AIIMS 1980 Biology Question Paper with Answer and Solution

(C) In the family $Solanaceae$,the ovary is bicarpellary and syncarpous.
Due to the shifting of the axis,the carpels are placed obliquely at an angle of about $45^o$ to the median plane.
This oblique placement occurs because the posterior carpel turns to the right and the anterior carpel turns to the left.

(C) The genetic code (triplet code) is universal,meaning it is common to all living organisms.
Energy storage via high-energy phosphate bonds (like $ATP$) is a fundamental mechanism in all life forms.
Ribosomes are universally present as the site of protein synthesis in all cells.
However,the specific types of proteins present in the body vary significantly between different species and even between individuals,as they are determined by the specific genetic makeup and evolutionary history of the organism. Therefore,this is not a common characteristic.

(B) In succulent plants ($CAM$ plants),the stomata open at night to take in $CO_2$.
During the night,carbohydrates are incompletely oxidized to form organic acids,primarily malic acid,through the process of dark fixation of $CO_2$.
This accumulation of organic acids increases the osmotic concentration of the guard cells,leading to the influx of water and the opening of stomata.
During the day,these organic acids are broken down to release $CO_2$ for photosynthesis,and the stomata close to prevent water loss.

(B) In plants like $Lotus$ (Nelumbo),the leaves are $epistomatic$,meaning stomata are present only on the upper surface of the leaves.
Coating the upper surface with wax will block these stomata,thereby inhibiting gaseous exchange ($CO_2$ uptake for photosynthesis and $O_2$ release) and stomatal transpiration.
Since $Hydrilla$,$Pistia$,and $Vallisneria$ are aquatic plants with different stomatal distributions or adaptations (often $hypostomatic$ or submerged),the impact of waxing the upper surface is most critical in $Lotus$ where the upper surface is the primary site for gas exchange.

(B) $Mg$ is a central constituent of the chlorophyll molecule,forming the core of the porphyrin ring,which is essential for its structure and function.
$Fe$ (Iron) is not a structural component of chlorophyll,but it is absolutely essential for the biosynthesis of chlorophyll molecules as it acts as a cofactor for enzymes involved in the pathway.
Therefore,both $Fe$ and $Mg$ are required for the synthesis and maintenance of chlorophyll in plants.

(D) The correct answer is $(d)$.
Necrosis,or the death of plant tissues,is a common symptom caused by the deficiency of mineral elements such as $P, K, Ca$ and $Mg$.
This condition typically manifests as the appearance of dead necrotic areas on leaves,often starting at the tips or margins.
Necrosis usually occurs following chlorosis (the loss of chlorophyll) and is a clear indicator of nutrient stress in the plant.

(B) Chlorosis is the loss of chlorophyll leading to the yellowing of leaves,which is a primary symptom of nitrogen deficiency.
Nitrogen is a mobile element within the plant.
When nitrogen is deficient,the plant translocates nitrogen from older tissues to younger,actively growing tissues to support their development.
Consequently,the deficiency symptoms,such as yellowing,first manifest in the older leaves before spreading to the younger ones.

(C) The synthesis of hexose sugars (glucose) occurs during the biosynthetic phase (Calvin cycle) of photosynthesis,which takes place in the stroma of the chloroplast.
Thylakoids are primarily responsible for the light-dependent reactions (photophosphorylation),where $ATP$ and $NADPH$ are produced.
The enzymes required for the Calvin cycle (such as $RuBisCO$) are located in the stroma,not within the thylakoids.
When thylakoids are removed and placed in a medium with $CO_2$ and $H_2O$,the light-dependent reactions may occur,but the enzymes necessary for carbon fixation and sugar synthesis are absent in the medium. Therefore,hexose sugars are not formed.

(C) The most appropriate reason for storing green apples at low temperatures is that it slows down the metabolic activities of the fruit.
Specifically,the rate of respiration is significantly reduced at low temperatures.
Since respiration is a catabolic process that consumes stored sugars (like galactose) to provide energy for ripening and senescence,reducing this rate helps the fruit remain fresh for a longer duration.

