CSIR NET Life Sciences Mock Test - 4 - UGC NET MCQ

As per the given data,

Let the number be x

Let the error percentage be y%

⇒ (7x/5 – 5x/7)/ (7x/5) = (y/100)

⇒ 7/5 – 5/7 = 7y/500

⇒ 24/35 = 7y/500

⇒ (24 × 500)/(35 × 7) = y

⇒ y = 48.98 %

Since PA ∶ PB = 3 ∶ 2 and AB is diameter which is equal to 10 cm,
⇒ PA = 3/5 × 10 = 6 cm and PB = 10 - 6 = 4 cm
Since AO is the radius of the circle,
∴ OP = 6 - 5 = 1 cm
In the same way,
ND = 6 cm and NC = 4 cm
∴ NO = 6 - 5 = 1 cm
Now applying Pythagoras theorem in ΔOSN,
⇒ OS² = NS² + NO²
Since OS is the radius of the circle, OS = 5 cm
⇒ 25 = NS² + 1
⇒ NS = 2√6
∴ SM = (2√6 - 1) cm

Given:

Speed of Train 1 = 108 km/hr

Length of Train1 = 120 m

Time taken by Train 1 to cross the platform = 10 sec

Speed of Train 2 = 198 km/hr

Time taken by Train 2 to cross the platform = 6 sec

Concept Used:

Speed of train 1 in m/sec = 108 × 5 / 18 = 30 m/sec

Distance traveled by Train 1 in 10 sec = 30 × 10 = 300 m

Speed of train 2 in m/sec = 198 × 5 / 18 = 55 m/sec

Distance traveled by Train 2 in 6 sec = 55 × 6 = 330 m

Calculation:

Length of the platform = Distance traveled by Train 1 in 10 sec - Length of Train1

Length of the platform

⇒300 - 120

⇒ 180

Length of the platform = 180 m.

Now,

Length of Train 2 = Distance traveled by Train 2 in 6 sec - Length of the platform

Length of Train 2

⇒ 330 - 180

⇒ 150

Length of Train 2 = 150 m

∴ The length of the other train is 150 m.

​The TGF-β signaling pathway culminates in the nuclear translocation of a complex comprising phosphorylated R-SMADs and co-SMAD (SMAD4), which then influences gene transcription.

• The TGF-β signaling pathway begins when TGF-β ligands bind to serine/threonine kinase receptors on the cell surface. These receptors are categorized into two types: Type II (TβRII) and Type I (TβRI) receptors.
• The initial step involves TGF-β ligands binding to Type II receptors. This is a crucial initiation event that sets the rest of the pathway in motion.
• After TGF-β binds to TβRII, the next step involves the recruitment of Type I receptors. TβRII phosphorylates TβRI, leading to the activation of TβRI. This phosphorylation is key to transmitting the signal downstream.
• Activated TβRI then phosphorylates receptor-regulated SMADs (R-SMADs). Common R-SMADs involved in TGF-β signaling are SMAD2 and SMAD3.
• Once phosphorylated, R-SMADs form a complex with a common-mediator SMAD, known as co-SMAD (SMAD4).
• The complex of phosphorylated R-SMADs and SMAD4 translocates to the nucleus. This movement into the nucleus is critical for the pathway's role in regulating gene expression.
• Inside the nucleus, the SMAD complex binds to specific DNA sequences and interacts with other transcription factors to regulate the transcription of target genes that control cellular processes such as growth, differentiation, apoptosis, and immune system regulation.

Fig- TGF- β signalling pathway

Key Points

• Statement A: Incorrect: The statement does not suggest that SMAD4 is directly activated by the ligand-receptor interaction. SMAD4 is activated through the formation of a complex with phosphorylated R-SMADs, not directly by the TGF-β receptor interaction.
• Statement B: Incorrect: The initiation of signaling does not begin primarily with Type I receptors; it starts with the TGF-β ligand binding to Type II receptors (TβRII), which in turn recruit and phosphorylate Type I receptors (TβRI).
• Statement C: Correct: This option accurately describes the sequence of events in the TGF-β signaling pathway. It emphasizes the formation of a complex between phosphorylated R-SMADs and SMAD4, their nuclear translocation, and their role in influencing gene transcription, which is in line with the statement's description.
• Statement D: Incorrect: The statement does not mention that the phosphorylation of TβRI by TβRII leads to their degradation. Instead, it highlights the phosphorylation as a step towards activating the SMAD signaling cascade.

Conclusion: Therefore, the correct answer is Option 3

The correct answer is Cell fusion

The immortalization of hybridomas for the production of an unlimited supply of monoclonal antibodies is achieved through a specific process.

