CSIR NET Life Sciences Mock Test - 3 - UGC NET MCQ

Calculation:

Let's number the given ball.

Moving ball 5 below the ball, in line with ball 2 along with moving ball 1 below ball 6, in line with ball 3.

We get

The minimum number of moves required is 2.

Concept:-

1. Basic concept of counting.

2. Concept of 'AND', when two things have to happen, we multiply the number of ways both can happen individually.

Calculation:-

Number of ways to take a train from Gwalior to Bhopal = 18

Number of ways to reach Raisen from Bhopal = 3

Since both cases have to happen, concept of 'AND' will come in this.

Total number of ways = 18 × 3 = 54

Given

Principal (P) = ₹20,000

Rate of interest (R) = 25%

Annual part payment = ₹2,500

Formula Used

Amount after n years = P(1 + R/100)ⁿ - Part payment using the structure of compound interest

Calculation

For the first year:

⇒ Amount after 1 year = 20,000 × (1 + 25/100)¹ - 2,500

⇒ A1 = 20,000 × 1.25 - 2,500

⇒ A1 = 25,000 - 2,500

⇒ A1 = 22,500

For the second year ⇒ P2 = 22,500

⇒ Amount after 2nd year = 22,500 × (1 + 25/100)¹ - 2,500

⇒ A2 = 22,500 × 1.25 - 2,500

⇒ A2 = 28,125 - 2,500

⇒ A2 = 25,625

The amount owed after two installments is ₹25,625.

The correct answer is A lag phase will be observed between the two exponential phases.

Concept:

• When E. coli is cultured in a medium containing both a preferred carbon source (glucose) and a non-preferred carbon source (lactose), the bacteria will initially consume the preferred carbon source (glucose).
• This results in a biphasic growth curve, known as diauxic growth.
• During the first exponential phase, E. coli grows rapidly utilizing glucose.
• Once glucose is depleted, a lag phase occurs where the bacteria adjust their metabolism to utilize lactose as the carbon source.
• Following this lag phase, a second exponential phase occurs as the bacteria begin to grow using lactose.

Explanation:

• Option 1: Growth curve will be the same as when grown in the presence of only glucose. This is incorrect because the growth curve with both glucose and lactose will show a lag phase after glucose depletion, which does not occur when only glucose is present.
• Option 2: Growth curve will be the same as when grown in the presence of only lactose. This is incorrect because the presence of glucose will initially suppress the utilization of lactose, resulting in a different growth pattern.
• Option 3: A lag phase will be observed between the two exponential phases. This is correct because after glucose is exhausted, E. coli will experience a lag phase while adjusting to metabolize lactose.
• Option 4: Two lag phases will be observed between the two exponential phases. This is incorrect because there is typically only one lag phase observed between the consumption of glucose and the utilization of lactose.

During vertebrate limb development, the regulation of apoptosis is crucial for the proper formation of digits. Apoptosis, or programmed cell death, allows for the removal of cells to shape developing structures, such as separating the fingers and toes. The Bcl-2 family of proteins plays a critical role in this process. Members of the Bcl-2 family include both pro-apoptotic and anti-apoptotic proteins that tightly regulate the apoptotic pathways, ensuring that cell death occurs in a controlled manner during digit formation.

1. Bcl-2 family: The Bcl-2 protein family is directly involved in the intrinsic (mitochondrial) pathway of apoptosis, managing the balance between cell survival and cell death through interactions between its pro-apoptotic and anti-apoptotic members.

   • During limb development: The controlled apoptosis mediated by the Bcl-2 family ensures proper digit formation by selectively removing the cells in the inter-digital regions.

2. Caspase family: Caspases are enzymes that execute the apoptosis process once it is initiated, playing a downstream role in the pathway.

   • During limb development: Though crucial in executing cell death, caspases are not the primary regulators in the context of specific tissue patterning like digit separation.

3. TGF-beta family: The TGF-beta family is involved in regulating cell growth, differentiation, and embryonic development.

   • During limb development: While important in various developmental processes, TGF-beta is not primarily recognized for regulating apoptosis in the context of digit formation.

4. Wnt family: Wnt proteins are crucial for regulating cell-to-cell interactions during embryogenesis.

   • During limb development: They play a key role in the regulation of limb patterning and outgrowth, but not specifically in apoptosis for digit formation.

The correct answer is Option 4 i.e.Because water molecules have a high affinity for ions, reducing the solvent availability for proteins

In the salting-out process, the high affinity of water molecules for the added salt ions results in the ions preferentially interacting with water. This reduces the number of water molecules available to solvate the proteins, leading to decreased protein solubility and eventual precipitation.

Explanation:

The process of salting out is a crucial principle in biochemistry, particularly in the field of protein purification and characterization. It involves the addition of high concentrations of salt to a solution, leading to the selective precipitation of proteins. The core mechanism behind this phenomenon is the interaction between the added salt ions, water molecules, and the proteins present in the solution.

High Affinity of Water for Ions:-

• Water is a polar solvent, composed of molecules with a slight positive charge on the hydrogen atoms and a slight negative charge on the oxygen atom. This polar nature enables water to form hydration shells around ions, which are dispersed when salts are dissolved in water. The negative end of water molecules is attracted to cations (+), while the positive end is attracted to anions (-), leading to the formation of a structured layer of water molecules around each ion. This interaction is energetically favorable and results in the stabilization of the ions in solution.

