AIPMT 1990 Biology Question Paper with Answer and Solution

(A) $Vibrio$ $cholerae$ is a curved,rod-shaped bacterium that possesses a single polar flagellum.
Bacteria with a single flagellum at one end are classified as $Monotrichous$.
Therefore,$Vibrio$ $cholerae$ belongs to the $Monotrichous$ group.

(D) The $III^{rd}$ cranial nerve is the oculomotor nerve.
It is responsible for controlling most of the extrinsic eye muscles,which manage ocular movements.
It also carries parasympathetic fibers that control the constriction of the pupil (sphincter pupillae) and the ciliary muscles responsible for accommodation.
Therefore,damage to this nerve results in the loss of ocular movements,dilation of the pupil (due to loss of parasympathetic control),and loss of accommodation.

(B) The kingdom $Monera$ consists of prokaryotic organisms, primarily bacteria.
$Escherichia$ (e.g., $Escherichia coli$) is a well-known bacterium belonging to the kingdom $Monera$.
$Amoeba$ is a protist (kingdom $Protista$).
$Gelidium$ is a red alga (kingdom $Protista$).
$Spirogyra$ is a green alga (kingdom $Plantae$).
Therefore, the correct option is $B$.

(A) The primary difference between Gram-positive and Gram-negative bacteria is the composition and structure of their cell wall.
Gram-positive bacteria possess a thick layer of peptidoglycan (murein) in their cell wall,which retains the crystal violet stain.
Gram-negative bacteria have a much thinner layer of peptidoglycan and an additional outer membrane containing lipopolysaccharides $(LPS)$,which prevents the retention of the crystal violet stain.

(A) $Trypanosoma$ is a genus of kinetoplastids,a monophyletic group of unicellular parasitic flagellate protozoa.
It is known to be polymorphic,meaning it exhibits different morphological forms during its life cycle (e.g.,trypomastigote,epimastigote,amastigote).
It is digenetic,requiring two hosts (an insect vector and a vertebrate host) to complete its life cycle.
It is an obligate parasite,meaning it cannot complete its life cycle without a host.
It is highly pathogenic,causing diseases such as African sleeping sickness and Chagas disease.
Therefore,the most distinct characteristic among the given options is that it is polymorphic.

(C) The malaria parasite,$Plasmodium$,is a member of the kingdom $Protista$.
Within the group of protozoan protists,$Plasmodium$ is classified under the class $Sporozoa$ (also known as $Apicomplexa$).
These organisms are characterized by having an infectious spore-like stage in their life cycle.
Therefore,the correct option is $C$.

(A) $Paramecium$ exhibits nuclear dimorphism,meaning it possesses two types of nuclei: the micronucleus and the macronucleus.
The micronucleus is diploid and contains the germline genetic information,which is responsible for reproduction and genetic recombination during conjugation.
The macronucleus is polyploid and controls the metabolic activities and vegetative functions of the cell.
Therefore,the primary genetic information for inheritance is stored in the micronucleus.

(B) Absorptive heterotrophic nutrition is a characteristic feature of $Fungi$.
$Fungi$ are heterotrophic organisms that obtain their nutrients by absorbing dissolved organic compounds from their environment.
They secrete digestive enzymes into their surroundings,which break down complex organic matter into simpler substances,which are then absorbed through their cell walls.
In contrast,$Algae$,$Bryophytes$,and $Pteridophytes$ are primarily autotrophic,as they contain chlorophyll and perform photosynthesis.

(B) The life cycle of mosses (Bryophyta) includes a juvenile stage known as the protonema.
$1$. The protonema is a creeping,green,branched,and frequently filamentous stage.
$2$. It develops directly from a spore.
$3$. Among the given options,$Funaria$ is a moss,whereas $Riccia$ is a liverwort,and $Chlamydomonas$ and $Spirogyra$ are algae.
$4$. Therefore,the protonema is characteristic of the life cycle of $Funaria$.

