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

Since all have attended exactly two meeting therefore, number of people attended only one meeting or all three meetings is equals to zero.

Given,
21 professors attended Mumbai meeting,
27 attended Delhi meeting and
30 attended Chennai meeting
Let x Professor attended Mumbai and Chennai meeting, then 30 − x will attend Chennai and Delhi meeting. (Since, total 30 have attended Chennai meeting.)
Similarly since total 21 people have attended Mumbai Meeting, So, 21 − x will attend Delhi and Mumbai meeting.
Since 27 people have attended Delhi meeting so,
(21 – x) + (30 – x) = 27
51 – 2x = 27
x = 12
So, people who have attended Chennai and Delhi meeting = 30 – 12 = 18.

Calculation:

The number is always divisible by 3.

∵ The sum of smallest 3 consecutive number = 2 + 3 + 4 = 9

Taking another example;

82 + 83 + 84 = 249.

∵ 249 is a multiple of 3, thus it is divisible by 3.

From the given data, we get

When T = 0 hr, A has 2 liters and B has 0 liters

When T = 1 hr, A has 1 liter and B has 1 liter

When T = 1.5 hr, A has 1.5 liters and B has 0.5 liters

When T = 2 hr, A has 0.5 liters and B has 1.5 liters

When T = 2.5 hr, A has 1 liter and B has 1 liter

When T = 3 hr, A has 0 liters and B has 2 liters

Concept:

Calculation:
Given,
Two cubes numbered 1 to 6.
So, Total outcomes = {1, 2, 3, 4, 5, 6} = 6
Possibility of even number = Favourable outcomes = {2, 4, 6} = 3
The probability that an even number is rolled out for Cube 1 and Cube 2,

The probability that an even number is rolled out on each dice is,

Calculation :

Given, Rohan sells two commodities for Rs. 19,800 each, one sold at a profit of 10% and the other at a loss of 10%.

So, as the selling price of both the items is the same, therefore there would always be an overall loss.

Thus, the loss percent = = 1% loss

To find the amount of loss we have to find the cost price of each commodity.

Let CP₁ and CP₂ be the cost prices of the two commodities.

So, CP₁ × (11/10) = 19800 ⇒ CP₁ = Rs.18,000

and, CP₂ × (9/10) = 19800 ⇒ CP₂ = Rs.22,000

So, total cost price = 18000 + 22000 = Rs.40,000

⇒ Loss amount = (loss% × total CP)/100 = 40000/100 = Rs.400

Important Points

In a case like this when the SP of two articles are the same and one is sold at a profit of x % and the other at a loss of x %, there would always be a loss of %

However, make sure that the selling price is the same and that that profit percentage and loss percentage are numerically equivalent.

On the other hand, in a case when the CP of two articles are the same and one is sold at a profit of x % and the other at a loss of x %, there would be neither overall profit nor loss.

The correct answer is an odd number and a perfect square
Given:
For any four consecutive decimal digits, the largest value of the product of the sum of any two and the sum of the other two is:

Let's consider four consecutive decimal digits: 1, 2, 3, and 4.

The possible sums of any two digits are:

• 1 + 2 = 3
• 1 + 3 = 4
• 1 + 4 = 5
• 2 + 3 = 5
• 2 + 4 = 6
• 3 + 4 = 7

Now, we need to find the product of the sum of any two and the sum of the other two.

Let's consider all possible combinations:

• (1 + 2) * (3 + 4) = 3 * 7 = 21
• (1 + 3) * (2 + 4) = 4 * 6 = 24
• (1 + 4) * (2 + 3) = 5 * 5 = 25

Among these, the largest value is 25, which is an odd number and a perfect square.

Therefore, the correct answer is an odd number and a perfect square

The correct answer is Option 3 i.e.Signal-based sorting utilizes specific amino acid sequences or motifs within proteins to direct them to their proper cellular locations, often involving recognition by sorting receptors.

Signal-based sorting is a crucial mechanism in eukaryotic cells that directs proteins and other molecules to their appropriate locations within or outside the cell.