• Cell fusion: Hybridomas are created by fusing a specific antibody-producing B-cell with a myeloma cell (a type of cancerous cell). The myeloma cell provides the hybridoma with the ability to grow indefinitely due to its cancerous nature, while the B-cell provides the ability to produce a specific monoclonal antibody. This combination results in a cell line that is both immortal and capable of producing the desired antibody.
• Ectopic expression of telomerase: While the expression of telomerase can help to maintain telomere length and extend cell lifespan, this is not the standard or primary method used for creating hybridomas.
• Transformation by a retrovirus: This method involves using a retrovirus to alter the genetic material of cells, which can lead to immortalization in some contexts. However, it is not the standard technique used in the creation of hybridomas for monoclonal antibody production.
• Introduction of oncogenes: Introducing oncogenes can drive cells to proliferate uncontrollably and can lead to immortalization. Although this is a potential technique for immortalizing cells, it is not the usual method employed for creating hybridomas.

The correct answer is Rice

Concept:

• A unique symbiotic association between a pteridophyte and a cyanobacterium is utilized in agriculture, particularly for the cultivation of rice.
• Pteridophytes, such as Azolla, form a symbiotic relationship with cyanobacteria like Anabaena, which helps in nitrogen fixation, enriching the soil with essential nutrients.
• This association is especially beneficial for rice paddies, where it can significantly enhance soil fertility and reduce the need for chemical fertilizers.

Explanation:

• Jute: Jute is primarily grown for its fiber and does not benefit significantly from the pteridophyte-cyanobacterium symbiotic association. Jute cultivation relies more on well-drained soils and organic matter rather than nitrogen fixation.
• Rice: The symbiotic relationship between Azolla (a pteridophyte) and Anabaena (a cyanobacterium) is widely used in rice fields. The cyanobacterium fixes atmospheric nitrogen, which is then utilized by the rice plants, enhancing growth and yield.
• Soybean: Soybean cultivation relies heavily on symbiotic nitrogen fixation, but it involves a different association with Rhizobium bacteria in root nodules, not with pteridophytes and cyanobacteria.
• Cotton: Cotton cultivation does not typically benefit from the symbiotic association between pteridophytes and cyanobacteria. Cotton plants require a different set of nutrients and soil conditions for optimal growth.

The correct answer is 1.9

Concept:

• Bacterial growth characterized by four phases: lag phase, exponential (log) phase, stationary phase, and death phase.
• During the exponential (log) phase, bacteria divide at a constant rate, leading to exponential growth in their numbers.
• The doubling time is the time required for the bacterial population to double in number during the exponential growth phase.
• The formula to calculate the doubling time (T_d) during exponential growth is given by:

where:

• T is the time period (3 hours in this case),
• Nt is the final concentration (30,000 cells/ml),
• N0 is the initial concentration (10,000 cells/ml).

Explanation:

Given: Initial concentration (N0) = 10,000 cells/ml, Final concentration (Nt) = 30,000 cells/ml, and Time (t) = 3 hours.

Using the formula:

• 
• 
• T_d = 0.903 / 0.477 = 1.9 hrs

Therefore, the correct answer is 1.9 hours.

The correct answer is It can stabilize DNA-histone interactions.

Concept:

• Chromatin is the complex of DNA and histone proteins that packages eukaryotic DNA into a more compact, dense form.
• Nucleosomes, the basic unit of chromatin, consist of a segment of DNA wrapped around a histone octamer (two copies each of H2A, H2B, H3, and H4).
• For various cellular processes such as transcription, DNA repair, and replication, the chromatin structure needs to be dynamic. Specifically, the nucleosomes must be repositioned or restructured to allow access to DNA. Chromatin remodeling complexes, like SWI/SNF, are essential in modulating the structure and organization of chromatin.

The SWI/SNF complex uses the energy from ATP hydrolysis to change nucleosome positioning.

• SWI2/SNF2 Subunit: This subunit has ATPase activity and is homologous to DNA helicases, enabling the complex to exert mechanical force necessary for remodeling the nucleosomes.
• Disruption of DNA-Histone Interactions: The force generated by ATP hydrolysis is used to break DNA-histone contacts, allowing the nucleosome to slide along the DNA or be evicted entirely.
• By displacing or repositioning nucleosomes, the SWI/SNF complex exposes promoter regions and other regulatory elements to transcription factors and RNA polymerase II, facilitating the initiation of transcription.

Explanation:

Tumor Suppressor Role: True

• The SWI/SNF complex play role in tumor suppression. Mutations or alterations in SWI/SNF subunits have been associated with a variety of cancers, indicating its role in maintaining proper cell cycle regulation and preventing oncogenesis.

Co-activator of Transcription: True

• The SWI/SNF complex serves as a co-activator of transcription by remodeling chromatin in a way that makes DNA more accessible to transcription factors and the transcriptional machinery, thereby facilitating the initiation of transcription.

Homology to DNA Helicases: True

• One of the key subunits of the SWI/SNF complex, SWI2/SNF2, has homology to DNA helicases. This subunit provides the ATPase activity necessary for the chromatin remodeling function of the SWI/SNF complex, enabling the repositioning or eviction of nucleosomes.