Reduction in Solvent Availability:-

• As salt concentration increases, more and more water molecules are sequestered away to form hydration shells around the ions, significantly reducing the number of free water molecules available to solvate other solutes, such as proteins. Proteins in solution are surrounded by a layer of structured water molecules that stabilize their conformation and keep them soluble. This structured water layer is crucial for maintaining the delicate balance of forces that govern protein solubility, including hydrophobic interactions, hydrogen bonding, and electrostatic interactions.

Impact on Protein Solubility:-

• When the availability of water molecules to form hydration shells around proteins decreases, the solubility of the proteins decreases accordingly. The proteins cannot maintain their solvation layer, leading to an increased propensity for intermolecular interactions. As a result, proteins begin to aggregate, eventually precipitating out of the solution. The phenomenon of salting out capitalizes on the principle that different proteins will precipitate at different salt concentrations, allowing for the selective purification of specific proteins from a complex mixture.

Conclusion:

In summary, salting out is driven by the preferential interaction between water molecules and salt ions, which reduces the availability of water for protein solvation. This leads to decreased protein solubility and precipitation, a process that is widely utilized in biochemistry for protein purification. This selective precipitation is not just a matter of removing proteins from solution but a fine method to isolate specific proteins based on their unique solubility characteristics under various salt concentrations.

Concept:

• Numerous sorts of cell movements are carried out by actin filaments, frequently in conjunction with myosin.
• The first molecular motor, myosin, is a protein that transforms chemical energy in the form of ATP into mechanical energy to produce force and movement.

Explanation:

• Muscle contraction is the most notable example of such movement, and it serves as a model for studying actin-myosin interactions and the motor activity of myosin molecules.
• Actin and myosin interactions, however, play a crucial function in cell biology as they are responsible for a variety of movements of non-muscle cells, including cell division, in addition to muscle contraction.
• Furthermore, actin-myosin interactions and actin polymerization appear to be the primary drivers of the crawling movements of cells across a surface, which are regulated by the actin cytoskeleton.​

Fig 1: General structure of myosin

HMM: Heavy meromyosin
LMM: Light meromyosin
S1 refers to the motor domain plus the lever arm, which binds the essential and regulatory light chains (ELC and RLC respectively)
​Hence correct answer is option 1

Double circulation is a key feature of the cardiovascular systems in mammals and birds, which ensures more efficient separation and management of oxygenated and deoxygenated blood. It comprises two circuits:

• Pulmonary Circulation: Blood travels from the heart to the lungs, where it is oxygenated, and then back to the heart. This circuit ensures that blood is oxygen-rich before it is pumped to the rest of the body.
• Systemic Circulation: Oxygenated blood is pumped from the heart to all organs and tissues of the body, delivers oxygen and nutrients, and then returns deoxygenated blood back to the heart.

The cardiovascular organization called double circulation provides vigorous flow of blood to the brain, muscles, and other organs.

• This assertion remains true. Double circulation in mammals ensures efficient oxygenation and nutrient delivery to various organs, including the brain and muscles, by separating oxygenated and deoxygenated blood streams and ensuring the blood is re-pressurized before being sent to the systemic circulation.

The blood is pumped a second time after it loses pressure in the capillary beds of the lungs or skin.

• This reason is also true. After blood goes through the capillaries of the lungs where it is oxygenated, it returns to the heart where it is pumped again at high pressure to the rest of the body. This re-pressurization is crucial for maintaining the high flow rate needed for effective systemic circulation. The inclusion of "skin" may be irrelevant or misleading in the context of discussing double circulation but does not negate the truth of the re-pressurization in the lungs and its impact on blood flow. Skin does not play a role in the re-oxygenation or re-pressurization of blood. However, capillaries do reach the skin and all other organs, where they deliver oxygen and nutrients.

Therefore, Both [a] and [r] are true and [r] is the correct reason for [a]

Concept:

• Botanical nomenclature is the system of rules and procedures for assigning names to plants.
• This is important in the field of botany, as it provides a standardized way of referring to different plant species, which is necessary for accurate communication and research.
• The rules of botanical nomenclature are governed by the International Code of Nomenclature for algae, fungi, and plants (ICN).
• This is a set of guidelines that have been developed and updated over time by an international group of experts in plant taxonomy.
• One of the key principles of botanical nomenclature is that each plant species should have only one accepted name.
• This helps to avoid confusion and ensure that different researchers are referring to the same species.
• The ICN provides rules for the formation and spelling of plant names, as well as guidelines for the priority of names and how to handle cases where multiple names have been proposed for the same species.
• Another important aspect of botanical nomenclature is the way in which plant names are cited in scientific literature.
• When a new plant species is described, it is given a name that is usually composed of a genus name and a specific epithet.
• If the species is later moved to a different genus by a subsequent author, the original author's name is included in parentheses after the new combination.
• These rules help to ensure that plant names are accurately and consistently cited in scientific literature.

Explanation:

Option 1: x et y

• When a name is published by two authors, their names must be linked by "et" or "&".
• Example - Didymopanax gleasonii Briton et Wilson (i.e., x et y format).
• Therefore, this option would be incorrect.

Option 2: x ex y

• When a name is suggested by an author but is published validly by another author, the latter author name is connected to the first author name by "ex".
• Example - Acalypha racemosa Wall. ex Baill (x ex y format).
• Therefore, this option would be incorrect.

Option 3: (x) y

• This option would be the correct answer.
• According to the rules of botanical nomenclature, when a species is reclassified by a subsequent author, the new combination should be cited with the original author's name in parentheses after the new combination.
• Example - Citrus grandis (L.) Osbeck.
• This option correctly follows this rule.

