirements.
However,in the context of standard multiple-choice questions regarding the average short-lived $WBC$s,the most appropriate answer is less than $10$ days.
Granulocytes (neutrophils,eosinophils,and basophils) typically have a short life span,circulating in the blood for $4-8$ hours and residing in tissues for $4-5$ days.
Monocytes have a life span of $10-20$ hours in the blood before migrating into tissues to become macrophages.
Lymphocytes can live for days,months,or even years depending on the immunological memory requirements.
However,in the context of standard multiple-choice questions regarding the average short-lived $WBC$s,the most appropriate answer is less than $10$ days.
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19
BiologyMediumMCQAIPMT · 1997
In an animal cell,protein synthesis takes place in:
A
Only in the cytoplasm
B
In the cytoplasm as well as in mitochondria
C
In the nucleolus as well as in the cytoplasm
D
Only on ribosomes attached to the nucleus
Solution
(B) Protein synthesis in animal cells occurs primarily in the cytoplasm,where ribosomes translate mRNA into polypeptide chains.
Additionally,mitochondria contain their own $70S$ ribosomes,which are capable of synthesizing specific mitochondrial proteins encoded by the mitochondrial genome.
Therefore,protein synthesis occurs in both the cytoplasm and the mitochondria.
Additionally,mitochondria contain their own $70S$ ribosomes,which are capable of synthesizing specific mitochondrial proteins encoded by the mitochondrial genome.
Therefore,protein synthesis occurs in both the cytoplasm and the mitochondria.
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20
BiologyMediumMCQAIPMT · 1997
Centromere is part of
A
Ribosomes
B
Mitochondria
C
Chromosome
D
Endoplasmic reticulum
Solution
(C) The centromere is a specialized region of the $Chromosome$ that holds the two sister chromatids together.
During cell division ($Mitosis$ and $Meiosis$),the centromere serves as the attachment site for spindle fibers via the kinetochore complex.
It is also known as the primary constriction of the $Chromosome$.
During cell division ($Mitosis$ and $Meiosis$),the centromere serves as the attachment site for spindle fibers via the kinetochore complex.
It is also known as the primary constriction of the $Chromosome$.
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21
BiologyMediumMCQAIPMT · 1997
$A$ cofactor $(prosthetic\ group)$ is a part of a holoenzyme. It is:
A
$A$ loosely attached inorganic part
B
An accessory non-protein substance attached firmly
C
$A$ loosely attached organic part
D
None of these
Solution
(B) holoenzyme consists of an apoenzyme (protein part) and a cofactor (non-protein part).
Cofactors are classified into three types: prosthetic groups, co-enzymes, and metal ions.
Prosthetic groups are organic compounds that are distinguished from other cofactors in that they are tightly bound to the apoenzyme.
Therefore, a prosthetic group is an accessory non-protein substance that is attached firmly to the enzyme.
Cofactors are classified into three types: prosthetic groups, co-enzymes, and metal ions.
Prosthetic groups are organic compounds that are distinguished from other cofactors in that they are tightly bound to the apoenzyme.
Therefore, a prosthetic group is an accessory non-protein substance that is attached firmly to the enzyme.
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22
BiologyEasyMCQAIPMT · 1997
Which of the following structures will not be common to a mitotic cell of a higher plant?
A
Cell plate
B
Centromere
C
Centriole
D
Spindle fibre
Solution
(C) Centrioles are cylindrical organelles that play a crucial role in the formation of the spindle apparatus during nuclear division in animal cells.
In higher plants,cell division occurs without centrioles. Instead,they form spindle fibers using microtubule organizing centers (MTOCs) located in the cytoplasm.
Therefore,centrioles are absent in the mitotic cells of higher plants.
In higher plants,cell division occurs without centrioles. Instead,they form spindle fibers using microtubule organizing centers (MTOCs) located in the cytoplasm.
Therefore,centrioles are absent in the mitotic cells of higher plants.
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23
BiologyMediumMCQAIPMT · 1997
How many mitotic divisions are needed for a single cell to produce $128$ cells?
A
$7$
B
$14$
C
$28$
D
$32$
Solution
(A) In mitosis,one cell divides to form two daughter cells.
If $n$ is the number of mitotic divisions,the number of cells produced is given by the formula $2^n$.
We need to find $n$ such that $2^n = 128$.
Since $128 = 2^7$,it follows that $n = 7$.
Therefore,$7$ mitotic divisions are required to produce $128$ cells from a single cell.
If $n$ is the number of mitotic divisions,the number of cells produced is given by the formula $2^n$.
We need to find $n$ such that $2^n = 128$.
Since $128 = 2^7$,it follows that $n = 7$.
Therefore,$7$ mitotic divisions are required to produce $128$ cells from a single cell.
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24
BiologyMediumMCQAIPMT · 1997
During cell division in apical meristem,the nuclear membrane reappears in:
A
Interphase
B
Telophase
C
Prophase
D
$S$ phase
Solution
(B) During the process of mitosis,the $M$ phase consists of four stages: Prophase,Metaphase,Anaphase,and Telophase.
In Prophase,the nuclear envelope and nucleolus disappear.
In Telophase,which is the final stage of karyokinesis,the chromosomes cluster at opposite spindle poles and their identity is lost as discrete elements.
The nuclear envelope assembles around the chromosome clusters at each pole,and the nucleolus,Golgi complex,and endoplasmic reticulum reform,marking the reappearance of the nuclear membrane.
In Prophase,the nuclear envelope and nucleolus disappear.
In Telophase,which is the final stage of karyokinesis,the chromosomes cluster at opposite spindle poles and their identity is lost as discrete elements.
The nuclear envelope assembles around the chromosome clusters at each pole,and the nucleolus,Golgi complex,and endoplasmic reticulum reform,marking the reappearance of the nuclear membrane.
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25
BiologyMediumMCQAIPMT · 1997
When a cell is fully turgid, which of the following will be zero?
A
Wall pressure
B
Osmotic pressure
C
Turgor pressure
D
Water potential
Solution
(D) In a fully turgid cell, the cell wall exerts a pressure equal and opposite to the turgor pressure.
As a result, the net movement of water into the cell stops because the cell is in equilibrium with the surrounding water.
The water potential $(\Psi_w)$ of a cell is given by the equation $\Psi_w = \Psi_s + \Psi_p$.
In a fully turgid cell, the solute potential $(\Psi_s)$ is equal and opposite to the pressure potential $(\Psi_p)$, making the net water potential $(\Psi_w)$ equal to $0$.
As a result, the net movement of water into the cell stops because the cell is in equilibrium with the surrounding water.
The water potential $(\Psi_w)$ of a cell is given by the equation $\Psi_w = \Psi_s + \Psi_p$.
In a fully turgid cell, the solute potential $(\Psi_s)$ is equal and opposite to the pressure potential $(\Psi_p)$, making the net water potential $(\Psi_w)$ equal to $0$.
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26
BiologyMediumMCQAIPMT · 1997
Bidirectional translocation of minerals takes place in
A
Phloem
B
Xylem
C
Parenchyma
D
Cambium
Solution
(A) The translocation of minerals and organic nutrients in plants occurs through the $Phloem$.
Unlike $Xylem$,which primarily transports water and minerals in an upward (unidirectional) direction from roots to leaves,$Phloem$ is responsible for the bidirectional transport of food (sucrose) and minerals from the source (leaves) to the sink (roots,fruits,or growing regions).
Therefore,the correct answer is $Phloem$.
Unlike $Xylem$,which primarily transports water and minerals in an upward (unidirectional) direction from roots to leaves,$Phloem$ is responsible for the bidirectional transport of food (sucrose) and minerals from the source (leaves) to the sink (roots,fruits,or growing regions).
Therefore,the correct answer is $Phloem$.
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27
BiologyMediumMCQAIPMT · 1997
Which of the following is not caused by the deficiency of mineral nutrition?
A
Necrosis
B
Chlorosis
C
Etiolation
D
Shortening of internodes
Solution
(C) $Etiolation$ is a phenomenon observed in plants grown in the absence of light (darkness). It is characterized by long, weak stems, smaller leaves, and a pale yellow or white appearance due to the lack of chlorophyll. It is not caused by mineral deficiency, whereas $Necrosis$, $Chlorosis$, and $Shortening$ \text{ of internodes} are well-known symptoms of mineral nutrient deficiencies.
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28
BiologyMediumMCQAIPMT · 1997
$NADPH_2$ (or $NADPH + H^+$) is generated through:
A
Glycolysis
B
Photosystem-$I$
C
Photosystem-$II$
D
Anaerobic respiration
Solution
(B) In the light-dependent reactions of photosynthesis,the $Z$-scheme involves both Photosystem-$II$ $(PS-II)$ and Photosystem-$I$ $(PS-I)$.
During the electron transport chain,electrons are excited in $PS-I$ and passed through a series of carriers to the enzyme $NADP$ reductase.