Stabilizing DNA-Histone Interactions: Not True

• The primary function of the SWI/SNF complex is to destabilize DNA-histone interactions, not stabilize them.
• By using energy from ATP hydrolysis, the SWI/SNF complex disrupts histone-DNA contacts, leading to the repositioning or removal of nucleosomes, thus making DNA more accessible for transcription and other processes.
• Stabilizing DNA-histone interactions would be counterproductive to its main role in chromatin remodeling.

• Character displacement involves a divergence in traits (such as body size) to minimize competition when species are sympatric (living in the same geographic area).
• This divergence does not occur when the species are allopatric (living in separate geographic areas), as there is no direct competition for resources.
• Character displacement is the term used to describe an evolutionary change that occurs when two similar species inhabit the same environment.
• Under such conditions, natural selection favors a divergence in the characters--morphology, ecology, behavior, or physiology of the organisms.
• The best example of Character displacement is in the Galapagos Islands, a group of coexisting finches have distinctly different body and beak sizes as well as beak shapes.
• Moreover, character displacement in traits associated with resource use (ecological character displacement) has been studied largely independently of that in traits associated with reproduction (reproductive character displacement).
• The Ecological compression hypothesis predicts that when two species occur together in narrow sympatry, individuals in the overlap zone will use a smaller range of habitats or compress to reduce competition between species and a larger or unchanged range of prey than individuals in allopatry.
• This phenomena shows results in shorter timescale and does not involve heritable change.

Concept:

• Photoperiodism is an important process that regulates many aspects of plant growth and development, including germination, flowering, and dormancy. Plants use photoreceptors to sense light and to determine the duration of light and dark periods.
• These photoreceptors respond to specific wavelengths of light, such as red and far-red light, as well as blue light.
• Long-day plants, such as spinach and lettuce, require a certain minimum length of daylight to flower.
• They will only flower when the duration of daylight exceeds a certain threshold, which varies depending on the species.
• Short-day plants, such as soybeans and chrysanthemums, require a certain maximum length of daylight to flower.
• They will only flower when the duration of daylight is below a certain threshold.
• In long-day plants, exposure to red light during the dark period can induce flowering, while exposure to far-red light can inhibit flowering
• In short-day plants, exposure to red light can inhibit flowering, while exposure to far-red light can induce flowering

Explanation

• The phenomenon described in the question is called photoperiodism, which is the process by which plants use light to regulate their growth and development.
• Long-day plants are those that require a certain minimum length of daylight to flower, while short-day plants require a certain maximum length of daylight to flower.
• In the case of long-day plants, a flash of red light during the dark period can interrupt the plant's perception of the length of the night and induce flowering.
• This is because red light activates a pigment called phytochrome, which exists in two forms: Pr (the inactive form) and Pfr (the active form).
• Exposure to red light converts Pr to Pfr, which is responsible for the induction of flowering.
• On the other hand, a flash of far-red light during the dark period can reverse the effect of red light and inhibit flowering.
• This is because far-red light converts Pfr back to Pr, which reduces the level of the active form of phytochrome and cancels out the effect of the red light.

Therefore, the correct answer is option 1.

A restriction enzyme that recognizes a specific 4-base sequence is expected to cut DNA at points where this 4-base sequence appears. The probability of any given nucleotide being a specific base (A, T, C, or G) is 1/4, assuming a random distribution of nucleotides in the DNA sequence.

For a 4-base sequence, the probability of finding that specific sequence at any given location on the DNA is

This means, on average, this specific 4-base sequence would appear once in every 256 bases along the DNA molecule. For a bacteriophage with a genome size of 5,000 bp, we would expect this specific sequence to appear, and thus for the restriction enzyme to cut, approximately:

• 5,000 / 256 = 19.53

Since the bacteriophage genome can only be divided into an integer number of fragments, we would expect about 20 fragments to be produced on average by the restriction enzyme that recognizes a specific 4-base sequence. So, the answer is About 20

• Theory of metabolic difference (Riddle): This theory suggests that metabolic differences between male and female embryos determine sex.
• Theory of alternate dominance (Castle): This theory implies that sex is determined by the dominance of alternate alleles.
• Theory of heterozygosity (Correns): This theory posits that the sex determination is influenced by the presence of heterozygous individuals, where one sex is heterozygous and the other is homozygous.
• Quantitative theory (Goldschmidt): This theory suggests that the determination of sex is based on the quantity of specific factors or genes.

Statement 1: IL-2 is critical for the clonal expansion of T cells.

• IL-2 (Interleukin-2) is essential for the clonal expansion of T cells. This cytokine acts as a growth factor for T cells, promoting their proliferation after they have recognized an antigen. When T cells are activated by an antigen, IL-2 is produced and binds to its receptor on the surface of T cells, stimulating their growth and division. This process is vital for mounting an effective immune response.

Statement 2: IL-2 induces the differentiation of naïve T cells into cytotoxic T cells and helper T cells.

