BiologyQ1–24 of 24 questions
Page 1 of 1 · English
1
BiologyMediumMCQAIIMS · 1984
Multiflagellate male gametes (sperms) are found in
A
Chlamydomonas
B
Funaria
C
Dryopteris
D
Riccia
Solution
(C) The correct answer is $C$. In $Dryopteris$ (a fern),the antheridium produces numerous antherozoids. Each antherozoid is large,uninucleate,spirally coiled,and multiflagellate (possessing many flagella). In contrast,bryophytes like $Funaria$ and $Riccia$ typically produce biflagellate antherozoids.
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2
BiologyMediumMCQAIIMS · 1984
Sweet potato is a modification of
A
Leaf
B
Root
C
Stem
D
Flowering axis
Solution
(B) Sweet potato $(Ipomoea \text{ } batatas)$ is a modified adventitious root.
These roots become swollen and fleshy to store food, which is a characteristic feature of tuberous roots.
These roots become swollen and fleshy to store food, which is a characteristic feature of tuberous roots.
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3
BiologyMediumMCQAIIMS · 1984
Bicarpellary syncarpous gynoecium,parietal placentation,tetradynamous stamens and siliqua fruit are characteristic features of which family?
A
Cucurbitaceae
B
Cruciferae
C
Compositae
D
Solanaceae
Solution
(B) The correct answer is $B$ (Cruciferae).
$1$. Bicarpellary syncarpous gynoecium: The gynoecium consists of two carpels that are fused together.
$2$. Parietal placentation: The ovules are attached to the inner wall of the ovary,which is typically unilocular due to the formation of a false septum (replum).
$3$. Tetradynamous stamens: There are $6$ stamens in total,arranged in two whorls. The outer whorl has $2$ short stamens,and the inner whorl has $4$ long stamens.
$4$. Siliqua fruit: This is a characteristic dry dehiscent fruit that develops from a bicarpellary,syncarpous,and superior ovary.
$1$. Bicarpellary syncarpous gynoecium: The gynoecium consists of two carpels that are fused together.
$2$. Parietal placentation: The ovules are attached to the inner wall of the ovary,which is typically unilocular due to the formation of a false septum (replum).
$3$. Tetradynamous stamens: There are $6$ stamens in total,arranged in two whorls. The outer whorl has $2$ short stamens,and the inner whorl has $4$ long stamens.
$4$. Siliqua fruit: This is a characteristic dry dehiscent fruit that develops from a bicarpellary,syncarpous,and superior ovary.
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4
BiologyMediumMCQAIIMS · 1984
Angiosperms are characterized by the presence of which of the following?
A
Tracheids only
B
Vessels absent
C
Vessels present
D
Sieve tubes absent
Solution
(C) Angiosperms,or flowering plants,are characterized by the presence of vessels in their xylem tissue. Vessels are efficient water-conducting tubes that are a hallmark of angiosperm anatomy,distinguishing them from most gymnosperms which typically lack vessels and rely primarily on tracheids for water transport.
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5
BiologyMediumMCQAIIMS · 1984
The innermost layer of the cortex which shows Casparian thickenings in its cells is called as
A
Epidermis
B
Endodermis
C
Pericycle
D
Exodermis
Solution
(B) The correct answer is $B$.
The $Endodermis$ is the innermost layer of the cortex in roots.
It consists of a single layer of barrel-shaped,compactly arranged parenchymatous cells.
The radial and tangential walls of these cells contain water-impermeable,waxy material called $Suberin$,which forms the $Casparian$ strips.
The $Endodermis$ is the innermost layer of the cortex in roots.
It consists of a single layer of barrel-shaped,compactly arranged parenchymatous cells.
The radial and tangential walls of these cells contain water-impermeable,waxy material called $Suberin$,which forms the $Casparian$ strips.
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6
BiologyMediumMCQAIIMS · 1984
Singer and Nicholson's model of plasma membrane differs from Robertson's model in the
A
Number of lipid layers
B
Arrangement of lipid layers
C
Arrangement of proteins
D
Absence of protein in Singer and Nicholson's model
Solution
(C) The fluid mosaic model proposed by Singer and Nicholson $(1972)$ describes the plasma membrane as a mosaic of proteins embedded in a fluid phospholipid bilayer.
In contrast,Robertson's unit membrane model proposed that the membrane consists of a phospholipid bilayer sandwiched between two continuous layers of globular proteins.
Therefore,the primary difference between the two models lies in the arrangement of proteins,where Singer and Nicholson suggested that proteins are dispersed (integral and peripheral) within the lipid bilayer rather than forming continuous outer layers.
In contrast,Robertson's unit membrane model proposed that the membrane consists of a phospholipid bilayer sandwiched between two continuous layers of globular proteins.
