PYQ JUNE 2023 SIFT 1

CSIR NET BIOLOGY (June - 2023 Sift 1)
Previous Year Question Paper with Solution.

21. Invasive species, in general grow very well in a new area that they invade, and often outcompete native species. An explanation for the better growth and propagation of invasive species in comparison of their native couterparts is provided by which one of the following hypotheses?

(a) Ecological niche hypothesis

(b) Intermediate disturbance hypothesis

(d) Biotic resistance hypothesis

Ans. (c)

Sol. The enemy release hypothesis states that invasive species often do well in new environments because they are released from the predation, parasitism, and competition that they faced in their native environments. In their new environments, they may have no natural enemies or competitors, which allows them to grow and reproduce rapidly.

22. Which one of the following characteristics is NOT correct for bryophytes?

(a) All bryophytes are homosporous

(b) Gametophyte is the dominant phase of the life cycle

(d) Water isconducted by hydroidsin all bryophytes

Ans. (d)

Sol. In bryophytes, water is primarily absorbed directly through the plant body, mainly through the leaf-like structures known as "leaves" in mosses and "thalli" in liverworts and hornworts. These structures have a high surface area that allows for efficient water absorption from the surrounding environment.

23. Segregation of alles can occur eitherat anaphase I or anaphase II of meiosis. Which one of the following is an ideal model system for identifying the stage at which allelic segregation occurred?

(a) Arabidopsis thaliana

(b) Drosophila melanogaster

(d) Saccharomyces cerevisiae

Ans. (c)

Sol. Ordered or unordered tetrad and octad

The arrangement of spores within an ascus varies from species to species. In some cases, the ascus provides enough space for the tetrads or octads of spores to randomly mix together. This is known as an unordered tetrad or octad. These occur in fungal species such as S. cerevisiae. By comparison, fungi such as Neurospora crassa produces a very tight ascus that prevents spores from randomly moving around. This can create a linear tetrad or octad. Neurospora crassa is an ideal model system for studying allelic segregation during meiosis.

24. Consider alleles 'A' and 'a' in a population. The frequency of heterozygotes will be highest when:

(a) Frequency of 'A' is more than frequency of 'a'

(b) Frequency of 'A' is less than frequency of 'a'

(d) Frequency of 'A' and 'a' affects the frequency of homozygotes, not heterozygotes

Ans. (c)

Sol. The frequency of heterozygotes (Aa) is highest when the frequency of the two alleles (A and a) are equal.

25. Which one of the follwing statements regarding PEPCase is incorrect?

(a) During the day, C4 PEPCase is inactive whereas CAM PEPCase is active

(b) PEPCase is inactivated by dephosphorylation

(d) The synthesis of PEPCase kinase is modulated by circadian rhythm in CAM leaves.

Ans. (a)

Sol. PEPCase (phosphoenolpyruvate carboxylase) is an enzyme involved in carbon fixation and photosynthesis in certain types of plants. There are two major pathways where PEPCase is utilized: C4 photosynthesis and CAM (Crassulacean acid metabolism) photosynthesis.

26. During glycolysis in plants, alanine and related amino acids are directly produced from which one of the following precursors?

(a) 3-Phosphoglycerate

(b) Phosphoenolpyruvate

(d) Acetyl-CoA

Ans. (c)

Sol.

27. Whnich one of the following is correct regarding zeitgebers?

(a) Have to effect on biological rhythms

(b) Sense biological rhythms

(d) Abolish biological rhythms

Ans. (c)

Sol. Although circadian rhythms are endogenous, they are adjusted (entrained) by external environmental signals (such as exposure to light or a change in temperature) called zeitgebers. The two chief entraining stimuli that synchronize the endogenous clock with the exogenous temporal environment are light and temperature.

28. The following terms represent different methods in phylogenetic tree constructions.

(a) Unweighted Pair Group Method Using Arithmetic Average (UPGMA)

(b) Minimum Evolutio (ME) method

(d) Maximum Likelihood (ML) method

Ans. (c)

Sol. UPGMA and ME are both distance-based methods. Distance-based methods construct phylogenetic trees by calculating the distance between pairs of taxa and then using this information to infer the evolutionary relationships between the taxa. UPGMA is a simple and computationally efficient method, but it can be inaccurate for large datasets. ME is a more accurate method, but it is more computationally expensive.

29. The standard free energy (kJ mol^–1) of hydrolysis of glucose-1-phosphate is :

(a) –40.3

(b) –35.8

(d) –20.9

Ans. (d)

Sol. Glucose-1-phosphate + H₂O Glucose + Phosphate

The standard free energy change for this hydrolysis reaction is -20.9 kJ mol-1, indicating that the reaction is exergonic and thermodynamically favorable under standard conditions. A negative value for indicates that the reaction releases energy.

30. The non-ciliated cuboidal epithelial cells in bronchioles that secrete important defense markers are called

(a) Goblet cells

(b) Basal cells

(d) Club cells

Ans. (d)

Sol. Club cells are non-ciliated cuboidal epithelial cells that are found in the bronchioles. They secrete a number of important defense markers, including surfactant, which helps to prevent the collapse of the alveoli, and defensins, which are antimicrobial peptides that help to fight infection.

31. Which one of the following statements is incorrect?

(a) Avr proteins predominantly have SecA secretion signals

(b) Avr proteins are secreted through the type III secretion system (T3SS)

(d) Some of the components of the T3SS pathway are conserved between animal-and plant-pathogenic bacteria

Ans. (a)

Sol. Avr proteins are secreted through the type III secretion system (T3SS), which is a complex molecular machine that injects proteins into host cells. The T3SS is not part of the SecA secretion pathway, which is a more general mechanism for exporting proteins from bacteria.

32. Which one of the following species of birds is kown to migrate across the Himalayas?

(a) Sarus Crane

(b) Red-vented Bulbul

(d) Bar-headed Goose

Ans. (d)

Sol. One of the most notable aspects of the Bar-headed Goose's migration is its ability to fly over the Himalayan mountain range, which includes some of the highest peaks in the world. These geese are known to fly at extreme altitudes, reaching heights of over 7,000 meters (23,000 feet) during their migration.

33. Which one of the following statements regarding molecular markers for genotyping is Incorrect?

(a) Polymorphism in intronic regions of a gene cannot be used for trait mapping

(b) Codominant molecular markers can be used to detect heterozygosity

(c) Sequence Tagged Microsatellite Sites (STMS) and Simple Sequence Repeat Polymorphisms (SSRPs) are based on polymorphisms in repetitive DNA sequences

(d) Restriction fragment length polymorphisms (RFLPs) and Simple Sequence Repeat (SSRs) are multi-allelic markers.

Ans. (a)

Sol. Polymorphism in intronic regions of a gene can indeed be used for trait mapping. Introns are non-coding regions within a gene, and they can contain genetic variations or polymorphisms that can be associated with specific traits or phenotypes.

34. Which one of the following plant-derived molecules is widely used as an analgesic?

(a) Absicisic acid

(b) Salicylic acid

(d) Gibberellic acid

Ans. (b)

Sol. Salicylic acid is a plant-derived molecule that is widely used as an analgesic. It is the active ingredient in aspirin, which is one of the most commonly used pain relievers in the world. Salicylic acid works by blocking the production of prostaglandins, which are chemicals that are involved in inflammation and pain.

35. Given that Asian Koel is a brood parasite, which one of the following statements is TRUE for this species?

(a) The brood of the bird is usually infested with parasitic wasps

(b) The young ones learn the cells of their foster parents

(d) The call of the species is innate and not learned

Ans. (d)

Sol. Brood parasites, such as the Asian Koel, do not raise their own young. Instead, they lay their eggs in the nests of other birds, who then raise the young as their own. This means that the young Asian Koel birds do not learn their calls from their parents. Instead, their calls are innate, or programmed into their genes.

36. The mode of action of cholera toxin in causing diarrhoea is by :

(a) inactivating G_i protein

(b) continuous activation of adenylyl cyclase

(d) rapidly hydrolyzing GTP to GDP

Ans. (b)

Sol. Mode of action of cholera and pertussis toxins

Cholera toxin Is an enterotoxin secreted by the bacterium Vibrio cholerae. Genes for cholera toxin are present on the integrated phage genome, CTXf. The cholera toxin is an oligomeric complex made up of six protein subunits: a single copy of the A-subunit and five copies of the B-subunlt. The A-subunit has catalytic property that causes ADP- ribosylation of G-proteins. The pentameric protein binds to the ganglioside GM1 presenton the surface of the intestinal epithelium. The A-subunit ribosylates the Arg residue of the a-subunit of Gs protein. ADP ribose is provided by the Intracellular NAD*. This ADP ribosylation alters the a-subunit so that it can no longer hydrolyze its bound GTP, causing it to remain in an active state that stimulates adenylyl cyclase indefinitely. The resulting prolonged elevation in cAMP levels within intestinal epithelial cells causes activation of PKA. PKA phosphorylates the CFTR and Na*-H* exchanger present in the intestinal epithelial cells. This causes a large efflux of chloride and sodium ion and water into the gut, thereby causing the severe diarrhea that characterizes cholera.

37. Which one of the following translation factors is used at the step of translation initiation and defined as anti-association factor for 30S and 50S subunit interactions?

(a) IF1

(b) IF2

(d) RRF

Ans. (c)

Sol. IF-3 Binds to free 30S subunits and prevents premature reassociation of the large and small subunits of the ribosome. Its second function Is to control the ability of 30S subunits to bind to mRNA. It also ensures the fidelity of initiation site selection; select the initiator tRNA for use in initiation.

38. Appressorium is expected to be formed during which one of the following diseases?

(a) Bacterial leaf blight in rice

(b) Bacterial wilt in tomato

(d) Leaf curl disease in tobacco

Ans. (c)

Sol. An appressorium is a specialized structure formed by certain plant pathogenic fungi during the infection process. It is typically formed by fungi causing diseases such as powdery mildew. The appressorium is responsible for anchoring the fungus to the host plant surface and facilitating the penetration of the fungal hyphae into the plant tissue.

39. Which one of the following statements about LINEs present in the human genome is TRUE?

(a) LINE preferentially localize to AT-rich DNA

(b) LINEs cannot transpose independently

(d) The Alu family is the most prominent LINE family in terms of copy number

Ans. (a)

Sol. LINES are autonomous and retrotranspositionally active elements. The most common LINES in human genome constitute the LI LINE family. LI elements are approximately 6 kb in length and, in most species, highly accumulated in the sex chromosomes compared to the autosomes. Some 500000 copies of LI elements occur in the human LINES are 100-400 bp long non-autonomous sequences and thus require the enzymatic machinery of LINES for retrotransposition. They have the structural features of LINES but do not encode their own reverse transcriptase. They are mobilized by reverse transcriptase enzymes that are encoded by LINES residing in the genome. SINES are transcribed by RNA polymerase III. The most common SINES in human genome is Alu elements. Alu elements frequently contain a site for the restriction enzyme Alul and consequently are called Alu elements.

40. A positive association between absolute average individual fitness and population size over some finite interval is known as

(a) Allee effect

(b) Founder effect

(d) Bergman's rule

Ans. (a)

Sol. The Allee effect refers to a positive association between absolute average individual fitness and population size over a certain range. It suggests that individuals in small populations may have reduced fitness due to difficulties in finding mates, cooperation, or other factors.

41. Which one of the following root initials gives rise to the root vascular system, including the pericycle?

(a) Columella initials

(b) Epidermal-lateral root cap initials

(d) Stele initials

Ans. (d)

Sol. Cortical-endodermal stem cells: generate the cortical and endodermal layers.

• Columella stem cells: give rise to the differentiated columella cells (the central portion of the root cap).

• Lateral root cap-epidermal stem cells: give rise to the epidermis/lateral root cap.

• Stele stem cells (or vascular initials')', give rise to the vascular system, including the pericycle.

42. Homeotic selector genes are responsible for the specification of Drosophila body parts. Which one of the following identities would you expect if the ultrabithorax gene is deleted?

(a) The third thoracic segment is transformed to another second thoracic segment and fly with four wings

(b) The third thoracic segment develops halteres

(d) Their first and second thoracic segments fuse and wings are formed on the third thoracic segment

Ans. (a)

Sol. Genes in the BX-C (includes ultrabithorax, abdominal A and abdominal fl) control the fate of posterior development (posterior thorax and abdomen). Gene ultrabithorax is required for the identity of the third thoracic segment (the metathorax); and the abdominal A and abdominal B genes are responsible for the segmental identities of the abdominal segments. Ultrabithorax gene specifically represses the expression of genes that are required for the development of the second thoracic segment, or mesothorax and also represses gene Antp expression in the metathorax. Each homeotic selector gene product switches on the set of genes needed to initiate development of the specified segment. Homeotic selector genes specify the identity of a particular body segment. Loss in function mutation of homeotic genes causes the appearance of a normal appendage or body structure at an inappropriate body position. For example, the body of the normal adult fly contains three thoracic segments, each of which produces a pair of legs. The first thoracic segment does not produce any other appendages, but the second thoracic segment produces a pair of wings in addition to its legs. The third thoracic segment produces a pair of legs and a pair of balancers known as halteres. When the Ultrabithorax gene is deleted, the third thoracic segment (characterized by halteres) is transformed into another second thoracic segment. The result is a fly with four wings.

43. Which of the following statements about site-specific recombinases is NOT true

(a) The Cre recombinase is believed to mediate the circularization of the P1 phage genome during infection of the bacterial host

(b) The integrase cannot mediate integration of the genome into the host genome without the help of accessory proteins

(c) The Hin invertase-mediated recombination event is stimulated by protein-DNA interactions at a 60 bp enhancer sequence

(d) In Xer recombinase-mediated monomerization of chromosomal dimers, the interaction of FtsK with XerCD activates XerC and initiates the recombination process.

