AIPMT 1988 Biology Question Paper with Answer and Solution

127 QuestionsEnglishWith Solutions

BiologyQ5188 of 127 questions

Page 2 of 2 · English

51
BiologyEasyMCQAIPMT · 1988
The wood-boring mollusc/shipworm is .....
A
Chiton
B
Teredo
C
Limax
D
Patella

Solution

(B) $Teredo$,commonly known as the shipworm,is a genus of saltwater clams in the family $Teredinidae$.
These molluscs are famous for boring into submerged wood,such as ship hulls and piers,which is why they are referred to as wood-boring molluscs or shipworms.
$Chiton$ is a marine mollusc known as the coat-of-mail shell.
$Limax$ is a genus of slugs.
$Patella$ is a genus of sea limpets.
52
BiologyEasyMCQAIPMT · 1988
Typhlops is a/an ........
A
Sea snake
B
Glass snake
C
Blind snake
D
Grass snake

Solution

(C) $Typhlops$ is a genus of non-venomous snakes commonly known as blind snakes.
They belong to the family $Typhlopidae$.
These snakes are adapted for burrowing,having smooth scales and small,vestigial eyes covered by scales,which makes them appear blind.
Therefore,the correct option is $C$.
53
BiologyEasyMCQAIPMT · 1988
$Necturus$ is commonly known as:
A
Hellbender
B
Congoeel
C
Mudpuppy
D
Blind worm

Solution

(C) $Necturus$ is a genus of aquatic salamanders in the family $Proteidae$.
They are commonly known as $Mudpuppies$ because of the mistaken belief that they make a barking sound like a dog.
$Hellbender$ refers to $Cryptobranchus$ $alleganiensis$.
$Congoeel$ refers to $Amphiuma$.
$Blind$ $worm$ refers to $Anguis$ $fragilis$ (a legless lizard) or $Caecilians$ (amphibians).
54
BiologyEasyMCQAIPMT · 1988
The Fire-bellied toad is known as ........
A
Amphiuma
B
Bombinator
C
Necturus
D
Salamander

Solution

(B) The Fire-bellied toad belongs to the genus $Bombinator$ (also commonly referred to as $Bombina$). These toads are known for their bright red or orange coloration on their bellies,which serves as a warning signal to predators (aposematism). Therefore,the correct option is $B$.
55
BiologyEasyMCQAIPMT · 1988
The organ-pipe coral is known as .......
A
Tubipora
B
Astraea
C
Heliopora
D
Fungia

Solution

(A) The organ-pipe coral is scientifically known as $Tubipora$ $musica$. It belongs to the phylum $Cnidaria$ (or $Coelenterata$). It is named 'organ-pipe coral' because its skeletal structure consists of parallel tubes that resemble the pipes of a musical organ. Therefore,the correct option is $A$.
56
BiologyEasyMCQAIPMT · 1988
Which of the following is not a true amphibian?
A
Salamander
B
Toad
C
Turtle
D
Frog

Solution

(C) Amphibians belong to the class $Amphibia$ and are characterized by their ability to live in both aquatic and terrestrial habitats. They typically have moist skin,lack scales,and undergo metamorphosis.
$A$. Salamanders are amphibians (Order $Urodela$).
$B$. Toads are amphibians (Order $Anura$).
$C$. Turtles belong to the class $Reptilia$. They have dry,scaly skin and lay amniotic eggs on land,which distinguishes them from amphibians.
$D$. Frogs are amphibians (Order $Anura$).
Therefore,the turtle is not an amphibian.
57
BiologyEasyMCQAIPMT · 1988
The fruit of groundnut is a ..........
A
Legume
B
Caryopsis
C
Berry
D
Nut

Solution

(A) The fruit of groundnut $(Arachis hypogaea)$ is botanically classified as a $Legume$ (or $Pod$).
Although it develops underground due to geocarpy, it retains the structural characteristics of a legume, which is a dry, dehiscent fruit derived from a single carpel that splits along two sutures.
58
BiologyMediumMCQAIPMT · 1988
In which of the following are fibers produced?
A
Coconut
B
Oak
C
Teak
D
Sheesham

