AIPMT 1991 Biology Question Paper with Answer and Solution

(A) Resolving power is defined as the ability of an optical instrument,such as a microscope,to distinguish between two closely placed points or objects as separate entities. It determines the clarity and detail of the image produced.

(C) The growth of the cell wall occurs through two main processes: Apposition and Intussusception.
Apposition refers to the deposition of new wall material on the surface of the existing wall.
Intussusception refers to the process where new wall materials (such as cellulose microfibrils and matrix substances) are deposited within the existing cell wall structure,allowing it to expand.
Therefore,the addition of new molecules to the pre-existing cell wall is known as Intussusception.

(D) In a healthy individual,the glomerular filtrate contains substances like water,glucose,amino acids,and salts.
As the filtrate passes through the renal tubules,essential substances are reabsorbed back into the blood.
Glucose is a high-threshold substance that is completely reabsorbed from the proximal convoluted tubule $(PCT)$ under normal physiological conditions.
Therefore,it is not normally found in the urine of a healthy person.

(B) Ammonia is highly toxic and requires a large amount of water for its elimination. Tadpoles are aquatic and live in water,so they excrete nitrogenous waste as ammonia (ammonotelic). Adult frogs are semi-terrestrial and have developed kidneys to conserve water,so they excrete nitrogenous waste primarily as urea (ureotelic).

(B) $ADH$ (Antidiuretic Hormone),also known as Vasopressin,is synthesized by the hypothalamus and released by the posterior pituitary gland (neurohypophysis).
Its primary function is to act on the distal convoluted tubules and collecting ducts of the nephrons in the kidneys.
It increases the permeability of these tubules to water,thereby promoting the reabsorption of water from the glomerular filtrate back into the blood.
This process helps in reducing the loss of water through urine (diuresis) and maintaining the body's water balance.

(C) nucleotide is the basic structural unit of nucleic acids like $DNA$ and $RNA$.
It consists of three main components:
$1$. $A$ nitrogenous base (either a purine or a pyrimidine).
$2$. $A$ pentose sugar (deoxyribose in $DNA$ or ribose in $RNA$).
$3$. $A$ phosphate group.
Therefore,the correct composition of a nucleotide is a nitrogenous base,a sugar,and a phosphate group.

(D) Cytochromes are iron-containing hemoproteins that act as electron carriers in the electron transport chain. The central metal ion in the heme group of cytochromes is $Fe$ (Iron). While copper $(Cu)$ is present in cytochrome c oxidase (Complex $IV$),the fundamental prosthetic group of all cytochromes is the iron-porphyrin complex. Therefore,$Fe$ is the primary mineral associated with cytochromes.

(D) The basic structural unit of a nucleic acid (such as $DNA$ or $RNA$) is the $Nucleotide$.
Each $Nucleotide$ consists of three components: a nitrogenous base,a pentose sugar (ribose or deoxyribose),and a phosphate group.
While a $Nucleoside$ consists only of a nitrogenous base and a pentose sugar,the addition of a phosphate group converts it into a $Nucleotide$,which acts as the monomeric building block for nucleic acid polymers.

(B) Isoenzymes (or isozymes) are enzymes that differ in amino acid sequence and molecular structure but catalyze the same chemical reaction.
They often have different kinetic properties (such as different $K_m$ values) or are regulated differently.
For example,lactate dehydrogenase $(LDH)$ exists in different isoenzymic forms in various tissues.

(D) In $metaphase$, each chromosome consists of two sister chromatids attached at the centromere. Therefore, if a cell has $2n$ chromosomes, it contains $4n$ chromatids.
In $anaphase$, the centromeres split, and the sister chromatids separate. These separated chromatids are now considered individual chromosomes.
As a result, the number of chromosomes doubles (becomes $4n$) while the number of chromatids per chromosome becomes $1$, effectively halving the total number of chromatids per chromosome compared to the metaphase state, or more accurately, the chromatids are now independent chromosomes.

