CSIR NET Life Sciences Mock Test - 1 - UGC NET MCQ

Given:

In a firm the average salary of male employees = Rs. 520

in a firm the average salary of female employees = Rs. 420

the overall average salary = Rs. 500,

Formula used:

Average = (n₁ × X₁ + n₂ × X₂)/(n₁ + n₂)

Calculation:

500 = (M × 520 + F × 420)/(M + F)

⇒ 500M + 500F = 520M + 420F

⇒ 80F = 20M

⇒ M : F = 80 : 20

∴ The required ratio of Male and Female is 80 : 20.

Given:

A dishonest shopkeeper his goods at cost price but uses a false weight of 950 gms for 1kg.

Formula:

Gain % = [(Error)/(True value – Error)] × 100

Calculation:

According to the question,

Seller uses a weight of 800 gms for 1 kg.

1 kg = 1000 gms.

Error in weight = 1000 – 950 = 50 gms.

According to the formula,

Gain % = [50/(1000 – 50)] × 100

⇒ 50/950 × 100

⇒ 5.26%

∴ The gain percent is 5.26%.

Given:
In a cylic quadrilateral ABCD.
∠ADC = 130°

Concepts Used:

• The sum of the opposite angles of a cyclic quadrilateral is supplementary i.e. 180º.
• The angle at the circumference in a semicircle is always 90°

Explanation:

We know that:
∠ADC + ∠ABC = 180º
∴ ∠ABC = 180º - 130°
⇒ ∠ABC = 50°
We also know that :
∠ACB = 90º
Now in ΔABC:
Sum of all interior angles of a triangle = 180º
∠ABC + ∠ACB + ∠CAB = 180º
∠CAB + 50º + 90º = 180º
∠CAB = 180º - 50º - 90º
∠CAB = 40º
Hence the correct answer is "40º".

Concept:

Explanation:

In ΔABP
= tan30° =
x + y = √3h . . . (i)
In ΔQBP
h/y = tan45° = 1
h = y . . . (ii)
Substitute the value of h in equation (i), we have
x + y = √3y
x = (√3 - 1) y
Let speed is v m/s and the time to cover distance x is 20 seconds.
Therefore,  x/v = 20 ⟹ x = 20v
∴ 20v = (√3 - 1) y
Time taken by boat to cover distance y = y/v

Time taken by boat to cover distance y = 10(√3 +1) seconds
Hence, the correct answer is option (2).

Calculation:

The correct answer is China
To determine which country experienced the maximum change in forest area in absolute terms, we have to consider both the percentage change in forest cover and the total land area of each country.
Suppose the forest cover percentages and land areas for the countries are as follows:

• Tanzania: Forest cover changed from 40% to 35%, Land area = 0.945 million km²
• Russian Federation: Forest cover changed from 50% to 49%, Land area = 17.1 million km²
• India: Forest cover changed from 21% to 24%, Land area = 3.287 million km²
• China: Forest cover changed from 16% to 23%, Land area = 9.597 million km²

To find the absolute change in forest area, we calculate:

• Tanzania: (35% - 40%) * 0.945 million km² = -0.05 * 0.945 = -0.04725 million km²
• Russian Federation: (49% - 50%) * 17.1 million km² = -0.01 * 17.1 = -0.171 million km²
• India: (24% - 21%) * 3.287 million km² = 0.03 * 3.287 = 0.09861 million km²
• China: (23% - 16%) * 9.597 million km² = 0.07 * 9.597 = 0.67179 million km²

From the calculations, we observe that China experienced the maximum change in forest area in absolute terms, which is approximately 0.67179 million km².

Therefore, the correct answer is option D(China).

The correct answer is Shorter bands (around 2 kb) in both transformed and SV40 cells, longer bands (around 5 kb) in normal epithelial cells.

Concept:

• Southern Blot Analysis: Southern blotting is a technique used to detect specific DNA sequences within a DNA sample by hybridizing with a labeled probe.
• Histone Gene Sequences: The histone genes are highly conserved and can be used as probes in Southern blotting to analyze changes in DNA structure and modifications.
• Transformation and SV40 T Antigen: Transformation of cells and the introduction of the SV40 T antigen can lead to changes in the genomic DNA, including the integration of foreign DNA or rearrangements that can be identified as changes in band patterns on a Southern blot.

Explanation:

Histone genes are crucial for packing DNA into chromatin. They are relatively conserved and stable in normal cells. In transformed cells (cells that have undergone changes to become cancer-like) and cells transduced with SV40 T antigen (a viral protein known to induce transformation), the genome undergoes several alterations:

• Genomic Instability: Transformation can lead to chromosomal rearrangements, amplifications, or deletions.
• Telomere Shortening: Transformed cells often exhibit telomere shortening due to replicative stress. However, telomere alteration is more relevant to other types of sequences rather than histone genes

Normal Epithelial Cells: These cells maintain their genomic integrity. Therefore, the restriction enzyme digestion should produce consistent, longer bands (around 5 kb) due to the stable and organized chromatin structure.

Transformed Cells and SV40 T Antigen Transduced Cells: These cells would likely experience genomic instability and alterations. This can lead to:

• Changes in the accessibility of restriction sites.
• Generation of shorter DNA fragments (around 2 kb) because of DNA breakpoints or the introduction of new restriction sites through genomic reorganization.

Band pattern:

Longer bands (around 5 kb) for normal epithelial cells due to the stable, organized chromatin structure.
Shorter bands (around 2 kb) in both transformed and SV40-transduced cells due to the genomic instability and reorganization that affects the DNA restriction pattern.

