CSIR NET Life Sciences Mock Test - 5 - UGC NET MCQ

Given:

m + n + mn = 118

Calculations:

m + n + mn = 118

Adding 1 on both sides,

=> m + n + mn + 1 = 118 + 1

=> m + mn + n + 1 = 119

=> m(1 + n) + 1(1 + n) = 17 * 7

=> (m + 1)(1 + n) = 17 * 7

=> m + 1 = 17

=> m = 17 - 1

=> m = 16

1 + n = 7

=> n = 7 - 1

=> n = 6

m + n = 16 + 6

m + n = 22

∴ The answer is 22.

Concept:

Basic Concepts and Pipes & Cisterns formula:

• If a pipe can fill a tank in a hrs, then the part filled in 1 hr =1/a.
• If a pipe can empty a tank in b hrs, then the part of the full tank emptied in 1 hr = 1/b.
• If a pipe can fill a tank in a hrs and the another pipe can empty the full tank in b hrs, then the net part filled in 1 hr, when both the pipes are opened =[1/a - 1/b] ∴ Time taken to fill the tank, when both the pipes are opened = ab/(b - a)
• If a pipe can fill a tank in a hrs and another can fill the same tank in b hrs, then the net part filled in 1 hr, when both pipes are opened = [1/a + 1/b] ∴ Time taken to fill the tank = ab/(a + b)
• If a pipe fills a tank in a hrs and another fills the same tank in b hrs, but a third one empties the full tank in c hrs, and all of them are opened together, the net part filled in 1 hr = [1/a+ 1/b-1/c] ∴ Time taken to fill the tank =abc/(bc + ac – ab) hrs.
• A pipe can fill a tank in a hrs. Due to a leak in the bottom it is filled in b hrs. If the tank is full, the time taken by the leak to empty the tank = ab/(b - a) hrs.

Calculation:

P → 6 hrs (Fill), Q → 9 hrs (Fill), R → 12 hrs (Empty)

In 1 hr, (P + R) can fill: [1/6 - 1/12]

In 4 hrs, (P + R) can fill:

After 4 hrs, P is closed and Q is opened for 6 hours:

In 6 hrs, (R + Q) can fill:

Total tank filled in 10 hours (4 + 6 hrs):

After 6 more hours, R is closed. So the remaining 1/2 part of the tank will be filled by Q only.

To fill 1/2 part, Q will take: 9/2 = 4.5 hrs

Total time taken to fill the tank = 10 + 4.5 = 14.5 hrs

The pattern followed here is:

Today is Monday

What day will it be after 61 days.

Let's find the odd days = 61 ÷ 7 = 5 Odd days

⇒ Monday + 5 Days = Saturday

So, Saturday falls after 61 days.

Hence, the correct answer is "Saturday".

If,
Number of Row equals to Number of Column (R = C).
Then,
Total squares = 1² + 2² + 3² + 4² + - - - - - - - - - - n²
Here, n = Total number of column.
In given question,
In Square ABCD

Number of Row (R) = 4
Number of Column (C) = 4
Number of Row = Number of Column = 4
Then,
Total squares = 1² + 2² + 3² + 4²
Total squares = 1 + 4 + 9 + 16
Total squares in ABCD = 30
In square EFGH,

Number of Row (R) = 2
Number of Column (C) = 2
Number of Row = Number of Column = 2
Then,
Total squares = 1² + 2²
Total squares = 1 + 4
Total squares = 5
We count two squares in square ABCD so we have to subtract it from square EFGH
So,
Total squares in EFGH = 5 – 2
Total squares in EFGH = 3
Similarly,
In square IGFK,
Total squares in IGFK = 3
In square KLMN,
Total squares in KLMN = 3
In square GNOP,
Total squares in GNOP = 3
Now adding all the squares = Squares in ABCD + Squares in EFGH + Squares in IGFK + Squares in KLMN + Squares in GNOP
Total squares = 30 + 3 + 3 + 3 + 3
Total squares = 42
Hence, 42 is the correct answer.

Calculation:

⇒ The total number of red and blue ball = 8 + 12 = 20

⇒ The total number of after 5 balls replacement = 20 - 1 + 5 = 24

⇒ Probability to first ball is res colour = 8/20

⇒ Probabaility of second ball is blue colour = 12/24

⇒ The required probability = (8/20) × (12/24) = 1/5

∴ The required result will be 1/5.

Additional Information

The probability = (The required events)/(Total events).

Alternate MethodGiven:

A bag contains 8 red balls and 12 blue balls

One ball Is drawn at random and replaced with 5 green balls. A second ball was drawn without replacement.