(A) The peristome is a specialized structure found in the capsule of mosses (Bryophyta).
It consists of a ring of tooth-like appendages surrounding the opening of the capsule.
These teeth are hygroscopic,meaning they respond to changes in humidity.
When the air is dry,the peristome teeth bend outward,allowing the spores inside the capsule to be released and dispersed by the wind.
Therefore,the primary function of the peristome is the regulation of spore dispersal.

(A) The capsule of $Funaria$ is differentiated into three distinct regions: the $apophysis$,the $theca$,and the $operculum$.
$1$. The $apophysis$ is the sterile,basal (lower) part of the capsule that connects it to the $seta$.
$2$. The $theca$ is the middle,fertile part containing the $spore$ sac.
$3$. The $operculum$ is the upper,lid-like part of the capsule.
Therefore,the $apophysis$ is located at the lower part of the capsule.

(A) The life cycle of $Taenia$ $solium$ (pork tapeworm) involves two hosts. The primary host is human,and the secondary (intermediate) host is the pig.
In the human intestine,the eggs are released and passed out with feces. When the pig (secondary host) ingests these eggs,the embryos (oncospheres) hatch and migrate to the muscles.
In the muscles of the pig,the oncosphere develops into a bladder-like larva known as $Cysticercus$ $cellulosae$ (also called bladder worm).
Therefore,the transmission of $Taenia$ to the secondary host occurs in the form of the oncosphere,which then develops into the $Cysticercus$ stage.

(A) Malpighian tubules are the primary excretory organs found in insects and many other terrestrial arthropods.
They function by absorbing waste products from the hemolymph and discharging them into the digestive tract for excretion.
Therefore,the correct option is $A$.

(D) The pancreas is a heterocrine (mixed) gland,meaning it has both exocrine and endocrine functions.
$1$. The exocrine part consists of acini cells,which secrete pancreatic juice containing digestive enzymes.
$2$. The endocrine part consists of the Islets of Langerhans,which contain different types of cells (such as alpha and beta cells) that secrete hormones like glucagon and insulin.
Therefore,these secretions are produced by different types of cells.

(B) Ecdysis,also known as moulting,is the process of shedding the old exoskeleton or cuticle in arthropods and some other invertebrates.
In these organisms,the outer layer of the body,which is the epidermis (specifically the cuticle secreted by the epidermis),must be shed to allow for growth.
Therefore,ecdysis is the shedding of the epidermis or its derivatives (cuticle).

(D) In amphibians like the $Frog$ $(Rana \, tigrina)$, the skin is moist and highly vascularized, which allows for cutaneous respiration. This means they can exchange gases directly through their skin, especially when they are underwater or in moist environments. Other options like lizards, birds, and mammals primarily use lungs for respiration.

(B) $Pheretima$ $posthuma$ (earthworm) is known as the 'friend of farmers'.
They make the soil porous by burrowing,which helps in the penetration of air and water into the soil.
Additionally,they enrich the soil by mixing their nitrogenous excretory products (vermicompost) and decaying organic matter into the soil,which increases soil fertility.

(D) The blood of $Pheretima$ (earthworm) is red in color.
This red color is due to the presence of hemoglobin dissolved in the plasma.
In earthworms,red blood cells $(RBCs)$ are absent,and hemoglobin is found freely dissolved in the plasma.
Therefore,the correct option is $D$.

(D) Vivipary is a phenomenon where seeds germinate while still attached to the parent plant.
This is a specialized adaptation found in plants growing in saline,marshy,or swampy environments,known as halophytes (e.g.,Rhizophora).
In these environments,the high salt concentration and lack of oxygen in the soil make it difficult for seeds to germinate if they fall directly into the mud.
Therefore,the embryo grows out of the seed coat and fruit while attached to the parent tree,allowing the seedling to establish itself quickly once it drops into the soft mud.

(B) The $Asteraceae$ family (also known as the $Compositae$ family) is distinctively characterized by the $Capitulum$ or $Head$ type of inflorescence. In this type,the main axis becomes flattened or convex,and many small,sessile flowers called $florets$ are arranged on it. These are often surrounded by an involucre of $bracts$.

(B) The Casparian strips are characteristic features of the endodermis in the roots of vascular plants.
These are bands of suberin-impregnated cell wall material that prevent the apoplastic movement of water and solutes into the vascular cylinder.
Therefore,the correct option is $B$.