The correct answer is Option 2 i.e. It activates the transcription of genes involved in DNA repair and can induce cell cycle arrest.

p53, often dubbed the "guardian of the genome," is a crucial tumor suppressor protein that plays a significant role in preventing cancer by maintaining the integrity of the genome. Its activities are multifaceted, but one of its primary functions is to respond to cellular stress, particularly DNA damage. Here's how it contributes to cell cycle regulation upon DNA damage detection:

• Activation upon DNA damage: When DNA damage is detected, p53 is stabilized and activated through post-translational modifications, primarily by phosphorylation by specific kinases that are part of the DNA damage response (DDR) pathway. Under normal conditions, p53 levels are kept low through continuous degradation mediated by the E3 ubiquitin ligase MDM2. However, DNA damage leads to the disruption of this interaction, thus stabilizing and activating p53.
• Transcriptional regulation of target genes: Once activated, p53 functions as a transcription factor that binds to specific DNA sequences and regulates the expression of lot of target genes. These target genes are involved in various critical cellular processes, including DNA repair, cell cycle arrest, apoptosis (programmed cell death), and senescence (a state of permanent cell cycle arrest).

a. Cell cycle arrest: One way p53 helps repair DNA damage is by arresting the cell cycle, providing the cell with time to repair the damage before proceeding through critical phases like DNA replication (S phase) and cell division (M phase). p53 achieves this by upregulating the transcription of the CDK inhibitor p21 (WAF1/CIP1). p21 inhibits the activity of cyclin-CDK complexes, specifically cyclin E-CDK2 and cyclin A-CDK2, thus enforcing a G1/S checkpoint arrest. Additionally, p21 can inhibit cyclin B-CDK1, contributing to a G2/M arrest, thereby covering key transitions that are essential for DNA replication and mitosis.

b. DNA repair: p53 also induces the expression of several genes directly involved in DNA repair mechanisms, ensuring that any damaged DNA is repaired before the cell continues to divide.

• Apoptotic pathway activation: If the DNA damage is too severe and cannot be repaired, p53 can push the cell towards apoptosis, a programmed cell death, to prevent the propagation of damaged DNA, which could otherwise lead to tumorigenesis.

Conclusion:

p53's contribution to cell cycle regulation upon DNA damage is primarily through transcriptional activation of genes that are involved in halting the cell cycle (via the induction of p21 leading to cell cycle arrest) and promoting DNA repair pathways. It ensures that cells do not proceed to divide with damaged DNA, thereby maintaining genetic stability. Therefore, the correct answer is Option 2

Zellweger syndrome is a peroxisomal disorder. Peroxisomal disorders are a group of genetic diseases that affect the normal functioning of peroxisomes, which are cellular organelles involved in various metabolic pathways, including the breakdown of very long chain fatty acids, the synthesis of plasmalogens (important for the normal function of the myelin sheath of nerve cells), and the detoxification of hydrogen peroxide.

Zellweger syndrome is caused by mutations in genes responsible for the biogenesis and function of peroxisomes. These genes include PEX1, PEX2, PEX3, PEX5, and others, which encode peroxins, proteins necessary for the assembly of peroxisomes.
The mutation leads to impaired peroxisome assembly and function, resulting in multiple enzyme deficiencies and the accumulation of very long chain fatty acids and other metabolites that would normally be degraded in peroxisomes.

Other Options:

• Acute intermittent porphyria (AIP): It is a disorder of heme metabolism, specifically a hepatic porphyria, leading to an accumulation of porphyrin precursors. It is caused by mutations in the HMBS gene, which encodes the enzyme hydroxymethylbilane synthase.
• Maple syrup urine disease (MSUD): It is a metabolic disorder affecting the breakdown of branched-chain amino acids (leucine, isoleucine, and valine). It is caused by mutations in the BCKDHA, BCKDHB, or DBT genes, which encode components of the branched-chain alpha-keto acid dehydrogenase complex.
• Medium chain acyl-CoA dehydrogenase (MCAD) deficiency: It is a fatty acid oxidation disorder. It is caused by mutations in the ACADM gene, affecting the enzyme medium chain acyl-CoA dehydrogenase, crucial for the breakdown of medium-chain fatty acids.

The correct answer is Most primitive living angiosperm

Concept:

• Amborella trichopoda is a species of flowering plant that is widely considered to be the most primitive extant (living) angiosperm (flowering plant).
• It is endemic to New Caledonia and is the only species in the genus Amborella.
• The significance of Amborella lies in its evolutionary position; it represents the earliest branch of the angiosperm family tree.

Explanation:

1. Oldest known fossil of an angiosperm: This option is incorrect because the oldest known fossil of an angiosperm is not Amborella trichopoda. Fossils of angiosperms date back to the Early Cretaceous period, and Amborella is not a fossil but a living plant.
2. Most primitive living angiosperm: This is the correct answer. Amborella trichopoda is considered the most primitive living angiosperm due to its unique characteristics and its position in the phylogenetic tree of flowering plants.
3. Most primitive living vascular plant: This option is incorrect. Vascular plants include a wide variety of plant groups such as ferns, gymnosperms, and angiosperms. The most primitive vascular plants are thought to be ferns or fern-like plants, not Amborella.
4. Oldest known fossil of a seed plant: This option is also incorrect. The oldest known seed plants are gymnosperms, which appeared long before angiosperms. Amborella is an angiosperm, not a fossil of a seed plant.

