PYQ SEPTEMBER 2022 SIFT 2

CSIR NET BIOLOGY (Sept. - 2022 Sift 2)
Previous Year Question Paper with Solution.

21. The B-form double stranded DNA was invaded by a complementary RNA sequence to form an R loop structure. During this process,

(a) sugar puckering on the DNA strand that pairs with RNA will remain unchanged

(b) sugar puckering on the DNA strand that pairs with RNA will change

(c) sugar puckering on the DNA strand that pairs with RNA will remain unchanged but the number of base pairs per turn in the RNA-DNA hybrid will increase.

(d) sugar puckering on the DNA strand that pairs with RNA will change but the number of base pairs per turn in the RNA-DNA hybrid with remain unchanged

Ans. (b)

Sol. When RNA pairs with the DNA in an R-loop structure, there will be a change in the sugar puckering conformation of the DNA strand that pairs with RNA. This is because the ribose sugar in the RNA forms a different conformation compared to the deoxyribose sugar in DNA.

22. Which one of the following is NOT CORRECT in the context of protein structure and folding?

(a) -sheets are more common in the interiors of proteins than surfaces

(b) -sheets are less likely to form than -helices in the earliest stages of protein folding

(d) -helices are less likely to form than -sheets in the earliest stages of protein folding

Ans. (d)

Sol. -helices are actually more likely to form in the earliest stages of protein folding compared to -sheets.

23. Which of the following represents the most oxidized form of carbon?

(a) HCOOH

(b) HCHO

(d) CO₂

Ans. (d)

Sol. In carbon dioxide, carbon is in the +4 oxidation state since it has two double bonds to oxygen (each with an oxidation state of -2).

Since carbon is in the highest oxidation state (+4) in CO2, it represents the most oxidized form of carbon among the given compounds

24. Catabolic end product of purines is

(a) Xyloric acid

(b) Allantoin

(d) Uric acid

Ans. (d)

Sol. Purines are nitrogenous bases found in nucleotides, which are the building blocks of DNA and RNA. The catabolic breakdown of purines results in the production of uric acid. This process occurs as part of the purine degradation pathway.

Here is a brief overview of the breakdown process:

Purines (adenine and guanine) are converted to hypoxanthine and xanthine, respectively, through enzymatic reactions.

Hypoxanthine is further broken down to xanthine, and xanthine is finally converted to uric acid by the enzyme xanthine oxidase

25. CENP-A containing nuclesomes are found at the centromeric region of the chromosomes. CENP-A is a variant of which one of the following histones?

(a) H1

(b) H2A

(d) H4

Ans. (c)

Sol. CENP-A, also known as centromere protein A, is a variant of histone H3. It is a specialized histone protein that plays a crucial role in centromere formation and function. Centromeres are specific chromosomal regions essential for proper chromosome segregation during cell division. CENP-A is specifically localized to the centromeric regions and is involved in building the kinetochore, a protein structure that attaches the chromosomes to the mitotic spindle during cell division. This ensures the accurate distribution of genetic material to daughter cells.

26. Which one of the following is abundant in the plasma membranes of mammalian cells but is absent from most prokaryotic and plant cell membranes?

(a) Phosphoglycerides

(b) Ergosterol

(d) Cholesterol

Ans. (d)

Sol. Cholesterol is abundant in the plasma membranes of mammalian cells but is absent from most prokaryotic and plant cell membranes. Cholesterol is a type of lipid and is an essential component of animal cell membranes. It plays a vital role in regulating the fluidity and permeability of the membrane, which affects the overall membrane structure and function.

27. In Saccharomyces cerevisiae, DNA replication is tightly controlled, and DNA should replicate once per cell cycle. Choose the INCORRECT statement regarding why the cells do not re-replicate their DNA in the S-phase.

(a) Pre-replicative complex (Pre-RC) remains bound to the DNA in the S-phase and does not allow the re-replication

(b) Assembly of Pre-RC is inhibited by Cdk activity Cdk

(c) Assembly of Pre-RC is initiated at the end of mitosis, at the early G1 phase of the cell cycle (when the APC activity is high)

(d) Cdt1 that helps in the recruitment of MCM proteins in the G1phase is inactivated by geminin in the S-phase of the cell cycle Cdt1

Ans. (a)

Sol. The pre-replicative complex (Pre-RC) is a protein complex that forms at the origins of replication during the G1 phase of the cell cycle. It is responsible for initiating DNA replication. Once DNA replication starts in the S-phase, the Pre-RC components get displaced and are no longer present at the origins of replication. This ensures that each origin is only used once during a single cell cycle, preventing re-replication of the DNA.

28. To study the cell cycle progression for cultured mammalian cells, one would typically NOT utilize?

(a) Artificial thymidine analog BrdU

(b) Kinase inhibitor, LY294002

(d) Live cell imaging

Ans. (b)

Sol. Kinase inhibitor, LY294002: LY294002 is a specific inhibitor of phosphoinositide 3-kinases (PI3Ks), which are enzymes involved in various signaling pathways regulating cell growth, survival, and proliferation. Using a kinase inhibitor like LY294002 may interfere with specific signaling pathways, affecting cell cycle progression and complicating the interpretation of the results. Thus, it is not typically used when studying the normal cell cycle.

29. Which one of the following statements about DNA replication is INCORRECT?

(a) Once DNA replication commences, it always continues uninterrupted until the entire process is complete.

(b) Eukaryotic genomes replicate from multiple origins of replication

(d) Both, fully methylated as well as non-methylated oriC can initiate DNA replication, while hemi-methylated oriC does not.

Ans. (a)

Sol. DNA replication is a complex and highly regulated process, and there are several factors that can cause interruptions or stalls during replication. Some of the reasons for interruptions in DNA replication include:

DNA damage: If the replication machinery encounters damaged bases or lesions in the DNA template, it may stall or pause to allow repair enzymes to fix the damage before continuing.

Replication fork collapse: Replication forks can collapse due to various reasons, such as collisions with other cellular structures or obstacles on the DNA.

Stalling by certain DNA sequences: Certain sequences, known as replication fork barriers, can impede the progression of the replication machinery, causing pauses or interruptions.

Stalling by secondary structures: Secondary structures, such as hairpin loops, can form inthe DNA template, leading to replication stalling.

Replication termination: Replication must be precisely terminated once the entire DNA molecule is copied. This process is tightly regulated and can cause a temporary interruption before the process is completed.

30. The long DNA strand depicted below is serving as a template for lagging strand DNA synthesis. The short lines represent the newly synnesized Okazaki fragments.

At which positions among A, B, C and D would DNA primase act next?

(a) A

(b) B

(d) D

Ans. (d)

Sol. In the given template, DNA primase would act next at position D, which is the 3' end of the previous Okazaki fragment. The primase will synthesize a short RNA primer complementary to the template strand at position D, allowing DNA polymerase to initiate the synthesis of the next Okazaki fragment. This process continues as the lagging strand is synthesized in short fragments away from the replication fork.

31. The post-translational modifications in one or more core histones that are known to be associated with DNA repair pathways are:

(a) Phosphorylation at specific tyrosine residues

(b) Ubiquitination at specific lysine residues

(d) Methylation at specific serine residues

Ans. (b)

Sol. Ubiquitination is the process of adding a small protein called ubiquitin to a target protein. In the context of DNA repair, ubiquitination of histones can serve as a signal to recruit DNA repair proteins to damaged sites and facilitate the repair process. It can also regulate the chromatin structure to allow access to the DNA repair machinery.

32. The amino acid arginine is encoded by six codons : CGU, CGC, CGA, CGG, AGA and AGG. Assuming inosine is not an option in the tRNA anticodon, what is the minimum number of tRNAs (from theoptions given below) that would be sufficient to read these codons?

(a) Six

(b) Four

(d) Five

Ans. (c)

Sol. The minimum number of tRNAs required is determined by the number of unique anticodons necessary to recognize all six codons for arginine. Let's list the anticodons that can pair with the arginine codons:

CGU - Anticodon: GCA

CGC - Anticodon: GCG

CGA - Anticodon: GCU

CGG - Anticodon: GCC

AGA - Anticodon: UCU

AGG - Anticodon: UCC

33. Which one of the following proteins is NOT related to extracellular matrix?

(a) Cadherin

(b) Vitronectin

(d) Selectin

Ans. (c)

Sol. Lamin is a type of intermediate filament protein found in the nuclear lamina, which is a structure located inside the cell nucleus. It provides structural support to the nuclear envelope and is not directly related to the extracellular matrix (ECM).

34. Which one of the following statements about cancers is INCORRECT?

(a) The c-myc gene is translocated to one of the immunoglobulin loci in a majority of Burkitt's lymphomas.

(b) Viral integration into the cellular genome may convert a proto-oncogene into an oncogene

(d) Many cases of metastatic breast cancer display increased expression of human epidermal-growth-factor–like receptor 2(HER2).

Ans. (c)

Sol. The functions of p53 and Rb (Retinoblastoma) are actually inhibited by E6 and E7 proteins of human papillomavirus (HPV), not augmented. HPV is a group of viruses known to cause cervical cancer and is also associated with other types of cancers.

35. What is the nature of the successful anti-cancer Human Papilloma Virus (HPV) vaccine?

(a) Cyhemically inactivated virus

(b) Live attenuated mutant form of HPV HPV

(d) mRNA vaccine expressing viral L1 protein mixed with recomibnant viral proteins :

Ans. (c)

Sol. The L1 major capsid protein is the major structural protein of HPV, and when expressed in a recombinant system, it can self-assemble into VLPs. These VLPs resemble the authentic HPV virions, but since they lack the viral DNA or RNA, they cannot cause infection or replication. However, they can stimulate a strong immune response when introduced into the body.

36. An anti-idiotypic antibody with fluorescent tag was used for detection of immune cells in tissue sections fro a healthy individual specifically by cell surface labeling. Which of the following will have the highest chances of getting detected?

(a) Macrophages in lymph nodes

(b) Mature B cells in spleen

(d) Eosinophils in tonsils

Ans. (b)

Sol. In healthy individuals, mature B cells are present in the spleen, where they are involved in immune responses, including antibody production. When the anti-idiotypic antibody with the fluorescent tag is used for cell surface labeling, it will specifically detect and bind to the antigenic determinants on mature B cells in the spleen, resulting in their fluorescent labeling and detection.

37. Which one of the following best describes the ability of the cells to respond to a specific inducing signal?

(a) Potency

(b) Equivalence

(d) Specification

Ans. (c)

Sol. Competence refers to the ability of cells to respond to specific inducing signals or developmental cues. It means that the cells have reached a certain developmental stage where they are capable of responding to external signals and undergoing specific differentiation or developmental pathways.

38. The programmed cell death that separates the digits during a tetrapod limb development is dependent on which one of the following signaling pathways?

(a) BMP

(b) FGF

(d) Shh

Ans. (a)

Sol. BMP signaling promotes apoptosis in the regions between the digits, known as the interdigital regions, leading to the separation of the digits. This process ensures that the developing limb acquires its characteristic shape and structure with distinct fingers and toes.

39. Which one of the following mRNAs is a BMP inhibitor and can rescue the dorsal structures of ventralized Xenopus embryo when injected into it?

(a) beta-catenin

(b) Noggin

(d) Siamos

Ans. (b)

Sol. Noggin is an mRNA that encodes a bone morphogenetic protein (BMP) inhibitor. It acts as an antagonist of BMP signaling and can block the activity of BMPs. When Noggin mRNA is injected into ventralized Xenopus embryos (embryos with a loss of dorsal structures due to increased BMP signaling), it can rescue the dorsal structures by inhibiting BMP activity.

40. A type of regeneration in which the differentiated cells divide, maintaining their differentiated function without dedifferentiation and production of undifferentiated is known as

(a) Epimorphosis

(b) Morphallaxis

(d) Stem cell mediated regeneration

Ans. (c)

Sol. Compensatory regeneration is a type of regeneration in which differentiated cells proliferate and replace damaged or lost tissues, maintaining their differentiated function without dedifferentiation into stem cells or undifferentiated cells. This process allows the damaged tissue to be restored without the need for significant cell dedifferentiation or formation of a blastema, which is a mass of undifferentiated cells.