Option 4: (y) x

• This option would be incorrect.
• This option cites the original author's name in parentheses after the new combination, but this is not the correct order.
• The correct order is to cite the original author's name first, followed by the new author's name.

Actin's Role in the Cytoskeleton
Actin is one of the most abundant proteins in eukaryotic cells and plays a crucial role in various cellular processes, including muscle contraction, cell motility, cell division, and the maintenance of cell shape and structure. It exists in two forms: globular actin (G-actin), which polymerizes to form filamentous actin (F-actin) that makes up actin filaments, one of the three main components of the cytoskeleton—the others being microtubules and intermediate filaments.

1) α actins are found in various types of muscles

• Alpha (α) actins are specific isoforms of actin found in muscle cells, where they are essential components of the contractile apparatus. They are involved in the formation of the thin filament, interacting with myosin (a motor protein) in muscle fibers to facilitate muscle contraction—a process that is fundamental to all types of muscle tissue, including skeletal, cardiac, and smooth muscles. This specificity underlines actin's role in muscle function.

2) Polymerization of pure actin in vitro requires GTP

• This statement is incorrect and misunderstands the nucleotide requirement for actin polymerization.
• Actin monomers bind ATP before adding to the growing filament. Upon or shortly after the incorporation into the filament, the ATP is hydrolyzed to ADP.
• This ATP hydrolysis is crucial for actin dynamics, allowing actin filaments to undergo rapid assembly and disassembly- a process central to actin's function in cell motility and shape changes.
• GTP, on the other hand, is involved in the polymerization of tubulin, another cytoskeletal protein, into microtubules, not actin filaments.

3) Actin filaments have a slow-growing minus end and a fast-growing plus end

• Actin filaments display structural polarity with distinct ends: the slow-growing (or "pointed") minus end and the fast-growing (or "barbed") plus end.
• This polarity is essential for directional growth and is exploited by cells in processes such as cell motility, where the assembly of actin at the leading edge (plus end) of a cell pushes the membrane forward.
• The differential growth rates at these ends allow actin filaments to dynamically reorganize in response to cellular signals, enabling various actin-based structures and movements.

4) Cytochalasins are the inhibitors of actin polymerization

• Cytochalasins are a class of compounds that inhibit actin polymerization by binding to the fast-growing plus end of actin filaments, preventing the addition of new monomers.
• This inhibition effectively stalls the dynamic restructuring of the actin cytoskeleton, making cytochalasins useful tools in research to dissect the roles of actin in cellular processes.
• Their action demonstrates the critical nature of actin dynamics for cell function and the potential for pharmacological intervention in these processes.

Conclusion:-
The incorrect statement related to GTP's role in actin polymerization highlights the importance of ATP in actin dynamics, distinguishing it from other cytoskeletal elements like microtubules that do rely on GTP for polymerization. Therefore, the correct answer is Option 2.

Osmotic pressure and hydrostatic pressure influence fluid movement in the body, the key concept is the balance between these two forces within the capillary beds. This balance determines the direction of fluid movement between the capillaries and the interstitial space (the space between cells).

Osmotic pressure is predominantly generated by plasma proteins (notably albumin) and pulls water into the bloodstream from the surrounding tissue. During protein starvation, the concentration of plasma proteins can decrease, leading to a reduction in the osmotic pressure exerted by the blood.

Hydrostatic pressure is the pressure exerted by the fluid (blood) on the walls of the blood vessels. This pressure tends to push water out of the capillaries and into the interstitial space.

When the osmotic pressure on the venous side of capillary beds drops below the hydrostatic pressure (due to protein starvation, for example), the normal balance that maintains fluid levels in the tissues and blood vessels is disrupted. This imbalance means that the force pushing fluid out of the capillaries (hydrostatic pressure) is greater than the force pulling fluid back into the capillaries (osmotic pressure).

The correct answer to the question is fluids will tend to accumulate in tissues.

• This is because the decrease in osmotic pressure (as a result of protein starvation) compared to hydrostatic pressure will result in more fluid being pushed out of the capillaries and less being pulled back in, leading to fluid accumulation in the interstitial spaces, which can manifest as edema (swelling due to fluid accumulation in tissues).

The other options are not directly relevant to the change in osmotic and hydrostatic pressures described:

a. Hemoglobin's ability to release oxygen is primarily influenced by factors such as pH, temperature, and the concentration of 2,3-bisphosphoglycerate (2,3-BPG), not by the balance between osmotic and hydrostatic pressures.

c. The pH of interstitial fluids is more directly influenced by the concentrations of acids and bases in those fluids, not by the pressures mentioned.

d. The manner in which carbon dioxide is transported in the blood (dissolved in plasma, as bicarbonate ions, or bound to hemoglobin) is not directly affected by the balance between hydrostatic and osmotic pressures described in the question.