(A) Meiosis is a specialized type of cell division that reduces the chromosome number by half,resulting in the production of haploid daughter cells.
$1$. Segregation: During $Anaphase-I$,homologous chromosomes separate,ensuring that each daughter cell receives only one allele for each gene.
$2$. Independent Assortment: During $Metaphase-I$,the random alignment of homologous chromosome pairs leads to different combinations of chromosomes in the gametes.
$3$. Crossing Over: During $Prophase-I$,the exchange of genetic material between non-sister chromatids of homologous chromosomes creates new combinations of alleles.
These three processes collectively ensure that the daughter cells are genetically distinct from the parent cell and from each other.

(A) Water potential $(\Psi_w)$ is a measure of the potential energy of water in a system compared to pure water.
According to the standard plant physiology definition,the water potential of a cell is the sum of the solute potential $(\Psi_s)$ and the pressure potential $(\Psi_p)$.
Therefore,the formula is $\Psi_w = \Psi_s + \Psi_p$.

(A) Plants absorb water from the soil primarily through their roots.
Soil water exists in various forms:
$1$. Gravitational water: This water drains away deep into the soil due to gravity and is generally unavailable to plants.
$2$. Hygroscopic water: This water is held tightly by soil particles and is not available to plants.
$3$. Chemically bound water: This water is part of the chemical structure of soil minerals and is unavailable.
$4$. Capillary water: This water is held in the capillary spaces between soil particles against the force of gravity. It is the form of water that is readily available for absorption by plant roots.
Therefore,the correct answer is $A$.

(C) The Calvin cycle was discovered by Melvin Calvin and his colleagues.
They used radioactive $C^{14}$ isotope in algal photosynthesis studies to trace the path of carbon in the dark reaction of photosynthesis.
This technique is known as the radioactive isotope technique or carbon labeling.

(B) In the eastern states of India,the monsoon season is characterized by heavy cloud cover.
Photosynthesis is a light-dependent process.
During the monsoon,the intensity and duration of sunlight reaching the plants are significantly reduced due to persistent cloud cover.
Since light is a limiting factor for photosynthesis,the reduced availability of light leads to a decrease in the rate of photosynthesis,which ultimately results in lower rice production.

(A) Ferredoxin is an iron-sulfur protein that acts as an electron carrier in the photosynthetic electron transport chain.
In the light-dependent reactions of photosynthesis,electrons are excited in $PS-I$ (Photosystem $I$) and are transferred to ferredoxin.
Therefore,ferredoxin is functionally and structurally associated with the $PS-I$ complex.

(A) Photosynthetic pigments such as chlorophyll $a$,chlorophyll $b$,xanthophylls,and carotenoids are embedded within the thylakoid membranes of the chloroplast.
These pigments are organized into photosystems ($PS-I$ and $PS-II$) which are essential for capturing light energy during the light-dependent reactions of photosynthesis.

(B) The process of photosynthesis is divided into two main phases: the light-dependent reaction and the light-independent reaction (also known as the dark reaction or Calvin cycle).
$1$. The light-dependent reaction occurs in the thylakoid membranes of the chloroplast,where $ATP$ and $NADPH$ are produced.
$2$. The dark reaction (Calvin cycle) occurs in the stroma of the chloroplast,where the $ATP$ and $NADPH$ produced in the light reaction are used to fix $CO_2$ into carbohydrates.
Therefore,the correct location for the dark reaction is the stroma.

(B) The complete oxidation of one glucose molecule yields a net gain of $38 \,ATP$ in prokaryotes (or under specific shuttle conditions in eukaryotes).
$1$. Glycolysis produces $2 \,ATP$ directly and $2 \,NADH$.
$2$. The link reaction produces $2 \,NADH$.
$3$. The Krebs cycle produces $2 \,GTP$ (equivalent to $ATP$),$6 \,NADH$,and $2 \,FADH_2$.
$4$. In the Electron Transport System $(ETS)$,each $NADH$ produces $3 \,ATP$ and each $FADH_2$ produces $2 \,ATP$.
$5$. Total $ATP$ calculation: $2 \,ATP$ (glycolysis) + $2 \,GTP$ (Krebs) + $10 \,NADH \times 3 = 30 \,ATP$ + $2 \,FADH_2 \times 2 = 4 \,ATP$.
$6$. Thus,$2 + 2 + 30 + 4 = 38 \,ATP$.
$7$. Specifically,$4 \,ATP$ are produced via substrate-level phosphorylation (outside mitochondria in glycolysis and inside in Krebs),and $34 \,ATP$ are produced via oxidative phosphorylation in the mitochondria.

