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

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".

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 T-cell anergy

Concept:

• T cells are a type of lymphocyte that play a central role in cell-mediated immunity. They recognize antigens presented by other cells via the T-cell receptor (TCR).
• Co-stimulation is a crucial secondary signal provided by antigen-presenting cells (APCs) necessary for full T-cell activation.
• Without co-stimulation, T cells can become anergic, meaning they become non-responsive to the antigen and cannot proliferate or secrete cytokines effectively.

Explanation:

• Upregulation of B7.1: This is incorrect. B7.1 (CD80) is a co-stimulatory molecule expressed on antigen-presenting cells (APCs) that binds to CD28 on T cells to provide the necessary second signal for T-cell activation. Lack of co-stimulation would not result in upregulation of B7.1 on T cells.
• T-cell apoptosis: This is incorrect. While T-cell apoptosis (programmed cell death) can occur under certain conditions, the absence of co-stimulation typically leads to anergy rather than apoptosis.
• T-cell anergy: This is correct. In the absence of co-stimulation, T cells become anergic, meaning they are functionally inactivated and unable to respond to subsequent encounters with the antigen.
• Upregulation of B7.2: This is incorrect. Similar to B7.1, B7.2 (CD86) is another co-stimulatory molecule on APCs and is not upregulated on T cells in the absence of co-stimulation.

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.

• Nitric oxide (NO) is a small, nonpolar molecule, which allows it to rapidly diffuse across the lipid bilayer of the cell membrane without the need for assistance from membrane proteins.
• The permeability of substances through the plasma membrane is influenced by their chemical nature.
• Gases such as O₂ and CO₂ are very small and nonpolar, allowing them to easily diffuse through the lipid bilayer.
• Nonpolar molecules like steroids can also pass through the lipid bilayer relatively easily, although not as effortlessly as gases.
• Uncharged polar molecules (e.g., water, urea) can permeate the membrane, but their polarity makes this process slower compared to nonpolar substances.
• Charged polar molecules and ions (like Na⁺, K⁺, and Cl^-) are generally impermeable to the lipid bilayer due to their charge and require specific transport mechanisms to cross the membrane.
• Lastly, large charged molecules such as DNA and RNA are the least permeable due to their size and charge, necessitating specialized transport mechanisms for crossing the plasma membrane.
  Therefore, the molecule that is most likely to diffuse through the cell membrane is Nitric oxide (NO)

The extrinsic apoptotic pathway, also known as the death receptor pathway, is triggered by the binding of extracellular ligands to death receptors on the cell surface. One such receptor-ligand pair is the Fas receptor (CD95) and its ligand FasL. Upon binding of FasL to Fas, the receptor trimerizes and recruits the adaptor protein FADD (Fas-associated death domain).

The recruitment of FADD facilitates the formation of the death-inducing signaling complex (DISC). Within the DISC, procaspase-8 molecules are recruited and brought into close proximity, leading to their dimerization and subsequent activation. Active caspase-8 then triggers the apoptotic cascade by directly cleaving and activating executioner caspases, such as caspase-3, and by cleaving BID, which can amplify the apoptotic signal through the intrinsic pathway.

1. Caspase-8:
   • The primary initiator caspase activated in the extrinsic apoptotic pathway upon the binding of FasL to the Fas receptor.
   • This activation leads to the cascade of downstream effector caspases, such as caspase-3, resulting in apoptosis.
2. Caspase-9:
   • This caspase is primarily involved in the intrinsic (mitochondrial) apoptotic pathway and is activated by the apoptosome complex.
3. Caspase-3:
   • This is an executioner caspase activated downstream in both intrinsic and extrinsic pathways to dismantle the cell during apoptosis.
4. Caspase-1:
   • This caspase is primarily involved in the inflammatory response and is known for its role in the processing of pro-inflammatory cytokines.

Conclusion:

Among the options provided, caspase-8 is the initiator caspase primarily activated upon the binding of FasL to the Fas receptor in the extrinsic apoptotic pathway.

The theory of optimal foraging is based on the assumption that natural selection only favours the behaviour which minimizes the energy spent on looking for food and maximizes the energy that the food will provide. In short, the foraging behaviour in which the net energy gained is maximized is favoured by natural selection.