• IL 2 primarily supports the growth, proliferation, and survival of T cells. The differentiation of naïve T cells into cytotoxic T cells CD8+ and helper T cells CD4+ is primarily driven by signals received through the T cell receptor TCR upon antigen presentation and the specific cytokine environment created by other cells such as dendritic cells. While IL 2 does play a role in the proliferation and survival of these differentiated cells, it is not the primary cytokine responsible for the initial differentiation process.
• Other cytokines and signaling pathways influence the differentiation process more directly. For instance, the differentiation of naïve T cells into helper T cells (Th1, Th2, Th17) is directed by other cytokines such as IL-12, IL-4, and TGF-β, respectively.
• Meanwhile, the differentiation into cytotoxic T cells also involves signals from antigen-presenting cells and cytokines like IL-7 and IL-15.

Chalcone synthase is a key enzyme in the flavonoid biosynthesis pathway. It catalyzes the first step in the pathway, which is the conversion of p-coumaroyl-CoA and malonyl-CoA to chalcone, the precursor for various flavonoids.

1. Decreased flavonoid production: This statement is correct. A loss of function mutation in the gene encoding chalcone synthase would impede the production of chalcone, thereby reducing the overall production of flavonoids in the plant.
2. Increased flavonoid production: This statement is incorrect. A loss of function mutation would not increase flavonoid production; rather, it would decrease it due to the lack of chalcone necessary for subsequent reactions in the biosynthetic pathway.
3. Increased lignin content: This statement is incorrect. Lignin biosynthesis is a separate pathway from flavonoid biosynthesis. While both may share some common precursors, a loss of function mutation in chalcone synthase specifically impacts flavonoid production, not lignin production.
4. Increased anthocyanin content: This statement is incorrect. Anthocyanins are a type of flavonoid derived from the same biosynthetic pathway. Therefore, a loss of function mutation in chalcone synthase would also decrease anthocyanin content, not increase it.\

Conclusion:

The correct answer is Decreased flavonoid production, as a loss of function mutation in the chalcone synthase gene would reduce the production of flavonoids.

The bacteriophage with a genome composed of single-stranded circular DNA is ØX174.

ØX174:

• Genome Type: Single-stranded circular DNA (ssDNA).
• Family: Microviridae.
• The ØX174 bacteriophage has a unique genome that consists of a single-stranded circular DNA. It infects Escherichia coli and is one of the most well-studied bacteriophages in molecular biology, especially known for its pioneering use in the sequencing of viral DNA.

λ (Lambda):

• Genome Type: Double-stranded linear DNA (dsDNA), which becomes circular upon entry into the host.
• Family: Siphoviridae.
• Bacteriophage λ is well-known for its role in genetic research, particularly in the study of recombination and DNA integration in its lysogenic cycle.

T5:

• Genome Type: Double-stranded linear DNA (dsDNA).
• Family: Myoviridae.
• Bacteriophage T5 infects Escherichia coli and has a complex genome that replicates using a unique mechanism involving partially overlapping DNA fragments.

P1:

• Genome Type: Double-stranded circular DNA (dsDNA).
• Family: Myoviridae.
• Bacteriophage P1 has a relatively large genome that exists as a linear DNA molecule with terminal redundancy in the phage particle, but it circularizes upon infection of the bacterial host.

Fluorescence In Situ Hybridization (FISH) is a molecular cytogenetic technique that allows for the visualization of specific DNA or RNA sequences within cells or tissues using fluorescent probes.

This method involves fixing the cells or tissue samples, denaturing the DNA or RNA to make it single-stranded, and then applying probes that are complementary to the target sequences and labeled with fluorescent dyes. When the probes hybridize (bind) to their target sequences, they can be detected using fluorescence microscopy.

FISH is widely used in research and clinical diagnostics to identify genetic abnormalities, study gene expression, and map the spatial distribution of specific sequences within the genome, providing valuable insights into the genetic basis of diseases and developmental processes.

Key Points

• Fluorescence in situ hybridization (FISH) is a laboratory technique used to detect and locate a specific DNA sequence on a chromosome.
• In this technique, the full set of chromosomes from an individual is affixed to a glass slide and then exposed to a “probe”—a small piece of purified DNA tagged with a fluorescent dye.
• The fluorescently labeled probe finds and then binds to its matching sequence within the set of chromosomes.
• With the use of a special microscope, the chromosome and sub-chromosomal location where the fluorescent probe bound can be seen.
• Fluorescence in situ hybridization (FISH) is a molecular cytogenetic technique that allows the localization of a specific DNA sequence or an entire chromosome in a cell.
• It is utilized to diagnose genetic diseases, gene mapping, and identification of chromosomal abnormalities, and may also be used to study comparisons among the chromosomes' arrangements of genes of related species.
• FISH involves unwinding of the double helix structure and binding of the DNA of all probes attached to a fluorescent molecule with a specific sequence of sample DNA, which can be visualized under the fluorescent microscope.
• By creatively combining chromosome-specific probes with gene-specific probes and antibodies, investigators can use FISH to provide exciting new insights about nuclear architecture.