(C) Cytokinins are a class of plant growth substances (phytohormones) that promote cell division,or cytokinesis,in plant roots and shoots. They were discovered by Miller et al. $(1955)$ and the first natural cytokinin identified was zeatin,isolated from corn kernels. While auxins like $IAA$ and $NAA$ promote cell enlargement and gibberellins promote stem elongation,cytokinins are specifically responsible for stimulating cell division.

(A) Apical dominance is the phenomenon where the main central stem of the plant grows more dominantly than the side (lateral) stems.
This process is primarily controlled by the hormone $Auxin$,specifically $Indole-3-Acetic Acid$ $(IAA)$.
$IAA$ is synthesized in the apical meristem and transported downwards,where it inhibits the growth of lateral buds.
Therefore,the correct option is $A$.

(A) Abscisic Acid $(ABA)$ is a plant growth inhibitor.
Its primary physiological role is to induce stomatal closure in response to water stress,thereby reducing transpiration and conserving water.
It also promotes seed dormancy and abscission of leaves and fruits.

(C) Florigen is a hypothetical hormone or signaling molecule that is believed to be responsible for inducing flowering in plants. It is synthesized in the leaves in response to photoperiodic stimuli and then transported to the shoot apical meristem to initiate the transition from vegetative growth to reproductive growth (flowering).

(A) Protein digestion begins in the stomach,where the enzyme $pepsin$ acts on proteins to break them down into proteoses and peptones in the presence of $HCl$.
While digestion continues in the duodenum due to pancreatic enzymes like $trypsin$,$chymotrypsin$,and $carboxypeptidase$,the initial and significant site for protein digestion is the stomach.

(B) The pancreas is a heterocrine (mixed) gland that functions as both an exocrine and an endocrine gland.
As an exocrine gland,it secretes pancreatic juice containing three main types of digestive enzymes: $1.$ Trypsinogen (protease),$2.$ Pancreatic amylase (carbohydrase),and $3.$ Pancreatic lipase (lipase).
As an endocrine gland,it secretes two primary hormones from the Islets of Langerhans: $1.$ Insulin and $2.$ Glucagon.
Therefore,the pancreas produces three digestive enzymes and two hormones.

(C) Carbonic anhydrase is a zinc-containing enzyme that catalyzes the reversible reaction: $CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$. This enzyme is present in very high concentrations within the $R.B.C.$ (Red Blood Cells) and in smaller amounts in the blood plasma. However,the primary site of its action for rapid transport of $CO_2$ is the $R.B.C.$

(C) The wall of blood capillaries is extremely thin and is composed of a single layer of squamous epithelial cells,which are known as $Endothelial$ $cells$. These cells rest on a basement membrane. This structure allows for the efficient exchange of gases,nutrients,and waste products between the blood and the surrounding tissues.

(C) In the $ABO$ blood grouping system,the presence of antigens on the surface of red blood cells determines the blood group.
$1$. Individuals with blood group $A$ have antigen $A$ and antibody $b$ in their plasma.
$2$. Individuals with blood group $B$ have antigen $B$ and antibody $a$ in their plasma.
$3$. Individuals with blood group $AB$ have both antigens $A$ and $B$ on their red blood cells but do not have any antibodies ($a$ or $b$) in their plasma.
$4$. Individuals with blood group $O$ do not have any antigens on their red blood cells but have both antibodies $a$ and $b$ in their plasma.
Therefore,the blood group with both antigens present and no antibodies is $AB$.

(C) Deep black soil (also known as Regur soil) is highly productive because it is rich in clay and humus.
Clay particles have a high water-holding capacity,which is essential for plant growth.
Humus provides essential nutrients and improves the soil structure,making it fertile for agriculture.

(D) Plants absorb water from the soil primarily through their root hairs.
Soil water exists in various forms:
$1$. Gravitational water: This water drains away due to gravity and is not available to plants.
$2$. Hygroscopic water: This water is held tightly by soil particles and is unavailable to plants.
$3$. Capillary water: This water is held in the capillary spaces between soil particles by surface tension. It is the form of water that is readily available for absorption by plant roots.
Therefore,the correct answer is capillary water.

