AIPMT 2007 Biology Question Paper with Answer and Solution

114 QuestionsEnglishWith Solutions

BiologyQ1100 of 114 questions

Page 1 of 2 · English

1
BiologyEasyMCQAIPMT · 2007
$ICBN$ stands for
A
International Code of Botanical Nomenclature
B
International Congress of Biological Names
C
Indian Code of Botanical Nomenclature
D
Indian Congress of Biological Names

Solution

(A) $ICBN$ stands for International Code of Botanical Nomenclature.
It is a set of rules and recommendations dealing with the formal botanical names given to plants.
The foundations of $ICBN$ were established in the book written by $C. Linnaeus$ named $Philosophia Botanica$.
It is independent of zoological nomenclature.
The rank of species is basic,and the relative order of the ranks of taxa is: species,genus,tribe,family,order,series,class,division,and kingdom.
The different ranks or categories have specific endings for their names,such as division (-phyta),class (-ae),and family (-aceae).
2
BiologyMediumMCQAIPMT · 2007
The living organisms can be unexceptionally distinguished from the non-living things on the basis of their ability for
A
interaction with the environment and progressive evolution
B
reproduction
C
growth and movement
D
responsiveness to touch

Solution

(D) The correct answer is $D$. Responsiveness to touch (or consciousness) is considered a defining property of living organisms.
$1$. Consciousness is the ability of organisms to sense their environment and respond to environmental stimuli,which could be physical,chemical,or biological.
$2$. All organisms,from the prokaryotes to the most complex eukaryotes,can sense and respond to environmental cues.
$3$. While growth and reproduction are characteristics of living organisms,they are not 'defining' features because non-living things can also grow (e.g.,crystal growth) and some living organisms (like sterile worker bees or infertile human couples) do not reproduce.
$4$. Therefore,consciousness is the only feature that is present in all living organisms without exception.
3
BiologyEasyMCQAIPMT · 2007
Which one of the following is a slime mould?
A
Physarum
B
Thiobacillus
C
Anabaena
D
Rhizopus

Solution

(A) : Slime moulds are peculiar protists that normally take the form of amoebae,but under certain conditions,they develop fruiting bodies that release spores,which are superficially similar to the sporangia of fungi.
The order $Physarales$ includes $Physarum$ species.
The fruiting bodies (sporangia) are characterized by the presence of an abundant amount of calcium salt.
The order comprises $142$ species which are placed under $12$ genera.
$Physarum$ $polycephalum$ is the best-known species.
The somatic phase is a multinucleate,diploid,holocarpic plasmodium,which is the product of syngamy.
4
BiologyMediumMCQAIPMT · 2007
Which one of the following statements about mycoplasma is wrong?
A
They are pleomorphic.
B
They are sensitive to penicillin.
C
They cause diseases in plants.
D
They are also called $PPLO$.

Solution

(B) : Mycoplasma are small,unicellular,non-motile prokaryotic organisms.
They are pleomorphic,meaning they can change their shape.
Therefore,they are known as pleuropneumonia-like organisms $(PPLO)$.
They lack a cell wall.
They contain cytoplasm,ribosomes,and $DNA$.
They are inhibited by tetracyclines but are insensitive to penicillin because penicillin acts by inhibiting cell wall synthesis,and mycoplasma lack a cell wall.
They cause various diseases in both plants and animals.
5
BiologyMediumMCQAIPMT · 2007
Which pair of the following belongs to basidiomycetes?
A
Puffballs and Claviceps
B
Peziza and stink horns
C
Morchella and mushrooms
D
Birds nest fungi and puffballs

Solution

(D) $Cyathus$ is known as bird's nest fungi,and $Lycoperdon$ is called puffballs. Both these fungi belong to the group of club fungi or basidiomycetes.
These fungi produce spores inside club-shaped fruit bodies called basidium.
Typically,a basidium has $4$ basidiospores produced exogenously.
$Peziza$,$Morchella$,and $Claviceps$ belong to ascomycetes (which produce ascospores in ascocarps).
Mushrooms are also basidiomycetes,but the pair in option $D$ consists entirely of basidiomycetes.
6
BiologyMediumMCQAIPMT · 2007
In gymnosperms,the pollen chamber represents
A
a cavity in the ovule in which pollen grains are stored after pollination
B
an opening in the megagametophyte through which the pollen tube approaches the egg
C
the microsporangium in which pollen grains develop
D
a cell in the pollen grain in which the sperms are formed

Solution

(A) The correct answer is $A$. In gymnosperms,the pollen chamber is a cavity formed in the ovule near the micropyle,where pollen grains are deposited and stored after pollination. The pollen grains germinate within this chamber to produce pollen tubes that carry male gametes towards the egg cell.
7
BiologyMediumMCQAIPMT · 2007
Spore dissemination in some liverworts is aided by
A
indusium
B
calyptra
C
peristome teeth
D
elaters

Solution

(D) : An $elater$ is a cell (or structure attached to a cell) that is hygroscopic,and therefore will change shape in response to changes in moisture in the environment.
Elaters come in a variety of forms,but are always associated with plant spores.
In plants that do not have seeds,they function in dispersing the spores to a new location.
In the liverworts,$elaters$ are cells that develop in the sporophyte alongside the spores.
They are complete cells,usually with helical thickenings at maturity that respond to moisture content.
In most liverworts,the $elaters$ are unattached,but in some leafy species (such as $Frullania$) a few $elaters$ will remain attached to the inside of the sporangium (spore capsule).
The $elaters$ by hygroscopic movement help in spore dispersal.
8
BiologyMediumMCQAIPMT · 2007
Flagellated male gametes are present in all the three of which one of the following sets?
A
Zygnema,Saprolegnia and Hydrilla
B
Fucus,Marsilea and Calotropis
C
Riccia,Dryopteris and Cycas
D
Anthoceros,Funaria and Spirogyra

Solution

(C) The correct answer is $C$.
Flagellated male gametes (antherozoids) are typically found in lower plant groups such as algae,bryophytes,and pteridophytes,where water is required for fertilization.
In the set $Riccia$,$Dryopteris$,and $Cycas$:
$1$. $Riccia$ is a bryophyte that produces biflagellate antherozoids.
$2$. $Dryopteris$ is a pteridophyte that produces multiflagellate antherozoids.
$3$. $Cycas$ is a gymnosperm that produces large,multiciliated (flagellated) male gametes.
Other options contain plants like $Spirogyra$ (non-flagellated gametes) or $Calotropis$ (angiosperm,non-flagellated pollen tubes),making them incorrect.
9
BiologyMediumMCQAIPMT · 2007
If you are asked to classify the various algae into distinct groups,which of the following characters should you choose?
A
Nature of stored food materials in the cell
B
Structural organization of thallus
C
Chemical composition of the cell wall
D
Types of pigments present in the cell

Solution

(D) : Algae are a group of chlorophyllous,non-vascular plants with a thallose plant body. Different algae show different pigments present in the cell,such as chlorophyll $a, b$,xanthophylls,carotenes,etc. These pigments provide the primary basis for the classification of various groups of algae into different classes. For example,members of Chlorophyceae possess chlorophyll $a$ and $b$,Phaeophyceae contains fucoxanthin,and Rhodophyceae contains $r$-phycoerythrin.
10
BiologyMediumMCQAIPMT · 2007
In the prothallus of a vascular cryptogam,the antherozoids and eggs mature at different times. As a result
A
there is high degree of sterility
B
one can conclude that the plant is apomictic
C
self fertilization is prevented
D
there is no change in success rate of fertilization.

Solution

(C) The phenomenon where male and female reproductive organs mature at different times is known as dichogamy. In the prothallus of vascular cryptogams (like ferns),the antherozoids and eggs mature at different times. This temporal separation ensures that the sperm from one prothallus fertilizes the egg of another,thereby preventing self-fertilization and promoting cross-fertilization.
11
BiologyMediumMCQAIPMT · 2007
Which one of the following is a matching pair of a body feature and the animal possessing it?
A
Ventral central nervous system $\Rightarrow$ Leech
B
Pharyngeal gill slits absent in embryo $\Rightarrow$ Chamaeleon
C
Ventral heart $\Rightarrow$ Scorpion
D
Post-anal tail $\Rightarrow$ Octopus

Solution

(A) The correct matching pair is $A$.
$1$. Leech (Annelida) possesses a ventral central nervous system consisting of a nerve ring and a ventral nerve cord with ganglia.
$2$. Pharyngeal gill slits are present in the embryos of all chordates,including Chamaeleon (Reptilia).
$3$. Scorpion (Arthropoda) has a dorsal heart,not a ventral one.
$4$. $A$ post-anal tail is a characteristic feature of chordates; Octopus (Mollusca) is a non-chordate and lacks this feature.
12
BiologyMediumMCQAIPMT · 2007
What is common between parrot, platypus, and kangaroo?
A
Toothless jaws
B
Functional post-anal tail
C
Oviparity
D
Homoiothermy

Solution

(D) : Homoiothermy is the maintenance by an animal of its internal body temperature at a relatively constant value by using metabolic processes to counteract fluctuations in the temperature of the environment. Homoiothermy occurs in birds and mammals, which are described as endotherms. The heat produced by their tissue metabolism and the heat lost to the environment are balanced by various means to keep body temperature constant: $36-38^{\circ}C$ in mammals and $38-40^{\circ}C$ in birds. The hypothalamus in the brain monitors blood temperature and controls thermoregulation by both nervous and hormonal means. Thus, parrot (bird), platypus (mammal), and kangaroo (mammal) are all homoiothermic animals.
13
BiologyMediumMCQAIPMT · 2007
What is true about $Nereis$,scorpion,cockroach,and silver fish?
A
They all possess a dorsal heart.
B
None of them is aquatic.
C
They all belong to the same phylum.
D
They all have jointed paired appendages.