Diagram:-

The correct answer is: Estuary

Freshwater ecosystems include environments where the water has low salt concentration, typically less than 1%. The main types of freshwater ecosystems are:

• Lentic: Refers to standing or still water bodies like lakes, ponds, and reservoirs.
• Lotic: Refers to flowing water bodies such as rivers and streams.
• Wetland: Areas where water covers the soil, or is present either at or near the surface of the soil for at least part of the year, supporting aquatic plants.

An Estuary, however, is a coastal area where freshwater from rivers and streams meets and mixes with saltwater from the ocean. Estuaries are characterized by brackish water, which is a mixture of freshwater and saltwater, making them part of a coastal or marine ecosystem rather than a purely freshwater ecosystem.

Thus, the environment that does NOT belong to the freshwater ecosystem is: Estuary

The correct answer is It binds to GPCR and activates G protein to slow the heart rate

Concept:

• Acetylcholine (ACh) is a neurotransmitter released by the parasympathetic nervous system.
• In the heart, ACh plays a crucial role in regulating heart rate by affecting the heart's pacemaker cells, particularly in the sinoatrial (SA) node.
• The binding of acetylcholine to its receptors on the pacemaker cells leads to a series of events that ultimately result in the slowing down of the heart rate.

Explanation:

• Option 1: It binds to GPCR and activates G protein to slow the heart rate: Acetylcholine binds to muscarinic receptors, which are G protein-coupled receptors (GPCRs) on the pacemaker cells. This binding activates the G protein, which then activates a potassium channel. The resulting efflux of potassium ions hyperpolarizes the cell, making it less likely to fire, thereby slowing the heart rate. This mechanism is a key way the parasympathetic nervous system reduces heart rate.
• Option 2: It stimulates GABA-activated ion-channel coupled receptor to increase the heart rate: This option is incorrect because acetylcholine does not interact with GABA receptors in the heart. GABA (gamma-aminobutyric acid) is a neurotransmitter that primarily functions in the central nervous system and is not involved in heart rate regulation.
• Option 3: It binds to GPCR and inhibits G protein to slow the heart rate: This option is incorrect because acetylcholine does not inhibit the G protein. Instead, it activates the G protein when it binds to the muscarinic receptors, leading to the slowing of the heart rate.
• Option 4: It inhibits GABA-activated ion-channel coupled receptor to increase the heart rate: This option is incorrect for multiple reasons. Firstly, acetylcholine does not interact with GABA receptors. Secondly, inhibiting GABA receptors is not a mechanism for increasing heart rate. GABAergic mechanisms are not typically involved in heart rate regulation.

The correct answer is 30.8%

• In genetic mapping, a three-point cross is used to determine the order and distance between three genes on a chromosome.
• Recombination frequency is calculated by observing the offspring phenotypes and determining the proportion of recombinant offspring.

Recombination events include single crossovers and double crossovers, and these events help to map the relative positions of genes on chromosomes.

• Parental genotypes: These are the offspring that retain the original combination of alleles from the parents. In this problem, there are 352 parental genotypes.
• Single crossovers: These occur when there is an exchange of genetic material between homologous chromosomes at one point. There are 142 single crossovers in this case.
• Double crossovers: These occur when there are two exchanges of genetic material between homologous chromosomes. There are 6 double crossovers in this problem.

Total number of offspring: The total number of offspring is the sum of parental genotypes, single crossovers, and double crossovers. Therefore, the total is 352 + 142 + 6 = 500.

Number of recombination events = SCOs (142) + 2 × DCOs (6)
= 142 + 12
= 154 recombination events

Total offspring = 500

Recombination frequency:
Percentage recombination = (Number of recombination events / Total number of offspring) × 100
= (154 / 500) × 100
= 30.8%

The correct answer is cGMP

The visual signal transduction cascade is a biochemical process that occurs in the photoreceptor cells of the retina, enabling the conversion of light signals into electrical signals that can be interpreted by the brain. This process is crucial for vision and involves several steps, which are highly regulated and involve a series of molecular interactions and reactions. The degradation of cyclic guanosine monophosphate (cGMP) rather than its synthesis plays a central role in this cascade.

1. Absorption of Light by Rhodopsin

• The process begins when photons of light are absorbed by rhodopsin, a photoreceptive protein found in the membrane of rod cells (and cone cells for different wavelengths/colors). Rhodopsin consists of the opsin protein bound to a light-sensitive retinal molecule (a derivative of vitamin A).

2. Conformational Change and Activation of Transducin

• The absorption of light induces a conformational change in retinal from the 11-cis to all-trans isomer, which, in turn, triggers a change in the opsin protein structure, activating rhodopsin.
• The activated rhodopsin then interacts with and activates transducin, a G-protein located in the photoreceptor cell membrane.

3. Activation of Phosphodiesterase (PDE)

• Activated transducin binds to and activates phosphodiesterase (PDE), an enzyme that hydrolyzes cGMP into GMP (guanosine monophosphate).