41. Which one of the following statements is INCORRECT?

(a) Dehydrins are intrinsically disordered proteins

(b) Dehydrins have minimal secondary structure.

(d) Dehydrins are highly hydrophobic proteins.

Ans. (d)

Sol. Dehydrins are a class of proteins that are induced in response to various abiotic stresses, such as dehydration (water deficit), cold, and salinity. They are known for their protective role in plants under stressful conditions. Dehydrins are characterized by their high hydrophilicity (water-attracting) rather than hydrophobicity (water-repelling).

42. Which of the following domains is present in symbiosis receptor-like kinase (SYMRK) proteins?

(a) Nucleotide binding repeat

(b) Leucine-rich repeat region

(d) W-box

Ans. (b)

Sol. The leucine-rich repeat (LRR) region is a common domain present in SYMRK proteins.

LRRs are structural motifs that are involved in protein-protein interactions and are frequently found in receptor-like kinases (RLKs). The LRR region allows the receptor to recognize specific ligands, such as Nod factors from rhizobia or signals from mycorrhizal fungi, initiating the symbiotic signaling pathway.

43. Identify the correct site of action of DBMIB (2,5-dibromo-3-methyl-6-isopropyl-pbenzoquinone), an inhibitor of the chloroplast electron transport chain.

(a) Q_A Q₈

(b) Q_B Q₈

(d) cytb₆f PC

Ans. (c)

Sol. In the chloroplast electron transport chain, PQ (plastoquinone) is an essential mobile electron carrier that shuttles electrons between Photosystem II (PSII) and the cytochrome b6f complex (Cytb6f). The Cytb6f complex plays a crucial role in transferring electrons from PQ to plastocyanin (PC), which is then used to reduce Photosystem I (PSI) during photosynthesis.

44. The plant hormone gibberellins (GA) are a group of

(a) Monoterpenes (C₁₀)

(b) Diterpenes (C₂₀)

(d) Sequiterpenes (C₁₅)

Ans. (b)

Sol. Gibberellins (GAs) are a group of plant hormones that belong to the diterpene class of compounds. Diterpenes are a class of terpenes, which are naturally occurring organic compounds derived from the isoprene unit (C5H8). Diterpenes consist of 20 carbon atoms (C20) and are known for their diverse biological activities in plants.

45. Hemoglobin A1c (HbA1c) from diabetic mellitus individuals has a glucose molecule attached to which one of the terminal amino acid residues of globin chain?

(a) Lysine of each chain

(b) Lysine of each chain

(d) Valine of each chain

Ans. (d)

Sol. In particular, glucose attaches to the N-terminal valine of the chains (not -chains) of hemoglobin. This process, known as glycation or glycosylation, occurs when glucose in the blood reacts with the N-terminal valine, forming a stable adduct that remains for the lifespan of the red blood cell. The level of HbA1c in the blood reflects the average blood glucose levels over the previous 2-3 months, providing valuable information about long-term glucose control in individuals with diabetes.

46. The pericytes are found in

(a) myelin sheath

(b) surrounding coat of a skeletal muscle fibre

(d) lymph capillary wall

Ans. (c)

Sol. Pericytes are specialized contractile cells found in the walls of blood capillaries. They are located in close association with endothelial cells, which form the inner lining of blood vessels. Pericytes play essential roles in blood vessel stabilization, regulation of blood flow, and the maintenance of the blood-brain barrier.

47. The reabsorption of water and NaCl in kidneys is inhibited by the increased secretion of the following substances EXCEPT one:

(a) Urodilatin

(b) Uroguanylin

(d) Norepinephrine

Ans. (d)

Sol. Norepinephrine, on the other hand, has an antidiuretic effect. It acts on the kidneys to promote the reabsorption of sodium and water, reducing urine output and helping to conserve body fluids.

48. Mullerian-inhibiting substance (MIS), a homodimer that causes regression of the Mullerian duct by apoptosis, is secreted by which one of the following cells?

(a) Leydig cells

(b) Sertoli cells

(d) Placental cells

Ans. (b)

Sol. In males, Sertoli cells secrete MIS/AMH, which causes regression and inhibition of the development of the Mullerian ducts, the precursor structures to female reproductive organs such as the fallopian tubes, uterus, and upper part of the vagina. By inhibiting the Mullerian ducts, MIS/AMH promotes the development of male reproductive structures, including the epididymis, vas deferens, and seminal vesicles.

49. In which one of the following human disorders, parents or grandparents are said to carry premutations?

(a) Down syndrome

(b) Fragile X syndrome

(d) Alkaptonuria

Ans. (b)

Sol. Fragile X syndrome is a genetic disorder caused by an expansion of a specific DNA sequence, known as CGG repeats, in the FMR1 (Fragile X Mental Retardation 1) gene. This expansion leads to the inactivation of the FMR1 gene and the loss of the FMRP (Fragile X Mental Retardation Protein), which is important for normal brain development and function.

50. The AFLP technique generates polymorphic DNA fragments that are generally scored as dominant markers. However, a pair of DNA fragments (say a and b) generated by AFLP can be termed as co-dominant, if on analysis of a large progeny of doubled haploids (DH) derived from and F₁ (resi;tomg from, a cross between two parents one with fragment 'a' and the other with 'b'), it is observed that :

(a) 50% of the progeny has both 'a' and 'b' fragments and the rest have none.

(b) 50% of the progeny has fragment 'a' and the remaining have fragment 'b'

(d) 75% of the progeny has both the fragments, while 25% has either 'a' or 'b'.

Ans. (b)

Sol. In AFLP (Amplified Fragment Length Polymorphism) analysis, the technique generates polymorphic DNA fragments that are scored as dominant markers. This means that the presence or absence of a fragment is determined, but the intensity of the bands is not considered.

When analyzing a large progeny of doubled haploids (DH) derived from an F1 cross between two parents, one with fragment 'a' and the other with 'b', a pair of DNA fragments (a and b) generated by AFLP can be termed as codominant if both fragments are present in different individuals of the progeny.

51. If a gamete produced following non disjunction of a chromosome at second meiotic division was fertilized by a normal gamete, what is the expected frequency of trisomic progeny?

(a) ¼

(b) 2/4

(d) 1

Ans. (a)

Sol. Let's consider a specific scenario where non-disjunction occurs during the second meiotic division. In this case, one of the gametes produced will have an extra copy of a particular chromosome, making it trisomic (2n + 1), and the other gamete will have one less copy of that chromosome, making it monosomic (2n - 1).

Now, if the trisomic gamete is fertilized by a normal gamete (2n), the resulting zygote will be trisomic (2n + 1). On the other hand, if the monosomic gamete is fertilized by a normal gamete, the resulting zygote will not be viable and will likely lead to early embryonic loss.

52. How many complementation groups do the following mutants m1 to m6 come under?

(a) Two

(b) Four

(d) Three

Ans. (c)

Sol. m1 and m2 do not complement each other, so they are in the same complementation group.

m3 and m4 do not complement each other, so they are in the same complementation group.

m5 does not complement m3 or m4, so it is in a separate complementation group.

m6 does not complement m2, m3, or m4, so it is in a separate complementation group.

m6 complements m5, so they are in the same complementation group.

53. The black buck (Antilope cervicapra) has been traditionally protected by which one of the following communities?

(a) Bhils

(b) Jats

(d) Ahirs

Ans. (c)

Sol. The Bishnoi community has been traditionally known for their strong conservation and protection of wildlife, particularly the blackbuck (Antilope cervicapra) in India. The Bishnoi community is a religious sect within Hinduism, and they have a deep-rooted belief in protecting and preserving nature and wildlife.

54. Savannas are biomes where tree and grass vegetation coexist over large areas. Which one of the following statements does NOT explain the occurrence of savannas in the Indian subcontinent?

(a) Selective logging of forests opens up the canopy and grasses take over

(b) Low rainfall maintains low tree cover that helps grasses establish

(d) Browsing by herbivores limits tree establishment

Ans. (a)

Sol. In the Indian subcontinent, savannas are typically found in regions with specific climatic conditions and seasonal rainfall patterns that favor the coexistence of grasses and scattered trees. Selective logging, while it can have an impact on forest ecosystems, is not the primary reason for the occurrence of savannas in the region.

55. Diclofenac toxicity has been suggested to be the cause for population decline in which one of the following animals?

(a) Gyps vultures

(b) Olive Ridley turtles

(d) Oceanic sharks

Ans. (a)

Sol. In the Indian subcontinent, the population decline of several vulture species, especially the Gyps vultures (such as the Indian vulture, White-rumped vulture, and Slender-billed vulture), has been largely attributed to diclofenac toxicity. Vultures play a crucial role in the ecosystem by scavenging on animal carcasses and helping in nutrient recycling. The decline in their population has raised concerns about the impact on the ecosystem and public health, as their absence may lead to increased disease transmission.

56. An Indian bird species known to defend flowers is the

(a) Purple-throated hummingbird

(b) Jungle babbler

(d) Crescent honeyeater

Ans. (c)

Sol. The Purple-rumped sunbird (Leptocoma zeylonica) is an Indian bird species known to defend flowers. Sunbirds are small, colorful birds that are specialized in feeding on nectar from flowers. They have long, slender bills and brush-tipped tongues that allow them to extract nectar from flowers.

57. Which one of the following survivorship curves is typical of invasive insect pest species?

(a) Invasive insect pest species do not follow specific survivorship curves

(b) Type II

(d) Type I

Ans. (b)

Sol. Type II survivorship curve shows a relatively constant mortality rate across the lifespan of the individuals in the population. In other words, the probability of death remains relatively constant throughout their life. This type of curve is often observed in species where the mortality is not strongly age-dependent.

58. Which one of the following statements is NOT correct?

(a) Both alpha and gamma diversities measure the presence and abundance of species in a community

(b) Gamma diversity can be expressed as the product of alpha and beta diversities across sites

(d) Gamma diversity can be expressed as the sum of alpha and beta diversities across sites

Ans. (c)

Sol. Gamma diversity refers to the total diversity of species in a large and heterogeneous area, such as a region, landscape, or entire ecosystem. It represents the overall diversity at the regional scale and includes all the species found across multiple sites within that area.

59. Which one of these statements is NOT CORRECT with respect to ecotones?

(a) Itertidal zones and estuaries are two examples of ecotones

(b) They are transitional areas of vegetation between two different plant communities

(d) They harbour only K-selected species that can survive in changing havitats

Ans. (d)

Sol. They harbor only K-selected species that can survive in changing habitats: This statement is not correct. Ecotones may support a mixture of both K-selected (species that are adapted to stable environments with high competition) and r-selected (species that are adapted to rapidly changing environments with high reproductive rates) species. The presence of both types of species allows ecotones to be dynamic and responsive to environmental changes.

60. Which one of the following countries has contributed the maximum towards CO2 emissions over the last decade?

(a) India

(b) USA

(d) Russia

Ans. (c)

Sol. China has been the largest contributor to CO2 emissions over the last decade. As one of the world's most populous countries and a rapidly growing economy, China's energy consumption and industrial activities have significantly increased, leading to a substantial increase in CO2 emissions.

61. An ecologist studying molluscs concluded that there is a correlation between the thickness of the shell and weight of the mollusc. Based on this information, one can conclude that

(a) heavier molluscs are better defended from attacks by predators.

(b) heavier molluscs are poorly defended from attacks by predators.

(d) weight and thickness are variable traits in mollusc population

Ans. (c)

Sol. A correlation between the thickness of the shell and the weight of the mollusc suggests that there is a relationship between these two traits. Correlation means that as one trait changes, the other trait tends to change in a consistent manner. In this case, as the weight of the mollusc increases, the thickness of the shell also tends to increase.

62. Which of the following life history traits is most likely in a rodent species when snakes prefer to prey upon large, older individuals of the rodent species that grow continuously over their lifespan?

(a) Early reproduction and slow growth rate

(b) Delayed reproduction and fast growth rate

(d) Early reproduction and fast growth rate

Ans. (a)

Sol. When snakes prefer to prey upon large, older individuals of a rodent species that grow continuously over their lifespan, it creates a selective pressure for the rodent species to adopt certain life history traits that enhance their survival and reproductive success.