Solution

(A) $Nereis$,scorpion,cockroach,and silver fish are all invertebrates and possess a dorsal heart.
$Nereis$ is a marine animal,whereas the other animals mentioned are terrestrial.
$Nereis$ belongs to Phylum $Annelida$,while the rest of the animals belong to Phylum $Arthropoda$.
Jointed appendages are a characteristic feature of Phylum $Arthropoda$,which includes scorpion,cockroach,and silver fish,but these are absent in $Nereis$.
14
BiologyMediumMCQAIPMT · 2007
Which of the following is a flowering plant with nodules containing filamentous nitrogen-fixing micro-organism?
A
Crotalaria juncea
B
Cycas revoluta
C
Cicer arietinum
D
Casuarina equisetifolia

Solution

(D) $Casuarinaceae$ is the family of dicotyledonous flowering plants placed in the order $Fagales$.
$Casuarina$ is a member of this family,characterized by drooping equisetoid twigs,evergreen nature,and being monoecious or dioecious.
The roots of $Casuarina$ possess nitrogen-fixing nodules that contain the soil actinomycetes called $Frankia$,which is a filamentous bacterium.
15
BiologyMediumMCQAIPMT · 2007
Passage cells are thin-walled cells found in
A
phloem elements that serve as entry points for substances for transport to other plant parts
B
testa of seeds to enable emergence of growing embryonic axis during seed germination
C
central region of style through which the pollen tube grows towards the ovary
D
endodermis of roots facilitating rapid transport of water from cortex to pericycle

Solution

(D) : Endodermis is a single-layered structure that separates the cortex from the stele.
There are both thick-walled and thin-walled cells in the endodermis.
The thin-walled cells are known as passage cells or transfusion cells,which are located opposite the protoxylem groups.
These cells facilitate the rapid transport of water from the cortex to the pericycle.
16
BiologyMediumMCQAIPMT · 2007
For a critical study of secondary growth in plants,which one of the following pairs is suitable?
A
Teak and pine
B
Deodar and fern
C
Wheat and maiden hair fern
D
Sugarcane and sunflower

Solution

(A) Secondary growth is observed in dicots and gymnosperms. It is not observed in pteridophytes and is rarely observed in monocots.
Secondary growth results in an increase in the girth or diameter of the stem due to the formation of secondary tissues by the activity of lateral meristems (vascular cambium and cork cambium).
Teak is a dicotyledonous angiosperm,and pine is a gymnosperm; both exhibit significant secondary growth.
Therefore,for the study of secondary growth,teak and pine are the best-suited pair.
17
BiologyMediumMCQAIPMT · 2007
Which one of the following pairs of structures distinguishes a nerve cell from other types of cell?
A
Vacuoles and fibres
B
Flagellum and medullary sheath
C
Nucleus and mitochondria
D
Perikaryon and dendrites

Solution

(D) : $A$ neuron (nerve cell) is one of the basic functional units of the nervous system. $A$ neuron is a cell specialized to transmit electrical nerve impulses and thus carry information from one part of the body to another. Each neuron has an enlarged portion,the cell body ($Perikaryon$ or $Cyton$),containing the nucleus. From the cell body,several short,branched processes called $Dendrites$ extend,through which impulses enter the neuron. $A$ longer process,the axon,extends outward and carries impulses away from the cell body. $Perikaryon$ and $Dendrites$ are characteristic structures that distinguish a neuron from other types of cells.
18
BiologyMediumMCQAIPMT · 2007
In which one of the following preparations are you likely to come across cell junctions most frequently?
A
Thrombocytes
B
Tendon
C
Hyaline cartilage
D
Ciliated epithelium

Solution

(D) The correct answer is $D$. Cell junctions are specialized structures that provide contact between neighboring cells or between a cell and the extracellular matrix. They are most abundant in epithelial tissues, which form the lining of various body cavities and surfaces.
$1$. $Ciliated$ $epithelium$ is a type of epithelial tissue consisting of columnar or cuboidal cells. Because epithelial cells are tightly packed to form a continuous barrier, they require various types of cell junctions (such as tight junctions, adherens junctions, and gap junctions) to maintain structural integrity and regulate transport.
$2$. $Thrombocytes$ (platelets) are blood cells involved in clotting and do not form tissue layers with junctions.
$3$. $Tendon$ and $hyaline$ $cartilage$ are types of connective tissue. In connective tissues, cells are typically separated by a large amount of extracellular matrix, and therefore, cell junctions are far less frequent compared to epithelial tissues.
19
BiologyEasyMCQAIPMT · 2007
Which one of the following is not a constituent of cell membrane?
A
Glycolipids
B
Proline
C
Phospholipids
D
Cholesterol

Solution

(B) The correct answer is $(b)$.
Chemically,a biomembrane consists of lipids $(20-70\%)$,proteins $(20-70\%)$,carbohydrates $(1-5\%)$,and water $(20\%)$.
The primary lipids found in the cell membrane include phospholipids,sterols (such as cholesterol),glycolipids,and sphingolipids (such as sphingomyelin and cerebrosides).
Proline is an amino acid,which is a building block of proteins,but it is not a structural constituent of the cell membrane itself in the same way that lipids or membrane proteins are.
20
BiologyMediumMCQAIPMT · 2007
Select the wrong statement from the following.
A
Both chloroplasts and mitochondria have an internal compartment,the thylakoid space bounded by the thylakoid membrane.
B
Both chloroplasts and mitochondria contain $DNA$.
C
The chloroplasts are generally much larger than mitochondria.
D
Both chloroplasts and mitochondria contain an inner and an outer membrane.

Solution

(A) is the incorrect statement.
$1$. Mitochondria and chloroplasts are semi-autonomous organelles. They contain their own $DNA$,which encodes for $mRNA$,$tRNA$,and $rRNA$.
$2$. Both organelles possess their own ribosomes,allowing them to synthesize some of their own proteins.
$3$. The thylakoid space and thylakoid membrane are specific to chloroplasts,not mitochondria. Mitochondria have an inner membrane folded into cristae,enclosing the matrix.
$4$. Both organelles are double-membrane bound structures,and chloroplasts are typically larger than mitochondria.
21
BiologyMediumMCQAIPMT · 2007
Biological organisation starts with
A
cellular level
B
organismic level
C
atomic level
D
submicroscopic molecular level

Solution

(D) The correct answer is $D$. Biological organization begins at the submicroscopic molecular level.
Molecular assemblies are large,organized sets of molecular units that constitute parts of organelles.
For example,a common macromolecular assembly is the microtubule,which is essential for maintaining cell structure and facilitating cell movement.
The cell (plasma) membrane,which surrounds many organelles and the cell itself,is also a highly organized molecular assembly.
22
BiologyMediumMCQAIPMT · 2007
Two cells $A$ and $B$ are contiguous. Cell $A$ has osmotic pressure $10 \ atm$,turgor pressure $7 \ atm$ and diffusion pressure deficit $3 \ atm$. Cell $B$ has osmotic pressure $8 \ atm$,turgor pressure $3 \ atm$ and diffusion pressure deficit $5 \ atm$. The result will be
A
no movement of water
B
equilibrium between the two
C
movement of water from cell $A$ to $B$
D
movement of water from cell $B$ to $A$

Solution

(C) The correct answer is $C$.
Diffusion pressure deficit $(DPD)$ is the reduction in the diffusion pressure of water in a system compared to its pure state.
It is calculated using the formula: $DPD = O.P. - T.P.$
For cell $A$: $DPD = 10 \ atm - 7 \ atm = 3 \ atm$.
For cell $B$: $DPD = 8 \ atm - 3 \ atm = 5 \ atm$.
Water always moves from a region of lower $DPD$ to a region of higher $DPD$.
Since cell $A$ has a lower $DPD$ $(3 \ atm)$ and cell $B$ has a higher $DPD$ $(5 \ atm)$,water will move from cell $A$ to cell $B$.
23
BiologyEasyMCQAIPMT · 2007
$A$ plant requires magnesium for
A
protein synthesis
B
chlorophyll synthesis
C
cell wall development
D
holding cells together

Solution

(B) $Mg$ is an essential mineral element for plants. It serves as a vital constituent of the chlorophyll molecule,forming the central atom of the porphyrin ring. Without $Mg$,chlorophyll cannot be synthesized,which directly impacts the plant's ability to perform photosynthesis. Therefore,the correct answer is $B$.
24
BiologyMediumMCQAIPMT · 2007
In the leaves of $C_4$ plants,malic acid formation during $CO_2$ fixation occurs in the cells of
A
bundle sheath
B
phloem
C
epidermis
D
mesophyll