Calculation:

Probability of getting 1st ball as red = 8/(8 + 12) = 8/20

Probability of getting 2nd ball as blue = 12/(8 + 12 + 5 – 1) = 12/24

∴ Required probability = 8/20 × 12/24 = 1/5

LED bulbs sold by
Y in April = (2/5) × 15% = 6%
Y in May = (4/5) × 10% = 8%
Y in June = (11/20) × 10% = 5.5%
∴ Total bulbs sold by Y from April to June = 6 + 8 + 5.5 = 19.5%
Bulbs sold by X and Y firm from April to June = 35% of (total bulbs sold by X and Y from January to June) = 50000
∴ The number of bulbs sold by Y from April to June = (50000/35) × 19.5% = 27857.14

The correct answer is 7.8
Given:
A 1 m long rod having diameter of 12 mm weighs 880 g.
Calculation:
First, convert the dimensions to consistent units:

• Length of rod = 1 m = 100 cm
• Diameter of rod = 12 mm = 1.2 cm
• Radius of rod (r) = Diameter / 2 = 1.2 cm / 2 = 0.6 cm

Next, calculate the volume of the cylindrical rod:

• Volume (V) of cylinder = πr²h
• V = π × (0.6 cm)² × 100 cm
• V ≈ 3.14 × 0.36 cm² × 100 cm
• V ≈ 113.04 cm³

Now, calculate the density (ρ):

• Density (ρ) = Mass / Volume
• ρ = 880 g / 113.04 cm³
• ρ ≈ 7.78 g/cm³

Therefore, the density of the material of the rod is approximately 7.8 g/cm³.

The correct answer is 1/64

• Probability that an offspring will be of blood group A = 1/4
• Probability that an offspring will be of blood group AB = 1/2

Therefore, probability that an offspring will NOT be of blood group A or AB = 1 - (1/4 + 1/2)

= 1 - 3/4

= 1/4

For three children, the probability that NONE of them will be of blood group A or AB:

= (1/4) × (1/4) × (1/4)

= 1/64

Therefore, the required probability is 1/64.

The correct answer is Migration → Ecesis → Aggregation → Reaction → Stabilization

Concept:

• Succession is the process by which the structure of a biological community evolves over time.
• There are several stages in the process of succession, including migration, ecesis, aggregation, reaction, and stabilization.
• The sequence in which these stages occur is crucial for the development of a stable and mature ecosystem.

Explanation:

• Migration: This is the initial stage where species from surrounding areas move into the new habitat. These pioneer species are typically hardy and can survive in the harsh conditions of the new environment.
• Ecesis: Following migration, the species that have arrived must establish themselves in the new habitat. This involves germination, growth, and reproduction.
• Aggregation: As more individuals of the species establish themselves, they start to form a community. This aggregation leads to an increase in population density and interactions among the species.
• Reaction: The community of species begins to modify the environment. These modifications can include changes in soil composition, microclimate, and other abiotic factors, making the habitat more suitable for other species to invade and establish.
• Stabilization: In the final stage, the ecosystem becomes stable and mature. The species composition becomes relatively constant, and the community reaches a dynamic equilibrium.

The correct answer is Monophyletic taxon and Polyphyletic taxon

Concept:

• In biological classification, the terms "clade" and "grade" are used to describe different types of groupings of organisms.
• A "clade" refers to a group of organisms that consists of a common ancestor and all its descendants, forming a single branch on the tree of life. This is also known as a monophyletic group or monophyletic taxon.
• A "grade" refers to a group of organisms that share a similar level of organizational complexity or adaptation, but do not necessarily share a common ancestor. This is referred to as a polyphyletic group or polyphyletic taxon.

Explanation:

• Monophyletic taxon: A group of organisms that consists of all the descendants of a common ancestor. An example would be the class Mammalia, which includes all mammals that share a common ancestor.
• Polyphyletic taxon: A group of organisms that are grouped together based on similarities but do not share a recent common ancestor. An example would be the grouping of flying animals like bats, birds, and insects.

Other Options:

• Taxon and Category: These are general terms in taxonomy where "taxon" refers to any group of organisms in a classification system, and "category" refers to a rank or level in the classification system. They are not specifically related to "clade" and "grade".
• Division and Phylum: These terms are taxonomic ranks in the classification hierarchy. "Division" is commonly used in plant taxonomy, while "Phylum" is used in animal taxonomy. They do not relate directly to the concepts of "clade" and "grade".
• Synonyms and Tautonyms: These terms relate to naming conventions in taxonomy. "Synonyms" are different names for the same species, and "tautonyms" are scientific names where the genus and species names are identical. They are not related to the concepts of "clade" and "grade".

The malate-aspartate shuttle is a biochemical system that transports reducing equivalents (specifically electrons) from the cytosol into the mitochondrial matrix. This shuttle is crucial for facilitating the transfer of NADH produced during glycolysis, which occurs in the cytosol, into the mitochondria where it can be used in the electron transport chain for ATP production.

A) It generates FADH₂ during the transport of reducing equivalents across the mitochondrial membrane.

• False. The malate-aspartate shuttle does not generate FADH₂; rather, it helps transport electrons from NADH in the cytosol to NADH in the mitochondria, which is then used to produce ATP via oxidative phosphorylation.

B) It operates primarily in skeletal muscle and brain tissue.

• False. While the malate-aspartate shuttle does operate in various tissues, it is particularly prominent in the liver, heart, and kidneys. In contrast, the glycerol-3-phosphate shuttle is more commonly utilized in skeletal muscle.