63. Pick the statement that includes both a proximate and an ultimate explanation for the evolution of a given behaviour

(a) Elevated heart beat and higher levels of stress hormon

(b) Scent marking along boundaries of territories and high aggression

(d) Higher maternal fitness and increased offspring survival

Ans. (c)

Sol. Proximate explanation: "Social communication through odors." This explains the immediate mechanism or cause of the behavior. In this case, the behavior is driven by the use of odors for communication among individuals in the social group.

Ultimate explanation: "Increased group survival." This explains the evolutionary reason or adaptive significance of the behavior. The behavior of social communication through odors has evolved because it promotes increased group survival. Effective communication through odors helps individuals in the group coordinate activities, establish social hierarchies, avoid conflicts, and enhance group cohesion, which can lead to better protection from predators and access to resources, thus increasing the overall survival and fitness of the group.

64. Bioaugmentation refers to:

(a) Developing microbial strains through genetic engineering which can degrade pollutants and toxic compounds efficiently

(b) Ex- situ bioremediation of toxins from soil or any other contaminant site by addition of selected microbes to enhance biodegradation

(c) Addition of nutrients at contaminated sites to enhance growth of indigenous microflora which will in turn degrade pollutants

(d) Addition of selected microbes both archaea and bacteria to the polluted site so that biodegradation is enhanced.

Ans. (d)

Sol. Bioaugmentation is a bioremediation technique used to enhance the biodegradation of pollutants and toxic compounds in contaminated environments. It involves the addition of specific microorganisms (such as bacteria, archaea, or fungi) to a polluted site to improve the natural biodegradation process.

The selected microbes are chosen for their ability to efficiently degrade the specific pollutants present in the contaminated site. These microbes can produce enzymes or metabolic pathways that break down the pollutants into less harmful substances, thus aiding in the cleanup of the polluted environment.

65. Which of the following methods can be used to selectively lyse newly dividing cells?

(a) MTT (3-(4,5-Dimethylthiazol- 2-yl)-2,5-diphenyltetrazolium bromide) treatment of dividing cells followed by UVB irradiation MTT

(b) Treatment of dividing cells with caspase inducers

(d) Treatment of dividing cells with 51Cr and measuring its release over a period of time.

Ans. (c)

Sol. Bromodeoxyuridine (BrdU) is a thymidine analog that is incorporated into the DNA of actively dividing cells during the S phase of the cell cycle. By exposing the BrdU-labeled cells to light, the incorporated BrdU can undergo photolysis, causing DNA damage and selective lysis of the newly dividing cells. This method is used to selectively target and kill dividing cells in a population.

66. The information obtained by comparing a new diagnostic test with the gold standard is summarized in a two-by-two table given below

What is the sensitivity and specificity of the new test?

(a) Sensitivity = 76%; Specificity = 71%

(b) Sensitivity = 32%; Specificity = 22%

(d) Sensitivity = 34%; Specificity = 39%

Ans. (c)

Sol. Given the information from the two-by-two table:

True Positives = 68

False Positives = 22

False Negatives = 32

True Negatives = 78

Using the formulas:

Sensitivity = 68 / (68 + 32) = 68 / 100 = 0.68 or 68%

Specificity = 78 / (78 + 22) = 78 / 100 = 0.78 or 78%

67. What can you infer if the correlation coefficient, [Pearson correlation (r)], is close to -1 (minus 1) for two set of variables?

(a) There is no relationship between the two variables

(b) There is an exponential relationship between the two variables

(c) There is a linear relationship in which when there is a decrease in one variable, there is also a decrease in the second variable.

(d) There is a linear relationship in which, when there is an increase in one variable, there is a decrease in the second variable

Ans. (d)

Sol. The correlation coefficient ranges from -1 to +1. A correlation coefficient of -1 indicates a perfect negative correlation, where the two variables move in exact opposite directions with a linear relationship. As the correlation coefficient approaches -1, the strength of the negative linear relationship increases.

68. The distribution of heights of college students aged between 18 to 20 was found approximately normally distributed with an average

(a) 2.4

(b) 3.1

(d) 2.9

Ans. (a)

Sol. The distribution of heights of college students aged between 18 to 20 was found approximately normally distributed with an average of 2.4.

69. Emission maximum of a fluorophore is shifted to longer wavelength when compared to the wavelength of excitation. What is the reason?

(a) Non-radiative loss of excitation energy

(b) Partial absorbance of incident light

(d) Radiative loss of excitation energy

Ans. (a)

Sol. The phenomenon described in the question is known as the "Stokes shift." In fluorescence spectroscopy, when a fluorophore absorbs a photon of light at a specific wavelength (excitation), it gets excited to a higher energy state. However, when the fluorophore returnsto its ground state, it emits a photon of light at a longer wavelength than the excitation wavelength.

70. What will be the percentage transmission when absorbance is 1, 2 and 3, respectively?

(a) 10, 1, 0.1

(b) 1, 10, 100

(d) 20, 10, 0

Ans. (a)

Sol. The percentage transmission (T%) can be calculated using the formula:

T% = 10^(-Absorbance) * 100

Given the absorbance values:

(a) Absorbance = 1 T% = 10^(-1) * 100 = 0.1 * 100 = 10%

(b) Absorbance = 2 T% = 10^(-2) * 100 = 0.01 * 100 = 1%

So, the correct answer is (a) 10, 1, 0.1, which represents the percentage transmission when the absorbance is 1, 2, and 3, respectively.

71. The F₁ subunit of F₀F₁ ATP synthase synthesizes ATP from ADP in the mitochondrial inner membrane. Purified F₁ subunit hydrolyses ATP to ADP. Which one of the following reason explans the difference between the activities of the F₁ subunit in soluble and membrane bound form?

(a) A conformational change in the F₁ subunit between the two environments

(b) The lipid bilayer environment facilitates the synthesis of ATP by enhancing the rate of the dehydration reaction

(c) The ATP synthesis reaction is driven by coupling to an electrochemical potential across the inner mitochondrial membrane

(d) In the soluble form, the electrochemical potential drives the F₁ subunit to hydrolyze ATP

Ans. (c)

Sol. The difference in activities of the F1 subunit between the soluble and membrane-bound forms is due to the absence of coupling to the electrochemical potential in the soluble form, leading to hydrolysis of ATP, whereas in the membrane-bound form, the electrochemical potential drives ATP synthesis.

72.

The regions of phi, psi space occupied by well characterized protein secondary structrues are marked on a Ramachandran plot as shown above. Which of the following statements is CORRECT?

(a) A-right handed a helix, B- strand, C- left handed a hekix, D - collagen

(b) A- strand, B- right handed a helix, C-left handed a helix, D- collagen

(d) A-left handed a helix, B- strand, C- collagen, D-right handed a helix

Ans. (b)

Sol. These regions are based on the known phi and psi angles observed in experimentally determined protein structures. The Ramachandran plot is a useful tool to analyze the allowed backbone conformations for amino acids in proteins.

73. An enzyme has a K_m of 5 × 10^-5 M and a V_max of 100 µmoles.lit^-1.min^-1 (K_m is the Michaelis constant and V_max is the maximal velocity).

What is the velocity in the presence of 1 × 10^–4 M substrate and 2 × 10^–4 M competitive inhibitor, given that the K_i for the inhibitor is 2 × 10^–4 M?

(a) 0.005 µmoles. lit^–1. min^–1

(b) 50 µmoles. lit^–1. min^–1

(d) 500 µmoles. lit^–1. min^–1

Ans. (b)

Sol. To calculate the velocity (V) in the presence of a competitive inhibitor, we can use the following equation:

V = (Vmax * [S]) / (Km * (1 + [I] / Ki))

where: V = Velocity in the presence of the inhibitor Vmax = Maximal velocity of the

enzyme-catalyzed reaction [S] = Substrate concentration Km = Michaelis constant of the

enzyme-substrate complex [I] = Inhibitor concentration Ki = Inhibition constant of the inhibitor

Given the values: Vmax = 100 µmoles.lit^-1.min^-1 Km = 5 × 10^-5 M [S] = 1 × 10^-4 M

(substrate concentration) [I] = 2 × 10^-4 M (inhibitor concentration) Ki = 2 × 10^-4 M

Substituting these values into the equation:

V = (100 * 1 × 10^-4) / (5 × 10^-5 * (1 + 2 × 10^-4/2 × 10^-4)) V = (100 * 1 × 10^-4) / (5 × 10^-5 * (1 + 1)) V = 100 / (5 * 2) V = 100 / 10 V = 10 µmoles.lit^-1.min^-1.

So, the velocity in the presence of 1 × 10^-4 M substrate and 2 × 10^-4 M competitive inhibitor is 10 µmoles.lit^-1.min^-1.

74. Pyruvate kinase, the enzyme that catalyzes the conversion of PEP to pyruvate transfers the Pi from PEP to ADP to generate ATP. The standard free energies of the half-reactions are given below.

ADP + Pi = ATP = +30.5kJmol^–1

How is the free energy for genration of ATP from ADP derived in the reaction catalyzed by pruvate kinase?

(a) through coupling with keto-enol tautomerism where the enol form of pyruvate is converted to the keto form

(b) through condensation of Pi with ADP ADP

(d) through coupling with hydrolysis of PPi PPi

Ans. (a)

Sol. The coupling of the reaction with keto-enol tautomerism allows for the conversion of pyruvate from its enol form to the keto form. This conversion releases energy, which helps in driving the synthesis of ATP from ADP and Pi in the reaction catalyzed by pyruvate kinase.

75.

Which of the four molecules shown above are optically active?

(a) A, B, C and D

(b) B and D only

(d) B only

Ans. (b)

Sol. B: This molecule has a chiral center at the sulfur (S) atom, as it is bonded to four different groups: two hydrogen atoms, one carbon atom, and one selenium (Se) atom. Therefore, molecule B is optically active.

D: This molecule has a chiral center at the selenium (Se) atom, as it is bonded to four different groups: two hydrogen atoms, one carbon atom, and one sulfur (S) atom. Therefore, molecule D is optically active.

76. The actin-binding proteins regulate microfilament turnover in a eukaryotic system. Match the actin-binding protein (in column I) with their functions (in column II).

Choose the CORRECT combination from below:

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

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

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

Ans. (c)

Sol. A. Cofilin (ii) Binds preferentially to filament containing ADP-actin

B. Profilin (i) Binds ADP-G-actin and catalyzes the exchange of ADP for ATP

C. Thymosin (iv) Binds to ATP-G-actin and inhibits addition of actin subunit to filament

D. Formin (iii) Assembles unbranched filament

77. Given below are a few steps in clathrin-coated vesicle formation in the secretory pathway.

(A) Receptor-ligand recognition and binding

(B) Recruitment of adapter protein and clathrin

(D) Uncoating of clathrin coats

Choose the option that correctly identifies the sequence of events in making a clathrincoated vesicle.

(a) A, B, C, D

(b) B, A, C, D

(d) B, A, D, C

Ans. (a)

Sol. (A) Receptor-ligand recognition and binding

(B) Recruitment of adapter protein and clathrin

(D) Uncoating of clathrin coats

78. To ensure proper segregation of chromosomes during mitosis, the sister chromatid pairs must be stably bi-oriented on the mitotic spindle. In animal cells, after nuclear envelope breakdown (NEBD), chromosomes glide along the microtubules' length with the help of the motor proteins. When the chromosomes reach the plus-end of microtubules, the kinetochores attach to the microtubules. Which one of the following is the correct option for the kinetochore-microtubules attachment configuration that ensures proper chromosome segregation?

(a) Monotelic

(b) Merotelic

(d) Syntelic

Ans. (c)

Sol. In amphitelic attachment, each sister chromatid of a chromosome is attached to microtubules from opposite spindle poles. This bi-orientation ensures that tension is generated at the kinetochores, which is required for proper chromosome alignment and segregation during mitosis. This configuration is essential for the accurate and faithful distribution of genetic material to the daughter cells.