Conclusion:

Therefore, the correct answer is Option 2

• Wilson's disease is a genetic disorder characterized by the body's inability to properly eliminate copper, leading to its accumulation in tissues, particularly in the liver and brain. The condition is most often associated with mutations that affect the protein ceruloplasmin, which plays a key role in copper transportation and metabolism.
• Wilson's disease is an autosomal recessive genetic disorder, which means that an individual must inherit one defective gene from each parent to be affected.
• This condition is caused by mutations in the ATP7B gene, which provides instructions for making a protein that plays critical roles in transporting copper and incorporating it into ceruloplasmin, a copper-carrying protein in the blood.
• When the ATP7B protein is not functioning correctly due to mutations, copper is not properly exported out of liver cells and fails to be incorporated into ceruloplasmin. As a result, copper begins to accumulate in the liver and, over time, in other organs, especially the brain.
• In the liver, excessive copper can cause liver damage, leading to symptoms such as fatigue, abdominal pain, jaundice, or signs of liver failure.
• If the liver's capacity to store copper is overwhelmed, copper starts to accumulate in other tissues, particularly the brain, eyes (forming a characteristic ring called Kayser-Fleischer ring), kidneys, and heart, leading to a range of neurological and psychiatric symptoms.
• These may include movement disorders (like tremors, involuntary movements, and poor coordination), speech and swallowing difficulties, cognitive impairments, mood swings, depression, and, in some cases, psychosis.

The correct answer is ​ABA biosynthesis

• NCED (9-cis epoxycarotenoid dioxygenase) is involved in the biosynthesis of abscisic acid (ABA).
• ABA is a plant hormone that plays a crucial role in many aspects of plant growth and development, including seed dormancy, control of organ size, stress responses (like drought tolerance), and stomatal closure.
• NCED is a key enzyme in the ABA biosynthesis pathway. It catalyzes the cleavage of 9-cis epoxycarotenoids to produce xanthoxin, which is a precursor of abscisic acid.
• NCED's action represents a rate-limiting step in the production of ABA, making it crucial for the regulation of this hormone's levels in plants, especially in response to water-deficit stress.
• Through the action of NCED and subsequent steps in the pathway, the accumulation of ABA leads to physiological responses that help the plant cope with stress conditions.

Fig: ABA biosynthetic pathways.9‐cis‐epoxycarotenoid dioxygenases (NCEDs) are attractive targets for chemical modulation in the ABA biosynthesis pathway.

The correct answer is posterior compartment of each segment.​

• The "engrailed" gene in Drosophila melanogaster (fruit fly) is a classic example of a gene involved in the patterning of embryonic segments, which plays a crucial role in the development of the segmented body plan characteristic of insects and many other animals.
• In the model organism Drosophila melanogaster engrailed acts as a segment-polarity gene in early embryonic development.
• Engrailed is a homeobox (Hox) gene that encodes for a transcription factor. Transcription factors are proteins that help turn specific genes on or off by binding to nearby DNA. In the case of Engrailed, the protein it encodes plays a pivotal role in specifying the identity of the posterior compartment in each of the segments that make up the body of the fruit fly.
• The position of engrailed staining illuminates the compartment and segment boundaries along the anterior-posterior axis. Once proper segments form, engrailed expressing cells are found in the posterior-most region of each segment.
• During early development in Drosophila, the embryo is divided into a series of segments. Each segment is further partitioned into an anterior and a posterior compartment, which are cellularly and genetically distinct regions that give rise to different parts of each segment in the mature organism.
• The anterior compartment of a segment forms under the influence of other genes and developmental signals, which confer an anterior identity to the cells in this compartment.
• The posterior compartment, where Engrailed is expressed, is defined by the activity of the Engrailed gene. The expression of Engrailed marks the cells in this compartment and actively contributes to their identity as "posterior." This distinction is crucial for the correct formation of the segmental structures and for the proper arrangement of limbs and other appendages in the mature fly.

Fig: In the first instar larva, the anterior-posterior axis has been formed and can be recognized by the expression of the engrailed gene in the posterior compartment. Engrailed, a transcription factor, activates the hedgehog gene.(Source:Gilbert 12^th edition)

Conclusion:

Therefore, the correct answer is Engrailed expression defines the posterior compartment of each segment in Drosophila melanogaster.

The correct answer is C and D

Concept:

• Peptide bond is amide linkage that is formed when unshared electron pair of α-amino acid of one amino acid attacks the carboxyl carbon of another amino acid.
• This is a nucleophilic acyl substitution reaction, where one molecule of water is released as the by-product.
• Peptide bond have 40% partial double bond characters because of which the 6-atom molecules are in rigid planar configuration and rotation of peptide bond is restricted.
• Angle of rotation of peptide bond is given by ω and it usually have a value of 180°(trans) and occasionally its value is ψ = 0°.

• However rotation is permitted along the N -Ca and Ca - C  bonds.
• The rotation along N - Ca. is called ϕ while the rotation along the Ca - C  is called ψ.
• ψ and ϕ can have value ranging from +180° to -180°, but most of the value are prohibited due to steric hinderence between the atoms in the backbone of polypeptide and side chains of amino acids.
• G. N. Ramachandran, was the first to determine the permitted values and this permitted values is indicated in the ϕ- ψ plane and it is called Ramachandran

plot.

​

• In the Ramachandran plot, the white region corresponds to region of strictly disallowed conformation, this region is strictly disallowed for amino acid accept glycine.
• The dark region corresponds to regions where there is no steric interference, hence this is allowed region.
• The outlined region corresponds to conformation having outer limit of Van der Waals distances, i.e, it allowed atoms to come close together.
• The Ramachandran Principle says that most likely confirmation are alpha helices, beta strands, and turns in the polypeptide chains as most of the other conformations are impossible because of steric collisions between atoms.

Explanation:

• Plot A: Shows a large cluster in the beta-sheet region. This protein is predominantly beta-sheet.
• Plot B: Also shows a cluster in the beta-sheet region, indicating this protein is also predominantly beta-sheet.
• Plot C: Shows a tight cluster in the alpha-helical region (around Φ = -60° and Ψ = -45°), indicating this protein is predominantly alpha-helical.
• Plot D: Also shows a cluster in the alpha-helical region, indicating this protein is predominantly alpha-helical as well.