The Cre-Lox system is a powerful technology used widely in molecular biology for genome editing, including the insertion, deletion, or modification of DNA within an organism's genome.

a. It uses Conservative site-specific recombination by Tyr recombinase

• The Cre-Lox system involves a type of recombination known as conservative site-specific recombination, and it involves a tyrosine (Tyr) recombinase enzyme.
• In this system, Cre (Causes recombination) is the Tyr recombinase that catalyzes the recombination events between two LoxP (Locus of X-over P1) sites.
• This enzyme mediates the precise cutting and rejoining of DNA strands, which does not involve the gain or loss of nucleotides at the site of recombination, hence the term "conservative". So, the correct choice based on the description given is a.

b. It uses Conservative site-specific recombination by Ser recombinase

• While Ser recombinase also mediates site-specific recombination in some biological contexts, the Cre enzyme specifically belongs to the family of Tyr recombinases. Thus, this statement is incorrect as it pertains to the Cre-Lox system.

c. Present in both prokaryotes and eukaryotes

• While the Cre-Lox system was originally identified in bacteriophage P1 (a prokaryotic context), it has been harnessed and is widely used in genetic engineering across a range of organisms, including eukaryotes. The system itself is not naturally present in eukaryotes but has been adopted and utilized in eukaryotic research for genome engineering.
• Thus, while the technology is applied in both prokaryotes and eukaryotes, the system naturally originates from a prokaryote.

d. Lox enzyme is a Tyr- recombinase recognize specific Cre site in bacterial genome so that bacteriophage DNA integrate into the genome of the bacteria

• This choice contains several inaccuracies. Firstly, there is no "Lox enzyme"; LoxP sites are specific DNA sequences recognized by the Cre recombinase enzyme.
• Additionally, the Cre-Lox system's primary role is not the integration of bacteriophage DNA into a bacterial genome but rather to mediate recombination between LoxP sites which can lead to DNA modifications such as deletions, insertions, or inversions.

Conclusion: The most accurate statement regarding the Cre-Lox recombination system is: a. It uses Conservative site-specific recombination by Tyr recombinase.

The correct answer is 1 g/l

Concept:

• Michaelis-Menten kinetics describes the rate of enzymatic reactions by relating reaction velocity to substrate concentration.
• According to the Michaelis-Menten equation:
  
  where:
  is the reaction velocity, V_max is the maximum velocity,[S] is the substrate concentration, and Km is the Michaelis constant (substrate concentration at which the reaction velocity is half of
• The reaction velocity doubles when the substrate concentration increases from 0.5 g/l to 1 g/l.

Explanation:

• When the substrate concentration is significantly higher than K_m the reaction velocity approaches V_max Conversely when the substrate concentration is much lower than K_m the reaction velocity is proportional to the substrate concentration.
• Given that the reaction velocity doubles when the substrate concentration increases from 0.5 g/l to 1 g/l, this suggests that the initial substrate concentration (0.5 g/l) is around the same magnitude as K_m.
• If K_m were much lower than 0.5 g/l (e.g., 0.001 g/l or 0.01 g/l), the reaction velocity would have already been close to V_max at 0.5 g/l and would not double by increasing the substrate concentration to 1 g/l.
• If K_m were much higher than 0.5 g/l (e.g., 1 g/l), the reaction velocity at 0.5 g/l would be significantly less than half of V_max and doubling the substrate concentration to 1g/l would result in a noticeable increase in reaction velocity.
• Therefore, the K_m value that fits this scenario is approximately 1g/l.

Key Points

• Jasmonates form a family of oxylipins arising from the enzymatic oxygenation of 16 and 18-carbon triunsaturated fatty acids.
• Oxylipins are oxygenated fatty acids containing one or more oxygen atoms other than those in the carboxyl group.
• The best-known jasmonates are jasmonic acid (JA), methyl jasmonate and jasmonic acid conjugated to some amino acids such as leucine (JA-leucine) and isoleucine (JA-isoleucine).
• JA is the best-known and best-characterized member of the Jasmonate family.
• Jasmonates play anti-herbivory and antimicrobial functions.
• Mutant plants that don't produce or respond to jasmonates are far more susceptible to insect or necrotrophic pathogen attacks.
• Besides their role in defense, jasmonates also participate in reproductive and vegetative development.

Explanation:

• Jasmonates are key players in plant defense mechanisms, widely accepted in plant biology.
• They are involved in responses to environmental stress and damage, including herbivore attack.
• This hormone activates the expression of defense-related genes, leading to the production of compounds that deter herbivores.

Hence the correct answer is option 4

The correct answer is ATP will inhibit nucleoside diphosphate reductase

• Nucleoside diphosphate reductase is a crucial enzyme in the synthesis of deoxyribonucleotides from ribonucleotides, which is vital for DNA replication and repair.
• Its activity is not directly inhibited by ATP; in fact, ATP can serve as a substrate or in some systems act as an allosteric activator for the enzyme, enhancing its activity.
• The regulation of nucleoside diphosphate reductase is complex and involves balance between different forms of ribonucleotide triphosphates (rNTPs) and deoxyribonucleotide triphosphates (dNTPs), but the statement that ATP will inhibit this enzyme is inaccurately generalized and therefore false.
• The control over nucleotide synthesis, especially in the context of needing more guanine nucleotides, involves several layers of regulation but not the inhibition of nucleoside diphosphate reductase by ATP

Key PointsA. Glutamine-PRPP amidotransferase will not be fully inhibited.

• This statement is true within the given scenario. Glutamine-phosphoribosyl-pyrophosphate (PRPP) amidotransferase is a key regulatory enzyme in purine nucleotide synthesis, catalyzing the first committed step in the pathway.
• It is subject to feedback inhibition by the end products, including both AMP and GMP (adenine and guanine nucleotides, respectively), to regulate purine nucleotide synthesis according to the cell’s needs.
• If the cell has an adequate supply of adenine nucleotides (e.g., AMP, ATP) but requires more guanine nucleotides (e.g., GMP, GDP, GTP), this enzyme would not be fully inhibited by GMP alone, allowing for the continued synthesis of the common precursor IMP (inosine monophosphate), which can then be directed towards the production of more guanine nucleotides.

B. AMP will be a feedback inhibitor of the condensation of IMP with aspartate.

• The specific feedback inhibition by AMP in purine synthesis relates to AMP's ability to regulate the enzyme adenylosuccinate synthetase in its own synthesis pathway from IMP but does not specifically target the condensation of IMP with aspartate, which is an earlier step in the purine synthesis pathway.
• The feedback inhibition mechanisms are designed to balance the levels of purine nucleotides (AMP and GMP) in the cell, focusing on their specific biosynthetic pathways from IMP.