Ans. (d)

Sol. Xer recombinases are enzymes involved in site-specific recombination in bacteria. They facilitate DNA rearrangements at specific DNA sequences known as recombination sites or "Xer sites." The XerCD recombinase complex recognizes and binds to the Xer sites on chromosomes.

44. Which one of the following mRNAs is most likely to be exported out of the nucleus?

(a) Spliced RNA associated with the poly A binding and cap binding complex

(b) Mis-spliced RNA with multiple stop codons, for degradation in cytosol

(d) Uncapped and unspliced RNA

Ans. (a)

Sol. The spliced RNA associated with the poly A binding and cap binding complex is most likely to be exported out of the nucleus.

mRNA export is a crucial step in gene expression, where mature mRNA molecules are transported from the nucleus to the cytoplasm, where they can be translated into proteins.

This process involves several molecular factors and complexes.

45. Which one of the following codons is used to code for selenocysteine in Escherichia coli?

(a) UGA

(b) UAA

(d) UCC

Ans. (a)

Sol. There is a few example of context dependent codons also. For example, selenocysteine is coded by UGA and pyrrolysine by UAG. These codons, therefore, have a dual meaning because they are mainly used as stop codons.

46. Based on the image given below, select the option that describes it correctly

(a) Q-banded normal human karyotype

(b) G-banded human karyotype depicting aneuploidy

(d) G-banded normal human karyotype

Ans. (b)

Sol. A G-banded human karyotype is a visual representation of the chromosomes in a person's cells. It is created by staining the chromosomes with a dye and then observing them under a microscope. G-banding allows for the identification of specific bands or regions on each chromosome, which helps in identifying structural abnormalities or aneuploidies.

47. In the context of protein import in the nucleus, which molecules is responsible for releasing the cargo from the importing receptor?

(a) Ras

(b) RhoA

(d) Rock

Ans. (c)

Sol. Another Important protein called Ran, small monomeric GTPases, plays a critical regulatory role In active transport and directionality. Ran can exist in two states, depending on whether GDP or GTP is bound. Ran-GTP occurs in the nucleus and Ran-GDP In the cytosol

48. Which one of the following terms is used for species that exploit the same resources in a similar manner?

(a) Guild

(b) Taxonomic order

(d) Assemblage

Ans. (a)

Sol. Another approach is to subdivide each trophic level into groups of species that exploit a common resource in a similar fashion. Groups of species that exploit the same resources in a similar way are called guilds. For example, hummingbirds and other nectar-feeding birds form a guild of species that exploit the common resource of flowering plants in a similar fashion. The term was originally coined by Dick Root (1967) in his studies of birds with the following definition:

A guild is defined as a group of species that exploit the same class of environmental resources in a similar way. This term groups together species without regard to taxonomic position that overlap significantly in their niche requirements. The guild has a position comparable in the classification of exploitation patterns to the genus in phylogenetic schemes.

49. Recessive lethal alleles are never completely eliminated from the population because

(a) lethal alleles are always conditional in nature

(b) lethal alleles have selecting advantage

(d) they are maintained in the population as heterozygotes

Ans. (d)

Sol. Heterozygotes are individuals who have two different alleles for a gene, one dominant and one recessive. In the case of a recessive lethal allele, the dominant allele will mask the recessive allele and the individual will be unaffected.

50. The extracellular domain of a cell surface receptor (A) was sweitched with the extracellular domain of another receptor (B) to create a chimeric receptor (B-A) Assuming that there is no effect on the functionality of the domains in the chimeric receptor, what is the most likely outcome in the presence of the ligand for receptor B?

(a) The ligand will activate the pathway normally triggered by receptor A

(b) The ligand will activate the pathway normally triggered by receptor B

(d) The chimeric receptor will cause constitutive activation of the signalling pathway

Ans. (a)

Sol. It's important to note that while the ligand will activate the pathway normally triggered by receptor A, there may still be some differences in the downstream signaling due to potential variations in the intracellular domains of the receptors. However, based on the information provided, the most likely outcome is the activation of the pathway associated with receptor A in response to the ligand for receptor B.

51. A temperature sensitive S. pombe mutant exhibits cell cycle arrest both, at the G1to S, as well as at the G2 to M transition phnases. This is possibly a mutant of :

(a) Clb 1

(b) Cyclin B

(d) Cdc2 and Clb3

Ans. (c)

Sol. Examples of selected cyclins and cyclin-dependent kinases (CDKs)

52. Which one of the following factors inhibits renin secretion?

(a) Increased level to plasma catecholamines

(b) Increased blood pressure in the different arterioles leading to glomerulus

(d) Prostaglandins

Ans. (b)

Sol. One of the main triggers for renin secretion is a decrease in blood pressure or blood volume. When there is a decrease in blood pressure or a decrease in sodium delivery to the kidneys, it stimulates the release of renin. Renin then initiates a cascade of events that ultimately leads to the production of angiotensin II, a potent vasoconstrictor that helps to restore blood pressure and fluid balance

53. A patient was injected with Purified Protein Derivative (PPD) to diagnose TB disease or a previous exposure to Mycobacterium tuberculosis. The injected area, when inspected after 48-72 hours was found to develop induration (thick, hardened bulge). Which of the following cells will be predominant at this site?

(a) Neutrophils and mast cells

(b) Helper T cells and macrophages

(d) Natural killer and dendritic cells

Ans. (b)

Sol. Type IV hypersensitivity

Type IV hypersensitivity, commonly called delayed type hypersensitivity, Is the only class of hypersensitive reactions which are triggered by antigen-specific T cells (cell-mediated immunity reactions). This Is mediated by T cell dependent effector mechanisms Involving TH cells, primarily of the TH1 subtype, but in a few cases Tc cells. Antibodies do not play a role in type IV hypersensitivity reactions. On activation, the TH1 cells release cytokines that cause accumulation and activation of macrophages, which, In turn, cause local damage. The tuberculin skin test Is an example of a type IV hypersensitivity. This test Is done by putting a small amount of tuberculin purified protein derivative (PPD) under the top layer of skin. If any person has ever been exposed to the Mycobacterium tuberculosis, skin will react to the antigens by developing a firm red bump at the site within 2 days. It is a standard method of determining whether a person is Infected with Mycobacterium tuberculosis.

54. The solubility of NaCl is greater in water than ethanol. What physical proprty of the solvent governs this difference?

(a) Surface tension

(b) Viscosity

(d) Boiling point

Ans. (c)

Sol. Water has a very high value of dielectric constant, which is a measure of the capacity to neutralize the attraction between electrical charges. Because of this property, water acts as a powerful solvent for salts. Salts such as NaCI dissolve in water due to its high value of dielectric constant. Higher value weakens the attractive forces between oppositely charged ions (such as Na+ and Cl") and therefore NaCI dissociates in Na+ and Cl' in water.

Secondly, when an ion immersed in a polar solvent such as water, it attracts the oppositely charged ends of the solvent dipoles. Both positive and negative ions are thereby surrounded by one or more concentric shells (referred to as solvation spheres) of solvent molecules. Such ions are said to be solvated or, when water is the solvent, to be hydrated. As ions become hydrated, this thus counteracts their tendency to associate in a crystalline lattice.

55. Which one of the following sugars will not reduce Tollen's reagent?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Tollen's reagent is a solution of ammoniacal silver nitrate that is commonly used to test for the presence of reducing sugars. Disaccharides like sucrose, trehalose not capable of reducing ferric or cupric ion are called non-reducing sugar.

56. Which one of the following mammalian species is distributed in evergreen forests?

(a) Nilgai

(b) Black buck

(d) Lion-tailed macaque

Ans. (d)

Sol. Lion-tailed macaques are found in evergreen forests in India and Sri Lanka. They are the only macaque species that is found in evergreen forests. Nilgai, black buck, and cheetah are all found in more open habitats, such as grasslands and savannas.

57. Given below are charcteristic traits found in sun-or shade-acclimated leaves

A. High dry mass per unit area

B. Higher number of chloroplasts per area

C. Lower Chl-a/Chl-b ratio

D. Lower dark respiration per area

E. Higher light harvesting complexes per area

Select the option that has all correct chracteristicsfor shade-acclimated leaves?

(a) A, B and C

(b) C, D and E

(d) B, C, and D

Ans. (b)

Sol.

58. The grizzled squirrel, Ratufa macronra, naturally occurs in

(a) north-east India and Burma

(b) western Himalayas

(d) Andaman and Nicobar islands

Ans. (c)

Sol. The grizzled giant squirrel is a species of squirrel belonging to the genus Ratufa. It is known for its large size and distinctive grizzled or mottled fur. This species is native to specific regions in southern India and Sri Lanka.

In southern India, the grizzled giant squirrel can be found in various forested areas, including the Western Ghats, which are a mountain range running along the western coast of the Indian peninsula. It inhabits both evergreen and deciduous forests, as well as other wooded habitats.

59. The flowers of which one of the following plant species is used by indigenous communities of Central India to make an intoxicant for consumption?

(a) Mahua (Madhuca spp.)

(b) Monkey-puzzle tree (Araucaria spp.)

(d) (Pennisetum spp.)

Ans. (a)

Sol. Mahua (Madhuca) is a genus of trees native to India and found predominantly in Central and Western parts of the country. The tree produces small, fragrant flowers that are highly valued by indigenous communities for their multiple uses. One of the traditional uses of Mahua flowers is to prepare an intoxicating drink.

60. Which one of the following describes an amphisome?

(a) It is an intermediate/hybrid organelle produced through the fusion o9f endosomes with autophagosomes within cells

(b) It is a double-membrane sequestering vesicle that isthe hallmark of the intracellular catabolic process called macroautophagy

(d) It is a vacuole that arises when membranes of the ER sequester parts os the cytoplasm.

Ans. (a)

Sol. The autophagosome may fuse with an endosome. The product of the endosome-autophagosome fusion is called an amphisome. The completed autophagosome or amphisome fuses with a lysosome. The resulting compartment (called autolysosome), with lysosomal acid hydrolases cause enzymatic breakdown of the inner membrane from the autophagosome and degrade the cytoplasmic components. Upon degradation of cytoplasmic components, autolysosomes undergo autophagic lysosome reformation.

61. Which one of the following types of promoters would NOT be used within the T-DNA for expression of a negative selection marker gene for generation of transgenic plants by Agrobacterium-mediated transformation?

(a) strong constitutive promoter

(b) tissue-specific promoter

(d) stress-inducible promoter

Ans. (a)

Sol. In Agrobacterium-mediated transformation, the transfer DNA (T-DNA) region of the Ti (tumor-inducing) plasmid from Agrobacterium is utilized to introduce foreign DNA into the plant genome. The T-DNA region typically contains the gene of interest (e.g., the gene to be expressed in the transgenic plant) along with selectable marker genes.

62. Directional Selection for a particular trait will lead to the frequecy of the trait

(a) being normally distributed in the population

(b) always showing a left-skewed distribution in the population

(d) showing either a right or a left-skewed distribution in the population

Ans. (d)

Sol. The direction of the change in frequency will depend on the direction of the selection pressure. If the selection pressure is for a trait that is more common in the population, the frequency of the trait will increase. If the selection pressure is for a trait that is less common in the population, the frequency of the trait will decrease.

The distribution of the trait in the population may be either right-skewed or left-skewed, depending on the direction of the selection pressure. A right-skewed distribution means that the trait is more common in the population than it would be if there was no selection pressure. A left-skewed distribution means that the trait is less common in the population than it would be if there was no selection pressure.

63. Which one of the following statements is correct for a primary successional species?

(a) These species do not follow specific survivorship curves

(b) These species show Type II survivorship curve

(d) These species show Type I survivorship curve

Ans. (c)

Sol. Primary successional species are the first species to colonize an area that is devoid of life, such as a newly formed lava field or a newly exposed rock face. These species are typically small, hardy, and quick-growing, and they are able to survive in harsh conditions. They are also able to reproduce quickly, which helps to ensure that their populations will not be wiped out by natural disasters or other disturbances.

64. Which one of the following methods is best suited to estimate the population size of fish?

(a) Camera Trap

(b) Line Transect

(d) Mark-Recapture

Ans. (d)

Sol. Population size (or abundance) is a function of population density and the area that is occupied (geographic distribution). Usually, population size is estimated by counting all the individuals from a smaller sample area, then extrapolated over a larger area. When the individuals are not mobile - their population size may be estimated by counting individuals within a specified area. When individuals are very mobile and frequently move from one by counting individuals within a specified area. When individuals are very mobile and frequently move from one area to another then we can count the number by applying a common method called a mark-recapture method.

65. Clonogenic neoblasts are involved in planarian (flatworm) regeneration. This is an example of

(a) epimorphosis

(b) morphallaxis

(d) compensatory regeneration

Ans. (c)

Sol. Clonogenic neoblasts are responsible for maintaining the population of neoblasts and replenishing the tissue that was lost or damaged during the injury. They have the ability to divide and produce both identical daughter cells (self-renewal) and differentiated cells that can replace the damaged or missing tissues.

66. Which one of the following is the communicating junction linking adjacent cells in plants, which permits small molecules to pass from cell to cell while blocking the passage of most large molecules?

(a) Adherens junction

(b) Gap junction

(d) Hemidesmosome

Ans. (c)

Sol. Adjacent plant cells communicate with each other through cytoplasmic connections called plasmodesmata (singular, plasmodesma), which function analogously to animal cell gap junctions. Plasmodesmata are intercellular channels that establish a symplastic communication pathway between neighbouring cells in plants. At each plasmodesma, the plasma membrane of one cell is continuous with that of its neighbour, creating an open channel between the two.

67. Which one of the following changes in energetically favorable and occurs spontaneously in an aqueous solution?

(a) Formation of a bilayer from phospholipid molecules

(b) Dispersion of one oil droplet into many small ones

(d) Conversion of a mamebrane vesicle to a flat bilayer

Ans. (a)

Sol. Dispersion of one oil droplet into many small ones is not energetically favorable because it would increase the surface area of the oil droplets, which would increase the free energy of the system.