Solution

(D) In $C_4$ plants,the process of $CO_2$ fixation occurs in two stages involving two different cell types.
$1$. The primary $CO_2$ fixation occurs in the mesophyll cells,where $CO_2$ is accepted by Phosphoenolpyruvate $(PEP)$ to form a $4$-carbon compound,oxaloacetic acid $(OAA)$.
$2$. This $OAA$ is then converted into malic acid (or aspartic acid) in the mesophyll cells.
$3$. Malic acid is then transported to the bundle sheath cells,where it undergoes decarboxylation to release $CO_2$ for the Calvin cycle.
Therefore,the formation of malic acid occurs in the mesophyll cells.
25
BiologyMediumMCQAIPMT · 2007
The first acceptor of electrons from an excited chlorophyll molecule of photosystem $II$ is
A
iron-sulphur protein
B
ferredoxin
C
quinone
D
cytochrome

Solution

(C) The correct answer is $C$. In the light-dependent reactions of photosynthesis,the reaction center of photosystem $II$ $(P680)$ gets excited by absorbing light energy and releases electrons. These high-energy electrons are immediately captured by the primary electron acceptor,which is a molecule of pheophytin (a chlorophyll derivative),followed by plastoquinone $(PQ)$. Among the given options,quinone acts as the primary electron acceptor in the electron transport chain of photosystem $II$.
26
BiologyMediumMCQAIPMT · 2007
The overall goal of glycolysis,Krebs' cycle and the electron transport system is the formation of
A
$ATP$ in one large oxidation reaction
B
sugars
C
nucleic acids
D
$ATP$ in small stepwise units

Solution

(D) The correct answer is $D$.
Respiration is an energy-releasing,enzymatically controlled,multistep catabolic process involving the stepwise breakdown of organic substances (hexose sugars) inside living cells.
Aerobic respiration includes three major processes: glycolysis,Krebs' cycle,and the electron transport chain.
During these processes,the substrate is completely broken down to form $CO_2$ and water.
$A$ large amount of energy is released in a stepwise manner in the form of $ATP$ molecules,rather than in a single large reaction,which prevents the loss of energy as heat.
27
BiologyMediumMCQAIPMT · 2007
All enzymes of the $TCA$ cycle are located in the mitochondrial matrix, except for one which is located in the inner mitochondrial membrane in eukaryotes and in the cytosol in prokaryotes. This enzyme is
A
isocitrate dehydrogenase
B
malate dehydrogenase
C
succinate dehydrogenase
D
lactate dehydrogenase

Solution

(C) The correct answer is $C$.
Most enzymes of the $TCA$ cycle (Krebs cycle) are found in the mitochondrial matrix.
However, the enzyme $succinate$ $dehydrogenase$ is an exception.
In eukaryotes, it is embedded in the inner mitochondrial membrane, where it also functions as Complex $II$ of the electron transport chain.
In prokaryotes, since they lack mitochondria, this enzyme is located in the cytosol or associated with the plasma membrane.
28
BiologyEasyMCQAIPMT · 2007
The wavelength of light absorbed by the $P_r$ form of phytochrome is (in $nm$)
A
$680$
B
$720$
C
$620$
D
$640$

Solution

(A) $P_r$ is the inactive form of phytochrome,which is blue in color.
It specifically absorbs red light in the range of $660\ nm$ to $680\ nm$.
Upon absorption of red light,it is converted into the active $P_{fr}$ form.
The $P_{fr}$ form absorbs far-red light at approximately $730\ nm$ and converts back to the $P_r$ form.
Therefore,the correct wavelength absorbed by the $P_r$ form is $680\ nm$.
29
BiologyMediumMCQAIPMT · 2007
Which one of the following pairs is not correctly matched?
A
Gibberellic acid - Leaf fall
B
Cytokinin - Cell division
C
$IAA$ - Cell wall elongation
D
Abscisic acid - Stomatal closure

Solution

(A) : Gibberellic acid is a plant growth hormone that promotes cell elongation in stems and leaves,and internodal growth in dwarf plants.
It is a growth-promoting hormone and does not induce leaf abscission.
Leaf abscission (leaf fall) is primarily associated with Abscisic acid $(ABA)$,not Gibberellic acid.
30
BiologyMediumMCQAIPMT · 2007
Opening of floral buds into flowers is a type of
A
autonomic movement of variation
B
paratonic movement of growth
C
autonomic movement of growth
D
autonomic movement of locomotion

Solution

(C) The correct answer is $C$.
Movements in plants can be classified into movements of locomotion and movements of curvature.
Movements of curvature involve the movement of individual parts of the plant in relation to other parts.
Curvature movements are further classified into growth movements and turgor movements.
Growth movements are caused by differential growth in different parts of an organ.
The opening of floral buds into flowers is a classic example of an autonomic growth movement,as it is driven by internal physiological processes related to the development of the floral organs.
While some floral movements are nastic (like thermonasty or photonasty) and influenced by external stimuli,the fundamental process of bud opening is primarily an autonomic growth-related phenomenon.
31
BiologyMediumMCQAIPMT · 2007
Which one of the following is a fat-soluble vitamin and its related deficiency disease?
A
Retinol-Xerophthalmia
B
Cobalamine-Beri-beri
C
Calciferol-Pellagra
D
Ascorbic acid-Scurvy

Solution

(A) The correct answer is $A$. Retinol (vitamin $A$) is a fat-soluble vitamin,and its deficiency leads to xerophthalmia (dryness and thickening of the cornea).
$B$. Cobalamine (vitamin $B_{12}$) is a water-soluble vitamin,and its deficiency causes pernicious anaemia,not beri-beri. Beri-beri is caused by the deficiency of vitamin $B_1$ (thiamine).
$C$. Calciferol (vitamin $D$) is a fat-soluble vitamin,but its deficiency causes rickets (in children) or osteomalacia (in adults),not pellagra. Pellagra is caused by the deficiency of vitamin $B_3$ (niacin).
$D$. Ascorbic acid (vitamin $C$) is a water-soluble vitamin,and its deficiency causes scurvy. Since the question asks for a fat-soluble vitamin and its related deficiency disease,only option $A$ is correct.
32
BiologyMediumMCQAIPMT · 2007
$A$ drop of each of the following is placed separately on four slides. Which of them will not coagulate?
A
Blood serum
B
Sample from the thoracic duct of lymphatic system
C
Whole blood from pulmonary vein
D
Blood plasma

Solution

(A) : Blood serum is blood plasma from which the fibrinogen and other clotting factors have been removed.
Since clotting factors are essential for the coagulation process,their absence prevents blood serum from clotting.
In contrast,whole blood,lymph (which contains clotting factors),and blood plasma (which contains fibrinogen) all have the potential to coagulate under appropriate conditions.
33
BiologyDifficultMCQAIPMT · 2007
$A$ person who is on a long hunger strike and is surviving only on water,will have
A
less amino acids in his urine
B
more glucose in his blood
C
less urea in his urine
D
more sodium in his urine.

Solution

(C) : $A$ person who is on a long hunger strike and is surviving only on water,will have less urea in his urine. Urea is an organic chemical compound which is the waste product produced when the body metabolizes protein. It is manufactured in the liver from the breakdown of amino acids and ammonia,and the kidneys transfer urea from the blood to the urine. The average person excretes about $30 \ g$ of urea a day. During starvation,the body must rely on its own tissues to provide fuel. While the primary fuel is stored fat,the body also needs a continuous supply of glucose. Since glycogen stores are depleted within $36$ hours,the body performs gluconeogenesis,where it makes glucose from amino acids derived from body proteins. However,in the long term,the body minimizes protein breakdown to preserve vital tissues,leading to a reduction in the overall nitrogenous waste production compared to a normal diet,resulting in less urea in the urine.
34
BiologyMediumMCQAIPMT · 2007
Which one of the following items gives its correct total number?
A
Types of diabetes - $3$
B
Cervical vertebrae in humans - $8$
C
Floating ribs in humans - $4$
D
Amino acids found in proteins - $16$

Solution

(C) The correct answer is $C$.
$1$. Types of diabetes: There are primarily $2$ main types of diabetes mellitus ($Type-I$ and $Type-II$), though gestational diabetes is also recognized.
$2$. Cervical vertebrae in humans: There are $7$ cervical vertebrae in humans.
$3$. Floating ribs in humans: There are $2$ pairs ($4$ ribs in total) of floating ribs. These are the $11^{th}$ and $12^{th}$ pairs of ribs, which do not attach to the sternum.
$4$. Amino acids found in proteins: There are $20$ standard amino acids that are commonly found in proteins.
35
BiologyMediumMCQAIPMT · 2007
In the human body,which one of the following is anatomically correct?
A
Collar bones $\Rightarrow 3$ pairs
B
Salivary glands $\Rightarrow 1$ pair
C
Cranial nerves $\Rightarrow 10$ pairs
D
Floating ribs $\Rightarrow 2$ pairs

Solution

(D) The correct anatomical counts in the human body are as follows:
$1$. Collar bones (Clavicles): There are $2$ collar bones in total,which form $1$ pair.
$2$. Salivary glands: There are $3$ pairs of salivary glands (parotid,submandibular,and sublingual).
$3$. Cranial nerves: There are $12$ pairs of cranial nerves originating from the brain.
$4$. Floating ribs: The $11^{th}$ and $12^{th}$ pairs of ribs are not connected ventrally to the sternum and are known as floating ribs,totaling $2$ pairs.
Therefore,the statement regarding floating ribs is anatomically correct.
36
BiologyMediumMCQAIPMT · 2007
During the transmission of a nerve impulse through a nerve fibre,the potential on the inner side of the plasma membrane has which type of electric change?
A
First positive,then negative and continue to be negative
B
First negative,then positive and continue to be positive
C
First positive,then negative and again back to positive
D
First negative,then positive and again back to negative.

