5/5 - (1 vote)

Part B

  1. Chromatin Immunoprecipitation [Option ID = 1881]
  2. Northern Blotting [Option ID = 1882]
  3. ELISA [Option ID = 1883]
  4. Real time PCR [Option ID = 1884]

4. Real time PCR

  • Real-time PCR, also known as quantitative PCR (qPCR), is a laboratory technique used in molecular biology to amplify and simultaneously quantify a targeted DNA molecule. It enables both detection and quantification of a specific sequence in a DNA sample. The procedure follows the general principle of polymerase chain reaction, but its key feature is that the amplified DNA is detected as the reaction progresses in real time, not at its end, as in conventional PCR. This is achieved by using fluorescent dyes that increase their fluorescence upon binding to the DNA. The fluorescence can then be measured after each amplification cycle, allowing the DNA quantity to be calculated.
  1. Both Statement I and Statement II are correct.
    [Option ID = 1885]
  2. Both Statement I and Statement II are incorrect.
    [Option ID = 1886]
  3. Statement I is correct but Statement II is incorrect.
    [Option ID = 1887]
  4. Statement I is incorrect but Statement II is correct.
    [Option ID = 1888]
  1. Replication protein A [Option ID = 1889]
  2. DNA helicase [Option ID = 1890]
  3. DNA topoisomerase [Option ID = 1891]
  4. DNA polymerase [Option ID = 1892]

The enzyme that catalyzes strand separation at the replication fork is DNA helicase. [Option ID = 1890]

Yes, DNA helicase is the enzyme that catalyzes the separation of DNA strands at the replication fork. It unwinds the DNA double helix by breaking the hydrogen bonds between the base pairs, allowing the DNA to be replicated.

  1. MicroRNAs (miRNAs)
    [Option ID = 1893]
  2. Short interfering RNA (siRNA)
    [Option ID = 1894]
  3. Heterochromatin siRNAs (hcsiRNAs)
    [Option ID = 1895]
  4. Trans‐acting siRNAs (tasiRNAs)
    [Option ID = 1896]

Heterochromatin siRNAs (hcsiRNAs) do not regulate gene expression by cleavage of mRNAs. Instead, they are involved in the formation of heterochromatin, a tightly packed form of DNA, which prevents gene expression.

  1. Lethal allele [Option ID = 1897]
  2. Co‐dominance [Option ID = 1898]
  3. Epistasis [Option ID = 1899]
  4. ABO genotype [Option ID = 1900]

3. Epistasis

  • Epistasis is a phenomenon in genetics where the effect of one gene is influenced by one or more other genes, which are sometimes called modifier genes. This can result in a masking effect, where one gene suppresses the phenotypic expression of another gene at a different locus.
  • The ABO blood group system is an example of both multiple allele and co-dominance. In this system, there are three alleles of the gene: IA, IB, and i. The IA and IB alleles are co-dominant, and both express themselves fully in a phenotype. The i allele is recessive to both IA and IB.
  • Epistasis can affect the ABO genotype by modifying the expression of these alleles. For example, a person may have the genotype IAIA or IAi, which would typically result in type A blood. However, if an epistatic gene is present that suppresses the production of A antigens, the person’s blood type may be O, despite their ABO genotype.
  1. Both Statement I and Statement II are correct
    [Option ID = 1901]
  2. Both Statement I and Statement II are incorrect
    [Option ID = 1902]
  3. Statement I is correct but Statement II is incorrect
    [Option ID = 1903]
  4. Statement I is incorrect but Statement II is correct
    [Option ID = 1904]

1. Both Statement I and Statement II are correct

  1. Terpenoids [Option ID = 1905]
  2. Flavonoids [Option ID = 1906]
  3. Alkaloids [Option ID = 1907]
  4. Brassinosteroids [Option ID = 1908]

The class of Nitrogen containing plant secondary metabolites is Alkaloids. [Option ID = 1907]

  • The class of Nitrogen containing plant secondary metabolites is Alkaloids [Option ID = 1907]. Alkaloids are a group of naturally occurring chemical compounds that contain mostly basic nitrogen atoms. This group also includes some related compounds with neutral and even weakly acidic properties.
  • Terpenoids [Option ID = 1905] are made from five-carbon isoprene units and do not typically contain nitrogen.
  • Flavonoids [Option ID = 1906] are a class of plant secondary metabolites based on the flavan nucleus, which is a 15-carbon skeleton, and they also do not typically contain nitrogen.
  • Brassinosteroids [Option ID = 1908] are a class of poly-hydroxysteroids that have been recognized as a sixth class of plant hormones; they do not contain nitrogen.
  1. 8 [Option ID = 1909]
  2. 128 [Option ID = 1910]
  3. 64 [Option ID = 1911]
  4. 32 [Option ID = 1912]
  • The frequency of the A1A2 genotype can be calculated using the Hardy-Weinberg principle, which states that the frequency of two alleles in a gene pool will remain constant from generation to generation in the absence of other evolutionary influences.
  • The frequency of the A1A2 genotype is given by 2pq, where p is the frequency of the A1 allele and q is the frequency of the A2 allele.
  • Given that p = 0.8 and q = 0.2, the frequency of the A1A2 genotype is 2pq = 2 * 0.8 * 0.2 = 0.32.
  • To find the number of plants with the A1A2 genotype in a population of 200 plants, multiply the frequency of the A1A2 genotype by the total population size: 0.32 * 200 = 64.
  • So, the number of plants with the A1A2 genotype is 64.
  • The correct answer is 3. 64 [Option ID = 1911].
  1. Symmetric hybrid [Option ID = 1913]
  2. Asymmetric hybrid [Option ID = 1914]
  3. Cybrid [Option ID = 1915]
  4. Homokaryons [Option ID = 1916]

Such a product will be called a Cybrid.

  • The product will be called a Cybrid [Option ID = 1915].
  • A cybrid is a cell or organism that contains the cytoplasm from one species and the nucleus from another species. In the case of protoplast fusion, the cytoplasm is contributed by both parents, but the nucleus (containing the genetic material) is from only one parent due to chromosome elimination.
  • The other options are not correct because:
    • 1. A symmetric hybrid [Option ID = 1913] is a hybrid organism that has equal genetic contribution from both parents, which is not the case here.
    • 2. An asymmetric hybrid [Option ID = 1914] is a hybrid organism that has unequal genetic contribution from both parents, but it still has genetic material from both parents, which is not the case here.
    • 3. Homokaryons [Option ID = 1916] are cells or organisms that contain nuclei that are genetically identical, which is not the case here as there is only one nucleus.
  1. Thousand [Option ID = 1917]
  2. Million [Option ID = 1918]
  3. Lakh [Option ID = 1919]
  4. Hundred [Option ID = 1920]
  • To create a representative genomic library, you need to ensure that every part of the genome is included at least once. However, to allow for a redundancy of at least one order of magnitude, you need to have each part of the genome represented 10 times.
  • Step 1: Calculate the number of vectors needed to cover the genome once. The genome size is 1 billion base pairs (bp) and each vector can carry 10 thousand base pairs (kbp). So, you divide the total genome size by the size each vector can carry: 1,000,000,000 bp / 10,000 bp = 100,000 vectors
  • Step 2: To allow for a redundancy of one order of magnitude (10 times), multiply the number of vectors needed to cover the genome once by 10: 100,000 vectors x 10 = 1,000,000 vectors
  • So, you would need 1 million vectors to create a representative genomic library of a plant with a genome of a billion base pairs, with a redundancy of at least one order of magnitude. Therefore, the correct answer is 2. Million [Option ID = 1918]
  1. Both Statement I and Statement II are true.
    [Option ID = 1921]
  2. Both Statement I and Statement II are false.
    [Option ID = 1922]
  3. Statement I is true but Statement II is false.
    [Option ID = 1923]
  4. Statement I is false but Statement II is true.
    [Option ID = 1924]

4. Statement I is false but Statement II is true.

The virulence trait of Agrobacterium tumefaciens is not borne on genomic DNA, but on a plasmid known as the Ti (tumor-inducing) plasmid. The Ti-plasmid is indeed used to introduce genes into dicot plants.

  1. It is rich in chlorophyll a. [Option ID = 1925]
  2. It is rich in beta–carotene. [Option ID = 1926]
  3. It is rich in chlorophyll b. [Option ID = 1927]
  4. It is rich in phycobilins. [Option ID = 1928]

Golden rice is pale yellow in color because it is rich in beta-carotene. [Option ID = 1926]

  • Golden rice is pale yellow in color because it is rich in beta-carotene [Option ID = 1926]. Beta-carotene is a carotenoid that gives a yellow-orange color to certain foods like carrots and sweet potatoes.
  • Option 1 and 3 are incorrect because chlorophyll a and b give plants a green color, not yellow.
  • Option 4 is incorrect because phycobilins are pigments found in red algae and cyanobacteria, not in rice.
  1. Phenotyping [Option ID = 1929]
  2. Genotyping [Option ID = 1930]
  3. Karyotyping [Option ID = 1931]
  4. Gene Silencing [Option ID = 1932]

Simple sequence repeats (SSRs) are best utilized for genotyping.