Solution

(A) Fibers are a type of sclerenchyma tissue that provide mechanical support to plants.
In the case of $Cocos$ $nucifera$ (coconut),the husk is composed of abundant sclerenchymatous fibers,which are commercially known as coir.
These fibers are highly lignified and provide strength to the fruit covering.
59
BiologyMediumMCQAIPMT · 1988
In which of the following is the cortex not differentiated into the hypodermis and general cortex?
A
Monocot stem
B
Dicot stem
C
Monocot root
D
Dicot root

Solution

(A) In a $Monocot$ stem,the ground tissue consists of a mass of parenchyma cells with no clear differentiation into hypodermis,cortex,endodermis,pericycle,and pith. The vascular bundles are scattered throughout the ground tissue. In contrast,$Dicot$ stems and roots show clear differentiation of the cortex into various zones like hypodermis,general cortex,and endodermis.
60
BiologyMediumMCQAIPMT · 1988
The cork cambium (phellogen) is formed from:
A
Phellogen
B
Vascular cambium
C
Phloem
D
Xylem

Solution

(A) The cork cambium,also known as $phellogen$,is a secondary meristematic tissue.
It is formed by the dedifferentiation of the cortical cells (parenchyma cells) located just below the epidermis during secondary growth in dicot stems and roots.
Therefore,the correct origin of the cork cambium is the cortical cells (part of the ground tissue system).
Among the given options,$Phellogen$ is the tissue itself,but the question asks for the origin. However,in the context of standard multiple-choice questions where the options are limited,$Phellogen$ is often cited as the layer involved in the periderm formation. Given the options provided,$A$ is the most appropriate choice as it refers to the tissue layer responsible for cork production.
61
BiologyEasyMCQAIPMT · 1988
The $Tunica-Corpus$ theory is associated with ..........
A
Root apex
B
Root cap
C
Shoot apex
D
Secondary growth

Solution

(C) The $Tunica-Corpus$ theory was proposed by $Schmidt$ in $1924$.
This theory explains the organization and development of the $Shoot$ $Apex$ in angiosperms.
According to this theory,the shoot apex consists of two distinct zones:
$1$. $Tunica$: The peripheral layer$(s)$ that divide anticlinally to increase the surface area.
$2$. $Corpus$: The central mass of cells that divide in various planes to increase the volume of the shoot apex.
Therefore,the correct option is $C$.
62
BiologyEasyMCQAIPMT · 1988
Which meristematic tissue is responsible for increasing the girth of the plant?
A
Lateral meristem
B
Intercalary meristem
C
Promeristem
D
Apical meristem

Solution

(A) The lateral meristem is responsible for secondary growth in plants,which leads to an increase in the diameter or girth of the stem and root. Examples include the vascular cambium and cork cambium. Apical meristems are responsible for primary growth (increase in length),while intercalary meristems are found at the base of leaves or internodes and help in elongation.
63
BiologyMediumMCQAIPMT · 1988
According to the fluid mosaic model,the cell membrane consists of:
A
Phospholipids and oligosaccharides
B
Phospholipids and hemicellulose
C
Phospholipids and integral proteins
D
Phospholipids,extrinsic proteins,and integral proteins

Solution

(D) The fluid mosaic model,proposed by $Singer$ and $Nicolson$ in $1972$,describes the cell membrane as a mosaic of various components.
These components primarily include a phospholipid bilayer,which provides the fluid structure,and proteins that are embedded within or attached to the surface.
The proteins are classified into two types: integral (or intrinsic) proteins,which are embedded in the bilayer,and peripheral (or extrinsic) proteins,which lie on the surface of the membrane.
Therefore,the cell membrane is composed of phospholipids,extrinsic proteins,and integral proteins.
64
BiologyMediumMCQAIPMT · 1988
In Hammerling's nucleocytoplasmic experiments,$Acetabularia$ is used,which is a:
A
Unicellular fungus
B
Multicellular fungus
C
Unicellular,uninucleate green alga
D
Unicellular,multinucleate green alga