Conclusion:

• C and D both have Ramachandran plots indicative of alpha-helical structure, making Option 4 the correct answer.

The correct answer is G = 18.5%, C = 24.1%, A = 32.8%, U = 24.6%
Given Base Composition of the DNA Template Strand:

• G (Guanine) = 24.1%
• C (Cytosine) = 18.5%
• A (Adenine) = 24.6%
• T (Thymine) = 32.8%

RNA Transcription:
During transcription, the RNA polymerase synthesizes RNA based on the DNA template strand, following specific base pairing rules:

• Adenine (A) in DNA pairs with Uracil (U) in RNA.
• Thymine (T) in DNA pairs with Adenine (A) in RNA.
• Cytosine (C) in DNA pairs with Guanine (G) in RNA.
• Guanine (G) in DNA pairs with Cytosine (C) in RNA.

Corresponding RNA Base Composition:

1. From G (24.1% in DNA) → C in RNA: C = 24.1%
2. From C (18.5% in DNA) → G in RNA: G = 18.5%
3. From A (24.6% in DNA) → U in RNA: U = 24.6%
4. From T (32.8% in DNA) → A in RNA: A = 32.8%

Final RNA Base Composition:

• G = 18.5%
• C = 24.1%
• A = 32.8%
• U = 24.6%

Conclusion: The correct answer is G = 18.5%, C = 24.1%, A = 32.8%, U = 24.6%, based on the accurate transcription rules from the DNA template strand.

The correct answer is Option 4

Concept:

C3 Photosynthesis:

• Pathway: In C3 plants, the first step of photosynthesis involves the enzyme Rubisco (Ribulose-1,5-bisphosphate carboxylase oxygenase) fixing CO₂ from the atmosphere directly into a 3-carbon compound, 3-phosphoglycerate (3-PGA). This reaction occurs in the stroma of chloroplasts where the Calvin cycle takes place. The Calvin cycle then processes 3-PGA to produce glucose.
• Environment: C3 plants thrive in cooler, more temperate regions where water is relatively abundant, and the rate of photorespiration is not excessively high. Photorespiration is a wasteful pathway that occurs when Rubisco fixes O₂ instead of CO₂, prevalent in high light intensities and temperatures.
• Examples: Wheat, rice, and soybeans.

C4 Photosynthesis:

• Pathway: C4 plants have evolved a mechanism to efficiently capture CO₂ in mesophyll cells and then transport it to bundle-sheath cells where the Calvin cycle is concentrated. In the mesophyll cells, CO₂ is initially fixed into a 4-carbon compound, oxaloacetate, which is then converted to malate or aspartate and transported to bundle-sheath cells. This compartmentalized process reduces photorespiration by concentrating CO₂ around Rubisco.
• Environment: This adaptation is beneficial in high-temperature environments, where photorespiration can significantly reduce the efficiency of photosynthesis in C3 plants.
• Examples: Maize, sugarcane, and sorghum.

CAM Photosynthesis:

• Pathway: CAM plants take in CO₂ at night and fix it into organic acids (such as malate), storing it in vacuoles. During the day, when stomata are closed to conserve water, the stored CO₂ is released for use in the Calvin cycle. This temporal separation of CO₂ uptake and its fixation in the Calvin cycle helps to minimize water loss in arid conditions.
• Environment: CAM plants are adapted to extremely arid environments where water conservation is paramount.
• Examples: Cacti, succulents, and some orchids.

Explanation
Option 4 accurately summarizes the key mechanisms and ecological adaptations of each pathway:

• C3 plants perform direct carbon fixation into a 3-carbon compound under cooler, moist conditions and are susceptible to photorespiration under higher temperatures and light intensities.
• C4 plants minimize photorespiration and enhance carbon fixation efficiency through a spatial separation of carbon fixation and the Calvin cycle, suited for hot and dry environments.
• CAM plants exhibit a temporal separation of carbon fixation (at night) from the Calvin cycle (during the day) to drastically reduce water loss in arid conditions.

Random Distribution:

• Mean =
• Estimated population = 7.5 x 4 = 30

Clumped Distribution:

• Mean =
• Estimated population = 6.5 x 4 = 26

Uniform Distribution:

• Mean =
• Estimated population = 12 x 4 = 48

Effectiveness of Sampling Based on Distribution Type:-

Regular dispersion

• In regular dispersion (also called a uniform or even distribution), the individuals are more or less spaced at an equal distance from one another.
• Individuals are more evenly spaced than expected by chance.
• This is rare in nature but is common in managed systems like croplands.
• A regular distribution pattern may be the result of competition or social interactions.

Random dispersion,

• In random dispersion, the position of one individual is unrelated to the positions of other individuals.
• This kind of distribution occurs where the environment is very uniform and there is no tendency to aggregate.
• This is also relatively rare in nature.

Clumped dispersion

• In clumped dispersion (also called a contagious or aggregated dispersion) individuals are aggregated into groups of varying size.Most populations exhibit this kind of dispersion.
• A clumped distribution pattern often results from the uneven distribution of a resource.
• Human population show Clumped distribution due to their social behaviours, economic condition and geographic factors.

Conclusion:

The uniform distribution provides the most accurate and stable estimate for the population size as it minimizes the chance of sampling error and variance among units. In contrast, the clumped distribution, though common in natural settings, requires careful consideration of sampling strategy to avoid biased estimates.