  • Simple Sequence Repeats (SSRs) are best utilized for genotyping [Option ID = 1930]. SSRs, also known as microsatellites, are short, repetitive DNA sequences that are highly variable in length among individuals. This makes them ideal for genotyping, which is the process of determining differences in the genetic make-up (genotype) of an individual by examining the individual’s DNA sequence.
  • SSRs are not typically used for phenotyping [Option ID = 1929] because phenotypes are the observable characteristics of an organism, which are influenced by both its genetic makeup and environmental factors. SSRs alone cannot provide information about an organism’s phenotype.
  • Karyotyping [Option ID = 1931] involves the visualization of an organism’s complete set of chromosomes to detect gross genetic changes, such as large-scale deletions or duplications, which is not the level of detail provided by SSRs.
  • Gene silencing [Option ID = 1932] is a technique used to reduce or eliminate the expression of a specific gene. While SSRs can occur in gene regions and potentially influence gene expression, they are not a tool used to intentionally silence genes.
  1. Both Statement I and Statement II are true.
    [Option ID = 1933]
  2. Both Statement I and Statement II are false.
    [Option ID = 1934]
  3. Statement I is true but Statement II is false.
    [Option ID = 1935]
  4. Statement I is false but Statement II is true.
    [Option ID = 1936]

The correct answer is: Statement I is false and Statement II is true.

Statement I is false because the fragments of DNA generated after cleavage with a restriction enzyme have a phosphate group on their 5′ ends and a -OH group on their 3′ ends, not the other way around. Statement II is true because in the context of PCR, the term “reanneal” is used to describe the process where the separated DNA strands pair up again during the annealing step.

  1. Both Statement I and Statement II are true.
    [Option ID = 1937]
  2. Both Statement I and Statement II are false.
    [Option ID = 1938]
  3. Statement I is true but Statement II is false.
    [Option ID = 1939]
  4. Statement I is false but Statement II is true.
    [Option ID = 1940]

Statement I is true and Statement II is false

  • Statement I is correct. Immobilization of plant cells can induce stress, which in turn stimulates the production of secondary metabolites. These metabolites are often produced as a response to stress or injury.
  • Statement II is incorrect. Cells derived from different explants (tissue samples) in culture do not always produce the same set of secondary metabolites. The type and quantity of secondary metabolites produced can vary depending on the specific plant species, the type of tissue used for explant, and the conditions under which the cells are cultured.
  1. Both A and R are correct and R is the correct explanation of A.
    [Option ID = 1941]
  2. Both A and R are correct but R is not the correct explanation of A.
    [Option ID = 1942]
  3. A is correct but R is not correct.
    [Option ID = 1943]
  4. A is not correct but R is correct.
    [Option ID = 1944]

The correct answer is (B) Both A and R are true but R is not the correct explanation of A.

  • Assertion A is true because a linear DNA molecule with three recognition sites for a restriction enzyme will indeed generate multiple DNA fragments upon partial digestion. However, Reason R, while also true, is not the correct explanation for Assertion A. Partial digestion does lead to a heterogeneous population of DNA fragments, but this is due to the fact that not all recognition sites are cut during partial digestion, not because of the number of recognition sites.
  • Restriction enzymes are proteins that cut DNA at specific sequences, known as recognition sites. If a linear DNA molecule has three recognition sites for a particular restriction enzyme, it can potentially be cut into four fragments.
  • However, during partial digestion, the enzyme doesn’t cut at all the recognition sites. Instead, it cuts at some sites and leaves others intact. This results in a mixture of DNA fragments of different sizes, hence a heterogeneous population of DNA fragments.
  • The number of recognition sites determines the maximum number of fragments that can be produced, but it’s the incomplete cutting during partial digestion that leads to the heterogeneity. Therefore, while both Assertion A and Reason R are true, Reason R is not the correct explanation for Assertion A.
  1. A and C Only
    [Option ID = 1945]
  2. B and D Only
    [Option ID = 1946]
  3. B and C Only
    [Option ID = 1947]
  4. A and D Only
    [Option ID = 1948]
  1. B and C Only [Option ID = 1946]

The correct statements are B and C. B: A higher auxin: cytokinin ratio would indeed induce the production of roots under culture conditions. C: Embryos can be obtained from somatic tissues of plants under in vitro culture conditions. This process is known as somatic embryogenesis.

  1. A‐I, B‐III, C‐IV, D‐II
    [Option ID = 1949]
  2. A‐IV, B‐I, C‐II, D‐III
    [Option ID = 1950]
  3. A‐III, B‐I, C‐IV, D‐II
    [Option ID = 1951]
  4. A‐IV, B‐III, C‐I, D‐II
    [Option ID = 1952]
  1. A‐I, B‐III, C‐IV, D‐II
    [Option ID = 1949]
  • A. Northern Hybridization – I. Transcriptome analysis
  • B. Next generation sequencing – III. Analysis of splice variants
  • C. Sanger’s DNA sequencing – IV. Capillary sequencers
  • D. In-situ immunolocalization – II. Detection of proteins in specific cell types
  1. A‐III, B‐I, C‐IV, D‐II
    [Option ID = 1953]
  2. A‐IV, B‐III, C‐II, D‐I
    [Option ID = 1954]
  3. A‐III, B‐IV, C‐I, D‐II
    [Option ID = 1955]
  4. A‐II, B‐I, C‐IV, D‐III
    [Option ID = 1956]

4. A‐II, B‐I, C‐IV, D‐III
[Option ID = 1956]

  • A. Bar – II. Herbicide resistance. The bar gene provides resistance to the herbicide Basta.
  • B. Meristem Culture – I. Virus-free plants. Meristem culture is a method used in plant tissue culture to produce virus-free plants.
  • C. Chalcone Synthase – IV. Flower color. Chalcone Synthase is an enzyme involved in the biosynthesis of flavonoids, which contribute to the color of flowers.
  • D. gai – III. Dwarf wheat. The gai gene is involved in the regulation of plant growth and development, including the development of dwarf varieties of plants such as wheat.
  1. Both Statement I and Statement II are true.
    [Option ID = 1957]
  2. Both Statement I and Statement II are false.
    [Option ID = 1958]
  3. Statement I is true but Statement II is false.
    [Option ID = 1959]
  4. Statement I is false but Statement II is true.
    [Option ID = 1960]

4. Statement I is false but Statement II is true.
[Option ID = 1960]

  • Statement I is false: Capping does occur by the addition of a methylated guanosine triphosphate to the 5′ end of the hnRNA , addition of guanosine occur at capping not the adenosine.
  • Statement II is true: During tailing, a poly(A) tail consisting of about 200-300 adenylate residues is added to the 3′ end of the hnRNA. This process is known as polyadenylation.
  1. Bacterial Artificial Chromosome (BAC) [Option ID = 1961]
  2. Agrobacterium [Option ID = 1962]
  3. T4 phage [Option ID = 1963]
  4. Cosmid [Option ID = 1964]

2.Agrobacterium [Option ID = 1962]. The vector most commonly used in crop improvement is Agrobacterium

  1. Helix turn helix [Option ID = 1965]
  2. Helix loop helix [Option ID = 1966]
  3. Leucine zipper [Option ID = 1967]
  4. Zinc finger [Option ID = 1968]

The lac repressor of E. coli contains 1. Helix turn helix DNA binding motifs.

  • The lac repressor of E. coli contains Helix-turn-helix DNA binding motifs. This motif is a specific structure found in proteins that can bind to the major groove of DNA. The motif is characterized by two alpha helices connected by a short strand of amino acids and is found in many proteins that regulate gene expression.
  • The other options, Helix-loop-helix, Leucine zipper, and Zinc finger, are also types of DNA-binding motifs, but they are not found in the lac repressor of E. coli.
    • The Helix-loop-helix motif is a protein structural motif characterized by two alpha helices connected by a loop. It is not found in the lac repressor.
    • The Leucine zipper is a common three-dimensional structural motif in proteins and it is not found in the lac repressor.
    • The Zinc finger motif is a small protein structural motif that is characterized by the coordination of one or more zinc ions in order to stabilize the fold. It is not found in the lac repressor.
  1. chemically mutagenizing a mouse and selecting for mutant offspring. [Option ID = 1969]
  2. creating a chimera by fusing cells from two different cell lines. [Option ID = 1970]
  3. infecting the mouse with a retrovirus. [Option ID = 1971]
  4. transfecting embryonic stem cells with an inactivated gene. [Option ID = 1972]

4.transfecting embryonic stem cells with an inactivated gene. [Option ID = 1972]

  • Knockout mice are usually created by transfecting embryonic stem cells with an inactivated gene [Option ID = 1972]. This process involves replacing or disrupting a specific gene in mouse embryonic stem cells, then inserting these cells into a mouse blastocyst. The resulting mouse is a “knockout” for the targeted gene, meaning that gene’s function is lost or altered.
  • Option 1 is incorrect because chemical mutagenesis is a random process and does not allow for the targeted inactivation of a specific gene.
  • Option 2 is incorrect because creating a chimera by fusing cells from two different cell lines does not necessarily result in the inactivation of a specific gene.
  • Option 3 is incorrect because infecting a mouse with a retrovirus does not specifically target and inactivate a particular gene. Retroviruses can insert their genetic material into the host genome, but this process is not precise or targeted.
  1. ssDNA [Option ID = 1973]
  2. dsDNA [Option ID = 1974]
  3. ssRNA [Option ID = 1975]
  4. dsRNA [Option ID = 1976]