Therefore,the primary difference between the two models lies in the arrangement of proteins,where Singer and Nicholson suggested that proteins are dispersed (integral and peripheral) within the lipid bilayer rather than forming continuous outer layers.
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7
BiologyMediumMCQAIIMS · 1984
An autonomic genome system is present in
A
Ribosomes and chloroplasts
B
Mitochondria and ribosomes
C
Mitochondria and chloroplasts
D
Golgi bodies and mitochondria
Solution
(C) Mitochondria and chloroplasts are known as semi-autonomous organelles.
They contain their own circular $DNA$ molecules and $70S$ ribosomes.
This allows them to synthesize some of their own proteins and replicate independently of the cell's nuclear genome.
They contain their own circular $DNA$ molecules and $70S$ ribosomes.
This allows them to synthesize some of their own proteins and replicate independently of the cell's nuclear genome.
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8
BiologyMediumMCQAIIMS · 1984
In which part of mitochondria is $ATP$ generated?
A
Matrix
B
Cristae
C
Outer membrane
D
$F_1$ particles (oxysomes)
Solution
(D) The generation of $ATP$ in mitochondria occurs through the process of oxidative phosphorylation. This process takes place in the inner mitochondrial membrane,specifically at the $F_1$ particles,also known as oxysomes or $ATP$ synthase complexes. These particles project into the mitochondrial matrix and are responsible for the synthesis of $ATP$ from $ADP$ and inorganic phosphate using the proton gradient established across the inner membrane.
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9
BiologyMediumMCQAIIMS · 1984
All are membrane-bound cell organelles except: $OR$ Which of the following cell organelles lacks a unit membrane?
A
Mitochondria
B
Lysosomes
C
Sphaerosomes
D
Ribosomes
Solution
(D) Ribosomes are non-membrane-bound organelles found in both prokaryotic and eukaryotic cells.
They are composed of $RNA$ and proteins.
Since they lack a lipid bilayer membrane,they are not considered membrane-bound organelles.
In contrast,mitochondria,lysosomes,and sphaerosomes are all surrounded by a unit membrane.
They are composed of $RNA$ and proteins.
Since they lack a lipid bilayer membrane,they are not considered membrane-bound organelles.
In contrast,mitochondria,lysosomes,and sphaerosomes are all surrounded by a unit membrane.
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10
BiologyEasyMCQAIIMS · 1984
The most widely accepted explanation for the ascent of sap in trees is:
A
Capillarity
B
Role of atmospheric pressure
C
Pulsating action of living cells
D
Transpiration cohesion theory of Dixon
Solution
(D) The most widely accepted theory for the ascent of sap is the $Transpiration$ $Cohesion-Tension$ theory,proposed by $Dixon$ and $Joly$ in $1894$.
According to this theory,the ascent of sap is driven by the transpiration pull generated at the leaf surface due to the evaporation of water.
This pull is transmitted down the water column in the $Xylem$ due to the cohesive and adhesive properties of water molecules,which prevent the column from breaking under tension.
According to this theory,the ascent of sap is driven by the transpiration pull generated at the leaf surface due to the evaporation of water.
This pull is transmitted down the water column in the $Xylem$ due to the cohesive and adhesive properties of water molecules,which prevent the column from breaking under tension.
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11
BiologyMediumMCQAIIMS · 1984
$C_4$ plants are adapted to:
A
Hot and dry climate
B
Temperate climate
C
Cold and dry climate
D
Hot and humid climate
Solution
(A) $C_4$ plants,such as maize and sugarcane,have evolved a specialized leaf anatomy known as $Kranz$ anatomy.
This adaptation allows them to minimize photorespiration,which is a wasteful process that occurs in $C_3$ plants under high temperatures.
By concentrating $CO_2$ around the enzyme $RuBisCO$ in the bundle sheath cells,$C_4$ plants can maintain high rates of photosynthesis even in hot and dry environments where water conservation is necessary.
This adaptation allows them to minimize photorespiration,which is a wasteful process that occurs in $C_3$ plants under high temperatures.
By concentrating $CO_2$ around the enzyme $RuBisCO$ in the bundle sheath cells,$C_4$ plants can maintain high rates of photosynthesis even in hot and dry environments where water conservation is necessary.
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12
BiologyEasyMCQAIIMS · 1984
$ATP$ synthesis occurs on the
A
Outer membrane of mitochondrion
B
Inner membrane of mitochondrion
C
Matrix
D
None of the above
Solution
(B) $ATP$ synthesis is an endergonic,oxidative process that occurs on the inner mitochondrial membrane,specifically involving the cristae in eukaryotes. In prokaryotes,this process occurs on the inner side of the cell membrane or mesosomes.