(B) The vascular cambium is a lateral meristem that is responsible for secondary growth in dicotyledonous plants.
During secondary growth,the cells of the vascular cambium divide periclinally.
The cells produced towards the inner side differentiate into secondary xylem,while the cells produced towards the outer side differentiate into secondary phloem.
Therefore,the vascular cambium produces secondary xylem and secondary phloem.

(B) In dicotyledonous roots,the pericycle is the layer of cells located inside the endodermis.
It is responsible for the initiation of lateral roots and part of the vascular cambium during secondary growth.
Therefore,the pericycle gives rise to lateral roots.

(C) In monocot leaves,the mesophyll is not differentiated into palisade and spongy parenchyma. Instead,it consists of undifferentiated parenchyma cells. Additionally,monocot leaves often possess bulliform cells (motor cells) in the upper epidermis,which help in leaf rolling during water stress. Reticulate venation is a characteristic of dicot leaves,while monocot leaves typically show parallel venation.

(A) Monocot leaves,specifically the leaves of grasses,possess $Intercalary$ $meristem$.
This type of meristem is located between mature tissues and is responsible for the growth in length of the leaves and internodes.
It allows the plant to regenerate parts that have been removed by grazing animals.

(C) The intrafascicular cambium and interfascicular cambium are responsible for secondary growth in plants.
Together,they form a complete ring known as the 'cambium ring'.
Since these tissues increase the thickness or girth of plant organs,they are classified as 'lateral meristems'.

(C) Collenchyma is a type of simple permanent tissue that provides mechanical support to the growing parts of the plant,such as young stems and petioles of leaves.
It is characteristically absent in monocot stems and roots.
It is primarily found in the hypodermis of dicot stems and the petioles of dicot leaves.
Therefore,collenchyma is found in the stem and petiole of dicot plants.

(A) Smooth muscle tissue,also known as involuntary or non-striated muscle,consists of cells that are fusiform (spindle-shaped) in structure.
These cells are unbranched and possess a single central nucleus (uninucleate).
They lack the striations (light and dark bands) found in skeletal or cardiac muscles.
Their contraction is controlled by the autonomic nervous system,making them involuntary in nature.
Therefore,the correct characteristics are fusiform,unbranched,non-striated,uninucleate,and involuntary.

(C) Joachim Hammerling performed experiments on the unicellular green alga $Acetabularia$ to demonstrate the role of the nucleus in heredity.
He used two species, $Acetabularia$ $crenulata$ and $Acetabularia$ $mediterranea$.
In his experiments, he performed grafting by exchanging the $stalk$ (which contains the cytoplasm) and the $rhizoid$ (which contains the nucleus) between the two species.
He observed that the characteristics of the cap that regenerated were determined by the species from which the $rhizoid$ (nucleus) was taken, not the $stalk$.
Therefore, the parts exchanged in his experiments were the $rhizoids$ and the $stalks$.

(B) The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illumination $(Resolving Power \propto 1/\lambda)$.
Visible light has a relatively long wavelength,limiting the resolution of light microscopes.
In contrast,electron beams behave like waves with extremely short wavelengths (much smaller than visible light).
Because the wavelength of an electron beam is significantly shorter,the electron microscope can resolve much smaller details,resulting in a much higher resolving power.

(A) The total magnification $(M)$ of a compound microscope is given by the formula: $M = (L/f_o) \times (D/f_e)$,where $L$ is the tube length,$f_o$ is the focal length of the objective lens,$f_e$ is the focal length of the eyepiece,and $D$ is the least distance of distinct vision.
Numerical aperture $(NA)$ is a measure of the light-gathering ability and resolving power of the lens,defined as $NA = n \sin(\theta)$.
While magnification depends on the focal lengths of the lenses and the tube length,the numerical aperture is primarily related to the resolution of the microscope,not the magnification itself.

(C) The $Proximal \text{ Convoluted Tubule } (PCT)$ of the nephron is lined by simple cuboidal brush border epithelium.
This brush border consists of numerous microvilli that significantly increase the surface area for the reabsorption of water, electrolytes, and nutrients from the filtrate.
Therefore, the brush border is a characteristic feature of the $PCT$.