• Optimal foraging theory (OFT) suggests that when animals search for food, they do so in a way that maximizes their net energy intake per unit of time. This means they evolve strategies that balance the calories gained from food against the energy expended in finding and consuming it. The theory is built on the assumption that natural selection has shaped foraging behaviors to enable animals to optimize this balance, thus enhancing their survival and reproductive success.
• While the size of the prey can affect the energy gain from a foraging effort, it's the overall energy efficiency (i.e., the energy gained versus the energy spent in the process, which includes the time and risk involved) that is most critical according to OFT. Hence, statement 2 is not the primary assumption of OFT.
• Statement 3 is also incorrect as OFT does not primarily concern the genetics of an individual but rather the behavior that evolution has shaped across generations in response to environmental pressures.
• Statement 4 combines statements 1 and 2, and since 1 is the correct assumption and 2 is not, statement 4 is also incorrect.

• CRISPR-Cas9 technology leverages the natural defense mechanisms of bacteria, using a guided RNA molecule to identify specific sequences of DNA and the Cas9 enzyme to create precise double-stranded breaks (DSBs) in the DNA at these locations.
• Once the DSB is introduced, the cell's native double-stranded DNA break repair pathways namely the non-homologous end joining (NHEJ) and homology-directed repair (HDR) are activated.
• NHEJ can lead to insertions or deletions (indels) at the cut site, which can disrupt or delete genes, while HDR, when supplied with a repair template, can introduce specific DNA sequence changes.
• This capability makes CRISPR-Cas9 a powerful tool for gene editing, allowing researchers to knock out genes, introduce specific mutations, and study gene functions in a precise and controlled manner, with wide-ranging applications in genetics, biotechnology, and medicine.

Key Points

• The CRISPR/Cas bacterial immune system is used as a simple, RNA-guided platform for highly efficient and targeted genome editing. CRISPR genome editing technology takes advantage of DNA repair. crRNAs guide Cas9 proteins to target sequences, which are recognized through their complementarity to the crRNA. The Cas9 nuclease then introduces a double-strand break in the target DNA.
• CRISPR/Cas9 (Clustered Regularly Interspaced Short Palindromic Repeats/CRISPR-associated protein 9) is a technology that has revolutionized the field of genetic engineering by providing a simple, efficient method for making precise, targeted changes to the genome of almost any organism.
• CRISPR/Cas9 is based on a natural defense mechanism found in bacteria, which uses CRISPR sequences and Cas proteins like Cas9 to defend against viral attacks by slicing up the invading DNA. The Cas9 protein is combined with a guide RNA (gRNA) that has been engineered to match the sequence of the targeted gene. This system can be used for a variety of genetic modifications, such as knockouts, where the targeted gene is disabled, as well as more precise edits, like changing a single base pair.

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 Option 3

• Strigolactones (SLs) represent a class of plant hormones with diverse roles in plant growth, development, and interaction with the environment. They were initially discovered due to their role in stimulating the germination of parasitic plant species like Striga and Orobanche but have since been recognized as fundamental hormonal signals within plants themselves.
• One of the critical aspects of strigolactone biology is their mechanism of action, specifically how they regulate plant architecture and root development. This action primarily involves a receptor protein named DWARF14 (D14).
• Biosynthesis: Strigolactones are synthesized from carotenoids via a series of enzymatic reactions. These hormones are primarily synthesized in the roots but can also be produced in other parts of the plant, including the stems and leaves. Their biosynthesis involves the carotenoid cleavage dioxygenases (CCDs), particularly CCD7 and CCD8, which play crucial roles in the initial steps of strigolactone production.
• Transport: After synthesis, strigolactones can move both upwards and downwards within the plant. They utilize the xylem for transport from roots to shoots and can also move through the phloem, facilitating their multifaceted roles in the plant. This dual transport mechanism allows strigolactones to function both locally and at distant sites from their synthesis point.

Mode of Action:-

• Receptor Binding: The primary mode of action of strigolactones involves their perception by the receptor protein DWARF14 (D14), which is an α/β hydrolase. D14 is capable of binding to strigolactones, and this binding is critical for initiating the strigolactone signaling pathway.
• Signal Transduction: Upon binding to strigolactones, D14 interacts with other proteins, such as the F-box protein D3 in rice (or MAX2 in Arabidopsis), leading to the ubiquitination and subsequent degradation of proteins that negatively regulate the strigolactone signaling pathway, such as DWARF53 in rice and SMXLs in Arabidopsis. This degradation releases the suppression on downstream signaling pathways, enabling the plant to alter its growth and development according to its needs.
• Biological Effects: The consequence of strigolactone signaling includes the inhibition of shoot branching, which allows plants to allocate their resources more efficiently under nutrient-poor conditions. Additionally, strigolactones promote root growth and enhance the establishment of symbiotic relationships with arbuscular mycorrhizal fungi, which help the plant in nutrient acquisition, particularly phosphorus.