79. Following statements were made about mitochondria:

A. The D loop of the mitochondrial genome is required for replication, but not for the regulation of transcription.

B. The L strand of mitochondrial genome possesses more cytosine.

C. In plants, most mitochondrial tRNAs are encoded by the nuclear genome and then imported into the mitochondrion.

D. Cycloheximide inhibits protein synthesis by mitochondrial ribosomes, but does not affect eukaryotic cytosolic ribosomes.

E. Some organisms have been found to carry linear mitochondrial DNA.

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

(a) A, B, C

(b) B, D, E

(d) B, C, E

Ans. (d)

Sol. B. The L strand of mitochondrial genome possesses more cytosine.

C. In plants, most mitochondrial tRNAs are encoded by the nuclear genome and then imported into the mitochondrion.

E. Some organisms have been found to carry linear mitochondrial DNA.

80. From a newly fertilized mouse egg, maternal pronucleus was removed and replaced with a second paternal pronucleus. Following observations/ statements were made:

A. This will result in formation of an androgenetic embryo.

B. This will result in formation of a gynogenetic embryo.

C. The embryo will not survive beyond mid-gestation since parental genomes serve distinct complementary functions due to variable imprinting pattern.

D. The adult originating from the embryo will be a clone of the father.

E. The embryo will develop as adult but will die early due to rapid shortening of the telomeres.

Which one of the following represents correct combination of above statements?

(a) A, B, C

(b) B, C, E

(d) A and E only

Ans. (c)

Sol. The removal of the maternal pronucleus and replacement with a second paternal pronucleus will result in an androgenetic embryo, which means it contains only paternal genetic material. However, such embryos usually do not survive beyond mid-gestation because parental genomes serve distinct complementary functions due to genomic imprinting, which is the differential expression of genes depending on whether they are inherited from the mother or the father. The lack of the maternal genome in the embryo would lead to imbalanced gene expression and result in developmental abnormalities leading to early embryonic lethality.

81. The following statements are made with respect to merodiploids of the lac operon, where 'l' is the lac repressor. 'O' is the ac operator. "Z" is the lacZ gene encoding beta-galactosidase and "Y" is the lac Y gene encoding permease

A. In I^SO⁺Z⁺Y⁺/I⁺O^cZ^–Y⁺ the lacZ is inducible and lacY is constitutively expressed

B. In I⁺ O^CZ⁺ Y^–/I^–O⁺Z^–Y⁺ the lacZ and lacY are both inducible

C. In I⁺O^cZ⁺Y^–/I^SO⁺Z^–Y⁺ the lacZ is constitutively expressed and lacY is inducible

D. I^SO⁺Z⁺Y⁺/I⁺O^CZ^–Y⁺ the lacZ is inducible and lacY is constitutively expressed

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

(a) B and D only

(b) A and B only

(d) B, C and D

Ans. (b)

Sol. In statement A, the lacZ gene is inducible, and the lacY gene is constitutively expressed. In statement B, both lacZ and lacY genes are inducible.

82. The following statements refer to the E.coli repucative DNA polymerase

A. DNA Pol 1 displays very limited processivily and possesses 3' 5' exonuclease activity allowing fidelity of DNA replication

B. DNA Pol III is suitable for leading strand DNA synthesis due to its high processivity and 5' 3' exonuclease activity that removes incorrect nucleotides incorporated during DNA synthesis.

C. DNA Pol 1 possesses 5' 3' exonuclease activity which allows removal of the RNA primer while its 5' 3' polymerase activity allows it to fill the gap created by removal of the RNA primer

D. DNA Pol III is suitable for lagging strand DNA synthesis due to its low processivity and 5' 3' exonuclease activity.

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

(a) D only

(b) B and C

(d) A and C

Ans. (b)

Sol. B. DNA Pol III is suitable for leading strand DNA synthesis due to its high processivity and 5' 3' exonuclease activity that removes incorrect nucleotides incorporated during DNA synthesis.

C. DNA Pol 1 possesses 5' 3' exonuclease activity, which allows removal of the RNA primer while its 5' 3' polymerase activity allows it to fill the gap created by the removal of the RNA primer.

83. The 5' UTR of ferritin mRNA forms a stem-loop structure called the iron regulatory element [IRE]. The Iron Regulatory Binding Protein [IRBP] binds this IRE.

The following statements were made with reference to IRBP- IRE interaction:

A. IRBP-IRE interaction prevents eIF4A from resolving the stem-loop structure, thus preventinginitiation of translation of ferritin genes.

B. IRBP-IRE interaction recruits eIF4A to the 5' UTR, thus promoting translation initiation.

C. In presence of ferrous ions IRBP is unable to bind the IRE.

D. eIF4A binds directly at the 5' UTR and disrupts the stem-loop structure, thus promoting translation initiation.

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

(a) B only

(b) A and D

(d) B and C

Ans. (c)

Sol. A. IRBP-IRE interaction prevents eIF4A from resolving the stem-loop structure, thus preventing initiation of translation of ferritin genes.

C. In the presence of ferrous ions, IRBP is unable to bind the IRE

84. An in vitro translation system capable of incorporating ~8 amino acids s-1 was programmed to translate a single mRNA that codes for an alanine-rich (~35% alanine with uniform distribution of alanine) protein of 275 amino acids (~30kDa) including a hexa-histidine tag at the C-terminal end of the protein. The protein possesses three methionine residues at amino acid positions 1, 135 and 230 and generates polypeptides of ~15 kDa, ~10 kDa and ~5 kDa upon degradation with cyanogen bromide. The translation reaction was initiated and the ongoing reaction was supplemented with 14C Ala after 5 min. Soon after addition of 14C Ala, aliquots were drawn at 2, 20, and 200 s, and reactions in the aliquots were instantaneously stopped. The translated proteins were purified on Ni-NTA columns, processed for degradation by CNBr, resolved on SDS-PAGE, and visualized by nonquantitative autoradiography. Which of the following autoradiograms represents the expected pattern of the bands?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. This autoradiogram shows the progressive increase in intensity of the bands corresponding to 15 kDa, 10 kDa, and 5 kDa polypeptides over time, indicating the incorporation of 14C-labeled alanine into the growing protein chain.

85. Some of the steps in the process of eukaryotic DNA replication mentioned below require hydrolysis of ATP.

A. Phosphodiester bond formation

B. DNA strand separation by helicase

C. Clamp-loader association with clamp and DNA

D. Joining of Okazaki fragments

Choose the following option that correctly identifies all the steps utilizing ATP hydrolysis

(a) A, B and D only

(b) B, C and D only

(d) B and D only

Ans. (d)

Sol. Both of these steps involve energy-consuming processes that utilize ATP hydrolysis to drive the necessary reactions.

86. The sequence below represents part of the coding strand of the bacterial gene Z. The arrow indicates the transcription start site

The following statements were made with reference to transcription and translation of the stand

A. Insertion of an 'A' nucleotide after position +8 increases the length of the transcript by 1 nucleotide and changes the amino acid sequence of the protein being translated.

B. Substitution of the T and position 22 changes the primary structure of the protein without altering transcript length

C. Insertion of an 'A' after position 26 changes the primary structure of the protein and results in synthesis of a truncated protein

D. Delection of 'A' at position 9 creates the STOP codon that prevents translation of the protein

Which one of the options below represents the combination of all correct statement?

(a) C only

(b) A and D

(d) A and B

Ans. (a)

Sol. Insertion of an 'A' after position +26 would cause a frameshift mutation and change the entire amino acid sequence, likely resulting in a truncated and non-functional protein.

87. Which one of the following statements relating to the mechanism of color development in response to LacZ expression in Escherichia coli is INCORRECT?

(a) E. coli growth on LB agar with X-gal results in blue colored colonies because LacZ produced in the cell hydrolyses X-gal present in the medium into a blue colored product.

(b) When the membranes of the cells harboring LacZ are permeabilized and cells incubated in a buffer with ONPG, the solution turns yellow because LacZ encoded protein hydrolyzes ONPG.

(c) E. coli growth on MacConkey agar results in pink colored colonies because LacZ encoded protein produced in the cell hydrolyzes the neutral red dye present in the medium into a pink colored product.

(d) E. coli growth on MacConkey agar results in pink colored colonies due to shift in pH of the medium MacConkey

Ans. (c)

Sol. E. coli growth on MacConkey agar results in pink or red colored colonies due to the fermentation of lactose. MacConkey agar contains lactose as a carbohydrate source and neutral red as a pH indicator. If E. coli can ferment lactose, it produces acid by-products, which lowers the pH of the medium, causing the neutral red indicator to turn pink or red. LacZ (beta-galactosidase) hydrolyzes lactose, enabling the fermentation process and resulting in the color change of the colonies on the MacConkey agar. The color change is not due to LacZ hydrolyzing neutral red.

88. For an experiment, the hapten DNP was conjugated with the carrier protein BSA or with the carrier protein OVA. A set of mice were primed with either DNP-BSA conjugate or with OVA which was not conjugated to DNP. The following experiments were then performed.

A. X-ray irradiated syngeneic mice were injected with spleen cells from both DNP-BSA– primed mice and OVA-primed mice and then challenged with DNP-OVA conjugate.

B. X-ray irradiated syngeneic mice were injected with T-cell depleted spleen cells from DNP-BSA–primed mice and spleen cells from OVA-primed mice and then challenged with DNP-OVA conjugate.

C. X-ray irradiated syngeneic mice were injected with spleen cells from DNP-BSA–primed miceand T-cell depleted spleen cells from OVA-primed-mice and then challenged with DNP-OVA conjugate.

Which one of the following options of mice will generate secondary anti-hapten response to DNP?

(a) The mice in experiment A only

(b) The mice in experiment B only

(d) The mice in experiments A and B

Ans. (d)

Sol. Both experiments involve the transfer of B cells that can recognize and respond to the DNP hapten, leading to a secondary anti-hapten response when the mice are challenged with the DNP-OVA conjugate. Therefore, the mice in experiments A and B will generate a secondary anti-hapten response to DNP.

89. Consider the defects in human macrophage cell lines (antigen presenting cells, Column A) and their possible consequence on T cell activation (Column B).

Select the option that represents all the correct matches

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

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

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

Ans. (c)

Sol. A. 2-microglobulin deficiency (i) - Cannot activate CD4+ or CD8+ T cells.

B. TL24 knockout macrophages (ii) - Cannot activate CD4+ T cells.

C. Macrophages with HLA region for DP, DQ, and DR deleted (iii) - Cannot activate CD8+ T cells.

D. B7 knockout macrophages (iv) - Can activate CD4+ or CD8+ T cells.

90. Consider the cancer types in Column X and the cancer related descriptions (Column Y)

Select the option that represent all the correct matches

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

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

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

Ans. (a)

Sol. A. Leukemia (i) - Hematopoietic cell cancer that does not grow as a solid tumor.

B. Sarcoma (ii) - A tumor that has arisen from endodermal tissue.

C. Carcinoma (iii) - A tumor that develops from mesodermal connective tissue.

D. Melanoma (iv) - Cancer that develops from the pigment-producing cells of the skin.

91. Cervical cancer cells were unireated (–), or treated (+) with compound 'X', a putative anti-cancer drug. The cell extracts were analyzed by immunoblotting for the levels of specific markers as indicated by the band thickness. The following results were obtained.

Which one of the following options best describes the action of compound 'X'?

(a) Compound 'X' induced cell death via the intrinsic pathway by activating caspase 8 and apoptosis was p53 independent

(b) Compound 'X' induced cell death via the extrinsic pathway by inducing the Fas ligand associated death domain (FADD) and apoptosis was p53 dependent

(c) Compound 'X' induced cell death by reducing the expression of Bax in a p53-dependent manner and consequently increasing the expression of caspase 9

(d) Compound 'X' induced cell death by activating the death domain together with increasing the expression of the pro-apoptotic protein in a p53 independent manner

Ans. (b)

Sol. The immunoblotting results show increased levels of FADD, cleaved caspase 8, and cleaved caspase 3 in the "+X" treated cells, indicating activation of the extrinsic pathway of apoptosis. The involvement of FADD and caspase 8 suggests that the extrinsic pathway is being triggered by an external death receptor, likely the Fas ligand. Additionally, the action of compound 'X' appears to be p53 dependent, as indicated by changes in the levels of p53 and apoptosis-related markers.

