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NEET Biology Class 12 - Molecular Basis of Inheritance Ultra-Hard Question Bank

Duration: 240 minutesTotal Marks: 960Questions: 240Negative Marking: -1

Instructions:

  1. Each question has four options (1), (2), (3), (4). Choose the most correct answer.
  2. Each correct answer carries 4 marks.
  3. Each wrong answer will deduct 1 marks.
  4. Unanswered questions will not be penalised.
1.Assertion (A): Erwin Chargaff's rule states that for a double-stranded DNA, the ratios
between Adenine and Thymine, and Guanine and Cytosine, are constant and equal to
one.\nReason (R): The two strands of a double-stranded polynucleotide chain always run in
anti-parallel directions, meaning one has 5′ → 3′ polarity and the other has 3′ → 5′ polarity.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
2.Consider the following statements regarding the polynucleotide chain:\nI. A nitrogenous
base is linked to the OH of 5′-C pentose sugar through an N-glycosidic linkage to form a
nucleoside.\nII. Uracil is present in RNA at the 2′-position of the ribose sugar.\nIII. In RNA,
every nucleotide residue has an additional --OH group present at the 2′-position of the ribose
sugar.\nIV. Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a
dinucleotide.\nV. Thymine is also chemically known as 5-methyl uracil.\n\nWhich of the
combinations represents ONLY INCORRECT statements?
(1)II and IV
(2)I and II
(3)I, II, and III
(4)III, IV, and V
3.Which of the following statements about DNA packaging in prokaryotes is/are
CONSISTENT with NCERT?\nI. In prokaryotes like E. coli, a defined nucleus is absent, and
DNA is scattered throughout the cell.\nII. DNA is held with some basic proteins that have
positive charges in a region termed as 'nucleoid'.\nIII. The DNA in the nucleoid is organized in
large loops held by basic proteins.\nIV. Prokaryotic DNA packaging involves histone octamers
that form nucleosomes.\n\nChoose the correct option:
(1)II and III
(2)I and II
(3)II, III, and IV
(4)I, II, and III
4.Assertion (A): Heterochromatin is described as transcriptionally active chromatin, whereas
euchromatin is transcriptionally inactive.\nReason (R): Euchromatin is loosely packed and
stains light, while heterochromatin is densely packed and stains dark.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
5.Identify the INCORRECT statements regarding Frederick Griffith's experiments
(1928):\nI. Griffith used Streptococcus pneumoniae, which causes pneumonia in mice.\nII. S-
strain bacteria possess a smooth, shiny mucus (polysaccharide) coat and are virulent.\nIII. R-
strain bacteria lack the coat and do not cause pneumonia.\nIV. Mice injected with a mixture of
heat-killed S-strain and live R-strain died, and Griffith recovered living S-strain bacteria from
them.\nV. Griffith concluded that the transforming principle was DNA, which transformed the
R-strain into S-strain.\n\nChoose the incorrect statements:
(1)I, II, and III
(2)I and IV
(3)V only
(4)IV and V
6.Assertion (A): In the biochemical characterization of Griffith's transforming principle,
digestion with proteases and RNases did not affect transformation, while digestion with DNase
inhibited it.\nReason (R): Proteases and RNases degrade proteins and RNA respectively, which
are not the transforming substances, whereas DNase degrades DNA, proving that DNA is the
transforming substance.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
7.Regarding the components of a nucleotide chain, which of the following statements are
CORRECT?\nI. Purines include Adenine and Guanine, while Pyrimidines include Cytosine,
Uracil, and Thymine.\nII. Cytosine is common to both DNA and RNA, whereas Thymine is
present in DNA and Uracil is present in RNA.\nIII. Adenosine, Deoxyadenosine, Guanosine, and
Deoxythymidine are all examples of nucleotides.\nIV. When a phosphate group is linked to the
3′-OH of a nucleoside, a nucleotide is formed.\nV. The backbone of a polynucleotide chain is
formed due to sugar and phosphates.
(1)I, III, and IV
(2)I, II, and V
(3)I, II, and III
(4)II, IV, and V
8.Assertion (A): In the double-helix structure of DNA, a purine always comes opposite to a
pyrimidine on the complementary strands.\nReason (R): This specific pairing generates an
approximately uniform distance between the two strands of the helix.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
9.Consider the following dimensions and features of the DNA double helix. Identify the
CORRECT statements:\nI. The two chains are coiled in a right-handed fashion.\nII. The pitch
of the helix is 34 nm.\nIII. There are roughly 10 base pairs in each turn of the helix.\nIV. The
distance between a base pair in a helix is approximately 0.34 nm.\nV. The plane of one base pair
stacks over the other in the double helix, which confers stability in addition to H-
bonds.\n\nChoose the correct option:
(1)II, III, and V
(2)I, II, and IV
(3)All of the above
(4)I, III, IV, and V
10.Assertion (A): Friedrich Meischer in 1869 identified DNA as an acidic substance present in
the nucleus and named it 'Nuclein'.\nReason (R): The double helix structure of DNA was
proposed immediately after Meischer's discovery because of the advanced technical capabilities
available at that time.
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
11.Identify the CORRECT statement(s) regarding the Central Dogma of Molecular
Biology:\nI. It was proposed by James Watson and Francis Crick jointly.\nII. It states that
genetic information flows from DNA → RNA → Protein.\nIII. In some viruses, the flow of
information is in the reverse direction, from RNA to DNA.\nIV. The process of reverse genetic
information flow is catalyzed by DNA-dependent RNA polymerase.\n\nChoose the correct
option:
(1)II and III
(2)I and IV
(3)II, III, and IV
(4)I, II, and III
12.Regarding the calculation of DNA length in mammalian and bacterial cells, which of the
following is/are CORRECT?\nI. The distance between two consecutive base pairs is taken as
0.34 × 10-9 m.\nII. The total number of base pairs in a diploid mammalian cell is 6.6 × 109
bp.\nIII. The calculated length of DNA in a typical diploid mammalian cell is approximately 2.2
metres.\nIV. The dimension of a typical eukaryotic cell nucleus is approximately 10-6 m.\nV. If
the length of E. coli DNA is 1.36 mm, its genome has approximately 4.0 × 106 bp.\n\nChoose the
correct option:
(1)II, IV, and V
(2)I, III, and V
(3)All of the above
(4)I, II, III, and IV
13.Assertion (A): Histones are basic, positively charged proteins because they are rich in basic
amino acid residues carrying positive charges in their side chains.\nReason (R): Histones are
rich in lysine and arginine, both of which carry a positive charge in their side chains at
physiological pH.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)(A) is true but (R) is false.