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13
BiologyMediumMCQAIIMS · 1984
Pruning makes the hedge plant dense because
A
Injury induces dense growth
B
Apical dominance is removed
C
Root sprouts additional branches
D
Pruning removes shade and allows germination of new seedlings to impart a dense growth
Solution
(B) Pruning involves the removal of the apical buds of the plant.
Apical buds produce auxins,which are responsible for apical dominance,a phenomenon where the main central stem grows more dominantly than the side branches.
When the apical bud is removed,the source of auxin is eliminated,which breaks the apical dominance.
This allows the lateral (axillary) buds to grow,resulting in the development of more branches and making the hedge plant appear dense.
Apical buds produce auxins,which are responsible for apical dominance,a phenomenon where the main central stem grows more dominantly than the side branches.
When the apical bud is removed,the source of auxin is eliminated,which breaks the apical dominance.
This allows the lateral (axillary) buds to grow,resulting in the development of more branches and making the hedge plant appear dense.
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14
BiologyMediumMCQAIIMS · 1984
The leaves of $Mimosa$ $pudica$ (Sensitive Plant) droop down on touch because
A
Plants have a nervous system
B
The leaves are very tender
C
The leaf tissues are injured
D
The turgor pressure of the leaf base changes
Solution
(D) The movement of leaves in $Mimosa$ $pudica$ upon touch is known as seismonasty.
This phenomenon occurs due to a rapid change in the turgor pressure within the specialized cells at the base of the leaflets,known as the pulvinus.
When touched,water moves out of the cells of the pulvinus,causing them to lose turgidity and collapse,which results in the drooping of the leaves.
This phenomenon occurs due to a rapid change in the turgor pressure within the specialized cells at the base of the leaflets,known as the pulvinus.
When touched,water moves out of the cells of the pulvinus,causing them to lose turgidity and collapse,which results in the drooping of the leaves.
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15
BiologyEasyMCQAIIMS · 1984
Which one of the following does not produce any digestive enzyme?
A
Intestinal mucosa
B
Gastric mucosa
C
Pancreas
D
Liver
Solution
(D) The $Liver$ does not produce any digestive enzymes.
It produces bile juice, which contains bile salts and bile pigments.
Bile juice is essential for the emulsification of fats, which increases the surface area for the action of lipase, but it does not contain any enzymes itself.
In contrast, the intestinal mucosa, gastric mucosa, and pancreas all secrete various digestive enzymes (such as peptidases, pepsin, and pancreatic amylase/lipase/proteases, respectively).
It produces bile juice, which contains bile salts and bile pigments.
Bile juice is essential for the emulsification of fats, which increases the surface area for the action of lipase, but it does not contain any enzymes itself.
In contrast, the intestinal mucosa, gastric mucosa, and pancreas all secrete various digestive enzymes (such as peptidases, pepsin, and pancreatic amylase/lipase/proteases, respectively).
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16
BiologyEasyMCQAIIMS · 1984
At times,the ligaments are overstretched or torn. This condition is called:
A
Dislocation
B
Fracture
C
Sprain
D
Tension
Solution
(C) $Sprain$ is an injury to a ligament,which is the fibrous tissue that connects bones to other bones. It occurs when a ligament is overstretched or torn due to sudden movement or trauma. In contrast,a $Fracture$ refers to a broken bone,$Dislocation$ refers to the displacement of bones at a joint,and $Tension$ is a general term for physical stress.
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17
BiologyMediumMCQAIIMS · 1984
In case of disputed parentage,the blood group analysis of the mother,child,and alleged father can:
A
Definitely prove a man to be the father
B
Only prove that he cannot be the father
C
Not be of any use
D
None of the above
Solution
(B) Blood group analysis is based on the inheritance of $ABO$ blood group alleles $(I^A, I^B, i)$.
While it can exclude a man from being the father if his blood group is incompatible with the child's genotype,it cannot definitively prove paternity.
This is because many individuals in the population share the same blood group phenotypes.
Therefore,blood group analysis can only prove that a man cannot be the father,but it cannot confirm paternity with absolute certainty.
While it can exclude a man from being the father if his blood group is incompatible with the child's genotype,it cannot definitively prove paternity.
This is because many individuals in the population share the same blood group phenotypes.
Therefore,blood group analysis can only prove that a man cannot be the father,but it cannot confirm paternity with absolute certainty.
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18
BiologyMediumMCQAIIMS · 1984
$A$ person meets with an accident and a great loss of blood has occurred. There is no time to analyze his blood group. It is safe to transfuse blood of group:
A
$AB, Rh^+$
B
$AB, Rh^-$
C
$O, Rh^-$
D
$O, Rh^+$
Solution
(C) The $O$ blood group is known as the universal donor because it lacks both $A$ and $B$ antigens on the surface of red blood cells.