This enzyme facilitates the reduction of $NADP^+$ to $NADPH + H^+$ (often referred to as $NADPH_2$) using electrons derived from the electron transport chain and protons from the stroma.
Therefore,the generation of $NADPH_2$ is specifically associated with the activity of $PS-I$ in the non-cyclic photophosphorylation pathway.
During the electron transport chain,electrons are excited in $PS-I$ and passed through a series of carriers to the enzyme $NADP$ reductase.
This enzyme facilitates the reduction of $NADP^+$ to $NADPH + H^+$ (often referred to as $NADPH_2$) using electrons derived from the electron transport chain and protons from the stroma.
Therefore,the generation of $NADPH_2$ is specifically associated with the activity of $PS-I$ in the non-cyclic photophosphorylation pathway.
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29
BiologyEasyMCQAIPMT · 1997
The core metal of chlorophyll is
A
$Fe$
B
$Mg$
C
$Ni$
D
$Cu$
Solution
(B) The core metal of chlorophyll is $Mg$ (Magnesium).
Chlorophyll is a porphyrin derivative with a central magnesium atom coordinated to four nitrogen atoms of the porphyrin ring.
This central $Mg^{2+}$ ion is essential for the light-harvesting function of chlorophyll molecules during photosynthesis.
If the central $Mg$ atom is replaced by $Fe$ (Iron),the molecule is converted into a different pigment,such as those found in cytochromes,which are involved in electron transport rather than light absorption.
Chlorophyll is a porphyrin derivative with a central magnesium atom coordinated to four nitrogen atoms of the porphyrin ring.
This central $Mg^{2+}$ ion is essential for the light-harvesting function of chlorophyll molecules during photosynthesis.
If the central $Mg$ atom is replaced by $Fe$ (Iron),the molecule is converted into a different pigment,such as those found in cytochromes,which are involved in electron transport rather than light absorption.
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30
BiologyMediumMCQAIPMT · 1997
The chlorophyll $a$ molecule has which of the following at its carbon atom $3$ of the pyrrole ring $II$?
A
Aldehyde group
B
Methyl group
C
Carboxylic group
D
Magnesium
Solution
(B) Chlorophyll $a$ is a pigment molecule consisting of a porphyrin ring with a central magnesium atom.
In the structure of chlorophyll $a$,the pyrrole ring $II$ contains a methyl group $(-CH_3)$ attached to its carbon atom $3$.
In contrast,chlorophyll $b$ has an aldehyde group $(-CHO)$ at the same position.
In the structure of chlorophyll $a$,the pyrrole ring $II$ contains a methyl group $(-CH_3)$ attached to its carbon atom $3$.
In contrast,chlorophyll $b$ has an aldehyde group $(-CHO)$ at the same position.
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31
BiologyEasyMCQAIPMT · 1997
What are the end products of fermentation when sugars are used as raw material?
A
Alcohol and acetic acid
B
$CO_2$
C
Alcohol and $CO_2$
D
Alcohol
Solution
(C) Fermentation is a type of anaerobic respiration that occurs in microorganisms like yeast.
In this process, sugars (like glucose) are incompletely oxidized in the absence of oxygen.
The chemical equation for alcoholic fermentation is: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2 + \text{Energy}$.
Therefore, the end products are alcohol (ethanol) and carbon dioxide $(CO_2)$.
In this process, sugars (like glucose) are incompletely oxidized in the absence of oxygen.
The chemical equation for alcoholic fermentation is: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2 + \text{Energy}$.
Therefore, the end products are alcohol (ethanol) and carbon dioxide $(CO_2)$.
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32
BiologyMediumMCQAIPMT · 1997
The mechanism of $ATP$ formation in both chloroplasts and mitochondria is explained by:
A
Relay pump theory of Godlewski
B
Cholodny-Went model
C
Chemiosmotic theory
D
Munch's hypothesis (mass flow model)
Solution
(C) The mechanism of $ATP$ synthesis in both chloroplasts (photophosphorylation) and mitochondria (oxidative phosphorylation) is explained by the Chemiosmotic theory,proposed by Peter Mitchell.
This theory states that $ATP$ synthesis is linked to the development of a proton gradient across the membrane (thylakoid membrane in chloroplasts and inner mitochondrial membrane in mitochondria).
The movement of protons $(H^+)$ from the intermembrane space or thylakoid lumen back into the matrix or stroma through the $F_0-F_1$ $ATP$ synthase complex provides the energy required for the phosphorylation of $ADP$ to $ATP$.
This theory states that $ATP$ synthesis is linked to the development of a proton gradient across the membrane (thylakoid membrane in chloroplasts and inner mitochondrial membrane in mitochondria).
The movement of protons $(H^+)$ from the intermembrane space or thylakoid lumen back into the matrix or stroma through the $F_0-F_1$ $ATP$ synthase complex provides the energy required for the phosphorylation of $ADP$ to $ATP$.
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33
BiologyMediumMCQAIPMT · 1997
The link between glycolysis and the Kreb's cycle is:
A
Citric acid
B
Acetyl $CoA$
C
Succinic acid
D
Oxaloacetic acid
Solution
(B) The link between glycolysis and the Kreb's cycle is Acetyl $CoA$.
Glycolysis occurs in the cytoplasm and produces pyruvate.
This pyruvate is then transported into the mitochondria,where it undergoes oxidative decarboxylation to form Acetyl $CoA$.
Acetyl $CoA$ acts as a connecting link because it is the substrate that enters the Kreb's cycle ($TCA$ cycle) for further oxidation.
Glycolysis occurs in the cytoplasm and produces pyruvate.
This pyruvate is then transported into the mitochondria,where it undergoes oxidative decarboxylation to form Acetyl $CoA$.
Acetyl $CoA$ acts as a connecting link because it is the substrate that enters the Kreb's cycle ($TCA$ cycle) for further oxidation.
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34
BiologyMediumMCQAIPMT · 1997
The correct sequence of electron acceptors in $ATP$ synthesis is:
A
$Cyt$ $a, a_3, b, c$
B
$Cyt$ $b, c, a, a_3$
C
$Cyt$ $b, c, a_3, a$
D
$Cyt$ $c, b, a, a_3$
Solution
(B) In the Electron Transport System $(ETS)$,electrons are transferred through a series of cytochromes arranged in order of their increasing redox potential.
This ensures that electron flow occurs in a stepwise manner from a more electronegative compound to the final electron acceptor,which is oxygen $({O_2})$.
The correct sequence of electron transport through the cytochromes is $Cyt$ $b \rightarrow Cyt$ $c \rightarrow Cyt$ $a \rightarrow Cyt$ $a_3$.
This ensures that electron flow occurs in a stepwise manner from a more electronegative compound to the final electron acceptor,which is oxygen $({O_2})$.
The correct sequence of electron transport through the cytochromes is $Cyt$ $b \rightarrow Cyt$ $c \rightarrow Cyt$ $a \rightarrow Cyt$ $a_3$.
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35
BiologyMediumMCQAIPMT · 1997
In $Kreb's$ cycle, $FAD$ acts as an electron acceptor during the conversion of:
A
Succinyl $CoA$ to succinic acid
B
$alpha$-ketoglutarate to succinyl $CoA$
C
Fumaric acid to malic acid
D
Succinic acid to fumaric acid
Solution
(D) In the $Kreb's$ cycle (also known as the Citric Acid Cycle), the conversion of succinic acid to fumaric acid is catalyzed by the enzyme succinate dehydrogenase.
During this specific reaction, $FAD$ (Flavin Adenine Dinucleotide) acts as an electron acceptor and is reduced to $FADH_2$.
This is the only step in the $Kreb's$ cycle where $FAD$ is used as an electron carrier instead of $NAD^+$.
Therefore, the correct option is $D$.
During this specific reaction, $FAD$ (Flavin Adenine Dinucleotide) acts as an electron acceptor and is reduced to $FADH_2$.
This is the only step in the $Kreb's$ cycle where $FAD$ is used as an electron carrier instead of $NAD^+$.
Therefore, the correct option is $D$.
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36
BiologyMediumMCQAIPMT · 1997
Gibberellic acid induces flowering in:
A
In some gymnospermic plants only
B
In long-day plants under short-day conditions
C
In short-day plants under long-day conditions
D
In day-neutral plants under dark conditions
Solution
(B) Gibberellins are plant hormones that can substitute for the requirement of long days in many long-day plants.
When long-day plants are grown under short-day conditions,they fail to flower because the endogenous production of gibberellins is insufficient.
However,if these plants are treated with exogenous gibberellic acid,they can be induced to flower even under non-inductive short-day conditions.
When long-day plants are grown under short-day conditions,they fail to flower because the endogenous production of gibberellins is insufficient.
However,if these plants are treated with exogenous gibberellic acid,they can be induced to flower even under non-inductive short-day conditions.