4. dsRNA [Option ID = 1976]

  • RNAi, or post-transcriptional gene silencing, is a response to the presence of dsRNA (double-stranded RNA). This process does not respond to ssDNA (single-stranded DNA), dsDNA (double-stranded DNA), or ssRNA (single-stranded RNA) because it specifically involves the breakdown of dsRNA into small interfering RNAs (siRNAs).
  • These siRNAs then guide the RNA-induced silencing complex (RISC) to degrade messenger RNAs (mRNAs) with complementary sequences, thereby silencing the gene from which the mRNA was transcribed.
  1. DNA Ligase [Option ID = 1977]
  2. DNA helicase [Option ID = 1978]
  3. DNA topoisomerase [Option ID = 1979]
  4. DNA polymerase [Option ID = 1980]
  1. DNA Ligase [Option ID = 1977]
  • The protein required to seal a nick during Okazaki fragment maturation is 1. DNA Ligase [Option ID = 1977]. DNA Ligase is responsible for joining the Okazaki fragments together to form a continuous DNA strand.
  • As for why not the others:
    • 2. DNA helicase [Option ID = 1978] is involved in unwinding the DNA double helix during replication, not in sealing nicks.
    • 3. DNA topoisomerase [Option ID = 1979] relieves the tension in the DNA molecule that is created by the unwinding, but it does not seal nicks.
    • 4. DNA polymerase [Option ID = 1980] synthesizes the new DNA strand but does not have the ability to join Okazaki fragments together.

Self splicing introns are an example of:[Question ID = 496][Question Description = 186_GATBS1_SECTIONB_APR22_Q86]

  1. Spliceosomes [Option ID = 1981]
  2. Ribozymes [Option ID = 1982]
  3. Mitrons [Option ID = 1983]
  4. Inteins [Option ID = 1984]
  1. Ribozymes [Option ID = 1982]
  • Self-splicing introns are an example of Ribozymes. Ribozymes are RNA molecules that have the ability to catalyze specific biochemical reactions, similar to the action of protein enzymes. The term “ribozyme” comes from ribonucleic acid enzyme. Self-splicing introns are segments of the RNA that can remove themselves and join the remaining portions together.
  • Spliceosomes, on the other hand, are complex structures that assist in the removal of introns from pre-mRNA. They are not self-splicing and require additional proteins for their function.
  • Mitrons and Inteins are not related to the process of self-splicing. Mitrons are a type of small RNA molecule, while Inteins are protein segments that can excise themselves and join the remaining portions together, similar to self-splicing introns but in proteins, not RNA.
  1. 5’ AAGG 3’
    3’ TTCC 5’ [Option ID = 1985]
  2. 3’ AGTC 5’
    5’ TCAG 3’ [Option ID = 1986]
  3. 5’ GGCC 3’
    3’ CCGG 5’ [Option ID = 1987]
  4. 5’ ACCA 3’
    3’ TGGT 5’ [Option ID = 1988]

3. 5’ GGCC 3’
3’ CCGG 5’ [Option ID = 1987]

  • Option 3, 5’ GGCC 3’ 3’ CCGG 5’, can be a restriction site for a type II restriction enzyme. These enzymes typically recognize palindromic sequences, which read the same forwards and backwards on both strands of the DNA molecule.
  • The other options are not palindromic and therefore would not be recognized by a type II restriction enzyme.
  1. none of the proteins in the cell will contain phenylalanine. [Option ID = 1989]
  2. lysine will partially replace phenylalanine at certain positions. [Option ID = 1990]
  3. the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine‐specifying anticodons. [Option ID = 1991]
  4. lysine will replace phenylalanine at all amino acids positions. [Option ID = 1992]

In a bacterial cell with a defective aminoacyl-tRNA synthetase that attaches lysine to tRNAs with the anticodon AAA instead of the normal phenylalanine, the consequence during protein synthesis would be:

2. Lysine will partially replace phenylalanine at certain positions. [Option ID = 1990]

This means that some of the proteins synthesized in the cell will have lysine incorporated in place of phenylalanine at specific amino acid positions, leading to altered protein structures and functions12.

  1. 81
    [Option ID = 1993]
  2. 123
    [Option ID = 1994]
  3. 256
    [Option ID = 1995]
  4. 4096
    [Option ID = 1996]
  • Each base pair has occupied 1/4 part if the all content are equal , means content of bases are 25%(0.25) each.
  • So, A=1/4, T=1/4, C=1/4, G=1/4, in case of equal content.
  • But here in this case A+T = 0.40, it indicates A= 0.20, T= 0.20
  • hence the GC content what would be in total is equal to 0.60 but further G and C divided into 0.30 each.
  • HaeIII ezyme generated fragment GGCC = 0.30 x 0.30 x 0.30 x 0.30 = 0.0081
  • Inverse the value to get the average size of fragment = 1/0.0081 = 10000/81 =123.45
  • So, the correct answer is 123.45 [Option ID = 1994]
  1. Both A and R are true and R is the correct explanation of A
    [Option ID = 1997]
  2. Both A and R are true but R is not the correct explanation of A
    [Option ID = 1998]
  3. A is true but R is false
    [Option ID = 1999]
  4. A is false but R is true
    [Option ID = 2000]
  1. Both A and R are true and R is the correct explanation of A [Option ID = 1997]

both Assertion A and Reason R are correct, and Reason R is the correct explanation for Assertion A. In bacteria, transcription and translation are coupled, meaning they occur simultaneously. This allows for attenuation of the trp operon. When tryptophan levels are high, the ribosome quickly translates regions 1 and 2 of the leader sequence, causing it to stall at region 2. This allows region 3 to pair with region 4, forming a transcription termination loop that halts further transcription of the trp operon.

  1. A→ B→C
    [Option ID = 2001]
  2. C→ B→A
    [Option ID = 2002]
  3. C→ A→B
    [Option ID = 2003]
  4. A→ C→B
    [Option ID = 2004]
  1. A→ B→C
    [Option ID = 2001]

The correct order of progressively higher levels of chromatin organization is: A. Nucleosome, B. 30 nm chromatin fibre, C. Looped domain.

  1. Both Statement I and Statement II are correct
    [Option ID = 2005]
  2. Both Statement I and Statement II are incorrect
    [Option ID = 2006]
  3. Statement I is correct but Statement II is incorrect
    [Option ID = 2007]
  4. Statement I is incorrect but Statement II is correct
    [Option ID = 2008]

3. Statement I is correct but Statement II is incorrect
[Option ID = 2007]

  • Statement I: True. After digestion with certain restriction enzymes, DNA fragments with cohesive (sticky) ends are formed. These ends can pair with complementary sequences.
  • Statement II: False. The cohesive ends of DNA fragments will have phosphate groups at the 5′ end and a hydroxyl group at the 3′ end. This is due to the nature of DNA structure and the way restriction enzymes cut the DNA.
  1. Both Statement I and Statement II are true [Option ID = 2009]
  2. Both Statement I and Statement II are false [Option ID = 2010]
  3. Statement I is true but Statement II is false [Option ID = 2011]
  4. Statement I is false but Statement II is true [Option ID = 2012]

3. Statement I is true but Statement II is false [Option ID = 2011]

  • double digestion may give 4 fragments instead of 5 fragments .
  1. DNA polymerases selectively cleave rNTPs. [Option ID = 2013]
  2. rNTPs are cyclized by DNA polymerase. [Option ID = 2014]
  3. Steric exclusion of rNTPs from the DNA polymerase active site. [Option ID = 2015]
  4. dNTPs are more stable than rNTPs. [Option ID = 2016]

3. Steric exclusion of rNTPs from the DNA polymerase active site. [Option ID = 2015]

Steric exclusion of rNTPs from the DNA polymerase active site occurs when an active site residue, usually one with a bulky side chain, collides with the 2′-OH group on the ribose ring of an incoming rNTP. This prevents rNTPs from binding stably in the active site of the dPol. 

  1. operon is expressed at low level to allow lactose entry. [Option ID = 2017]
  2. lactose binds with the inducer to activate the operon. [Option ID = 2018]
  3. expression of operon is blocked by an active repressor. [Option ID = 2019]
  4. lactose needs a transporter to enter the cell. [Option ID = 2020]

The regulation of the lac operon is considered negative because expression of operon is blocked by an active repressor. [Option ID = 2019]

The regulation of the lac operon is considered negative because the repressor protein, when active, binds to the operator region of the operon and prevents transcription. This effectively “turns off” the operon. When lactose is present, it binds to the repressor and changes its shape, making it unable to bind to the operator. This allows transcription to proceed, effectively “turning on” the operon. This type of regulation is called negative because the operon is turned off by the active form of the regulatory molecule (the repressor).