C) It transports reducing equivalents from the cytosol to the mitochondrial matrix by converting oxaloacetate to malate.

• True. In this shuttle, oxaloacetate in the cytosol is converted to malate by the enzyme malate dehydrogenase, using NADH as a reducing agent. The malate is then transported into the mitochondria, where it is converted back to oxaloacetate, regenerating NADH in the mitochondrial matrix.

D) It generates NADH when oxaloacetate is converted back to malate in the cytosol.

• False. When oxaloacetate is converted to malate in the cytosol, NADH is consumed (not generated) because the conversion is catalyzed by malate dehydrogenase using NADH as a reducing equivalent.

Conclusion: The correct statement that describes the malate-aspartate shuttle is: It transports reducing equivalents from the cytosol to the mitochondrial matrix by converting oxaloacetate to malate.

The correct answer is Glutamate

Concept:

• Eye is a photosensor that can even detect a single photon and transmit that signal to the brain.
• Cells of retina are the visual photoreceptors of the eyes.
• The visual cell contains two main parts: outer segment and inner segment

1. Outer segment - it consists of the light absorbing visual pigment.
2. Inner segment - it consists of nucleus, mitochondrial and other organelles that support the functions of the outer segment.

• Cilium or ciliary process connects the inner and outer segments of the retina.
• Terminals present in the inner segment have synapses with the horizontal cells and bipolar cells which in turn have synapses with the ganglion and amacrine cells.
• Visual cells are of two types- rod cells and cone cells.

1. Rod cells - it consists of an elongated outer segment. It contains rhodopsin pigment which is responsible for dim-light vision also called scotopic vision.
2. Cone cells - it consists of a dome-shaped outer segment. It has photoreceptors for day-light vision also called photopic vision.

Explanation:

• Rods and Cones Mechanism: In darkness, both types of cells are depolarized, which is contrary to many other neurons that are depolarized (activated) by stimuli. This depolarization occurs because, in the absence of light, a cyclic GMP (cGMP)-dependent mechanism keeps sodium channels open, allowing the influx of sodium ions.
• Release of Glutamate: While depolarized in darkness, rods and cones continuously release the neurotransmitter glutamate into the synaptic gap between themselves and the next layer of retinal neurons (bipolar cells and horizontal cells). Glutamate is a major neurotransmitter in the central nervous system, including the retina, where it acts as the primary excitatory neurotransmitter.
• Effect of Light: When light strikes the photopigments in the outer segments of rods and cones, it triggers a biochemical cascade that leads to the conversion of cGMP into GMP, closing the sodium channels and hyperpolarizing the cell. As a result, the release of glutamate is reduced. This decrease in glutamate alters the activity of bipolar cells and horizontal cells, leading to the generation of visual signals that are processed by the brain.
• Glutamate's Role in Visual Processing: The modulation of glutamate release by light conditions is crucial for visual processing. By reducing glutamate release when light is present, rods and cones enable a dynamic response to varying light conditions, which is essential for visual acuity and contrast detection.

Loss-of-function mutations in tumor suppressor genes lead to uncontrolled cell division because these genes normally act to inhibit cell growth and prevent tumor formation. When a tumor suppressor gene is mutated and loses its function, the regulation of the cell cycle is compromised, potentially leading to cancer. For example, well-known tumor suppressor genes include TP53, RB1, and PTEN. These genes are crucial for controlling various processes like DNA repair, cell cycle arrest, and apoptosis. Mutations that inactivate tumor suppressors eliminate these important control mechanisms, allowing cells to proliferate uncontrollably.

• Oncogene: These are genes that, when mutated or overexpressed, promote cell division and survival. Oncogenes result from the activation (gain-of-function) of proto-oncogenes, not from loss-of-function mutations.
• Proto-oncogene: These are normal genes that can become oncogenes when mutated or overexpressed. Proto-oncogenes typically promote cell growth and division. Loss-of-function mutations in these genes do not lead to cancer; instead, their overactivation does.
• DNA repair gene: These genes are involved in the correction of DNA damage. While mutations in DNA repair genes can contribute to cancer by allowing genomic instability, they are not typically classified as causing uncontrolled cell division directly.

Conclusion:

The correct answer is Tumor suppressor gene since loss-of-function mutations in these genes remove critical regulatory controls on cell division, leading to uncontrolled proliferation and cancer development.

The correct answer is Genes for mannopine synthesis and mannopine catabolism

Concept:

• Agrobacterium rhizogenes, a soil bacterium, is known for its ability to transfer part of its DNA (T-DNA) into plant cells, leading to the formation of hairy roots.
• The Ri (Root-inducing) plasmid contains genes that facilitate this transformation, with specific genes located within and outside the T-DNA region.
• Genes located within the T-DNA region typically include those responsible for the synthesis of plant growth regulators like auxins and cytokinins, which alter the plant's hormone balance to support the growth of the hairy roots.
• Outside the T-DNA region, the plasmid contains genes that benefit the bacterium, such as those involved in opine catabolism.