Solution

(D) The correct answer is $D$.
$1$. At the resting state,the inner side of the axonal membrane is negatively charged relative to the outer side due to the differential permeability of the membrane to $K^+$ and $Na^+$ ions.
$2$. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing an influx of $Na^+$. This leads to depolarization,where the inner side becomes positively charged.
$3$. Subsequently,the membrane becomes permeable to $K^+$ ions,which move out,leading to repolarization,where the inner side returns to its original negative state.
$4$. Thus,the potential changes from negative to positive and back to negative during the transmission of an action potential.
37
BiologyEasyMCQAIPMT · 2007
Bowman's glands are located in the
A
Anterior pituitary
B
Female reproductive system of cockroach
C
Olfactory epithelium of our nose
D
Proximal end of uriniferous tubules.

Solution

(C) : Bowman's glands,also known as olfactory glands,are branched tubuloalveolar glands found in the mucous membrane of the olfactory region of the nasal cavity.
These glands produce mucus that serves to moisten the olfactory epithelium and dissolve odor-containing gases,which is essential for the sense of smell.
38
BiologyMediumMCQAIPMT · 2007
Feeling the tremors of an earthquake,a scared resident of the seventh floor of a multistory building starts climbing down the stairs rapidly. Which hormone initiated this action?
A
Adrenaline
B
Glucagon
C
Gastrin
D
Thyroxine

Solution

(A) : Adrenaline (epinephrine),also known as the emergency hormone,is produced by the adrenal medulla. It prepares the body for '$fright, flight, or fight$' responses by increasing heart rate,improving muscle power,and increasing the rate and depth of breathing. It also inhibits non-essential processes like digestion during an emergency. These effects are similar to the stimulation of the sympathetic nervous system.
39
BiologyMediumMCQAIPMT · 2007
$A$ person is having problems with calcium and phosphorus metabolism in his body. Which one of the following glands may not be functioning properly?
A
Parotid
B
Pancreas
C
Thyroid
D
Parathyroid

Solution

(D) : Parathormone $(PTH)$ is secreted by the chief cells of the parathyroid gland,which regulates the metabolism of calcium and phosphate in the body.
It increases calcium absorption from the intestine and also enhances calcium resorption from the nephrons of the kidneys.
Therefore,a deficiency or improper functioning of the parathyroid gland leads to an imbalance in calcium and phosphorus levels,specifically resulting in low blood calcium levels $(hypocalcemia)$.
40
BiologyMediumMCQAIPMT · 2007
Two plants can be conclusively said to belong to the same species if they
A
have more than $90$ percent similar genes
B
look similar and possess identical secondary metabolites
C
have same number of chromosomes
D
can reproduce freely with each other and form seeds.

Solution

(D) The correct answer is $(D)$.
According to the biological species concept,organisms are classified as belonging to the same species if they can interbreed freely under natural conditions to produce fertile offspring.
In the case of plants,if two individuals can reproduce freely with each other and form viable seeds,they are considered to belong to the same species.
While genetic similarity,morphological appearance,and chromosome numbers are important,the ability to interbreed is the definitive criterion for species classification.
41
BiologyMediumMCQAIPMT · 2007
Increased asthmatic attacks in certain seasons are related to
A
eating fruits preserved in tin containers
B
inhalation of seasonal pollen
C
low temperature
D
hot and humid environment

Solution

(B) : Increased asthmatic attacks in certain seasons are related to the inhalation of seasonal pollen.
Pollens are microscopic grains produced by plants for the purpose of reproduction.
Pollen allergy is a hypersensitive immune reaction to specific pollen grains present in the air.
Pollen-induced allergic reactions often manifest as extrinsic asthma,allergic rhinitis,and bronchitis.
42
BiologyMediumMCQAIPMT · 2007
Which of the following is an example of negative feedback control in humans?
A
Constriction of skin blood vessels and shivering of skeletal muscles during extreme cold.
B
Secretion of tears when sand particles enter the eye.
C
Salivation in the mouth upon seeing delicious food.
D
Secretion from sweat glands and dilation of skin blood vessels during extreme heat.

Solution

(A) Negative feedback is a regulatory mechanism in which a stimulus causes an opposite output in order to maintain homeostasis.
$A$ is an example of negative feedback because when the body temperature drops (stimulus),the body responds by constricting blood vessels and shivering (response) to increase heat production and conserve body temperature.
$B$ and $C$ are examples of reflex actions or conditioned responses,not homeostatic negative feedback.
$D$ is incorrect because dilation of blood vessels (not constriction) occurs during heat to release heat.
43
BiologyMediumMCQAIPMT · 2007
Living organisms can be distinguished from non-living things on the basis of which ability,excluding exceptions?
A
Interaction with the environment and progressive evolution
B
Reproduction
C
Growth and movement
D
Response to touch

Solution

(D) Living organisms exhibit several characteristics such as growth,reproduction,metabolism,cellular organization,and consciousness.
Among these,consciousness (the ability to sense their environment and respond to environmental stimuli) is considered the defining property of living organisms.
While growth and reproduction are characteristics of living beings,they are not 'defining' properties because non-living objects can also grow (e.g.,mountains,boulders) and some living organisms (e.g.,sterile worker bees,infertile human couples) cannot reproduce.
Therefore,the ability to sense the environment and respond to stimuli is the most distinct feature that separates living from non-living things.
44
BiologyMediumMCQAIPMT · 2007
The organization of living organisms starts with which level?
A
Cellular level
B
Organismal level
C
Atomic level
D
Sub-microscopic molecular level

Solution

(D) The biological organization of living organisms begins at the sub-microscopic molecular level. Atoms combine to form molecules,which then form macromolecules,organelles,and eventually cells. Therefore,the fundamental basis of life's organization starts at the sub-microscopic molecular level.
45
BiologyEasyMCQAIPMT · 2007
Ergot of rye is caused by a species of which of the following?
A
Phytophthora
B
Uncinula
C
Ustilago
D
Claviceps

Solution

(D) Ergot of rye is a plant disease caused by the fungus $Claviceps \text{ purpurea}$.
This fungus belongs to the class Ascomycetes.
It infects the ovaries of rye and other cereal grasses, replacing the grain with a dark, hard fungal mass known as an ergot or sclerotium.
These sclerotia contain toxic alkaloids that can cause ergotism in humans and animals if consumed.
46
BiologyMediumMCQAIPMT · 2007
In which of the following sets are flagellated male gametes found?
A
Anthoceros,Funaria and Spirogyra
B
Zygnema,Saprolegnia,Hydrilla
C
Fucus,Marsilea,Calotropis
D
Riccia,Dryopteris,Cycas

Solution

(D) Flagellated male gametes (antherozoids) are typically found in groups that require water for fertilization,such as Bryophytes,Pteridophytes,and some Gymnosperms.
$1$. $Riccia$ is a Bryophyte,which produces flagellated antherozoids.
$2$. $Dryopteris$ is a Pteridophyte,which also produces flagellated antherozoids.
$3$. $Cycas$ is a Gymnosperm,which is unique among seed plants for having flagellated,motile male gametes.
Therefore,the set $Riccia, Dryopteris, Cycas$ contains organisms that all possess flagellated male gametes.
47
BiologyMediumMCQAIPMT · 2007
Which of the following pairs are correct?
Animal Morphological Feature
$(i)$ Crocodile $4$-chambered heart
$(ii)$ Sea cucumber Echinodermata (Water vascular system)
$(iii)$ Obelia Metagenesis
$(iv)$ Lemur Thecodont
A
$(ii), (iii)$ and $(iv)$
B
$(i), (iii)$ and $(iv)$
C
Only $(i)$ and $(ii)$
D
Only $(i)$ and $(iv)$

Solution

(B) Let us analyze each pair:
$(i)$ Crocodile: It belongs to the class Reptilia. Unlike most reptiles, crocodiles possess a $4$-chambered heart. This is correct.
$(ii)$ Sea cucumber: It belongs to the phylum Echinodermata. The term 'Abicharno' (in the original query) is incorrect; the characteristic feature is the water vascular system. Thus, this pair is incorrect.
$(iii)$ Obelia: It exhibits metagenesis (alternation of generations between polyp and medusa forms). This is correct.
$(iv)$ Lemur: It is a mammal. Mammals possess thecodont dentition (teeth embedded in sockets of the jaw bone). This is correct.
Therefore, pairs $(i), (iii),$ and $(iv)$ are correct.
48
BiologyMediumMCQAIPMT · 2007
Which of the following mammalian cells are not capable of metabolizing glucose to $CO_2$ aerobically?
A
White blood cells
B
Smooth muscle cells
C
Liver cells
D
Red blood cells