14.Regarding the structure of a nucleosome, which of the following statements is/are
CORRECT?\nI. A histone octamer is composed of eight molecules of histones representing four
different types.\nII. The positively charged DNA is wrapped around the negatively charged
histone octamer.\nIII. A typical nucleosome contains 200 bp of DNA helix.\nIV. Nucleosomes
constitute the repeating unit of chromatin, which appears as 'beads-on-string' under an electron
microscope.\n\nChoose the correct option:
(1)II and III
(2)All of the above
(3)I, III, and IV
(4)I and III only
15.Read the following statements regarding higher-level chromatin packaging and identify
the CORRECT ones:\nI. The 'beads-on-string' structure in chromatin is packaged to form
chromatin fibers.\nII. Chromatin fibers are further coiled and condensed at the prophase stage
of cell division to form chromosomes.\nIII. The packaging of chromatin at higher levels requires
an additional set of proteins collectively called Non-histone Chromosomal (NHC) proteins.\nIV.
Euchromatin is densely packed and stains dark, while heterochromatin is loosely packed and
stains light.\n\nChoose the correct option:
(1)III and IV
(2)I, II, and III
(3)II and III
(4)I and III
16.Assertion (A): In a human diploid cell, there are approximately 3.3 × 107 nucleosomes
present in the chromatin.\nReason (R): A typical nucleosome contains 200 bp of DNA, and the
haploid content of human DNA is 3.3 × 109 bp.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
17.Regarding the strains of Streptococcus pneumoniae used by Griffith, which of the following
is/are INCORRECT?\nI. Smooth (S) strain bacteria form smooth, shiny colonies because they
possess a mucous (polysaccharide) coat.\nII. Rough (R) strain bacteria form rough colonies
because they lack the polysaccharide coat.\nIII. R-strain bacteria are virulent and cause
pneumonia in infected mice.\nIV. S-strain bacteria are non-virulent and do not develop
pneumonia.\nV. The virulence of the S-strain is directly attributed to the presence of its mucus
coat.\n\nChoose the incorrect statements:
(1)III and IV
(2)I, II, and V
(3)II and III
(4)IV and V
18.Read the following experimental setups in Griffith's transforming experiment and identify
the CORRECT combinations:\nI. S-strain (live) → Inject into mice → Mice die\nII. R-strain
(live) → Inject into mice → Mice live\nIII. S-strain (heat-killed) → Inject into mice → Mice
live\nIV. S-strain (heat-killed) + R-strain (live) → Inject into mice → Mice die\nV. Living R-
strain bacteria were recovered from the dead mice in setup IV.\n\nChoose the correct option:
(1)All of the above
(2)I, II, III, and IV
(3)I, II, and V
(4)I, III, and IV
19.Assertion (A): Griffith concluded that the transforming principle was DNA, which enabled
the R strain to synthesize a smooth polysaccharide coat and become virulent.\nReason (R):
Prior to the biochemical characterization by Avery, MacLeod, and McCarty, genetic material
was widely thought to be a protein.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
20.Regarding the biochemical characterization of the transforming principle (1933-44),
identify the CORRECT statements:\nI. Avery, MacLeod, and McCarty purified proteins, DNA,
RNA, etc., from heat-killed S-strain cells.\nII. They found that addition of proteases completely
inhibited the transformation of live R cells.\nIII. Digestion with RNase did not affect
transformation, proving that RNA is not the transforming substance.\nIV. Digestion with DNase
did inhibit transformation, suggesting that DNA caused the transformation.\nV. All biologists
immediately accepted their conclusion that DNA is the hereditary material.\n\nChoose the
correct option:
(1)I, III, and IV
(2)All of the above
(3)I, II, and IV
(4)III, IV, and V
21.Assertion (A): A polynucleotide polymer has a free phosphate moiety at the 5′-end of the
sugar, which is referred to as the 5′-end of the chain.\nReason (R): At the opposite end, the
sugar has a free --OH group at the 3′-C, which is referred to as the 3′-end of the polynucleotide
chain.
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
22.Which of the following represents the correct set of features of the DNA Double Helix
according to Watson and Crick?\nI. It is made of two polynucleotide chains where the backbone
is constituted by bases, and sugars project inside.\nII. The two chains run in parallel directions
with both strands having 5′ → 3′ polarity.\nIII. Adenine forms two hydrogen bonds with
Thymine, and Guanine forms three hydrogen bonds with Cytosine.\nIV. The helical coiling
generates a constant uniform distance between the strands because a purine always comes
opposite to a pyrimidine.\n\nChoose the correct option:
(1)I, II, and III
(2)II and IV
(3)I and IV
(4)III and IV
23.Assertion (A): In some viruses, the flow of genetic information occurs in the reverse
direction, that is, from RNA to DNA.\nReason (R): Tobacco Mosaic Virus (TMV) and QB
bacteriophage are examples of viruses that replicate via a DNA intermediate during reverse
transcription.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
24.Regarding eukaryotic histones, which of the following statements is/are CORRECT?\nI. A
protein acquires charge depending upon the abundance of amino acids residues with charged
side chains.\nII. Histones are rich in the basic amino acid residues Lysine and Arginine.\nIII.
Both lysine and arginine carry negative charges in their side chains.\nIV. Histones are organized
to form a unit of eight molecules called a histone octamer.\n\nChoose the correct option:
(1)I and II only
(2)I, II, and IV
(3)II, III, and IV
(4)All of the above
25.Read the following statements about chromatin and nucleosomes:\nI. A typical nucleosome
contains 400 bp of DNA helix.\nII. Nucleosomes constitute the repeating unit of a structure in
the nucleus called chromatin.\nIII. Chromatin appears as thread-like stained (coloured) bodies
seen in the nucleus.\nIV. Under an electron microscope, the nucleosomes in chromatin are seen
as 'beads-on-string' structure.\n\nWhich of the statements is/are INCORRECT?