Explanation:

• Auxin Synthesis Genes: These genes within the T-DNA promote root initiation and elongation but are not involved in providing a nitrogen source directly to Agrobacterium.
• Cytokinin Synthesis Genes: These genes within the T-DNA promote cell division and shoot formation but are similarly not involved in nitrogen provision for Agrobacterium.
• Mannopine Synthesis and Mannopine Catabolism Genes:
  • The synthesis genes are located within the T-DNA region and induce the plant to produce mannopine and agropine, which are types of opines.
  • The catabolism genes, located outside the T-DNA region, allow Agrobacterium to utilize these opines as a source of carbon and nitrogen, providing a growth advantage in the niche created by the transformed plant tissue.
• Auxin and Cytokinin Synthesis Genes: These genes contribute to plant hormone alterations but do not directly provide nitrogen to Agrobacterium.

Additional Information:

• Opines: Opines are low molecular weight compounds synthesized by transformed plant cells under the influence of T-DNA genes. Different types include mannopine, nopaline, and octopine, which serve as nutrient sources for Agrobacterium.
• Ecological Niche Advantage: The ability to catabolize opines gives Agrobacterium a competitive advantage over other soil microbes, as it can exploit these unique nutrient sources created by its own genetic transformation of the plant.

• Electrophoresis is a laboratory technique used to separate DNA, RNA, or protein molecules based on their size and electrical charge.
• In this method, an electric field is applied to move the molecules through a gel matrix.
• Pulsed-field gel electrophoresis (PFGE) is a specific type of electrophoresis designed to separate very large DNA segments, which are often too large to be separated by standard gel electrophoresis techniques.
• PFGE is used to separate large DNA fragments that cannot be resolved by conventional gel electrophoresis.
• It works by periodically changing the direction of the electric field, allowing large DNA molecules to reorient and migrate through the gel more effectively.
• This technique is particularly useful in the analysis of large genomes, such as those of bacteria and yeast, and in DNA fingerprinting for epidemiological studies.
• Two-dimensional electrophoresis:
  • This technique separates proteins in two steps: first by isoelectric focusing (based on charge) and then by SDS-PAGE (based on size).
  • It is primarily used for protein analysis and is not suitable for separating large DNA segments.
• High-resolution electrophoresis:
  • This term generally refers to electrophoresis techniques that provide high resolution for separating small differences in molecule size or charge.
  • It is not specifically designed for very large DNA segments.
• Polyacrylamide gel electrophoresis (PAGE):
  • PAGE is commonly used for separating small to medium-sized proteins and nucleic acids.
  • It is not effective for resolving very large DNA segments due to the limitations in gel pore size.

• Affinity chromatography is a method of separating biochemical mixtures based on a highly specific interaction between antigen and antibody, enzyme and substrate, receptor and ligand, or protein and nucleic acid.
• It is a type of chromatographic laboratory technique used for purifying biological molecules within a mixture by exploiting specific interactions between the molecule of interest and a binding material (immobilized ligand) attached to a column.
• The principle behind affinity chromatography is that the target molecule to be purified will bind to the immobilized ligand based on specific interactions, while other components of the mixture will not.
• To elute, or release, the bound protein from the column, you can use a buffer containing a high concentration of free ligand. The free ligand competes with the immobilized ligand for binding to the target protein. Because the free ligand is present in high concentration, it will effectively outcompete the immobilized ligand, leading to the detachment of the target protein from the column.
• 0.1 mM NaCl: This would not specifically disrupt the protein-ligand interaction in most affinity chromatography setups.
• Acetonitrile: It is commonly used in reversed-phase chromatography as a part of the mobile phase to elute proteins based on their hydrophobicity, not in affinity chromatography which relies on specific biological interactions.
• Janmobilized ligand: This seems to be a typographical error. Nonetheless, neither an immobilized ligand nor modifications of this concept would directly apply to eluting proteins in the context described for affinity chromatography.

Conclusion:

Thus, free ligand is the correct answer for how proteins bound to an affinity column are commonly eluted by exploiting specific competitive interactions between the immobilized and free forms of the ligand.

Gastrulation is a critical phase of early embryonic development across many organisms, marking the period when the embryo begins to form distinct layers (germ layers) which will later differentiate into specific tissues and organs. In zebrafish, gastrulation involves complex movements and reshaping of cells that set the foundation for the organism's body plan, including the establishment of the anterior-posterior and dorsal-ventral axes.

Epiboly: Epiboly is a specific process during zebrafish gastrulation characterized by the spreading and thinning of the blastoderm (the layer of cells surrounding the yolk) to eventually envelope the yolk cell completely. This movement begins around the sphere stage of zebrafish development and involves several types of cell movements and shape changes, such as radial intercalation (where cells rearrange themselves from multiple layers into fewer layers, thereby covering more surface area) and deep cell epiboly (involvement of deep, more interior cells in the movement).

• The blastoderm itself is initially positioned at the animal pole of the embryo, covering only a portion of the yolk cell. As epiboly progresses, the blastoderm extends over and around the yolk, eventually encapsulating it entirely by the end of gastrulation. This process ensures the correct positioning of the embryonic germ layers- ectoderm, mesoderm, and endoderm relative to each other and to the yolk, which serves as a nutrient source for the developing embryo.