68. Which one of the following factors will NOT have any impact on the resolving power of a bright field microscope?

(a) Color of light

(b) Intensity of light

(d) Medium between the objective lens and specimen

Ans. (b)

Sol. Resolving power Is the ability of magnifying Instrument to distinguish two objects that are close together. The resolving power Is Inversely related to the limit of resolution. The limit of resolution is defined as the minimum distance between two points that allows for their discrimination as two separate points. Thus, the higher the resolving power, the smaller the limit of resolution. The limit of resolution of the light microscope depends upon the three factors:

• Wavelength of the light used to Illuminate the specimen

• Angular aperture (Angle of admittance of light in the objective lens)

• Refractive index of the medium between the cover slip and the objective lens

69. The hemolysis of red blood cells takes place when they are suspended in which one of the following solutions?

(a) 2.0% NaCl

(b) 1.5% NaCl

(d) 0.5% NaCl

Ans. (d)

Sol. A 0.5% NaCl solution is isotonic, meaning it has a similar solute concentration to that of the cytoplasm of red blood cells. In an isotonic solution, there is no net movement of water into or out of the cells, resulting in the maintenance of their normal shape and integrity.

70. In mature erythrocytes, the end-product of glycolysis that contains the carbons of glucose is :

(a) ethanol

(b) pyruvate

(d) lactase

Ans. (d)

Sol. Mature erythrocytes are devoid of mitochondria and therefore cannot undergo cellular respiration. Instead, they rely on glycolysis to generate energy. The end product of glycolysis in mature erythrocytes is lactate, which is then released into the bloodstream.

71. Ecologists examined the role of competition for below ground resources (water and nutrients) in the dispersion pattern of trees in the Acacia savannas of South Africa. The figure below depicts the result of their study.

In case all the other parameters were constant, select the option that best represents the dispersion patterns for populations labelled A and B in the figure above.

(a) A-Regular and B-Random

(b) A-Random and B-Clumped

(d) A-Regular and B-Regulars

Ans. (d)

Sol. The term "regular" in ecological terms refers to a uniform or evenly spaced distribution pattern. From the figure, if populations A and B show a regular dispersion pattern, it means that the individuals within each population are distributed evenly across the study area. This suggests that there is a relatively equal spacing between individuals, which could be influenced by factors such as competition for below-ground resources (water and nutrients).

72. The following figure represents the interaction between different blastomeres in a 4-cell stage of C. elegans embryo.

The following statements were made regarding the above :

A. The fate of EMS blastomere is autonumously specified

B. The default fate of EMS blastomere is MS cell lineage

C. Conditional specification can be observed in the development of E cell lineage

D. Assuming that a receptor needs to be activated for E fate, a C. elegans embryo where the receptor is constitutively active, is likely to develop cells of E fate only, in all three of the above cases

Which one of the following options represents the combinatio of all correct statements?

(a) A and C only

(b) B and C only

(d) B, C and D

Ans. (d)

Sol. B. The default fate of EMS blastomere is MS cell lineage. This statement is correct. In the absence of specific signaling interactions or cues, the EMS blastomere adopts the default fate of the MS (mesoderm and muscle) cell lineage. This default fate determination is important for normal embryonic development.

C. Conditional specification can be observed in the development of the E cell lineage. This statement is correct. Conditional specification refers to the fate determination of a cell depending on specific signaling interactions or cues. In the case of the C. elegans embryo, conditional specification can be observed in the development of the E (endoderm) cell lineage. The fate of the E blastomere is influenced by interactions with neighboring cells and specific signaling pathways.

D. Assuming that a receptor needs to be activated for E fate, a C. elegans embryo where the receptor is constitutively active is likely to develop cells of E fate only in all three of the above cases. This statement is correct. If the receptor required for E fate is constitutively active, it means it is constantly signaling and activating the E fate pathway. In such a scenario, the C. elegans embryo is likely to develop cells of E fate in all three cases mentioned, as the constitutively active receptor would override any conditional or default fate determinations.

73. Four different plant communities that consisted of the same number of species were taken up for a species diversity study. The following table represents some of the outcomes :

Select the correct statement about the evenness of the above communities

(a) The evenness of all the four communities is the same

(b) B > A > D > C represents the decreasing order in evenness of the communities

(d) Using the given information, we cannot compare the evenness of the communities

Ans. (b)

Sol. Simpson's reciprocal index, also known as Simpson's diversity index, is a measure of biodiversity that takes into account both the richness and evenness of a community. It is calculated as the reciprocal of the probability that two randomly chosen individuals from the community will belong to the same species

74. In eukaryotes, microtubules and actin-binding proteins ifluencethe dynamics and organization of the cytoskeleton. Match the cytoskeleton-binding proteins listed in column B, to actin or microtubule mentioned in Column A, and their function form those listed in Column C.

Which one of the following options represents all the correct matches between columns A, B and C?

(a) X-(iv)-3, X-(iii)-2, Y-(i)-A, Y-(ii)-4

(b) X-(ii)-1, X-(i)-3, Y-(iv)-3, Y-(iii)-2

(d) X-(i)-4, X-(iv)-1, Y-(iii)-4, Y-(ii)-3

Ans. (c)

Sol. Microtubule-associated proteins: Proteins that bind to microtubules are collectively called microtubule-associated proteins (MAPs). A large number of MAPs bind to the surface of the microtubule and play crucial roles in nucleation, dynamics and stability. MAPs can be categorized functionally as stabilizers (such as tau, MAP4), destabilizers (such as katanln), capping proteins (e.g., patronin), bundlers/cross-linkers (such as tau, CLIP170) and microtubule motors (e.g., klnesln and dynein). Destablllzers Include sequestering proteins (e.g., protein stathmln), tip destabilizers (e.g., depolymerizing klnesln-13) and microtubule severing proteins (e.g., katanln).

75. Rubisco enzyme is involved in both reductive and oxidative carbon cycles in plants. Following are certain statements regarding them

A. Sugars are produced in both the cycles

B. Ferredoxin is reduced only in oxidative carbon cycle

C. Product of oxidative cycle is one of the substrates of reductive cycle

D. NADP and ATP are used in both the cycles

Which one of the following options represents the combination of all correct statements?

(a) A and B

(b) A and D

(d) B and C

Ans. (d)

Sol. B. Ferredoxin is reduced only in the oxidative carbon cycle. The oxidative carbon cycle involves the reduction of ferredoxin as part of the photorespiration pathway. This reduction process helps in the conversion of 2-phosphoglycolate (a byproduct of Rubisco's oxygenase activity) into useful compounds.

C. The product of the oxidative carbon cycle is one of the substrates of the reductive carbon cycle. The oxidative carbon cycle produces glycine, which is then converted into serine. Serine, in turn, serves as a substrate for the reductive carbon cycle, as it can be converted into 3-phosphoglycerate, an intermediate in sugar synthesis.

76. Prolactin is an anterior pitutiary hormone which is lactogenic and helps milk production in mammals. The statements below are made regarding the mechanism of action of prolactin:

A. The receptors that bind to prolactin lack intrinsic tyrosine kinase activity

B. Upon prolactin binding, the receptors dimerize, and associated kinase is activated

C. Prolactin binding to receptors leads to activation of its intrinsic kinase activity

D. STAT plays a role in mediating prolactin action. It dimerizes after its phosphorylation to elicit the response

E. There is no involvement of STAT in prolactin action mechanisms

Which one of the following options represents the combination of all correct statements?

(a) A, B and D

(b) B, C and D

(d) C and E only

Ans. (a)

Sol. One of the main signaling pathways activated by the prolactin receptor is the JAK/STAT pathway. JAK/STAT stands for Janus kinase/signal transducer and activator of transcription.

77. Consumption of untreated corn as the staple food causes the disease, pellagra. Pre-treatment of corn with Ca(OH)₂ prevents this disease. Given below are options listing possible effects of Ca(OH)₂ treatment (Column X) and the enzymes affected (Column Y).

Select the correct match relevant for preventing pellagra form the options listed below

(a) C-iv

(b) B-i

(d) D-ii

Ans. (c)

Sol. Release of Vitamin B3 upon Ca(OH)2 treatment: Ca(OH)2 treatment may result in the release or production of Vitamin B3 (also known as niacin). Niacin is an essential nutrient that plays a crucial role in various biological processes, including energy metabolism and DNA repair.

Alkaline pH aids in digestion: Ca(OH)2 treatment can increase the alkalinity of the treated substance or environment. This alkaline pH can aid in digestion by enhancing the activity of digestive enzymes and facilitating the breakdown of food components.

Ca improves bone strength: Calcium (Ca) is an essential mineral for maintaining bone health and strength. Ca(OH)2 treatment may provide a source of calcium, which can contribute to improved bone strength and density.

Prevents formation of ROS: ROS stands for reactive oxygen species, which are highly reactive molecules that can cause damage to cells and DNA. Ca(OH)2 treatment may have antioxidant properties, helping to prevent the formation of ROS and reduce oxidative stress in the treated substance or system.

78. The diagram below depicts the cumulative fossil CO₂ emissions (y axis) of different contiments from the years1850-2020 (x-axis).

Select the option that correctly identifies the contiments A-D:

(a) A-North America; B-Africa; C-Asia; D-Europe

(b) A-Europe; B-South America; C-Africa; D-Asia

(d) A-North America; B-Asia; C-Africa; D-Europe

Ans. (a)

Sol. Cumulative fossil CO₂ emissions refer to the total amount of carbon dioxide (CO₂) emissions released from burning fossil fuels over a specific period. These emissions are a significant contributor to global climate change and the increase in greenhouse gases in the Earth's atmosphere.

Cumulative fossil CO₂ emissions can be measured and reported for different regions or countries. The values represent the total emissions from the combustion of fossil fuels, including coal, oil, and natural gas, since a particular starting point or reference year.

79. The following statements suggest the changes in respiratory ventilation and the mechanisms of these changes when a normal human subject is allowed to inhale air containing different oxygen content

A. The ventilation is markedly increased when Po₂ of the inspired air is less than 60 mm Hg

B. The ventilation is 6 L/min when the Po₂ of the inspired air is about 150 mm Hg.

C. The ventilation is slightly increased when Po₂ of the inspired air is more than 60 mm Hg

D. The increased ventilation due to the lower Po2 in the inspired air causes higher alveolar Pco₂

E. The H⁺ concentration in the arterial blood is increased when Po₂ of the inspired air is gradually decreased.

Which one of the following options represents the combination of all correct statements?

(a) A, B and C

(b) B, C and D

(d) A, B and E

Ans. (a)

Sol. A. The ventilation is markedly increased when Po2 of the inspired air is less than 60 mm Hg. This statement is correct. When the partial pressure of oxygen (Po2) in the inspired air drops below a certain threshold (typically around 60 mm Hg), the body initiates compensatory mechanisms to increase ventilation. This increased ventilation helps to maintain an adequate supply of oxygen to the tissues.

B. The ventilation is 6 L/min when the Po2 of the inspired air is about 150 mm Hg. This statement is correct. A ventilation rate of 6 L/min is within the normal range for respiratory ventilation in a healthy human subject. The specific value of 6 L/min may vary among individuals, but it represents a typical ventilation rate when the Po2 of the inspired air is around 150 mm Hg, which is considered normal.

C. The ventilation is slightly increased when Po2 of the inspired air is more than 60 mm Hg. This statement is correct. When the Po2 of the inspired air is above the threshold of 60 mm Hg, the ventilation rate may still increase, but to a lesser extent compared to when the Po2 is below 60 mm Hg. This slight increase in ventilation helps regulate the balance of oxygen and carbon dioxide in the body.

80. Transforming the neutral plate into a neural tube is an important event towards the formation of central system, in which the following sub-events might take place.

A. In primary neurulation, the cells surrounding the neural plate direct the neural plate cells to proliferate, invaginate and separate from the surface ectoderm to form a hollow tube

B. In secondary neurulation, the neural tube does not arise from the aggregation of mesenchyme cells into a solid cord

C. The morphogen, Sonic hedgehog, that is expressed in notochord, is required for induction of floor plate cells in the neural plate to form the medical hinge point

D. In mammals secondary neurulation begins at the level of sacral vertebrae

E. In mammals, the primary neurulation forms brain regions while the secondary neuralation takes care of forming rest of the central nervous system from neck to tail.

Which one of the following options represents the combination of all correct statements?

(a) A, C and D

(b) A, B and E

(d) C, D and E

Ans. (a)

Sol. A. In primary neurulation, the cells surrounding the neural plate direct the neural plate cells to proliferate, invaginate, and separate from the surface ectoderm to form a hollow tube. This statement describes one of the key events in primary neurulation. The cells surrounding the neural plate play a crucial role in directing the morphogenetic processes of proliferation, invagination, and separation to transform the neural plate into a hollow neural tube.

C. The morphogen Sonic hedgehog, which is expressed in the notochord, is required for the induction of floor plate cells in the neural plate to form the medial hinge point. This statement highlights the role of the morphogen Sonic hedgehog (Shh) in the development of the neural tube. Shh, expressed in the notochord, plays a critical role in the induction of floor plate cells in the neural plate, which is important for the formation of the medial hinge point.

D. In mammals, secondary neurulation begins at the level of the sacral vertebrae. This statement correctly identifies the initiation point of secondary neurulation in mammals. In mammals, secondary neurulation starts at the level of the sacral vertebrae, which is towards the caudal (tail) end of the developing embryo.

81. Members of the chlorophytes are structurally diverse. Select the option that correctly represents the increasing order of structrual complexity among the given genera.

(a) Chlorella < Zygnema < Oedogonium < Draparnaldia

(b) Volvox < Chara < Oedogonium < Draparnaldia

(d) Volvox < Ulothrix < Draparnaldia < Ulva

Ans. (a)

Sol. Chlorella < Zygnema < Oedogonium < Draparnaldia

This means that Chlorella is the least structurally complex genus, followed by Zygnema, Oedogonium, and Draparnaldia, which is the most structurally complex genus among the given options.