Solution

(C) $Acetabularia$ is a genus of single-celled green algae that is famous for its large size and complex shape.
It is a unicellular organism that contains only one nucleus,which is located in the rhizoid (base).
Joachim Hammerling used $Acetabularia$ in his experiments to demonstrate the influence of the nucleus on the development and morphology of the cell,proving that the nucleus controls cellular activities.
65
BiologyMediumMCQAIPMT · 1988
Intercostal muscles are found in ...........
A
Rat
B
Thigh
C
Ribs
D
Diaphragm

Solution

(C) Intercostal muscles are a group of muscles that run between the ribs and help form and move the chest wall. They are primarily involved in the mechanical aspect of breathing by helping to expand and shrink the size of the chest cavity. Therefore,they are associated with the ribs.
66
BiologyEasyMCQAIPMT · 1988
Acute vision is found in ................
A
Eagle
B
Shark
C
Bat
D
Frog

Solution

(A) Acute vision or high visual acuity is a characteristic feature of birds of prey, such as the $Eagle$.
These birds possess a high density of photoreceptor cells (cones) in their retina, specifically in the fovea, which allows them to detect small prey from great distances with high precision.
67
BiologyMediumMCQAIPMT · 1988
In the intermediate lobe of the pituitary gland,$MSH$ is responsible for:
A
Pigmentation of skin in lower vertebrates.
B
Lightening of skin in lower vertebrates.
C
Both $(a)$ and $(b)$.
D
Pigmentation of skin in humans.

Solution

(A) $MSH$ stands for Melanocyte Stimulating Hormone,which is secreted by the intermediate lobe (pars intermedia) of the pituitary gland.
In lower vertebrates,$MSH$ acts on melanocytes and stimulates the synthesis of melanin,which results in the darkening (pigmentation) of the skin.
This process helps these organisms adapt to their environment by changing their skin color.
68
BiologyMediumMCQAIPMT · 1988
Which hormone exhibits an anti-insulin effect?
A
Cortisol
B
Calcitonin
C
Oxytocin
D
Aldosterone

Solution

(A) The hormone that exhibits an anti-insulin effect is $Cortisol$.
Insulin is a hypoglycemic hormone that lowers blood glucose levels by promoting the uptake of glucose into cells.
In contrast,$Cortisol$ is a glucocorticoid that acts as a hyperglycemic agent.
It stimulates gluconeogenesis (the production of glucose from non-carbohydrate sources) and inhibits the cellular uptake and utilization of glucose,thereby increasing blood glucose levels.
Thus,$Cortisol$ opposes the action of insulin.
69
BiologyMediumMCQAIPMT · 1988
Meiosis-$I$ is a reductional division. Meiosis-$II$ is an equational division because ...........
A
Pairing of homologous chromosomes occurs.
B
Crossing over takes place.
C
Separation of chromatids occurs.
D
Separation of homologous chromosomes occurs.

Solution

(C) Meiosis-$I$ is called a reductional division because the number of chromosomes is reduced to half in the daughter cells.
Meiosis-$II$ is called an equational division because it is similar to mitosis,where the number of chromosomes remains the same in the daughter cells as in the parent cell.
This happens because,during Meiosis-$II$,the sister chromatids of each chromosome separate and move to opposite poles,similar to the separation of chromatids in mitosis.
70
BiologyMediumMCQAIPMT · 1988
Phenylmercuric acetate $(PMA)$ results in which of the following effects?
A
Decreased photosynthesis
B
Decreased transpiration
C
Decreased respiration
D
Destruction of plants

Solution

(B) Phenylmercuric acetate $(PMA)$ is a well-known anti-transpirant chemical.
It acts by inducing the partial closure of stomata in plants.
By reducing the size of the stomatal aperture,it significantly decreases the rate of transpiration without affecting photosynthesis to the same extent.
Therefore,the application of $PMA$ leads to a reduction in water loss through transpiration.
71
BiologyMediumMCQAIPMT · 1988
The opening and closing of stomata are caused by:
A
Circadian rhythm
B
Genetic clock
C
Pressure of gases inside the leaves
D
Turgor pressure in guard cells