Solution

(D) Mature mammalian red blood cells $(RBCs)$ lack mitochondria and other organelles.
Since mitochondria are the sites of aerobic respiration (the Krebs cycle and electron transport chain),$RBCs$ cannot perform aerobic respiration to metabolize glucose into $CO_2$.
Instead,they rely entirely on anaerobic glycolysis to produce $ATP$ for their energy requirements.
49
BiologyMediumMCQAIPMT · 2007
Which of the following pairs of structures is distinct from the other types of cells found in neurons?
A
Vacuoles and filaments
B
Flagella and myelin sheath
C
Nucleus and mitochondria
D
Perikaryon and dendrites

Solution

(B) Neurons are specialized cells that contain typical cell organelles like the $Nucleus$ and $Mitochondria$ (found in most cells). $Perikaryon$ (cell body) and $Dendrites$ are characteristic structures of neurons. $Vacuoles$ and $Filaments$ are also present in various cell types. However,the $Myelin$ $sheath$ is a specialized insulating layer formed by glial cells (Schwann cells or oligodendrocytes) around the axon,and $Flagella$ are typically absent in mature neurons. Among the given options,the pair $Flagella$ and $Myelin$ $sheath$ is the most distinct because $Myelin$ $sheath$ is a unique feature of nerve fibers,and $Flagella$ are not standard components of neuronal structure.
50
BiologyEasyMCQAIPMT · 2007
About $98\%$ of living organisms are composed of elements including carbon,hydrogen,nitrogen,oxygen,and:
A
Sulfur and magnesium
B
Magnesium and scandium
C
Phosphorus and sulfur
D
Calcium and phosphorus

Solution

(C) Living organisms are primarily composed of elements such as carbon,hydrogen,nitrogen,oxygen,phosphorus,and sulfur. These elements constitute the bulk of biological macromolecules like proteins,nucleic acids,lipids,and carbohydrates. Phosphorus is essential for $DNA$,$RNA$,and $ATP$,while sulfur is a critical component of certain amino acids like cysteine and methionine.
51
BiologyEasyMCQAIPMT · 2007
Male gametes in angiosperms are formed by the division of
A
generative cell
B
vegetative cell
C
microspore mother cell
D
microspore

Solution

(A) In the pollen sac $(microsporangium)$ of the anther,haploid microspores are formed by meiosis.
Each microspore undergoes mitosis to produce a two-celled pollen grain consisting of a small generative cell and a large vegetative cell.
The generative cell then undergoes a second mitotic division to form two male gametes.
Therefore,the male gametes are formed by the division of the generative cell.
52
BiologyMediumMCQAIPMT · 2007
Which one of the following is surrounded by a callose wall?
A
Male gamete
B
Egg
C
Pollen grain
D
Microspore mother cell

Solution

(D) The correct answer is $(D)$.
In the anther,the primary sporogenous tissue differentiates into microspore mother cells $(MMC)$.
These $MMC$ are diploid and are initially connected by plasmodesmata.
Each $MMC$ is surrounded by a callose wall,which is deposited between the plasma membrane and the primary cell wall.
This callose wall is synthesized by the tapetum.
Following the deposition of the callose wall,the $MMC$ undergoes meiosis to form a tetrad of microspores.
After meiosis,the callose wall is degraded by the enzyme callase,allowing the microspores to separate into individual pollen grains.
53
BiologyMediumMCQAIPMT · 2007
Which part of the ovary in mammals acts as an endocrine gland after ovulation?
A
Stroma
B
Germinal epithelium
C
Vitelline membrane
D
Graafian follicle

Solution

(D) The correct answer is $D$. After ovulation,the ruptured Graafian follicle transforms into a temporary endocrine gland known as the corpus luteum.
During the menstrual cycle,the Graafian follicle protects the developing oocyte. Upon ovulation,the secondary oocyte is released,and the remaining follicular cells (zona granulosa and theca cells) undergo luteinization to form the corpus luteum.
The corpus luteum secretes large amounts of progesterone,which is essential for maintaining the endometrium during pregnancy,and also produces some estrogen.
54
BiologyMediumMCQAIPMT · 2007
In the human female,menstruation can be deferred by the administration of
A
combination of $FSH$ and $LH$
B
combination of estrogen and progesterone
C
$FSH$ only
D
$LH$ only

Solution

(B) The menstrual cycle is regulated by the levels of hormones like estrogen and progesterone.
During the cycle,the withdrawal of these hormones leads to the shedding of the uterine endometrium,which is known as menstruation.
By administering synthetic estrogen and progesterone,the hormonal levels in the blood are maintained artificially,which prevents the shedding of the endometrium and thus defers menstruation.
This is the principle behind the use of oral contraceptive pills.
Therefore,the correct option is $B$.
55
BiologyMediumMCQAIPMT · 2007
$A$ human male produces sperms with the genotypes $AB, Ab, aB,$ and $ab$ pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person?
A
$AaBB$
B
$AABb$
C
$AABB$
D
$AaBb$

Solution

(D) The production of four types of gametes $(AB, Ab, aB, ab)$ in equal proportions indicates that the individual is heterozygous for both genes.
According to the law of independent assortment,a dihybrid individual with the genotype $AaBb$ produces four types of gametes in equal frequency $(1:1:1:1)$.
Therefore,the genotype of the person must be $AaBb$.
56
BiologyMediumMCQAIPMT · 2007
In pea plants,yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant,what ratio of yellow and green seeded plants would you expect in $F_1$ generation?
A
$9 : 1$
B
$1 : 3$
C
$3 : 1$
D
$50 : 50$

Solution

(D) In pea plants,yellow seed color $(Y)$ is dominant over green seed color $(y)$.
$A$ heterozygous yellow seeded plant has the genotype $(Yy)$.
$A$ green seeded plant must have the homozygous recessive genotype $(yy)$.
When these two are crossed: $(Yy) \times (yy)$.
The gametes produced by the $(Yy)$ parent are $(Y)$ and $(y)$,while the $(yy)$ parent produces only $(y)$ gametes.
The resulting offspring genotypes are $(Yy)$ (yellow) and $(yy)$ (green) in a $1:1$ ratio.
Therefore,the ratio of yellow to green seeded plants in the $F_1$ generation is $50:50$ (or $1:1$).
57
BiologyMediumMCQAIPMT · 2007
Inheritance of skin colour in humans is an example of
A
point mutation
B
polygenic inheritance
C
codominance
D
chromosomal aberration

Solution

(B) : Polygenic (or quantitative) inheritance is a type of inheritance in which the expression of a trait is controlled by two or more genes. In this,each dominant allele contributes a unit fraction to the trait,and the total phenotypic expression is the cumulative effect of all dominant alleles of the polygenes.
Human skin colour is a classic example of polygenic inheritance,controlled by three pairs of genes: $A, B$,and $C$.
Dark skin colour (Negro) results from the presence of all six dominant alleles $(AABBCC)$.
Very light or white skin colour results from the presence of all six recessive alleles $(aabbcc)$.
58
BiologyMediumMCQAIPMT · 2007
In the hexaploid wheat, the haploid $(n)$ and basic $(x)$ numbers of chromosomes are
A
$n = 21$ and $x = 21$
B
$n = 21$ and $x = 14$
C
$n = 21$ and $x = 7$
D
$n = 7$ and $x = 21$

Solution

(C) Hexaploid wheat $(Triticum \text{ } aestivum)$ has a total chromosome number of $2n = 42$.
The haploid number $(n)$ is half of the diploid number, so $n = 42 / 2 = 21$.
The basic chromosome number $(x)$ represents the number of chromosomes in a single genome set. Since hexaploid wheat is $6x = 42$, the basic number is $x = 42 / 6 = 7$.
Therefore, $n = 21$ and $x = 7$.
59
BiologyMediumMCQAIPMT · 2007
$A$ common test to find the genotype of a hybrid is by
A
crossing of one $F_2$ progeny with female parent
B
studying the sexual behaviour of $F_1$ progenies
C
crossing of one $F_1$ progeny with recessive parent
D
crossing of one $F_2$ progeny with male parent

Solution

(C) The correct method to determine the genotype of a hybrid (or an individual showing a dominant phenotype) is a test cross.
In a test cross,an individual of unknown genotype is crossed with a homozygous recessive parent.
Therefore,crossing an $F_1$ hybrid with its recessive parent allows us to identify the genotype of the $F_1$ individual.
Option $C$ is the most accurate description of this process in the context of genetic analysis.
60
BiologyMediumMCQAIPMT · 2007
Which one of the following statements is correct?
A
There is no evidence of the existence of gills during embryogenesis of mammals.
B
All plant and animal cells are totipotent.
C
Ontogeny repeats phylogeny.
D
Stem cells are specialized cells.