  1. grow on the galactose‐rich medium [Option ID = 2021]
  2. grow on lactose rich medium [Option ID = 2022]
  3. develop resistance to antibiotics [Option ID = 2023]
  4. produce color in the presence of a chromogenic substrate [Option ID = 2024]

4. produce color in the presence of a chromogenic substrate [Option ID = 2024]

  • Recombinants with insertional inactivation of β‐galactosidase are identified by their inability to produce color in the presence of a chromogenic substrate.
  • This is because β‐galactosidase is an enzyme that cleaves certain bonds in chromogenic substrates, such as X-gal, producing a colored product.
  • If the gene for β‐galactosidase is disrupted by the insertion of a DNA fragment, the enzyme will not be produced, and the bacteria will not be able to cleave the chromogenic substrate, hence no color will be produced.
  1. number of transformants obtained. [Option ID = 2025]
  2. nutrient utilization rate. [Option ID = 2026]
  3. the level of resistance to antibiotic. [Option ID = 2027]
  4. Western blotting. [Option ID = 2028]

The level of expression of recombinant protein can be gauged by 4. Western blotting.

Western blotting is a technique used to detect specific proteins in a sample. It involves separating proteins by gel electrophoresis, transferring them to a membrane, and then using antibodies to identify the protein of interest. The intensity of the signal from the antibody-protein interaction can give an estimate of the level of expression of the recombinant protein.

  1. 10% [Option ID = 2029]
  2. 80% [Option ID = 2030]
  3. 30% [Option ID = 2031]
  4. 70% [Option ID = 2032]

4. 70% [Option ID = 2032]

  • To calculate the percentage of sequence identity, you need to compare the two sequences and count the number of identical positions.
  • The sequences are:
    • 1. VKSFLWTQAL
    • 2. VPSFRWTQSL
    • Comparing these two sequences, we see that the identical positions are: V, S, F, W, T, Q, and L. That’s 7 identical positions.
    • The total number of positions in each sequence is 10.
    • To calculate the percentage of sequence identity, divide the number of identical positions by the total number of positions and multiply by 100.
    • So, (7/10) x 100 = 70% Therefore, the percentage of sequence identity for the given sequences is 70%.
  1. Polypeptide chain, m‐RNA, t‐RNA [Option ID = 2033]
  2. t‐RNA, m‐RNA, Polypeptide chain [Option ID = 2034]
  3. Polypeptide chain, t‐RNA, m‐RNA [Option ID = 2035]
  4. m‐RNA, Polypeptide chain, t‐RNA [Option ID = 2036]

The correct sequence in which they leave the ribosome at the time of completion of protein synthesis is: Polypeptide chain, t‐RNA, m‐RNA [Option ID = 2035].

  1. eliminates incomplete cDNA from a gene library. [Option ID = 2037]
  2. create expression libraries based on genes that are currently being expressed. [Option ID = 2038]
  3. identify and construct new probes for southern hybridization. [Option ID = 2039]
  4. identify sets of genes that are only expressed under certain conditions. [Option ID = 2040]

Subtractive hybridization is useful to 4. identify sets of genes that are only expressed under certain conditions.

  • subtractive hybridization is a technique used in genetics to identify unique sequences of DNA or RNA, such as genes that are expressed in one cell type but not in another.
  • This method is particularly useful for identifying genes that are only expressed under certain conditions, such as in response to a specific stimulus or at a particular stage of development.
  • The process involves removing or “subtracting” common sequences between two or more DNA samples, leaving behind only the unique sequences.
  1. intracellular bacteria reservoirs. [Option ID = 2041]
  2. cells with upregulated MHC‐I molecules. [Option ID = 2042]
  3. cells with downregulated MHC‐I molecules. [Option ID = 2043]
  4. cells with self‐marker on their surface. [Option ID = 2044]

3. cells with downregulated MHC‐I molecules. [Option ID = 2043]

  • Natural Killer (NK) cells are a type of lymphocyte, or white blood cell, that play a crucial role in the immune system’s response to infections and cancer. They are particularly significant because they have the ability to recognize and kill cells that have downregulated Major Histocompatibility Complex class I (MHC-I) molecules.
  • MHC-I molecules are found on the surface of all nucleated cells in the body and present internal proteins to the immune system. This allows the immune system to monitor the health of these cells. If a cell is infected with a virus or becomes cancerous, the MHC-I molecules will present abnormal proteins, alerting the immune system to the problem.
  • However, some viruses and cancer cells have evolved to evade the immune system by downregulating, or decreasing the expression of, MHC-I molecules. This prevents the immune system from recognizing that these cells are infected or abnormal. This is where NK cells come in. They have receptors that can detect the absence or reduced presence of MHC-I molecules on a cell.
  • If the MHC-I molecules are downregulated, the NK cells recognize this as a sign that the cell is infected or cancerous, and they respond by killing the cell. This makes NK cells a crucial line of defense against infections and cancer.
  1. binding peptide antigens for recognition by antigen‐specific receptors on T‐cells. [Option ID = 2045]
  2. mediation of T‐independent B‐cell responses. [Option ID = 2046]
  3. helping in endocytosis of antigens by phagocytic cells. [Option ID = 2047]
  4. aid opsonization of foreign particle. [Option ID = 2048]
  1. binding peptide antigens for recognition by antigen‐specific receptors on T‐cells. [Option ID = 2045]

The primary function of the class I and class II MHC molecules is binding peptide antigens for recognition by antigen-specific receptors on T-cells.

  1. The neo gene produces an antibiotic that kills unwanted cells.
    [Option ID = 2049]
  2. The neo gene is the right size for disabling other genes.
    [Option ID = 2050]
  3. The neo gene provides a selectable marker.
    [Option ID = 2051]
  4. The neo gene produces a repressor that inhibits transcription of the target gene.
    [Option ID = 2052]

The correct answer is 3. The neo gene provides a selectable marker.

This means that cells which have incorporated the neo gene can be identified and selected for, usually because they have gained resistance to a certain antibiotic.

  1. Negative selection [Option ID = 2053]
  2. MHC restriction [Option ID = 2054]
  3. Affinity maturation [Option ID = 2055]
  4. Lineage commitment [Option ID = 2056]

The selection of cells whose T-cell receptors respond to self-MHC is known as: 2. MHC restriction [Option ID = 2054]

  • The selection of cells whose T-cell receptors respond to self-MHC is known as MHC restriction [Option ID = 2054]. This process ensures that T cells recognize antigens only when they are presented on the body’s own MHC molecules.
  • Negative selection [Option ID = 2053] refers to the elimination of T cells that react strongly with self-antigens, which is a different process.
  • Affinity maturation [Option ID = 2055] is a process that occurs in B cells, not T cells, and involves the increase in affinity of the B cell receptor for a specific antigen over time.
  • Lineage commitment [Option ID = 2056] refers to the process by which a cell becomes committed to a specific developmental path, which is not specific to the interaction of T-cell receptors with self-MHC.
  1. A‐II, B‐I, C‐III, D‐IV
    [Option ID = 2057]
  2. A‐II, B‐III, C‐I, D‐IV
    [Option ID = 2058]
  3. A‐III, B‐I, C‐IV, D‐II
    [Option ID = 2059]
  4. A‐II, B‐III, C‐IV, D‐I
    [Option ID = 2060]

4. A‐II, B‐III, C‐IV, D‐I
[Option ID = 2060]

  1. Inactivated Vaccines (A): These vaccines are based on the killed or altered disease-causing germ.
  2. Live Attenuated Vaccines (B): These vaccines are based on the weakened form of the pathogen.
  3. Conjugate Vaccines ©: These vaccines are based on the combination of the weak antigen coat of the pathogen and strong carrier proteins.
  4. Toxoid Vaccines (D): These vaccines are based on the poisonous protein made by the pathogen.
  1. Fibroblasts [Option ID = 2061]
  2. Neutrophils [Option ID = 2062]
  3. Dendritic cells [Option ID = 2063]
  4. Epithelial cells [Option ID = 2064]

3. Dendritic cells [Option ID = 2063]

3. Dendritic cells are Professional Antigen Presenting Cells. They capture, process, and present antigens to T cells in the immune system, initiating an immune response.

  1. Both A and R are correct and R is the correct explanation of A
    [Option ID = 2065]
  2. Both A and R are correct but R is not the correct explanation of A
    [Option ID = 2066]
  3. A is correct but R is not correct
    [Option ID = 2067]
  4. A is not correct but R is correct
    [Option ID = 2068]
  1. Both A and R are correct and R is the correct explanation of A
    [Option ID = 2065]

Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A. Polyclonal antibodies are produced by different B cell clones, each recognizing a different epitope on the same antigen, which allows them to bind to the same antigen but recognize different epitopes.

Identify the statement that is NOT applicable to enzyme catalyzed reaction.
[Question ID = 518][Question Description = 208_GATBS1_SECTION‐B_APR22_Q108]

  1. The reaction proceeds with the conversion of the substrate to a higher energy transition state.
    [Option ID = 2069]
  2. Enzymes alter the equilibrium constant of the reaction.
    [Option ID = 2070]
  3. Enzymes decrease the energy of activation required for the reaction.
    [Option ID = 2071]
  4. Some enzymes involve multiple steps of electron transfer.
    [Option ID = 2072]

2. Enzymes alter the equilibrium constant of the reaction. [Option ID = 2070]

The statement that is NOT applicable to enzyme catalyzed reaction is: “Enzymes alter the equilibrium constant of the reaction.” Enzymes speed up the rate of a reaction but do not change the equilibrium constant.