(1)I only
(2)II and III
(3)None of the above
(4)I and IV
26.Assertion (A): Euchromatin stains light and is transcriptionally active, whereas
heterochromatin stains dark and is transcriptionally inactive.\nReason (R): Euchromatin is
densely packed, whereas heterochromatin is loosely packed within the eukaryotic nucleus.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
27.Match the organisms with their correct genomic DNA sizes as per NCERT and choose the
correct combination:\nI. Bacteriophage φ × 174 - 5386 nucleotides\nII. Bacteriophage lambda -
48502 base pairs\nIII. Escherichia coli - 4.6 × 106 base pairs\nIV. Human DNA (haploid content)
- 3.3 × 109 base pairs\n\nChoose the correct option:
(1)I and IV only
(2)All of the above are correct
(3)I, II, and III only
(4)II and IV only
28.Assertion (A): Injecting heat-killed S-strain Streptococcus pneumoniae into mice does not
cause pneumonia and the mice survive.\nReason (R): Heating kills the S-strain bacteria,
rendering them unable to replicate, but it does not destroy the transforming principle present in
them.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
29.Which of the following statements about the biochemical characterization of the
transforming principle is/are CORRECT?\nI. Avery, MacLeod, and McCarty discovered that
DNA alone from S-bacteria caused R-bacteria to become transformed.\nII. They found that
proteases and RNases inhibited transformation.\nIII. They showed that digestion with DNase
did not inhibit transformation.\nIV. Their work shifted the consensus of all biologists to accept
DNA as the genetic material immediately in 1944.\n\nChoose the correct option:
(1)I, III, and IV
(2)I and II
(3)None of the above
(4)I only
30.Assertion (A): If the sequence of bases in one strand of a double-stranded DNA is known,
the sequence in the other strand can be predicted.\nReason (R): The base pairing between the
two polynucleotide strands is complementary, meaning they have identical base sequences.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
31.In the Hershey-Chase experiment (1952) to prove DNA is the genetic material, which of the
following statements is/are CORRECT?\nI. Bacteriophages grown on radioactive phosphorus
contained radioactive DNA because DNA contains phosphorus.\nII. Bacteriophages grown on
radioactive sulfur contained radioactive protein because protein contains sulfur.\nIII. DNA
contains sulfur but protein does not, so radioactive sulfur labeled DNA.\nIV. Protein contains
phosphorus but DNA does not, so radioactive phosphorus labeled protein.\n\nChoose the correct
option:
(1)I and II
(2)II and IV
(3)III and IV
(4)I, II, and III
32.Assertion (A): In the Hershey-Chase experiment, after infecting E. coli cells with
radioactive phages, blending was used to separate the heavier bacterial cells from lighter viral
capsids.\nReason (R): Centrifugation separates mixtures based on density, which forces the
lighter viral particles into the pellet and heavier bacteria into the supernatant.
(1)Both (A) and (R) are false.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
33.Regarding the final observations of the Hershey-Chase experiment, identify the
CORRECT statements:\nI. Bacteria infected with phages containing radioactive DNA (P32)
were found to be radioactive.\nII. Bacteria infected with phages containing radioactive protein
(S35) were not radioactive.\nIII. Radioactivity was detected in the supernatant in the S35 setup,
showing viral protein coats did not enter the bacteria.\nIV. Radioactivity was detected in the
pellet in the P32 setup, indicating that DNA entered the bacteria.\n\nChoose the correct option:
(1)I, II, and III only
(2)II and IV only
(3)All of the above
(4)I and IV only
34.A molecule that acts as a genetic material must fulfill all of the following criteria
EXCEPT:\nI. It should be able to generate its replica (Replication).\nII. It should be stable
chemically and structurally.\nIII. It should provide scope for rapid changes (mutations) to allow
fast adaptation.\nIV. It should be able to express itself in the form of 'Mendelian
Characters'.\n\nChoose the incorrect statement(s):
(1)III only
(2)I and III
(3)III and IV
(4)II and IV
35.Assertion (A): Because of the rules of base pairing and complementarity, both DNA and
RNA have the ability to direct their own duplications.\nReason (R): Other major biomolecules
in living systems, such as proteins, fail to fulfill this first criterion of replication.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
36.Identify the CORRECT statements explaining why DNA is chemically more stable than
RNA:\nI. The 2′-OH group present at every nucleotide in RNA is a highly reactive group
making it easily degradable.\nII. DNA lacks the reactive 2′-OH group in its deoxyribose sugar,
making it less reactive.\nIII. RNA is catalytic, which makes it chemically reactive and
unstable.\nIV. The presence of Uracil in DNA instead of Thymine confers additional
stability.\nV. Complementary strands of DNA, if separated by heating, can come together under
appropriate conditions.\n\nChoose the correct option:
(1)I, II, III, and V
(2)All of the above
(3)I, II, and IV
(4)II, III, and IV
37.Assertion (A): Viruses having an RNA genome and shorter life span mutate and evolve at a
faster rate.\nReason (R): RNA is chemically highly stable, allowing mutations to accumulate
without degrading the genome.
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
38.According to the 'RNA World' hypothesis in NCERT, which of the following statements
is/are CORRECT?\nI. RNA was the first genetic material.\nII. Essential life processes like
metabolism, translation, and splicing evolved around RNA.\nIII. RNA acted as a genetic
material as well as a protein-degrading enzyme.\nIV. DNA evolved from RNA with chemical
modifications that made it more stable.\nV. Double-stranded DNA resists changes by evolving a
process of repair.\n\nChoose the correct option:
(1)II, III, and IV
(2)All of the above
(3)I, II, and III
(4)I, II, IV, and V
39.Assertion (A): DNA is preferred for the storage of genetic information, whereas RNA is
better for the transmission of genetic information.\nReason (R): DNA is chemically less reactive
and structurally more stable, whereas RNA can directly code for proteins and express
characters easily.
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
40.In Meselson and Stahl's experiment (1958) proving semiconservative replication, which of
the following is/are CORRECT?\nI. E. coli was grown in a medium containing 15NH4Cl as the
sole nitrogen source.\nII. 15N is a heavy, stable isotope of nitrogen and is not radioactive.\nIII.
15N could be distinguished from normal 14N by density gradient centrifugation in cesium
chloride (CsCl).\nIV. 15N is a radioactive isotope of nitrogen that can be detected using
autoradiography.\n\nChoose the correct option:
(1)I, II, and III
(2)II, III, and IV
(3)I and IV
(4)All of the above
41.Assertion (A): In the Meselson-Stahl experiment, DNA extracted from E. coli cells one
generation after transfer from 15N to 14N medium had a hybrid density.\nReason (R):
Semiconservative replication dictates that the hybrid DNA molecule contains one heavy 15N
parental strand and one light 14N newly synthesized strand.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)(A) is true but (R) is false.
42.Suppose E. coli is allowed to grow in a 14N-containing medium for 80 minutes after being
transferred from a 15N medium. What will be the proportions of Light, Hybrid, and Heavy
DNA molecules respectively?\nI. The total number of generations completed is 4
generations.\nII. The ratio of Hybrid to Light DNA molecules will be 1:7.\nIII. The proportion
of hybrid DNA molecules will be 12.5 percent.\nIV. The proportion of light DNA molecules will
be 87.5 percent.\nV. The proportion of heavy DNA molecules will be 0 percent.\n\nChoose the
correct option:
(1)II, III, and V only
(2)I and IV only
(3)All of the above
(4)I, II, and IV only
43.Regarding Taylor's experiments in 1958, which of the following statements is/are
CORRECT?\nI. Taylor used radioactive thymidine to detect the distribution of newly
synthesized DNA in chromosomes.\nII. The experiments were performed on Vicia faba (faba
beans).\nIII. The experiments proved that DNA in chromosomes replicates
semiconservatively.\nIV. The experiment was conducted on Escherichia coli using radioactive
sulfur.\n\nChoose the correct option:
(1)All of the above
(2)II and IV
(3)I, II, and III
(4)I and III only
44.Assertion (A): DNA polymerase in E. coli must polymerize nucleotides at an average rate
of approximately 2000 base pairs per second.\nReason (R): E. coli has 4.6 × 106 bp and
completes its replication process within 18 minutes.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
45.Deoxyribonucleoside triphosphates (dNTPs) serve dual purposes during DNA replication.