• This process utilizes specific amino acid sequences or motifs known as signal sequences or targeting signals present in the proteins.
• Signal Sequences or Motifs are short stretches of amino acids within a protein that are recognized by the cell's sorting machinery. Different signals target proteins to different locations, such as the nucleus, mitochondria, endoplasmic reticulum (ER), and others.
• These signals are recognized by sorting receptors, which then facilitate the transport of these proteins to specific destinations such as the endoplasmic reticulum, Golgi apparatus, lysosomes, mitochondria, peroxisomes, or plasma membrane.
• The recognition and subsequent sorting are highly selective, ensuring that proteins are correctly localized to maintain cellular function.

Vesicle-based sorting, on the other hand, involves the encapsulation of proteins into membrane-bound vesicles for transport within the cell.

• This process is crucial for the transport of proteins between different organelles, such as from the endoplasmic reticulum to the Golgi apparatus or from the Golgi apparatus to other destinations like lysosomes or the plasma membrane.
• Vesicle-based sorting can recognize specific signal sequences on proteins, facilitated by coat proteins (e.g., COPI, COPII, and clathrin) and adaptors that select cargo for packaging into vesicles based on these signals.
• Certain proteins in the ER have exit signals that sort them into budding vesicles coated with coat proteins such as COPII. These vesicles then detach and transport the cargo proteins to the Golgi apparatus.
• Selective Cargo Packaging: Specific cargo receptors in the vesicle coat interact with cargo proteins that have the appropriate signal sequences, ensuring that only proteins meant for transport get packaged into vesicles.
• Vesicle Targeting and Fusion: After budding, vesicles are targeted to specific destinations. Each vesicle has specific markers that are recognized by target membranes, ensuring that vesicles fuse only with the correct organelle. For instance, vesicles from the ER fuse with the cis-Golgi, and vesicles from the trans-Golgi can fuse with the plasma membrane or endosomes.
• Therefore, vesicle-based sorting is not a non-selective process; it plays a key role in distinguishing between different proteins for targeted transport based on specific signals.

Conclusion:

Statement 1 and 2 misconstrue the mechanisms and specificity involved in signal-based and vesicle-based sorting, while 4 inaccurately portrays the selectivity and purpose of vesicle-based sorting.Therefore,correct answer is Statement 3

Concept:

• Five different genes are known to specify floral organ identity in Arabidopsis: APETALA1 (AP1), APETALA2(AP2), APETALA3 (AP3), PISTILLATA (PI), and AGAMOUS (AG).
• The organ identity genes initially were identified through mutations that dramatically alter the structure and thus the identity of the floral organs produced in two adjacent whorls.
• Because mutations in these genes change floral organ identity without affecting the initiation of flowers, they are homeotic genes.
• These homeotic genes fall into three classes—types A, B, and C—defining three different kinds of activities

​Explanation:

Fig 1: ABC model for the acquisition of floral organ identity

• The acquisition of floral organ identity is based on the interactions of three different types of activities of floral homeotic genes: A, B, and C.
• In the first whorl, expression of type A (AP2) alone results in the formation of sepals.
• In the second whorl, expression of both type A (AP2) and type B (AP3/PI) results in the formation of petals.
• In the third whorl, the expression of B (AP3/PI) and C (AG) causes the formation of stamens.
• In the fourth whorl, activity C (AG) alone specifies carpels.
• In addition, activity A (AP2) represses activity C (AG) in whorls 1 and 2, while C represses A in whorls 3 and 4.

Hence the correct answer is option 3

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.

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.

α-Keratins are a major component of the intermediate filaments found in epithelial cells and are essential constituents of structures such as hair, nails, and the outer layer of skin. α-Keratin has a chemical composition that is notably rich in hydrophobic residues.

Role of Hydrophobic Residues: Hydrophobic amino acids, such as alanine, valine, leucine, isoleucine, methionine, and phenylalanine, play a significant role in the structure of α-keratin. These residues contribute to the protein's ability to form the α-helical coiled coil structure, which is a defining feature of intermediate filaments. The hydrophobic interactions between amino acid side chains are critical for stabilizing this helical structure, allowing it to resist solvation by water and thus providing structural integrity in the harsh environments of skin, hair, and nails.