  1. A‐IV, B‐III, C‐II, D‐I
    [Option ID = 2073]
  2. A‐IV, B‐III, C‐I, D‐II
    [Option ID = 2074]
  3. A‐III, B‐I, C‐IV, D‐II
    [Option ID = 2075]
  4. A‐II, B‐III, C‐IV, D‐I
    [Option ID = 2076]

2. A‐IV, B‐III, C‐I, D‐II
[Option ID = 2074]

  • The correct matches between the enzymes in List I and the ions in List II that are associated with them are:
    • A. Cytochrome oxidase – IV. Copper ions
    • B. DNA polymerase – II. Magnesium ions
    • C. Nitrate reductase – I. Molybdenum ions
    • D. Urease – II. Nickel ions
  1. A, B and C only [Option ID = 2077]
  2. A, C and E only [Option ID = 2078]
  3. C and E only [Option ID = 2079]
  4. B and D only [Option ID = 2080]

The correct answer is “A, C and E only” [Option ID = 2078].

Isoleucine (A), Tryptophan (C), and Phenylalanine (E) are both glucogenic and ketogenic. Glucogenic amino acids can be converted into glucose through gluconeogenesis, while ketogenic amino acids can be converted into ketone bodies. Isoleucine, Tryptophan, and Phenylalanine are unique because they can be converted into both glucose and ketone bodies, making them both glucogenic and ketogenic. Serine (B) and Proline (D) are only glucogenic, not ketogenic.

  1. length and degree of unsaturation of hydrocarbon chain [Option ID = 2089]
  2. degree of unsaturation only [Option ID = 2090]
  3. length of hydrocarbon chain only [Option ID = 2091]
  4. Number of side and branch chains present [Option ID = 2092]
  • The melting point of fatty acids in plant-derived oils is influenced by both the length and degree of unsaturation of the hydrocarbon chain [Option ID = 2089].
  • The length of the hydrocarbon chain affects the melting point because longer chains have more surface area that can be in contact with each other, leading to stronger intermolecular forces and a higher melting point.
  • The degree of unsaturation affects the melting point because unsaturated fatty acids (those with double bonds) have kinks in their structure that prevent them from packing together as tightly as saturated fatty acids (those without double bonds). This results in weaker intermolecular forces and a lower melting point. Therefore, both factors – the length of the hydrocarbon chain and the degree of unsaturation – influence the melting point of fatty acids in plant-derived oils.
  1. Chondroitin
    [Option ID = 2093]
  2. Inulin
    [Option ID = 2094]
  3. Hyaluronic acid
    [Option ID = 2095]
  4. Heparin
    [Option ID = 2096]

2. Inulin
[Option ID = 2094]

  • Inulin is NOT a mucopolysaccharide.
  • It is a type of polysaccharide, but it is not classified as a mucopolysaccharide.
  • The other options, Chondroitin, Hyaluronic acid, and Heparin, are all examples of mucopolysaccharides
  1. Both Statement I and Statement II are true
    [Option ID = 2097]
  2. Both Statement I and Statement II are false
    [Option ID = 2098]
  3. Statement I is true but Statement II is false
  4. [Option ID = 2099]
  5. Statement I is false but Statement II is true
    [Option ID = 2100]

4. Statement I is false but Statement II is true
[Option ID = 2100]

  • Statement I is incorrect. Fatty acid synthase is not a tetramer, but a dimer of a multienzyme polypeptide.
  • Statement II is correct. Each polypeptide in fatty acid synthase does harbor seven independent enzymatic functions.
  1. 2 Km
    [Option ID = 2101]
  2. 4 Km
    [Option ID = 2102]
  3. 0.8 Km
    [Option ID = 2103]
  4. Km
    [Option ID = 2104]

2. 4 Km
[Option ID = 2102]

  • In Michaelis-Menten kinetics, the velocity of an enzymatic reaction (v) is given by the equation: v = Vm[S] / (Km + [S]) where Vm is the maximum velocity, [S] is the substrate concentration, and Km is the Michaelis constant.
  • If we want the velocity to be 80% of Vm,
  • we can set up the equation as follows: 0.8Vm = Vm[S] / (Km + [S])
  • Solving for [S] gives: [S] = 0.8Km / (1 – 0.8) = 4Km
  • So, the substrate concentration required to obtain a velocity equal to 80% Vm is 4 times the Michaelis constant, or 4Km.
  • Therefore, the correct answer is 4 Km [Option ID = 2102].
  1. resolution is poor.
    [Option ID = 2105]
  2. resolved structure is not reliable.
    [Option ID = 2106]
  3. difficult to interpret the data for larger proteins (>50 KDa).
    [Option ID = 2107]
  4. difficult to interpret the data for smaller proteins (<15 KDa).
    [Option ID = 2108]
  1. B and D Only
    [Option ID = 2109]
  2. A and D Only
    [Option ID = 2110]
  3. A and C Only
    [Option ID = 2111]
  4. C and D Only
    [Option ID = 2112]

The correct answer is 2. A and D Only [Option ID = 2110]. Immobilization of enzymes can increase the specificity of the enzyme in batch reactions and decrease the operational cost of the industrial process.

  1. Epinephrine [Option ID = 2113]
  2. Insulin [Option ID = 2114]
  3. Estrogen [Option ID = 2115]
  4. Progesterone [Option ID = 2116]

The hormone that activates the enzyme adenylate cyclase is Epinephrine.

The rate of sedimentation of a particle is dependent on the: [Question ID = 530][Question Description =220_GATBS1_SECTION‐B_APR22_Q120]

  1. density of the particle only. [Option ID = 2117]
  2. size of the particle only. [Option ID = 2118]
  3. viscosity of the medium only. [Option ID = 2119]
  4. size and density of the particle along with the viscosity of the medium. [Option ID = 2120]

4. The rate of sedimentation of a particle is dependent on the size and density of the particle along with the viscosity of the medium.

  1. the focal length of the electron microscope is significantly larger. [Option ID = 2121]
  2. the contrast is enhanced by staining with atoms of heavy metal. [Option ID = 2122]
  3. electron beams have much shorter wavelengths than visible light. [Option ID = 2123]
  4. the electron microscope has a much greater ratio of image size to real size. [Option ID = 2124]

3. electron beams have much shorter wavelengths than visible light. [Option ID = 2123]. The resolution power of a transmission electron microscope is of the sub-nanometer level because electron beams have much shorter wavelengths than visible light.

  1. A→ B→C→D
    [Option ID = 2125]
  2. A→ B→D→C
    [Option ID = 2126]
  3. A→ C→B→D
    [Option ID = 2127]
  4. B→ A→C→D
    [Option ID = 2128]

2. A→ B→D→C
[Option ID = 2126]

  • The correct sequence of the MAP kinase signal transduction pathway is: A. Ras, B. Raf, D. Mek, C. ERK.
  • The MAP kinase signal transduction pathway is a series of protein interactions that transmit a signal from a receptor on the surface of a cell to the DNA in the nucleus of the cell.
  • The correct sequence is:
    • 1. Ras: This is a small G protein that gets activated when a signal binds to the cell surface receptor.
    • 2. Raf: Activated Ras then activates Raf, a kinase enzyme.
    • 3. Mek: Raf phosphorylates and activates Mek, another kinase enzyme.
    • 4. ERK: Mek then phosphorylates and activates ERK (also known as MAPK), the final kinase in the pathway.
    • Once activated, ERK can enter the nucleus and regulate gene expression by phosphorylating various transcription factors.
  1. B→C→E→D→A
    [Option ID = 2129]
  2. A→D→C→E→B
    [Option ID = 2130]
  3. C→A→E→B→D
    [Option ID = 2131]
  4. D→E→B→C→A
    [Option ID = 2132]

Option ID = 2130: A→D→C→E→B

  • This translates to:
    • 1. Parasite feeds on the hemoglobin in the RBCs (A)
    • 2. The food vacuole of the parasite increases in size (D)
    • 3. The nucleus of the parasite moves on one side (C)
    • 4. The haemozoin granules are formed (E)
    • 5. Parasite forms a schizont by multiple fissions (B)
  1. A and B Only
    [Option ID = 2133]
  2. B Only
    [Option ID = 2134]
  3. B and C Only
    [Option ID = 2135]
  4. C Only
    [Option ID = 2136]

The correct answer is “B and C Only [Option ID = 2135]”.

Cyclin-dependent kinase (Cdk) is present throughout the cell cycle and it is an enzyme that attaches phosphate groups to other proteins. However, Cdk is not inactive in the presence of cyclin B. In fact, it is activated by binding to cyclin B.