Which of the following describes these purposes?\nI. They act as substrates for DNA polymerase
to build the new strand.\nII. They provide energy for the polymerization reaction.\nIII. The
energy is released by releasing the terminal phosphate groups, similar to ATP hydrolysis.\nIV.
They act as transcription factors to initiate replication.\n\nChoose the correct option containing
the dual purposes:
(1)I and II only
(2)I and IV
(3)II and IV
(4)I, II, and III
46.Assertion (A): Long DNA molecules replicate within a small opening of the DNA helix,
referred to as the replication fork.\nReason (R): The two strands of a long DNA molecule cannot
be separated in their entire length due to very high energy requirements.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
47.Consider the following statements regarding replication directionality at the replication
fork:\nI. DNA-dependent DNA polymerase catalyzes polymerization only in the 5′ → 3′
direction.\nII. On the template strand with 3′ → 5′ polarity, replication is continuous.\nIII. On
the template strand with 5′ → 3′ polarity, replication is discontinuous.\nIV. The discontinuously
synthesized fragments are later joined by the enzyme DNA helicase.\n\nChoose the incorrect
statement(s):
(1)IV only
(2)I and IV
(3)None of the above
(4)II and III
48.Assertion (A): In recombinant DNA technology, a foreign piece of DNA must be integrated
into a vector to propagate in host cells.\nReason (R): DNA polymerase cannot initiate replication
randomly at any site; it requires a specific sequence called the origin of replication, which the
vector provides.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
49.Which of the following statements about eukaryotic DNA replication timing and
coordination is/are CORRECT?\nI. DNA replication in eukaryotes takes place at the G1-phase
of the cell cycle.\nII. Eukaryotic DNA replication and cell division must be highly
coordinated.\nIII. A failure in cell division after DNA replication results in polyploidy.\nIV.
Polyploidy is a chromosomal anomaly.\n\nChoose the correct option:
(1)II, III, and IV
(2)II and III only
(3)All of the above
(4)I, II, and III
50.Which of the following is/are CONSISTENT with the NCERT text regarding the initiation
of DNA replication?\nI. DNA polymerase can initiate the replication process on its own given
enough ATP.\nII. Replication can initiate randomly at any place in E. coli DNA.\nIII. There is a
definite region in E. coli DNA where replication originates, termed origin of replication.\nIV.
Replication requires RNA primers to initiate synthesis.\n\nChoose the correct option:
(1)I and III
(2)II and III
(3)III and IV
(4)III only
51.In Hershey and Chase's experiment, what was the chemical rationale behind using
radioactive Phosphorus (32P) and radioactive Sulfur (35S)?\nI. Phosphorus is a component of
DNA nucleotides, specifically in the sugar-phosphate backbone.\nII. Phosphorus is not found in
amino acid side chains of viral proteins.\nIII. Sulfur is present in proteins, specifically in certain
amino acid residues like methionine and cysteine.\nIV. Sulfur is a key constituent of nitrogenous
bases of DNA, specifically cytosine.\n\nWhich of the rationale statements is/are CORRECT?
(1)I, II, and III
(2)II and IV
(3)All of the above
(4)I and III only
52.Assertion (A): RNA is highly reactive, labile, and easily degradable compared to
DNA.\nReason (R): In RNA, every nucleotide residue has an additional --OH group present at
the 2′-position of the ribose sugar.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
53.Which of the following is/are CONSISTENT with NCERT regarding the catalytic nature
of RNA?\nI. RNA is now known to be catalytic, hence reactive.\nII. Catalytic RNA molecules
are highly stable compared to non-catalytic RNA.\nIII. DNA chemically is less reactive and
structurally more stable than RNA.\nIV. Ribozyme is an example of a protein enzyme that has
RNA-like folding.\n\nChoose the correct option:
(1)I, II, and III
(2)II and IV
(3)All of the above
(4)I and III
54.Assertion (A): The presence of Thymine at the place of Uracil confers additional chemical
stability to DNA.\nReason (R): Thymine is also chemically known as 5-methyl uracil, and the
methyl group acts as a structural stabilizer.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
55.Which of the following statements about protein synthesis and genetic expression is/are
CORRECT?\nI. RNA can directly code for protein synthesis and express characters easily.\nII.
DNA is directly translated into proteins without any RNA intermediates.\nIII. DNA is dependent
on RNA for protein synthesis.\nIV. The protein-synthesizing machinery in cells has evolved
around DNA.\n\nChoose the correct option:
(1)II and IV
(2)I and III
(3)All of the above
(4)I, II, and III
56.Assertion (A): Watson and Crick proposed a scheme for semiconservative replication
immediately after proposing the double helix structure of DNA.\nReason (R): The
complementary base pairing they postulated immediately suggested a copying mechanism for
the genetic material.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
57.Regarding the centrifugation in Meselson-Stahl's experiment, which of the following is/are
CORRECT?\nI. Separation of 15N-DNA and 14N-DNA is based on density differences.\nII. The
density gradient was prepared using cesium chloride (CsCl).\nIII. The density gradient
separates molecules based on their radioactive decay rates.\nIV. Centrifugal force drives denser
molecules (15N-DNA) to sediment faster or further than lighter ones (14N-DNA).\n\nChoose the
correct option:
(1)I and II only
(2)All of the above
(3)I, II, and IV
(4)II, III, and IV
58.Assertion (A): DNA extracted from the E. coli culture 40 minutes after transfer to 14N
medium is composed of equal amounts of hybrid and light DNA.\nReason (R): After two
generations of replication in a light medium, the ratio of 15N-14N hybrid DNA molecules to
14N-14N light DNA molecules is 1:1.\n
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
59.Regarding the primary enzyme of replication, DNA-dependent DNA polymerase, which of
the following is/are CORRECT?\nI. It uses a DNA template to catalyze the polymerization of
deoxynucleotides.\nII. It is highly efficient and catalyzes reactions with a high degree of
accuracy.\nIII. Any mistake during replication would result in mutations.\nIV. It can synthesize
DNA strands in both 5′ → 3′ and 3′ → 5′ directions.\n\nChoose the correct option:
(1)II, III, and IV
(2)I and II only
(3)I, II, and III
(4)All of the above
60.Assertion (A): On the template strand with 5′ → 3′ polarity, replication is discontinuous,
producing Okazaki fragments.\nReason (R): DNA ligase binds these discontinuously synthesized
fragments together by catalyzing phosphodiester bond formation.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
61.Consider the following transcription unit sequence:\n3′-
ATGCATGCATGCATGCATGCATGC-5′ (Template Strand)\n5′-
TACGTACGTACGTACGTACGTACG-3′ (Coding Strand)\n\nWhich of the following is the
CORRECT sequence of the transcribed RNA as per NCERT rules?\nI. 5′-
UACGUACGUACGUACGUACGUACG-3′\nII. 5′-TACGTACGTACGTACGTACGTACG-3′
\nIII. 3′-UACGUACGUACGUACGUACGUACG-5′\nIV. The RNA sequence is identical to the
coding strand, with Thymine replaced by Uracil.\n\nChoose the correct option:
(1)I and IV
(2)II and IV
(3)I only
(4)III and IV
62.Assertion (A): The strand of DNA that is displaced during transcription and does not code