C. ATP will stimulate the production of GMP from IMP.

• This statement is correct. In the biosynthesis of guanine nucleotides from the precursor IMP, specific steps require energy input and are regulated by allosteric effectors.
• ATP serves as an allosteric activator in the conversion of IMP to xanthosine monophosphate (XMP), which is subsequently converted to GMP.
• This regulation allows the cell to increase the synthesis of guanine nucleotides when ATP levels are high and more guanine nucleotides are needed, indicating a balanced supply of energy resources and nucleotide precursors for DNA and RNA synthesis.

Therefore, the correct answer is Option 4

Concept:

• In taxonomy, a lectotype is a specimen selected from among a group of syntypes to serve as the name-bearing type specimen for a taxon when no holotype was designated at the time of the original description.

Type specimen:

• The specimen (or specimens) that is designated as the name-bearing reference for a particular taxon (e.g., species).

There are several types of type specimens, including holotypes, syntypes, and lectotypes.

• Holotype: The single specimen upon which the original description of a new taxon is based and specimen is designated at the time of the original description, and serves as the name-bearing type for the taxon.
• Syntype: One of two or more specimens cited in the original description of a taxon when no holotype was designated.
• Lectotype:
  • A specimen selected from among the syntypes to serve as the name-bearing type when no holotype was designated.
  • When a taxon is originally described and no holotype is designated, the author may instead designate two or more syntypes (i.e., specimens) to serve as the reference for the new taxon.
  • Later, if it becomes necessary to clarify which specimen is the name-bearing type (e.g., because the original syntypes are lost or destroyed), a lectotype can be designated by a subsequent author or authority.
  • The lectotype is selected from among the original syntypes and becomes the name-bearing type for the taxon.

Explanation:

option 1: A type specimen that is selected after the original type specimen is lost or destroyed

• This option describes the definition of a lectotype in botanical nomenclature.
• When the original type specimen designated for a taxon is lost, destroyed, or insufficient, a new type specimen may be designated to serve as the type.
• The new type specimen is chosen from among the remaining original material, and is referred to as a lectotype.
• The lectotype provides a clear reference point for the taxon when the original type specimen is not available for study.

Option 2: A type specimen that is selected from among the syntypes to serve as the single type for a taxo

• This option describes the definition of a lectotype in zoological nomenclature, which is slightly different from botanical nomenclature.
• In zoology, a syntype is one of two or more original specimens that were simultaneously designated as types for a taxon.
• If the syntypes are found to represent more than one taxon or if one syntype is found to be a better representative of the taxon than the others, a lectotype may be selected from among the syntypes to serve as the single type for the taxon.

Option 3: A type specimen that is designated to replace an existing type specimen that is found to be inadequate

• This option is not the definition of a lectotype, but instead describes the concept of a neotype.
• A neotype is a new type specimen that is designated to replace an existing type specimen that is found to be inadequate, for example if it is lost, destroyed, or in poor condition.
• A neotype can also be designated if the original type specimen is determined to be a composite of multiple taxa or is found to represent a taxon other than the one originally described.

Option 4: A type specimen that is selected to serve as the type for a taxon when the original description is unclear or insufficient

• This option describes the definition of a type specimen known as a epitype.
• An epitype is a new type specimen that is designated to clarify the identity of a taxon when the original description is unclear or insufficient to determine the taxon's identity.
• The epitype is chosen from among the remaining original material and serves as the primary reference specimen for the taxon.

The correct answer is Chromatofocusing

• MALDI-TOF (Matrix-Assisted Laser Desorption/Ionization - Time of Flight): This is a mass spectrometry technique that can be used to determine the molecular mass of proteins, peptides, and other macromolecules. The protein molecules are ionized without fragmentation, and their mass-to-charge ratio (m/z) is measured. The molecular mass can be directly deduced from the m/z ratio, making MALDI-TOF a suitable method for determining molecular mass.
• Gel Filtration Chromatography: Also known as size-exclusion chromatography, this technique separates molecules based on size. Smaller molecules enter the pores of the stationary phase and elute later than larger molecules, which are excluded from these pores. While gel filtration chromatography is useful for purification and size separation, it does not directly provide the molecular mass of proteins. Instead, it allows the estimation of molecular mass based on calibration with standards of known molecular weight, which means it’s not as direct or accurate for exact mass determination as mass spectrometry.
• Chromatofocusing: This is a type of ion-exchange chromatography that separates proteins based on their isoelectric point (pI), the pH at which a molecule carries no net charge. While useful for separating proteins with different isoelectric points, it does not provide direct information about the molecular mass of proteins.
• SDS-PAGE (Sodium Dodecyl Sulfate - Polyacrylamide Gel Electrophoresis): This technique separates proteins primarily based on their molecular mass under denaturing conditions. Proteins are treated with SDS, an anionic detergent, which denatures secondary and tertiary structures and applies a negative charge to the proteins in proportion to their mass. When subjected to an electric field, smaller proteins migrate faster through the gel matrix than larger ones. While SDS-PAGE is widely used for estimating the molecular mass of proteins by comparing their migration to that of molecular mass standards, it doesn’t directly measure molecular mass but rather allows for estimation based on relative mobility.

Conclusion:

Therefore, the method that cannot be used to determine the molecular mass of a protein directly and accurately is Chromatofocusing

The correct answer is remove the antibiotic cassette

Concept:

• Genome engineering is a process that involves the modification of an organism's genetic information to achieve desired traits.
• Flippase is an enzyme that recognizes specific DNA sequences known as FRT (Flippase Recognition Target) sites and catalyzes recombination between these sites.
• Antibiotic cassettes are often used in genome engineering to select for cells that have successfully incorporated the desired genetic modification. However, these cassettes need to be removed after selection to avoid interference with subsequent modifications or the organism's normal functions.