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37
BiologyMediumMCQAIPMT · 1997
If a tree flowers thrice in a year $(October, January, \text{and } July)$ in Northern India, it is said to be
A
Photosensitive but thermo-insensitive
B
Thermosensitive but photo-insensitive
C
Photo and thermo-insensitive
D
Photo and thermosensitive
Solution
(B) The phenomenon of flowering in response to the duration of light and dark periods is known as photoperiodism.
If a plant flowers multiple times throughout the year under varying day lengths (photoperiods), it indicates that the plant is not dependent on a specific critical day length for flowering.
Therefore, such a plant is considered photo-insensitive.
Since the question implies that the plant flowers regardless of the changing photoperiods across different seasons in Northern India, option $(b)$ is the most appropriate description in the context of typical plant physiology questions where such plants are classified as thermo-sensitive but photo-insensitive.
If a plant flowers multiple times throughout the year under varying day lengths (photoperiods), it indicates that the plant is not dependent on a specific critical day length for flowering.
Therefore, such a plant is considered photo-insensitive.
Since the question implies that the plant flowers regardless of the changing photoperiods across different seasons in Northern India, option $(b)$ is the most appropriate description in the context of typical plant physiology questions where such plants are classified as thermo-sensitive but photo-insensitive.
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38
BiologyMediumMCQAIPMT · 1997
What will be the effect on phytochrome in a plant subjected to continuous red light?
A
Level of phytochrome will decrease
B
Phytochrome will be destroyed
C
Phytochrome synthesis will increase
D
None of these
Solution
(D) Phytochrome exists in two interconvertible forms: $P_{660}$ (red light-absorbing form) and $P_{730}$ (far-red light-absorbing form).
When a plant is exposed to red light,$P_{660}$ is converted into the physiologically active form $P_{730}$.
Continuous red light maintains the phytochrome in the $P_{730}$ form.
Since both forms are interconvertible and the total amount of phytochrome remains relatively stable under these conditions,none of the options suggesting destruction or decrease in level are correct.
Therefore,the correct answer is $(d)$.
When a plant is exposed to red light,$P_{660}$ is converted into the physiologically active form $P_{730}$.
Continuous red light maintains the phytochrome in the $P_{730}$ form.
Since both forms are interconvertible and the total amount of phytochrome remains relatively stable under these conditions,none of the options suggesting destruction or decrease in level are correct.
Therefore,the correct answer is $(d)$.
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39
BiologyMediumMCQAIPMT · 1997
If the pancreas is removed,the compound which remains undigested is
A
Carbohydrates
B
Fats
C
Proteins
D
All of these
Solution
(D) The pancreas is a vital digestive gland that secretes pancreatic juice into the small intestine.
Pancreatic juice contains various enzymes such as trypsin and chymotrypsin for protein digestion,pancreatic amylase for carbohydrate digestion,and pancreatic lipase for fat digestion.
It also contains nucleases that act on nucleic acids.
If the pancreas is removed,these essential enzymes are no longer secreted into the digestive tract.
Consequently,the digestion of carbohydrates,fats,and proteins is severely impaired or ceases entirely.
Therefore,all of these compounds remain undigested.
Pancreatic juice contains various enzymes such as trypsin and chymotrypsin for protein digestion,pancreatic amylase for carbohydrate digestion,and pancreatic lipase for fat digestion.
It also contains nucleases that act on nucleic acids.
If the pancreas is removed,these essential enzymes are no longer secreted into the digestive tract.
Consequently,the digestion of carbohydrates,fats,and proteins is severely impaired or ceases entirely.
Therefore,all of these compounds remain undigested.
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40
BiologyMediumMCQAIPMT · 1997
What is common among amylase,rennin,and trypsin?
A
These all are proteins
B
These all are proteolytic enzymes
C
These are produced in the stomach
D
These act at a $pH$ lower than $7$
Solution
(A) All enzymes are biological catalysts that are essentially proteinaceous in nature. Amylase,rennin,and trypsin are all enzymes; therefore,they are all proteins. While they differ in their specific substrates and sites of action,their fundamental chemical composition as proteins is the common factor.
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41
BiologyEasyMCQAIPMT · 1997
Which one of the following vitamins can be synthesized by bacteria inside the gut?
A
$B_1$
B
$A$
C
$D$
D
$K$
Solution
(D) The human gut microbiome,particularly in the large intestine,contains symbiotic bacteria such as $Escherichia \ coli$.
These bacteria are capable of synthesizing certain essential vitamins,most notably Vitamin $K$ and various $B$ vitamins (such as $B_{12}$,$B_7$,and $B_9$).
Among the given options,Vitamin $K$ is the most well-known vitamin synthesized by gut flora.
Note: While some $B$ vitamins are synthesized,$B_1$ (thiamine) is primarily obtained through diet,making $K$ the most accurate answer in the context of typical biological curriculum.
These bacteria are capable of synthesizing certain essential vitamins,most notably Vitamin $K$ and various $B$ vitamins (such as $B_{12}$,$B_7$,and $B_9$).
Among the given options,Vitamin $K$ is the most well-known vitamin synthesized by gut flora.
Note: While some $B$ vitamins are synthesized,$B_1$ (thiamine) is primarily obtained through diet,making $K$ the most accurate answer in the context of typical biological curriculum.
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42
BiologyMediumMCQAIPMT · 1997
In lungs,the air is separated from the venous blood through:
A
Squamous epithelium + endothelium of blood vessel
B
Squamous epithelium + tunica media blood vessel
C
Transitional epithelium + tunica external blood vessel
D
None of these
Solution
(A) The diffusion membrane of the lungs is composed of three layers: the thin squamous epithelium of the alveoli,the endothelium of the alveolar capillaries,and the basement substance in between them.
Therefore,the air in the alveoli is separated from the blood in the capillaries by the squamous epithelium of the alveoli and the endothelium of the blood vessel.
Therefore,the air in the alveoli is separated from the blood in the capillaries by the squamous epithelium of the alveoli and the endothelium of the blood vessel.
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43
BiologyMediumMCQAIPMT · 1997
The basic functional unit of the human kidney is:
A
Nephron
B
Pyramid
C
Nephridia
D
Henle's loop
Solution
(A) The $Nephron$ is the structural and functional unit of the kidney, which extracts waste products from the blood.
Each kidney contains approximately one million nephrons.
It is composed of two main parts: the $Bowman's$ capsule and the renal tubule.
Therefore, the correct option is $A$.
Each kidney contains approximately one million nephrons.
It is composed of two main parts: the $Bowman's$ capsule and the renal tubule.
Therefore, the correct option is $A$.
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44
BiologyMediumMCQAIPMT · 1997
In ureotelic animals,urea is formed by
A
Ornithine cycle
B
Cori cycle
C
Krebs cycle
D
$EMP$ pathway
Solution
(A) In ureotelic animals,such as mammals and many terrestrial amphibians,ammonia produced by metabolism is converted into urea in the liver.
This process is known as the Ornithine cycle,also called the Urea cycle or Krebs-Henseleit cycle.
In this cycle,ammonia and carbon dioxide are combined to form urea,which is less toxic and requires less water for excretion compared to ammonia.
This process is known as the Ornithine cycle,also called the Urea cycle or Krebs-Henseleit cycle.
In this cycle,ammonia and carbon dioxide are combined to form urea,which is less toxic and requires less water for excretion compared to ammonia.
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45
BiologyEasyMCQAIPMT · 1997
Nissl's granules are present in the ....... and are made up of ....... respectively.
A
Muscle cells and deoxyribonucleic acid
B
Mast cells and $RNA$
C
Osteocytes and $DNA$
D
Neuron and $RNA$
Solution
(D) Nissl's granules are characteristic structures found in the cell body (cyton) and dendrites of neurons.
These granules are composed of rough endoplasmic reticulum $(RER)$ and free ribosomes.
Since ribosomes are primarily composed of ribosomal $RNA$ $(rRNA)$ and proteins,they are essentially sites of protein synthesis.
Therefore,Nissl's granules are present in neurons and are made up of $RNA$ and proteins.
These granules are composed of rough endoplasmic reticulum $(RER)$ and free ribosomes.
Since ribosomes are primarily composed of ribosomal $RNA$ $(rRNA)$ and proteins,they are essentially sites of protein synthesis.
Therefore,Nissl's granules are present in neurons and are made up of $RNA$ and proteins.
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46
BiologyEasyMCQAIPMT · 1997
In the chemistry of vision in mammals,the photosensitive substance is called
A
Selerotin
B
Retinol
C
Rhodopsin
D
Melanin
Solution
(C) $Rhodopsin$ is a visual purple pigment (formed by vitamin $A$ and opsin protein) that is sensitive to dim light.
It plays a crucial role in scotopic vision,which is vision in low-light conditions.
It plays a crucial role in scotopic vision,which is vision in low-light conditions.