92. Physical attachment between cells and extracellular matrix is critical in both animals and plants because it imparts rigidity and strength to tissues and organs. However, junctions between cell-cell or between cell-matrix are diverse in structure and play roles beyond providing physical support. Column "X" lists some of the cell junctions and column "Y" lists their characteristic functions

Select the option with all correct matches

(a) A – i; B – ii; C – iv; D – iii

(b) A – ii; B – iii; C – iv; D – i

(d) A – iv; B – i; C – ii; D – iii

Ans. (b)

Sol. A. Tight junctions seal gaps between epithelial cells.

B. Gap junctions allow passage of small water-soluble molecules from cell to cell in animal tissues.

C. Plasmodesmata allow passage of small molecules but not macromolecules (with some exceptions) in plants.

D. Desmosomes connect intermediate filaments in one cell to those in the next cell.

93. Inbred mouse strains with different MHC haplotypes (homozygous H-2^m MHC haplotype and homozygous for the H-2ⁿ haplotype) were mated resulting in F1 progeny (H-2^m/n). Skin transplantation experiments were performed between these mouse strains (parents with H-2^m and H-2ⁿ MHC haplotypes and progeny with H-2^m/n MHC haplotype). Which one of the following statements with respect to acceptance and rejection of the skin graft is correct?

(a) Skin graft from the progeny will be accepted by any of the parents (recipient)

(b) Skin graft from one parent (donor) will be accepted by the other parent (recipient).

(c) Skin graft from progeny will be accepted by the parent homozygous for the H-2^m haplotype, but not by the parent homozygous for the H-2ⁿ haplotype.

(d) Skin graft from any of the parents will be accepted by the progeny recipient.

Ans. (d)

Sol. In this scenario, the F1 progeny (H-2m/n) have inherited one MHC haplotype from each parent (H-2m from one parent and H-2n from the other). As a result, the F1 progeny will have both MHC types in their system. This makes them tolerant to both MHC haplotypes, and they will not mount an immune response against skin grafts from either parent. Therefore, skin grafts from any of the parents will be accepted by the F1 progeny.

94. Tbx4 and Tbx5 are critical in the specification of hindlimbs and forelimbs, respectively. The following statements were made regarding experiments involving expression of Tb × 4 or Tb × 5 genes and their probable outcomes:

A. When chick embryo was made to express Tbx4 throughout the flank tissue, limbs induced in the anterior region often become legs instead of wings.

B. Loss of TbX4 function in the hindlimb field completely inhibits leg initiation and growth.

C. Loss of Tbx5 gene in chick results in complete failure of forelimb formation which includes even the most proximal shoulder/girdle structure.

Which one of the following options represents all correct statements as made above?

(a) A only

(b) A and B only

(d) A, B and C

Ans. (d)

Sol. A. When the chick embryo is made to express Tbx4 throughout the flank tissue, limbs induced in the anterior region often become legs instead of wings. This shows that Tbx4 is critical for specifying hindlimb identity.

B. Loss of Tbx4 function in the hindlimb field completely inhibits leg initiation and growth.

This statement indicates that Tbx4 is necessary for the formation and growth of hindlimbs.

C. Loss of Tbx5 gene in the chick results in complete failure of forelimb formation, including even the most proximal shoulder/girdle structure. This demonstrates that Tbx5 is essential for specifying forelimb identity and the development of the entire forelimb, including the shoulder region.

95. The following statements were made regarding the patterning of anterior-posterior body plan of Drosophila:

A. Microinjection of bicoid mRNA in the middle of a bicoid-deficient embryo leads to formation of 'head' in the middle and telson at the two ends.

B. Nanos protein inhibits the translation of caudal mRNA at the posterior half of the embryo.

C. The Bicoid protein activates the zygotic expression of the hunchback gene.

D. The segment polarity genes are expressed in segments of the embryo.

Which one of the following options represents all correct statements as made above?

(a) A and B only

(b) A and C only

(d) B, C and D

Ans. (b)

Sol. A. Microinjection of bicoid mRNA in the middle of a bicoid-deficient embryo does not lead to the formation of 'head' in the middle and telson at the two ends. Instead, it leads to the formation of duplicated anterior structures (two heads) in the middle of the embryo.

C. The Bicoid protein does not activate the zygotic expression of the hunchback gene directly. Instead, it acts as a morphogen and forms a concentration gradient along the anterior-posterior axis, which then regulates the expression of hunchback and other downstream genes.

96. The group of 6 cells (P3.p to P8.P) called vulval precursor cells (VPCs) of C. elegans form an equivalence group. The following statements were made as evidence that VPCs form an equivalence group:

A. If the anchor cell is destroyed the VPCs contribute to the formation of hypodermal tissues.

B. If the 3 central cells (P5.p to P7.p) are destroyed the remaining cells can generate vulval cells.

C. If expression of lin-3 is increased VPCs contributing to the secondary lineage can form cells of primary lineage.

D. Ectopic expression of let-23 in P5.p and P7.p VPCs converts them to primary cell lineage.

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

(a) A and B only

(b) B and C only.

(d) B, C and D

Ans. (b)

Sol. Statement B is correct because if the central cells (P5.p to P7.p) are destroyed, the remaining cells (P3.p, P4.p, and P8.p) can still generate vulval cells, indicating that they have the potential to adopt the vulval cell fate.

Statement C is correct because if the expression of lin-3 is increased, VPCs contributing to the secondary lineage can form cells of the primary lineage, suggesting that the fate of VPCs can be altered by changes in the expression of lin-3.

97. The mammalian genital ridge is bipotential. Which one of the following statements regarding determination of the fate of genital ridge is INCORRECT?

(a) The activation of Sox9 gene promotes testis determining pathway.

(b) The accumulation of â-catenin is critical for activating ovarian development

(d) Though Wnt4 is expressed in the bipotential gonads, it is an important factor in ovary determination

Ans. (c)

Sol. R-spondin 1 (Rspo1) is not involved in promoting the testis determining pathway. Instead, R-spondin 1 is critical for ovarian development. It acts to stabilize -catenin, which in turn activates the Wnt/-catenin signaling pathway, leading to the development of ovaries. The testis determining pathway, on the other hand, is promoted by the activation of the Sox9 gene.

98. In mammals, autophagy is involved in specific cytosolic rearrangements needed for proliferation and differentiation during embryogenesis and postnatal development. Embryos have the ability to activate general protective strategy against many stressinducing conditions. Which one of the following statements DOES NOT conform to the role of autophagy during early development?

(a) Autophagy is a process of cytosolic renovation, crucial for cell fate decisions

(b) Autophagy plays a dual role both in adaptation to stress and starvation during morphogenesis and in cell elimination along with apoptosis.

(c) Functional characterization of the autophagy regulatory genes indicates that autophagy is definitely not an evolutionarily conserved process

(d) Defects in autophagy during early embryogenesis can be lethal for the organism

Ans. (c)

Sol. Autophagy is indeed an evolutionarily conserved process found in a wide range of organisms, from yeast to mammals. The core machinery of autophagy and many of its regulatory genes are highly conserved throughout evolution. Autophagy is essential for cell survival and homeostasis and plays a crucial role during embryogenesis and postnatal development.

99. The table below summarizes the key signaling pathways that orchestrate development, their receptors, transcription effectors and output.

Which of the above pathways is correctly depicted in one of the options given below?

(a) A, C and D

(b) Only B and E

(d) Only C and E

Ans. (d)

Sol. In option C, the RTK (Receptor Tyrosine Kinase) signaling pathway is correctly depicted with EGFR (Epidermal Growth Factor Receptor) as the receptor and Pointed/Yan as the transcriptional effectors. The output of this pathway is morphogenesis.

In option E, the JNK (c-Jun N-terminal Kinase) signaling pathway is correctly depicted with TNF (Tumor Necrosis Factor) as the receptor and Jun/Fos as the transcriptional effectors. The output of this pathway is migration.

100. Members of the WUSCHEL RELATED HOMEOBOX (WOX) transcription factor family play an important role during zygote elongation and division in Arabidopsis. Following are certain statements regarding the expression of different members of WOX gene family during zygote elongation.

A. WOX2 and WOX8 are present in both the egg cell and the zygote.

B. WOX2 is present in the apical and basal cell.

C. WOX8 along with WOX9 regulates the development of basal lineage.

D. WOX8 and WOX9 are directly activated in the zygote by the transcription factor WRKY2.

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

(a) A, B and C

(b) A, B and D

(d) B, C and D

Ans. (c)

Sol. Statement A is correct as WOX2 and WOX8 are present in both the egg cell and the zygote. Statement C is correct as WOX8, along with WOX9, regulates the development of the basal lineage.

Statement D is correct as WOX8 and WOX9 are directly activated in the zygote by the transcription factor WRKY2.

101. Phytochrome photoreceptors exist in two isoforms, P_R and P_FR. Following are certain statements regarding the function of P_FR:

A. P_FR form induces phosphorylation and ubiquitin linked degradation of PIFs transcription factor.

B. P_FR mediated degradation of PIFs inhibits photomorphogenesis.

C. P_FR inhibits the activity of COPI.

D. P_FR increases the stability of transcription factors HFR 1, HY5 and LAF1.

Which one of the following combinations is correct?

(a) A, B and C only

(b) A, C and D only

(d) A, B and D only

Ans. (b)

Sol. Statement A is correct as PFR form induces phosphorylation and ubiquitin-linked degradation of PIFs (Phytochrome Interacting Factors) transcription factors.

Statement C is correct as PFR inhibits the activity of COPI (Coatomer protein I), which is involved in vesicle trafficking.

Statement D is correct as PFR increases the stability of transcription factors HFR1, HY5, and LAF1, which are involved in various photomorphogenesis processes.

102. A student listed following combinations of enzymes and their involvement in different phases of Calvin-Benson cycle:

A. Phosphoglycerate kinase – Reduction phase

B. Glyceraldehyde-3-phosphate dehydrogenase – Regeneration phase

C. Triose-phosphate isomerase – Reduction phase

D. Phosphoribulokinase – Regeneration phase

Which one of the following combinations is correct?

(a) .A, B and C

(b) B and C only

(d) A and D only

Ans. (d)

Sol. Glyceraldehyde-3-phosphate dehydrogenase and triose-phosphate isomerase are enzymes involved in the Calvin-Benson cycle, but their involvement is not correctly matched in the options provided.

103. The following statements were made regarding submergence tolerance in plants.

A. Wetland plants have structural barrier to prevent O₂ diffusion into soil.

B. Dryland plants have structural barrier to prevent O₂ diffusion into soil.

C. Lowering of cytosolic Ca²⁺ prevents aerenchyma formation.

D. Activation of ethylene signal transduction pathway prevents aerenchyma formation.

Which one of the following options has all correct statements?

(a) A and C only

(b) B and C only

(d) B and D only

Ans. (a)

Sol. A. Wetland plants have structural barriers to prevent O2 diffusion into the soil, which is an adaptation for submergence tolerance.

C. Lowering of cytosolic Ca2+ prevents aerenchyma formation, which is another adaptation for submergence tolerance in plants.

104. The first common enzyme in the biosynthesis of the branched-chain amino acids (Leu, Ile and Val) is acetohydroxyacid synthase (AHAS). Following statements are made about the enzyme:

A. AHAS requires thiamine diphosphate as cofactor.

B. The plant AHAS comprises a large catalytic subunit and a smaller regulatory subunit.

C. The large subunit alone is sensitive to inhibition by Leu, Ile and Val in plants.

D. Most of the bacterial and fungal AHAS enzymes are sensitive to inhibition by Val only.

Select the option with all correct statements.

(a) A, B and C

(b) A, C and D

(d) A, B and D

Ans. (d)

Sol. A. AHAS requires thiamine diphosphate as a cofactor, which is a correct statement.

B. The plant AHAS comprises a large catalytic subunit and a smaller regulatory subunit, which is also true.

D. Most of the bacterial and fungal AHAS enzymes are sensitive to inhibition by Val only, which is also a correct statement.