(D) Multiple allelism refers to the existence of more than two alleles for a single gene locus within a population.
In humans,the $ABO$ blood grouping system is a classic example of multiple allelism.
The gene $I$ controls the $ABO$ blood grouping,which has three alleles: $I^A$,$I^B$,and $i$.
Since there are three alleles,there are more than two possible phenotypes and genotypes in the population.

(A) Let the allele for brown eyes be $B$ (dominant) and the allele for blue eyes be $b$ (recessive).
Since the man has brown eyes but his mother was blue-eyed $(bb)$,he must have inherited one $b$ allele from her. Thus,his genotype is $Bb$.
The woman is blue-eyed,so her genotype must be $bb$.
When we cross the man $(Bb)$ with the woman $(bb)$:
$Bb \times bb \rightarrow Bb, Bb, bb, bb$.
This results in $50\%$ brown-eyed $(Bb)$ and $50\%$ blue-eyed $(bb)$ children,which is a ratio of $1:1$.

(A) Color blindness is an $X$-linked recessive trait. For a girl to be color-blind,she must inherit the recessive allele $(X^c)$ from both parents. This means her father must be color-blind $(X^cY)$ and her mother must be either a carrier $(X^CX^c)$ or color-blind $(X^cX^c)$. Since the question implies the rarity of the condition,the most specific scenario is that the father is color-blind and the mother is at least a carrier. However,looking at the provided options,the condition where a girl is color-blind requires the father to be color-blind and the mother to carry the gene. Option $A$ is the most accurate description of the genetic requirement.

(D) dihybrid condition refers to an organism that is heterozygous for two different traits or genes.
In the given options,$Tt\, Rr$ represents an individual that is heterozygous for both the gene controlling height $(T/t)$ and the gene controlling seed shape $(R/r)$.
Therefore,$Tt\, Rr$ is the correct representation of a dihybrid genotype.

(C) Mendel proposed three main laws of inheritance based on his experiments with pea plants:
$1$. Law of Dominance: States that in a heterozygote,one allele masks the presence of another.
$2$. Law of Segregation: States that alleles separate during gamete formation.
$3$. Law of Independent Assortment: This is the third and final law proposed by Mendel. It states that the alleles of two (or more) different genes get sorted into gametes independently of one another. This law is observed in dihybrid crosses.

(B) Gregor Johann Mendel is known as the father of genetics. He performed hybridization experiments on garden peas ($Pisum$ $sativum$) for seven years $(1856-1863)$ and proposed the laws of inheritance in living organisms.

(B) In genetics,$Mendel$ referred to the contrasting forms of a character as 'contrasting traits'. These traits are controlled by different forms of a gene,which are known as alleles (or allelomorphs). Therefore,a pair of contrasting traits is represented by a pair of alleles.

(D) In genetics,a factor (or allele) that fails to express its phenotype in the presence of a dominant allele is known as a $Recessive$ factor. According to Mendel's Law of Dominance,in a heterozygous condition,the allele that expresses itself is called the $Dominant$ allele,while the one that remains masked is called the $Recessive$ allele.

(B) In $Antirrhinum$ (snapdragon),the inheritance of flower color shows incomplete dominance.
When a homozygous red-flowered plant $(RR)$ is crossed with a homozygous white-flowered plant $(WW)$,the $F_1$ generation produces pink-flowered plants $(RW)$.
This occurs because the dominant allele $(R)$ is not completely dominant over the recessive allele $(W)$,resulting in an intermediate phenotype (pink) in the heterozygote.

(B) The process of copying genetic information from one strand of $DNA$ into $RNA$ is known as $Transcription$.
In this process,only a segment of $DNA$ and only one of the two strands is copied into $RNA$.
$Replication$ is the process of duplicating $DNA$.
$Translation$ is the process of polymerizing amino acids to form a polypeptide based on the sequence of $mRNA$.
$Translocation$ refers to the movement of the ribosome along the $mRNA$ during protein synthesis.