Explanation:

• Remove the antibiotic cassette: This is the correct answer. Flippase enzyme recognizes the FRT sites flanking the antibiotic resistance gene (antibiotic cassette) and catalyzes the recombination between these sites, thereby excising the antibiotic cassette from the genome. This step is crucial for subsequent rounds of genetic modifications and for the proper functioning of the organism.
• Add FRT sequence: This is incorrect. FRT sequences are added to the genome through molecular cloning techniques before the Flippase enzyme can act on them. Flippase does not add FRT sequences; it only recognizes and acts upon them.
• Remove FRT sequence: This is incorrect. Flippase does not remove FRT sequences; it facilitates recombination between FRT sites. The FRT sequences generally remain in the genome after the recombination event.
• Add the antibiotic cassette: This is incorrect. The addition of an antibiotic cassette is typically done through cloning and transformation techniques prior to the Flippase-mediated recombination process. Flippase is not involved in adding antibiotic cassettes.

The correct answer is p53.

Apaf-1 (Apoptotic protease activating factor-1) plays a crucial role in the intrinsic (mitochondrial) pathway of apoptosis. It is involved in the formation of the apoptosome, a complex necessary for the activation of caspase-9, which subsequently activates caspase-3 to carry out apoptosis.

In a cell line deficient in Apaf-1, the intrinsic pathway of apoptosis would be disrupted due to the inability to form the apoptosome, leading to resistance to apoptosis triggered by internal signals such as UV radiation. Despite this deficiency, other proteins that are part of or interact with different pathways may still function normally.

1. Cytochrome c: Cytochrome c is involved in the intrinsic pathway and its release from mitochondria leads to apoptosome formation. In the absence of Apaf-1, even if cytochrome c is released, it cannot promote apoptosis effectively.
2. Caspase-9: Caspase-9 is an initiator caspase that is activated by the apoptosome. Without Apaf-1, caspase-9 activation would be impaired.
3. Caspase-3: Caspase-3 is an executioner caspase activated by caspase-9. With Apaf-1 deficiency, caspase-9 is not activated, and thus caspase-3 activation would be compromised.
4. p53: p53 is a tumor suppressor protein that regulates the cell cycle and promotes apoptosis in response to DNA damage, including that caused by UV radiation. p53 functions upstream of the mitochondrial pathway and its activity does not rely directly on Apaf-1. Hence, p53 is most likely still functioning normally in the Apaf-1 deficient cell line

Key Points

• Cytochrome c, caspase-9, and caspase-3 are directly involved in the Apaf-1-dependent intrinsic pathway of apoptosis and would be affected by Apaf-1 deficiency.
• p53 functions upstream and independently of Apaf-1, so its normal function is not directly impacted by Apaf-1 deficiency.

Conclusion:

The correct answer is Option d, because p53 is most likely still functioning normally in the cell line deficient in Apaf-1.

The correct answer is Indohyus.
Indohyus is an extinct genus of a small deer-like, herbivorous animal from the family Raoellidae, which lived about 48 million years ago. In 2007, scientists discovered fossils of Indohyus in the Kashmir region of India, and these fossils provided critical evidence linking it to cetaceans (the group that includes modern whales, dolphins, and porpoises).
Key evidence suggesting that Indohyus was a close terrestrial ancestor of whales includes:

• Bone structure: It had dense bones, similar to those seen in modern aquatic animals that spend a lot of time in water, suggesting it was semi-aquatic.
• Ear structure: The structure of the ear bones in Indohyus resembled that of modern whales, indicating an evolutionary relationship.
• Lifestyle: It is believed to have waded in water and perhaps fed on aquatic plants, which shows an adaptation to aquatic environments, a key transition towards full aquatic life seen in modern whales.

This discovery was significant because it helped scientists understand the evolutionary shift from land-dwelling mammals to fully aquatic cetaceans.
The other options you mentioned are unrelated to the ancestry of whales:

• Jainosaurus: A genus of titanosaurs, a type of herbivorous dinosaur.
• Rajasaurus: A carnivorous dinosaur from India.
• Indosuchus: A genus of theropod dinosaurs, also from India.

Thus, the correct fossil that is considered a close ancestor of whales is Indohyus.

The correct answer is:Several species are involved in coevolutionary interactions.

Concept:

Guild coevolution, also known as diffuse coevolution, refers to coevolutionary interactions that involve multiple species rather than just two. In this type of coevolution, several species interact and exert selective pressures on each other in a more generalized way, rather than in a tightly coupled, pairwise interaction. These species belong to the same guild—a group of species that exploit the same kinds of resources or face similar ecological pressures.

• For example, a group of plants may evolve defenses like toxins or thorns to protect against a guild of herbivores (e.g., insects, mammals), and in response, the herbivores may evolve mechanisms to overcome those defenses. These coevolutionary dynamics occur not between just one plant and one herbivore but across multiple species.

Explanation:

• "One species uses the other as a resource": This describes a trophic interaction like predation or parasitism, not coevolution, which involves reciprocal evolutionary changes.
• "Two species coevolve reciprocally, but only to each other": This describes pairwise coevolution, where two species (such as a specific predator and prey or plant and pollinator) directly influence each other’s evolution. Diffuse coevolution, on the other hand, involves multiple species.
• "A species escapes association from a predator and diversifies. Later, a different predator adapts to the host and diversifies.": While this describes an evolutionary process, it suggests a sequence of events involving specific species, not the broader network of interactions typical of guild coevolution.

Thus, the statement "Several species are involved in coevolutionary interactions" best describes guild coevolution.