Furthermore,$Rh^-$ blood is considered safer for emergency transfusions because it lacks the $Rh$ antigen,preventing the risk of anti-$Rh$ antibody formation in an $Rh^-$ recipient.
Therefore,$O, Rh^-$ blood is the safest choice for emergency transfusion when the patient's blood group is unknown.
Furthermore,$Rh^-$ blood is considered safer for emergency transfusions because it lacks the $Rh$ antigen,preventing the risk of anti-$Rh$ antibody formation in an $Rh^-$ recipient.
Therefore,$O, Rh^-$ blood is the safest choice for emergency transfusion when the patient's blood group is unknown.
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19
BiologyMediumMCQAIIMS · 1984
The function of the allantois in mammals is:
A
Respiration only
B
Excretion
C
Nutrition,excretion,and respiration
D
Protection from shocks
Solution
(C) The allantois is one of the extra-embryonic membranes in mammals. Its primary functions include the storage of nitrogenous waste (excretion),participation in the exchange of gases (respiration),and it also plays a role in the transport of nutrients from the placenta to the embryo (nutrition). Therefore,option $(c)$ is the correct answer.
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20
BiologyMediumMCQAIIMS · 1984
Pure homozygous offsprings in a dihybrid cross in the $F_2$ generation will be
A
$1/2$
B
$1/4$
C
$1/8$
D
$1/16$
Solution
(C) In a dihybrid cross,the $F_2$ generation produces $16$ possible combinations of genotypes.
Out of these $16$ combinations,only $2$ are completely homozygous: the double dominant $(AABB)$ and the double recessive $(aabb)$.
Therefore,the proportion of pure homozygous offspring is $2/16$,which simplifies to $1/8$.
Thus,the correct option is $1/8$.
Out of these $16$ combinations,only $2$ are completely homozygous: the double dominant $(AABB)$ and the double recessive $(aabb)$.
Therefore,the proportion of pure homozygous offspring is $2/16$,which simplifies to $1/8$.
Thus,the correct option is $1/8$.
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21
BiologyMediumMCQAIIMS · 1984
The reason why some harmful mutations are not eliminated from the gene pool is that:
A
They have future survival value
B
They are recessive and carried by heterozygous individuals
C
They are dominant and show up more frequently
D
Genetic drift occurs because of a small population area
Solution
(B) Harmful mutations that are recessive are not eliminated from the gene pool because they are masked by the dominant allele in heterozygous individuals. Since the phenotype is not expressed in the heterozygote,natural selection cannot act upon it to remove it from the population. These alleles remain hidden in the gene pool and can be passed on to future generations.
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22
BiologyEasyMCQAIIMS · 1984
Barbara McClintock is famous for her work on
A
Wheat
B
Rice
C
Maize
D
Pisum
Solution
(C) Barbara McClintock discovered jumping genes (transposons) in $1940$ in maize $(Zea \text{ } mays)$.
She referred to these as controlling elements or mobile genetic elements.
For her pioneering work on these genetic elements, she was awarded the Nobel Prize in $1983$.
She referred to these as controlling elements or mobile genetic elements.
For her pioneering work on these genetic elements, she was awarded the Nobel Prize in $1983$.
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23
BiologyEasyMCQAIIMS · 1984
$BOD$ stands for:
A
Biological organism death
B
Biochemical organic matter decay
C
Biotic oxidation demand
D
Biochemical oxygen demand
Solution
(D) $BOD$ stands for Biochemical Oxygen Demand. It refers to the amount of dissolved oxygen that is required by aerobic microorganisms in a body of water to break down the organic material present in a given water sample at a specific temperature over a specific time period. Higher $BOD$ values indicate higher levels of water pollution.
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24
BiologyMediumMCQAIIMS · 1984
If a biochemical analysis of mitochondria is to be done,the best procedure would be
A
Grind the cell and filter out the mixture and take the debris
B
Subject the cells to cell fractionation $(centrifugation)$ and obtain mitochondria
C
Select cells which have a large number of mitochondria
D
Plasmolyse the cell,filter it and take the debris
Solution
(B) To perform a biochemical analysis of mitochondria,one must first isolate them from the rest of the cellular components.
This is achieved through the process of cell fractionation.
First,the cells are broken down (homogenized) to release the organelles.
Then,the mixture is subjected to differential centrifugation,where organelles are separated based on their size and density.
Mitochondria can be isolated in the pellet fraction at specific speeds of centrifugation,allowing for a pure biochemical analysis.
This is achieved through the process of cell fractionation.
First,the cells are broken down (homogenized) to release the organelles.
Then,the mixture is subjected to differential centrifugation,where organelles are separated based on their size and density.
Mitochondria can be isolated in the pellet fraction at specific speeds of centrifugation,allowing for a pure biochemical analysis.
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