for any RNA molecule is paradoxically referred to as the coding strand.\nReason (R): All
reference points while defining a transcription unit's coordinates are made with respect to the
coding strand.\n
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
63.Regarding the promoter and terminator in a transcription unit, which of the following
statements is/are INCORRECT?\nI. The promoter is located towards the 5′-end (upstream) of
the structural gene, defined with respect to the template strand.\nII. The promoter provides the
binding site for RNA polymerase.\nIII. The presence of a promoter defines the template and
coding strands.\nIV. The terminator is located towards the 3′-end (downstream) of the coding
strand.\nV. Switching the positions of promoter and terminator reverses the definition of
template and coding strands.\n\nChoose the incorrect statements:
(1)None of the above
(2)I and IV
(3)II and V
(4)I only
64.Why are both strands of DNA not transcribed simultaneously? Identify the CORRECT
explanations according to NCERT:\nI. If both strands acted as templates, they would code for
RNA molecules with identical sequences.\nII. Coding for RNA with different sequences would
result in proteins with different amino acid sequences, complicating translation.\nIII. The two
complementary RNA molecules produced simultaneously would base-pair to form double-
stranded RNA.\nIV. Double-stranded RNA would prevent translation, making the transcription
process futile.\n\nChoose the correct option:
(1)All of the above
(2)I and III only
(3)II, III, and IV
(4)I, II, and III
65.Assertion (A): Eukaryotic structural genes are described as split genes because their
coding sequences are interrupted by non-coding intervening sequences.\nReason (R): In mature
mRNA, introns are expressed sequences that code for polypeptides, while exons are removed
during splicing.\n
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
66.Match eukaryotic RNA polymerases with their specific transcript targets. Identify the
CORRECT matches:\nI. RNA Polymerase I transcribes 28S, 18S, and 5.8S rRNAs.\nII. RNA
Polymerase II transcribes tRNA and snRNAs.\nIII. RNA Polymerase III transcribes 5S rRNA,
tRNA, and snRNAs.\nIV. RNA Polymerase II transcribes heterogeneous nuclear RNA (hnRNA),
the precursor of mRNA.\n\nChoose the correct option:
(1)II and IV only
(2)I, III, and IV
(3)All of the above
(4)I and III only
67.Regarding eukaryotic post-transcriptional modifications, which of the following
statements is/are CORRECT?\nI. Splicing removes introns and joins exons in a defined
order.\nII. Capping involves adding methyl guanosine triphosphate to the 5′-end of
hnRNA.\nIII. Tailing involves adding 200--300 adenylate residues to the 3′-end of hnRNA.\nIV.
Tailing occurs in a template-dependent manner using DNA template instructions.\nV. Fully
processed hnRNA is called mature tRNA.\n\nChoose the correct option:
(1)I, II, and IV
(2)I, II, and III
(3)All of the above
(4)II, III, and V
68.Assertion (A): The split-gene arrangement represents an ancient feature of the eukaryotic
genome, and splicing represents the dominance of the RNA world.\nReason (R): The presence of
introns is reminiscent of antiquity, indicating that genomes were originally composed of non-
coding regions.\n
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
69.Regarding transcription in bacteria, which of the following statements is/are CORRECT?
\nI. There is a single DNA-dependent RNA polymerase that catalyzes transcription of all types
of RNA.\nII. RNA polymerase requires separate enzymes for mRNA, tRNA, and rRNA
transcription.\nIII. The single RNA polymerase initiates, elongates, and terminates transcription
on its own without other factors.\nIV. The structural genes in bacteria are mostly
polycistronic.\n\nChoose the correct option:
(1)II and III
(2)I, III, and IV
(3)I and IV
(4)All of the above
70.Assertion (A): In bacteria, transcription and translation can be coupled, meaning
translation can begin before mRNA is fully transcribed.\nReason (R): Bacterial mRNA does not
require any processing to become active, and transcription and translation take place in the
same compartment.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
71.Regarding the steps of bacterial transcription, which of the following is/are CORRECT?
\nI. RNA polymerase is capable of catalyzing all three steps: initiation, elongation, and
termination.\nII. RNA polymerase is only capable of catalyzing the process of elongation.\nIII.
RNA polymerase associates transiently with initiation-factor (σ) and termination-factor (ρ) to
initiate and terminate transcription.\nIV. The association with factors alters the specificity of the
RNA polymerase.\n\nChoose the correct option:
(1)I and III only
(2)I, III, and IV
(3)II and III only
(4)II, III, and IV
72.Assertion (A): George Gamow argued that the genetic codon must constitute a
combination of three nucleotides to code for all 20 amino acids.\nReason (R): A doublet code
(combinations of two bases) would only generate 16 codons, which is insufficient to code for 20
amino acids.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
73.Match the scientists with their correct methodology/contribution in deciphering the genetic
code:\nI. Har Gobind Khorana - Developed chemical method for synthesizing homopolymers
and copolymers of RNA.\nII. Marshall Nirenberg - Developed a cell-free system for protein
synthesis, which helped decipher the code.\nIII. Severo Ochoa - Discovered an enzyme that
polymerizes RNA in a template-dependent manner.\nIV. George Gamow - Suggested the triplet
nature of codons based on mathematical permutations.\n\nChoose the correct option:
(1)I, II, and IV
(2)All of the above
(3)II and IV only
(4)I, II, and III
74.Regarding the salient features of the genetic code, which of the following is/are
CORRECT?\nI. Out of 64 codons, 61 codons code for amino acids, and 3 do not code for any
amino acids.\nII. Some amino acids are coded by more than one codon, hence the code is
unambiguous.\nIII. One codon codes for only one amino acid, hence the code is degenerate.\nIV.
The codon is read in mRNA in a contiguous fashion with periodic punctuations.\nV. The genetic
code is nearly universal, with UUU coding for Phenylalanine from bacteria to
humans.\n\nChoose the correct option:
(1)II, IV, and V
(2)I, II, and IV
(3)I and V
(4)I, III, and V
75.Assertion (A): The genetic code is universal, meaning a specific codon codes for the exact
same amino acid in all living organisms.\nReason (R): Some exceptions to the universality of the
genetic code have been identified in mitochondrial codons and in some protozoans.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
76.Which of the following statements regarding the codon AUG is/are CORRECT?\nI. It has
dual functions.\nII. It codes for Methionine (met).\nIII. It acts as an initiator codon.\nIV. It is
also one of the three stop codons.\n\nChoose the correct option:
(1)I and II only
(2)All of the above
(3)II and IV
(4)I, II, and III
77.Assertion (A): Codons UAA, UAG, and UGA do not code for any amino acids and are
referred to as stop terminator codons.\nReason (R): These codons bind to specific tRNAs that
lack amino acid attachments, stopping translation.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
78.Regarding mutations and the genetic code, which of the following is/are CORRECT?\nI.