105. Plant nodulation genes encode proteins with receptor-like-features. Following are the list of some nodulation proteins (Column X) and their possible domain characteristics (Column Y):

Which of the following is the correct match?

(a) A – i, B – ii, C – iii

(b) A – ii, B – iii, C – i

(d) A – i, B – iii, C – ii

Ans. (b)

Sol. A – ii, B – iii, C – i

106. Following statements were made regarding gibberellins (GA) biosynthesis in plants and fungi.

A. Two separate enzymes are involved in synthesis of ent-kaurene from GGDP in plants

B. Only a single bifunctional enzyme catalyses the synthesis of ent- kaurene from GGDP in fungi

C. GA-biosynthesis genes are mostly clustered on a single chromosome in fungi.

D. GA-biosynthesis genes are randomly located on chromosomes in fungi.

Which one of the following combination of statements is correct?

(a) A, B and C only

(b) A, B and D only

(d) A and D only

Ans. (a)

Sol. A. In plants, the synthesis of ent-kaurene from GGDP involves two separate enzymes. B. In fungi, the synthesis of ent-kaurene from GGDP is catalyzed by a single bifunctional enzyme.

C. GA-biosynthesis genes are mostly clustered on a single chromosome in fungi, allowing coordinated regulation.

107. Insulin is a polypeptide hormone that reduces blood glucose levels in human. Following statements are made for insulin synthesis and structure:

A. It is synthesized in rough endoplasmic reticulum of the B cells of islets of Langerhans.

B. It is synthesized in cytosol on free ribosomes of the B cells of islets of Langerhans.

C. Insulin has an AB heterodimer structure with one intrachain (A8-A13) and two interchain disulfide bridges (A6-B10 and A21-B18)

D. Insulin has an AB heterodimer structure with one intrachain (A6-A11) and two interchain disulfide bridges (A7-B7 and A20-B19).

E. The gene for insulin is located on the long arm of chromosome 11 and has two introns and three exons.

F. The gene for insulin is located on the short arm of chromosome 11 that has two introns and three exons.

Which one of the following combination of statements is correct?

(a) B, D and E

(b) A, C and F

(d) A, D and F

Ans. (d)

Sol. A. Insulin is synthesized in the rough endoplasmic reticulum of the B cells of islets of Langerhans.

D. Insulin has an AB heterodimer structure with one intrachain (A6-A11) and two interchain disulfide bridges (A7-B7 and A20-B19).

F. The gene for insulin is located on the short arm of chromosome 11 and has two introns and three exons.

108. Plasma proteins have vital roles in the body ranging from maintaining osmolarity to transport of hormones. Certain statements are given below for the functions of selected plasma proteins:

A. Von Willebrand factor is normally synthesized in the liver.

B. Ceruloplasmin is a copper carrier protein.

C. Genetic deficiency of -antiproteinase causes emphysema.

D. Most plasma proteins including albumin are covalently glycosylated.

E. -acid glycoprotein (AGP) level increases during body's response to inflammation.

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

(a) .A, B and C only

(b) B, C and D only

(d) A, D and E only

Ans. (c)

Sol. B. Ceruloplasmin is a copper carrier protein.

C. Genetic deficiency of á1-antiproteinase causes emphysema.

E. -acid glycoprotein (AGP) level increases during the body's response to inflammation.

109. A potential difference of about -70 mV between inside and outside of a single axonal membrane in resting condition may be recorded by suitable electrodes and amplifier. The physico-chemical and biological basis of the origin of this resting membrane potential (RMP) are suggested below:

(A) The RMP is close to the equilibrium of Na⁺ ion.

(B) There must be an unequal distribution of diffusible ions across the axonal membrane for the RMP.

(D) The concentration gradient of Na+ and K+ ions across the axonal membrane required for the RMP, is dependent on the activity of Na+, K+-ATPase.

(E) Impermeable proteins in the axoplasm do not affect the distribution of diffusible ions across the axonal membrane which is required for the RMP.

(F) Na+, K+-ATPase pump in the axonal membrane which is essential for the RMP, is not electrogenic.

Choose all correct statements from the following options:

(a) A, B and C only

(b) B, C and D only

(d) D, E and F only D

Ans. (b)

Sol. (B) There must be an unequal distribution of diffusible ions across the axonal membrane for the RMP. The resting membrane potential (RMP) is established due to the uneven distribution of ions across the axonal membrane. This is maintained by the selective permeability of the membrane to different ions through ion channels.

(C) The axonal membrane must be permeable to one or more species of ions for the RMP. The resting membrane potential is dependent on the membrane's permeability to ions. The ion channels in the membrane allow specific ions to move in and out, thus influencing the resting potential.

(D) The concentration gradient of Na+ and K+ ions across the axonal membrane required for the RMP is dependent on the activity of Na+, K+-ATPase. The Na+, K+-ATPase pump actively maintains the concentration gradients of sodium (Na+) and potassium (K+) ions across the axonal membrane. This pump helps establish the ionic basis of the resting membrane potential.

110. Iron deficiency is a common problem in humans worldwide. The homeostasis of iron in the body is maintained using various proteins (column- X) and their function (column- Y):

Choose the correct option from below that most appropriately matches in column X with that of column Y.

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

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

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

Ans. (b)

Sol. A. Ferritin is an intramucosal cell iron binding protein. It stores iron in a non-toxic form within cells.

B. Ferroportin is the protein that allows iron to leave mucosal cells and enter the bloodstream.

C. Transferrin is the plasma iron-binding protein. It transports iron in the bloodstream to various tissues.

D. Hepcidin is a peptide hormone that regulates iron metabolism by controlling the amount of iron released from cells into the bloodstream. Hypoxia is known to reduce its synthesis.

111. Sound waves are transmitted from the external environment to the cochlea through the middle ear during hearing. The functions of the middle ear in hearing are suggested below:

(A) During the transmission of sound waves through the middle ear, the movement of the head of stapes induces a piston like movement on the oval window.

(B) The tympanic membrane functions as a resonator that reproduces the vibration of sound source.

(C) The sound pressure on the tympanic membrane is increased 1.3 times on the oval window by the lever system of malleus and incus.

(D) The area of tympanic membrane is greater than that of the footplate of stapes, and hence sound pressure on tympanic membrane is increased on oval window.

(E) The contraction of tensor tympani muscle causes the manubrium of the malleus to be pulled outward.

(F) The footplate of the stapes is pulled inward by the contraction of stapedius muscle.

Choose the option with all CORRECT statements

(a) A, B and C

(b) B, C and D

(d) D, E and F

Ans. (b)

Sol. B. The tympanic membrane functions as a resonator and amplifies the vibration of the sound source.

C. The lever system of the malleus and incus increases the sound pressure on the oval window by about 1.3 times.

D. The area of the tympanic membrane is greater than that of the footplate of the stapes, which results in an increase in sound pressure on the oval window.

112. In fever caused by bacterial infection, the set-point of the thermoregulatory mechanism is changed to a new point above 37°C. The following statements were proposed by a reserarcher to explain the pathogenesis of this fever

A. The infection induced cytokines inhibit PGE₂ in the hypothalamus, and that increases the body temperature

B. The increased levels of circulating and IL after infection are not able to induce fever

C. The endotoxins of bacteria act on the macrophages and monocytes of the infected person to initiate the process that results in the rise of body temperature

D. The cytokines produced from the macrophages by endotoxins act as endogenous pyrogens in the infected person

E. The inffection induced circulating cytokines act on the organum basculosum of lamina terminalis (OVLT) which activates pre-optic area of hypothalamus resulting in the increase of body temperature

F. The inhibition of COX2 gene expression by the increased level of ciculating cytokines causes the rise of body temperature in the infected person

Choose the option with all correct statements :

(a) A, B and C

(b) B, C and D

(d) E and F D

Ans. (c)

Sol. C. This statement is correct. Endotoxins of bacteria act on macrophages and monocytes to produce cytokines, such as TNF-á and IL-1â, which act as endogenous pyrogens to induce fever.

D. This statement is correct. The cytokines produced from macrophages and monocytes by endotoxins act as endogenous pyrogens in the infected person, leading to fever.

E. This statement is correct. Infection-induced circulating cytokines act on the organum vasculosum of the lamina terminalis (OVLT), which activates the pre-optic area of the hypothalamus, resulting in an increase in body temperature.

113.. Fertilization between two mating types (P1 and P2) of Neurospora, led to a diploid ascus cell which gave rise to ascus containing 8 haploid ascospores. A set of DNA markers representing two linked loci was analyzed in P1, P2 and the octads labeled 01 to 08 arranged from the tip to the base of the ascus. The observed profile is represented below

Which one of the following is a correct conciusion of the above observation?

(a) Bands labeled (a) and (c) are allelic and segregation occurred during meiosis II

(b) Bands labeled (b) and (d) are allelic and segregation occurred during meiosis II

(d) Bands labeled (c) and (d) are allelic and segregation occurred during meiosis I

Ans. (b)

Sol. In Neurospora, as represented in the observed profile, the bands labeled (b) and (d) are allelic, indicating that they represent different alleles of the same gene. The segregation of these alleles occurred during meiosis II when the diploid ascus cell underwent meiosis to form haploid ascospores. The segregation of allelic markers during meiosis II is consistent with the formation of 8 haploid ascospores in the ascus.

114. In a plant r+ and a+ genes encode for a regulatory and a structural protein, respectively. These genes are responsible for blue color of flower. Mutation in either of the genes leads to white flowers, which is a recessive character. The two genes assort independently.

When two homozygous white flowered plants are crossed, the F₁ plants have blue colored flowers. If the F₁ plant is backcrossed, the progeny will have plants with blue and white flowers in the ratio of :

(a) 9 : 7

(b) 1 : 1

(d) 1 : 0

Ans. (b)

Sol. Parent 1: r+r+ a+a+ Parent 2: r+r+ a+a+

All offspring (F1 generation) will be heterozygous for both genes and have the genotype r+r+ a+a+.

When the F1 plant is backcrossed (crossed with one of the homozygous white-flowered parents), the possible combinations for the offspring are:

r+r+ a+a+ (blue flower) x r+r+ a+a+ (white flower) = r+r+ a+a+ (blue flower) (1/4 probability)

r+r+ a+a+ (blue flower) x r+ r+ a+ a+ (white flower) = r+r+ a+a+ (blue flower) (1/4 probability)

r+r+ a+a+ (blue flower) x r+r+ a+a+ (white flower) = r+r+ a+a+ (blue flower) (1/4 probability)

r+r+ a+a+ (blue flower) x r+ r+ a+ a+ (white flower) = r+r+ a+a+ (white flower) (1/4 probability)

Therefore, the ratio of plants with blue flowers to plants with white flowers in the progeny is 1:1 (option b).

115. During development, many gene products are provided by the females to the eggs which are needed for normal development of the zygote. Such genes are called as maternal-effect genes. The following are a set of crosses between parents carrying a recessive mutant allele (m) and the offspring obtained:

Which of the above cross(es) is/are indicative that the mutation is in a maternal-effect gene?

(a) Cross III only

(b) Cross V only

(d) Cross II and V

Ans. (b)

Sol. Cross V is indicative that the mutation is in a maternal-effect gene. In this cross, when two parents carrying one copy of the recessive mutant allele (m/+) are crossed, both normal and mutant phenotypes are observed in the offspring. This suggests that the genotype of the mother determines the phenotype of the offspring, indicating the presence of maternal-effect genes. The maternal gene products provided by the mother to the eggs play a critical role in determining the development of the zygote.

116. The pedigree below is in reference to Angelman syndrome (AS), which is caused by a mutation in the UBE3A gene on chromosome 15. The gene is also patemally imprinted individuals showing AS, have not been indicated in the given pedigree. Individual 1-1 does not have AS. individuals marked with dots are carries for UBE3A mutation.

Which one of the following options lists individuals all of whom are likely to show AS?

(a) II-1, III-1 and IV-1 II-1, III-1

(b) III-1, III-2 and IV-2 III-1, III-2

(d) II-1 and II-V II-1

Ans. (b)

Sol. pedigree, individuals who are likely to show Angelman syndrome (AS) would be those who have inherited two copies of the mutated UBE3A gene (one from the mother, which is imprinted and not functional, and one from the father, which is functional).