(C) According to Chargaff's rule for $DNA$,the amount of adenine $(A)$ is equal to the amount of thymine $(T)$,and the amount of cytosine $(C)$ is equal to the amount of guanine $(G)$.
Given: $A = 120$,therefore $T = 120$.
Given: $C = 120$,therefore $G = 120$.
The total number of nitrogenous bases is $A + T + C + G = 120 + 120 + 120 + 120 = 480$.
Since each nitrogenous base is attached to a sugar and a phosphate group to form a nucleotide,the total number of nucleotides in the $DNA$ fragment is equal to the total number of nitrogenous bases,which is $480$.

(D) In $DNA$,the four nitrogenous bases are Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
In $RNA$,Thymine is replaced by Uracil $(U)$.
Therefore,the base pairing in $RNA$ involves Adenine,Guanine,Cytosine,and Uracil.

(D) $DNA$ $(Deoxyribonucleic acid)$ is a polymer of $Deoxyribonucleotides$.
Each $Deoxyribonucleotide$ consists of three components: a nitrogenous base, a pentose sugar $(Deoxyribose)$, and a phosphate group.
These units link together via phosphodiester bonds to form the long polynucleotide chains that constitute the $DNA$ double helix.

(A) The study of fossils is known as $Paleontology$. It involves the examination of plant and animal remains preserved in the Earth's crust to understand the history of life on Earth. $Herpetology$ is the study of amphibians and reptiles, $Saurology$ is the study of lizards, and $Organic \text{ } Evolution$ refers to the process of gradual change in organisms over time.

(C) According to the Oparin-Haldane hypothesis,the early atmosphere of the Earth was a reducing atmosphere. It contained gases like $CH_4$,$NH_3$,$H_2$,and water vapor,but it lacked free molecular oxygen $(O_2)$. Oxygen was produced later through the process of photosynthesis by cyanobacteria.

(C) The geological time scale is divided into eras representing major stages in Earth's history.
$1$. The Archeozoic and Proterozoic are part of the Precambrian eon.
$2$. The Phanerozoic eon is divided into three main eras: Paleozoic,Mesozoic,and Cenozoic.
$3$. The Paleozoic era (ancient life) occurred first,followed by the Mesozoic era (middle life),and finally the Cenozoic era (recent life).
Therefore,the correct chronological sequence is Paleozoic $\to$ Mesozoic $\to$ Cenozoic.

(A) Mutualism is a type of population interaction in which both interacting species benefit from each other. In many cases,this relationship is obligate,meaning the organisms are essential for each other's survival (e.g.,lichens,mycorrhizae). Commensalism is an interaction where one species benefits and the other is neither harmed nor benefited. Therefore,the correct term for a mutually beneficial interaction essential for survival is Mutualism.

(C) food chain represents the sequence of transfer of energy from one trophic level to another.
In option $A$,the sequence is incorrect because a chameleon (carnivore) cannot be eaten by an insect.
In option $B$,the sequence is incorrect because a fox (carnivore) does not eat grass.
In option $C$,the sequence is $Phytoplankton$ (producer) $\to$ $Zooplankton$ (primary consumer) $\to$ $Fish$ (secondary consumer). This represents a correct aquatic food chain.
In option $D$,the sequence is incorrect as it describes a detritus food chain,but the flow from bacteria to insect larvae is not the standard representation of energy transfer in this context.

(B) The food chain that begins with dead organic matter and involves microorganisms (decomposers) breaking down this matter is called the $Detritus$ $food$ $chain$ $(DFC)$.
In this chain,the primary source of energy is dead organic matter (detritus) rather than living producers.
Microorganisms like bacteria and fungi act as decomposers,breaking down the complex organic compounds into simpler inorganic substances.
Therefore,the correct option is $B$.

(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.

(C) The greenhouse effect is a natural phenomenon that warms the Earth's surface. When the Sun's energy reaches the Earth's atmosphere,some of it is reflected back to space and the rest is absorbed and re-radiated by greenhouse gases. The primary greenhouse gases include $CO_2$,methane,and water vapor. An increase in the concentration of $CO_2$ in the atmosphere traps more heat,leading to global warming. Therefore,the greenhouse effect is considered a warning due to the rise in global temperatures caused by the accumulation of these gases.

(B) Household waste primarily consists of organic materials such as food scraps,vegetable peels,and paper,which can be broken down by microorganisms. Therefore,it is classified as biodegradable pollution.