Solution

(C) The correct statement is $(c)$.
Ernst Haeckel $(1866)$ proposed the Biogenetic Law, also known as the Recapitulation Theory.
This theory states that "Ontogeny repeats phylogeny," meaning the developmental history of an individual organism (ontogeny) reflects the evolutionary history of its species (phylogeny).
For example, during the embryonic development of vertebrates, embryos show features like gill slits that are characteristic of their ancestors.
61
BiologyMediumMCQAIPMT · 2007
The concept of chemical evolution is based on
A
interaction of water,air and clay under intense heat
B
effect of solar radiation on chemicals
C
possible origin of life by combination of chemicals under suitable environmental conditions
D
crystallization of chemicals

Solution

(C) The concept of chemical evolution,also known as the Oparin-Haldane hypothesis,suggests that life originated from non-living organic molecules.
These organic molecules were formed through the combination of simpler inorganic chemicals under the primitive Earth's reducing environmental conditions.
Therefore,it explains the possible origin of life by the combination of chemicals under suitable environmental conditions.
62
BiologyMediumMCQAIPMT · 2007
The finches of Galapagos islands provide an evidence in favour of
A
evolution due to mutation
B
retrogressive evolution
C
biogeographical evolution
D
special creation

Solution

(C) The finches of the Galapagos islands,famously studied by Charles Darwin,provide evidence for adaptive radiation,which is a type of biogeographical evolution. These birds evolved from a single ancestral species that migrated to the islands and subsequently diversified into different species to exploit various ecological niches. This process demonstrates how geographical isolation and environmental pressures drive evolutionary change.
63
BiologyMediumMCQAIPMT · 2007
When two species of different genealogy come to resemble each other as a result of adaptation,the phenomenon is termed
A
microevolution
B
co-evolution
C
convergent evolution
D
divergent evolution

Solution

(C) The correct answer is $C$.
Convergent evolution is the development of superficially similar structures in unrelated organisms,usually because the organisms live in the same kind of environment.
This phenomenon occurs when different groups of organisms evolve similar adaptive features to perform the same function due to similar environmental pressures.
Examples include the wings of insects,birds,and bats,which serve the same function of flight but evolved independently.
Another example is the streamlined body shape of whales (mammals) and fish,which evolved independently to facilitate movement in an aquatic environment.
64
BiologyMediumMCQAIPMT · 2007
Adaptive radiation refers to
A
evolution of different species from a common ancestor
B
migration of members of a species to different geographical areas
C
power of adaptation in an individual to a variety of environments
D
adaptations due to geographical isolation.

Solution

(A) : Adaptive radiation (divergent evolution) is the process of evolution of different species from a common ancestor in a given geographical area,starting from a point and radiating to other areas of geography.
As the original population increases in size,it spreads out from its center of origin to exploit new habitats and food sources.
In time,this results in a number of populations,each adapted to its particular habitat.
Eventually,these populations will differ from each other sufficiently to become new species.
65
BiologyMediumMCQAIPMT · 2007
If you suspect a major deficiency of antibodies in a person,to which of the following would you look for confirmatory evidence?
A
Haemocytes
B
Serum globulins
C
Fibrinogen in plasma
D
Serum albumins

Solution

(B) : Serum globulins are proteins that include gamma globulins (antibodies) and a variety of enzymes and carrier/transport proteins.
Serum protein electrophoresis $(SPEP)$ is used to analyze the profile of these globulins by separating them based on size and charge.
There are four major groups identified: alpha-$1$ globulins,alpha-$2$ globulins,beta globulins,and gamma globulins.
Since the gamma fraction primarily consists of immunoglobulins (antibodies),a deficiency in antibodies is directly reflected by a low level of serum globulins.
66
BiologyEasyMCQAIPMT · 2007
Lysozyme present in perspiration,saliva,and tears,destroys:
A
certain types of bacteria
B
all viruses
C
most virus-infected cells
D
certain fungi

Solution

(A) $Lysozyme$ is an antibacterial enzyme with natural antibiotic properties.
It is present in tears,nasal mucus,milk,and saliva in most animals and acts as part of the body's first line of innate defense.
$Lysozyme$ functions by degrading the polysaccharide protective cell wall of certain bacteria,thereby causing their lysis.
It acts as a non-specific innate opsonin by binding to the bacterial surface,which reduces the negative charge and facilitates phagocytosis by white blood cells.
Therefore,it primarily targets and destroys certain types of bacteria.
67
BiologyMediumMCQAIPMT · 2007
Compared to a bull,a bullock is docile because of:
A
higher levels of cortisone
B
lower levels of blood testosterone
C
lower levels of adrenaline/noradrenaline in its blood
D
higher levels of thyroxine

Solution

(B) : $A$ bullock is a castrated bull. Bulls are castrated to make them more meek and docile. Castration is any action,surgical,chemical,or otherwise,by which a male loses the functions of the testes. Castration involves the removal or destruction of one or both testicles,which results in sterility,decreased sexual desire,and the inhibition of secondary sex characteristics. This procedure is performed to improve the quality of meat and decrease the aggressiveness of farm animals. Since testosterone is the primary hormone responsible for aggression and sexual drive in males,its reduction through castration leads to a more docile temperament.
68
BiologyMediumMCQAIPMT · 2007
In the cloning of cattle,a fertilized egg is taken out of the mother's womb and:
A
in the eight-cell stage,cells are separated and cultured until small embryos are formed,which are then implanted into the womb of other cows.
B
in the eight-cell stage,the individual cells are separated under an electrical field for further development in culture media.
C
from this,up to eight identical twins can be produced.
D
the egg is divided into $4$ pairs of cells which are implanted into the womb of other cows.

Solution

(A) The correct process for cattle cloning via embryo transfer involves taking a fertilized egg from the donor cow. At the $8$-cell stage,the blastomeres are separated. These individual cells are then cultured in vitro until they develop into small embryos. Finally,these embryos are implanted into the wombs of surrogate cows (recipient cows) to complete their development. This technique allows for the production of multiple offspring from a single high-quality embryo.
69
BiologyEasyMCQAIPMT · 2007
Which one of the following is a viral disease of poultry?
A
Coryza
B
Newcastle disease
C
Pasteurellosis
D
Salmonellosis

Solution

(B) : Newcastle disease is a highly contagious viral disease affecting many domestic and wild avian species.
The causal agent,Newcastle disease virus $(NDV)$,is a negative-sense single-stranded $RNA$ virus.
Transmission occurs by exposure to faecal and other excretions from infected birds,and through contact with contaminated feed,water,equipment,and clothing.
Signs of infection with $NDV$ vary greatly depending on factors such as the strain of virus and the health,age,and species of the host.
Symptoms include respiratory signs (gasping,coughing),nervous signs (depression,inappetence,drooping wings,paralysis),swelling of the eyes and neck,diarrhoea,rough or thin-shelled eggs,and reduced egg production.
70
BiologyMediumMCQAIPMT · 2007
In maize,hybrid vigour is exploited by
A
crossing of two inbred parental lines
B
harvesting seeds from the most productive plants
C
inducing mutations
D
bombarding the seeds with $DNA$

Solution

(A) : Hybridisation or heterosis or hybrid vigour is defined as the superiority of the hybrid over its parents. It has been commercially exploited in different commercial crops like maize,sorghum,and bajra. The main steps include: selection of parents,selfing of parents to create inbred lines,emasculation,bagging,crossing of desired and selected inbred parents,and finally seed setting and harvesting.
71
BiologyEasyMCQAIPMT · 2007
Probiotics are
A
cancer inducing microbes
B
new kind of food allergens
C
live microbial food supplement
D
safe antibiotics

Solution

(C) : Probiotics are dietary supplements containing potentially beneficial bacteria or yeast,with lactic acid bacteria $(LAB)$ as the most common microbes used. $LAB$ have been used in the food industry for many years,because they are able to convert sugars (including lactose) and other carbohydrates into lactic acid. They not only provide the characteristic sour taste of fermented dairy foods such as yogurt,but also act as a preservative by lowering the $pH$ and creating fewer opportunities for spoilage organisms to grow.
72
BiologyMediumMCQAIPMT · 2007
Which one of the following pairs is wrongly matched?
A
Yeast - Ethanol
B
Streptomycetes - Antibiotic
C
Coliforms - Vinegar
D
Methanogens - Gobar gas

Solution

(C) : Coliform bacteria are commonly used as a bacterial indicator of the sanitary quality of foods and water. They are not involved in the production of vinegar. Vinegar is produced by the bacterium $Acetobacter$ $aceti$.
73
BiologyMediumMCQAIPMT · 2007
$A$ genetically engineered microorganism used successfully in bioremediation of oil spills is a species of
A
Trichoderma
B
Xanthomonas
C
Bacillus
D
Pseudomonas

Solution

(D) : Bioremediation is the process of using living microorganisms to clean up a contaminated site.
Microorganisms perform this by removing toxins from materials.
They decompose these compounds by using enzymes,which are specific proteins that control reactions in living cells.
Organisms that produce enzymes capable of degrading petroleum are useful in cleaning up oil spills.
Some common examples that break down oil are genetically engineered species of $Pseudomonas$ and $Azotobacter$.
Bioremediation accounts for $5$ to $10$ percent of all pollution treatment and has been used successfully in cleaning up leaking underground gasoline storage tanks.
Bioremediation has many applications,from the ordinary garden compost to the removal of selenium and other toxic metals from waste.
The best agents for bioremediation are those that can break down contaminants without becoming contaminated or harmful themselves.
74
BiologyMediumMCQAIPMT · 2007
Geometric representation of age structure is a characteristic of
A
population
B
landscape
C
ecosystem
D
biotic community

Solution

(A) The correct answer is $A$.
Population possesses several attributes that are characteristics of the group as a whole,rather than of an individual.
Age distribution is one such attribute,which refers to the number or percentage of individuals in a population across different age groups.
This distribution is represented geometrically in the form of an age pyramid,which helps in understanding the growth status of the population.
75
BiologyMediumMCQAIPMT · 2007
The population of an insect species shows an explosive increase in numbers during the rainy season,followed by its disappearance at the end of the season. What does this show?
A
The food plants mature and die at the end of the rainy season.
B
Its population growth curve is of $J$-type.
C
The population of its predators increases enormously.
D
$S$-shaped or sigmoid growth of this insect.