(i) High levels of gene flow from other populations: High gene flow generally introduces new genetic variation, which can be beneficial for adaptation.
(ii) Low levels of genetic variation within the population: Low genetic variation reduces the ability of the population to adapt to environmental changes and increases extinction risk.
(iii) Frequent environmental disturbances: Environmental disturbances can challenge the population, but their impact depends on the population's resilience and genetic diversity.
(iv) High mutation rate introducing deleterious alleles: High mutation rates can increase the genetic load, reducing fitness and increasing extinction risk.
(v) Strong selective pressure favoring a single phenotypic trait: This can reduce genetic diversity and adaptability, making the population more vulnerable to changes in environmental conditions.

A. Conservation: Biosphere reserves are designated for the conservation of biodiversity and ecosystems.
B. Education: They serve as centers for environmental education and awareness.
C. Human habitation allowed: Unlike traditional protected areas, biosphere reserves often allow human habitation in certain zones, promoting sustainable development.
E. Strong legal back-up: These areas are legally protected and have specific regulations governing their management.
G. Research: Biosphere reserves are important for scientific research and monitoring of ecosystems.

Incorrect Options:

D. Human habitation not allowed:

• Biosphere reserves are designed to balance conservation and sustainable human development. They are divided into zones:
  • Core zone: Strictly protected, no human activity allowed.
  • Buffer zone: Limited human activity for research and education.
  • Transition zone: Human activities like agriculture, forestry, and settlements are permitted, but must be sustainable.

F. No supporting act:

• Biosphere reserves are legally recognized and protected by national and international laws. They have specific regulations and management plans to ensure their conservation and sustainable use.

Key Points

Binary Fission: This is the primary method of bacterial propagation.

In this process, a single bacterial cell doubles its content and then divides into two identical daughter cells.

• Bacterial cell division often occurs through a process known as binary fission. In binary fission, a single bacterial cell, termed the "parent," makes a copy of its DNA and then divides itself into two new cells, termed "daughter cells."

• In a balanced growth, the growth rate of bacteria is proportional to their current number, which makes bacterial growth an example of the first-order reaction.

• E. coli follows the typical growth cycle in liquid nutrient broth with four distinct phases: lag phase, log or exponential phase, stationary phase, and death or decline phase.

• Bacteria do not divide by "sexual reproduction"; rather, they predominantly divide via asexual reproduction through a process known as binary fission.
• They can exchange genetic material through processes known as conjugation, transformation, and transduction, but these are not forms of sexual reproduction as they do not result in the splitting of cells.

Hence the correct answers is option 4

• Halophile: Organisms that require high salt concentrations for growth. Found in environments like salt lakes, salt mines, and marine environments. These bacteria thrive in environments with high levels of salt.
• Piezophile (Barophile): Organisms that grow optimally at high hydrostatic pressure. Typically found in deep-sea environments where the pressure is exceedingly high. They have adapted to survive and thrive under high-pressure conditions.
• Mesophile: Organisms that grow best at moderate temperatures. Common in environments that are not extreme, such as soil, water, and human bodies. Their optimal growth temperature range is between 20°C and 45°C.
• Xerophile: Organisms that grow best in conditions with very low water activity. Found in extremely dry environments like deserts, dried foods, and some salted products. They can survive and proliferate under very dry conditions.

The correct answer is All three statements are correct

Explanation:

The three statements pertain to the principles of phylogenetic analysis and the application of the principle of parsimony in determining ancestral sequences based on observed SNP differences in scorpion species Mesobuthus tamulus.

Statement 1: "In phylogenetic analysis of scorpion species Mesobuthus tamulus, single nucleotide polymorphisms (SNPs) were used to infer ancestral relationships."

• This statement is correct. SNPs are commonly used in phylogenetic analysis to infer evolutionary relationships among species or within species because they provide information on genetic variation.

Statement 2: "The principle of parsimony suggests that the ancestral nucleotide at a given node will minimize the number of changes required to explain observed SNP differences in descendants."

• This statement correctly describes the principle of parsimony used in phylogenetic analysis. The principle of parsimony seeks the simplest explanation or path with the fewest evolutionary changes, which is consistent with standard phylogenetic methods.

Statement 3: "If three individuals show A, A, and T at a specific locus, the most parsimonious ancestral nucleotide at the corresponding internal node is likely to be A."

• This statement logically follows from the principle of parsimony. If there are two instances of the nucleotide 'A' and one instance of 'T', it would require fewer mutations to assume the ancestral nucleotide is 'A' to explain the observed data in the descendants than if the ancestral nucleotide were 'T'.

Key Points:

• Phylogenetic Analysis: SNPs are commonly used in phylogenetic studies to infer relationships among species by comparing genetic differences.
• Principle of Parsimony: This fundamental evolutionary concept prefers the simplest phylogenetic tree with the fewest changes or mutations.
• Evaluating Ancestral Nucleotides: In estimating ancestral states, the nucleotide requiring the fewest changes when observed among descendants is chosen, aligning with principles of parsimony.

Key Points

• Isoelectric focusing IEF is a technique used to separate proteins based on their isoelectric point pI, which is the pH at which a protein carries no net electrical charge.
• In the given scenario, the IEF gel showed multiple sharp bands corresponding to different pI values, indicating the presence of proteins with different charges.