  1. the cell cycle is completed bypassing the M phase. [Option ID = 2137]
  2. the cell cycle is completed bypassing the G2 checkpoint. [Option ID = 2138]
  3. completion of cell cycle will depend on G2 checkpoint. [Option ID = 2139]
  4. the cell will next enter M phase bypassing the G2 checkpoint. [Option ID = 2140]
  • The correct answer is 3. Completion of the cell cycle will depend on the G2 checkpoint. [Option ID = 2139]
  • Explanation:
    • After the G1 checkpoint, the cell enters the S phase where DNA replication occurs.
    • Following the S phase, the cell enters the G2 phase, where it prepares for mitosis.
    • The G2 checkpoint ensures that all DNA has been correctly replicated and that the cell is ready to proceed to the M phase (mitosis).
    • If the cell does not pass the G2 checkpoint, it will not proceed to the M phase.
  • Why not others:
    • 1. The cell cycle is not completed by bypassing the M phase. [Option ID = 2137] The M phase is crucial for cell division and cannot be bypassed.
    • 2. The cell cycle is not completed by bypassing the G2 checkpoint. [Option ID = 2138] The G2 checkpoint ensures that the cell is ready for mitosis, and cannot be bypassed.
    • 4. The cell will not next enter the M phase bypassing the G2 checkpoint. [Option ID = 2140] The G2 checkpoint ensures that the cell is ready for mitosis, and cannot be bypassed.
  1. 1000 [Option ID = 2141]
  2. 10000 [Option ID = 2142]
  3. 100000 [Option ID = 2143]
  4. 1000000 [Option ID = 2144]
  1. 1000 [Option ID = 2141]
  • The probability of mutation per base pair per replication is 5 x 10^-10.
  • The genome length of the bacteria is 5 Mb (megabases), which is 5 x 10^6 base pairs.
  • The average number of mutations per replication cycle is the product of the mutation rate and the genome length, which is (5 x 10^-10) x (5 x 10^6) = 0.0025.
  • To have one mutation on average per replication cycle, we need to divide 1 by the average number of mutations per replication cycle. So, 1 / 0.00025 = 4000.
  • Therefore, the number of bacterial cells required so that there is one mutation on an average per replication cycle is 4 x 1000.
  1. reduced internal pore diffusion. [Option ID = 2145]
  2. Reduced external film diffusion. [Option ID = 2146]
  3. decrease in energy of activation for the enzyme. [Option ID = 2147]
  4. enhanced enzyme activity due to mechanical shear. [Option ID = 2148]

2. Reduced external film diffusion. [Option ID = 2146]

  • The behavior can be attributed to 2. Reduced external film diffusion. [Option ID = 2146].
  • This is because increasing the agitation speed reduces the thickness of the external film around the enzyme, thereby enhancing the mass transfer rate and increasing the fractional conversion.
  1. remains unchanged.
    [Option ID = 2149]
  2. increases.
    [Option ID = 2150]
  3. decreases.
    [Option ID = 2151]
  4. increases initially followed by a decrease.
    [Option ID = 2152]

3. apparent viscosity of the fluid decreases.

  • In the power law model for fluid rheology, the flow behavior index (n) determines the relationship between shear stress and shear rate.
  • If n < 1, the fluid is shear-thinning or pseudoplastic.
  • This means that as shear stress increases, the apparent viscosity of the fluid decreases.
  • This is because the fluid’s resistance to flow decreases with increasing shear stress.
  • In the power law model, the apparent viscosity (μ) can be expressed as a function of the shear rate the consistency index (k), and the flow behavior index (n). The relationship is given by:
  • This equation shows that the apparent viscosity is not a constant, but depends on the shear rate. If n < 1 (as in the case of shear-thinning or pseudoplastic fluids), the apparent viscosity decreases with increasing shear rate. This is because these fluids become less resistant to flow (i.e., less viscous) under higher shear conditions.
  • Therefore, the correct answer is: 3. decreases. [Option ID = 2151]
  1. diauxic growth. [Option ID = 2153]
  2. exponential growth. [Option ID = 2154]
  3. linear growth. [Option ID = 2155]
  4. lag phase followed by log phase growth. [Option ID = 2156]
  1. diauxic growth. [Option ID = 2153]
  • When a growth medium contains two different carbon sources, one of which is preferentially utilized, the resulting growth pattern is known as diauxic growth.
  • In diauxic growth, microorganisms initially consume the preferred carbon source until it is depleted. Afterward, they switch to utilizing the secondary carbon source, leading to a characteristic biphasic growth curve. The phases include an initial exponential growth phase (associated with the preferred carbon source) followed by a second exponential growth phase (associated with the secondary carbon source). This phenomenon reflects the adaptation of microorganisms to changing nutrient availability12.

Therefore, the correct answer is Option ID 2153: diauxic growth. 🌱

4. A‐IV, B‐I, C‐II, D‐III
[Option ID = 2160]

  • The correct matches between List I and List II are:
  • A. Sedimentation coefficient – IV. Centrifugation
  • B. Partition Coefficient – I. Aqueous two-phase extraction
  • C. Rejection coefficient – II. Ultrafiltration
  • D. Activity coefficient – This term is typically used in the context of chemical thermodynamics to describe the deviation of a mixture from ideal behavior, and does not directly correspond to any of the methods listed in List II.
  1. A‐I, B‐III, C‐II, D‐IV
    [Option ID = 2162]
  • The correct matches for the enzymes and their industries are:
    • A. Penicillinase – I. Pharmaceutical
    • B. Pectinase – III. Wine
    • C. Trypsin – II. Leather
    • D. Renin – IV. Dairy
  1. for supplying CO for the metabolic reaction. [Option ID = 2165]
  2. to control pH by generating carbonic acid. [Option ID = 2166]
  3. to control pH by generating bicarbonate. [Option ID = 2167]
  4. to aid photosynthetic reaction. [Option ID = 2168]

In animal cell culture, a CO2 enriched atmosphere in the incubation chamber is used 3. to control pH by generating bicarbonate.[Option ID = 2167]

  1. 0.80 [Option ID = 2169]
  2. 1.25 [Option ID = 2170]
  3. 0.05 [Option ID = 2171]
  4. 0.45 [Option ID = 2172]
  1. 0.80 [Option ID = 2169]
  • The yield of product based on substrate consumed is calculated by dividing the specific product formation rate by the specific rate of substrate utilization.
  • So, Yield = Specific product formation rate / Specific rate of substrate utilization
  • Substituting the given values: Yield = 0.20 h / 0.25 h = 0.80
  • So, the yield of product based on substrate consumed is 0.80 [Option ID = 2169].
  1. reduces by 50% [Option ID = 2173]
  2. remains unchanged [Option ID = 2174]
  3. increases by 100% [Option ID = 2175]
  4. increases by 300% [Option ID = 2176]

4. increases by 300% [Option ID = 2176]

  • When you break a large bubble into smaller ones, you create more surface area. This is because each small bubble has its own surface, and when you add up all these surfaces, it’s more than the surface of the original large bubble.
  • So, if you halve the diameter of a bubble, you end up with four times as many bubbles of that size fitting in the same space, which means four times the surface area. That’s why the gas-liquid interfacial area increases by 300%.
  • If you halve the diameter of a bubble, you actually create eight times as many bubbles of that size from the original bubble (since volume is proportional to the cube of the radius).
  • However, the surface area of each of these smaller bubbles is only one-fourth of the original. So, if you multiply the eight times more bubbles by the one-fourth surface area each one has, you get twice the original surface area.
  • This is a 100% increase. But, because we’re dealing with a gas-liquid interface, both sides of the bubble contribute to the interfacial area.
  • So, we double the 100% increase to account for both sides of the bubble, leading to a 200% increase. Then, because the original surface area is considered 100%, the total increase is 300% of the original.
  • Finally, we add the original 100% (the original surface area before the increase), which gives us a total of 300%.
  1. Both A and R are true and R is the correct explanation of A
    [Option ID = 2085]
  2. Both A and R are true but R is not the correct explanation of A
    [Option ID = 2086]
  3. A is true but R is false
    [Option ID = 2087]
  4. A is false but R is true
    [Option ID = 2088]

3. A is true but R is false
[Option ID = 2087]

  • The Assertion A is correct because the major source of energy for the basic functioning of the cell is indeed derived from oxidative metabolism. This process occurs in the mitochondria, where glucose and other substrates are oxidized to produce ATP, the cell’s main energy currency.
  • However, the Reason R is incorrect. Mitochondria do not transfer high energy electrons from molecular oxygen to glucose. Instead, during the process of cellular respiration, glucose is oxidized and its high-energy electrons are transferred to molecular oxygen, producing carbon dioxide (CO2) and water (H2O) as byproducts. This process also generates ATP. Therefore, Reason R is a misrepresentation of the process of oxidative metabolism.
  1. 60 [Option ID = 2081]
  2. 100 [Option ID = 2082]
  3. 125 [Option ID = 2083]
  4. 150 [Option ID = 2084]
  1. 60 [Option ID = 2081]
  • The average molecular weight of the aromatic amino acids (tryptophan, phenylalanine, and tyrosine) is approximately 181.33 g/mol.
  • If the peptide has a molecular weight of 11,000 g/mol, then the likely number of amino acid residues in the peptide can be calculated by dividing the total molecular weight by the average weight of an amino acid.
  • 11,000 g/mol ÷ 181.33 g/mol ≈ 60.67
  • Therefore, the peptide likely contains around 60 amino acid residues. So, the correct answer is 1. 60 [Option ID = 2081].
  1. 3 fold [Option ID = 2177]
  2. 9 fold [Option ID = 2178]
  3. 27 fold [Option ID = 2179]
  4. 81 fold [Option ID = 2180]
  • In Michaelis-Menten kinetics, the reaction velocity (v) is given by the equation: v = Vmax[S]/(Km + [S]) where Vmax is the maximum reaction velocity, [S] is the substrate concentration, and Km is the Michaelis constant.
  • The reaction velocity is 10% of Vmax when [S] = 0.1Km and 90% of Vmax when [S] = 9Km.
  • Therefore, to increase the reaction velocity from 10% of Vmax to 90% of Vmax, the substrate concentration has to be increased from 0.1Km to 9Km.
  • This is an increase by a factor of 9Km/0.1Km = 90, which is a 9-fold increase.
  • So, the correct answer is: 9 fold [Option ID = 2178].
  1. have cell wall with mycolic acid [Option ID = 2181]
  2. thrive well in acidic environment [Option ID = 2182]
  3. are fastidious bacteria [Option ID = 2183]
  4. cause acid depletion in soil [Option ID = 2184]
  1. Acid Fast bacteria have cell walls with mycolic acid.
  2. Yes, acid-fast bacteria, such as Mycobacterium tuberculosis, have a unique cell wall structure. Their cell walls contain a high concentration of mycolic acid, a type of fatty acid. This mycolic acid layer makes the cell wall waxy and resistant to staining by Gram stain. Instead, a special staining technique called acid-fast staining is used to identify these bacteria.
  1. positive sense single stranded RNA virus [Option ID = 2185]
  2. negative sense single stranded RNA virus [Option ID = 2186]
  3. double stranded RNA virus without segmented genome [Option ID = 2187]
  4. double stranded RNA virus with segmented genome [Option ID = 2188]