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47
BiologyMediumMCQAIPMT · 1997
Hormones thyroxin,adrenaline and the pigment melanin are formed from
A
Tryptophan
B
Glycine
C
Tyrosine
D
Proline
Solution
(C) The amino acid $Tyrosine$ serves as a precursor for the synthesis of several important biological molecules.
$1$. $Thyroxine$ $(T_4)$ and $Triiodothyronine$ $(T_3)$ are thyroid hormones synthesized from $Tyrosine$ residues in the protein $Thyroglobulin$.
$2$. $Adrenaline$ (epinephrine) and $Noradrenaline$ (norepinephrine) are catecholamines synthesized from $Tyrosine$ via the pathway involving $L-DOPA$ and $Dopamine$.
$3$. $Melanin$,the pigment responsible for skin,hair,and eye color,is also synthesized from $Tyrosine$ through a series of enzymatic reactions involving $Tyrosinase$.
$1$. $Thyroxine$ $(T_4)$ and $Triiodothyronine$ $(T_3)$ are thyroid hormones synthesized from $Tyrosine$ residues in the protein $Thyroglobulin$.
$2$. $Adrenaline$ (epinephrine) and $Noradrenaline$ (norepinephrine) are catecholamines synthesized from $Tyrosine$ via the pathway involving $L-DOPA$ and $Dopamine$.
$3$. $Melanin$,the pigment responsible for skin,hair,and eye color,is also synthesized from $Tyrosine$ through a series of enzymatic reactions involving $Tyrosinase$.
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48
BiologyEasyMCQAIPMT · 1997
Iodine is associated with which of the following hormones?
A
Thyroxin
B
Calcitonin
C
Oxytocin
D
Secretin
Solution
(A) Iodine is an essential trace element required for the synthesis of thyroid hormones,specifically Thyroxin $(T_4)$ and Triiodothyronine $(T_3)$.
Thyroid follicles produce these hormones by incorporating iodine atoms into the amino acid tyrosine.
Deficiency of iodine leads to the enlargement of the thyroid gland,a condition known as goiter.
Thyroid follicles produce these hormones by incorporating iodine atoms into the amino acid tyrosine.
Deficiency of iodine leads to the enlargement of the thyroid gland,a condition known as goiter.
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49
BiologyMediumMCQAIPMT · 1997
Which of the following is a free-living,aerobic,and non-photosynthetic nitrogen-fixing bacterium?
A
Rhizobium
B
Nostoc
C
Azospirillum
D
Azotobacter
Solution
(D) $Azotobacter$ is a free-living,aerobic,and non-photosynthetic nitrogen-fixing bacterium found in the soil.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.
$Nostoc$ is a photosynthetic cyanobacterium that can fix nitrogen.
$Azospirillum$ is an associative symbiotic nitrogen-fixing bacterium.
$Rhizobium$ is a symbiotic nitrogen-fixing bacterium.
$Nostoc$ is a photosynthetic cyanobacterium that can fix nitrogen.
$Azospirillum$ is an associative symbiotic nitrogen-fixing bacterium.
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50
BiologyMediumMCQAIPMT · 1997
The site of respiration in bacteria is ................
A
Episome
B
Mesosome
C
Ribosome
D
Microsome
Solution
(B) In prokaryotic cells like bacteria,membrane-bound organelles such as mitochondria are absent.
Mesosomes are specialized membranous structures formed by the extensions of the plasma membrane into the cell in the form of vesicles,tubules,and lamellae.
These structures play a crucial role in cellular respiration,$DNA$ replication,and the secretion process.
Therefore,the mesosome acts as the site of respiration in bacteria.
Mesosomes are specialized membranous structures formed by the extensions of the plasma membrane into the cell in the form of vesicles,tubules,and lamellae.
These structures play a crucial role in cellular respiration,$DNA$ replication,and the secretion process.
Therefore,the mesosome acts as the site of respiration in bacteria.
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51
BiologyMediumMCQAIPMT · 1997
Which alga is rich in protein?
A
Chlorella
B
Spirogyra
C
Oscillatoria
D
Ulothrix
Solution
(A) $Chlorella$ $vulgaris$ is a unicellular green alga that is highly rich in protein.
It is widely used as a food supplement by space travelers and as a single-cell protein $(SCP)$ source due to its high nutritional value.
It belongs to the class $Chlorophyceae$.
It is widely used as a food supplement by space travelers and as a single-cell protein $(SCP)$ source due to its high nutritional value.
It belongs to the class $Chlorophyceae$.
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52
BiologyEasyMCQAIPMT · 1997
During adverse seasons,therophytes survive by:
A
Bulbs
B
Corms
C
Rhizomes
D
Seeds
Solution
(D) Therophytes are annual plants that complete their life cycle within a single growing season. During adverse conditions (such as extreme cold or drought),they survive in the form of dormant seeds. These seeds germinate when favorable conditions return,allowing the plant to complete its life cycle.
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53
BiologyMediumMCQAIPMT · 1997
If an angiospermic male plant is diploid $(2n)$ and the female plant is tetraploid $(4n)$,what will be the ploidy level of the endosperm?
A
Haploid
B
Triploid
C
Tetraploid
D
Pentaploid
Solution
(D) In angiosperms,the endosperm is formed by the process of double fertilization,specifically through the fusion of one male gamete $(n)$ with two polar nuclei ($n+n$ or $2n$ depending on the female parent's ploidy).
Given that the male plant is diploid $(2n)$,its gametes are haploid $(n)$.
Given that the female plant is tetraploid $(4n)$,its embryo sac is formed from a megaspore mother cell $(4n)$ via meiosis,resulting in haploid megaspores $(2n)$.
Thus,the polar nuclei in the central cell of the embryo sac will each be diploid $(2n)$.
The endosperm is formed by the fusion of one male gamete $(n)$ and two polar nuclei $(2n + 2n)$.
Therefore,the ploidy of the endosperm is $n + 2n + 2n = 5n$,which is Pentaploid.
Given that the male plant is diploid $(2n)$,its gametes are haploid $(n)$.
Given that the female plant is tetraploid $(4n)$,its embryo sac is formed from a megaspore mother cell $(4n)$ via meiosis,resulting in haploid megaspores $(2n)$.
Thus,the polar nuclei in the central cell of the embryo sac will each be diploid $(2n)$.
The endosperm is formed by the fusion of one male gamete $(n)$ and two polar nuclei $(2n + 2n)$.
Therefore,the ploidy of the endosperm is $n + 2n + 2n = 5n$,which is Pentaploid.
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54
BiologyMediumMCQAIPMT · 1997
Which plant will lose its economic value if its fruits are produced by induced parthenocarpy?
A
Grape
B
Pomegranate
C
Orange
D
Banana
Solution
(B) Parthenocarpy is the development of fruit without fertilization, resulting in seedless fruits.
In $Pomegranate$ $(Punica \text{ } granatum)$, the edible part of the fruit is the succulent $testa$ (the outer layer of the seed).
If $Pomegranate$ fruits are produced by induced $parthenocarpy$, they will be seedless.
Since the seed (specifically the $testa$) is the part consumed by humans, the absence of seeds renders the fruit economically useless.
In contrast, for fruits like $Grape$, $Orange$, and $Banana$, seedlessness is often a desirable trait that increases their market value.
In $Pomegranate$ $(Punica \text{ } granatum)$, the edible part of the fruit is the succulent $testa$ (the outer layer of the seed).
If $Pomegranate$ fruits are produced by induced $parthenocarpy$, they will be seedless.
Since the seed (specifically the $testa$) is the part consumed by humans, the absence of seeds renders the fruit economically useless.
In contrast, for fruits like $Grape$, $Orange$, and $Banana$, seedlessness is often a desirable trait that increases their market value.
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55
BiologyEasyMCQAIPMT · 1997
Fertilizin is a chemical substance produced from
A
Mature eggs
B
Acrosome
C
Polar bodies
D
Middle piece of sperm
Solution
(A) Fertilizin is a chemical substance secreted by the mature egg (specifically the jelly coat or vitelline membrane).
It is a glycoprotein or mucopolysaccharide in nature.
Its primary function is to attract and facilitate the binding of compatible sperm to the egg surface during the process of fertilization.
It is a glycoprotein or mucopolysaccharide in nature.
Its primary function is to attract and facilitate the binding of compatible sperm to the egg surface during the process of fertilization.
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56
BiologyMediumMCQAIPMT · 1997
After crossing two plants,the progenies are found to be male sterile. The phenomenon is found to be maternally inherited and is due to some genes which reside in
A
Nucleus
B
Chloroplast
C
Mitochondria
D
Cytoplasm
Solution
(D) The correct answer is $D$. Male sterility in plants is often a result of cytoplasmic male sterility $(CMS)$. This phenomenon is maternally inherited because the organelles responsible for this trait,such as mitochondria and chloroplasts,are inherited through the egg cytoplasm. Since these genes reside in the cytoplasm (specifically within the organellar $DNA$),they do not follow Mendelian inheritance patterns.