4. Decrease in cGMP Levels Leads to Closing of Ion Channels

• In the resting state, cGMP levels are high inside photoreceptor cells, and cGMP binds to and opens cyclic nucleotide-gated (CNG) ion channels, allowing a steady influx of Na+ and Ca2+ ions, thereby keeping the cell partially depolarized.
• The activation of PDE and subsequent degradation of cGMP lead to a rapid decrease in intracellular cGMP levels. As a result, CNG ion channels close, reducing the influx of Na+ and Ca2+ ions.

5. Hyperpolarization and Signal Transmission

• The closure of CNG ion channels causes the cell membrane to hyperpolarize. In the photoreceptors, hyperpolarization leads to a decrease in the release of the neurotransmitter glutamate at the synapses with bipolar and horizontal cells.
• This change in neurotransmitter release alters the activity of bipolar cells, which then transmit the signal to retinal ganglion cells. The axons of these ganglion cells converge and form the optic nerve, carrying visual information to the brain for processing.

Importance of cGMP Degradation
The reduction in cGMP levels and the subsequent steps of this cascade highlight the importance of the precise control of second messengers like cGMP in the visual system. This mechanism allows for a rapid and sensitive response to changes in light intensity, enabling vision in a wide range of lighting conditions. The entire process is a remarkable example of how molecular events translate into physiological responses, such as the sensation of sight.

Conclusion:

In the process, light activates rhodopsin, which then activates the G-protein transducin. This, in turn, activates the enzyme phosphodiesterase, which degrades cGMP. As a result, the reduced concentration of cGMP causes cyclic nucleotide-gated ion channels to close, leading to hyperpolarization of the photoreceptor cells and the transmission of visual signals to the brain.Therefore, the correct answer is Option 3 i.e. cGMP

The correct answer is Sf9

• Cell lines are cultures of cells that can be propagated repeatedly and indefinitely in vitro. They are used in various biological and medical research applications.
• Insect cell lines are derived from insect tissues and are commonly used in research, particularly for the expression of recombinant proteins.
• Sf9 is a well-known insect cell line derived from the ovarian cells of the fall armyworm, Spodoptera frugiperda.

Other Options:

• HEK 293: This is a human embryonic kidney cell line. It is widely used in cell biology and biotechnology but is not an insect cell line.
• DH5α: This is a strain of the bacterium Escherichia coli, used as a host for plasmid DNA cloning. It is not a cell line, and certainly not an insect cell line.
• CHO: This stands for Chinese Hamster Ovary cell line. It is a mammalian cell line widely used in biotechnology for the production of therapeutic proteins, but it is not an insect cell line.

Concept:

• Small chemicals and ions known as second messengers transmit signals from cell-surface receptors to effector proteins.
• They comprise a wide range of chemical species and possess a variety of characteristics that enable them to signal within membranes (for example, hydrophobic molecules like lipids and lipid derivatives), within the cytoplasm (for example, polar molecules like nucleotides and ions), or between the two (e.g., gases and free radicals).
• When cells are activated, second messengers, which are generally present in low concentrations in resting cells, can be created or released very quickly.
• Second messenger levels are precisely regulated in both time and space, and they are enhanced during signaling by enzymatic processes or the activation of ion channels.
• These messengers then rapidly disperse from the source and bind to the target proteins, changing their activity, location, stability, and other characteristics in order to spread signals.

​Explanation:

• Cyclic GMP, diacylglycerol, and inositol triphosphate are all examples of second messengers that are involved in intracellular signal transduction pathways.
• These molecules are produced in response to the binding of a ligand (such as a hormone or neurotransmitter) to a cell surface receptor and serve as intermediaries that help to transmit the signal from the cell surface to the interior of the cell.
• The production of inositol triphosphate and diacylglycerol depends on the phospholipid phosphatidylinositol, which is found in the plasma membrane of cells.
• The fact that it is not created in response to the attachment of a ligand to a cell surface receptor prevents it from being regarded as a second messenger, despite the fact that it is essential to intracellular signal transduction pathways.

Therefore, the correct answer is option 4.

The correct answer is Operant conditioning

Concept:

• Operant conditioning: This is a method of learning that occurs through rewards and punishments for behavior. Through operant conditioning, an association is made between a behavior and a consequence for that behavior.
• B.F. Skinner, a renowned psychologist, conducted extensive research on operant conditioning. One of his famous experiments involved placing a rat in a box, now known as a "Skinner box." The rat learned to press a lever to receive food, thereby reinforcing the behavior of pressing the lever through positive reinforcement.
• Key components of operant conditioning include reinforcement (positive and negative), punishment (positive and negative), and extinction.