(B) The nephron is the structural and functional unit of the kidney.
It consists of a renal corpuscle (glomerulus and Bowman's capsule) and a renal tubule.
The renal tubule is further divided into the proximal convoluted tubule $(PCT)$,the loop of Henle,and the distal convoluted tubule $(DCT)$.
Therefore,both the proximal and distal convoluted tubules are integral parts of the nephron.

(C) In almost all mammals,the number of cervical vertebrae is $7$. This is a characteristic feature of the class $Mammalia$. Whether it is a camel,a rabbit,a horse,or a whale,they all possess $7$ cervical vertebrae. Therefore,the number of cervical vertebrae in a camel is equal to that of a rabbit,a horse,and a whale. Among the given options,the statement that the number is the same as that of a whale is correct.

(D) The deltoid tuberosity is a rough,triangular area on the lateral surface of the shaft of the $Humerus$ bone.
It serves as the insertion point for the deltoid muscle,which is responsible for the abduction of the arm at the shoulder joint.

(D) The parasympathetic nervous system is responsible for 'rest and digest' activities.
It promotes functions that conserve energy.
Constriction of the pupil (miosis) is a characteristic parasympathetic response.
In contrast,contraction of hair muscles (piloerection),stimulation of sweat glands,and increase in heart rate are functions associated with the sympathetic nervous system.

(D) The brain contains several interconnected cavities known as ventricles,which are filled with cerebrospinal fluid $(CSF)$.
$(1)$ The first and second ventricles are the lateral ventricles,located in the cerebral hemispheres,and are called paracoels.
$(2)$ The third ventricle is located in the diencephalon and is known as the diocoel.
$(3)$ The fourth ventricle is located in the hindbrain (medulla oblongata) and is known as the metacoel.
$(4)$ Rhinocoele refers to the cavities within the olfactory lobes.
Therefore,the third ventricle is the diocoel.

(B) The hormone $Thyroxine$ is essential for the metamorphosis of tadpoles into adult frogs. The presence of iodine in the water is crucial because it is a key component required for the synthesis of $Thyroxine$ by the thyroid gland. If iodine or $Thyroxine$ is added to the water,it accelerates the process of metamorphosis,causing the tadpoles to transform into adult frogs more rapidly.

(C) Insulin is a peptide hormone produced by the beta cells of the Islets of Langerhans in the pancreas. It plays a crucial role in regulating blood glucose levels by facilitating the uptake of glucose into cells.

(C) The root system in plants is highly plastic and responds to the availability of nutrients in the soil. When there is a deficiency of mineral salts (nutrients) in the soil,the plant directs more resources toward root growth to explore a larger volume of soil for nutrient uptake. This is a physiological adaptation to ensure survival under nutrient-limited conditions.

(B) In angiosperms,the xylem is the principal water-conducting tissue.
It consists of tracheids and vessels,which are the main elements responsible for the long-distance transport of water and minerals from roots to other parts of the plant.
Sieve tubes are part of the phloem,which is responsible for the translocation of food (organic solutes).
Since both tracheids and vessels are involved in water conduction in angiosperms,the most accurate answer is that water conduction occurs through these xylem elements. However,in the context of multiple-choice questions where 'all of the above' is provided,it is important to note that sieve tubes do not conduct water. Therefore,the question implies the primary xylem elements. Given the options,'Vessels' are the most specialized and efficient water-conducting structures in angiosperms.

(D) The process being demonstrated is $Imbibition$.
$Imbibition$ is a special type of diffusion where water is absorbed by solids (colloids),causing them to increase in volume.
When dry or soaked seeds are placed in a confined space like a glass jar and allowed to absorb water,they swell significantly.
This swelling exerts a tremendous amount of pressure,known as $Imbibition$ pressure.
In this scenario,the pressure exerted by the swelling mustard seeds exceeded the structural strength of the glass jar,leading to its breakage.