Incorrect statements:

1) Strigolactones are not exclusively biosynthesized in the leaves; they are also synthesized in roots. In fact, strigolactones were first recognized for their role in promoting the germination of parasitic plants seeds, like Striga spp., which suggested their synthesis and release from roots.

2) While strigolactones are primarily synthesized in the roots and transported to other parts of the plant, stating that their transportation occurs via the xylem "relying solely on transpiration pull" oversimplifies their transport mechanisms. Strigolactones, like many signaling molecules, may be transported through various means, including both xylem and phloem, depending on their destination and function.

4) This statement is incorrect in suggesting that strigolactones move through the plant via phloem only and do not interact or affect xylem-based transport systems. Strigolactones can be transported through both xylem and phloem. Their systemic movement is complex and tailored to their roles in different plant processes.

Conclusion:

Therefore, the correct statement is Option 3

The correct answer is 2´, 3´ dideoxy nucleotides

• DNA sequencing is defined as the process of determining the sequence of nucleotide bases in a given DNA sample.
• The chain termination method also known as the Sanger method is developed by Fred Sanger. This forms the basis of automated 'cycle' sequencing reactions.
• In the 1980s, two important developments allowed researchers to believe in the quick and accurate production of DNA.
• Applied Biosystems has the first commercialised Sanger method for DNA sequencing. Even in the Human Genome Project, the Sanger sequencing method is used to sequence
• It is also called the chain termination method because it uses dideoxynucleotides ddNTPs. It is an artificial molecule that lacks a hydroxyl group at the 3' position in the ribose sugar.

The following are the requirements of the Sanger sequencing method:

• DNA polymerase enzyme
• Four DNA nucleotides i.e., dATP, dTTP, dCTP and dGTP.
• Primer to start the DNA synthesis.
• Template DNA
• Dideoxynucleotides of all four nucleotides each having conjugated with different dye. Modified nucleotides that are used to terminate DNA synthesis. Unlike normal nucleotides, ddNTPs lack a hydroxyl group (-OH) at the 3' carbon of the sugar ring.

In normal DNA synthesis, nucleotides are added to the growing DNA strand by forming a phosphodiester bond between the 3' hydroxyl group of the last nucleotide in the strand and the 5' phosphate group of the incoming dNTP. This reaction is catalyzed by DNA polymerase.

• Dideoxynucleotides (ddNTPs) used in Sanger sequencing lack this crucial 3' hydroxyl group. When a ddNTP is incorporated into the DNA strand, it prevents the addition of any further nucleotides because there is no 3' hydroxyl group to form a bond with the 5' phosphate group of the next nucleotide. This results in the termination of chain elongation.

Conclusion:

The correct answer is 2', 3' dideoxy nucleotides because it is these modified nucleotides that lack the essential 3' hydroxyl group required for chain elongation, allowing them to act as terminators of DNA synthesis during Sanger sequencing.

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 (1) - (b), (2) - (c), (3) - (d), (4) - (a)

1. Inhibitor of mitochondrial ETC - Rotenone is known to inhibit the electron transport chain in mitochondria.
2. Gravity sensor - Statoliths, in plant cells, are involved in sensing gravity.
3. Stress responsive protein - Osmotin is a plant protein that is induced under stress conditions.
4. Inhibitor of Non-cyclic electron transport - DCMU (3-(3,4-dichlorophenyl)-1,1-dimethylurea) is an inhibitor that blocks electron transport from photosystem II to plastoquinone in photosynthesis .

Corrected table

The correct answer is 20

• Amylase is an enzyme that catalyzes the hydrolysis of starch into sugars such as glucose. The activity of amylase can be measured by the amount of glucose produced over a period of time.
• One unit of amylase activity is defined as the amount of enzyme required to produce 1 µmole of glucose per minute.

Given: 100 µL of extract produces 6 µmoles of glucose in 30 minutes.

• Amylase activity (in units) = amount of glucose produced (µmoles) / incubation time (minutes).
• For 100 µL of extract: Amylase activity = 6 µmoles / 30 minutes = 0.2 units.

Since this activity is for 100 µL of extract, we need to scale it up to the total volume of extract (100 mL).

• Activity in 100 mL of extract = 0.2 units × (100 mL / 0.1 mL) = 200 units.

Given that 10 g of plant material was used to prepare the 100 mL extract, the amylase activity per gram of plant material is calculated as follows:

• Amylase activity (units/g) = Total activity (units) / Amount of plant material (g).
• Amylase activity (units/g) = 200 units / 10 g = 20 units/g.