117. Four different Hfr strains of E. coli were mated with F^— recipients, and the time of entry of various donor markers were found to be as below:

Hfr 1: met [15 min] thr [30 min] phe [42 min] mal [57 min]

Hfr 2: bio [50 min] thy [51 min] his [60 min] mal [77 min]

Hfr 3: cys [10 min] phe [26 min] his [58 min]

Hfr4: his [12 min] bio [22 min] azi [27 min] thi [44min]

Based upon the above observations, the following statements were made assuming met to be at 0 min and thr at 15 min:

A. his is located at 59 min

B. azi is located at 74 min

C. cys is located at 11 min

D. mal is located at 76 min

Which one of the following options represents all correct statements?

(a) A and D only

(b) B, C and D

(d) C and D only

Ans. (c)

Sol. Let's analyze the information given:

Hfr 1: met [15 min] thr [30 min] phe [42 min] mal [57 min]

Hfr 2: bio [50 min] thy [51 min] his [60 min] mal [77 min]

Hfr 3: cys [10 min] phe [26 min] his [58 min] Hfr 4: his [12 min] bio [22 min] azi [27 min] thi [44 min]

We assume that met is at 0 min and thr is at 15 min.

A. his is located at 59 min - Correct, based on Hfr 2 (60 min) and Hfr 3 (58 min).

B. azi is located at 74 min - Correct, based on Hfr 4 (27 min) and Hfr 2 (77 min).

C. cys is located at 11 min - Correct, based on Hfr 3 (10 min).

118. The table lists information about different classes of retroelements :

Which one of the following options has all correct matches between column X and Y?

(a) A- (ii), B- (i), C – (iii)

(b) A- (i), B- (iii), C – (ii)

(d) A- (iv), B- (iii), C – (i)

Ans. (a)

Sol. A. LTR retrotransposons - (ii) Copia elements

B. Non-LTR retroposons- (i) 7-21 bp target sequence

C. SINEs- (iii) Alu elements

D. L1- (iv)

119. The Western Ghats (WG) is a 1600 km mountain chain along the west coast of peninsular India, which intercepts the south-west monsoon winds. Monsoon starts in the southern WG and moves progressively north and retreats in the reverse direction. The southern WG also receives some rainfall from the north-east monsoon. Based on this information, which one of the following statements is most likely to be INCORRECT

(a) Vegetation in the southern WG experiences a more seasonal climate

(b) Vegetation in the northern WG experiences a more seasonal climate

(c) Generally, less seasonal areas tend to have higher plant diversity, so tree diversity will decrease from south to north in the WG

(d) Tree species in the northern WG will have to handle longer dry seasons than species found in the southern WG

Ans. (a)

Sol. The Western Ghats intercept the south-west monsoon winds, which results in a more pronounced monsoon season in the southern region. The southern WG also receives some rainfall from the northeast monsoon. Therefore, the vegetation in the southern WG may experience a more consistent and less seasonal climate compared to the northern WG. The northern WG, on the other hand, may experience more seasonal fluctuations in rainfall and temperature due to the monsoon's progressive movement from south to north and its retreat in the reverse direction.

120. Which one of the following statements about corals is NOT CORRECT?

(a) Corals possess special stinging cells called nematocytes in their tentacles for capturing prey

(b) Several corals have mutualistic interactions with microorganisms called zooxanthellae that photosynthesize and pass some of the food to their hosts

(d) All corals grow only in the photic zones as they need sunlight for their growth

Ans. (d)

Sol. While many corals do indeed require sunlight for their growth and are found in the photic zones (shallow waters where sunlight can penetrate), there are also deep-sea corals that can thrive in deeper, darker waters where little to no sunlight reaches. These deep-sea corals have adapted to survive in low-light conditions and do not rely on photosynthesis like the shallow-water corals with symbiotic zooxanthellae.

121.

Select the correct arrangement of characters that are being described by A to D.

(a) A = Bony skeleton, B = Four limbs, C = Hair, D = Amniotic egg

(b) A = Vertebrate, B = Bony skeleton, C = Amniotic egg, D = Hair

(d) A = Amniotic egg, B = Four limbs, C = Vertebrate D = Hair

Ans. (b)

Sol. A = Vertebrate, B = Bony skeleton, C = Amniotic egg, D = Hair

122. Select the option that correctly identifies all organisms that are included in the International Code of Nomenclature (Shenzhen Code, 2017) along with plants:

(a) Prokaryotes together with all algae and fungi, except their fossils

(b) All algae and fungi along with their fossils, except Microsporidia.

(d) Photosynthetic algae and fungi.

Ans. (b)

Sol. (b) correctly identifies all the organisms included in the ICN along with plants. The ICN covers all algae and fungi, including their fossils, except for Microsporidia. Microsporidia are a group of intracellular parasites that were originally classified as fungi but have been reclassified as a separate group within the kingdom Fungi. As such, they are not covered by the ICN. All other algae and fungi, along with their fossils, fall under the rules and recommendations of the ICN for scientific naming.

123. Match the Indian Biosphere reserves (Column P) with the key fauna (Column Q) they are intended to protect. Which one of the following options has all correct matches between column P and Q?

Which one of the following options has all correct matches between column P and Q?

(a) A-(iii); B- (i); C- (ii); D- (v); E- (iv)

(b) A- (ii); B- (iv); C- (v); D- (i); E- (iii)

(d) A- (v); B- (iii); C- (ii); D- (iv); E- (i)

Ans. (a)

Sol. (a) A-(iii); B- (i); C- (ii); D- (v); E- (iv)

124. Table below shows the protected areas, their description and the protected area types.

Select the option that is NOT CORRECT based on the information provided in the Table

(a) Beas Conservation Reserve, D3, T1

(b) Kaziranga National Park, D1, T2

(d) Manas Wildlife Sanctuary, D4, T1

Ans. (d)

Sol. The Manas Wildlife Sanctuary is described as "Home to many endangered species, including tiger, pygmy hog, Indian rhinoceros, and Asian elephant." However, the protected area type listed for Manas Wildlife Sanctuary is T1, which is incorrect. There is no information about T1 in the table, so the protected area type for Manas Wildlife Sanctuary should be listed as N/A (Not Applicable) or unspecified.

125. Fragmentation breaks up contiguous tracts of natural habitats into smaller patches. In a fragmented landscape where a previously large forest has become a mosaic of patches of different sizes, the following statements can be made about the fragment size and its species diversity.

A. Smaller fragments will always have lower species richness than larger fragments.

B. Species richness will depend on fragment size.

C. Species richness will depend on physical connectivity between fragments.

D. Species richness cannot be compared between large and small fragments.

Select the option where both the statements are correct

(a) A and B

(b) B and C

(d) B and D

Ans. (b)

Sol. Statement B is correct: Species richness will depend on fragment size. Generally, larger fragments are likely to support higher species richness compared to smaller fragments. Statement C is correct: Species richness will depend on physical connectivity between fragments. Physical connectivity allows for the movement of species between fragments, promoting gene flow and maintaining higher species richness in the landscape.

126. The Montreal Protocol and its subsequent amendments have resulted in reduced ozone depletion. It is also observed that ozone depletion over the South Pole is much more severe than over the North Pole. In this regard, consider the following statements.

A. A polar vortex is formed around the North Pole.

B. Stratospheric temperatures over the South Pole are much lower compared to the North Pole.

C. Emissions of ozone depleting substances are higher in the southern hemisphere compared to northern hemisphere.

D. More extensive formation of polar stratospheric clouds over the South Pole compared to the North Pole.

Select the option which includes the correct combination of statements that explain the difference in the ozone depletion between the poles.

(a) A and D

(b) B and D

(d) A and C

Ans. (b)

Sol. Statement B is correct: Stratospheric temperatures over the South Pole are much lower compared to the North Pole. The extremely cold temperatures in the stratosphere over the South Pole create conditions that favor the formation of polar stratospheric clouds (PSCs), which play a crucial role in ozone depletion.

Statement D is correct: More extensive formation of polar stratospheric clouds over the South Pole compared to the North Pole. The more extensive formation of PSCs over the South Pole provides a surface for chemical reactions that release chlorine and bromine, which are responsible for ozone depletion. This is one of the reasons why ozone depletion is more severe over the South Pole.

127. The intensity of compelition can be inferred from knowing the carrying capacity (K) and the population size (N) in the equation below :

Assume that populations have the same intrinsic growth rates(r) and carrying capacities (K). Then, at which one of the following values of the second term (K–N)/K in the euqation, is the intraspecific competition likely to be the highest?

(a) 0.001

(b) 0.009

(d) 0.015

Ans. (a)

Sol. To determine at which value of the second term (K-N)/K the intraspecific competition is likely to be highest, we need to analyze the equation dn/dt = rN(K-N)/K.

The intensity of intraspecific competition is highest when the value of (K-N)/K is smallest, i.e., when N is closest to K.

So, let's examine the options:

(a) (K-N)/K = 0.001 (b) (K-N)/K = 0.009 (c) (K-N)/K = 0.15 (d) (K-N)/K = 0.015

Among these options, (a) has the smallest value for (K-N)/K, which means that N is closest to K in option (a). Therefore, intraspecific competition is likely to be highest when (K-N)/K = 0.001.

128. The graph below shows the accumulation of species in two sites A and B as more plots are sampled.

Based on this graph, following statements were made

A. In both sites sampling more plots will not add any more species

B. Sampling more plots will add more species in Site B but not Site A.

C. Sites A and B are likely to have similar species richness

D. Site B is likely to have higher species richness than Site A

Which one of the following options contains both statements that are INCORRECT?

(a) A and C

(b) A and B

(d) C and D

Ans. (a)

Sol. A. In both sites, sampling more plots will not add any more species.

This statement is incorrect because both Site A and Site B show an increasing trend in the number of species with an increasing number of sampled plots. More plots will likely add more species in both sites.

C. Sites A and B are likely to have similar species richness.

This statement is correct as the accumulation curves for both sites are reaching a plateau, indicating that further sampling is less likely to add significantly more species.

129. Study the global ecosystem data provided in the following table.

Based on the data provided in the table, choose the correct option that represents ecosystems with the highest global, primary production and the highest relative NPP, respectively.

(a) Tropical rainforest and tropical rainforest

(b) Swamp and marsh, and tropical rainforest

(d) Open ocean and open ocean.

Ans. (d)

Sol. Let's calculate the TPP and Relative NPP for each ecosystem:

Tropical rainforest: TPP = 17 * 2000 = 34,000 gm/yr Relative NPP = 2000 / 44 45.45 gm/kg

Swamp and marsh: TPP = 2 * 2500 = 5000 gm/yr Relative NPP = 2500 / 15 166.67 gm/kg

Cultivated land: TPP = 14 * 644 = 9,016 gm/yr Relative NPP = 644 / 1.1 585.45 gm/kg

Open ocean: TPP = 332 * 127 = 42,164 gm/yr Relative NPP = 127 / 0.003 42,333.33 gm/kg

Now, let's compare the values:

The highest global primary production is in the open ocean (42,164 gm/yr).

The highest relative NPP is also in the open ocean (42,333.33 gm/kg).

Therefore, the correct option is (d) Open ocean and open ocean.

130. Interacting plant (A-J) and insect herbivore (P-Y) species in a community are deplected in the network below.

Consider the following statements about the network drawn above

A. Insects are more specialised than plants

B. There are no obligate interactions in this network

C. The community is modular

D. Missing links always represent the absence of an interaction

Given the network which one of the options below is correct?

(a) A only

(b) A and C only

(d) B and D only

Ans. (b)

Sol. A. Insects are more specialized than plants: This statement is correct because we can see that some insects (P, Q, R, S, and T) interact with only one plant species, while all plant species interact with more than one insect species. This indicates that the insects have more specialized interactions compared to plants.

C. The community is modular: This statement is correct. Modularity in ecological networks refers to the presence of tightly connected groups of species that interact more frequently within the group than with species outside the group. In this network, we can see that certain groups of plants and insects form tight clusters of interactions, indicating modularity in the community.

131. A researcher observed ants in contact with plant hoppers that were feeding on tree sap. Which of the following conclusions made by her would be correct?