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57
BiologyMediumMCQAIPMT · 1997
Genetic identity of a human male is determined by
A
Autosome
B
Nucleolus
C
Sex chromosome
D
Cell organelles
Solution
(C) In humans,the sex of an individual is determined by the presence of specific sex chromosomes. Females possess two $X$ chromosomes $(XX)$,while males possess one $X$ chromosome and one $Y$ chromosome $(XY)$. The presence of the $Y$ chromosome is the primary factor that triggers the development of male characteristics,thus determining the genetic identity of a human male.
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58
BiologyMediumMCQAIPMT · 1997
$A$ fruit fly is heterozygous for sex-linked genes and is mated with a normal female fruit fly. In what proportion will the male-specific chromosomes enter the egg cells (in $:1$)?
A
$1$
B
$2$
C
$3$
D
$7$
Solution
(A) In fruit flies ($Drosophila$ $melanogaster$),the sex determination is of the $XY$ type.
When a male is heterozygous for sex-linked genes,he possesses one $X$ chromosome and one $Y$ chromosome.
During meiosis,the $X$ chromosome and the $Y$ chromosome segregate into different gametes.
As a result,$50\%$ of the sperm cells receive the $X$ chromosome and $50\%$ receive the $Y$ chromosome.
Therefore,the ratio of male-specific chromosomes ($X$ and $Y$) entering the gametes is $1:1$.
When a male is heterozygous for sex-linked genes,he possesses one $X$ chromosome and one $Y$ chromosome.
During meiosis,the $X$ chromosome and the $Y$ chromosome segregate into different gametes.
As a result,$50\%$ of the sperm cells receive the $X$ chromosome and $50\%$ receive the $Y$ chromosome.
Therefore,the ratio of male-specific chromosomes ($X$ and $Y$) entering the gametes is $1:1$.
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59
BiologyMediumMCQAIPMT · 1997
Which of the following $RNAs$ picks up specific amino acid (from amino acid pool) in the cytoplasm to ribosome during protein synthesis?
A
$tRNA$
B
$mRNA$
C
$rRNA$
D
All of these
Solution
(A) The $tRNA$ (transfer $RNA$) acts as an adapter molecule.
It has many varieties,where each variety is specific for a particular amino acid.
During protein synthesis,$tRNA$ picks up its specific amino acid from the amino acid pool in the cytoplasm and transports it to the $mRNA$ template on the ribosomes.
This process ensures that the amino acids are added in the correct sequence as specified by the genetic code.
It has many varieties,where each variety is specific for a particular amino acid.
During protein synthesis,$tRNA$ picks up its specific amino acid from the amino acid pool in the cytoplasm and transports it to the $mRNA$ template on the ribosomes.
This process ensures that the amino acids are added in the correct sequence as specified by the genetic code.
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60
BiologyMediumMCQAIPMT · 1997
The codons causing chain termination are
A
$TAG, TAA, TGA$
B
$GAT, AAT, AGT$
C
$AGT, TAG, UGA$
D
$UAG, UGA, UAA$
Solution
(D) The codons that cause chain termination are known as stop codons or nonsense codons.
These codons do not code for any amino acid during the process of protein synthesis (translation).
The three stop codons are $UAG$,$UGA$,and $UAA$.
These codons do not code for any amino acid during the process of protein synthesis (translation).
The three stop codons are $UAG$,$UGA$,and $UAA$.
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61
BiologyMediumMCQAIPMT · 1997
Which of the following steps of translation does not consume a high-energy phosphate bond?
A
Translocation
B
Amino acid activation
C
Peptidyl transferase reaction
D
Aminoacyl $tRNA$ binding to $A$-site
Solution
(C) The correct answer is $C$.
$1$. Amino acid activation requires $ATP$ to form aminoacyl-$AMP$ complex.
$2$. Aminoacyl $tRNA$ binding to the $A$-site requires $GTP$ hydrolysis.
$3$. Translocation requires $GTP$ hydrolysis to shift the ribosome along the $m-RNA$.
$4$. The peptidyl transferase reaction is a ribozyme-catalyzed reaction that forms a peptide bond between the amino acid at the $A$-site and the polypeptide chain at the $P$-site. This reaction does not require the hydrolysis of $ATP$ or $GTP$ as the energy is derived from the high-energy ester bond between the amino acid and the $tRNA$.
$1$. Amino acid activation requires $ATP$ to form aminoacyl-$AMP$ complex.
$2$. Aminoacyl $tRNA$ binding to the $A$-site requires $GTP$ hydrolysis.
$3$. Translocation requires $GTP$ hydrolysis to shift the ribosome along the $m-RNA$.
$4$. The peptidyl transferase reaction is a ribozyme-catalyzed reaction that forms a peptide bond between the amino acid at the $A$-site and the polypeptide chain at the $P$-site. This reaction does not require the hydrolysis of $ATP$ or $GTP$ as the energy is derived from the high-energy ester bond between the amino acid and the $tRNA$.
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62
BiologyEasyMCQAIPMT · 1997
Genes located at the same locus of chromosomes are called
A
Polygenes
B
Oncogenes
C
Multiple alleles
D
None of these
Solution
(C) Multiple alleles are defined as a series of three or more alternative forms of a gene that occupy the same locus on a specific chromosome in a population.
Since they occupy the same locus,they represent different states of the same gene and do not undergo crossing over between them.
Since they occupy the same locus,they represent different states of the same gene and do not undergo crossing over between them.
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63
BiologyMediumMCQAIPMT · 1997
Common origin of man and chimpanzee is best shown by
A
Dental formula
B
Cranial capacity
C
Binocular vision
D
Chromosome number
Solution
(D) The common ancestry of humans and chimpanzees is best evidenced by the similarity in their genetic material,specifically the chromosome banding patterns. While humans have $46$ chromosomes and chimpanzees have $48$,the similarity is profound. It is widely accepted that human chromosome $2$ resulted from the fusion of two ancestral chromosomes that remain separate in chimpanzees,explaining the difference in chromosome count.
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64
BiologyMediumMCQAIPMT · 1997
Which one of the following statements is correct regarding the evolution of mankind?
A
Neanderthal man and Cro-Magnon man lived at the same time.
B
Australopithecus lived in Australia.
C
$Homo$ $erectus$ is preceded by $Homo$ $habilis$.
D
None of these.
Solution
(C) The correct statement is that $Homo$ $erectus$ is preceded by $Homo$ $habilis$.
$Homo$ $habilis$ lived approximately $2.0-1.5$ million years ago,while $Homo$ $erectus$ appeared around $1.5$ million years ago.
$Australopithecus$ lived in East Africa,not Australia.
Neanderthal man and Cro-Magnon man did not live at the same time; Cro-Magnon man appeared much later.
$Homo$ $habilis$ lived approximately $2.0-1.5$ million years ago,while $Homo$ $erectus$ appeared around $1.5$ million years ago.
$Australopithecus$ lived in East Africa,not Australia.
Neanderthal man and Cro-Magnon man did not live at the same time; Cro-Magnon man appeared much later.
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65
BiologyMediumMCQAIPMT · 1997
If a person shows production of interferons in his body, the chances are that he has got an infection of
A
Typhoid
B
Measles
C
Tetanus
D
Malaria
Solution
(B) Interferons are antiviral proteins produced by virus-infected cells to protect non-infected cells from further viral infection.
Among the given options, $Typhoid$ is caused by a bacterium $(Salmonella \text{ } typhi)$, $Tetanus$ is caused by a bacterium $(Clostridium \text{ } tetani)$, and $Malaria$ is caused by a protozoan $(Plasmodium \text{ } species)$.
$Measles$ is a viral disease caused by the $Rubeola \text{ } virus$.
Therefore, the production of interferons indicates a viral infection, making $Measles$ the correct answer.
Among the given options, $Typhoid$ is caused by a bacterium $(Salmonella \text{ } typhi)$, $Tetanus$ is caused by a bacterium $(Clostridium \text{ } tetani)$, and $Malaria$ is caused by a protozoan $(Plasmodium \text{ } species)$.
$Measles$ is a viral disease caused by the $Rubeola \text{ } virus$.
Therefore, the production of interferons indicates a viral infection, making $Measles$ the correct answer.
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66
BiologyEasyMCQAIPMT · 1997
Which of the following diseases is now considered nearly eradicated from India?
A
Plague
B
Kala-azar
C
Smallpox
D
Poliomyelitis
Solution
(C) Smallpox is an acute,highly communicable disease caused by the $Variola$ virus.
It has been officially declared eradicated from India and the world through global vaccination programs.
While other diseases like polio have been eliminated from India,smallpox is the classic example of a disease that has been completely eradicated globally.
It has been officially declared eradicated from India and the world through global vaccination programs.