Hydrophobic Interactions and Stability: The hydrophobic residues are positioned along the α-helices in such a way that they promote the interaction of two α-helices into a coiled-coil structure. The intertwining of these helices through the hydrophobic interactions contributes significantly to the mechanical strength and elasticity of keratin fibers. These properties are essential for the protective functions of skin and the flexibility and resilience of hair and nails.

Dimerization and Filament Formation: α-Keratin molecules dimerize to form coiled-coil dimers, which further assemble into intermediate filaments. The hydrophobic interactions facilitate this dimerization and are key to the filament assembly process, providing the structural framework necessary for the formation of the strong and insoluble keratin filaments found in epithelial cells.

Hydrophobic vs. Other Amino Acid Residues: These hydrophobic residues are fundamental in maintaining the structure and integrity of α-keratin through their contributions to the protein's secondary and tertiary structures.

Key PointsKeratins are a diverse group of fibrous proteins that form the key structural material in the outer layer of human skin, as well as in hair, nails, and various other epithelial tissues and organs. They are part of the larger family of intermediate filament proteins, which are critical components of the cytoskeleton in eukaryotic cells. The structure and function of keratins are essential not only for the mechanical resilience and protective capabilities of tissues but also for cellular integrity and the mediation of cell-cell interactions.

Types of Keratins

• Keratins can be broadly classified into two types: alpha-keratins and beta-keratins.
• Alpha-keratins are primarily found in mammals, while beta-keratins are present in birds and reptiles. The distinction between these two types lies in their secondary structure and functional properties.
• Alpha-keratins are characterized by their helical rod domains that form coiled-coil structures, whereas beta-keratins have a beta-sheet conformation.

Structure of Keratins

• Alpha-Helical Rod Domain: This central part of alpha-keratins forms a coiled-coil structure, essential for the dimerization of keratin monomers. These dimers further assemble into protofilaments, which then associate to form the intermediate filaments characteristic of the keratin cytoskeletal network.
• Head and Tail Domains: In addition to the rod domain, keratins also have non-helical terminal regions known as the head (amino-terminal) and tail (carboxy-terminal) domains. These regions are involved in regulating filament assembly and mediating interactions with other cellular components, including membrane proteins and signaling molecules

Conclusion:

In summary, α-keratin's chemical composition is notably rich in hydrophobic amino acid residues, which are key to its ability to form the coiled-coil structures essential for its structural role in epithelial cells and related tissues.

Activation (Turning On):

• The alpha subunit of a heterotrimeric G-protein is inactive when it is bound to GDP.
• The process of activation begins when a signal (such as a ligand binding to a G-protein-coupled receptor, GPCR) leads to the exchange of GDP for GTP on the alpha subunit. This exchange triggers a conformational change that activates the alpha subunit.
• The activation process can be summarized as the binding of GTP to the alpha subunit.

Deactivation (Turning Off):

• The alpha subunit remains active as long as it is bound to GTP.
• The intrinsic GTPase activity of the alpha subunit eventually hydrolyzes the bound GTP to GDP, resulting in the inactivation of the alpha subunit.
• The hydrolysis of GTP to GDP turns off the alpha subunit.

The initiation of DNA replication in prokaryotes involves a well-ordered sequence of molecular events to ensure accurate duplication of the DNA.

1. Helicase unwinding: The replication process begins when the enzyme helicase binds to the origin of replication and starts unwinding the double-stranded DNA, creating a replication fork. This exposes single-stranded DNA for further processing.
2. Binding of single-stranded binding proteins (SSBs): As helicase unwinds the DNA, SSBs quickly bind to the single-stranded DNA to prevent it from re-annealing or forming secondary structures, keeping the strands separated and accessible.
3. Primase synthesizing RNA primers: The enzyme primase then synthesizes a short RNA primer on each of the separated DNA strands. This RNA primer provides a free 3' hydroxyl group (-OH) necessary for DNA polymerase to begin synthesizing the new DNA strand.