Explanation of Incorrect Options:

• Classical conditioning: This type of learning was first described by Ivan Pavlov. It involves creating an association between a naturally occurring stimulus and a previously neutral one. Pavlov’s experiment with dogs, where he paired the sound of a bell with food, causing the dogs to salivate at the sound alone, is a prime example. This is not the correct answer because Skinner's experiment focused on the consequences of behavior, not the association between stimuli.
• Sensitization: This refers to an increased reaction to a stimulus after repeated exposure. It is a form of non-associative learning, where the response to a stimulus becomes stronger over time. Sensitization is not related to the reinforcement or punishment of behavior, making it an incorrect option here.
• Habituation: This is another form of non-associative learning where there is a decrease in response to a stimulus after repeated exposure. For example, a person might stop noticing the sound of a ticking clock after being exposed to it for some time. Like sensitization, habituation does not involve the reinforcement or punishment of behavior, so it is not the correct answer.

The correct answer is 20 mM/s

Michaelis-Menten kinetics describes the rate of enzymatic reactions by relating reaction rate v) to substrate concentration [S]. It is given by the equation:
v = (Vmax [S]) / (Km + [S])

• Vmax is the maximum reaction rate achieved by the enzyme, and Km is the substrate concentration at which the reaction rate is half of Vmax.
• Given in the problem:
  • K_m = 10 mM
  • V_max = 30 mM/s
  • [S] = 20 mM
• v = (30 mM/s x 20 mM) / (10 mM + 20 mM) = 600 / 30 = 20 mM/s

Therefore, the reaction velocity at 20 mM substrate concentration is 20 m M/s .

The correct answer is The lac operon is highly expressed only when both lactose and glucose are absent.

• The lac operon is controlled by both the Lac repressor and the activator protein, CAP: This statement is correct. The Lac repressor inhibits the operon in the absence of lactose, while CAP enhances expression when glucose is low.
• The lac operon is highly expressed only when both lactose and glucose are absent: This statement is incorrect. The lac operon is highly expressed when lactose is present and glucose is absent. In the presence of lactose, the repressor is inactivated, allowing transcription. However, the operon is not activated when both lactose and glucose are absent because CAP cannot stimulate transcription without the absence of glucose.
• The lac operon is highly expressed only when lactose is present and glucose is absent: This statement is correct. For maximum expression of the lac operon, lactose must be present to inactivate the repressor, and glucose must be absent to allow CAP to enhance transcription.
• In the presence of lactose, the repressor cannot bind to the operator: This statement is correct. Lactose binds to the Lac repressor, preventing it from binding to the operator and allowing transcription to occur.

Therefore, the incorrect statement is indeed The lac operon is highly expressed only when both lactose and glucose are absent because the lac operon is not highly expressed under those conditions; it requires lactose to be present and glucose to be absent for high expre

The correct answer is Mortality and reproduction are strongly dependent on population density.
r-selected species are characterized by traits that favor rapid reproduction and quick population growth, particularly in unstable or unpredictable environments. Here’s a breakdown of the traits:

1. Mortality and reproduction are strongly dependent on population density: This statement is NOT characteristic of r-selected species. Instead, r-selected species typically experience less regulation of their mortality and reproduction by population density. They often thrive in conditions where resources are abundant, allowing them to reproduce rapidly without the constraints of high competition or density-dependent factors.

2. Tend to occupy habitats that are unpredictable and/or ephemeral: This statement is characteristic of r-selected species. They often inhabit environments that are unstable, which require quick colonization and rapid life cycles.

3. Thrive in habitats where resource competition is low: This is also characteristic of r-selected species. They often exploit resources quickly and can thrive in environments with little competition.

4. Have superior capabilities to colonize new habitats: This is characteristic of r-selected species. They are typically adapted for rapid dispersal and colonization, allowing them to establish populations in new or disturbed areas quickly.

Conclusion: Therefore, the trait that is NOT characteristic of r-selected tree species is Mortality and reproduction are strongly dependent on population density.

The correct answer is 30

Explanation:

Environmental filtering refers to the process by which certain species from a regional species pool are selected to be part of a local community based on their ability to thrive in specific environmental conditions. This filtering often results in a reduced number of species in the local community compared to the broader regional pool.

Factors to Consider:

1. Regional Species Pool: This is the total number of species available in a larger geographic area, which in this case comprises 60 species.
2. Local Species Pool: This is the subset of the regional pool that can persist and reproduce in the local environment. Due to environmental filtering, not all species in the regional pool will be suitable for the local conditions.
3. Environmental Filtering: The process typically reduces species richness in the local pool. Depending on the environmental constraints (e.g., soil type, climate, competition), many species may be excluded from the local community.

Local Species Pool:

Given that environmental filtering tends to significantly reduce the number of species that can coexist in a specific habitat, it’s reasonable to expect that the local species pool would be much smaller than the regional species pool. In many ecological contexts, the local species pool could be around half or less of the regional pool, especially in more specialized environments.

In this case, a local species pool of 30 species represents a significant filtering from the 60 species available in the regional pool, aligning well with the principles of environmental filtering.

Therefore, the likely local species pool of this community is 30.