Mechanisms and Cellular Coordination: Epiboly in zebrafish is a highly coordinated event that requires interactions among various cell types, including the enveloping layer (EVL), which is the outermost layer of the blastoderm, and the deep cells, which form the bulk of the blastoderm. Marginal cells at the blastoderm edge play a pivotal role in initiating and maintaining the movement. Furthermore, changes in cell shape, adhesions between cells, and cytoskeletal reorganizations are essential for the forces driving epiboly. The precise regulation of these cell and molecular events ensures the successful completion of epiboly, setting the stage for further developmental processes.

• Triticum aestivum, commonly known as bread wheat, is a hexaploid species with the genomic composition AABBDD. It was formed through a series of hybridization events involving three different species:
• Triticum urartu contributed the 'A' genome. Triticum urartu is a diploid species with the genomic composition AA and is recognized as the primary donor of the A genome in hexaploid wheat.
• Triticum turgidum (a tetraploid species with the genomic composition AABB) was formed through hybridization of Triticum urartu (A genome donor) and a species with the B genome (believed to be related to Aegilops speltoides). This species subsequently contributed the A and B genomes to hexaploid wheat.
• Aegilops tauschii (or Aegilops squarrosa) contributed the 'D' genome during a later hybridization with Triticum turgidum, leading to the formation of hexaploid bread wheat.

The correct answer is Carbon, Nitrogen,Phosphorus

• The Redfield ratio is fundamentally an index of concentration that represents the ratio of carbon (C), nitrogen (N), and phosphorus (P) in marine phytoplankton.
• It typically exists in a molar ratio of 106:16:1 (C:N:P).
• This ratio is significant because it reflects the elemental composition most commonly found in oceanic phytoplankton and is used to predict the nutrient dynamics of the ocean.

Components of the Redfield Ratio:

• Carbon (C): An essential element for all living organisms, carbon forms the backbone of organic molecules.
• Nitrogen (N): A critical component of amino acids, proteins, and nucleic acids, nitrogen is essential for the growth and reproduction of phytoplankton and other marine organisms.
• Phosphorus (P): A key part of ATP (adenosine triphosphate), nucleic acids, and phospholipids, phosphorus is vital for energy transfer and genetic information in cells.

The correct answer is Option 1

Concept:

• Proteins are complex molecules that can unfold or denature when exposed to certain conditions, such as changes in temperature, pH, or the presence of chemicals like urea.
• Urea disrupts the hydrogen bonds in proteins, leading to the unfolding of the protein structure.
• The process of protein unfolding can be monitored by measuring changes in properties like absorbance, fluorescence, or circular dichroism as the concentration of urea increases.

Explanation:

• Option 1: This graph typically shows a sigmoidal curve representing the unfolding of a protein. Initially, the protein remains folded with low concentrations of urea. As the concentration of urea increases, the protein starts to unfold, leading to a sharp increase in the measured property. At high concentrations of urea, the protein is fully unfolded, and the curve plateaus.
• Option 2: This graph might show a linear relationship, which is incorrect because protein unfolding is a cooperative process and typically does not follow a linear pattern.
• Option 3: This graph could represent an immediate and complete unfolding at low urea concentrations, which is also incorrect as protein unfolding generally occurs over a range of urea concentrations.
• Option 4: This graph might show no significant change, suggesting that urea has no effect on the protein, which contradicts the known denaturing properties of urea.

Key Points

• When analyzing the expression of a hypothetical gene using Northern and Western blot hybridizations under control and induced conditions, the two techniques provide complementary information about gene expression at the RNA and protein levels, respectively.

Northern Blot Hybridization:

• Principle:
  1. Northern blotting is a technique used to detect and analyze specific RNA molecules in a sample.
  2. It involves the separation of RNA molecules using gel electrophoresis, followed by their transfer to a solid support membrane (typically a nitrocellulose or nylon membrane).
  3. The RNA molecules on the membrane are then hybridized with a labeled DNA or RNA probe specific to the target gene of interest.
• Analysis:
  1. By performing Northern blot hybridization, you can determine the relative abundance and size of the mRNA transcript of the hypothetical gene under control and induced conditions.
  2. The intensity of the hybridization signal corresponds to the amount of mRNA present in each condition.
  3. Additionally, the size of the hybridized bands indicates the transcript's molecular weight, which can provide insights into alternative splicing or processing events.

Western Blot Hybridization:

• Principle:
  1. Western blotting is a technique used to detect and analyze specific proteins in a sample.
  2. It involves the separation of proteins using gel electrophoresis, followed by their transfer to a solid support membrane, similar to Northern blotting.
  3. The transferred proteins on the membrane are then probed with specific antibodies that bind to the target protein of interest, which are subsequently detected using labeled secondary antibodies.
• Analysis:
  1. By performing Western blot hybridization, you can assess the presence, abundance, and potential changes in protein expression of the hypothetical gene under control and induced conditions.
  2. The intensity of the protein bands in the Western blot reflects the relative abundance of the protein in each condition. Additionally, the molecular weight of the protein bands provides information about the size of the protein product, which can be compared to the expected size of the hypothetical gene's protein product.
• By combining the results of Northern and Western blot hybridizations, you can gain a comprehensive understanding of the gene's expression pattern.
• Comparing the mRNA levels from the Northern blot with the corresponding protein levels from the Western blot allows you to assess the correlation between mRNA transcription and protein expression, providing valuable insights into gene regulation and potential post-transcriptional regulatory mechanisms.