While other diseases like polio have been eliminated from India,smallpox is the classic example of a disease that has been completely eradicated globally.
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67
BiologyMediumMCQAIPMT · 1997
Diphtheria is caused by
A
Bacteria
B
Virus
C
Nematodes
D
None of these
Solution
(A) Diphtheria is a serious infectious disease caused by the bacterium $Corynebacterium$ $diphtheriae$.
It commonly affects the mucous membranes of the nose,throat,and tonsils.
The symptoms include high fever,sore throat,and difficulty in breathing due to the formation of a thick grey coating in the throat.
It commonly affects the mucous membranes of the nose,throat,and tonsils.
The symptoms include high fever,sore throat,and difficulty in breathing due to the formation of a thick grey coating in the throat.
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68
BiologyMediumMCQAIPMT · 1997
Which of the following diseases is expected to be completely curable in the next two decades?
A
Cancer
B
Poliomyelitis
C
Tuberculosis
D
None of these
Solution
(A) The question refers to the historical context of medical advancements where $Cancer$ was often cited in academic literature as a disease that would see significant breakthroughs in treatment and potential curability within a few decades due to advancements in biotechnology and gene therapy. While $Poliomyelitis$ is largely controlled through vaccination and $Tuberculosis$ is already curable with antibiotics,$Cancer$ remains the primary focus of ongoing research for a definitive cure.
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69
BiologyMediumMCQAIPMT · 1997
Which of the following has been recently used for increasing the productivity of super milch cows?
A
Artificial insemination by a pedigreed bull only
B
Superovulation of a high-production cow only
C
Embryo transplantation only
D
$A$ combination of superovulation,artificial insemination,and embryo transplantation into a 'carrier cow' (surrogate mother)
Solution
(D) The correct method for increasing the productivity of super milch cows is the Multiple Ovulation Embryo Transfer $(MOET)$ technology.
$1$. Superovulation: Hormones (like $FSH$) are administered to induce the production of more than one egg (ova) per cycle.
$2$. Artificial Insemination: The cow is mated with an elite bull or artificially inseminated.
$3$. Embryo Transplantation: The fertilized eggs (embryos) at the $8-32$ cell stage are recovered non-surgically and transferred to surrogate mothers (carrier cows) to allow the genetic mother to be ready for another round of superovulation.
This combination allows for the rapid multiplication of high-yielding breeds.
$1$. Superovulation: Hormones (like $FSH$) are administered to induce the production of more than one egg (ova) per cycle.
$2$. Artificial Insemination: The cow is mated with an elite bull or artificially inseminated.
$3$. Embryo Transplantation: The fertilized eggs (embryos) at the $8-32$ cell stage are recovered non-surgically and transferred to surrogate mothers (carrier cows) to allow the genetic mother to be ready for another round of superovulation.
This combination allows for the rapid multiplication of high-yielding breeds.
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70
BiologyEasyMCQAIPMT · 1997
Honey is:
A
Acidic
B
Neutral
C
Alkaline
D
Basic after some days
Solution
(A) The $pH$ of honey typically ranges from $3.4$ to $6.1$,with an average value of approximately $3.9$.
Because the $pH$ is less than $7$,honey is acidic in nature.
Because the $pH$ is less than $7$,honey is acidic in nature.
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71
BiologyEasyMCQAIPMT · 1997
What is Agent Orange?
A
$A$ biodegradable insecticide
B
$A$ weedicide containing dioxin
C
Colour used in fluorescent lamps
D
$A$ hazardous chemical used in luminous plants
Solution
(B) Agent Orange is a powerful herbicide and defoliant that was used by the $U.S.$ military during the Vietnam War.
It is a mixture of two herbicides, $2,4-D$ and $2,4,5-T$.
The most significant concern regarding Agent Orange is that it was contaminated with a highly toxic byproduct called dioxin (specifically $TCDD$), which is known to cause serious health issues, including cancer and birth defects.
It is a mixture of two herbicides, $2,4-D$ and $2,4,5-T$.
The most significant concern regarding Agent Orange is that it was contaminated with a highly toxic byproduct called dioxin (specifically $TCDD$), which is known to cause serious health issues, including cancer and birth defects.
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72
BiologyMediumMCQAIPMT · 1997
Biofertilizers include:
A
Cow dung manure and farmyard waste
B
$A$ quick-growing crop ploughed back into the soil (Green manure)
C
$BGA$ (Blue-Green Algae),Anabaena,and Azolla
D
All of the above
Solution
(D) Biofertilizers are organisms that enrich the nutrient quality of the soil.
$1$. Cow dung manure and farmyard waste are organic fertilizers that provide essential nutrients to the soil.
$2$. Green manure involves growing quick-growing crops and ploughing them back into the soil to increase fertility.
$3$. $BGA$ (Blue-Green Algae) like Anabaena and the fern Azolla are well-known biofertilizers that fix atmospheric nitrogen,thereby enhancing soil fertility.
Since all these options contribute to improving soil nutrient status,the correct answer is $D$.
$1$. Cow dung manure and farmyard waste are organic fertilizers that provide essential nutrients to the soil.
$2$. Green manure involves growing quick-growing crops and ploughing them back into the soil to increase fertility.
$3$. $BGA$ (Blue-Green Algae) like Anabaena and the fern Azolla are well-known biofertilizers that fix atmospheric nitrogen,thereby enhancing soil fertility.
Since all these options contribute to improving soil nutrient status,the correct answer is $D$.
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73
BiologyMediumMCQAIPMT · 1997
The quantity of $CO_2$ in the atmosphere is about (in $\%$)
A
$0.003$
B
$0.03$
C
$0.3$
D
$3.0$
Solution
(B) The concentration of $CO_2$ in the Earth's atmosphere is approximately $0.03\%$ to $0.04\%$,which is equivalent to about $300$ to $400$ $ppm$ (parts per million).
Among the given options,$0.03\%$ is the standard value typically cited in biological and environmental science textbooks.
Among the given options,$0.03\%$ is the standard value typically cited in biological and environmental science textbooks.
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74
BiologyMediumMCQAIPMT · 1997
Which of the following acts as $''$nature's scavengers$''$?
A
Man
B
Animals
C
Insects
D
Micro-organisms
Solution
(D) The correct answer is $(d)$.
Micro-organisms,specifically detritivores and decomposers (such as bacteria,actinomycetes,and fungi),are referred to as $''$nature's scavengers$''$.
They are called scavengers because they break down complex organic matter into simpler substances,thereby cleaning the earth and recycling nutrients back into the ecosystem.
Micro-organisms,specifically detritivores and decomposers (such as bacteria,actinomycetes,and fungi),are referred to as $''$nature's scavengers$''$.
They are called scavengers because they break down complex organic matter into simpler substances,thereby cleaning the earth and recycling nutrients back into the ecosystem.
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75
BiologyMediumMCQAIPMT · 1997
Which of the following ecosystems has the highest gross primary productivity?
A
Grassland
B
Coral reef
C
Mangroves
D
Rain forest
Solution
(B) The correct answer is $B$. Coral reefs are among the most productive ecosystems on Earth. They are coastal regions characterized by maximum primary productivity and high biodiversity,where almost every group of marine algae and every animal phylum is represented. While tropical rain forests are highly productive on land,coral reefs exhibit higher gross primary productivity per unit area due to efficient nutrient cycling and high solar energy utilization.
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76
BiologyEasyMCQAIPMT · 1997
$MAB$ stands for
A
Man and biosphere
B
Man antibiotics and bacteria
C
Man and biotic community
D
Mayer,Anderson and Bishby
Solution
(A) The $MAB$ programme stands for Man and the Biosphere programme.
It was formally launched by $UNESCO$ in $1971$.
It is an interdisciplinary programme of research and training that emphasizes an ecological approach to the study of the interrelationship between man and his environment.
It was formally launched by $UNESCO$ in $1971$.
It is an interdisciplinary programme of research and training that emphasizes an ecological approach to the study of the interrelationship between man and his environment.
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77
BiologyEasyMCQAIPMT · 1997
Formation of ozone hole is maximum over
A
India
B
Antarctica
C
Europe
D
Africa
Solution
(B) The depletion of the ozone layer,commonly referred to as the ozone hole,is most pronounced over the continent of $Antarctica$.
This phenomenon occurs primarily due to the unique atmospheric conditions in the polar region,specifically the formation of polar stratospheric clouds during the extremely cold winter months.
These clouds provide a surface for chemical reactions that release active chlorine from chlorofluorocarbons $(CFCs)$,which then catalyze the destruction of ozone molecules $(O_3)$.
This phenomenon occurs primarily due to the unique atmospheric conditions in the polar region,specifically the formation of polar stratospheric clouds during the extremely cold winter months.
These clouds provide a surface for chemical reactions that release active chlorine from chlorofluorocarbons $(CFCs)$,which then catalyze the destruction of ozone molecules $(O_3)$.