Fig 1: Isoelectric focusing

Explanation:

• When the band corresponding to the pI of the protein of interest pI 5.2 was cut and subjected to SDS PAGE, three distinct bands were observed.
• SDS PAGE separates proteins based on their molecular weight, with smaller proteins migrating faster than larger ones.
• The presence of three distinct bands on SDS PAGE suggests the presence of different molecular weight forms or subunits of the protein.
• Therefore, based on the given observations, it can be inferred that the protein of interest may be composed of three subunits option c or may exist in different molecular weight variants.
• However, the inference that cannot be drawn is that the protein is composed of a single subunit option d.

The presence of multiple bands on SDS PAGE suggests the possibility of different subunits or molecular weight variants of the protein, and it cannot be concluded that the protein is composed of a single subunit based on these observations alone.

Thus the correct answer is option 4

The correct answer is Only A,B and C are correct.

Concept:

Photorespiration in C3 plants:

• C3 plants undergo photorespiration, a process where Rubisco (the enzyme responsible for carbon fixation in the Calvin cycle) catalyzes the addition of oxygen to ribulose-1,5-bisphosphate (RuBP) instead of carbon dioxide.
• This leads to the formation of a two-carbon compound, which undergoes a series of reactions known as the photorespiratory pathway.
• Photorespiration occurs when the concentration of oxygen in the leaf exceeds that of carbon dioxide, such as during hot and dry conditions.
• It results in the loss of fixed carbon and reduces the efficiency of photosynthesis.

CO₂ Compensation Point:

• The CO₂ compensation point is the point at which the rate of CO₂ uptake by photosynthesis equals the rate of CO₂ release by respiration. At this point, there is no net carbon gain or loss by the plant.
• In C3 plants, the CO₂ compensation point is higher compared to C4 plants.
• This higher CO₂ compensation point in C3 plants is primarily due to the presence of photorespiration.

Effect of Photorespiration on CO₂ Compensation Point:

• Photorespiration competes with photosynthesis for the same substrate (RuBP), leading to a decrease in the efficiency of carbon fixation.
• As a result, C3 plants require a higher atmospheric CO₂ concentration to compensate for the carbon losses due to photorespiration.
• The higher CO₂ compensation point reflects the greater reliance of C3 plants on atmospheric CO₂ for efficient photosynthesis compared to C4 plants.

Explanation:

Statement A. Correct. The CO₂ compensation point is generally lower in C4 plants compared to C3 plants. This is mainly due to the efficient initial fixation of CO₂ by PEP carboxylase in the mesophyll cells of C4 plants. PEP carboxylase has a much higher affinity for CO₂ and does not fix oxygen (O₂), which reduces photorespiration significantly compared to C3 plants. As a result, C4 plants can maintain photosynthesis with much lower internal CO₂ concentrations, hence their lower CO₂ compensation point.

Statement B. Correct. In C3 plants, an increase in temperature tends to enhance photorespiration, a process where oxygen is consumed and CO₂ is released. This process competes with CO₂ fixation by Rubisco, the enzyme responsible for carbon fixation in the Calvin cycle. Since higher temperatures favor the oxygenase activity of Rubisco over its carboxylase activity, more O₂ is consumed, and more CO₂ is released, thus increasing the CO₂ compensation point.

Statement C. Correct. The pH increase within the chloroplast stroma during the light reactions of photosynthesis favors the carboxylation activity of Rubisco. During the light reactions, the generation of ATP and NADPH is accompanied by the uptake of protons in the stroma, raising its pH. A higher stroma pH favors the binding of CO₂ over O₂ by Rubisco, which can lead to a reduced CO₂ compensation point under conditions that minimize photorespiration, such as high light intensity that promotes high rates of photosynthesis.

Statement D. Incorrect. The statement about C4 plants and the effect of leaf internal pH on the CO₂ compensation point is misleading. C4 plants carry out the initial fixation of CO₂ in the mesophyll cells via PEP carboxylase, which has a much higher affinity for CO₂ and is not directly affected by the internal pH in the manner described. The efficiency of PEP carboxylase in C4 plants is generally not inhibited by variations in internal pH under normal physiological conditions. The CO₂ concentration mechanism in C4 plants efficiently delivers CO2 to Rubisco in the bundle sheath cells, significantly reducing the impact of oxygen, thereby minimizing photorespiration and maintaining a low CO₂ compensation point regardless of minor pH variations.

Conclusion:

Therefore, Statements A, B, and C correctly reflect the factors influencing the CO₂ compensation point in C3 and C4 plants, making Option 2 the correct choice.

(P) Extensin: Extensin is a type of hydroxyproline-rich glycoprotein that is specific to plant cell walls. It plays a crucial role in cell wall architecture by providing structural support and cross-linking through its amino acid residues. Extensin is part of a larger family of structural proteins.
(Q) Lignin: Lignin is an organic polymer found in the cell walls of many plants, particularly in wood and bark. It is composed of phenolic compounds and is essential for providing rigidity and does not rot easily, thus increasing the plant's structural integrity and resistance to pathogens. Lignin is a phenolic polymer, which means it is made up of phenylpropanoid units.
(R) Pectin: Pectins are complex carbohydrates that are rich in galacturonic acid residues. They are polysaccharides with multiple negative charges due to the presence of carboxyl groups (-COOH) which can ionize to form -COO⁻. Pectins play a major role in plant cell adhesion by forming a gel-like matrix in which cellulose fibrils are embedded, and they are crucial for maintaining cell wall plasticity.
(S) Cellulose: Cellulose is the most abundant carbohydrate polymer on Earth and is composed of β(1→4) linked D-glucose units. It is a critical structural component of the primary cell wall of green plants, many forms of algae, and the oomycetes. The long chains of glucose units form strong fibrils that provide tensile strength and rigidity to the plant cell wall. Tensile strength is crucial for maintaining cell shape and resisting internal turgor pressure

Concept:

• There are numerous types of plants and animals in the living world. The structural and functional unit of life, cells constitute the foundation of the living world.
• With the help of carbon-based and related substances (during metabolism), they have the ability to change their shape and appearance in order to carry out growth and reproductive functions.