Statement A: They protect plants against being eaten by herbivores and against being infected by microbial pathogens.
Secondary metabolites serve various ecological functions, primarily defense mechanisms in plants. They help protect plants from herbivores, insects, and microbial pathogens through different strategies. For example:

• Alkaloids like nicotine and morphine can be toxic or deterrent to herbivores.
• Phenolic compounds like tannins can inhibit microbial infection and deter herbivores.
• Terpenes like pyrethroids have insecticidal properties. These compounds often have bitter tastes, toxic properties, or antimicrobial activities that reduce the likelihood of herbivory and infection.

Statement B: Terpenes, the largest class of secondary metabolites, are synthesized by the methylerythritol phosphate (MEP) pathway and shikimic acid pathway.
Terpenes are the largest class of secondary metabolites, are synthesized by two main pathways:

• Methylerythritol phosphate (MEP) pathway: Operates in the plastids and is responsible for the biosynthesis of monoterpenes, diterpenes, and certain other terpenoids.
• Mevalonate (MVA) pathway: Occurs in the cytosol and is responsible for the biosynthesis of sesquiterpenes, sterols, and other terpenoids.

The shikimic acid pathway is not involved in terpene biosynthesis. Rather, it is used for the production of aromatic amino acids (like phenylalanine, tyrosine, and tryptophan) and many phenolic compounds.

Statement C: The most abundant classes of phenolic compounds in plants are derived from phenylalanine.
Phenolic compounds, such as flavonoids, tannins, lignans, and lignin, are derived from phenylalanine. This amino acid undergoes deamination by the enzyme phenylalanine ammonia-lyase (PAL) to form cinnamic acid and subsequent phenylpropanoids. These phenylpropanoids serve as precursors for a wide variety of phenolic compounds, making phenylalanine a key building block in their biosynthesis.

Statement D: Alkaloids are nitrogen-containing secondary metabolites in plants.
Alkaloids are a diverse group of naturally occurring organic compounds that contain nitrogen atoms. They are known for their wide range of pharmacological effects on animals and humans. Examples include:

• Morphine and codeine (isolated from the opium poppy)
• Caffeine (found in coffee, tea, and other plants)
• Nicotine (from tobacco plants)
• Quinine (from the cinchona tree) These compounds often serve protective roles by deterring herbivores and inhibiting microbial growth.

Therefore, the correct statements are A,C and D

The correct answer is Cholinergic and non-adrenoceptive neurons

Concept:

The experiment aims to understand the role of specific neurotransmitter systems (cholinergic and adrenergic) in regulating function F. By stimulating or inhibiting these systems, researchers can determine their contribution to the overall function.

• Electrical Stimulation: Directly activates neurons in area A, leading to an increase in function F. The involvement of the adrenergic system is suggested by the inhibition of this effect by prazosin.
• Carbachol Injection: Directly activates cholinergic neurons in area A, also leading to an increase in function F. The lack of inhibition by prazosin indicates that the cholinergic pathway is independent of the adrenergic one.

Explanation:

(i) Electrical Stimulation of Brain Area (A)

• Increased Function (F): This indicates that electrical stimulation of area A activates certain neuronal pathways that lead to an increase in function F.
• Effect Prevented by Prazosin: Prazosin is an α1-adrenergic receptor antagonist. The ability of prazosin to prevent the increase in function F suggests that the increase is mediated through adrenergic signaling. Specifically, the stimulation probably activates adrenergic neurons or terminals in area A, leading to the observed effect.

(ii) Injection of Carbachol into Brain Area (A)

• Increased Function (F): Carbachol is a cholinergic agonist, mimicking the action of acetylcholine to stimulate cholinergic receptors. The fact that carbachol directly injected into area A also increases function F indicates the presence of cholinergic receptors involved in mediating this function.
• Not Prevented by Prazosin: Since prazosin does not prevent the carbachol-induced increase in function F, it indicates that the carbachol's effect is independent of adrenergic pathways. This shows the increase in function F is specifically due to activation of cholinergic receptors, not involving α1-adrenergic receptors blocked by prazosin.

The correct answer is Option 2

Concept:

Genetic maps and physical maps provide two different ways of understanding the locations of genes on chromosomes:

• Genetic maps are created based on genetic linkage information, which is derived from how often two genes are inherited together. Genetic linkage is influenced by the frequency of recombination between genes during meiosis.
• The unit of measurement is the centiMorgan (cM), which reflects the probability of recombination occurring between genes. One centiMorgan corresponds to a 1% chance that a marker at one genetic locus will be separated from a marker at another locus due to recombination in a single generation.
• Physical maps are created based on the direct measurement of the DNA sequence, giving the exact number of base pairs between genes. This is measured in kilobases (kb) or megabases (Mb), where 1 kb = 1,000 base pairs, and 1 Mb = 1,000,000 base pairs.

Explanation:

• Recombination does not occur uniformly across the genome. Some areas are more prone to recombination than others, often referred to as "recombination hotspots."
• Conversely, some regions may have very low recombination rates, sometimes due to structural features of the chromosome or the presence of DNA sequences that suppress recombination. Therefore, the recombination frequency, which determines the distances on a genetic map, does not correlate perfectly with the physical distances measured in kilobases on the DNA.
• This results in genetic maps that can be quite different from physical maps, particularly in regions where recombination rates are significantly higher or lower than average.
• The provided image likely shows that some genes, which appear close together on a physical map (short distance in kb), are farther apart on a genetic map (large distance in cM), or vice versa. This is a direct result of the recombination frequency variation across the chromosome.
• If recombination was uniform, the two maps would be more similar, with the distances on the genetic map increasing proportionally with the physical distances. However, due to the factors mentioned above, this is not the case.

Conclusion:

Thus, the discrepancy between the two maps is best explained by the fact that the recombination frequency varies and is not uniform across the chromosome. This affects the relative distances between genes on a genetic map compared to a physical map.

Each individual was tested using two different probes: one for the wild type allele (with its specific reporter dye) and one for the variant allele (with a different reporter dye).