Other Options:

• Binding of DNA polymerase, helicase unwinding, RNA primer synthesis by primase: This sequence is incorrect because DNA polymerase can only bind after RNA primers are synthesized by primase, not before.
• RNA primer synthesis by primase, helicase unwinding, binding of DNA polymerase: This sequence is also incorrect because helicase must unwind the DNA first before primase can synthesize RNA primers.
• Helicase unwinding, RNA primer synthesis by primase, ligation by DNA ligase: This is incorrect because DNA ligase is involved much later in the replication process, where it seals nicks in the DNA backbone during the final stages of replication, not during the initiation phase.

Thus, the correct sequence is helicase unwinding, followed by SSB binding, and then RNA primer synthesis by primase

The correct answer is synapomorphy

Concept:

• In phylogenetics, a shared derived character state refers to a trait that is shared by two or more taxa and is derived from their most recent common ancestor.
• This concept is crucial for constructing phylogenetic trees and understanding evolutionary relationships among different organisms.
• Characters can be classified based on their evolutionary origins and the groups they define.

Explanation:

• Synapomorphy: A synapomorphy is a shared derived character state that is present in an ancestral species and shared exclusively by its evolutionary descendants. This trait is used to define a clade, a group of organisms that includes an ancestor and all its descendants. For example, the presence of feathers in birds is a synapomorphy that defines the clade Aves.
• Apomorphy: An apomorphy is a derived character state that is unique to a particular taxon. It indicates evolutionary novelty. However, it is not necessarily shared with other taxa, making it less useful for defining shared evolutionary relationships.
• Symplesiomorphy: A symplesiomorphy is a shared ancestral character state. It is a trait that is found in the common ancestor of a group but is not unique to that group. This makes it less useful for defining specific evolutionary relationships. For example, the presence of a vertebral column is a symplesiomorphy for vertebrates.
• Plesiomorphy: A plesiomorphy is an ancestral character state. It is a trait that was present in the ancestor of a group but may have been altered or lost in some descendants. Plesiomorphies do not provide information about the derived relationships within a group.

The correct answer is Riboflavin

• Vitamins are organic compounds that are essential for normal growth and nutrition and are required in small quantities in the diet because they cannot be synthesized by the body.
• Some vitamins play crucial roles in cellular metabolism and energy production.
• Riboflavin: Riboflavin, also known as Vitamin B2, is a water-soluble vitamin that is part of the high-energy metabolite FAD (flavin adenine dinucleotide). FAD is a coenzyme involved in several important metabolic pathways, including the Krebs cycle (Citric Acid Cycle), which is essential for energy production in the cell. FAD serves as an electron carrier in the mitochondria, facilitating the production of ATP, the energy currency of the cell.

Other Options:

• Thiamine: Thiamine, or Vitamin B1, is another water-soluble vitamin that is crucial for carbohydrate metabolism. It is a part of the coenzyme thiamine pyrophosphate (TPP), which plays a role in the decarboxylation of pyruvate to acetyl-CoA in the Krebs cycle. However, it is not directly involved in the formation of high-energy metabolites like FAD.
• Pantothenate: Pantothenate, or Vitamin B5, is a component of coenzyme A (CoA), which is essential for fatty acid metabolism and the synthesis and oxidation of fatty acids. While CoA is vital for energy metabolism, it does not form a high-energy metabolite like FAD.
• Folate: Folate, or Vitamin B9, is important for DNA synthesis, repair, and methylation. It is crucial for cell division and growth, especially during periods of rapid growth such as pregnancy and infancy. Folate is not involved in forming high-energy metabolites but is essential for nucleic acid synthesis.

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.

Statement 1: The G1 checkpoint, also known as the restriction point, assesses cell size, nutrients, and DNA integrity before proceeding with DNA replication.

• The G1 checkpoint, or restriction point, is a critical juncture in the cell cycle where the cell evaluates various internal and external factors to determine if conditions are favorable for DNA synthesis (S phase).
• At this point, the cell assesses its size, the availability of nutrients, growth factors, and the integrity of its DNA.
• If any of these conditions are not met, the cell can enter a quiescent state known as G0, where it remains until conditions improve.
• Thus, this statement is accurate in describing the G1 checkpoint's functions.