Explanation:

Option 1: The definition of biological species was given by Ernst Mayr- CORRECT

• Ernst Mayr defined species as a “group of interbreeding natural populations that are reproductively isolated”.
• This is the most accepted species concept.

Option 2: Photoperiod does not affect reproduction in plants- INCORRECT

• Photoperiod is the duration of light for which an organism is illuminated with light.
• Plants show different physiological reactions in response to the length of light and dark periods.
• Photoperiod affects reproduction in both plants and animals. In plants, photoperiod affects flowering.

Option 3: Binomial nomenclature system was given by RH Whittaker- INCORRECT

• Binomial nomenclature was given by Carolus Linnaeus. He explained the rules of writing scientific names.
• Five kingdom classification was given by R. H. Whittaker. The five kingdom includes- Monera, Protista, Fungi, Plantae, and Animalia.

Option 4: In unicellular organisms, reproduction is synonymous with growth- CORRECT

• Growth is the characteristic of living organisms.
• Cell division in unicellular organisms is equivalent to reproduction.

So, the correct answer is option 3.

The correct answer is 30 - 40 residues

Step 1: Convert growth rate to seconds

• The bacterial cell grows at 1.2 µm per minute. First, convert the growth rate to micrometers per second:
• 1.2μm/min=60seconds/1.2μm​=0.02μm/s

Step 2: Convert micrometers to nanometers

• Since the length of one N-acetylglucosamine (NAG) residue is given in nanometers, convert the growth rate from micrometers per second to nanometers per second:
• 0.02μm/s=0.02×1000nm/s=20nm/s

Step 3: Calculate the number of NAG residues added per second

• Each NAG residue in peptidoglycan has a length of 0.5 nm.
• Number of residues per second= 20nm/s​ / 0.5 nm = 40 residues

The calculated number of NAG residues added per second is 40 residues.

A. Chemical Signals
Sieve tubes are crucial in the transport of chemical signals across the plant. These chemical signals include hormones (like auxins, gibberellins, cytokinins, etc.), which are instrumental in regulating plant growth, development, and responses to environmental stresses. The flow of these signals through the phloem allows for coordinated actions between different parts of the plant, ensuring that growth and responses to the environment are harmonized throughout the organism.

B. mRNAs
Recent research has shown that sieve tubes also function as conduits for the transport of messenger RNAs (mRNAs). This is significant because mRNAs carry genetic information from DNA to the machinery in the cell that synthesizes proteins. The ability to transport mRNAs allows for a systemic response to environmental cues and developmental signals. For instance, an mRNA produced in the leaves that encodes for a protein required to enhance stress resistance can be transported to other parts of the plant, enabling the entire plant to mount a coordinated response to stress conditions.

C. Electrical Signals
Electrical signaling in plants is a relatively less understood mechanism compared to animals. However, plants do use electrical signals to communicate rapid responses across long distances within the organism, such as in response to wounding or environmental threats. This communication happens through a variety of pathways, including the phloem. Sieve tubes can participate in the propagation of these signals. For example, when a leaf is damaged, electrical signals can be sent through the phloem to distant parts of the plant, triggering the release of defensive chemicals or other protective responses even in areas not directly affected by the damage.

D. Spheroplasts - Spheroplasts are not transmitted through sieve tubes. Spheroplasts refer to cells from which the cell wall has been almost entirely removed, commonly found in bacterial studies or in certain experimental setups involving plant cells. This choice is not relevant to the function of sieve tubes in plants

Conclusion
The functions of sieve tubes in the phloem extend beyond the mere transport of nutrients and photosynthates. They are integral to the plant's communication system, enabling it to react as a unified organism to internal developmental cues and external environmental stimuli. Chemical signals and mRNAs are conveyed through these sieve tubes, facilitating long-distance signaling that is crucial for plant growth, development, and survival. Additionally, the role of sieve tubes in the transmission of electrical signals highlights their importance in rapid response communications, cementing the phloem's role as a central communication hub in plants. This multifaceted communication network allows plants to adaptively respond and thrive in their environments.

Concept:

• Phylogeny is the study of ancestral relationship, lineages and evolutionary history of groups of organisms.
• Phylogenetic analysis is also known as cladistic analysis.
• A dendogram or branching diagram constructed by cladistic method is called as phylogenetic tree or cladogram.
• A cladogram serves the basis of phylogenetic classification.

Important Points

• Monophyletic group -
  • It includes a common ancestor and all the descendants of that ancestor.
• Paraphyletic group -
  • It includes a common ancestor and only some of the descendants.
  • It does not include all the known descendants.
• Polyphyletic group -
  • It consists of two or more ancestors.
  • Thus, it includes two or more separate groups, each having a separate common ancestor.

Explanation:

The given data suggests a paraphyletic group because:

• It has one common ancestor - species T.
• Species X, Y, Z, U and V are all descendants of the common ancestor T.
• The grouping does not include all the descendants as species U and V are excluded.