To calculate the amount of each primer required for a PCR reaction, let's follow these steps:

Determine the molecular weight of one primer:

• Each nucleotide has an average molecular weight of approximately 300 Da (Daltons).
• Each primer is 20 nucleotides long.
• Molecular weight of one primer = 20 nucleotides x 300 Da/nucleotide = 6000 Da

Convert the molecular weight to daltons per mole:

1 Da = 1 g/mol, so the molecular weight in g/mol is 6000 g/mol.
Convert molarity to mass:

• We need 20 pmoles (20 x 10^-12 moles) of the primer.
• Mass of primer required = Molecular weight x Amount in moles
• Mass of primer required = 6000 g/mol x 20 x 10^-12 mol
• Mass of primer required = 120 x 10^-9 g= 120 ng

A) Incorrect - Transmission electron microscopy (TEM) creates images using electrons transmitted through a sample. The resulting images are traditionally grayscale, showing differences in density and composition. Color can be added artificially during image processing, but it does not use fluorescent labeling. Fluorescent labeling is used in various types of fluorescence microscopy, not TEM.

B) Incorrect - Scanning Electron Microscopy (SEM) does not typically require sectioning of the sample. SEM images the surface of samples by scanning the surface with a focused beam of electrons. The electrons interact with atoms in the sample, producing signals that contain information about the sample's surface topography and composition. Samples may need to be coated with a thin layer of conductive material but sectioning is not a general requirement.

C) Correct - Confocal microscopy indeed uses optical methods to obtain images from a specific focal plane within the specimen and significantly reduces or excludes light (out-of-focus light) from other planes. This selective plane illumination improves the optical resolution and contrast of the images, allowing for the construction of three-dimensional images by stacking individual focal planes.

D) Correct - Differential-interference contrast (DIC) microscopy, relies on the interference of polarized light. It enhances contrast in unstained, transparent samples by detecting differences in the refractive index and thickness between different parts of the sample. The technique uses two beams of polarized light that pass through the sample and interfere with each other, producing an image with a pseudo-3D appearance and highlighting edges and boundaries within the cell or material.

E) Incorrect - Visualization in epifluorescence microscopy is typically achieved through the use of fluorescent dyes or markers, not staining by heavy metal atoms. Heavy metal staining is a technique used in electron microscopy to increase contrast in the images, not in epifluorescence microscopy, which relies on the excitation of fluorescent molecules and detection of the emitted light to visualize components of the specimen.

Conclusion:

From the provided statements, C (correct) and D (correct) are true, while A (incorrect) is an incorrect statement. Thus, the combination containing two correct and one incorrect statement from your options includes A (incorrect), C (correct), and D (correct).

Concept:

• Live-attenuated vaccines: it inject the live version of the pathogen that causes a diseases into the body. The germ is of weakened version so that it does not cjases any symptoms of the diseases and it is unable to reproduce in the body. It is most commonly used for generation of viral vaccines. It is able to produce immune response in a similar way as in the natural infection but individual is not able to pass on the viruses to another. Rotavirus, measles, yellow fever, chickenpox, etc vaccines are example of live-attenuated vaccines.
• Inactivated vaccines: it is developed by using bacteria or viruses that are killed by using some chemical or heat. It does not trigger a strong immune response as seen in teh case of live-attenuated vaccines but it is known to cause fewer side-effects as compared to that of live-attenuated vaccines. Hepatitis A, flu, polio, rabies, etc vaccines are examples of inactivated vaccines.
• Subunit vaccines: In the case where the antigen present on the surface of the viral or bacterial surface triggers the immune response than in that case the vaccines is prepared by isolating the specific antigen from the pathogens. This does not causes many side-effects as it is have great specifically targetted.
• Recombinant vaccines: these vaccines are prepared by using genetic engineering techniques. The gene that codes for the proteins that causes infection is isolated and inserted in the vector to produce recombinant proteins.
• Toxoids vaccines: This vaccines creates immunity against the specific parts of the pathogen that causes diseases and not the entire pathogen and the immune response is specific to this particular toxin. it does not provide life-long immunity.
• DNA vaccines: it is genes that code for the proteins of the pathogen. It consists of of a strong bacterial promoter, genes of interest and the polyadenylated termination sequences. The plasmid is grown in bacteria, purified, dissolved in salt and then injected in the host.
• RNA vaccines: It is mRNA of the pathogen that is injected in the host.

Explanation:

• The diphtheria vaccine is manufactured by inactivation of diphtheria toxin, this inactivated toxin is called a "toxoid." Once injected, the toxoid causes an immune response to the toxin, but, unlike the toxin, it doesn't cause disease.Hence, diphtheria vaccine is toxoids vaccine.
• Similarly, tatanus vaccine is also toxoid vaccine as it is generated by using the toxin.
• Influenze vaccine is generated by killing or inactivation of the virus, hence, influenza shots are conventional inactivated vaccines.
• Hepatitis B vaccine is produced by using genetic engineering. It contains non-infectious antigen of hepatisis B virus that is produced in yeast cells using recombinat DNA technology.
• Similarly foot and mouth disease vaccine is also generated using the genetic engineering technique.

Hence, the correct answer is option 3.