Important Points

Statement P: ​Control at transcription initiation.

• Transcription initiation is a critical step in gene expression regulation. Transcription factors and regulatory elements can bind to the promoter region of a gene and either activate or repress transcription. This control allows for the regulation of gene expression at the level of mRNA synthesis.

Statement Q: Alternative splicing (Q)

• Alternative splicing is a mechanism by which different combinations of exons within a pre-mRNA molecule can be included or excluded, leading to the production of multiple mRNA isoforms from a single gene.
• This process allows for the generation of different protein variants, expanding the diversity of the proteome.

Statement R: Control of translation initiation

• After mRNA is transcribed, it undergoes translation to synthesize proteins.
• The initiation step of translation can be regulated to control the rate of protein synthesis.
• Factors such as eukaryotic initiation factors (eIFs) and upstream open reading frames (uORFs) can influence translation initiation and affect gene expression.

Statement S: Protein stability

• The stability of proteins can be regulated by various mechanisms, including protein degradation pathways. Proteins can be targeted for degradation through processes like ubiquitination and proteasomal degradation, influencing their abundance and lifespan within the cell.

Conclusion:

Therefore, the correct statements about the regulation of gene expression are P,Q,R and S

A. This is correct. Central chemoreceptors, located in the medulla oblongata, are primarily sensitive to changes in the pH of cerebrospinal fluid, which is influenced by the levels of carbon dioxide.

• Central chemoreceptors located in the medulla oblongata are sensitive to changes in pH and CO₂ levels in cerebrospinal fluid, which indirectly reflect CO₂ levels in the blood. Increased CO₂ levels lead to increased respiratory rate and depth to expel CO₂ and regulate pH.

B. This is incorrect. Peripheral chemoreceptors, located in the carotid and aortic bodies, respond primarily to changes in the partial pressures of oxygen (PaO₂) and carbon dioxide (PaCO₂), and the pH of the blood, not directly to arterial blood pressure. They signal the respiratory center to increase breathing rate when oxygen levels are low.

C. This is correct. Peripheral chemoreceptors do respond to increases in carbon dioxide (as well as decreases in oxygen levels and pH), which then contribute to an increase in the breathing rate and depth.

D. This is incorrect. Hypoxemia (low oxygen levels in the blood) primarily increases the activity of peripheral chemoreceptors, not central chemoreceptors. Central chemoreceptors are more responsive to changes in carbon dioxide levels rather than oxygen levels.

Therefore, the correct statements are A and C

The correct answer is Option 3

Explanation:

The graph shows the levels of plasma growth hormone in ng/ml under four different conditions related to the dietary treatment of kwashiorkor, a form of malnutrition typically caused by protein deficiency:

• Protein deficiency (kwashiorkor) – baseline condition without treatment.
• Carbohydrate treatment for 3 days – short-term dietary intervention using carbohydrates.
• Protein treatment for 3 days – short-term dietary intervention using protein.
• Protein treatment for 25 days – prolonged dietary intervention using protein.

Key Points

• Statement 1 is incorrect because the graph shows a response to short-term carbohydrate treatment (a decrease in plasma growth hormone levels) and a less pronounced response to short-term protein treatment.
• After 3 days of protein treatment, there is not a significant decrease in plasma growth hormone levels. It's only after a prolonged period, specifically 25 days, that a significant reduction in these levels is observed. This points to a delayed response to the protein treatment in the context of kwashiorkor, where immediate changes may not be as pronounced or rapid, and that a sustained period of nutritional rehabilitation is needed to observe significant changes in this biomarker.
• In kwashiorkor, which is primarily caused by protein deficiency, immediate nutritional intervention may not instantly normalize the biomarkers of health such as plasma growth hormone levels because the body requires time to heal from the effects of prolonged malnutrition.
• The initial high levels of growth hormone could be due to the body's response to stress and an attempt to mobilize resources to promote growth and protein synthesis when adequate nutrition is not available. Once a consistent supply of protein is established, the body can then begin the process of recovery, resulting in the gradual normalization of growth hormone levels.

Bacterial transformation is a process by which bacterial cells take up naked DNA molecules. If these DNA molecules originate from a bacterium of a different genotype, and are integrated into the recipient bacterium's chromosome, they can bring about a genetic change, a process known as recombination.

Co-transformation is a phenomenon often seen in bacterial transformation where multiple genes can be transferred together if they are physically close on the donor's chromosome, i.e., they are linked.

The genes are co-transformed together because they're linked on the chromosome.