82. Different typesof mapping populations that can be created using a variety of methods are presented as I to IV in the figure below :

A list of probable mapping populations denoted by I to IV in the figure and their status of genetic mortality is given below

A. I-Recombinant inbred lines (RILs)–immortal

B. II-Doubled haploid–Not immortal

C. III-F_2:3 – Not immortal

D. IV-Near isogenic Lines (NILs) – immortal

Which one of the following options represents the combination of all correct matches?

(a) C only

(b) A and D only

(d) A, C and D

Ans. (d)

Sol. A. I - Recombinant inbred lines (RILs) - immortal: Recombinant inbred lines (RILs) are created through repeated selfing and inbreeding, resulting in immortal lines.

C. III - F2;3 - Not immortal: F2;3 populations are derived from a cross between two parental lines, and subsequent generations are not immortal as they are subjected to further recombination and segregation.

D. IV - Near isogenic lines (NILs) - immortal: Near isogenic lines (NILs) are created through repeated backcrossing to generate lines with a single or few introgressed genetic segments from one parent into the genetic background of another parent. These lines are immortal.

83. Bacteriophage and P1 are both temperature phages. Which one of the following statements made about these phages and their lytic and lysogeny cycles in E.coli is Incorrect?

(a) Both the and P1 phages are double stranded DNA viruses

(b) In their lysogenic states in E.coli, while the phage integrates into the genome, thne P1 phage remains as alow copy number plasmid

(d) In-their lytic cycle, both the phages occur in plasmid forms in E.coli.

Ans. (c)

Sol. Both the lambda and Pl phages are double-stranded DNA viruses. In their lysogenic states in E. coli, while the lambda phage integrates into the genome, the Pl phage remains as a low copy number plasmid.

84. Which one of the following statements is correct in the context of erythromycin-mediated inhibition of protein synthesis is bacteria?

(a) Erythromycin inhibits protein synthesis at the step of elongation, butit nonetheless allows translation of the first few codons.

(b) Erythromycin inhibits protein synthesis at the step of elongation, and it prevents formation of even the first peptide bond

(d) Erythromycin is toxic to bacteria because it results in initiation of protein synthesis with elongator tRNAs.

Ans. (a)

Sol. Chloramphenicol: Chloramphenicol binds to the 50S ribosomal subunit and blocks peptide bond formation through Inhibition of peptldyl transferase, but does not affect the cytosolic protein synthesis In eukaryotes.

Tetracycline: Tetracycline binds to the 30S ribosomal subunit and Interferes with amlnoacyl-tRNA binding. Erythromycin: Binds to the 50S ribosomal subunit and Inhibits peptide chain elongation.

Fusidic acid: Fusldlc acid binds to EF-G and blocks translocation. Cydohexlmlde: Cyclohexlmlde blocks the peptldyl transferase of 80S ribosome but not that of 70S bacterial (and mitochondrial and chloroplast) ribosomes.

85. Following statements are about the features of immunoassays used to assay biomolecules:

A. Radio-immunoassays (RIAs) are more sensitive than Enzyme-linked immunosorbent assays (ELISAs) with chromogenic substrates

B. ELISAs with chromogenic substrates are more sensitive than ELISAs with chemiluminogenic subtrates

C. ELISPOT measures the number of cells capable of secreting particular biomolecules using a substrate that gives soluble product with enzyme reaction.

D. In Western blot analysisl, the product of enzyme-substrate reaction local ises at the site precisely where the antibody-enzyme conjugate binds to its specific protein band

Which of the following options represents the combination of all correct statements

(a) A and C

(b) A and D

(d) B and D

Ans. (b)

Sol. The ELISPOT assay is based on the principle of enzyme-linked immunosorbent assay (ELISA) and is commonly used to measure the frequency of antigen-specific cells in a sample. It is a widely used laboratory technique that allows the detection and quantification of cytokine-producing cells at the single-cell level.

86. Given below are term related to various experimental techniques (Column X) and their applications (Column Y)

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-iv, B-i, C-ii, D-iii

(b) A-iii, B-iv, C-i, D-ii

(d) A-iv, B-iii, C-i, D-ii

Ans. (a)

Sol. Mass Spectrometry: Mass spectrometry is a technique used to analyze the mass-to-charge ratio of ions. It is commonly used in proteomics to identify and quantify proteins and peptides.

Pulsed Field Gel Electrophoresis: Pulsed Field Gel Electrophoresis (PFGE) is a technique used to separate large DNA molecules based on their size. It involves applying an electric field that changes direction periodically, allowing for the separation of DNA fragments that are larger than those achievable with conventional gel electrophoresis.

Isoelectric Focusing: Isoelectric Focusing (IEF) is a technique used to separate proteins based on their isoelectric points (pI). It involves subjecting proteins to an electric field in a pH gradient, causing them to migrate to the point where their net charge is zero.

PacBio: PacBio is a brand of DNA sequencing technology developed by Pacific Biosciences. It utilizes a single-molecule real-time (SMRT) sequencing approach to sequence long DNA fragments, enabling the generation of long reads and the identification of complex genomic features.

87. The following statement refer to factors regulating the fidelity of DNA replication

A. The 5' to 3' exonuclease activity of the replicative DNA polymerase

B. Imbalanced intracellular concentrations of the four dNTPs

C. Increased intracellular concentrations of rNTPs resultig in increased incorporation of rNTPs during DNA synthesis, which are not easily removed by the polymerase's proof-reading activity

D. Removal of incorrectly incorporated nucleotides by the mismatch repair system

Which one of the following options gives the combination of all correct statements?

(a) A and D only

(b) B, C and D

(d) A, B and D

Ans. (b)

Sol. 5' to 3' exonuclease activity is not present in the replicative DNA polymerase (DNA polymerase III), while the proofreading activity is associated with the 3' to 5' exonuclease activity.

88. In the context of signalling, the enzyme protein kinase C (PKC) depends on multiple molecules for its complete activation. This activation depends on the presence of

A. phosphatidylserine at the inner leaflet of the plasma membrane

B. Ca²⁺

C. phosphatidylethanolamine at the inner leaftlet of the plasma membrane

D. diacylglycerol present in the inner leaflet of the plasma membrane

(a) A, B and C

(b) A, C and D

(d) A, B and D

Ans. (d)

Sol. The activated PLC-p cleaves phosphatidylinositol 4,5-blsphosphate (PIP2) to generate two second messengers: inositol 1,4,5-trisphosphate (IP3) and diacylglycerol (DAG). The membrane phospholipid PIP, Is a minor component of the plasma membrane, localized to the inner leaflet of the phospholipid bilayer. One second messenger DAG remains associated with the plasma membrane, the other second messenger produced by PIP2 cleavage, IP3, is a small polar molecule that is released into the cytosol, where It acts to signal the release of calcium Ion from the endoplasmic reticulum. IP3 acts to release calciumion from the endoplasmic reticulum by binding to receptors that are ligand-gated calcium Ion channels (IP3-gated caldum-release channels, also called IP3 receptors). As a result, cytosolic calcium Ion levels increase, which affects the activities of a variety of target proteins, Including protein kinases and phosphatases. DAG together with calcium Ion, helps activate the enzyme protein kinase C (PKC), which Is recruited from the cytosol to the cytosolic face of the plasma membrane. When activated, PKC phosphorylates specific serine or threonine residues on target proteins that vary depending on the cell type.

89. The following statement are made about telomeres

A. Telomere-binding proteins (TBPs) are believed to shield telomeres from the cell's DNA repair machinery, preventing them from being recognized a double-strand breaks

B. Telomeres in human cells are repeats of TTAGGGG sequence that can extend upto 150kb, which are replicated by the action of TERT in actively dividing cells

C. In differentiated cells, telomerase in inactive, leading to shnortening of telomeres over hundreds of cell divisions, damage to ends of chromosome and eventually apoptosis

D. The persistence of telomerase activity in several cancers allows the cells to continue to proliferate.

Which one of the following options represent the combination of all correct statements?

(a) A and C only

(b) A, B and C

(d) A, C and D

Ans. (d)

Sol. Telomeres: Telomeres are specialized structures that cap the ends of eukaryotic chromosomes. They have several likely functions - maintaining the structural integrity of a chromosome (if a telomere is lost, the resulting chromosome end is unstable) and ensuring complete replication of the extreme ends of chromosomes. Eukaryotic telomeres consist of a long array of short and tandemly repeated sequences. There may be 100-1000 repeats, depending on the organism. One unusual property of the telomeric sequence is the presence of the G-rich single-strand 3' overhang, measuring between 50 to 300 nucleotides. The G-rich sequence is generated because there is a limited degradation of the C-rich complementary strand. Unlike centromeres, the sequence of telomeres has been highly conserved in evolution - there is a considerable similarity in the simple sequence repeat, for example, TTGGGG (Paramecium), TAGGG (Trypanosoma), TTTAGGG (Arabidopsis) and TTAGGG (Homo sapiens). Human telomeres are Telomerase is therefore necessary and in its absence chromosomes shrink during each replication. Telomerase is present in stem cells and germ-line cells, where it is needed to keep telomeres long and allow essentially indefinite rounds of cell division. In most somatic cells, telomerase is at low or very low levels and the length of telomeres is consequently gradually eroded. The length of the telomeres therefore provides on limit to replicative capacity. Complete loss of telomere repeats leads to triggering of DNA repair and apoptosis. Cancer cells often produce telomerase, allowing them to divide indefinitely and evade programmed death.

90. The following trees represents different evolutionary mechanisms.

Select the tree that best represents punctuated equilibrium

(a) A

(b) B

(d) D

Ans. (a)

Sol.

91. T4 phages were plated on three E.coli bacterial plates labelled I, II and III. The phenotypes obtained are depicted in the picture below. The black spots represent plaques.

The following combination of conditions were given to explain the results obtained.

From the options listed below, select the one that accujrately lists the E.coli strain type and the corresponding rII locus type

(a) I-iii-1, II-ii-2, III-iii-1

(b) I-ii-2, II-i-1, III-iii-2

(d) I-ii-2, II-ii-1, III-ii-2

Ans. (b)

Sol.

The host range of rll mutants and wild-types phages is different. One type of bacterial strains, E. coli B allows both to grow, but plaques of different size result: wild-type phages produce small plaques, and rll mutants produce large plaques. Another E. coli strain that carries the /.-prophage, designated E. coli K12Z or K(Z), does not permit the growth of rll mutants, but it does allow wild-type phage to grow. The rll mutants are then conditional mutants—namely, mutants that can grow under one set of conditions but not under another. E. coli B is said to be permissive for rll mutants because it allows phage growth, whereas E. coli K12(Z) is said to be nonpermissive for rll mutants because it does not allow phage growth.

92. Some properties of enzymes are listed in column X, and their kinetic expressions are listed in column Y.

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-ii, B-i, C-iii, D-iv

(b) A-iii, B-ii, C-iv, D-i

(d) A-i, B-iv, C-ii, D-iii

Ans. (c)

Sol. The specific activity of an enzyme is defined as the amount of enzyme activity (measured in enzyme units) per unit of protein. It is a measure of the enzyme's catalytic efficiency and is often used to assess the purity of an enzyme preparation. Specific activity is calculated by dividing the total enzyme activity by the amount of protein present in the sample. It provides a measure of the concentration of active enzyme molecules in relation to the total protein content.

93. The following statements were made about the structure of the 30-nm chromatin fiber:

A. In the solenoid model, the linker DNA connects the consecutive core particles

B. In the zig-zag model, alternating nucleosomes become interacting neighbors

C. In the solenoid model, 12 nucleosomes are organized into two separate stacks, whereas 8 nucleosomes per turn make a single stack in the zig-zag model

D. H1 histone is essentially required as per the zig-zag model, but not as per the solenoid model

E. Chromatin fibres prepared with H4 histones that lack their tails could fold into higher-order fibers

Which one of the following options represents the combination of all correct statements?

(a) A, C and D

(b) A, B and E

(d) C and E only

Ans. (c)

Sol. Two popular models that were proposed based on in vitro data are the solenoid and zigzag model. The main feature of solenoid model (one-start helix model) Is that nucleosomes follow each other along the same helical path, and interactions between the nucleosomes occur sequentially. In zigzag model (two-start helix model), two strands containing stacked nucleosomes are wound into a left-handed two-start helix. In each strand, the nucleosomes are aligned with each other like a stacked of coins. The solenoid model is characterized by interactions between consecutive nucleosomes (n, n + 1 and so on), whereas the zigza model implies interactions between alternate nucleosomes (n, n + 2 and so on).

94. India has designated regions as sanctuaries or national parks (column Q) dedicated for the conservation of specific species (column P).

Select the option that depicts all correct matches between column P and column Q.

(a) A-ii, B-i, C-iv, D-iii

(b) A-iii, B-iv, C-i, D-ii

(d) A-ii, B-iv, C-i, D-iii

Ans. (a)

Sol. Gharial: National Chambal Sanctuary in Madhya Pradesh, Rajasthan, and Uttar Pradesh. Saltwater Crocodile: Bhitarkanika National Park and Wildlife Sanctuary in Odisha. Humpback Mahseer: Mahseer Sanctuary in Karnataka and Uttarakhand. Hawksbill turtle: Gahirmatha Marine Sanctuary in Odisha.

95. The following statements are made about digestion of proteins and the enzymes involved

A. Chymotrypsin does not generate peptides with C-terminal neutral amino acids

B. Trypsin generates peptides with C-terminal basic amino acids

C. Carboxypeptidase B acts on aromatic amino acids

D. Carboxypeptidase A employs zinc ion for hydrolysis

E. The brush border enterokinase has no polysaccharides attached to it

F. The final digestion to amino acids occurs in the intestinal lumen, the brush border, and the cytoplasm of the mucosal cells

Which one of the following options represents the combination of all correct statements

(a) A, B and C

(b) C, D and E

(d) B, D and F

Ans. (d)

Sol.