(D) Enzymes,vitamins,and hormones are all essential biological molecules that play a critical role in the regulation of metabolic processes within living organisms.
Enzymes act as biological catalysts that speed up biochemical reactions.
Hormones act as chemical messengers that coordinate various physiological activities.
Vitamins act as co-factors or co-enzymes necessary for the proper functioning of enzymes.
Therefore,they are collectively involved in the regulation of metabolism.

(D) The glomerular filtrate is formed by the process of ultrafiltration in the glomerulus.
Due to the large molecular size of plasma proteins,they cannot pass through the filtration slits of the glomerular membrane.
Therefore,the glomerular filtrate is essentially plasma minus proteins.
Thus,the correct option is $D$.

(C) Adrenaline (epinephrine) is a hormone secreted by the adrenal medulla. It acts on the heart by binding to $\beta_1$-adrenergic receptors. This binding increases the heart rate (positive chronotropic effect) and the force of contraction (positive inotropic effect). When a physiological concentration is injected,it leads to a sustained increase in the heart rate to support the body's 'fight-or-flight' response.

(C) The amount of water retained by a unit weight of undisturbed soil after complete drainage under gravity is known as $Field \ capacity$.
When soil is thoroughly wetted and allowed to drain,the water that moves downward due to gravity is called $Gravitational \ water$.
Once the downward movement stops,the water held in the soil pores against gravity by capillary forces is known as $Capillary \ water$.
The total water content remaining in the soil at this point is defined as the $Field \ capacity$ of the soil.

(B) The condition where the testes fail to descend into the scrotum during development is known as $Cryptorchidism$. Since the temperature of the scrotum is lower than the body temperature,which is essential for spermatogenesis,the failure of descent leads to sterility in the individual.

(D) The dwarf pea plant is genetically homozygous recessive $(tt)$.
Gibberellic acid treatment only affects the phenotype (height) by promoting stem elongation but does not change the genotype $(tt)$.
When this treated plant $(tt)$ is crossed with a pure tall plant $(TT)$,
the cross is $TT \times tt$.
The resulting offspring in the $F_1$ generation will all have the genotype $Tt$.
According to Mendel's Law of Dominance,the $T$ allele is dominant over the $t$ allele.
Therefore,all offspring will exhibit the tall phenotype.

(B) The frequency of crossing over is directly proportional to the physical distance between two genes on a chromosome.
When two genes are situated very close to each other,the probability of a crossover event occurring between them is extremely low.
Therefore,hardly any crossovers are detected between such closely linked genes.
This phenomenon is known as tight linkage.

(C) The original sequence is $AAC\ GAC\ AGC\ GGC\ ACA\ AAA$.
If the first base $(A)$ is deleted,the sequence becomes $ACG\ ACG\ AGC\ GGC\ ACA\ AA$.
This is a frameshift mutation. However,in the context of this specific question,if we consider the deletion of the first base of the first triplet,the reading frame shifts entirely.
Option $C$ is the most scientifically accurate description of a frameshift mutation,as the deletion of a single base changes every subsequent triplet codon,leading to a completely different sequence of amino acids in the resulting polypeptide chain.

(D) $DNA$ replication is semiconservative,meaning each original strand serves as a template for a new strand.
Initially,we have $1$ $DNA$ molecule with $2$ radioactive strands $(R-R)$.
After the $1^{st}$ duplication: $2$ molecules are formed,each with $1$ radioactive and $1$ non-radioactive strand ($R-N$ and $R-N$).
After the $2^{nd}$ duplication: $4$ molecules are formed. The $2$ radioactive strands from the previous step are distributed into $2$ separate molecules,while the other $2$ molecules consist entirely of non-radioactive strands $(R-N, N-N, N-N, R-N)$.
After the $3^{rd}$ duplication: $8$ molecules are formed. The $2$ radioactive strands remain in $2$ separate molecules,while the remaining $6$ molecules are entirely non-radioactive.
Therefore,only $2$ $DNA$ molecules will contain radioactive thymidine.

(B) The process of copying genetic information from one strand of the $DNA$ into $RNA$ is termed as transcription. This process occurs in the nucleus where $DNA$ serves as a template for the synthesis of $mRNA$.

(B) The frequency of crossing over between two genes is directly proportional to the physical distance between them on the chromosome. When two genes are situated very close to each other,the probability of a crossover event occurring in the small segment between them is extremely low. Therefore,they exhibit strong linkage,and hardly any crossing over is detected between them.