Statement 2: During the S phase, DNA is replicated, and the replication machinery ensures that each daughter cell will receive an exact copy of the genome.

• The S phase (Synthesis phase) of the cell cycle is dedicated to replicating the cell's DNA. During this phase, the entire genome is duplicated to ensure that each daughter cell receives a complete set of chromosomes.
• The cell has mechanisms such as proofreading and error correction to minimize replication errors, ensuring the fidelity of the replication process.
• Therefore, this statement correctly characterizes the S phase.

Statement 3: The G2/M checkpoint ensures that DNA replication is completed successfully and checks for DNA damage. If the cell passes this checkpoint, it is committed to mitosis.

• After DNA replication, cells enter the G2 phase, during which they prepare for mitosis (M phase). The G2/M checkpoint assesses whether DNA replication has been completed accurately and whether the cell has repaired any DNA damage incurred during replication. Only when these criteria are satisfied does the cell proceed to mitosis.
• This checkpoint is crucial for preventing the propagation of damaged DNA to daughter cells. Hence, this statement accurately describes the purpose and significance of the G2/M checkpoint.

Statement 4: The spindle assembly checkpoint (SAC) occurs during the G2 phase and ensures that all chromosomes are properly aligned on the spindle before anaphase onset.

• The spindle assembly checkpoint (SAC), a crucial mechanism in ensuring the correct segregation of chromosomes, actually operates during the transition from metaphase to anaphase in mitosis, not the G2 phase.
• The SAC verifies that all chromosomes are correctly attached to the mitotic spindle and that the chromosomes are aligned correctly at the metaphase plate.
• If this checkpoint is satisfied, the cell proceeds with anaphase, ensuring accurate chromosome segregation.
• The incorrect element in this statement is the timing of the SAC, which does not occur in the G2 phase but rather during mitosis.

Conclusion:

The correct answer is Option A, which identifies statements 1, 2, and 3 as correct, describing essential checkpoints and processes accurately. Statement 4 contains the incorrect placement of the spindle assembly checkpoint in the cell cycle, which occurs during metaphase to anaphase transition in mitosis, not the G2 phase.

Winged seeds: These are lightweight seeds with wing-like structures that enable them to be dispersed by the wind. Examples of plants with winged seeds include maple trees (genus Acer).

Fleshy fruit: These are fruits that have a soft and edible flesh. Examples include fruits from the rose family (genus Rosa).

Capsular fruit: These are dry fruits that release seeds by splitting open when they mature, as seen in the genus Capsicum (peppers).

Caryopsis: This is a type of dry, one-seeded fruit where the seed coat is fused with the fruit wall, characteristic of grasses like maize (genus Zea).

Amide-linked myristoyl anchors involve the attachment of a myristoyl group (a type of fatty acid) to the protein through an amide bond formed with the nitrogen atom of the amino acid's side chain, commonly glycine. This anchor type embeds the protein in the membrane by inserting the hydrophobic tail of the myristoyl group into the lipid bilayer.

Thioester-linked fatty acyl anchors attach proteins to membranes through a thioester bond between a fatty acid (like palmitic acid) and a cysteine residue in the protein. As opposed to what option A suggests, this and other anchor types primarily facilitate embedding or associating with the membrane, but not necessarily attachment to the membrane's exterior surface exclusively.

GPI anchors are complex glycolipids that attach to the C-terminus of proteins. They anchor proteins to the external leaflet of the lipid bilayer, not the cytosolic side (the inner surface), contradicting what option C claims.

Thioether-linked prenyl anchors involve the attachment of prenyl groups (such as farnesyl or geranylgeranyl groups) to proteins via a thioether bond to a cysteine residue. Unlike what option D states, this bond involves the sulfur atom of the amino acid and not an ester linkage to the glycerol backbone of phospholipids.