When genes p⁺ and r⁺ are co-transformed, it implies that those genes are close together on the chromosome. The same reasoning is applied to genes q⁺ and s⁺, and t⁺ and s⁺.

Putting these together, we'll find that:

• r needs to between p and q
• q and s need to be together
• s and t need to be together

Thus, the order could be prqst

Key Points

• Endotherms are warm-blooded creatures that maintain a constant body temperature regardless of their environment.
• They create most of the heat they need internally and can increase metabolic heat production when it's cold to keep their body temperature constant.
• Notable examples of endotherms are birds and mammals.
• Yet, an endotherm's metabolic heat production comes at a cost.
• They generally have low production efficiencies due to the significant amount of energy spent maintaining their high body temperature.
• The high metabolism rates endotherms possess means that a larger part of their daily energy intake is being utilized simply for maintaining bodily functions, hence leaving less energy available for growth and reproduction, which in turn denotes lower production efficiency.
• Ectotherms, on the other hand, are cold-blooded animals like reptiles, fishes, and amphibians.
• They do not generate significant metabolic heat and instead, rely on external environmental conditions to regulate their body temperature.
• This allows them to have higher production efficiency since less energy is required to maintain body temperature, leaving more for growth and reproduction.
• Their mode of energy use allows them to survive in areas with lower food availability or in ecosystems with low primary productivity where endothermic creatures might have difficulty thriving due to higher energy needs.

Explanation

• it is known that ectotherms are more energy efficient, which suggests they have high production efficiency. This validates statement B.
• Endotherms, conversely, are less efficient, suggesting they have low production efficiency, thus supporting statement C.
• while endotherms maintain a constant high body temperature but have low production efficiency, ectotherms adjust their body temperature to environmental conditions and have high production efficiency.

Hence the correct answer is option 2

Concept:

• Adrenocorticotropic hormone (ACTH) is a tropic hormone produced by the anterior pituitary.
• The hypothalamic-pituitary axis controls it.
• ACTH regulates cortisol and androgen production.
• Diseases associated with ACTH include Addison disease, Cushing syndrome, and Cushing disease.

• The anterior pituitary produces ACTH.
• It is considered a tropic hormone.
• Tropic hormones indirectly affect target cells by first stimulating other endocrine glands.
• Corticotropin-releasing hormone (CRH) is released from the hypothalamus, which stimulates the anterior pituitary to release adrenocorticotropic hormone (ACTH).
• ACTH then acts on its target organ, the adrenal cortex.

• The adrenal cortex secretes glucocorticoids from the zona fasciculata and androgens from the zona reticularis.
• The secretion of glucocorticoids provides a negative feedback loop for inhibiting the release of CRH and ACTH from the hypothalamus and anterior pituitary, respectively.
• Stress stimulates the release of ACTH.

Explanation:

• ACTH works on G protein-coupled receptors on extracellular membranes on zona fasciculata and zona reticularis of the adrenal cortex.
• cAMP is the secondary messenger system.
• Activation of the g-couple receptor activates adenylyl cyclase, thus increase cAMP production.

• The circadian rhythm influences cortisol secretion.
• The highest levels of cortisol are seen in the early morning, and the lowest levels are in the evening.
• This concept is important for diagnostic testing.
• The immune/inflammatory (I/I) response is influenced by the brain in a major way.
• This is achieved via regulation of peripheral nervous system functions and endocrine responses.
• Among other pathways, the brain regulates this response through the hypothalamic–pituitary–adrenal (HPA) axis, which is activated during stress.
• On the other hand, receptors for a number of hormones, neurotransmitters, and neuropeptides are carried by cells of the immune system, leading to modulation of their responses by changes in neuroendocrine and/or autonomic activity..

The song sparrow's survivorship curve shows a mixed pattern. Initially, it has high juvenile mortality, which is characteristic of Type III curves where many young die off quickly but those that survive the early stage tend to live longer. As the sparrows mature, their survivorship curve becomes linear, which is typical of a Type II curve where the rate of mortality is more constant and not heavily dependent on age. This mixed pattern reflects the complexity of survival strategies and environmental interactions that influence the life history of the species.

Many survivorship curves show components of three generalized types at different times in the life history of a species.

• Options A and C are incorrect because they describe singular types of curves (Type I and Type II, respectively), which do not account for the initial high juvenile mortality followed by a linear decline observed in the song sparrow.
• Option D is incorrect as it suggests that the entire life of the song sparrow is characterized by a Type III curve, which does not transition into another pattern as actually observed.

The correct answer is A and B

Concept:

• Alkaline carbonate treatment at high pH (around 11.5) is commonly used to distinguish between peripheral and integral membrane proteins.
• Peripheral membrane proteins are typically released into the soluble fraction because they are loosely associated with the membrane, often through ionic interactions or hydrogen bonds.
• Integral membrane proteins, on the other hand, remain associated with the membrane (insoluble pellet) because they are embedded in the lipid bilayer through hydrophobic interactions.