Therefore, the correct answer is 20.

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: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.

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

Fold purification of an enzyme is calculated by comparing the specific activity (enzyme activity per amount of protein) at different stages of the purification process. The specific activity is calculated using the formula:

Step I: Cell-Free Extract

• Volume: 20 ml
• Total Protein: 100 mg
• Total Activity: 150 micromoles per min
• Specific activity at Step I:

Step II: Ni-NTA Chromatography

• Volume: 4 ml
• Total Protein: 10 mg
• Total Activity: 120 micromoles per min
• Specific activity at Step II:

Fold Purification: Fold purification is determined by comparing the specific activities of the enzyme at different purification steps.

Thus, the fold-purification of the enzyme after Ni-NTA chromatography is 8

The correct answer is Option 3 i.e. i, iv & v

Explanation:

Conservative site-specific recombination is a precise process that does not involve the loss or gain of nucleotides at the recombination site. Instead, it involves the cutting and rejoining of two defined DNA sequences in a way that either inverts or translocates a segment of DNA, depending on the orientation of the sequences recognized by the recombinase enzyme.

i. Recombination between inverted repeats leads to inversion of genes: This is a correct statement with respect to conservative site-specific recombination. When the recombination sites are in an inverted orientation relative to each other on the same piece of DNA, the enzyme-mediated recombination event can lead to the inversion of the DNA segment flanked by these sites. This process is precise and doesn’t involve the loss or gain of DNA but results in a change in orientation of the segment.

ii. Recombination between direct repeats leads to inversion of genes: This statement is incorrect in the specific context of conservative site-specific recombination between direct repeats. When the recombination sites are in the same orientation (direct repeats), recombination can lead to the deletion of the segment between the repeats if they are on the same molecule or duplication/integration into another molecule if they are on separate molecules. It does not lead to inversion.

iii. Conservation DNA synthesis by doubling of genes by ser-recombinase: This statement misinterprets the nature of conservative site-specific recombination. The term "conservative" in this context means that the overall process is precise and does not involve the net loss or gain of DNA nucleotides at the site of recombination. It does not refer to the doubling of genes. While serine recombinases (a type of recombinase enzyme characterized by its catalytic mechanism) do mediate conservative site-specific recombination, they do not inherently result in the duplication of genes through this process.

iv. Conservation of energy by recombinase: While energetically efficient,including those involved in site-specific recombination, do require energy to catalyze the cutting and rejoining of DNA strands, typically using ATP as an energy source. However, the statement about "conservation of energy" is vague and doesn't specifically elucidate a unique aspect of the conservative site-specific recombination process.

v. Involves both single-strand and double-strand breakage: This is correct regarding the mechanism of action of recombinase enzymes in conservative site-specific recombination. The recombination process can involve making both single-strand cuts (creating nicked intermediates) and double-strand breaks in the DNA as part of the mechanism to cut and rejoin the DNA at specific sites. The precise actions depend on the type of recombinase enzyme and the specific recombination event.

Therefore, the most accurate answer from the provided options, considering the conventions of conservative site-specific recombination, is i, iv & v

Concept:

• The Arabidopsis flower is one of the best understood plant developmental systems.
• The flower consists of four organ types arranged in concentric whorls.
• Identities of the different organ types are established by three classes of floral organ identity genes, the actions of which are best summarized by the ABC model of flower development.
• This model stipulates that these genes function in a combinatorial fashion to specify organ types, such that whorl 1 sepals are specified by A activity, whorl 2 petals by A and B activities, stamens in the third whorl by B and C activities, and the central carpels by C activity
• An example of the A class gene is APETALA1 (AP1), which also is a meristem identity gene.
• The B class genes consist of APETALA3 (AP3) and PISTILLATA (PI).
• These three organ identity genes encode MADS domain transcription factors and are expressed in spatially restricted patterns within the developing flower