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78
BiologyMediumMCQAIPMT · 1997
In coming years,skin-related disorders will be more common due to:
A
Pollutants in air
B
Use of detergents
C
Water pollution
D
Depletion of ozone layer
Solution
(D) The correct answer is $D$.
As the ozone layer in the stratosphere becomes thinner or develops holes,it allows a higher intensity of harmful ultraviolet $(UV)$ radiations to reach the Earth's surface.
These $UV$ radiations,particularly $UV-B$,are known to cause $DNA$ damage in skin cells,which can lead to skin-related disorders such as skin cancer (e.g.,melanoma) and premature aging of the skin.
As the ozone layer in the stratosphere becomes thinner or develops holes,it allows a higher intensity of harmful ultraviolet $(UV)$ radiations to reach the Earth's surface.
These $UV$ radiations,particularly $UV-B$,are known to cause $DNA$ damage in skin cells,which can lead to skin-related disorders such as skin cancer (e.g.,melanoma) and premature aging of the skin.
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79
BiologyMediumMCQAIPMT · 1997
Phosphate pollution is caused by
A
Phosphate rock only
B
Agricultural fertilizers only
C
Sewage and phosphate rock
D
Sewage and agricultural fertilizers
Solution
(D) Phosphate pollution is primarily caused by human activities that introduce excess phosphorus into aquatic ecosystems.
$(1)$ Agricultural fertilizers: Large quantities of superphosphates are used in farming,which leach into water bodies through runoff.
$(2)$ Sewage and detergents: Synthetic detergents and domestic sewage contain significant amounts of phosphates,which are discharged into water bodies,leading to eutrophication.
$(1)$ Agricultural fertilizers: Large quantities of superphosphates are used in farming,which leach into water bodies through runoff.
$(2)$ Sewage and detergents: Synthetic detergents and domestic sewage contain significant amounts of phosphates,which are discharged into water bodies,leading to eutrophication.
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80
BiologyMediumMCQAIPMT · 1997
Amniocentesis is a technique used to detect:
A
Deformity in the brain
B
Hereditary diseases
C
Deformity in the heart
D
All of these
Solution
(B) Amniocentesis is a prenatal diagnostic technique based on the chromosomal pattern in the amniotic fluid surrounding the developing fetus.
It is primarily used to detect chromosomal abnormalities and certain hereditary metabolic disorders (genetic diseases) in the fetus.
While it can sometimes provide clues about structural developments,its primary clinical application is the identification of genetic and hereditary conditions.
It is primarily used to detect chromosomal abnormalities and certain hereditary metabolic disorders (genetic diseases) in the fetus.
While it can sometimes provide clues about structural developments,its primary clinical application is the identification of genetic and hereditary conditions.
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81
BiologyMediumMCQAIPMT · 1997
The present population of the world is about
A
$500$ million
B
$100$ million
C
$15$ trillion
D
$6$ billion
Solution
(D) The correct answer is $D$. The United Nations designated Fatima,a female infant born in Sarajevo (Bosnia and Herzegovina) on October $14, 1999$,as the world's $6$ billionth child. By $2001$,the global population reached approximately $6.2$ billion. While the current world population has since exceeded $8$ billion,$6$ billion is the standard reference point often used in historical educational contexts regarding population milestones.
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82
BiologyMediumMCQAIPMT · 1997
Foetal sex can be determined by examining cells from the amniotic fluid by looking for
A
Barr bodies
B
Autosomes
C
Chiasmata
D
Kinetochore
Solution
(A) Amniotic fluid contains foetal cells that can be examined to determine the presence of sex chromatin,known as $Barr$ bodies.
The presence of a $Barr$ body indicates that the developing foetus is female,as it possesses two $X$ chromosomes ($XX$ genotype),where one $X$ chromosome is inactivated and condensed into a $Barr$ body.
Male foetuses ($XY$ genotype) do not possess $Barr$ bodies.
The presence of a $Barr$ body indicates that the developing foetus is female,as it possesses two $X$ chromosomes ($XX$ genotype),where one $X$ chromosome is inactivated and condensed into a $Barr$ body.
Male foetuses ($XY$ genotype) do not possess $Barr$ bodies.
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83
BiologyMediumMCQAIPMT · 1997
Amniocentesis is a process to
A
Determine any disease in heart
B
Determine any hereditary disease in the embryo
C
Know about the disease of brain
D
All of these
Solution
(B) Amniocentesis is a prenatal diagnostic technique used to determine the sex and health condition of the foetus.
In the early stages of pregnancy,a small amount of amniotic fluid is extracted using a surgical needle.
This fluid contains foetal cells,which are then cultured and analyzed to detect chromosomal abnormalities or hereditary diseases in the embryo.
In the early stages of pregnancy,a small amount of amniotic fluid is extracted using a surgical needle.
This fluid contains foetal cells,which are then cultured and analyzed to detect chromosomal abnormalities or hereditary diseases in the embryo.
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84
BiologyEasyMCQAIPMT · 1997
The genetic material in the bacterium $Escherichia$ $coli$ $(E. coli)$ is:
A
Single-stranded $DNA$
B
Deoxyribose sugar
C
Double-stranded $DNA$
D
Single-stranded $RNA$
Solution
(C) $E. coli$ is a prokaryotic organism.
Prokaryotes possess a circular,double-stranded $DNA$ molecule as their primary genetic material,which is located in the nucleoid region.
Therefore,the genetic material in $E. coli$ is double-stranded $DNA$.
Prokaryotes possess a circular,double-stranded $DNA$ molecule as their primary genetic material,which is located in the nucleoid region.
Therefore,the genetic material in $E. coli$ is double-stranded $DNA$.
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85
BiologyMediumMCQAIPMT · 1997
Which of the following algae is rich in protein?
A
Spirogyra
B
Ulothrix
C
Oscillatoria
D
Chlorella
Solution
(D) Chlorella and Spirulina are unicellular algae that are rich in proteins and are used as food supplements by space travelers. Therefore,$Chlorella$ is the correct answer.
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86
BiologyDifficultMCQAIPMT · 1997
Which plant loses its economic utility if its fruits are produced by induced $..........$?
A
Grapes
B
Pomegranate
C
Banana
D
Orange
Solution
(B) The correct answer is $B$ (Pomegranate).
Economic utility of fruits like grapes,bananas,and oranges is often enhanced by parthenocarpy (seedless fruits).
However,in the case of pomegranate,the edible part is the juicy seed coat (aril).
If parthenocarpy is induced in pomegranate,the fruits will be seedless,thereby destroying the edible part and rendering the fruit economically useless.
Economic utility of fruits like grapes,bananas,and oranges is often enhanced by parthenocarpy (seedless fruits).
However,in the case of pomegranate,the edible part is the juicy seed coat (aril).
If parthenocarpy is induced in pomegranate,the fruits will be seedless,thereby destroying the edible part and rendering the fruit economically useless.
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87
BiologyMediumMCQAIPMT · 1997
What do the genes found on mitochondrial $DNA$ exhibit?
A
Generally maternal inheritance
B
They always exhibit paternal inheritance.
C
They exhibit inheritance from both parents like nuclear genes.
D
They are not inherited.
Solution
(A) Mitochondria are inherited through the cytoplasm of the egg cell during fertilization. Since the sperm contributes almost no cytoplasm to the zygote,the mitochondrial $DNA$ $(mtDNA)$ is passed down exclusively from the mother to the offspring. This phenomenon is known as maternal inheritance.
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88
BiologyEasyMCQAIPMT · 1997
The human egg is ...
A
Alecithal
B
Microlecithal
C
Mesolecithal
D
Polylecithal
Solution
(A) The human egg (ovum) is classified as $Alecithal$ or $Microlecithal$ because it contains a very negligible or almost no amount of yolk. Since the human embryo receives nutrition from the mother through the placenta,the egg does not require a large amount of stored yolk.
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89
BiologyMediumMCQAIPMT · 1997
Which of the following secretes fertilizin?
A
Immature egg
B
Mature egg
C
Sperm
D
Polar bodies
Solution
(B) Fertilizin is a chemical substance (a glycoprotein) secreted by the mature egg (ovum) to attract sperm. It interacts with the antifertilizin present on the surface of the sperm,facilitating the recognition and binding of the sperm to the egg during the process of fertilization.
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90
BiologyEasyMCQAIPMT · 1997
In a $28$-day human ovarian cycle,ovulation occurs on:
A
Day $1$
B
Day $5$
C
Day $14$
D
Day $28$
Solution
(C) The human ovarian cycle typically lasts for $28$ days.
Ovulation is the process where a mature ovarian follicle ruptures and releases an egg (ovum).
In a standard $28$-day cycle,ovulation is triggered by a surge in Luteinizing Hormone $(LH)$ and typically occurs at the midpoint of the cycle,which is the $14$th day.