Large deletions and rearrangements in DNA can result in the loss or gain of a gene and
function.\nII. Sickle cell anemia is a classical example of a point mutation.\nIII. In sickle cell
anemia, a single base-pair change in the beta-globin gene results in the substitution of
Glutamate by Valine.\nIV. In sickle cell anemia, Valine is replaced by Glutamate in the beta-
globin chain.\n\nChoose the correct option:
(1)I and II only
(2)II and IV
(3)All of the above
(4)I, II, and III
79.Consider the following statements regarding frameshift mutations:\nI. Insertion or
deletion of one or two bases changes the reading frame from the point of insertion or
deletion.\nII. Such mutations are referred to as frameshift insertion or deletion mutations.\nIII.
Insertion or deletion of three or multiple bases inserts or deletes one or multiple codons, and the
reading frame remains unaltered from that point onwards.\nIV. Frameshift mutations prove
that the genetic code is read in a contiguous, non-overlapping triplet fashion.\n\nChoose the
correct option containing all CORRECT statements:
(1)I, II, and III only
(2)II, III, and IV only
(3)I and III only
(4)All of the above
80.Assertion (A): Insertion of a single base 'B' in the codon sequence shifts the reading frame
of all subsequent codons downstream of the mutation.\nReason (R): The genetic code is
degenerate, meaning multiple codons code for the same amino acid.\n
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
81.Read the following statements about cistrons and structural genes:\nI. A cistron is defined
as a segment of DNA coding for a polypeptide.\nII. Monocistronic structural genes are found
mostly in eukaryotes.\nIII. Polycistronic structural genes are found mostly in bacteria or
prokaryotes.\nIV. Polycistronic structural genes are found mostly in eukaryotes, while
monocistronic genes are prokaryotic.\n\nChoose the correct option:
(1)I, II, and III
(2)III and IV
(3)I and IV
(4)I and II only
82.Assertion (A): Regulatory DNA sequences are sometimes loosely defined as regulatory
genes even though they do not code for any RNA or protein.\nReason (R): These regulatory
sequences affect the inheritance of a character by controlling the expression of structural
genes.\n
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
83.In bacteria, three major types of RNAs are needed to synthesize a protein. Match their
functions as per NCERT:\nI. mRNA - provides the template.\nII. tRNA - brings amino acids and
reads the genetic code.\nIII. rRNA - plays structural and catalytic roles during translation.\nIV.
mRNA - plays structural and catalytic roles, while rRNA provides the template.\n\nWhich of the
combinations is/are CORRECT?
(1)I and II only
(2)IV only
(3)I, II, and III
(4)II, III, and IV
84.Assertion (A): Eukaryotes exhibit higher complexity in transcription compared to bacteria,
including the division of labor among multiple nuclear RNA polymerases.\nReason (R):
Eukaryotes contain at least three distinct RNA polymerases in the nucleus, in addition to
organelle-specific RNA polymerases.\n
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
85.Regarding the universality of the genetic code, which of the following is/are CORRECT?
\nI. UUU codes for Phenylalanine (phe) in both bacteria and humans.\nII. The genetic code is
absolutely universal without any exceptions.\nIII. Exceptions to the universality have been
found in mitochondrial codons.\nIV. Exceptions to the universality have been found in some
protozoans.\n\nChoose the correct option:
(1)II and III
(2)All of the above
(3)I and II
(4)I, III, and IV
86.Assertion (A): The codon UUU codes exclusively for Phenylalanine, representing the
unambiguous nature of the genetic code.\nReason (R): Phenylalanine is coded by UUU and
UUC, representing the degenerate nature of the genetic code.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
87.Deciphering the genetic code required interdisciplinary collaboration. Identify the correct
fields of the contributing scientists as per NCERT:\nI. George Gamow - Physicist\nII. Har
Gobind Khorana - Organic Chemist\nIII. Marshall Nirenberg - Biochemist\nIV. Severo Ochoa -
Geneticist\n\nChoose the correct option containing all CORRECT matches:
(1)I, II, and III
(2)II and IV
(3)I and II
(4)All of the above
88.Assertion (A): Severo Ochoa's enzyme was helpful in polymerizing RNA with defined
sequences in a template-independent manner.\nReason (R): Polynucleotide phosphorylase
catalyzes the template-free enzymatic synthesis of RNA polymers.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
89.Why is the primary transcript in eukaryotes (hnRNA) non-functional? Identify the
CORRECT reasons according to NCERT:\nI. It contains both exons and introns.\nII. Exons are
non-coding sequences that disrupt the gene.\nIII. Splicing is required to remove introns and join
exons in a defined order.\nIV. It lacks capping and tailing modifications, making it
unstable.\n\nChoose the correct option:
(1)I, III, and IV
(2)I and III only
(3)All of the above
(4)II and IV
90.Assertion (A): Capping adds methyl guanosine triphosphate to the 5′-end of hnRNA, while
tailing adds 200--300 adenylate residues to the 3′-end.\nReason (R): These additions are
catalyzed by RNA polymerase II in a template-dependent manner reading DNA instructions.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
91.Regarding the structure of tRNA, which of the following statements is/are CORRECT?\nI.