Solution

(B) The correct answer is $(B)$.
$J$-shaped growth pattern is observed in populations where resources are abundant for a short period,leading to exponential growth,followed by a sudden crash due to environmental resistance or seasonal changes.
In this case,the insect population increases rapidly during the rainy season and disappears when the season ends,which is a characteristic feature of $J$-type growth curves.
This pattern is common in insects,algae blooms,and annual plants.
76
BiologyMediumMCQAIPMT · 2007
$A$ high density of elephant population in an area can result in
A
intraspecific competition
B
interspecific competition
C
predation on one another
D
mutualism

Solution

(A) : Competition is rivalry for obtaining the same resource.
Competition is of two types: intraspecific and interspecific.
Intraspecific competition is the competition amongst members of the same species for a common resource,such as food,space,and mates.
Therefore,if the density of an elephant population in an area increases,it will lead to intraspecific competition.
This competition results in the establishment of territories,which may force individuals to move out to secure shelter,mates,and food for the remaining population.
77
BiologyMediumMCQAIPMT · 2007
Which one of the following pairs of organisms are exotic species introduced in India?
A
Lantana camara, water hyacinth
B
Water hyacinth, Prosopis cineraria
C
Nile perch, Ficus religiosa
D
Ficus religiosa, Lantana camara

Solution

(A) The correct answer is $A$.
In India, a large variety of exotic animal and plant species have been introduced from other parts of the world over the ages.
Some exotic plants have turned into invasive weeds, multiplying rapidly and causing significant harm to the native ecosystem.
Examples of such invasive exotic species include $Lantana$ $camara$ and water hyacinth $(Eichhornia$ $crassipes)$.
78
BiologyMediumMCQAIPMT · 2007
Identify the odd combination of the habitat and the particular animal concerned.
A
Sunderbans - Bengal Tiger
B
Periyar - Elephant
C
Rann of Kutch - Wild Ass
D
Dachigam National Park - Snow Leopard

Solution

(D) : Dachigam National Park is located $22 \ km$ from Srinagar,Jammu and Kashmir. It is famous for the Hangul or Kashmiri stag,which is the most endangered species of red deer in the world. Snow leopards are typically found in high-altitude regions like Hemis National Park,not Dachigam. Therefore,this is the odd combination.
79
BiologyEasyMCQAIPMT · 2007
One of the endangered species of Indian medicinal plants is that of
A
Ocimum
B
Garlic
C
Nepenthes
D
Podophyllum

Solution

(D) : An endangered species is a population of an organism which is at risk of becoming extinct because it is either few in number or threatened by changing environmental or predation parameters.
$Podophyllum$ is an endangered species of Indian medicinal plants.
They contain podophyllotoxin and podophyllin,which are used as a purgative and as a cytostatic agent.
They are also grown as ornamental plants for their attractive foliage and flowers.
Extracts of these plants are used for treating genital warts and some types of skin cancers.
80
BiologyMediumMCQAIPMT · 2007
In which one of the following have the $BOD$ (Biochemical Oxygen Demand) of sewage $(S)$,distillery effluent $(DE)$,paper mill effluent $(PE)$,and sugar mill effluent $(SE)$ been arranged in ascending order?
A
$SE < PE < S < DE$
B
$PE < S < SE < DE$
C
$S < DE < PE < SE$
D
$SE < S < PE < DE$

Solution

(B) The $BOD$ (Biochemical Oxygen Demand) is a measure of the organic matter present in water. Higher organic content leads to higher $BOD$.
- Paper mill effluent $(PE)$ typically has a lower $BOD$ compared to the others listed.
- Sewage $(S)$ has a moderate $BOD$.
- Sugar mill effluent $(SE)$ is rich in sugars,leading to a high $BOD$.
- Distillery effluent $(DE)$ contains high concentrations of organic waste from fermentation processes,resulting in the highest $BOD$ among these options.
Therefore,the correct ascending order is $PE < S < SE < DE$.
81
BiologyEasyMCQAIPMT · 2007
In a coal-fired power plant,electrostatic precipitators are installed to control the emission of:
A
$NO_X$
B
$SPM$
C
$CO$
D
$SO_2$

Solution

(B) The correct answer is $B$.
Electrostatic precipitators are used in coal-fired power plants to remove $SPM$ (Suspended Particulate Matter) from the exhaust gases.
$SPM$ consists of fine particles like dust,soot,and smoke that remain suspended in the air for long periods.
These devices use an electric field to charge the particles,which are then collected on grounded plates,preventing them from being released into the atmosphere.
82
BiologyMediumMCQAIPMT · 2007
Which one of the following statements is correct?
A
Both $Azotobacter$ and $Rhizobium$ fix atmospheric nitrogen in root nodules of plants.
B
Cyanobacteria such as $Anabaena$ and $Nostoc$ are important mobilizers of phosphates and for plant nutrition in soil.
C
At present it is not possible to grow maize without chemical fertilizers.
D
Extensive use of chemical fertilizers may lead to eutrophication of nearby water bodies.

Solution

(D) : Eutrophication is the excessive growth of algae, plants, and animals in water bodies due to nutrient enrichment, particularly with nitrogen and phosphate.
Nutrients present in sewage, agricultural wastes, and fertilizers cause dense growth of plants and planktonic algae.
The excessive growth of planktonic algae that causes coloration of water is called algal bloom, which is toxic to animals and humans.
Eutrophic water bodies also support the excessive growth of floating plants.
Algal blooms and floating plants cut off light from submerged plants, resulting in their death.
There is a drastic decrease in oxygen replenishment inside the water.
Non-availability of oxygen results in the death of aquatic animals such as fish, which further adds to the organic loading of water.
Decomposition is replaced by putrefaction, which is anaerobic, leading to the absence of oxygen in water and the death of aquatic animals.
83
BiologyMediumMCQAIPMT · 2007
Which one of the following is not a bioindicator of water pollution?
A
Blood-worms
B
Stone flies
C
Sewage fungus
D
Sludge worms

Solution

(B) Bioindicators are organisms whose presence,absence,or abundance provides information about the quality of the environment.
Blood-worms,sewage fungus,and sludge worms are well-known bioindicators of organic water pollution,as they thrive in polluted waters with low oxygen levels.
Stone flies,on the other hand,are highly sensitive to water pollution and require clean,well-oxygenated water to survive. Therefore,they are typically used as indicators of clean water rather than indicators of water pollution.
84
BiologyMediumMCQAIPMT · 2007
Why are bullocks more docile than bulls?
A
Higher levels of thyroxine
B
Higher levels of cortisone
C
Lower levels of testosterone in the blood
D
Lower levels of adrenaline and noradrenaline in the blood

Solution

(C) Bulls are intact male cattle that possess high levels of the hormone $Testosterone$,which is responsible for aggressive behavior and secondary sexual characteristics. Bullocks (steers) are castrated male cattle. Castration removes the primary source of $Testosterone$ production (the testes). Consequently,the significantly lower levels of $Testosterone$ in the blood of bullocks make them much more docile and easier to manage compared to bulls.
85
BiologyMediumMCQAIPMT · 2007
In human females,the menstrual cycle can be inhibited by the administration of:
A
Only $LH$
B
Combination of $FSH$ and $LH$
C
Combination of estrogen and progesterone
D
Only $FSH$

Solution

(C) The menstrual cycle is regulated by the feedback mechanism of hormones involving the hypothalamus,pituitary gland,and ovaries.
High levels of estrogen and progesterone exert a negative feedback effect on the anterior pituitary and hypothalamus,inhibiting the secretion of $FSH$ and $LH$.
Since $FSH$ and $LH$ are required for follicular development and ovulation,their inhibition prevents the progression of the menstrual cycle.
This principle is the basis for hormonal contraceptive pills,which typically contain a combination of synthetic estrogen and progesterone to prevent ovulation.
86
BiologyMediumMCQAIPMT · 2007
In maize plants,genes $R$ and $y$ are located very close to each other on the chromosome map. When a cross is made between $RRYY$ and $rryy$ genotypes,what will the $F_2$ analysis show?
A
Higher number of recombinant types
B
Segregation in the expected $9:3:3:1$ ratio
C
Segregation in a $3:1$ ratio
D
Higher number of parental types

Solution

(D) When two genes are located very close to each other on the same chromosome,they exhibit the phenomenon of linkage.
Linkage refers to the physical association of genes on a chromosome,which results in the tendency of these genes to be inherited together.
Because the genes $R$ and $y$ are closely linked,the frequency of crossing over between them is very low.
Consequently,the gametes produced will predominantly be of the parental type ($RY$ and $ry$).
In the $F_2$ generation,this results in a significantly higher proportion of offspring showing parental phenotypes compared to recombinant phenotypes,deviating from the independent assortment ratio of $9:3:3:1$.
87
BiologyMediumMCQAIPMT · 2007
Telomeric repetitive $DNA$ sequences control the function of eukaryotic chromosomes because they...
A
Act as replicons.
B
Are $RNA$ transcription initiators.
C
Help in chromosome pairing.
D
Prevent chromosome shortening.