Solution

(D) The opening and closing of stomata are primarily regulated by the turgor pressure within the guard cells.
When the guard cells take up water through osmosis,they become turgid and bulge outward,causing the stomatal pore to open.
Conversely,when water leaves the guard cells,they become flaccid,causing the stomatal pore to close.
This process is influenced by factors like light,$CO_2$ concentration,and potassium ion $(K^+)$ flux.
72
BiologyMediumMCQAIPMT · 1988
The primary mechanism for the opening of stomata is:
A
Exosmosis
B
Endosmosis
C
Decrease in the concentration of cytoplasm
D
Plasmolysis of guard cells

Solution

(B) The opening and closing of stomata are regulated by the turgor pressure of the guard cells.
When guard cells take up water through the process of endosmosis,they become turgid.
This increase in turgor pressure causes the outer thin walls of the guard cells to bulge outwards,which in turn pulls the inner thick walls apart,leading to the opening of the stomatal pore.
Therefore,endosmosis is the fundamental process responsible for the opening of stomata.
73
BiologyEasyMCQAIPMT · 1988
The translocation of minerals from roots to leaves occurs through:
A
Xylem
B
Phloem
C
Sieve tubes
D
None of the above

Solution

(A) The translocation of minerals and water from the roots to the aerial parts of the plant,such as leaves,occurs through the $Xylem$ tissue. $Xylem$ is a complex permanent tissue responsible for the upward conduction of water and dissolved minerals. While $Phloem$ is responsible for the translocation of organic nutrients (food) from leaves to other parts of the plant,$Xylem$ is the primary pathway for mineral transport.
74
BiologyMediumMCQAIPMT · 1988
In which process is $NADP^+$ reduced to $NADPH$?
A
$PS-I$
B
$PS-II$
C
Calvin cycle
D
Non-cyclic photophosphorylation

Solution

(D) During the light-dependent reactions of photosynthesis,specifically in non-cyclic photophosphorylation (also known as the $Z$-scheme),electrons are transferred from water to $PS-II$,then to $PS-I$,and finally to $NADP^+$. The enzyme $FNR$ (Ferredoxin-$NADP^+$ reductase) facilitates the reduction of $NADP^+$ to $NADPH$ using electrons derived from $PS-I$. Therefore,$NADPH$ is produced during non-cyclic photophosphorylation.
75
BiologyMediumMCQAIPMT · 1988
In the photosynthetic pathway,carbon dioxide is fixed during:
A
$PS-I$
B
$PS-II$
C
Light reaction
D
Dark reaction

Solution

(D) In photosynthesis,the process is divided into two main stages: the light-dependent reaction and the light-independent reaction (also known as the dark reaction or Calvin cycle).
During the light reaction,light energy is captured to produce $ATP$ and $NADPH$.
Carbon dioxide fixation occurs during the dark reaction (Calvin cycle),where the $ATP$ and $NADPH$ produced in the light reaction are utilized to convert $CO_2$ into glucose.
Therefore,carbon dioxide is fixed during the dark reaction.
76
BiologyEasyMCQAIPMT · 1988
The size of the chlorophyll molecule is:
A
Head $15 \times 15 \mathring A$,Tail $25 \mathring A$
B
Head $20 \times 20 \mathring A$,Tail $25 \mathring A$
C
Head $15 \times 15 \mathring A$,Tail $20 \mathring A$
D
Head $10 \times 12 \mathring A$,Tail $25 \mathring A$