Coronavirus is a 1. positive sense single stranded RNA virus [Option ID = 2185]

  1. β ‐lactamase [Option ID = 2189]
  2. Phosphodiesterase [Option ID = 2190]
  3. Transesterase [Option ID = 2191]
  4. Glycopeptide transpeptidase [Option ID = 2192]

Penicillin inhibits cell wall synthesis by inhibiting 4. Glycopeptide transpeptidase [Option ID = 2192].

  • Penicillin works by inhibiting the enzyme glycopeptide transpeptidase, which is involved in the synthesis of the bacterial cell wall. This prevents the bacteria from properly forming a cell wall, which is crucial for their survival and growth.
  • As a result, the bacteria become vulnerable and can easily be destroyed by the immune system or die due to environmental stress.
  • The correct matches between the organisms and the diseases they cause are:
  • A. Rickettsia prowazekii – II. Epidemic typhus
  • B. Variola major – III. Small pox
  • C. Varicella zoster Virus – IV. Shingles
  • D. Borrelia recurrentis – I. Relapsing fever
  1. B only
    [Option ID = 2197]
  2. A, C and D only
    [Option ID = 2198]
  3. B and C only
    [Option ID = 2199]
  4. D and E only
    [Option ID = 2200]
  1. B only
    [Option ID = 2197]
  • The incorrect mechanism of antibiotic action is B. Rifampicin: Blocks correct positioning of A‐site aminoacyl-tRNA for peptidyl transfer reaction.
  • Rifampicin actually inhibits DNA-dependent RNA polymerase, thereby preventing transcription. The other options correctly describe the mechanisms of action for those antibiotics.
  • The other options correctly describe the mechanisms of action for those antibiotics.
  • Here’s why:
  • A. Tetracycline: It does inhibit the binding of aminoacyl-tRNA to the A-site of the ribosome, preventing the addition of new amino acids to the growing peptide chain during protein synthesis.
  • C. Chloramphenicol: It inhibits the function of EF-Tu, a protein involved in protein synthesis. EF-Tu delivers aminoacyl-tRNA to the ribosome, and by inhibiting this, Chloramphenicol prevents protein synthesis.
  • D. Ciprofloxacin: It inhibits bacterial topoisomerases, enzymes that are necessary for DNA replication. By doing this, it prevents the bacteria from replicating its DNA and thus, from reproducing.
  • E. Puromycin: It mimics the 3’ end of aminoacyl-tRNA and inserts itself into the A-site of the ribosome, causing premature chain termination during protein synthesis.
  • Each of these antibiotics has a different target within the bacterial cell, and they all inhibit some aspect of protein synthesis or DNA replication, which are vital processes for the bacteria. This is why they are effective as antibiotics
  1. Both statement I and statement II are correct
    [Option ID = 2201]
  2. Both statement I and statement II are incorrect
  3. [Option ID = 2202]
  4. Statement I is correct but statement II is not correct
    [Option ID = 2203]
  5. Statement I is not correct but statement II is correct
    [Option ID = 2204]

5. Statement I is not correct but statement II is correct
[Option ID = 2204]

Statement I is incorrect. The recombination frequency of genes varies from 0% to 50%, not 50% to 80%. Statement II is correct. Recombination can occur anywhere along the length of the chromosome.

  • A. Silent mutation – IV. Amino acid sequence is unchanged.
  • B. Missense mutation – III. Amino acid sequence is changed in one amino acid.
  • C. Nonsense mutation – I. Amino acid codon is changed to a stop codon.
  • D. Frameshift mutation – II. Amino acid sequence beyond the site of mutation is changed.
  1. Both Statement I and Statement II are correct
    [Option ID = 2209]
  2. Both Statement I and Statement II are incorrect.
    [Option ID = 2210]
  3. Statement I is correct but Statement II is incorrect.
    [Option ID = 2211]
  4. Statement I is incorrect but Statement II is correct.
    [Option ID = 2212]
  1. Both Statement I and Statement II are correct

Sickle cell trait does provide some protection against malaria, and it can be transmitted if both parents are heterozygous for the β‐globin gene.

Certainly! Let’s analyze the given genotypes of the lac operon and determine which one leads to constitutive enzyme synthesis:

  1. Genotype A: I+O+Z+Y+/I+O+Z+Y+
    • This genotype represents the wild-type operon.
    • The expression of the operon will be inducible, not constitutive.
  2. Genotype B: I−OCZ+Y+/I−OCZ+Y+
    • This genotype also represents the constitutive operon.
    • The expression of the operon will be constitutive.
  3. Genotype C: I−O−Z+Y+/I−O−Z+Y+
    • This genotype represents the wild-type operon.
    • The expression of the operon will be inducible, not constitutive.
  4. Genotype D: I−O+Z+Y+/I−O+Z+Y+
    • This genotype involves a constitutive mutation.
    • The lac operon will be expressed continuously, even in the absence of lactose.
    • Therefore, the enzyme synthesis would be constitutive.

Therefore, the correct answer is option 3: B and D Only. Genotype C (I−O+Z+Y+/I−O+Z+Y+) leads to constitutive enzyme synthesis 12.

  1. 32 cm [Option ID = 2217]
  2. 29 cm [Option ID = 2218]
  3. 19 cm [Option ID = 2219]
  4. 25.4 cm [Option ID = 2220]

2. 29 cm [Option ID = 2218]

  • The base height of the plant is 20 cm. Each additive allele contributes equally to the height. The maximum height is 38 cm, which would be achieved if all alleles were present (AA, BB, CC). This means that each allele contributes (38-20)/6 = 3 cm to the height (6 because there are two copies of each of three genes).
  • The genotype AAbbCc has three additive alleles . Therefore, the height of the plant would be the base height plus the contribution from the three additive alleles, which is 20 cm + 3*3 cm = 29 cm.

2. Salamanders are a sister group to the group containing lizards, goats, and humans.

  1. 4 [Option ID = 2225]
  2. 8 [Option ID = 2226]
  3. 6 [Option ID = 2227]
  4. 12 [Option ID = 2228]

3. 6 [Option ID = 2227]

In the human ABO blood group system, there are three common alleles represented by the letters A, B, and O. Let’s explore the possible genotypes:

  1. AA: Homozygous for the A allele.
  2. AO: Heterozygous, carrying one A allele and one O allele.
  3. BB: Homozygous for the B allele.
  4. BO: Heterozygous, carrying one B allele and one O allele.
  5. AB: Heterozygous, carrying both A and B alleles.
  6. OO: Homozygous for the O allele.

Therefore, there are a total of six different genotypes in the human ABO genetic locus 12. These genotypes correspond to the four common blood types:

  • Type A: Carries the A antigen.
  • Type B: Carries the B antigen.
  • Type AB: Carries both A and B antigens.
  • Type O: Does not carry either A or B antigens.

So, the correct answer is 6 genotypes 12.

  1. there are no tRNAs corresponding to stop codons. [Option ID = 2229]
  2. the corresponding tRNA do not have an amino acid. [Option ID = 2230]
  3. the corresponding tRNA cannot bind to ribosome binding site. [ ID = 2231]
  4. stop codon codes for pseudo amino acid. [Option ID = 2232]
  1. The stop codons lead to termination of translation because there are no tRNAs corresponding to stop codons. [Option ID = 2229]
  1. some of the daughters can be normal. [Option ID = 2233]
  2. all the daughters will be affected. [Option ID = 2234]
  3. all the daughters and none of the sons will be affected. [Option ID = 2235]
  4. all the sons and some of the daughters will be affected. [Option ID = 2236]

3. all the daughters and none of the sons will be affected. [Option ID = 2235] So, both “all the daughters will be affected” and “all the daughters and none of the sons will be affected” are correct.