Fluorescence is generated when the Taq polymerase cleaves the probes during the PCR, separating the reporter dye from the quencher, thus allowing the dye to fluoresce.

Individual I shows maximum fluorescence for the dye attached to the wild type probe.

• This means that Individual I has the wild-type allele because the probe specific to the wild-type allele is binding to the DNA and being cleaved by the Taq polymerase, generating strong fluorescence for the wild type dye.
• Therefore, Individual I is homozygous for the wild type allele.

Individual II shows maximum fluorescence for the dye attached to the variant probe.

• This means that Individual II has the variant allele because the probe specific to the variant allele is binding to the DNA and being cleaved by the Taq polymerase, generating strong fluorescence for the variant dye.
• Therefore, Individual II is homozygous for the variant allele.

Individual III exhibits equal fluorescence for both the dyes.

• This means that Individual III has both the wild type and the variant alleles because both probes are binding to the DNA and being cleaved by the Taq polymerase, leading to equal fluorescence from both the wild type and variant dyes.
• Therefore, Individual III is heterozygous, having one wild type allele and one variant allele.

A. RNA molecules can exhibit catalytic activity and act as enzymes, known as ribozymes.

• This statement is correct. Ribozymes are RNA molecules capable of catalyzing specific biochemical reactions, including RNA splicing in gene expression, without the need for additional proteins or enzymes.
• This discovery was initially surprising because, before ribozymes were known, proteins were thought to be the only molecules capable of acting as enzymes.
• Ribozymes highlight the versatile nature of RNA beyond its roles in storing and transmitting genetic information. They also provide evidence supporting the hypothesis of an "RNA world" early in the evolution of life, where RNA served both as genetic material and as a catalyst for biochemical reactions.

B. All DNA polymerases require a primer to initiate DNA synthesis, which can be either DNA or RNA.

• This statement is correct. DNA polymerases cannot start synthesizing DNA de novo; they need a short primer sequence that is hydrogen-bonded to the template strand to provide a free 3' hydroxyl group to which new nucleotides can be added.
• DNA polymerases are enzymes that synthesize DNA molecules from deoxyribonucleotides, the building blocks of DNA.
• They can't start a new DNA chain from scratch but can only add nucleotides to an existing strand or primer.
• Therefore, DNA synthesis is always initiated by a short primer sequence that binds to the template strand.
• This primer provides the free 3'-OH group necessary for the addition of the first nucleotide.
• In biological systems, the primer can be made of either RNA or DNA. For example, in DNA replication, RNA primers are synthesized by the enzyme primase and then extended by DNA polymerase.

C. A-DNA is the biologically active form that exists under physiological conditions.

• This statement is incorrect. B-DNA is considered the predominant and biologically active form of DNA under physiological conditions, which include an aqueous environment and normal cellular ionic strength.
• B-DNA has a right-handed double helix with about 10 nucleotides per helical turn. This structure is stable and functional in most biological contexts.
• On the other hand, A-DNA is a more compact right-handed helical structure that can occur under dehydrated conditions or in high-salt conditions, but it's less common in cells under normal physiological conditions.

D. DNA methylation in eukaryotes occurs predominantly on cytosine bases adjacent to guanine bases, a pattern known as CpG methylation.

• This statement is correct. This is a key epigenetic mechanism involved in regulating gene expression.
• In eukaryotic genomes, methylation typically happens on cytosine residues that are followed by guanine residues, known as CpG sites.
• This methylation can repress gene activity by altering the DNA structure or by inhibiting the binding of transcription factors. CpG methylation plays vital roles in various biological processes, including development, genomic imprinting, X-chromosome inactivation, and the suppression of repetitive elements.

E. 5' capping and 3' polyadenylation are modifications specific to eukaryotic tRNA to stabilize it and prevent degradation.

• This statement is incorrect. These modifications are characteristic of eukaryotic mRNA, not tRNA.
• The 5' cap and 3' poly-A tail are added to mRNA molecules to protect them from degradation, help in their export from the nucleus, and aid in translation initiation. tRNA undergoes different modifications.
• These modifications protect mRNA from degradation, assist in the export of mRNA from the nucleus to the cytoplasm, and are involved in the initiation of translation. tRNA, in contrast, undergoes a different set of modifications, such as the addition of the CCA sequence at its 3' end, which is important for amino acid attachment and recognition by the ribosome.

Conclusion:

Based on the detailed explanation of each statement, the correct combination of all correct statements is A,B and D

Flippase

• Flippase specifically moves phospholipids from the outer leaflet of the phospholipid bilayer to the inner leaflet.
• This movement is ATP-dependent, meaning it requires energy to carry out this function. This is because flippase often works against a concentration gradient, actively transporting phospholipids to maintain or create lipid asymmetry between the two leaflets of the cell membrane.
• The activity of flippase is crucial for the proper distribution of specific lipids within the cellular membrane. For example, maintaining a higher concentration of phosphatidylethanolamine and phosphatidylserine on the inner leaflet is essential for cellular signaling, membrane curvature, and the initiation of certain processes like apoptosis. Flippase helps preserve this asymmetry, which is vital for the cell's normal functioning.

Floppase

• Floppase transfers phospholipids from the inner leaflet to the outer leaflet of the membrane.
• Similar to flippase, floppase's action is also dependent on ATP for energy. This dependence highlights the active transport mechanism for adjusting the lipid composition actively and maintaining the membrane's functional asymmetry.
• This enzyme plays a key role in ensuring that certain lipids are available on the outer leaflet of the cell membrane when needed. This can be important for cell identification, signal transduction, and interactions with the extracellular environment.
• Floppase contributes to the dynamic nature of the cell membrane, enabling the cell to adapt its membrane composition in response to various signals or stress.