Solution

(C) The chlorophyll molecule consists of a porphyrin head and a phytol tail.
The porphyrin head is a square-shaped structure with dimensions of approximately $15 \times 15 \mathring A$.
The phytol tail is a long hydrocarbon chain attached to the head,which is approximately $20 \mathring A$ in length.
Therefore,the correct dimensions are Head $15 \times 15 \mathring A$ and Tail $20 \mathring A$.
77
BiologyMediumMCQAIPMT · 1988
Where is $NADP^+$ reduced to $NADPH$?
A
$HMP$ pathway
B
Calvin cycle
C
Glycolysis
D
$EMP$ pathway

Solution

(A) In the process of photosynthesis,specifically during the light-dependent reactions,$NADP^+$ is reduced to $NADPH$ by the enzyme $NADP^+$ reductase.
However,in the context of metabolic pathways provided in the options,the $HMP$ (Hexose Monophosphate) pathway,also known as the Pentose Phosphate Pathway $(PPP)$,is the primary pathway where $NADP^+$ is reduced to $NADPH$.
$NADPH$ produced in this pathway serves as a crucial reducing agent for biosynthetic reactions,such as fatty acid synthesis.
Therefore,the correct option is $A$.
78
BiologyMediumMCQAIPMT · 1988
What does $R.Q.$ stand for?
A
$C/N$
B
$N/C$
C
$CO_2/O_2$
D
$O_2/CO_2$

Solution

(C) $R.Q.$ stands for Respiratory Quotient.
It is defined as the ratio of the volume of $CO_2$ evolved to the volume of $O_2$ consumed during respiration.
The formula is: $R.Q. = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}}$.
Therefore, the correct option is $C$.
79
BiologyMediumMCQAIPMT · 1988
The partial oxidation of glucose into pyruvic acid through a series of intermediate steps is known as?
A
$TCA$ cycle
B
Glycolysis
C
$HMS$ pathway
D
Krebs cycle

Solution

(B) Glycolysis is the process of partial oxidation of glucose into two molecules of pyruvic acid.
It occurs in the cytoplasm of the cell and does not require oxygen.
It involves a series of ten enzyme-catalyzed reactions that convert one molecule of glucose $(C_6H_{12}O_6)$ into two molecules of pyruvate $(CH_3COCOOH)$.
80
BiologyEasyMCQAIPMT · 1988
Which hormone is primarily associated with cell division?
A
$IAA$
B
$NAA$
C
Cytokinin/Zeatin
D
Gibberellic acid

Solution

(C) Cytokinins are a class of plant growth substances (phytohormones) that promote cell division,or cytokinesis,in plant roots and shoots. They were discovered by Miller et al. $(1955)$ and the first natural cytokinin identified was zeatin,isolated from corn kernels. While auxins like $IAA$ and $NAA$ promote cell enlargement and gibberellins promote stem elongation,cytokinins are specifically responsible for stimulating cell division.
81
BiologyMediumMCQAIPMT · 1988
Which chemical must be applied to the roots to keep the leaves of a plant green?
A
Gibberellin
B
Cytokinin
C
Auxin
D
Ethylene

Solution

(B) Cytokinins are plant hormones that promote cell division and delay the senescence (aging) of leaves. By applying cytokinin to the roots,the transport of nutrients to the leaves is maintained,which helps in preventing the breakdown of chlorophyll and keeps the leaves green for a longer period. This phenomenon is known as the Richmond-Lang effect.
82
BiologyEasyMCQAIPMT · 1988
What causes the movement of leaves in the $Mimosa$ $pudica$ plant?
A
Thermonasty
B
Seismonasty
C
Hydrotropism
D
Chemonasty

Solution

(B) The movement of leaves in $Mimosa$ $pudica$ (touch-me-not plant) upon being touched is known as seismonasty.
Seismonasty is a type of nastic movement in plants that occurs in response to mechanical stimuli such as touch,vibration,or shaking.
This movement is caused by a sudden loss of turgor pressure in the cells of the pulvinus (the swollen base of the leaf stalk).
When the plant is touched,electrical and chemical signals cause water to move out of the pulvinus cells,leading to the drooping of the leaves.
83
BiologyEasyMCQAIPMT · 1988
The duct arising from the parotid gland and opening into the vestibule is known as:
A
Haversian duct
B
Stensen's duct
C
Wolffian duct
D
Intra-orbital duct