Additional Information

Lipid Anchors and Membrane Proteins: Membrane proteins are integral to cell function, involved in processes such as signaling, transport, and cellular recognition. Many of these proteins are anchored to the membrane not by transmembrane domains but by lipid anchors covalently attached to specific sites on the protein. The nature of the lipid anchor influences not only the protein's position in the membrane but also its mobility, accessibility to other molecules, and functional role within cellular processes.

a) Amide-linked Myristoyl Anchors:
Myristoylation is a form of lipid anchoring where a myristoyl group (a 14-carbon saturated fatty acid) is covalently attached via an amide bond to the nitrogen atom of an internal glycine residue of a protein. This modification typically occurs co-translationally. The hydrophobic nature of the myristoyl chain helps to anchor the protein to the membrane by embedding it into the lipid bilayer. This form of anchoring can affect the protein's signaling functions and interactions with other cellular components.

b) Thioester-linked Fatty Acyl Anchors:
Palmitoylation is an example of this type of anchor, where a fatty acid, such as palmitate, is attached to cysteine residues on a protein through a thioester bond. This reversible modification allows for dynamic regulation of protein-membrane association, influencing protein localization, trafficking, and function within the cell. Unlike myristoylation, palmitoylation can occur post-translationally and often targets proteins to specific membrane microdomains, such as lipid rafts.

c) Thioether-linked Prenyl Anchors:
Prenylation involves the attachment of prenyl groups (either farnesyl or geranylgeranyl groups) to the sulfur atom of cysteine residues near the C-terminus of proteins via a thioether bond. This modification serves to anchor the protein to membranes and plays a crucial role in the membrane association of small GTPases, which are vital for intracellular signaling pathways. Prenylation increases the hydrophobicity of the protein, facilitating its insertion into the lipid bilayer and affecting its localization and function.

d) Amide-linked Glycosylphosphatidylinositol (GPI) Anchors
GPI anchoring involves attaching proteins to a complex glycolipid, glycosylphosphatidylinositol, which is then inserted into the outer leaflet of the plasma membrane. This post-translational modification happens in the endoplasmic reticulum. GPI-anchored proteins play diverse roles, including enzyme activities, receptors, and adhesion molecules. The presence of the carbohydrate moiety in the GPI anchor also contributes to the formation of the glycocalyx, a carbohydrate-rich zone on the cell surface that is involved in cell-cell recognition, communication, and protection.

Conclusion:

Therefore, the correct answer is Option 2.

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.

• 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.

III. Mevalonic acid is converted to mevalonic acid pyrophosphate.

• This step is one of the early stages in the mevalonate pathway, which is responsible for the synthesis of isoprenoids or terpenoids. Mevalonic acid conversion to mevalonic acid pyrophosphate involves a kinase enzyme, which phosphorylates mevalonic acid.

V. Isopentenyl pyrophosphate is converted to dimethylallyl pyrophosphate.

• Isopentenyl pyrophosphate (IPP) is an important intermediate in the synthesis of all isoprenoids. IPP is converted to dimethylallyl pyrophosphate (DMAPP) by an isomerase. Both IPP and DMAPP are five-carbon molecules known as isoprenes, which serve as the building blocks for all terpenoids.

IV. Dimethylallyl pyrophosphate is converted to geranyl pyrophosphate.

• This step is the synthesis of the ten-carbon molecule, geranyl pyrophosphate (GPP), from the five-carbon precursors DMAPP (dimethylallyl pyrophosphate) and IPP (isopentenyl pyrophosphate) through a head-to-tail condensation reaction. GPP serves as a precursor for the synthesis of monoterpenes and for further elongation to larger terpenoids.

I. Geranyl pyrophosphate is converted to farnesyl pyrophosphate.

• Farnesyl pyrophosphate (FPP) is a fifteen-carbon molecule made from the condensation of one GPP (ten carbons) and one additional IPP (five carbons). This step is crucial for the synthesis of sesquiterpenes and also serves as a precursor for further biosynthesis of larger terpenoids.

II. Geranylgeranyl pyrophosphate is converted to copalyl pyrophosphate.

• Geranylgeranyl pyrophosphate (GGPP) is a twenty-carbon molecule, and its formation involves the head-to-tail condensation of one GPP (ten carbons) and one IPP (five carbons) twice. The conversion of GGPP to copalyl pyrophosphate is a key step in the biosynthesis of diterpenes. Copalyl pyrophosphate can then be further modified to produce various diterpenoid compounds.