The secondary structure of tRNA looks like a clover-leaf.\nII. The actual three-dimensional
structure of tRNA looks like an inverted L.\nIII. tRNA has an anticodon loop that has bases
complementary to the code.\nIV. tRNA has an amino acid acceptor end to which it binds to
amino acids.\nV. There is a specific tRNA for initiation, referred to as initiator tRNA.\n\nChoose
the correct option:
(1)I, II, and III only
(2)All of the above
(3)II, III, and V only
(4)I, III, and IV only
92.Assertion (A): During translation, there is a specific initiator tRNA for the start codon, but
there are no tRNAs for the stop codons.\nReason (R): Stop codons do not code for any amino
acid and are recognized by protein release factors to terminate translation.\n
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
93.Regarding the first phase of translation, charging of tRNA, which of the following is/are
CORRECT?\nI. Amino acids are activated in the presence of ATP.\nII. Activated amino acids
are linked to their cognate tRNA, a process called aminoacylation of tRNA.\nIII. If two such
charged tRNAs are brought close together, peptide bond formation between them is
energetically favored.\nIV. The presence of a catalyst increases the rate of peptide bond
formation.\n\nChoose the correct option:
(1)II and IV only
(2)I and II only
(3)I, II, and III only
(4)All of the above
94.The ribosome is the cellular factory responsible for synthesizing proteins. Identify the
CORRECT statements regarding the ribosome as per NCERT:\nI. The ribosome consists of
structural RNAs and about 80 different proteins.\nII. In its inactive state, it exists as two
subunits: a large subunit and a small subunit.\nIII. When the small subunit encounters an
mRNA, the process of translation of the mRNA to protein begins.\nIV. There are two sites in the
large subunit for subsequent amino acids to bind to and be close to each other.\nV. In bacteria,
23S rRNA acts as a ribozyme catalyst for peptide bond formation.\n\nChoose the correct option:
(1)I, II, and III only
(2)All of the above
(3)I, III, and IV only
(4)II, III, and V only
95.Assertion (A): The ribosome is a key example of an organelle where RNA acts as a catalyst
rather than just a structural scaffold.\nReason (R): In bacteria, the 23S rRNA acts as a
ribozyme catalyst for peptide bond formation during translation.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
96.A translation unit in mRNA consists of several regions. Identify the CORRECT statements
regarding these regions as per NCERT:\nI. A translation unit is flanked by the start codon
(AUG) and the stop codon.\nII. It codes for a polypeptide sequence.\nIII. Untranslated Regions
(UTRs) are sequences in mRNA that are not translated into protein.\nIV. UTRs are present only
at the 5′-end (before the start codon).\nV. UTRs are present only at the 3′-end (after the stop
codon).\nVI. UTRs are present at both the 5′-end (before the start codon) and the 3′-end (after
the stop codon), and are required for efficient translation.\n\nChoose the correct option:
(1)I, II, and IV
(2)I, II, and III
(3)I, II, and VI
(4)II, III, and V
97.Assertion (A): Untranslated Regions (UTRs) of mRNA do not code for any amino acid but
are still essential for the survival and growth of cells.\nReason (R): UTRs are required for the
efficient translation process, ensuring the ribosome correctly binds and processes the mRNA.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
98.Read the following statements about translation steps and identify the CORRECT
ones:\nI. For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is
recognized only by the initiator tRNA.\nII. During elongation, complexes composed of an amino
acid linked to tRNA sequentially bind to the appropriate codon in mRNA by complementary
base pairing.\nIII. The ribosome moves from codon to codon along the mRNA in the 3′ → 5′
direction.\nIV. As the ribosome moves, amino acids are added one by one, translated into
polypeptide sequences.\nV. At the end, a release factor binds to the stop codon, terminating
translation and releasing the complete polypeptide.\n\nChoose the correct option:
(1)All of the above
(2)II, III, and V
(3)I, II, IV, and V
(4)I, II, and III
99.Assertion (A): In eukaryotes, regulation of gene expression can be exerted at four distinct
levels.\nReason (R): The four levels include: (i) transcriptional level (ii) processing level (iii)
transport of mRNA from nucleus to cytoplasm, and (iv) translational level.\n
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
100.The Lac Operon in E. coli contains three structural genes. Identify the CORRECT
matches between genes and their protein products:\nI. z gene - codes for beta-galactosidase (β-
gal).\nII. y gene - codes for permease.\nIII. a gene - codes for transacetylase.\nIV. i gene - codes
for an inducer protein that activates transcription.\n\nChoose the correct option:
(1)I, II, and III
(2)All of the above
(3)I and II only
(4)II, III, and IV
101.Assertion (A): In the absence of lactose, the repressor protein binds to the promoter
region of the operon, preventing RNA polymerase from transcribing the structural
genes.\nReason (R): The repressor protein is synthesized constitutively from the i gene of the lac
operon.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is true but (R) is false.
102.Regarding the inducer of the lac operon, which of the following statements is/are
CORRECT?\nI. Lactose or allolactose acts as the inducer for the operon.\nII. In the presence of
the inducer, the repressor is inactivated by binding to the inducer.\nIII. This inactivation allows
RNA polymerase access to the promoter, and transcription proceeds.\nIV. Glucose or galactose
can also act as inducers for the lac operon if lactose is absent.\n\nChoose the correct option:
(1)I, II, and III
(2)II, III, and IV
(3)All of the above
(4)I and II only
103.Assertion (A): A very low level of expression of the lac operon must be present in the E.
coli cell all the time.\nReason (R): Lactose cannot enter the cell unless permease, a product of
the operon, is present to transport it.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
104.The Human Genome Project utilized two major methodologies. Which of the following
is/are CORRECT matches regarding these methodologies?\nI. Expressed Sequence Tags (ESTs)
- focused on identifying all the genes that are expressed as RNA.\nII. Sequence Annotation -
involved sequencing the entire genome (coding and non-coding) and later assigning functional
regions.\nIII. Sequence Annotation - focused on identifying only the intron sequences of the
genome.\nIV. Expressed Sequence Tags (ESTs) - involved sequencing the entire genomic DNA
first, then selecting the exons.\n\nChoose the correct option:
(1)I, II, and IV
(2)I and II
(3)II and III
(4)All of the above
105.Regarding the cloning and sequencing workflow of HGP, identify the CORRECT
statements:\nI. Total DNA from a cell is isolated and converted into random fragments of
relatively smaller sizes.\nII. These fragments are cloned in host cells like bacteria and yeast
using vectors called BAC and YAC.\nIII. BAC stands for Bacterial Artificial Chromosome; YAC
stands for Yeast Artificial Chromosome.\nIV. Sequencing was done using automated DNA
sequencers based on the method developed by Frederick Sanger.\nV. Sanger is also credited with
developing methods for the sequencing of amino acids in proteins.\n\nChoose the correct option:
(1)I, II, and III only
(2)All of the above
(3)I, II, and IV only
(4)II, III, and IV only
106.Assertion (A): In HGP, the sequenced genomic fragments were aligned using specialized
computer-based programs rather than manual alignment.\nReason (R): These sequences were
generated as random fragments, and alignment was only possible based on overlapping regions
present in them.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
107.The salient features of the human genome are highly quantitative. Identify the
INCORRECT statements as per NCERT:\nI. The human genome contains 3164.7 million base
pairs.\nII. The average gene consists of 30,000 bases, with the largest known gene being
dystrophin with 2.4 million bases.\nIII. The total number of genes is estimated at 30,000.\nIV.
Almost all (99.9 percent) nucleotide bases are exactly the same in all people.\nV. The functions
are unknown for over 50 percent of the discovered genes.\nVI. Less than 2 percent of the
genome codes for proteins.\n\nChoose the incorrect statements:
(1)I, III, and V
(2)II and VI
(3)II only
(4)None of the above
108.Regarding the chromosomal statistics of HGP, identify the CORRECT statements:\nI.
Chromosome 1 was the last of the 24 human chromosomes to be completely sequenced.\nII.