• AP1 plays a key role in specifying the identity of the floral meristem and acts redundantly with LFY to activate the expression of the downstream B class floral homeotic genes AP3 and PI
• AP1 is expressed throughout the developing flower from floral stage 1 to early stage 3.
• Later in development, its expression is downregulated in the central dome and is restricted to the outer two whorls.
• In contrast, AP3 and PI start to be expressed during floral stage 3 in a subset of AP1-expressing cells.
• Although it has been shown that the expression of AP1 at high levels in the epidermal layer causes cell autonomous activation of AP3, ubiquitous expression of AP1 under the control of the 35S promoter does not alter floral morphology and therefore is unlikely to alter AP3 and PI expression.
• Together, these results suggest that the activity of the AP1 protein must be regulated during early flower development so that the expression domains of AP3 and PI are narrower than those of AP1.
• The β-glucuronidase (GUS) reporter gene system is an important technique with versatile uses in the study of flower development.
• Transcriptional and translational GUS fusions are used to characterize gene and protein expression patterns, respectively, during reproductive development.
• Additionally, GUS reporters can be used to map cis-regulatory elements within promoter sequences and to investigate whether genes are regulated posttranscriptionally.
• The EMBRYONIC FLOWER (EMF) genes EMF1 and EMF2 are required to maintain vegetative development and repress flower development.
• EMF1 encodes a putative transcriptional regulator, and EMF2 encodes a Polycomb group protein homolog.
• The floral induction pathway in Arabidopsis has been under intense investigation, and multiple floral signal transduction pathways have been identified
• The research study in the query above suggests that AP1 is involved in stimulation of flowering in the emf background

​Hence correct answer is option 4

Substrate-level phosphorylation during glycolysis involves the direct transfer of a phosphate group to ADP to form ATP, bypassing the need for an electron transport chain. The molecules directly involved in substrate-level phosphorylation during glycolysis are:1, 3-bisphosphoglycerate & Phosphoenolpyruvate

A. 1,3-bisphosphoglycerate (1,3-BPG)

• In the glycolytic pathway, 1,3-BPG is generated from glyceraldehyde 3-phosphate (G3P) in a reaction catalyzed by the enzyme glyceraldehyde 3-phosphate dehydrogenase. Subsequently, 1,3-BPG donates a phosphate group to ADP to form ATP and 3-phosphoglycerate (3-PG) in a reaction catalyzed by phosphoglycerate kinase. This is one of the steps where substrate-level phosphorylation occurs.

B. Glucose 6-phosphate (G6P)

• Glucose 6-phosphate is the product of the first step of glycolysis, where glucose is phosphorylated by hexokinase or glucokinase using ATP. This step does not involve substrate-level phosphorylation but rather consumes ATP.

C. Phosphoenolpyruvate (PEP)

• In one of the final steps of glycolysis, PEP is converted to pyruvate by the enzyme pyruvate kinase. During this conversion, a phosphate group from PEP is transferred to ADP to form ATP. This represents another instance of substrate-level phosphorylation within the glycolytic pathway

D. Fructose 1,6-bisphosphate (F1,6BP)

• Fructose 1,6-bisphosphate is formed from fructose 6-phosphate in a reaction catalyzed by the enzyme phosphofructokinase-1 (PFK-1). This is a key regulatory step of glycolysis that consumes ATP and does not result in the production of ATP through substrate-level phosphorylation.

Diagram:

Conclusion:

Therefore, the correct statements are A and C

Statement A: RNase H cleaves the RNA strand in RNA-DNA hybrids, facilitating the synthesis of complementary DNA (cDNA): Incorrect

• FEN (Flap Endonuclease) has a specific role in DNA replication and repair. It primarily participates in removing RNA primers and processing Okazaki fragments during lagging strand synthesis in DNA replication. However, it does not directly facilitate the synthesis of cDNA from RNA-DNA hybrids.
• RNase H is an endonuclease that specifically degrades the RNA strand of RNA-DNA hybrids. This activity is essential during reverse transcription, where it removes the RNA template strand after the synthesis of complementary DNA (cDNA) by reverse transcriptase.

Statement B: DNA polymerase I from E. coli has 3' to 5' exonuclease activity but lacks 5' to 3' exonuclease activity: Incorrect
DNA polymerase I from E. coli possesses both:

• 5' to 3' exonuclease activity for removing RNA primers during replication.
• 3' to 5' exonuclease activity for proofreading.
• Therefore, the statement is incorrect because DNA polymerase I does have 5' to 3' exonuclease activity.

Statement C: T4 DNA polymerase is commonly used for end-repair of DNA fragments during library preparation for next-generation sequencing (NGS): Correct
T4 DNA polymerase has both 3' to 5' exonuclease activity and polymerase activity. It is widely used in molecular biology to convert DNA overhangs into blunt ends by:

• Removing single-stranded overhangs (via exonuclease activity).
• Filling in recessed 3' ends (via polymerase activity).

This step is crucial during DNA library preparation for NGS.

Statement D: Cas9 is an RNA-guided endonuclease that introduces double-strand breaks at specific locations in the genome: Correct

• Cas9, derived from the CRISPR system, uses a guide RNA (gRNA) to target complementary DNA sequences. It introduces precise double-strand breaks at these locations, enabling genome editing.

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