(a) This is an example of ants being predatory

(b) This is an example of ants upsetting the ecological balance of nature.

(d) This is an example of the tree attracting ants to get rid of plant hoppers.

Ans. (d)

Sol. The observed interaction between ants and plant hoppers, where the ants are in contact with the plant hoppers that are feeding on tree sap, suggests a mutualistic relationship known as "ant-plant mutualism" or "myrmecotrophy."

In this mutualistic interaction, certain tree species have evolved mechanisms to attract and maintain ant colonies on their branches or trunks. These ants, in turn, protect the tree from herbivores like plant hoppers by actively removing or preying on them. The tree provides food resources (sap, nectar, or specialized structures) and shelter to the ants, while the ants provide protection to the tree by reducing herbivore damage.

132. The following statements explain various evolutionary outcomes:

A. Within a lineage, organisms show a constant rate of extinction.

B. Even in the absence of changing interactions, organisms are constantly evolving.

C. Organisms with novel genotypes are at a selective disadvantage.

D. Coevolution between two interacting species act to maintain genetic variation through time.

Which of the following combinations of the above statements are supported by the 'Red Queen hypothesis'?

(a) A and D

(b) A and B

(d) C and D

Ans. (a)

Sol. Statement A: "Within a lineage, organisms show a constant rate of extinction." This statement is supported by the Red Queen hypothesis because it implies that in a dynamic and changing environment, species must constantly evolve and adapt to avoid extinction.

Statement D: "Coevolution between two interacting species act to maintain genetic variation through time." The Red Queen hypothesis predicts that coevolution between species will drive continuous adaptations and counter-adaptations. This constant interaction and reciprocal selection pressure will maintain genetic variation within the species over time.

133. In a particular population A, individuals are under stress and they produce smaller offspring. Based on this, one may conclude that

(a) stress in a population affects offspring size but not the number of offspring

(b) stressed adults prefer to produce smaller offspring that require less food

(d) stress in a population directly affects offspring size.

Ans. (c)

Sol. Stress in a population may be linked to offspring size, meaning that when individuals are under stress, they tend to produce smaller offspring. However, the given information does not provide direct evidence or information about the number of offspring or the reasons why stressed adults produce smaller offspring.

134. Sexually reproducing organisms employ signals to attract mates. If such signals honestly reflect an individual's quality, then which of the following is expected?

(a) Organisms in poor metabolic condition signal more

(b) Organisms in poor metabolic condition signal less

(d) Organisms in good metabolic condition will signal less.

Ans. (b)

Sol. Option (b) is expected in sexually reproducing organisms if signals honestly reflect an individual's quality. In this scenario, organisms in poor metabolic condition are expected to signal less, while organisms in good metabolic condition are likely to signal more. This is because individuals in good condition can afford to invest more in signaling, while those in poor condition may have limited resources to allocate to signaling efforts. Honest signaling helps potential mates accurately assess the quality of individuals and make optimal mating choices.

135. A plant species with unisexual flowers has the following traits: floral longevity = 12 hours, pollen:ovule = 10:1, male and female flowers with synchronized anthesis. Given these, which of the following mutations would be most detrimental to seed set in this plant species?

(a) The pollen:ovule ratio drops to 3:1

(b) Longevity of male and female flowers increases to 16 hours

(d) The pollen:ovule ratio increases to 15:1.

Ans. (a)

Sol. The most detrimental mutation to seed set in this plant species would be option (a), where the pollen:ovule ratio drops to 3:1. In a unisexual plant with synchronized anthesis and a pollen:ovule ratio of 10:1, the reduction in the number of pollen grains relative to ovules would significantly decrease the chances of successful pollination and subsequent seed set. A lower pollen:ovule ratio would mean that fewer pollen grains are available for fertilizing the ovules, leading to reduced seed production.

136. In the cladograms given below each nucleotide change is indicated by a black bar

Which one of the following options represents two equally most parsimonious frees?

Which one of the following options represents two equally most parsimontous trees?

(a) A and B

(b) B and C

(d) A and D

Ans. (a)

Sol. both A and B are equally most parsimonious trees.

137. PBMCs from the blood collected from a tubercubsis (TB) patient were given to four lab techniques to perform ELISOPT assay for interferon (IFN). While all steps recommended for ELISPOT assay followed, the first step was performed differently by the four lab techniques, as detailed below.

A. Lab technician 1 coated each well with 250,000 formaldehyde-treated cells and stimulated the cells with TB-specific antigen

B. Lab technician 2 coated each well with 250,000 cells and did not stimulate the cells with TB-specific antigen

C. Lab technician 3 depleted T cells from PBMCs completely, coated the wells with monocyte-enriched PBMCs, and stimulated them with TB-specific antigen

D. Lab technician 4 coated each well with 250,000 cells and stimulated the cells with TB-specific antigens.

Which of the lab technicians assays will yield a correct ELISPOT result for interteron ?

(a) Lab technician 1

(b) Lab technician 2

(d) Lab technician 4

Ans. (d)

Sol. The correct lab technician assay that will yield a correct ELISPOT result for interferon (IFN) is (d) Lab technician 4. In this case, the technician coated each well with 250,000 cells and stimulated the cells with TB-specific antigens. This is the standard procedure for ELISPOT assay to measure IFN production in response to specific antigens.

138. In a modified version of ELISA, a student first incubated antibody against the Pseudomonas aeruginosa exotoxin A (Pa-exotoxin A) with culture samples in a 0.5mL tube to check for Pseudomonas contamination. Each antibody-culture mixture was then added to a microtiter plate whose wells were coated with Pa-exotoxin A. This was followed by removing the antibody-culture mix from the wells, washing the wells, adding enzyme-conjugated secondary antibody specific for the isotype of the primary antibody, and then detection with enzyme- specific substrate reaction absorbance at 450 nm. The values of absorbance at 450 nm for each of four samples A-D is given below:

Select the option that arranges the samples from having highest to least contamination.

(a) C, D, A, B

(b) B, A, D, C

(d) B, A, C, D

Ans. (a)

Sol. The absorbance values indicate the level of contamination, and higher absorbance values correspond to higher contamination levels. Therefore, sample C has the highest contamination, followed by sample D, then sample A, and finally sample B with the least contamination.

139. The table below lists terms used in bioremediation (column X) and explanations for the terms (column Y).

Which one of the following options is a correct match between terms in column X and explanations in column Y?

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

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

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

Ans. (d)

Sol. A. Bioventing (iii) It is a technique to add oxygen directly to a site of contamination in an unsaturated zone which stimulates in situ aerobic degradation.

B. Natural attenuation (i) Indigenous level of containment degradation without any treatment.

C. Air spraying (ii) It is a technique of adding oxygen to the saturated zone below the water table to stimulate degradation.

D. Biostimulation (iv) Modification of environmental conditions by adding nutrients to enhance the biodegradation process.

140. The functions of some components used for magnetic resonance imaging (MRI) technique are proposed in the following statements:

(A) The static magnetic field used by MRI causes all the magnetically sensitive particles to align themselves in same direction.

(B) The pulse sequence used by MRI is an oscillating magnetic field which causes perturbation of static magnetic field.

(C) The receiver coil placed near a portion of subject's body is a radiofrequency coil that records the relaxation time of protons.

(D) Various parameters of pulse sequence cannot be adjusted to maximize the ability to image certain substances.

(E) The signal intensity received by the receiver coil can provide the location of brain from which it is coming

Choose the option with all INCORRECT statements.

(a) A and B

(b) B and C

(d) D and E

Ans. (d)

Sol. Statement (D) is incorrect because various parameters of the pulse sequence can be adjusted to maximize the ability to image certain substances. Different pulse sequences can be used to highlight specific tissues or substances of interest in MRI imaging.

Statement (E) is incorrect because the signal intensity received by the receiver coil does not provide the location of the brain from which it is coming. MRI signals provide information about the intensity of protons in tissues, but additional spatial encoding techniques are required to determine the location of the signal within the body. These spatial encoding techniques are applied during data acquisition and processing to generate the final MRI images.

141. Given below are terms related to Genome-editing tools in Column A and their feature in Column B.

Which one of the following options is the most appropriate match between terms of Column A and Column B?

(a) A – iv, B – iii, C – ii, D – i

(b) A – iii, B – i, C – iv, D – ii

(d) A – iii, B – iv, C – ii, D – i

Ans. (b)

Sol. ZFN (Zinc Finger Nuclease) is a fusion of the Zinc finger DNA binding domain with the endonuclease domain of the FokI restriction enzyme.

Meganuclease is a homing endonuclease (I-SceI) with a repeat of 35 amino acids, where each amino acid binds to a specific DNA base in the target sequence. CRISPR/Cas9 uses guide RNA for target specificity.

TALEN (Transcription Activator-Like Effector Nuclease) uses TAL effectors, which are repeat domains of approximately 35 amino acids each that bind to specific DNA bases in the target sequence.

142. Given below are plots of the linear derivation of Michaelis-Menten kinetic equation and statements related to the variables (initial velocity – V₀ and substrate concentration-[S]) used.

A. In plot (i) both x and y axes have dependent variables

B. In plot (ii) neither x nor y axis has independent varibles

C. In plot (i) only y-axis has a dependent variable

D. In both the plots, x axis has an independent variable

Select the option that has all the correct statements

(a) A, B and D only

(b) A and C only

(d) B and D only

Ans. (c)

Sol. Plot (i) represents a linear derivation of the Michaelis-Menten kinetic equation, where the x-axis represents the substrate concentration ([S]), which is the independent variable, and the y-axis represents the initial velocity (V0), which is the dependent variable. So, statement C is correct for plot (i).

Plot (ii) is a double-reciprocal plot, also known as a Lineweaver-Burk plot, where both x-axis and y-axis have dependent variables. The x-axis represents 1/[S], which is the dependent variable, and the y-axis represents 1/V0, which is also the dependent variable. So, statement B is correct for plot (ii).

143. Given below are a list of statistical terms in Column A and associated properties/features/descriptors in Column B.

Which one of the options given below is the most appropriate match between entries of Column A with those of Column B?

(a) A – ii, B – i, C – iv, D – iii

(b) A – ii, B – iii, C – iv, D-i

(d) A – iv, B – iii, C – ii, D – i

Ans. (c)

Sol. ANOVA (A) is a statistical test used for the comparison of means of two or more samples.

So, the correct match is A – iii.

Poisson distribution (B) is used to quantify errors in count data. So, the correct match is B – i.

Standard error (C) is a measure of the dispersion of repeated sample means around the true value. So, the correct match is C – iv.

Kurtosis (D) is a descriptor that measures the pointedness of a frequency distribution. So, the correct match is D – ii.

144. Three reactions were performed to detect a 150 bp DNA fragment rich in GC content, using PCR amplification method and the following radiolabeled material (i) 5^'32 P-labelled primers (ii) -³²P-labelled dCTP and, (iii) -³²P-labelled dATP. All the reactions had the remaining components for a successful PCR amplification. After PCR amplification the samples were run on a 2% Agarose gel. The gel was then exposed to radiographic film. From the radiographs given below, which is the correct representation of the reactions (i), (ii) and (iii) in lanes A, B and C respectively.

(a)

(b)

(c)

(d)

Ans. (b)

Sol. When the GC content is too low or too high, it can cause issues with PCR amplification, but 40-60% is just right.

145. Given below are radio-imaging technologies with the type of radiation/ radioisotope that is used for the same.

A. Computed tomography scanner uses UV-rays

B. Magnetic resonance imaging [MRI] uses non-ionization radiation

C. Thyroid scintigraphy uses Iodine-123 (I¹²³)

D. Phase-contrast radiography uses X-rays

E. Fluoroscopy uses X-rays

Which of the options represents all correct statements?

(a) B and E only

(b) A, C and D only

(d) B, C, D and E only

Ans. (d)

Sol. Option (d) includes all the correct matches between the radio-imaging technologies and the type of radiation or radioisotope used for each.

CSIR NET BIOLOGY (Sept. - 2022 Sift 2)
Previous Year Question Paper with Solution.

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Previous Year Question Paper with Solution.