Ovulation is the process where a mature ovarian follicle ruptures and releases an egg (ovum).
In a standard $28$-day cycle,ovulation is triggered by a surge in Luteinizing Hormone $(LH)$ and typically occurs at the midpoint of the cycle,which is the $14$th day.
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91
BiologyMediumMCQAIPMT · 1997
Amniocentesis is a process used for:
A
Determining any heart disease.
B
Determining any genetic disorder in the fetus.
C
Obtaining information about brain diseases.
D
All of the above.
Solution
(B) Amniocentesis is a prenatal diagnostic technique used to determine genetic disorders in the fetus.
It involves taking a sample of the amniotic fluid surrounding the developing fetus.
This fluid contains fetal cells,which can be analyzed for chromosomal abnormalities,metabolic disorders,or genetic mutations.
Therefore,the correct option is $B$.
It involves taking a sample of the amniotic fluid surrounding the developing fetus.
This fluid contains fetal cells,which can be analyzed for chromosomal abnormalities,metabolic disorders,or genetic mutations.
Therefore,the correct option is $B$.
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92
BiologyMediumMCQAIPMT · 1997
$A$ male individual with $47$ chromosomes due to the addition of an $X$ chromosome suffers from a condition called:
A
Down syndrome
B
Super female
C
Turner syndrome
D
Klinefelter syndrome
Solution
(D) Klinefelter syndrome is a genetic disorder in males caused by the presence of an extra $X$ chromosome,resulting in a karyotype of $47, XXY$.
This condition occurs due to the non-disjunction of chromosomes during meiosis.
Individuals with this syndrome exhibit masculine development but also show feminine characteristics such as gynaecomastia (development of breast tissue) and are sterile.
This condition occurs due to the non-disjunction of chromosomes during meiosis.
Individuals with this syndrome exhibit masculine development but also show feminine characteristics such as gynaecomastia (development of breast tissue) and are sterile.
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93
BiologyMediumMCQAIPMT · 1997
After crossing two plants,the male progeny is found to be sterile. This phenomenon is inherited maternally,and the cause lies in genes located in the ...
A
Nucleus
B
Chloroplast
C
Mitochondria
D
Cytoplasm
Solution
(C) The phenomenon where a trait is inherited exclusively from the mother is known as cytoplasmic inheritance or maternal inheritance.
In plants,male sterility is often associated with genes located in the mitochondrial $DNA$.
This is known as Cytoplasmic Male Sterility $(CMS)$.
Since mitochondria are inherited through the egg cytoplasm during fertilization,the trait follows maternal inheritance patterns.
Therefore,the correct option is $C$.
In plants,male sterility is often associated with genes located in the mitochondrial $DNA$.
This is known as Cytoplasmic Male Sterility $(CMS)$.
Since mitochondria are inherited through the egg cytoplasm during fertilization,the trait follows maternal inheritance patterns.
Therefore,the correct option is $C$.
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94
BiologyMediumMCQAIPMT · 1997
Some mutations occur at the same locus of a chromosome. They produce ..........
A
Pseudoalleles
B
Polygenic
C
Oncogenes
D
Multiple genes
Solution
(A) Pseudoalleles are genes that are functionally related and are located very close to each other on the same chromosome. They appear to be alleles of the same gene because they occupy the same locus or closely linked loci and affect the same trait. Mutations occurring at these closely linked sites result in the formation of pseudoalleles.
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95
BiologyMediumMCQAIPMT · 1997
In humans,the genetic identity of a male is determined by which of the following?
A
Autosomes
B
Nucleolus
C
Sex chromosomes
D
Cell organelles
Solution
(C) In humans,sex determination is based on the presence of specific sex chromosomes.
Females have two $X$ chromosomes $(XX)$,while males have one $X$ chromosome and one $Y$ chromosome $(XY)$.
The presence of the $Y$ chromosome,which carries the $SRY$ gene,triggers the development of male characteristics.
Therefore,the genetic identity of a male is determined by the sex chromosomes.
Females have two $X$ chromosomes $(XX)$,while males have one $X$ chromosome and one $Y$ chromosome $(XY)$.
The presence of the $Y$ chromosome,which carries the $SRY$ gene,triggers the development of male characteristics.
Therefore,the genetic identity of a male is determined by the sex chromosomes.
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96
BiologyMediumMCQAIPMT · 1997
$A$ mutation in a single base pair of a gene results in the production of a non-functional protein. This type of mutation is called .............
A
Frame-shift mutation
B
Missense mutation
C
Nonsense mutation
D
Reverse mutation
Solution
(B) mutation where a single base pair substitution results in the replacement of one amino acid with another is called a $Missense$ $mutation$. If this change occurs at a critical site in the protein,it can lead to the production of a non-functional protein. $Nonsense$ $mutation$ results in a premature stop codon,while $Frame-shift$ $mutation$ involves the insertion or deletion of bases,changing the entire reading frame. Therefore,the most appropriate term for a single base change causing loss of function is a $Missense$ $mutation$.
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97
BiologyMediumMCQAIPMT · 1997
The sex of a fetus can be determined by examining the cells of the amniotic fluid for the presence of:
A
Barr bodies
B
Autosomes
C
Swastika
D
Kinetochore
Solution
(A) The sex of a fetus can be determined by amniocentesis,a procedure where amniotic fluid is collected.
In female mammals,one of the two $X$ chromosomes is inactivated and condensed into a structure known as a $Barr$ body,which can be observed in the nuclei of somatic cells.
Since males have only one $X$ chromosome,they do not possess $Barr$ bodies.
Therefore,the presence of $Barr$ bodies in the cells of the amniotic fluid indicates a female fetus,while their absence indicates a male fetus.
In female mammals,one of the two $X$ chromosomes is inactivated and condensed into a structure known as a $Barr$ body,which can be observed in the nuclei of somatic cells.
Since males have only one $X$ chromosome,they do not possess $Barr$ bodies.
Therefore,the presence of $Barr$ bodies in the cells of the amniotic fluid indicates a female fetus,while their absence indicates a male fetus.
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98
BiologyMediumMCQAIPMT · 1997
$A$ fruit fly is heterozygous for a sex-linked gene. It is crossed with a normal female fruit fly. The specific chromosome for the male enters the egg in the ratio of ............. (in $: 1$)
A
$1$
B
$2$
C
$3$
D
$7$
Solution
(A) In fruit flies $(Drosophila)$,sex determination follows the $XY$ system where the female is $XX$ and the male is $XY$.
If a male is heterozygous for a sex-linked gene,it means the gene is located on the $X$ chromosome.
When a heterozygous male $(X^A Y)$ is crossed with a normal female $(X^A X^A)$,the gametes produced by the male are $X^A$ and $Y$ in a $1:1$ ratio.
Since the question asks for the ratio in which the specific chromosome for the male ($Y$ chromosome) enters the egg (or rather,the ratio of gametes containing $X$ vs $Y$),the segregation of sex chromosomes during meiosis results in $50\%$ $X$-bearing sperm and $50\%$ $Y$-bearing sperm.
Thus,the ratio is $1:1$.
If a male is heterozygous for a sex-linked gene,it means the gene is located on the $X$ chromosome.
When a heterozygous male $(X^A Y)$ is crossed with a normal female $(X^A X^A)$,the gametes produced by the male are $X^A$ and $Y$ in a $1:1$ ratio.
Since the question asks for the ratio in which the specific chromosome for the male ($Y$ chromosome) enters the egg (or rather,the ratio of gametes containing $X$ vs $Y$),the segregation of sex chromosomes during meiosis results in $50\%$ $X$-bearing sperm and $50\%$ $Y$-bearing sperm.
Thus,the ratio is $1:1$.
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99
BiologyMediumMCQAIPMT · 1997
In a bacterial chromosome,the genes are held together by which of the following?
A
Histones
B
Basic proteins
C
Acidic proteins
D
Actin
Solution
(B) In prokaryotes like bacteria,the $DNA$ is not organized into complex chromatin structures with histones as seen in eukaryotes. Instead,the bacterial $DNA$ is organized in large loops held together by certain non-histone proteins,which are often basic in nature (such as polyamines). These proteins help in folding and compacting the circular $DNA$ into a structure known as the nucleoid.
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100
BiologyMediumMCQAIPMT · 1997
Which codons signal the termination of the polypeptide chain?
A
$TAG, TAA, TGA$
B
$GAT, AAT, AGT$
C
$AGT, TAG, UGA$
D
$UAA, UAG, UGA$
Solution
(D) In the process of translation,the termination of the polypeptide chain is signaled by stop codons (also known as nonsense codons).
These codons do not code for any amino acid.
The three stop codons in the genetic code are $UAA$,$UAG$,and $UGA$.
Therefore,the correct option is $D$.
These codons do not code for any amino acid.
The three stop codons in the genetic code are $UAA$,$UAG$,and $UGA$.
Therefore,the correct option is $D$.
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