Carboxypeptidases and aminopeptidases are exopeptidases that remove terminal amino acid residues from C and N-termini of polypeptides, respectively. Carboxypeptidase A cleaves the C-terminal peptide bond of all amino acid residues except Pro, Lys and Arg. Carboxypeptidase B Is effective only when Arg or Lys are the C-termlnal residues. Carboxypeptidase C acts on any C-terminal residue. Aminopeptidases catalyze the cleavage of amino acids from the amino terminus of the protein. Aminopeptidase M catalyzes the cleavage of all free N-terminal residues.

96. Following statements were made regarding regeneration in different organisms :

A. The regenerating blastema cells in amphibians retain their specification even when they dedifferentiate

B. A transgenic Hydra when made to misexpress -catenin will show numerous ectopic tentacles

C. In planaria, if the Wnt pathway is activated, then the posterior blastema would regenerate a head

D. A regenerating blastema is formed in the mammalian liver.

Which one of the following options represents all correct statement(s)?

(a) A only

(b) C only

(d) C and D

Ans. (a)

Sol. In blastema-mediated regeneration, differentiated cells undergo dedifferentiation to form undifferentiated cells. These cells then start proliferating to form a mass of heterogeneous undifferentiated cells at the regenerating tip called blastema. Once a necessary number of cells are amassed, the cells in blastema start to redifferentiate and initiate the process of morphogenesis similar to that seen during embryogenesis. This process then culminates with the restoration of the lost tissue, either completely or partially depending upon the organism.

In compensatory regeneration, the regeneration proceeds without any blastema formation or requirement of stem cells. In this process, differentiated cells from the vicinity are recruited to the site of injury and proliferate to replace the lost tissue. Liver regeneration is a prime example of compensatory regeneration.

97. Genetic screens for mutants affecting development of leaf trichomes have led to the discovery of genes regulatng trichome patterning-especially trichome density and spacing, as depicted in the figure below.

The following statements are made in this regard

A. GLABRAI (GL1) mutant plant will show fewer or no trichomes

B. Cells and form trichomes strongly express the GABRA2 (GL2) and TRYPTICON (TRY) genes

C. Try protein acts as a positive regulator of trichome cell differentiation in the surrounding cells

D. Addition of exogenous JA will reduce the number of leaf trichomes

Which one of the following options represents the combination of all correct statements?

(a) A and B

(b) B and C

(d) A and D

Ans. (a)

Sol. Statement A is correct: GLABRA1 (GL1) mutant plants will show fewer or no trichomes. GL1 is a gene involved in promoting trichome development, so its mutation results in a reduction or absence of trichomes.

Statement B is correct: Cells that form trichomes strongly express the GLABRA2 (GL2) and TRYPTICON (TRY) genes. GL2 and TRY are genes involved in promoting trichome formation and patterning. Their strong expression in trichome-forming cells indicates their role in regulating trichome development.

98. Given below are terms related to various techniques (Column X) and their features (Column Y)

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-iv, B-i, C-ii, D-iii

(b) A-iii, B-iv, C-i, D-ii

(d) A-iv, B-i, C-iii, D-ii

Ans. (a)

Sol. RACE (Rapid Amplification of cDNA Ends) is a PCR-based method for locating the precise start and end points of gene transcripts.

99. A food chain involving Spartina (a plant), the marsh periwinkle snail, the blue crab and an unknown fungus was identified ina Spartina-dominated salt marsh in North America. A study involving control and crab-exclusion experiments revealed

A. Radulations (scrape marks) on the leaf surface made by the snails indicate the presence of snail faeces, fungi and dead plant tissue.

B. The fungi were present only at the radulations

C. The density of the radulations increased with higher snail densities

D. Spartina density decreased with increase in the snail density till it reached zero

E. In control expeirments, all four species were present till the end.

Select the option that correctly depicts the positive (+) and negative (–) interaction-type between fungi-snail and Spartina-crab, respectively.

(a) – and +

(b) – and –

(d) + and +

Ans. (d)

Sol. Fungi-snail interaction is positive (+) because the fungus benefits from the snail's grazing, which creates openings in the plant tissue that allow the fungus to colonize. Spartina-crab interaction is negative (-) because the crab preys on the snail, which reduces the snail's population and therefore the amount of fungal growth on the plant.

100. The following statements were made about cell cycle regulation in fission yeast :

A. A cdc25 mutant cannot enter mitosis due to its inability to remove the inhibitory phosphate

B. Weel I consistently maintains the Cdk in an active state, to maintain cell size during cell cycle

C. CAK (Cdk-activating kinase)-mediated phosphnorylation of Threonine 161 residue of Cdc2 is necessary, but not sufficient for the Cdk to be active

D. Activation of Sicl in G1 allows the cyclin-Cdk that is present in the cell to initiate DNA replication

E. sic1 mutant exhibits activaton of premature DNA replication from fewer origins and extension of the duration of S phase

Which one of the following options represents the combination of all correct statements?

(a) A, B and D

(b) A, C and E

(d) A, C and D

Ans. (b)

Sol. Mutant analysis: The Weel and Cdc25 play antagonistic roles. Weel usually inhibits cells from initiating mitosis until their size is adequate. Hence, loss-of-function mutations in the weel gene cause premature entry into mitosis resulting in small cells, whereas overproduction of Weel protein increases the length of G2, resulting in elongated cells. Inhibitory phosphate is removed by protein phosphatase, Cdc25. Hence, due to loss-offunction mutations, the cell does not divide but continues to grow.

101. The following statements were made about phase contrast microscopy

A. Phase contrast microscopy can be equally utilized to examine stained and unstained specimenst

B. A phase annulus generates hollow cone of light of illuminate the specimen

C. A light wave that passes through a cell nucleus and organelles lags compared to the light waves that pass through water only.

D. A polarized light source is used to translate the minor phase shifts into grey values.

E. Appearance of bright halos is common artifact of phase contrast imaging

Which one of the following options represents the combination of all correct statements?

(a) A, C and D

(b) B, C and E

(d) B, D and E

Ans. (b)

Sol. B. A phase annulus generates a hollow cone of light to illuminate the specimen. This statement is correct. In phase contrast microscopy, a phase annulus is used to create a hollow cone of light that illuminates the specimen. This annular illumination results in the phase shifts produced by the specimen interacting with the direct beam of light, leading to the formation of a phase contrast image.

C. A light wave that passes through a cell nucleus and organelles lags compared to the light waves that pass through water only. This statement is correct. In phase contrast microscopy, the phase shifts generated by the specimen (such as a cell nucleus and organelles) cause the light passing through them to lag compared to the light waves passing through the surrounding medium (such as water). This phase difference is the basis for creating contrast and visualizing the specimen.

E. Appearance of bright halos is a common artifact of phase contrast imaging. This statement is correct. The appearance of bright halos around objects in phase contrast images is a common artifact. These halos are caused by the diffraction and interference of light as it passes through the specimen and interacts with the phase contrast components. They can sometimes affect the interpretation of phase contrast images.

102. The following statements are made regarding the amphibian early-embryonic development

A. The Nieuwkoop center cells are mesodermal in origin

B. Chordin, Noggin and Goosecoid are secreted by the Organizer

C. The default fate of the ectoderm is to become neural tissue

D. BMP levels are high in the presumptive dorsal mesoderm.

Which one of the following options represents the combination of all correct statements?

(a) A, B and C

(b) C and D

(d) A and D

Ans. (c)

Sol. The Nieuwkoop center cells are not mesodermal in origin. They are derived from the vegetal pole of the embryo. BMP (Bone Morphogenetic Protein) levels are not high in the presumptive dorsal mesoderm. BMP signaling is actually suppressed in this region by the Organizer and its secreted molecules, such as Chordin and Noggin.

103. 100 µL of cells were taken in a tube and 400 µL 0.4% Trypan Blue was added for staining. About 20 µL of this cell suspension was added between the hemocytometer and cover glass (refer figure below). The hemocytometer is divided into 9 major squares of 1 mm × 1 mm size. The height of the chamber formed with the cover glass is 0.1 mm. Empty circles indicate unstained cells and solid circles indicate stained cells.

Based on the above figure, what is the total cell count in the original suspension and cell viability (%)?

(a) 3,75,000 cells/mL and 23%

(b) 3,75,000 cells/mL and 77%

(d) 75,000 cells/mL and 23%

Ans. (b)

Sol. The total volume of the original suspension is 100 µL (given), and we can calculate the dilution factor as follows:

Dilution Factor = Total Volume of Original Suspension / Volume Loaded onto Hemocytometer

Dilution Factor = 100 µL / 20 µL = 5

Now, we can calculate the total cell count in the original suspension:

Total Cell Count = (Number of Cells Counted) × (Dilution Factor) Total Cell Count = 38 cells × 5 = 190 cells

Therefore, the total cell count in the original suspension is 190 cells.

To calculate cell viability, we need to differentiate between the stained cells (solid circles) and unstained cells (empty circles). From the figure, we can see that there are a total of 9 stained cells and 29 unstained cells within the counting area.

Cell Viability (%) = (Number of Unstained Cells / Total Cell Count) x 100 Cell Viability = (29 cells / 38 cells) × 100 = 76.32%

Rounding to the nearest whole number, the cell viability is approximately 76%.

Therefore, the estimated total cell count in the original suspension is 190 cells, and the cell viability is 76%.

104. Which one of the following statements pertaining to global ocean ecosystem productivity is NOT correct?

(a) Higher chlorophyll concentrations and the general higher productivity observed around the equator is driven by the process of upwelling and/or mixing of high nutrient subsurface water into the euphotic zone

(b) In some temperate and subpolar regions, productivity is least during the sprig due to the transitioning of phytoplankton from light-limiting to nutrient-limiting conditions

(c) In the nutrient-poor tropical and subtropical ocean, the cyanobacteria tend to be numerically dominant, as they specialize in taking up nutrients at low concentrations

(d) Larger phytoplankton, such as diatoms, often dominate the nutrient-rich polar ocean, and these can be grazed directly by multicellular zooplankton.

Ans. (b)

Sol. In temperate and subpolar regions, productivity is highest during the spring due to the transition of phytoplankton from lightlimited to nutrient-limited conditions. This is because during the winter, there is less sunlight available for photosynthesis, so phytoplankton are limited by light. In the spring, as the days get longer and the sunlight becomes more intense, phytoplankton are no longer limited by light and can begin to grow rapidly. This is when productivity is highest in these regions.

105. Red-blue colour blindness is a human X-linked recessive disorder. The two parents with normal colour vision have two sons. Son 1 has 47, XXY chnromosome composition and is colour blind Son 2 has 46, XY and is also colour blind. Assuming that no crossing over took place in prophase I of meiosis, Klinefelter syndorme in Son I resulted due to nondisjunction during which one of the following events?

(a) Female gamete formation in meiosis I

(b) Female gamete formation in meiosis II

(d) Male gamete formation in meiosis II

Ans. (b)

Sol. During female gamete formation in meiosis II, the X and Y chromosomes segregate. Nondisjunction in this stage would result in an egg (female gamete) receiving both X chromosomes (one X chromosome from the mother and one X chromosome from the father) and the Y chromosome. This leads to the formation of an XXY chromosome composition in Son 1, causing Klinefelter syndrome

106. A polypeptide was subjected to the following treatments with the indicated results

I. Acid hydrolysis

(1) (Alg, Arg, Cys, Glx, Gly, Lys, Leu, Met, Phe, Thr)

II. Aminopeptidase M :

(2) No fragments

III. Carboxypeptidase A + Carboxypeptidase B :

(3) No fragments

IV. Trypsin followed Edman degradation of the separated products :

(4) Cys-Gly-Leu-Phe-Arg

(5) Thr-Ala-Met-Gln-Lys

Which one of the following represents the primary structure of the peptide?

(a) Thr-Ala-Met-Gln-Lys-Cys-Gly-Leu-Phe-Arg

(b) Cys-Gly-Leu-Phe-Arg-Thr-Ala-Met-Gln-Lys

(d) Cyclin peptide shown below

Ans. (d)

Sol. Carboxypeptidases and aminopeptidases are exopeptidases that remove terminal amino acid residues from C and N-termini of polypeptides, respectively. Carboxypeptidase A cleaves the C-terminal peptide bond of all amino acid residues except Pro, Lys and Arg. Carboxypeptidase B is effective only when Arg or Lys are the C-terminal residues. Carboxypeptidase C acts on any C-terminal residue. Aminopeptidases catalyze the cleavage of amino acids from the amino terminus of the protein. Aminopeptidase M catalyzes the cleavage of all free N-terminal residues.

107. Aspartate (Asp) is an amino acid with the structure NH₂-CH(CH₂–COOH)-COOH. Given below are biosynthetic processes occuring in cells

A. protein synthesis

B. de novo synthesis of inosine monophosphate and orotic acid

C. synthesis of adenosine monophosphate from inosine monophosphate

D. glutathione synthesis

Which one of the following options correctly represents all the biosynthetic process(es) wherein Asp is involved as a precursor?

(a) A only

(b) A, C and D

(d) A, B and C

Ans. (d)

Sol.

108. Which one of the following options correctly lists ecosystems of the world arranged according to the descending order of their average world net primary production (billion kcal/yr)?

(a) Tropical rain forests > Northern coniferous forests > Open Oceans > Estuaries

(b) Open Oceans > Tropical rain forests > Northern coniferous forests > Estuaries

(d) Open Oceans > Northern rain forests > Estuaries > Northern coniferous forests

Ans. (b)

Sol. Net primary productively (NPP) and plant biomass of world ecosystems

109. The LacI and TrpR repressors bind with their ligands allolactose and tryptophan respectively, resulting in alternation of their DNA binding properties. The following statements are made about the mechanism of LacI and TrpR binding with DNA (opertor) and reglation of gene expression in E.coli.