Solution

(D) Telomeres are specialized structures at the ends of linear eukaryotic chromosomes consisting of repetitive $DNA$ sequences and associated proteins.
During $DNA$ replication,the enzyme $DNA$ polymerase cannot replicate the extreme ends of the lagging strand because it requires an $RNA$ primer to initiate synthesis.
If the ends of chromosomes consisted of essential genes,they would be lost during each round of cell division.
Telomeres act as protective caps that prevent the loss of vital genetic information by allowing the chromosome to shorten slightly without affecting coding regions,thus preventing chromosome shortening and degradation.
88
BiologyMediumMCQAIPMT · 2007
The molecular basis of cell differentiation depends on the regulation of:
A
$RNA$ polymerase
B
Ribosomes
C
Transcription factors
D
Anticodons

Solution

(C) Cell differentiation is the process by which a less specialized cell becomes a more specialized cell type. This process is primarily controlled by the differential expression of genes. The regulation of gene expression,which determines which genes are turned on or off in a specific cell type,is largely mediated by $Transcription$ $factors$. These proteins bind to specific $DNA$ sequences to regulate the transcription of genetic information from $DNA$ to messenger $RNA$ $(mRNA)$,thereby controlling the protein synthesis profile of the cell.
89
BiologyMediumMCQAIPMT · 2007
The length of a $DNA$ molecule is much greater than the diameter of the nucleus in a eukaryotic cell. How is $DNA$ packaged?
A
By removing non-essential genes
B
By super-coiling into nucleosomes
C
By digesting $DNA$ with enzymes
D
By removing repetitive $DNA$

Solution

(B) In eukaryotic cells,the $DNA$ molecule is extremely long. For example,in humans,the total length of $DNA$ is approximately $2.2 \ m$,while the diameter of the nucleus is only about $10^{-6} \ m$. To fit this long $DNA$ into the small nucleus,it undergoes a complex packaging process. The $DNA$ is negatively charged and wraps around positively charged histone protein octamers to form structures called nucleosomes. These nucleosomes further coil and condense into chromatin fibers,a process known as super-coiling,which allows the $DNA$ to be compactly organized within the nucleus.
90
BiologyMediumMCQAIPMT · 2007
The differentiation of organs and tissues in a developing organism is associated with which of the following?
A
Developmental mutations
B
Differential expression of genes
C
Harmful mutations
D
Loss of genes

Solution

(B) In a developing organism,all cells contain the same set of genomic $DNA$.
However,different cells differentiate into specific tissues and organs because they express different sets of genes.
This process is known as the differential expression of genes,which is regulated by various transcription factors and epigenetic mechanisms.
Therefore,the correct answer is $B$.
91
BiologyMediumMCQAIPMT · 2007
In $DNA$ strand replication, Okazaki fragments are:
A
Result of replication.
B
Polymerase works in $3' \to 5'$ direction and forms a replication fork.
C
Prove the semi-conservative nature of $DNA$ replication.
D
Polymerization occurs in $5' \to 3'$ direction and explains replication in $3' \to 5'$ direction.

Solution

(D) During $DNA$ replication, the two strands of the double helix are antiparallel. $DNA$ polymerase can only synthesize new $DNA$ in the $5' \to 3'$ direction.
One strand, the leading strand, is synthesized continuously.
The other strand, the lagging strand, is synthesized discontinuously in short segments known as Okazaki fragments.
These fragments are synthesized in the $5' \to 3'$ direction, which allows the overall replication of the template strand that runs in the $3' \to 5'$ direction.
92
BiologyEasyMCQAIPMT · 2007
The $1$ gene $1$ enzyme relationship was first established in .......... .
A
Neurospora crassa
B
Salmonella typhimurium
C
Escherichia coli
D
Diplococcus pneumoniae

Solution

(A) The $1$ gene $1$ enzyme hypothesis was proposed by George Beadle and Edward Tatum in $1941$.
They conducted experiments on the bread mold $Neurospora$ $crassa$.
By inducing mutations using $X$-rays, they observed that specific mutations led to the loss of specific enzymatic activities, thereby proving that each gene is responsible for the synthesis of a specific enzyme.
93
BiologyMediumMCQAIPMT · 2007
Human gene expression is regulated when steroid molecules bind to:
A
Transfer $RNA$
B
Messenger $RNA$
C
Specific $DNA$ sequences
D
Ribosomes

Solution

(C) Steroid hormones are lipophilic molecules that can easily cross the plasma membrane of the cell.
Once inside the cytoplasm or nucleus,they bind to specific intracellular receptor proteins.
This hormone-receptor complex then acts as a transcription factor by binding to specific $DNA$ sequences (hormone response elements) in the promoter region of target genes.
This binding regulates the transcription of these genes,thereby controlling gene expression.
94
BiologyMediumMCQAIPMT · 2007
In $DNA$,the two polynucleotide chains are . . . . . . .
A
Parallel
B
Discontinuous
C
Antiparallel
D
Semi-conservative

Solution

(C) The $DNA$ double helix model,proposed by $Watson$ and $Crick$,states that the two polynucleotide chains run in opposite directions.
One strand runs in the $5' \rightarrow 3'$ direction,while the other runs in the $3' \rightarrow 5'$ direction.
This orientation is referred to as antiparallel.
95
BiologyMediumMCQAIPMT · 2007
Which of the following is a result of geographical isolation?
A
No changes in the animals of the separated region
B
Prevention of speciation
C
Speciation through reproductive isolation
D
Random production of new species

Solution

(C) Geographical isolation refers to the physical separation of populations of a species by barriers such as mountains,rivers,or oceans.
When populations are geographically isolated,they cannot interbreed.
Over time,due to different environmental pressures and genetic drift,these populations accumulate distinct genetic variations.
Eventually,these genetic differences become significant enough that even if the populations were to come into contact again,they would be unable to interbreed.
This process leads to reproductive isolation,which is a key mechanism in the formation of new species,known as speciation.
96
BiologyMediumMCQAIPMT · 2007
Which of the following human ancestors had a brain capacity of more than $1000 \ cc$?
A
Homo neanderthalensis
B
Homo erectus
C
Ramapithecus
D
Homo habilis

Solution

(A) The brain capacity of various human ancestors is as follows:
$1$. $Homo \ habilis$: $650-800 \ cc$.
$2$. $Homo \ erectus$: $800-1100 \ cc$.
$3$. $Homo \ neanderthalensis$: $1400 \ cc$.
$4$. $Ramapithecus$: Had a smaller brain capacity compared to the genus $Homo$.
Since $Homo \ neanderthalensis$ has a brain capacity of $1400 \ cc$,which is greater than $1000 \ cc$,it is the correct answer.
97
BiologyMediumMCQAIPMT · 2007
The phenomenon of industrial melanism in the peppered moth $(Biston \text{ } betularia)$ demonstrates that:
A
Dark melanin color is produced due to random mutations.
B
Dark-colored moths have a selective advantage over light-colored moths in industrial areas.
C
Light-colored moths have no selective advantage in either polluted or non-polluted areas.
D
Darkness is a trait induced by pollution.

Solution

(B) Industrial melanism is a classic example of natural selection. Before industrialization in England, the light-colored form of the peppered moth was more abundant as it could camouflage against lichen-covered tree trunks. With the onset of industrialization, soot covered the trees, making the dark-colored (melanic) form better camouflaged against predators. This provided the dark-colored moths with a selective advantage in polluted industrial areas, leading to an increase in their population frequency. Thus, it demonstrates that natural selection favors traits that provide a survival advantage in a changing environment.
98
BiologyMediumMCQAIPMT · 2007
Select the correct statement from the given options.
A
Darwinian variations are small and directionless.
B
Fitness is the ability to adapt and be selected by nature.
C
All mammals except whales and camels have seven cervical vertebrae.
D
Mutations are random and directional.

Solution

(B) The correct statement is that fitness is the ability to adapt and be selected by nature.
According to Darwin,fitness is the ultimate measure of an organism's success in an environment,defined by its ability to survive and reproduce,which leads to natural selection.
Option $A$ is incorrect because Darwinian variations are small and directional.
Option $C$ is incorrect because almost all mammals,including whales and camels,have seven cervical vertebrae.
Option $D$ is incorrect because mutations are random and directionless.
99
BiologyMediumMCQAIPMT · 2007
The finches of Galapagos islands provide evidence in favor of:
A
Speciation
B
Evolution due to mutation
C
Retrogressive evolution
D
Biogeographical evolution

Solution

(A) The finches of the Galapagos islands,famously studied by Charles Darwin,are a classic example of adaptive radiation.
Adaptive radiation is a process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats).
This process leads to the formation of new species from a common ancestor,which is known as speciation.
Therefore,the Galapagos finches provide evidence in favor of speciation.
100
BiologyMediumMCQAIPMT · 2007
When two species with different genetic backgrounds show similarities due to adaptation,this phenomenon is called:
A
Divergent evolution
B
Microevolution
C
Co-evolution
D
Convergent evolution

Solution

(D) When two or more unrelated or distantly related species evolve similar traits as a result of having to adapt to similar environments or ecological niches,this process is known as $Convergent \ evolution$.
This leads to the development of analogous organs,which perform similar functions but have different evolutionary origins.
For example,the wings of butterflies and birds,or the flippers of dolphins and penguins,show convergent evolution.

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