Explanation:

• Peripheral proteins (or extrinsic proteins) are bound to membranes primarily by electrostatic interactions and hydrogen bonds and do not interact with the hydrophobic core of the phospholipid bilayer. They are usually bound to the membrane indirectly by interactions with integral membrane proteins or directly by interactions with lipid polar head groups. Most peripheral proteins are soluble in an aqueous solution.

i. Proteins A and B are fractionated into soluble components.

• Proteins A and B are peripheral membrane proteins.
• A and B proteins are found in the soluble fraction, they are likely peripheral membrane proteins, which can be released from the membrane by treatments that disrupt ionic interactions.

ii. Protein C is fractionated into insoluble pellet components.

• Protein C is an integral membrane protein.
• Since protein C remains in the insoluble pellet fraction, it indicates that C is an integral membrane protein, firmly embedded in the membrane.

Genomic imprinting is a fascinating aspect of genetics wherein the expression of genes depends on whether the gene is inherited from the mother or the father. This form of epigenetic regulation leads to the expression of one allele (variant of a gene) while the other allele is silenced. This means that for imprinted genes, the organism relies on a single working copy instead of the usual two (one from each parent.

Genomic Imprinting Basics
At the core of genomic imprinting is the epigenetic modification of DNA and histones (proteins around which DNA is wrapped). These modifications do not change the DNA sequence but influence gene activity. Methylation of DNA is a common form of epigenetic modification involved in imprinting, usually leading to the silencing of the gene.

Bi-Parental Requirement & Disorders
In most cases, having two copies of each gene (one from each parent) provides a form of genetic backup. However, for imprinted genes, because one copy is epigenetically silenced, there is no backup. This monoallelic expression is crucial for normal development for certain genes. If the active copy is lost or mutated, there’s no secondary copy to compensate, leading to specific disorders depending on which gene is affected.

Prader-Willi Syndrome and Angelman Syndrome
Prader-Willi syndrome and Angelman syndrome are examples of disorders that occur due to deletions affecting imprinted genes on chromosome 15. Crucially, whether an individual develops PWS or AS depends on which parent contributes the chromosome with the deletion.

• Prader-Willi Syndrome (PWS): This condition arises when the deletion occurs on the paternal chromosome 15. The specific region contains several genes that are normally only active on the paternal copy. If deleted, the individual lacks these gene functions because the maternal copy in this region is imprinted (silenced).
• Angelman Syndrome (AS): AS occurs when the deletion is on the maternal chromosome 15. A critical gene affected in AS is UBE3A, which is only active on the maternal copy in certain brain regions. Hence, a deletion in the maternal chromosome means no functional UBE3A protein is produced in those brain areas, as the paternal UBE3A gene is silenced.

Consequences of Imprinting and Deletions
The phenomenon of imprinting explains why the loss of a small region on chromosome 15 does not behave as a recessive trait. In classical recessive inheritance, both alleles need to be mutated for a disorder to manifest, because one functional allele is sufficient for normal health. However, in the case of imprinted genes, there is effectively only one functional allele to start with. If that allele is lost due to a deletion, diseases like PWS and AS can occur because the "backup" gene is silenced by imprinting.

Implications
Genomic imprinting highlights the importance of epigenetic modifications and their role in human health and development. It disrupts traditional Mendelian genetics by introducing a layer where the parent-of-origin significantly influences the expression of genetic traits and susceptibility to certain genetic disorders.

Conclusion:

So, the accurate explanation is: "The copy of chromosome 15 from the other parent has genes in the region of the deletion that are imprinted, and thus inactive; in the absence of any active copies of these genes, development cannot proceed normally." Genomic imprinting is a crucial mechanism behind the phenomenon of these disorders and illustrates how not all genetic material is equally expressed from both parents, leading to unique patterns of inheritance for certain conditions.

A. Aldosterone increases sodium reabsorption and potassium excretion in the distal tubules, thereby increasing blood volume and GFR.

• Aldosterone, a mineralocorticoid hormone, acts on the distal convoluted tubule and collecting ducts to promote Sodium reabsorption and Potassium excretion.
• Indirectly, water retention (since water follows sodium osmotically). This increases blood volume, which can enhance renal perfusion and GFR.

B. Vasopressin (ADH) acts on the collecting ducts to increase water reabsorption, leading to concentrated urine and a decrease in GFR.

• Vasopressin (antidiuretic hormone or ADH) increases the permeability of the collecting ducts to water by upregulating aquaporin-2 channels. This results in water reabsorption and concentrated urine.
• However, ADH does not directly reduce GFR. Instead, its effect is more focused on water balance and urine concentration.

C. Angiotensin II directly constricts efferent arterioles to maintain GFR under conditions of low renal perfusion.

• Angiotensin II is part of the renin-angiotensin-aldosterone system (RAAS). It constricts efferent arterioles more than afferent arterioles, maintaining glomerular pressure and filtration even when renal blood flow is low.

D. Atrial natriuretic peptide (ANP) reduces renal vascular resistance, leading to increased GFR and enhanced sodium excretion.

• ANP is released in response to atrial stretch caused by increased blood volume. It dilates afferent arterioles and constricts efferent arterioles, increasing renal blood flow and GFR. It enhances sodium and water excretion (natriuresis).