In a family where the father is affected by an X-linked dominant trait and the mother is normal, the inheritance pattern leads to specific outcomes for their children:

  1. Some of the daughters can be normal.
    • This is because X-linked dominant traits are passed from the father to all of his daughters. However, not all daughters will necessarily express the trait, as it depends on whether they inherit the affected X chromosome or the normal X chromosome from their father.
  2. All the daughters will inherit the condition.
    • Since the father carries the affected X chromosome, he will pass it on to all of his daughters. As a result, all daughters will inherit the X-linked dominant trait.
  3. All the daughters and none of the sons will be affected.
    • Sons, on the other hand, inherit their father’s Y chromosome and not the X chromosome carrying the trait. Therefore, sons will not be affected by the X-linked dominant trait.
  4. All the sons and some of the daughters will be affected.
    • This statement is incorrect for X-linked dominant traits. Sons will not be affected, but daughters have a chance of being affected, depending on whether they inherit the dominant or recessive X chromosome from their father.

In summary, X-linked dominant traits result in all daughters inheriting the condition, while sons remain unaffected. The mother’s status does not directly impact the sons’ inheritance of the trait 123.

  1. M‐fold [Option ID = 2237]
  2. Alpha‐fold [Option ID = 2238]
  3. PDB [Option ID = 2239]
  4. Swiss‐Model [Option ID = 2240]

The protein prediction server based on artificial intelligence is Alpha-fold [Option ID = 2238].

  • AlphaFold is an artificial intelligence program developed by DeepMind that predicts the 3D structure of proteins based on their amino acid sequence. It uses machine learning algorithms to predict the way in which proteins will fold, which is crucial in understanding their function and behavior.
  • This technology has the potential to greatly accelerate research in fields such as drug discovery and bioengineering.
  1. Both Statement I and Statement II are correct
    [Option ID = 2241]
  2. Both Statement I and Statement II are incorrect
    [Option ID = 2242]
  3. Statement I is correct but Statement II is incorrect
    [Option ID = 2243]
  4. Statement I is incorrect but Statement II is correct
    [Option ID = 2244]

3. Statement I is correct but Statement II is incorrect
[Option ID = 2243]

  • Statement I is correct. BLAST (Basic Local Alignment Search Tool) is a tool that compares protein and nucleotide sequences to identify areas of similarity.
  • Statement II is incorrect. GenBank is a database of all publicly available nucleotide sequences and their protein translations, not a repository of data on the phenotypic results of gene knockouts in humans.
  • Therefore, the correct answer is “Statement I is correct and Statement II is incorrect.”
  1. replication machinery [Option ID = 2245]
  2. transcriptional machinery [Option ID = 2246]
  3. RNA capping machinery [Option ID = 2247]
  4. protein translation machinery [Option ID = 2248]

All are correct, 1. replication machinery [Option ID = 2245]

RNA viruses, which include a diverse group of pathogens like the common cold virus, influenza, and HIV, indeed rely on their host cells for various essential processes. Let’s break down their dependency:

  1. Replication Machinery: RNA viruses lack their own replication machinery. Instead, they hijack the host cell’s machinery to replicate their genetic material. This involves creating multiple copies of their RNA genome, which is essential for the virus to propagate.
  2. Transcriptional Machinery: RNA viruses also depend on the host cell’s transcriptional machinery. Once inside the cell, they utilize the host’s enzymes to transcribe their RNA genome into messenger RNA (mRNA). This mRNA serves as a template for protein synthesis.
  3. RNA Capping Machinery: RNA viruses do not have their own specialized machinery for adding a protective cap to their mRNA. Instead, they rely on the host cell’s capping enzymes to modify their mRNA. The cap is crucial for stability and efficient translation of viral proteins.
  4. Protein Translation Machinery: RNA viruses lack ribosomes and other components necessary for protein synthesis. Therefore, they utilize the host cell’s ribosomes to translate their viral mRNA into proteins. These proteins are essential for various viral functions, including replication and assembly.

In summary, RNA viruses are entirely dependent on their host cells for these critical processes, highlighting their intricate relationship with the cellular machinery 123. 🦠🔬

  1. Cytoplasm [Option ID = 2249]
  2. Nucleus [Option ID = 2250]
  3. Mitochondria [Option ID = 2251]
  4. Late endosomes [Option ID = 2252]
  1. Cytoplasm [Option ID = 2249]

RNA viruses generally replicate in the cytoplasm (Option 1) because they carry their own replication machinery, which includes RNA-dependent RNA polymerase. This enzyme is not found in the host cell’s nucleus, so the virus replicates in the cytoplasm where it can use its own machinery to replicate its RNA genome.

  1. Binds specifically to a cellular receptor.
    [Option ID = 2253]
  2. Cannot bind to neutralizing antibodies.
    [Option ID = 2254]
  3. Interacts with the cellular receptor leading to activation of endocytosis.
    [Option ID = 2255]
  4. Is composed of unique structural domains.
    [Option ID = 2256]

2. Cannot bind to neutralizing antibodies. [Option ID = 2254]

  • The statement that is NOT true for the receptor binding site of a virus is: “Cannot bind to neutralizing antibodies.” [Option ID = 2254]. This is incorrect because neutralizing antibodies can indeed bind to the receptor binding sites of viruses, preventing them from attaching to host cells.
  • The other options are true:
    • 1. “Binds specifically to a cellular receptor.” [Option ID = 2253] – Viruses have specific proteins that can bind to specific receptors on the host cell.
    • 3. “Interacts with the cellular receptor leading to activation of endocytosis.” [Option ID = 2255] – When a virus binds to a receptor on a host cell, it can trigger endocytosis, a process where the cell engulfs the virus.
    • 4. “Is composed of unique structural domains.” [Option ID = 2256] – The receptor binding site of a virus is composed of unique structural domains that allow it to bind to specific receptors on the host cell.
  1. Both Type I and Type III hypersensitivity [Option ID = 2257]
  2. Only Type I hypersensitivity [Option ID = 2258]
  3. Both Type I and Type II hypersensitivity [Option ID = 2259]
  4. Both Type II and Type III hypersensitivity [Option ID = 2260]

The correct answer is “Both Type II and Type III hypersensitivity” [Option ID = 2260].

  • Type II hypersensitivity reactions involve IgG and IgM antibodies binding to antigens on the surface of cells, leading to complement activation and cell lysis.
  • Type III hypersensitivity reactions involve the formation of immune complexes (antigen-antibody complexes) that can deposit in various tissues and activate the complement system, leading to inflammation and tissue damage.
  • Type I hypersensitivity does not involve complement activation. It is mediated by IgE antibodies and involves mast cells and basophils, leading to immediate allergic reactions.
  1. Histiocyte
    [Option ID = 2261]
  2. Kupffer cell
    [Option ID = 2262]
  3. Langerhans cell
    [Option ID = 2263]
  4. Dendritic cell
    [Option ID = 2264]

4. Dendritic cell
[Option ID = 2264]

The dendritic cell is not a tissue-specific macrophage. While histiocytes, Kupffer cells, and Langerhans cells are all types of tissue-specific macrophages, dendritic cells are a different type of immune cell. They are antigen-presenting cells, not macrophages. They capture, process, and present antigens to T-cells in the immune system, initiating an immune response.

  1. These are cancerous B cells
    [Option ID = 2265]
  2. They can divide indefinitely in a culture
    [Option ID = 2266]
  3. Their HGPRT gene is non‐functional
    [Option ID = 2267]
  4. They can produce antibodies
    [Option ID = 2268]

4. They can produce antibodies
[Option ID = 2268]

The statement “They can produce antibodies” is NOT TRUE about myeloma cells used in the hybridoma technique for the production of monoclonal antibodies. Myeloma cells are cancerous B cells that can divide indefinitely in culture, and their HGPRT gene is non-functional. However, they cannot produce antibodies on their own. They need to be fused with an antibody-producing B cell to form a hybridoma, which can then produce monoclonal antibodies.

  1. Covishield [Option ID = 2269]
  2. Sputnik V [Option ID = 2270]
  3. Covaxin [Option ID = 2271]
  4. Moderna mRNA‐1273 [Option ID = 2272]

Sputnik V is an inactivated vaccine.

  • Sputnik V [Option ID = 2270] is an inactivated vaccine. The other vaccines work differently: Covishield and Moderna mRNA-1273 are both mRNA vaccines, which use a piece of the virus’s genetic material to stimulate an immune response.
  • Covaxin is an inactivated vaccine, but it uses a different method of inactivation than Sputnik V.
  1. Shoot apical meristem [Option ID = 2273]
  2. Internode [Option ID = 2274]
  3. Leaf disc [Option ID = 2275]
  4. Root tip [Option ID = 2276]
  1. Shoot apical meristem [Option ID = 2273]

Scientists should choose the Shoot apical meristem for culturing. This is because the meristematic cells in the shoot apical meristem are undifferentiated and have the ability to divide and differentiate into various types of cells. Additionally, the shoot apical meristem is often virus-free even in infected plants, making it a good choice for tissue culture and micropropagation of a plant species nearing extinction due to virus infection.

  1. increased bioactive compounds.
    [Option ID = 2277]
  2. fortified Fe and Zn contents.
    [Option ID = 2278]
  3. enhanced shelf‐life.
    [Option ID = 2279]
  4. bigger and pulpier fruits.

3. The ‘Flavr Savr’ tomato is known for its enhanced shelf-life.