(A) In $C_4$ plants,the primary $CO_2$ acceptor is a $3$-carbon molecule called Phosphoenolpyruvate $(PEP)$.
This reaction is catalyzed by the enzyme $PEP$ carboxylase $(PEPCase)$ in the mesophyll cells.
The $CO_2$ combines with $PEP$ to form a $4$-carbon compound,Oxaloacetic acid $(OAA)$,which is the first stable product of the $C_4$ cycle.

(A) Kranz anatomy is a specialized structure found in the leaves of $C_4$ plants (such as maize,sugarcane,and sorghum).
In this anatomy,the mesophyll cells are arranged in a ring-like manner around the bundle sheath cells.
This structural arrangement facilitates the $C_4$ photosynthetic pathway,which helps in minimizing photorespiration and increasing the efficiency of carbon fixation in high-temperature and high-light conditions.

(B) The $EMP$ pathway,also known as Glycolysis,involves the breakdown of one molecule of glucose into two molecules of pyruvic acid.
During this process,$4$ molecules of $ATP$ are produced by substrate-level phosphorylation,and $2$ molecules of $ATP$ are consumed in the preparatory phase.
Additionally,$2$ molecules of $NADH + H^+$ are produced.
In aerobic respiration,each $NADH$ molecule yields $3\, ATP$ via the electron transport chain (in most eukaryotic cells),contributing $6\, ATP$.
However,the question specifically asks for the net gain from the $EMP$ pathway itself. The net gain of $ATP$ directly produced in glycolysis is $2\, ATP$ ($4$ produced - $2$ consumed).
Given the options provided,$8\, ATP$ is the standard answer representing the total energy yield ($2\, ATP$ net + $6\, ATP$ from $2\, NADH$) produced during the conversion of glucose to pyruvate in aerobic conditions.

(B) Glycolysis is the process of the breakdown of glucose into two molecules of pyruvic acid.
It occurs in the cytoplasm of the cell and does not require oxygen.
The overall reaction is: $Glucose + 2NAD^+ + 2ADP + 2Pi \rightarrow 2 \text{Pyruvic acid} + 2NADH + 2H^+ + 2ATP$.
Therefore, the end product of glycolysis is pyruvic acid.

(D) Glycolysis converts glucose into pyruvate,which is then converted into Acetyl $CoA$ by the link reaction.
$
$Fatty acids undergo $\beta$-oxidation to produce Acetyl $CoA$.
$
$Acetyl $CoA$ acts as the common substrate that enters the Krebs cycle ($TCA$ cycle) for further oxidation.
$
$Therefore,Acetyl $CoA$ serves as the metabolic bridge or link between carbohydrate metabolism (glycolysis) and fat metabolism ($\beta$-oxidation) and the Krebs cycle.

(B) Abscisic acid $(ABA)$ is a plant growth inhibitor.
It plays a crucial role in the abscission of leaves and fruits.
It also induces dormancy in seeds and buds,which helps plants survive unfavorable environmental conditions.
Therefore,$ABA$ is primarily associated with the regulation of leaf abscission and dormancy.

(A) $Auxin$ is a plant hormone that primarily promotes cell elongation and division. It is synthesized in the shoot and root apices (growing tips) of plants. Because it is produced in these regions to facilitate growth,the highest concentration of $Auxin$ is consistently found in the growing apices of the plant.

(D) Plant growth regulators (PGRs) are small,simple molecules of diverse chemical composition. They are synthesized by plants in very small amounts and are responsible for regulating various physiological processes such as growth,development,and responses to environmental stimuli. Therefore,they are defined as chemical substances synthesized by the plant that regulate its physiological processes.

(C) The gonads (testes in males and ovaries in females) are derived from the intermediate mesoderm during embryonic development.
Specifically,they develop from the gonadal ridges,which are thickenings of the mesothelium and the underlying mesenchyme of the mesoderm.

(B) Down syndrome is a chromosomal disorder caused by the presence of an additional copy of chromosome number $21$ (trisomy of $21$).
This condition affects the autosomes,not the sex chromosomes.
Therefore,the sex chromosome complement in a male child with Down syndrome remains normal,which is $XY$.