Conclusion:

Therefore, he correct sequence in biosynthetic pathway is III, V, IV, I, II

Natural Killer (NK) cells play a crucial role in the immune response against tumor cells.

1. Tumor cells often downregulate MHC- I molecules: This is correct. Many tumor cells downregulate MHC-I molecules as an adaptive mechanism to evade detection by cytotoxic T lymphocytes (CTLs), which typically recognize antigens presented by MHC-I molecules.
2. NK cells recognize the absence of MHC-I molecules on target cells: This is correct. NK cells have inhibitory receptors that typically interact with MHC-I molecules on the surface of normal cells. When MHC-I molecules are absent or downregulated (often the case with tumor cells), the inhibitory signal is not triggered, thereby activating NK cells to kill the target cell.
3. NK cells recognize tumor antigens presented on MHC-II molecules: This is incorrect. NK cells do not recognize antigens presented by MHC-II molecules, which is primarily the mechanism used by helper T cells. Instead, NK cells focus on the presence or absence of MHC-I molecules.
4. NK cells kill tumor cells through the release of perforins and granzymes: This is correct. NK cells induce apoptosis in target cells by releasing cytotoxic substances such as perforins and granzymes. Perforins create pores in the target cell membrane, allowing granzymes to enter the cell and trigger programmed cell death.

Key Points

• NK cells are vital in the immune response by targeting cells that lack MHC-I molecules.
• MHC-I downregulation is a common evasion strategy of tumor cells to avoid CTLs.
• Inhibitory receptors on NK cells detect the absence of MHC-I molecules, triggering NK cell activation.

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. ABA promotes leaf senescence independent of ethylene

• ABA is a plant hormone known for its role in stress responses, especially in response to drought.
• It is involved in the regulation of leaf senescence, a process of aging in leaves.
• While ABA can promote leaf senescence, ethylene is also a significant regulator of this process. The key point here is that ABA can promote leaf senescence independent of ethylene, meaning it has its own pathways and mechanisms for promoting senescence without relying on ethylene.
• However, in many cases, ABA and ethylene can work together synergistically to regulate leaf senescence. The complexity of hormone interactions in plants makes this a nuanced topic.

B. ABA promotes shoot growth and inhibits root growth at low water potential

• This statement is incorrect with respect to the typical roles played by ABA during water stress.
• ABA's primary function during drought stress is to help the plant conserve water. One way it does this is by causing the stomata (pores on the leaf surface) to close, reducing water loss through transpiration.
• In terms of growth, ABA typically inhibits shoot growth to reduce the plant's overall water demand.
• Contrarily, there is evidence that ABA can promote root growth under certain conditions, possibly to enhance the plant's ability to source water from the soil.
• Thus, the traditional view is that ABA inhibits shoot growth and can potentially promote root growth under stress conditions, contrary to what the statement suggests.

C. ABA inhibits gibberellin-induced enzyme production

• ABA and gibberellins (GAs) often have antagonistic effects in plant development processes, particularly in seed germination.
• During germination, GAs promote the synthesis of enzymes that break down stored starches and nutrients in the seed, making them available for the developing seedling.
• ABA acts as a counter-regulatory hormone, inhibiting this enzyme production and thereby suppressing germination. This balance between ABA and GA is crucial in seed dormancy and germination regulation, making this statement accurate.

D. Seed dormancy is controlled by the ratio of ABA and gibberellin

• The interplay between ABA and gibberellin is a key factor in managing seed dormancy and the initiation of germination.
• High levels of ABA are associated with maintaining dormancy, as it suppresses the processes necessary for germination.
• On the other hand, gibberellins promote germination by stimulating processes such as enzyme production for nutrient mobilization.
• The transition from dormancy to germination, therefore, is often triggered by a shift in the balance between these two hormones, with a decreased ABA to GA ratio favoring germination.
• This mechanism enables seeds to "decide" when environmental conditions are favorable for germination and subsequent seedling growth, underscoring the accuracy of statement D.

Conclusion:

Statements A, C and D accurately reflect the roles of ABA in inhibiting gibberellin-induced enzyme production and controlling seed dormancy.

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.