Solution

(B) The parotid glands are the largest salivary glands located in the cheeks.
The duct that carries saliva from the parotid gland into the oral cavity (specifically the vestibule near the second upper molar) is called Stensen's duct (or the parotid duct).
Therefore,the correct option is $B$.
84
BiologyEasyMCQAIPMT · 1988
Wharton's duct is associated with ............. .
A
Sublingual salivary gland
B
Parotid salivary gland
C
Submandibular salivary gland
D
Brunner's gland

Solution

(C) The human body has three pairs of salivary glands: the parotid,submandibular,and sublingual glands.
$1$. The parotid glands are the largest and their secretion is drained by Stensen's duct.
$2$. The submandibular (or submaxillary) glands are located below the jaw,and their secretion is drained by Wharton's duct.
$3$. The sublingual glands are the smallest and their secretion is drained by the ducts of Rivinus.
Therefore,Wharton's duct is associated with the submandibular salivary gland.
85
BiologyMediumMCQAIPMT · 1988
What is the primary product of hemoglobin degradation?
A
Bilirubin
B
Iron
C
Biliverdin
D
Calcium

Solution

(A) When red blood cells reach the end of their lifespan (approximately $120$ days),they are broken down in the spleen,liver,and bone marrow. The hemoglobin molecule is degraded into globin chains and heme. The heme group is further broken down into iron and biliverdin. Biliverdin is then rapidly converted into bilirubin,which is the primary pigment excreted by the liver into the bile. Therefore,bilirubin is the final major product of hemoglobin degradation.
86
BiologyEasyMCQAIPMT · 1988
$RBC$ are not found in .......
A
Frog
B
Cow
C
Camel
D
Cockroach

Solution

(D) In insects like the cockroach,the circulatory system is of the open type. The blood (hemolymph) does not contain respiratory pigments like hemoglobin and lacks $RBC$ (erythrocytes). Instead,oxygen is transported directly to tissues via the tracheal system. In contrast,frogs,cows,and camels are vertebrates and possess $RBC$ in their blood.
87
BiologyMediumMCQAIPMT · 1988
Erythroblastosis fetalis (hemolytic disease of the newborn) is observed in marriages between:
A
$Rh^{+ve}$ male and $Rh^{+ve}$ female
B
$Rh^{+ve}$ male and $Rh^{-ve}$ female
C
$Rh^{-ve}$ male and $Rh^{-ve}$ female
D
$Rh^{-ve}$ male and $Rh^{+ve}$ female

Solution

(B) Erythroblastosis fetalis occurs when an $Rh^{-ve}$ mother carries an $Rh^{+ve}$ fetus.
During the first pregnancy,the mother's blood is exposed to the $Rh$ antigens of the fetus during delivery,causing her body to produce anti-$Rh$ antibodies.
In subsequent pregnancies,if the fetus is again $Rh^{+ve}$,these maternal antibodies can cross the placenta and destroy the fetal red blood cells,leading to severe anemia or jaundice.
Therefore,this condition is observed in marriages between an $Rh^{+ve}$ male and an $Rh^{-ve}$ female.
88
BiologyMediumMCQAIPMT · 1988
What are the systolic and diastolic blood pressures of a healthy adult human?
A
$80 \, mm \, Hg$ and $80 \, mm \, Hg$
B
$70 \, mm \, Hg$ and $120 \, mm \, Hg$
C
$120 \, mm \, Hg$ and $80 \, mm \, Hg$
D
$50 \, mm \, Hg$ and $80 \, mm \, Hg$

Solution

(C) In a healthy adult human,the systolic blood pressure is the pressure exerted on the arterial walls during ventricular contraction,which is typically $120 \, mm \, Hg$.
The diastolic blood pressure is the pressure exerted during ventricular relaxation,which is typically $80 \, mm \, Hg$.
Therefore,the standard blood pressure reading is expressed as $120/80 \, mm \, Hg$.

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