(A) Colorblindness is an $X$-linked recessive disorder.
Let $X^C$ be the allele for colorblindness and $X$ be the allele for normal vision.
The husband has normal vision,so his genotype is $XY$.
The wife has normal vision,but since her father was colorblind,she must be a carrier,so her genotype is $XX^C$.
When we cross $XY \times XX^C$,the possible genotypes for their children are:
$1. XX$ (Normal daughter)
$2. XX^C$ (Carrier daughter)
$3. XY$ (Normal son)
$4. X^CY$ (Colorblind son)
Since colorblindness is $X$-linked recessive,a daughter can only be colorblind if she inherits the $X^C$ allele from both parents. In this case,the father is normal $(XY)$,so he cannot pass the $X^C$ allele to his daughter.
Therefore,the probability of their daughter being colorblind is $0\%$.

(D) Hemophilia is a sex-linked recessive disorder.
It is caused by a mutation in the genes located on the $X$ chromosome.
Since males have only one $X$ chromosome $(XY)$,a single recessive gene is sufficient to express the disease.
In contrast,females have two $X$ chromosomes $(XX)$,so they require two copies of the recessive gene to express the disease,making them less frequently affected than males.

(D) Phenylketonuria $(PKU)$ is an inborn error of metabolism that is inherited as an autosomal recessive trait.
It is caused by a mutation in the gene that codes for the enzyme phenylalanine hydroxylase,which leads to the accumulation of phenylalanine in the body.
Cataract,leprosy,and night blindness are generally not classified as genetic disorders in the context of Mendelian inheritance; leprosy is an infectious disease,while cataract and night blindness can be caused by environmental factors,aging,or nutritional deficiencies.

(D) The genetic code consists of $64$ codons,which are triplet sequences of nucleotides $(4^3 = 64)$.
Out of these $64$ codons,$61$ codons code for $20$ amino acids,while $3$ codons $(UAA, UAG, UGA)$ are stop codons (nonsense codons) that do not code for any amino acid.
Therefore,the statement that the genetic code is triplet is correct.

(A) According to Chargaff's rules for $DNA$,the amount of adenine $(A)$ is equal to thymine $(T)$,and the amount of cytosine $(C)$ is equal to guanine $(G)$. Thus,$A = T$ and $C = G$.
Option $(a)$ states $A = T, C = A$,which is incorrect because $C$ should equal $G$,not $A$.
Option $(b)$ is true because heating $DNA$ causes denaturation,which increases the volume and decreases the density.
Option $(c)$ is true because the ratio $(A + T) / (C + G)$ is species-specific and not a universal constant.
Since the question asks for what is $NOT$ true,and option $(a)$ contains an incorrect statement,it is the correct choice.

(A) The fundamental principle of embryonic development,often referred to as Von Baer's laws of embryology,was proposed by Karl Ernst von Baer.
These laws state that general characteristics of a group of animals appear in the embryo earlier than the specialized characteristics.
Therefore,the correct option is $A$.

(D) Parallel evolution (Parallelism) occurs when two independent species evolve together at the same time in the same environment and acquire similar characteristics. It is typically observed in closely related species that have diverged from a common ancestor but continue to evolve similar traits due to similar selective pressures. Therefore,it represents a form of divergent evolution occurring in closely related species.

(B) Analgesics are a class of drugs specifically designed to provide relief from pain. They act by blocking pain signals to the brain or by modifying the body's perception of pain. Therefore,the primary function of analgesic drugs is to relieve pain.

(C) In a pond ecosystem,the pyramid of numbers is upright.
$1$. At the base,there are a large number of producers (phytoplankton).
$2$. These are followed by a smaller number of primary consumers (zooplankton).
$3$. Then,there are even fewer secondary consumers (small fish).
$4$. Finally,there are very few tertiary consumers (large fish) at the top.
Thus,the number of organisms decreases as we move to higher trophic levels,resulting in an upright pyramid.

(D) Jet planes release gaseous emissions at high altitudes. The major aerosol pollutant released by jet planes is $CFCs$ (Chlorofluorocarbons). These $CFCs$ are released in the stratosphere,where they break down to release chlorine atoms,which subsequently deplete the ozone layer.