Scramblase

• Scramblase facilitates the bidirectional movement of phospholipids between the inner and outer leaflets of the cell membrane.
• Unlike flippase and floppase, scramblase operates without the need for ATP. This ATP-independence suggests that scramblase works to equilibrate phospholipids across the membrane, moving them in both directions, which does not actively create or maintain asymmetry but can disrupt it under certain conditions.
• Scramblase plays a significant role in certain physiological conditions, such as during cell apoptosis, where it helps externalize phosphatidylserine from the inner to the outer leaflet as an "eat-me" signal for phagocytes. Its activity can lead to a more randomized distribution of phospholipids, important for processes that require temporary disruption of membrane asymmetry.

Conclusion:

Flippase and floppase are critical for actively maintaining the asymmetric distribution of phospholipids across the cell membrane, an essential feature for cellular function that requires energy in the form of ATP. In contrast, scramblase acts independently of ATP, facilitating a non-selective, bidirectional movement of phospholipids, which can disrupt or alter this asymmetry under specific conditions.Therefore, the correct answer is Option 3 i.e.A-i,B-iii,C-ii

Option 1: Exposure to 660 nm converts Pr undefined Pfr

• This option is partially correct. The wavelength of 660 nm corresponds to red light, which indeed converts the inactive form of phytochrome (Pr) to its active form (Pfr). However, this statement alone does not fully address the influence on germination directly, which is the focus of the question.

Option 2: Exposure to 730 nm converts Pr undefined Pfr

• This statement is incorrect. The wavelength of 730 nm corresponds to far-red light, which actually converts the active form of phytochrome (Pfr) back to its inactive form (Pr), not the other way around. This process is essentially the reverse of what is described in option 1.

Option 3: Germination is influenced by red light

• This is the correct explanation. The sequence described in the question demonstrates that germination is initially promoted by exposure to red light (660 nm, converting Pr to Pfr), and then inhibited or reversed by exposure to far-red light (730 nm, converting Pfr back to Pr). A subsequent exposure to red light again promotes germination, clearly indicating that the germination process is influenced by red light through the phytochrome system.

Option 4: Germination is not influenced by red light

• This statement is incorrect based on the evidence provided. The experiment clearly shows that germination is influenced by red light, as initial germination occurs after exposure to 660 nm light, and the process can be reversed and then initiated again by alternating light exposures.

Summary: The question and its experimental setup illustrate the role of red light (660 nm) in promoting seed germination through the activation of the phytochrome system from its inactive form (Pr) to its active form (Pfr). The reversal of this process by far-red light (730 nm) and the reactivation by red light confirm that germination is indeed influenced by red light.

• PCR primers must anneal to the complementary strands of the DNA.
• Forward primer binds to the template strand, moving from 3' to 5'. It attaches to the start codon of the template DNA.
• Reverse primer binds to the coding strand, running from 5' to 3'. It attaches to the stop codon of the complementary strand of DNA.

• The primers must extend toward each other, ensuring amplification of the region between them.
• The forward primer matches the 5' end of the given strand.
• The reverse primer matches the reverse complement of the 3' end.

5' TCGGACTT 3' and 5' ATGCCGTA 3'

Forward primer: 5' TCGGACTT 3'

• This sequence matches the 5' end of the provided DNA strand, making it a valid forward primer.

Reverse primer: 5' ATGCCGTA 3'

• This sequence matches the reverse complement of the 3' end of the given strand (complement of 5' TACGGCAT 3').

Therefore, the correct answer is Option 2

A. Acetylcholine decreases heart rate by acting on muscarinic receptors.

• True. Acetylcholine, when released by the parasympathetic nervous system, primarily acts on muscarinic receptors in the heart, specifically M2 muscarinic receptors, leading to a decrease in heart rate.

B. Epinephrine increases heart rate by acting on β1-adrenergic receptors.

• True. Epinephrine (adrenaline), a hormone and neurotransmitter involved in the 'fight or flight' response, increases heart rate by acting on β1-adrenergic receptors located in the heart. Activation of these receptors increases cardiac output.

C. The vagus nerve increases heart rate.

• False. The vagus nerve, part of the parasympathetic nervous system decreases heart rate when activated. It releases acetylcholine, which binds to muscarinic receptors in the heart, thereby slowing down the heart rate.

D. Sympathetic stimulation decreases heart rate.

• False. Sympathetic stimulation actually increases heart rate. The sympathetic nervous system activates β1-adrenergic receptors in the heart through the release of norepinephrine and also through circulating epinephrine, which increases heart rate and cardiac output.

E. Thyroid hormones can increase heart rate.

• True. Thyroid hormones, such as thyroxine (T4) and triiodothyronine (T3), can increase heart rate by upregulating the expression of adrenergic receptors in the heart and enhancing the sensitivity of the heart to catecholamines such as epinephrine. They also directly affect heart muscle, leading to an increase in cardiac output.

Apoptosis, also known as programmed cell death, involves a series of characteristic morphological changes in cells. These include:

A. Formation of blebs on cell surface: This is one of the early signs of apoptosis, where the cell membrane shows irregular bulges known as blebs.

B. Swelling and bursting of cells: This description is more characteristic of necrosis, not apoptosis. In necrosis, cells swell and then burst, releasing their contents into the surrounding area, potentially causing inflammation.

C. Collapse of the cytoskeleton: The cytoskeleton, which helps maintain the cell's shape and internal organization, collapses during apoptosis due to the cleavage of structural proteins by caspases (a family of protease enzymes playing essential roles in programmed cell death).

D. Condensation and fragmentation of nuclear chromatin: Another hallmark of apoptosis is the condensation of the chromatin and the fragmentation of the nucleus, leading to the formation of apoptotic bodies that can be engulfed and digested by phagocytic cells without inducing inflammation.

Conclusion:

The characteristic morphological changes in cells undergoing apoptosis are the formation of blebs on the cell surface, collapse of the cytoskeleton, and condensation and fragmentation of nuclear chromatin.Therefore, the correct answer is A, C, and D