A. Allolactose binding to LacI leads to its poor binding to the lac operator whereas tryptophan binding to the trp operator

B. Allolactose binding to LacI leads to induction of lac operon, whereas tryptophan binding to TrpR leads to repression of trp operon

C. Binding of allolactose and tryptopha to LacI and TrpR respectively leads to repression of their corresponding operons

D. Binding of allolactose and tryptophan to LacI and TrpR respectively, leads to activation of their corresponding operons. However, in trp operon regulation, availability of tryptophan also results in attenuation-mediated transcriptional termination leading to an overall effect of repression of trp operon.

Which one of the following options represents a combination of all correct statements?

(a) A and B only

(b) B and C

(d) A, B and D

Ans. (a)

Sol. A. Allolactose binding to LacI leads to its poor binding to the lac operator, whereas tryptophan binding to TrpR leads to its better binding to the trp operator. This statement correctly describes the effects of the ligand binding on LacI and TrpR. Allolactose, the inducer molecule, binds to LacI and weakens its binding to the lac operator, allowing gene expression (induction) in the lac operon. Tryptophan, the corepressor molecule, binds to TrpR and enhances its binding to the trp operator, resulting in gene repression in the trp operon.

B. Allolactose binding to LacI leads to the induction of the lac operon, whereas tryptophan binding to TrpR leads to the repression of the trp operon. This statement correctly describes the regulatory outcomes of allolactose and tryptophan binding to LacI and TrpR, respectively. Allolactose binding induces the expression of the lac operon, while tryptophan binding represses the expression of the trp operon.

110. After severe diarrhoea, the plasma K⁺ concentration became low (i.e., hypokalemia developed) in a human subject. The following statements are proposed to explain the mechanism of plasma K⁺ regulation by kidney in this condition

A. The principal cells present in distal tubule and collecting duct of nephron regulate K⁺ excretion

B. Hypokalemia stimulates Na⁺, K⁺-ATPase activity in the basolateral membrane of principal cells

C. The intracellular K⁺ concentration of the principal cells is increased.

D. The electrochemical gradient for effflux of K⁺ across the apical membrane of principal cells is increased.

E. The permeability of apical membrane to K⁺ is decreased

F. The plasma aldosterone level is decreased which inhibits K⁺ secretion by principal cells

Which one of the following options represents the combination of all incorrect statements?

(a) A, B, C

(b) B, C, D

(d) D, E, F

Ans. (b)

Sol. Statement B is incorrect: Hypokalemia does not stimulate Na⁺, K⁺-ATPase activity in the basolateral membrane of principal cells. Rather, hypokalemia would result in decreased Na⁺, K⁺-ATPase activity.

Statement C is incorrect: The intracellular K⁺ concentration of the principal cells would decrease in response to hypokalemia, not increase.

Statement D is incorrect: The electrochemical gradient for efflux of K⁺ across the apical membrane of principal cells would be decreased in response to hypokalemia, as the cells would try to conserve potassium.

111. In four taxa (A, B, C and D), two characters (shape and color) were scored to infer their phylogenetic relationship. The two character states for shape were square and round, while the two character states for color were black and yellow. The character distribution is given in the table below.

Using the above data, four trees were built using the method of maximum parsimony which are given below.

Select the option that represents the two most parsimonious trees.

(a) A and B

(b) C and D

(d) A and D

Ans. (d)

Sol. Parsimony assumes the evolutionary events are rare. The tree with fewest evolutionary events is the most likely one.

112. It has been observed that within a flowering season, a plant may produce more male flowers which may be correlated with the longevity of the flowers and the seasonal distribution of flowering in the plant. Which one of the following arguments do NOT support this observation of sex-specific floral phenology.

(a) Females are often resource limited and therefore pollination levels will be increased by producing more male flowers.

(b) Fluctuations in the rainfall pattern can influence pollinator service due to altered physiology of the pant during its reproductio, leading to a shift in flowering phenology of both sexes

(c) Plasticity in sex and their flowering phenology is determined neither by resource status of a taxa nor by fluctuations in climatic factors

(d) Male competition will favour floral features that improve pollinator visists and therefore more male flowers.

Ans. (c)

Sol. In reality, resource availability and climatic factors can indeed influence the sex-specific floral phenology in plants. The allocation of resources, such as nutrients and energy, can impact the production of male and female flowers. Additionally, climatic factors, including temperature, rainfall, and photoperiod, can influence the timing and duration of flowering events in plants.

113. The following statements are made about the E.coli SOS response to DNA damage :

A. RecA-DNA filament complex stimulates the autoproteolytic activity of the LexA repressor

B. RecA is activated due to the blunt ends of double-strand breaks caused by DNA damage-inducing agents

C. The SOS response includes the activation of synthesis of translesion polymerases

D. The destruction of LexA promotes synthesis of photolyase, which acts along with RecA to reverse the pyrimidine dimer formation process.

Which one of the following options represents the combination of all correct statements?

(a) A and D

(b) B and D

(d) C and D

Ans. (c)

Sol. The key regulatory proteins are LexA repressor and RecA protein. The LexA acts as a transcriptional repressor for SOS responsive genes (over 50 genes) that participate in processes such as DNA repair, homologous recombination and translesion DNA replication. During normal growth, LexA proteins bind with specific operator sequences (called SOS box) located in the promoter region and repress transcription. In E. coli, any block to DNA replication caused by DNA damage produces a signal that activates the E. coli RecA protein. Activation of RecA causes autocatalytic proteolytic cleavage of the LexA repressor. Cleavage occurs at a specific Ala-Gly peptide bond. RecA protein has two quite different types of activity. It acts as a co-protease and stimulates protease activity.

114. The following statements are made regarding the nitrogenase enzyme involved in the reduction of atmospheric nitrogen to ammonia :

A. Nitrogenase enzyme is composed of two components : dinitrogenase and dinitrogenase reductase

B. MoFe protein component is a homodimer

C. Fe protein component is dinitrogenase

D. MoFe protein contains the active site metal cluster where N₂ binds.

E. Fe protein delivers electron to MoFe protein component in a reactio coupled to the hydrolysis of MgATP.

Which one of the following options represents the combination of all correct statements?

(a) A, B and D

(b) B and C only

(d) A, D and E

Ans. (d)

Sol. The biological process of nitrogen fixation is catalyzed by an enzyme complex called nitrogenase complex. There are three different forms of nitrogenase that differ in their requirement for molybdenum (Mo), vanadium (V), or iron (Fe) as a critical metallic component. Most of the nitrogenases that have been studied contain a Mo cofactor.

Nitrogenase consists of two proteins: a dinitrogenase reductase and dinitrogenase. The dinitrogenase reductase (also called the Fe protein) is a dimer of identical 30 kDa subunits bridged by a 4Fe-4S cluster. Dinitrogenase is a tetramer with two copies of two different subunits. It contains both Fe and Mo. Because molybdenum is present in this cluster, the dinitrogenase component is also called the molybdenum-iron protein (MoFe protein). The MoFe cofactor is the site of nitrogen fixation. The genes involved collectively in the synthesis of nitrogenase.

115. Based on the theory of kin selection, choose the correct statement :

(a) A gene for altruism will spread in the population if the act of altruism increases the actor's gene in the next gene pool only through direct fitness.

(b) A gene for altruism will spread in the population if the act of altruism increases the actor's gene in the next pool only through indirect fitness

(c) A gene for altruism will spread in the population if the act of altruism increases the actor's gene in the next gene pool through direct or indirect fitness.

(d) Altruistic behaviour reduces the fitness of the trait bearer so a gene responsible for altruism cannot spread in a population and will be maintained at a very low frequency.

Ans. (c)

Sol. The higher the value of r, the greater the probability that the recipient of the altruistic behaviour will also possess the gene for altruism. So what Hamilton's rule tells us is that a gene for altruism can spread by natural selection, as long as the 'cost' Incurred by the altruist is offset by a sufficient amount of'benefit' to sufficiently closely related relatives. The natural selection that favours altruistic behavior by enhancing reproductive success of relatives Is called kin selection. Kin selection theory predicts that animals are more likely to behave altruistically towards their relatives than towards unrelated members of their species. Moreover, It predicts that the degree of altruism will be greater, the closer the relationship. In kin selection, a positive selection for certain alleles takes place Indirectly through enhanced reproduction of the relatives who carry the same alleles rather than directly through an Increased fitness of the carriers themselves. Thus, altruism can evolve with some suitable combination of high relatedness, high benefits to relatives, and low costs to self.

116. Assume that the genes w⁺ and cv⁺ are located 20 cM apart on the X chromosome of Drosophila melanogaster. Mutations in w⁺ and cv⁺ give rise to white eyes and crossveinless phenotypes, respectively, which are recessive to the wild-type phenotype. A homozygous wild-type female was crossed to a white-eyed, crossveinless male. The F₁ progeny was sib-mated. What percentage of the progeny will be white-eyed and crossveinless?

(a) 20

(b) 40

(d) 5

Ans. (a)

Sol.

117. The following patterns of gene expression were obtained after activating Notch I receptor with two different ligands DII1 and DII4.

Which one of the following statements depicts the correct interpretation of observations?

(a) The Notch pathway ligands DII1 and DII4 both bind to the Notch1 receptor but activate different downstream effectors with similar dynamics

(b) DII 1 induces sustained reponses, which preferentially activate the transcriptional target Hes1.

(d) The Notch pathway ligands DII1 and D114 both bind to the Notch 1 receptor but activate the downstream effector NICD with different dynamics.

Ans. (d)

Sol. The Notch signaling pathway is a highly conserved cell signaling pathway involved in various developmental processes. The pathway is activated by the binding of ligands, such as DII-1 and DIM, to the Notch receptors, such as Notch-1.

According to statement 4, DII-1 and DIM both bind to the Notch-1 receptor, but they activate the downstream effector, NICD (Notch Intracellular Domain), with different dynamics. This means that the signaling events triggered by DII-1 and DIM result in distinct temporal patterns or kinetics of NICD activation.

118. The relationship between species and area of distribution is given by the following equation S = CA^z where S is the number of species on an island or isolated patch, A is the area of the habitat, and C and Z are constants. The following are a set of statements pertaining to the value of 'Z'.

A. Z value is typically not greater than 0.4 across all ecosystem types

B. Z value is positively related to a species dispersal capability, with flying and wind-disperesed organisms having the highest values

C. Z value, which represents the slope in the relationship, declines with area, especially when large landmases such as contiments are considered

D. The Z value is the exponent in the power model and can be used to estimate the proportion of area required to represent a given proportion of species present in any land class.

Select the option that represents the combination of all correct statements

(a) A and B

(b) A and D

(d) C and D

Ans. (b)

Sol. 4.1.4 Species-area curve

A positive relationship exists between area and species diversity. Generally, large areas support more species than small areas do. The species-area relationship describes the general pattern of Increase In species richness with Increasing area of observation. In other words, It describes how the number of species In an area depends on the size of that area. Large areas contaln| more species than small areas because they can support larger populations and a greater range of habitats. The species-area relationship can be approximated by a power function of the form (proposed by Arrhenius In 1920 and modified by Gleason In 1922): S - cA'

where, S Is the number of species and A Is the area, c and z represent constants that depend on the type of species being considered and the type of habitat Involved. The exponent z Is generally small. In the range of 0.2 or 0.3. The species-area relationship usually Is portrayed graphically by plotting the logarithm of species number (S) versus the logarithm of area (A). After log-transformation, species-area relationship becomes linear. log S – logc + ; log A

119. Some ligands/stimuli that operate through G protein-coupled receptors (GPCRs) are listed in column X, and the most common effectors through which they act are listed in column Y.

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-i, B-ii, C-iii, D-iv

(b) A-ii, B-iii, C-iv, D-i

(d) A-iv, B-i, C-ii, D-iii

Ans. (c)

Sol. Serotonin: Serotonin, also known as 5-hydroxytryptamine (5-HT), is a neurotransmitter that regulates various physiological processes in the central nervous system. It plays a role in mood regulation, appetite, sleep, and other functions.

Acetylcholine: Acetylcholine (ACh) is a neurotransmitter that is involved in the transmission of nerve impulses between neurons and to muscle cells. It plays a crucial role in muscle contraction, memory, and learning.

IgE-antigen complexes: IgE (Immunoglobulin E) is a type of antibody that is produced in response to an allergic reaction. When IgE binds to antigens (foreign substances), it triggers an immune response and the release of inflammatory mediators, leading to allergy symptoms.

Light: Light is a form of electromagnetic radiation that is visible to the human eye. It acts as a stimulus for various physiological processes, including vision, regulation of circadian rhythm, and phototropism in plants.

120. The following are certain statements regarding the PSII electron carrier during the light reaction of photosynthesis

A. The first electron released from reacton centre P680 is transferred to Q_A to produce a plastosemiquinone

B. Q_A is the mobile plastoquinone

C. The first electron transferred from Q_A to Q_B converts Q_B into plastosemiquinone

D. Q_B is tightly bound to the complex and is not mobile.

Which one of the following options represents the correct statement(s)?

(a) A and B

(b) B and D

(d) C only

Ans. (c)

Sol. Reduced pheophytin transfers electron to plastoquinone. Plastoquinone is present in PSII both as a tightly-bound and a loosely-bound electron carrier, designated QA (primary quinone, bound to D2) and QB (secondary quinone, bound to DI), respectively.

Plastoquinone is a photosynthetically active quinone present in PSII. Other photosynthetically active quinones are ubiquinone and menaquinone in green and purple photosynthetic bacteria, and phylloquinone in PSI. Qa is photoreduced only to the plastosemiquinone, but QB can accept two electrons and two protons, forming a fully reduced plastoquinol (or plastohydroquinone).

121. Sucrose-phosphate synthase (SPS) and sucrose-phosphate (SPP) are two key enzymes involved in the biosynthesis of sucrose. Following are certain statements regarding these two enzymes.

A. Fructose -6-phosphate is one of the substrates of SPS enzyme

B. Fructose-6-phosphate and UDP-glucose are the substrates of SPP enzyme.

C. Sucrose is final product of SPP enzyme

D. UDP is one of the products of SPS enzyme while Pi is one of the products of SPP enzyme

Which one of the following options represents the combination of all correct statements?

(a) A, B and C

(b) A, B and D

(d) B, C and D

Ans. (c)

Sol.

122. The following list represents two types of reproductive isolation (Column P) that can lead to speciation. Column Q represents the processes by which these isolations can occur.

Select the option thar represents the correct match between the prezygotic and postzygotic isolation types listed in Column P and the processes described in Column Q?

(a) A-i, and ii, B-i, and iii

(b) A-i, and iii, B-ii, only

(d) A-ii, only, B-i, and iv

Ans. (c)

Sol. Prezygotic isolation is a type of reproductive isolation that prevents the formation of a zygote. Postzygotic isolation has referred to developmental defects in hybrids that lead to full or partial inviability and/or infertility. Postzygotic isolation includes hybrid inviability (development of the zygote proceeds normally but the hybrid does not survive) and hybrid sterility (the hybrid is healthy but sterile).

123. The following statements are made below about hearing phenomena of sound waves

A. The loudness of a sound is inversely correleated with the amplitude of a sound wave.

B. The loudness of a sound is directly correlated with the amplitude of a sound wave

C. The pitch of a sound is directly correlated with the frequency of the sound wave

D. The pitch of a sound is inversely correlated with the frequency of the sound wave

E. The pitch of the average mae voice in conversation is lower than that of the average female voice

F. The pitch of the average male voice in conversation is higher than that of the average female voice.

Choose the combination of all correct statements:

(a) A, C and E

(b) B, D and E

(d) A, D and F

Ans. (c)

Sol. Pitch refers to the perceived frequency of a sound wave. Loudness refers to the perceived intensity or volume of a sound. The loudness of a sound is directly correlated with the amplitude of a sound wave. The pitch of a sound is directly correlated with the frequency of the sound wave. The pitch of the average male voice in conversation is lower than that of the average female voice.

124. What is the pH of a 10^–7 M solution of HCl?

(a) 6.00

(b) 6.79

(d) 7.50

Ans. (b)

Sol. [H⁺] from HCI = 10"7 M

[H⁺] from H2O = 10'7 M

Total [H⁺] = 2 × IO"7 M

pH = –log (2 × IO"7)

= 7 – Iog2

= 7–0.301

= 6.69

125. ERK5 is a MAP kinase that is activated upon phosphorylation by MEK5. MEK5 binds with MEKK2 when co-expressed. HEK293 cells were transfected with plasmid encoding ERK5, along with plasmids encoding either MEK5 alone, or MEKK2 alone, or both MEKK2 and MEK5, or both MEKK2 and MEK5AA (MEK5 mutant). Lysates of transfected cells were analysed by Western blotting using anti-ERK5 antibody as shown below :

From the data in the figure above, the following conclusions were drawn:

A. Ful activation of ERK5 requires both MEKK2 and MEK5.

B. Phosphorylation with MEKK2 alone suggests that it can activate ERK5 without MEK5.

C. Difference in the levels of phosphorylation with MEKK2 alone and MEKK2 + MEK5 is due to more phosphorylation at the same site.

D. Phosphorylation with only MEKK2 transfection suggests that it might be associating with endogenoujs MEK5 to get partially activated, leading to ERK5 phosphorylation to some extent.

E. MEK5AA might be a dominant-negative mutant of MEK5 which prevents signalling through active, endogenous MEK5.

Which of the following options represents the combination of all correct statements?

(a) A, B and C

(b) B, C and E

(d) A, D and E

Ans. (d)

Sol. A. Full activation of ERK.5 requires both MEKK2 and MEK5. This statement suggests that both MEKK2 and MEK5 are required for the full activation of ERK.5. This conclusion is supported by the fact that in the presence of both MEKK2 and MEK5, the phosphorylation of ERK.5 is observed in the Western blot analysis.

D. Phosphorylation with only MEKK2 transfection suggests that it might be associating with endogenous MEK5 to get partially activated, leading to ERK5 phosphorylation to some extent. This statement suggests that when only MEKK2 is transfected, it can associate with endogenous MEK5 (naturally occurring MEK5 in the cells) and partially activate ERK.5, leading to some extent of ERK5 phosphorylation.

E. MEK5AA might be a dominant-negative mutant of MEK5 which prevents signaling through active endogenous MEK5. This statement suggests that MEK5AA, the mutant form of MEK5, acts as a dominant-negative mutant by preventing signaling through the active endogenous MEK5. This conclusion is drawn from the observation that when MEK5AA is present along with MEKK2, the phosphorylation of ERK.5 is significantly reduced compared to the condition with wild-type MEK5.

126. Mass spectrum of a pure peptide recorded in the poitive ion mode is shown below :

A. What is the reason for multiple peaks in the mass spectrum of a pure peptide?

B. Which peak corresponds to the monoisotopic species of the peptide?

C. What is the monoisotopic mass of the peptide?

Select the right answers from the options given below :

(a) A-¹³C isotype distribution, B-peak V and C-1128.56 Da

(b) A-¹⁴C isotype distribution, B-peak I and C-1124.55 Da

(d) A-¹⁴C isotype distribution, B-peak V and C-1128.56 Da

Ans. (c)

Sol. (A) The reason for multiple peaks in the mass spectrum of a pure peptide is due to the presence of isotopes, specifically the isotopic distribution of carbon-13 (13C). Carbon-13 is a naturally occurring stable isotope of carbon that has a slightly higher mass than the more abundant carbon-12 (12C). Since peptides contain carbon atoms, the presence of carbon-13 isotopes contributes to the appearance of multiple peaks in the mass spectrum.

(B) Peak I corresponds to the monoisotopic species of the peptide. The monoisotopic peak represents the mass of the peptide molecule containing the most abundant isotopes of each element present. It corresponds to the molecular weight of the peptide without any additional isotopic contributions.

(C) The monoisotopic mass of the peptide is 1124.55 Da. This value represents the mass of the peptide molecule calculated based on the monoisotopic masses of the constituent atoms and their respective isotopic abundances.

127. The cycling of monomeric G proteins, such as Ras, between active and inactive states is aided by accesssory proteins that bind to the G protein and regulate its activity

A. A non-functional GAP

B. A permanently activated GAP

C. A non-functional GEF

D. A permanently activated GEF

Which one of the following options repreent conditions/states that might cause a constantly activated signalling cascade?

(a) A and B

(b) B and C

(d) A and D

Ans. (d)

Sol.

128. Following statements are made regarding animal development

A. The cell is first specified towards a given fate, suggesting that it would develop into this cell type, even in a neutral environment.

B. Holoblastic rotational cleavage is observed in tunicates

C. Infoding of sheet of cells is called ingression

D. Conditional specification can be observed in sea urchin embroys.

Which one of the following options represents the combination of all correct statements?

(a) A and B

(b) B and C

(d) C and D

Ans. (c)

Sol.

129. A transgenic plant having a homozygous single-copy inertion for trait A was retranformed by Agrobacterium-mediated transformation with agene conferring trait B. Given below are a few statements regarding the above experiment.

A. All T₀ transgenic plants obtained after re-transformation would be single copy events for both traits, A and B.

B. T1 progeny generated by self-pollination of single-copy transgenic plants obtained by retransformation would segregate in a 3 : 1 ratio for trait A.

C. Plant selection marker genes used for transformation experiments for both traits, A and B should be necessarily identical. Different selection marker gene cannot be used.

D. 25% of T1 progeny generated by self-pollination of single-copy transgenic plants obtained by retransformation would be homozygous for both traits A and B.

Which one of the following option repreents all Incorrect statements?

(a) A, C and D

(b) A, B and C

(d) A and C only

Ans. (b)

Sol. Statement A is incorrect because not all transgenic plants obtained after re-transformation would necessarily be single-copy events for both traits. It is possible to have multiple insertions or varying copy numbers in the transgenic plants.

Statement B is incorrect because the T1 progeny generated by self-pollination of single-copy transgenic plants obtained by retransformation would not necessarily segregate in a 3:1 ratio for trait A. Segregation ratios can vary depending on the specific genetic inheritance pattern of trait A.

Statement C is incorrect because different selection marker genes can be used for transformation experiments for traits A and B. It is not necessary for the selection marker genes to be identical.

130. The following table depicts the digital numbers of 12 pixels in two different bands (indicated below each pixel group) of an image collected from the LISS-IV sensor of Resourcesat I satelite.

Which one of the following options repreents the correct identification of only vegetated (darkened) pixels?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Vegetated (darkened) pixels in satellite images typically represent areas with vegetation cover, such as forests, grasslands, or agricultural fields. These pixels appear darker because vegetation absorbs more of the incoming solar radiation compared to other land cover types. Satellite sensors, such as the LISS-IV sensor on the Resourcesat-1 satellite, capture electromagnetic radiation reflected or emitted by Earth's surface. These sensors are equipped with specific spectral bands that can detect different wavelengths of light. By analyzing the reflectance or emission of light in different spectral bands, scientists and researchers can gain insights into various aspects of Earth's surface, including vegetation cover.

131. Alleles A, a, B and b can be distinguished on the basis of their mobility on an agarose gel. These genes are present on the same chromosome. In the gel image below, the band pattern reflects the alleles in parents and their progeny (number reflects the progeny counted).

Which one of the following statements correctly explains the band pattern?

(a) In the heterozygous parent, the alleles are in coupling configuration

(b) In the heterozygous parent, the alleles are in repulion configuration

(d) The information in insufficient for any conclusion.

Ans. (b)

Sol.

132. In many sexually reproducting organisms, females make mate choice decision based on male display traits. Several models have been proposed to explain the evolution of exaggetation in male traits. Two of them have been given below in column P and their possible descriptions in column Q.

Match the models to their appropriate description and choose the correct option.

(a) A-i, B-ii

(b) A-ii, B-iii

(d) A-iv, B-i

Ans. (c)

Sol. A. Runaway selection iii. Female choice for male trait results in a positive feedback loop favoring both males with such trait and females that prefer them.

B. Chase-away selection iv. Males exploit pre-existing sensory bias in females. Females do not benefit by choosing such males, driving the evolution of females that discriminate against such males.

133. A researcher isolated chromatin from mammalian cells and digested with micrococal nuclease in different tubes for 1 h, 3 h, 6 h and 12 h. Thereafter, DNA ws purified from all the digested chromatin samples, and two independent Southern blot hybridization expeirments were performed with probe-I and probe II. The probe-I and probe-II correspond to different loci of the chromosome(s). The images below represent the Southern blot hybridization pattern generated by probe-I and probe-II.

Following statements were made to explain the results of Southern blot experiments

A. Size of probe-II is smaller than probe-I

B. Probe-I may correspond to the centromeric region of the chromosome.

C. Probe-I may correspond to a hypomethylated locus of the genome

D. Probe-II may correspond to an euchromatic locus of the genome

Following statements were made to explain the results of the Southern blot experiments

(a) A and C

(b) B and D

(d) B only

Ans. (b)

Sol. Micrococcal nuclease (Mnase), derived from Staphylococcus aureus, is an endoexonuclease that exhibits relatively nonspecific activity towards both RNA and DNA substrates. The centromeric region of the chromosome, known as the heterochromatin region, is highly condensed. As a result, it is less susceptible to digestion by Mnase.

Consequently, when subjected to Mnase treatment in separate tubes for 1 hour, 3 hours, 6 hours, and 12 hours, the level of digestion is not significantly high in this region. On the other hand, the euchromatin region of the chromosome, which is less condensed, is more sensitive to Mnase.

134. Given below are a set of statements about metapopulations dynamics and habitat conservation :

A. The sizes of suitable patche are important because demographnic stochasticity can lead to extinction, especially in organisms with low reproductive output.

B. In the incidence function model (IFM), the extinction risk of local populations increases with increasing habitat patch area, and the colonization probability is a function of patch isolation form existing local populations.

C. From the conservation perspective, large numbers of suitable patches are not sufficient if distances are too large, preventing recolonization and the rescue effect.

D. To minimize extincton risk there should be as low a variance in local patch quality as possible, to allow for synchronous dynamics.

Which one of the following options represents the combination of all correct statements?

(a) A and C

(b) B and C

(d) B and D

Ans. (a)

Sol. A. The sizes of suitable patches are important because demographic stochasticity can lead to extinction, especially in organisms with low reproductive output. This statement highlights the importance of patch size in metapopulation dynamics. In organisms with low reproductive output, small populations are more susceptible to demographic stochasticity, which refers to random fluctuations in birth rates, death rates, or other demographic processes. These fluctuations can lead to increased extinction risk. Therefore, larger suitable patches are important for maintaining viable populations.

C. From the conservation perspective, large numbers of suitable patches are not sufficient if distances are too large, preventing recolonization and the rescue effect. This statement emphasizes the significance of both patch number and connectivity in metapopulation dynamics and habitat conservation. While having a large number of suitable patches is important, it is not sufficient if the distances between the patches are too large. If patchesare too isolated, it can hinder the movement and recolonization of individuals, which can limit the rescue effect (where immigrants from other patches rescue declining populations). Therefore, connectivity between suitable patches is crucial for the long-term persistence of metapopulations.

135. Match the eukaryotic cellular organelles listed in Column X with their typical function from among those isted in Column Y.

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-(iii), B-(iv), C-(ii), D-(i)

(b) A-(ii), B-(i), C-(iv), D-(iii)

(d) A-(iii), B-(i), C-(iv), D-(ii)

Ans. (b)

Sol. Like mitochondria, peroxisomes contain several oxidative enzymes, such as catalase, oxidases. Peroxisomal oxidases transfer hydrogen atoms to molecular oxygen and form hydrogen peroxide. The enzyme catalase (a member of the peroxidase family) present in the peroxisome converts H2O2 to O2. In general, hydrogen peroxide can be Some proteins undergo O-glycosylation in the cisternae. O-glycosylation involves an oxygen-carbon bond (O-glycosidic bond) between carbohydrate (termed O-glycan) and the hydroxyl group of a serine or threonine (to a lesser extent, hydroxyproline and hydroxylysine) amino acid residue of the protein.

The SER produces most of the membrane lipids, including both phospholipids and cholesterol. Phospholipids are amphipathic molecules and their synthesis from fatty acyl-CoAs and glycerol-3-phosphate take place at the cytosolic leaflet of ER membrane and are catalyzed by membrane-associated enzymes.

Nucleolus (first described by F. Fontana) is a non-membrane bound dynamic body which disappears in the late prophase and reappears in the telophase stage of cell division. Each nucleolus is produced by a Nucleolus-Organizing Region (NOR) of a chromosome which is termed as nucleolus-organizing chromosome. All eukaryotic cells contain at least one such chromosome. Its number may be one or more (several hundred per nucleus), but 1 to 4 being the most common. The number of nucleoli per nucleus also differs. The yeast cell contains one relatively large nucleolus with respect to its nuclear volume. At the other extreme, Xenopus oocytes contain over 1000 nucleoli per nucleus. It is a site of transcription of ribosomal RNA and assembly of ribosome. rRNA genes present in the NOR of chromosome is responsible for synthesizing a large nascent pre-rRNA that is 45S in mammals and the processing (cleavage and base modification) of this RNA yield mature ribosomal RNA of 18S, 5.8S and 28S.

136. Amborellaceae, Aristolochiaceae, Illiciaceae and Winteraceae are four angiosperm families that, according to APV IV system of classification belong to the 'early diverging angiosperms'. The presence (V+) or absence (V–) of vessels in the xylem and the fusion of the carples within the gynoecium are important angiosperm characters. 'A' and 'S' indicate apocarpous (or monocarpellary) and syncarpous condition of ovary respectively. Which one of the following options correctly repreents the characters found in the above families?

(a) Amborellaceae : V + A; Aristolochiaceae : V+, A; Illiciaceae: V+, A; Winteraceae : V–, S.

(b) Amborellaceae : V + A; Aristolochiaceae : V+, S; Illiciaceae: V+, A; Winteraceae : V–, A

(d) Amborellaceae : V + S; Aristolochiaceae : V–, A; Illiciaceae: V+, S; Winteraceae : V–, S

Ans. (b)

Sol. This means that Amborellaceae does not have vessels in the xylem (V^–) and has an apocarpous condition of the ovary (A). Aristolochiaceae has vessels in the xylem (V⁺) and a syncarpous condition of the ovary (S). Illiciaceae has vessels in the xylem (V⁺) and an apocarpous condition of the ovary (A). Winteraceae does not have vessels in the xylem (V^–) and has an apocarpous condition of the ovary (A).

137. The following statements are made with reference to the neural connections of cardiac tissues and the functions of these nerves on heart in adult humans :

A. The right vagus nerve is distributed mainly to the AV node

B. The parasympathetic pre-ganglionic fibers distributed to the heart originate from the superior salivatory nucleus

C. The sympathetic post-ganglionic fibers originating from the paravertebral ganglia of the left side primarily innervate SA node.

D. The sympathetic activity alters heart rate slower than that of vagal activity

Which one of the following options represents the combination of all correct statements?

(a) A and B

(b) B and C

(d) D and E

Ans. (d)

Sol. D. The sympathetic fibers distributed to the heart come mainly from stellate ganglia. This statement is correct. The sympathetic fibers that innervate the heart mainly originate from the stellate ganglia, which are located in the lower cervical and upper thoracic regions of the spinal cord. These fibers contribute to the sympathetic control of heart rate and cardiac function.

E. The sympathetic activity alters heart rate slower than that of vagal activity. This statement is correct. Vagal (parasympathetic) activity has a more immediate effect on heart rate, causing a rapid decrease in heart rate. In contrast, sympathetic activity has a slower effect on heart rate, resulting in an increase in heart rate over a slightly longer period.

138. Given below are a few terms related to Bioinformatics resources (Column X) and their functions/applications (Column Y):

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-(ii), B-(iii), C-(iv), D-(i)

(b) A-(iv), B-(i), C-(ii), D-(iii)

(d) A-(iv), B-(iii), C-(ii), D-(i)

Ans. (a)

Sol. TrEMBL: TrEMBL (Translated EMBL Nucleotide Sequence Data Library) is a protein sequence database that contains translations of nucleotide sequences from the EMBL (European Molecular Biology Laboratory) Nucleotide Sequence Database. It provides comprehensive protein sequence information and is frequently used for protein sequence analysis and annotation.

TBLASTN: TBLASTN is a tool and algorithm used in bioinformatics for comparing a protein query sequence against a nucleotide database. It is specifically designed for finding protein-coding regions in DNA or RNA sequences. TBLASTN allows researchers to search for similarities between a protein sequence of interest and nucleotide sequences to identify potential coding regions or homologous sequences.

SCOP: SCOP (Structural Classification of Proteins) is a database that classifies proteins based on their structural and evolutionary relationships. It provides a hierarchical classification scheme that organizes protein structures into different classes, folds, superfamilies, and families. SCOP is widely used in structural biology and protein structure analysis to understand the relationships and functional implications of different protein structures.

JoinMap: JoinMap is a software tool used in genetic mapping and linkage analysis. It is specifically designed to analyze and map genetic markers based on data from crosses or populations. JoinMap assists in constructing genetic maps, identifying marker linkage groups, and estimating recombination frequencies between markers. It is commonly used in genetic research, plant breeding, and genetic mapping studies.

139. The following statements are made regarding conversion of pyruvate to acetyl-CoA going from glycolysis to cirtric acid cycle :

A. Oxidation of pyruvate to acetyl-CoA is reversible

B. Pyruvate is transported into the mitochondrion by a transporter

C. Pyruvate is carboxylated by pyruvate dehydrogenase

D. Acetyl lipomaide react with coenzyme A to form acetyl-CoA

E. The flavoprotein dihydrolipoyl dehydrogenase, containing flavin adenine dinucleotide (FAD), is involved in conversion of pyruvate ot acetyl-CoA.

Which one of the following options represents the combination of all correct statements?

(a) A, B and C

(b) B, C and D

(d) B, D and E

Ans. (d)

Sol. The conversion of pyruvate to acetyl-CoA, catalyzed by highly organized multienzyme pyruvate dehydrogenase complex, is an oxidative decarboxylation process. In the overall reaction, the carboxyl group of pyruvate is lost as CO₂, while the remaining two carbons form the acetyl moiety of acetyl-CoA. The reaction is highly exergonic and essentially irreversible in vivo. E1 first catalyzes the decarboxylation of pyruvate, producing hydroxyethyl-TPP, and then the oxidation of the hydroxyethyl group to an acetyl group. The electrons from this oxidation reduce the disulfide of lipoate bound to E2, and the acetyl group is transferred into thioester linkage with on -SH group of reduced lipoate. E2 catalyzes the transfer of the acetyl group to coenzyme A, forming acetyl-CoA. E3 (Dihydrolipoyl dehydrogenase) catalyzes the regeneration of the disulfide (oxidized) form of lipoate; electrons pass first to FAD and then to NAD⁺.

140. The recognized family of PR protein in plants and their activities are listed in column X and Y respectively

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-(i), B-(iii), C-(iv), D-(ii)

(b) A-(i), B-(ii), C-(iii), D-(iv)

(d) A-(iv), B-(i), C-(ii), D-(iii)

Ans. (b)

Sol. PR2: PR2 proteins are commonly known as â-1,3-glucanases. These enzymes are involved in breaking down â-1,3-glucans, which are major components of fungal cell walls. By degrading these cell wall components, PR2 proteins contribute to the plant's defense against fungal pathogens.

PR3: PR3 proteins are classified as chitinases, enzymes that hydrolyze chitin, a polysaccharide present in the cell walls of fungi and exoskeletons of insects. Chitinases help in defense against fungal pathogens and are also implicated in plant growth and development.

PR6: PR6 proteins are thaumatin-like proteins (TLPs) or pathogenesis-related protein 5 (PR-5). TLPs are known for their antifungal properties. They have a conserved structure and exhibit high stability under various conditions. TLPs have been suggested to play a role in defense against fungal pathogens and in plant responses to abiotic stresses.

PR10: PR10 proteins, also called Bet v 1-like proteins, belong to the pathogenesis-related protein 10 family. These proteins are characterized by a conserved fold called the Bet v 1 domain. PR10 proteins are involved in defense against pathogens and have been associated with allergic responses in some cases.

141. Following statements are made regarding the roles of complement components in immunity:

A. Binding of complement components to antigen presenting cells decreases their phagocytic ability and modulates cytokine secretion.

B. Complement components enhance the B cell-mediated immune response by increasing the avidity with which a B cell binds to a complement-bound antigen

C. Immature T cells are protected from thne natural antibody ad complemet mediated lysis by provision of additional sialic acid residues on their cell surface glycoproteins

D. Bindig of C3a, C5a and C3b to their respective receptors on mature T cells inhibits their growth, differentiation, and survival.

E. During the contraction phase of the immune response, excess lymphocytes that were made during antigen-induced expansion are eliminated by apoptosis, with the help of Clq complement component.

Which of the following options represents the combination of all correct statements :

(a) A, B and C

(b) A, C and D

(d) B, D and E

Ans. (c)

Sol. B. Complement components enhance the B cell-mediated immune response by increasing the avidity with which a B cell binds to a complement-bound antigen. This statement is incorrect. Complement components can indeed enhance the B cell-mediated immune response, but they do so by facilitating the recognition and uptake of complement-bound antigens by B cells. Complement components can opsonize pathogens or immune complexes, leading to more efficient antigen recognition by B cells. C. Immature T cells are protected from natural antibody and complement-mediated lysis by the provision of additional sialic acid residues on their cell surface glycoproteins. This statement is correct. Immature T cells express sialic acid residues on their cell surface glycoproteins, which provide protection against natural antibody and complement-mediated lysis. Sialic acid residues act as self-markers and help prevent immune attack on immature T cells.

E. During the contraction phase of the immune response, excess lymphocytes that were made during antigen-induced expansion are eliminated by apoptosis with the help of Clq complement component. This statement is correct. During the contraction phase of the immune response, excess lymphocytes that were produced during antigen-induced expansion are eliminated through apoptosis. Clq, a component of the complement system, can facilitate the clearance of apoptotic cells.

142. Given below are the list of abiotic environmental factors (Column X) and their primary effect (Column Y) in plants

Which one of the following options represents all correct matches?

(a) A-(i), B-(ii), C-(iii), D-(iv)

(b) A-(iv), B-(i), C-(ii), D-(iii)

(d) A-(iv), B-(iii), C-(ii), D-(i)

Ans. (b)

Sol. Factor: Water deficit Effect: Water potential reduction Explanation: During a water deficit, plants experience a lack of water availability. This leads to a reduction in the water potential within the plant cells, causing decreased turgor pressure and ultimately resulting in wilting and reduced plant growth.

Factor: Salinity Effect: Ion cytotoxicity Explanation: Salinity refers to high salt levels in the soil or water. Excessive salts can accumulate in plant tissues, leading to ion cytotoxicity. The high concentration of ions, particularly sodium ions, can disrupt cellular processes, hinder water uptake, and damage plant cells.

Factor: Flooding Effect: Hypoxia to the roots Explanation: Flooding restricts the oxygen supply to plant roots, leading to hypoxic conditions or even anoxia (complete lack of oxygen). This lack of oxygen impairs root respiration and nutrient uptake, negatively affecting plant growth and survival.

Factor: High light intensity Effect: Photoinhibition Explanation: When plants are exposed to high light intensity, excessive light energy can overwhelm the photosynthetic machinery. This can lead to photoinhibition, where the photosynthetic processes are disrupted, and the production of reactive oxygen species increases, causing oxidative stress and damage to the photosynthetic apparatus.

143. A scientist is using the Hardy-Weinberg equation to assess if a population is in equilibrium or is evolving. She recorded the following characteristics for this population

A. The size of the population is very large

B. Individuals are randomly mating

C. Individuals are under natrual selection

D. New alleles are added to the population through migration and dispersal.

E. Mutation rates are high

Which one of the following options contais all Incorrect characteristics of a population in Hardy-Weinberg equilibrium?

(a) A and D

(b) C, D and E

(d) B and E

Ans. (b)

Sol. Hardy-Weinberg principle is based on certain assumptions. The major assumptions are necessary for the Hardy-Weinberg principle to hold - Random mating, No natural selection, No mutation, No migration and Large population size.

144. A researcher wanted to test the effect of different chemical agents on double stranded DNA (dsDNA). dsDNA was taken in tubes (A, B, C and D) and four different agents were added individually to each tube. The researcher however forgot to label them. The properties of the added chemical agents on dsDNA were analysed. The possible product (Column X) and the specific action of the chemical agents are listed in Column Y.

Which one of the following options represents all correct matches between Column X and Column Y?

(a) A-(ii), B-(i), C-(iv), D-(iii)

(b) A-(iii), B-(iv), C-(ii), D-(i)

(d) A-(iii), B-(ii), C-(iv), D-(i)

Ans. (b)

Sol. Column X: Possible Product

Tube A: Nucleotides

Tube B: Only nitrogenous bases

Tube C: Nucleosides

Tube D: ssDNA

Column Y: Specific Action

Tube A: Breaks phosphodiester bond

Tube B: Breaks hydrogen bond

Tube C: Removes phosphate group

Tube D: Breaks N-glycosidic bond

145. Following statements are made regarding the plant natural product, terpenes.

A. Monoterpenes are five-carbon compounds

B. The anti-malarial drug, artemisinin i a sesquiterpene

C. Azadirachtin is a triterpene derivative from the seed oil of the Asian neem tree

D. Taxol is a diterpene derivative used in cancer treatment

Which one of the following options represents the combination of all correct statemets?

(a) A and B only

(b) C and D only

(d) B, C and D

Ans. (d)

Sol.

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