Chromosome 1 sequencing was completed in May 2006.\nIII. Chromosome 1 has the most genes
(2968), and the Y chromosome has the fewest genes (231).\nIV. Chromosome 1 has the fewest
genes (231), and the Y chromosome has the most genes (2968).\n\nChoose the correct option:
(1)II and IV
(2)I, II, and III
(3)I and II only
(4)All of the above
109.DNA fingerprinting involves separating repetitive DNA. Which of the following
statements is/are CORRECT?\nI. Repetitive DNA is separated from bulk genomic DNA as
different peaks during density gradient centrifugation.\nII. The bulk DNA forms a major peak,
and other small peaks are referred to as satellite DNA.\nIII. Satellite DNA is classified into
micro-satellites, mini-satellites, etc., based on base composition, length of segment, and number
of repetitive units.\nIV. Satellite DNA codes for essential metabolic enzymes in the
cytoplasm.\n\nChoose the correct option:
(1)I, II, and III
(2)All of the above
(3)II, III, and IV
(4)I and II only
110.Assertion (A): Satellite DNA is used as a powerful tool in forensic science and DNA
fingerprinting.\nReason (R): These sequences normally do not code for any proteins, but they
show a high degree of polymorphism that is inherited.\n
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
111.The VNTR (Variable Number of Tandem Repeats) is a key marker in DNA fingerprinting.
Identify the CORRECT statements regarding VNTR as per NCERT:\nI. VNTR belongs to a
class of satellite DNA referred to as mini-satellite.\nII. A small DNA sequence is arranged
tandemly in many copy numbers.\nIII. The copy number varies from chromosome to
chromosome in an individual.\nIV. The size of VNTR varies in size from 0.1 to 20 kb.\nV. VNTR
belongs to a class of satellite DNA referred to as micro-satellite.\n\nChoose the correct option:
(1)All of the above
(2)II, III, and IV only
(3)II, IV, and V
(4)I, II, III, and IV
112.Identify the CORRECT sequence of steps involved in DNA Fingerprinting according to
NCERT:\nI. Isolation of DNA\nII. Hybridisation using labelled VNTR probe\nIII. Digestion of
DNA by restriction endonucleases\nIV. Detection of hybridised DNA fragments by
autoradiography\nV. Transferring (blotting) of separated DNA fragments to synthetic
membranes, such as nitrocellulose or nylon\nVI. Separation of DNA fragments by
electrophoresis\n\nChoose the correct sequence:
(1)I → III → V → VI → II → IV
(2)I → III → VI → II → V → IV
(3)I → III → VI → V → II → IV
(4)I → VI → III → V → II → IV
113.Assertion (A): DNA recovered from a single hair follicle or a tiny drop of blood at a crime
scene is sufficient to perform DNA fingerprinting analysis.\nReason (R): The sensitivity of the
DNA fingerprinting technique has been vastly increased by the use of polymerase chain reaction
(PCR).\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)(A) is true but (R) is false.
114.Regarding the autoradiogram bands generated in DNA fingerprinting, which of the
following is/are CORRECT?\nI. Hybridization with a VNTR probe yields many bands of
differing sizes.\nII. The banding pattern is unique for an individual DNA, differing from person
to person in a population.\nIII. Monozygotic (identical) twins share the exact same banding
pattern.\nIV. Dizygotic (fraternal) twins share the exact same banding pattern.\n\nChoose the
correct option:
(1)II, III, and IV
(2)I and II only
(3)All of the above
(4)I, II, and III
115.Assertion (A): Scientists have identified about 1.4 million locations where single-base
DNA differences (SNPs) occur in humans.\nReason (R): SNPs stands for Single Nucleotide
Polymorphisms, and mapping them is useful in tracing human history and disease-associated
sequences.\n
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
116.Regarding the regulation of transcription in bacteria, which of the following is/are
CORRECT?\nI. A polycistronic structural gene is regulated by a common promoter and
regulatory genes.\nII. Such arrangement is very common in bacteria and is referred to as an
operon.\nIII. Examples of operons include: lac operon, trp operon, ara operon, his operon, val
operon.\nIV. The operator region of an operon is adjacent to the promoter and is responsible for
binding regulatory proteins.\n\nChoose the correct option:
(1)I and II only
(2)II and IV only
(3)I, II, and III only
(4)All of the above
117.Assertion (A): E. coli cells grown in a medium containing both glucose and lactose will
prioritize metabolizing glucose first, and the lac operon will remain shut down.\nReason (R):
Glucose or galactose cannot act as inducers for the lac operon, and lactose is the substrate for
beta-galactosidase.\n
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
118.Regarding the elucidation of the lac operon, which of the following is/are CORRECT?\nI.
It was elucidated by Francois Jacob (a geneticist) and Jacques Monod (a biochemist)
jointly.\nII. They were the first to elucidate a transcriptionally regulated system.\nIII. The lac
operon is a monocistronic system regulated by a common promoter.\nIV. Francois Jacob was a
biochemist and Jacques Monod was a geneticist.\n\nChoose the correct option:
(1)II and IV
(2)I and II
(3)All of the above
(4)I, II, and III
119.Assertion (A): Regulation of the lac operon by the repressor protein is referred to as
negative regulation.\nReason (R): The lac operon is under negative control because repressor
binding to the operator shuts down transcription.\n
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
120.Regarding DNA polymorphism, which of the following statements is/are CORRECT?\nI.
Polymorphism (variation at genetic level) arises due to mutations.\nII. An inheritable mutation
observed in a population at high frequency (greater than 0.01) is referred to as DNA
polymorphism.\nIII. If an inheritable mutation is observed at a frequency less than 0.01, it is
still classified as DNA polymorphism.\nIV. DNA polymorphism is the basis of genetic mapping
of the human genome as well as DNA fingerprinting.\n\nChoose the correct option:
(1)II, III, and IV
(2)I, II, and IV
(3)All of the above
(4)I and II only
121.Which of the following statements correctly describe the chemical linkages in a
polynucleotide chain?\nI. The nitrogenous base is linked to the 1′-C of the pentose sugar via an
N-glycosidic linkage to form a nucleoside.\nII. A phosphate group is linked to the 5′-OH of a
nucleoside through a phosphoester linkage to form a nucleotide.\nIII. Two nucleotides are
linked through a 3′-5′ phosphodiester linkage to form a dinucleotide.\nIV. The phosphodiester
bond is formed between the phosphate group at the 5′-C of one nucleotide and the 3′-OH of the
next nucleotide pentose sugar.\n\nChoose the correct option:
(1)All of the above
(2)II and IV only
(3)I, II, and III only
(4)I and III only
122.Assertion (A): A double-stranded DNA segment containing 30\% Adenine will contain
20\% Cytosine.\nReason (R): In any double-stranded DNA, the ratio of Adenine to Thymine is
1:1, and the ratio of Guanine to Cytosine is 1:1, meaning A + G = T + C.
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
123.Regarding the pyrimidine bases present in nucleic acids, which of the following
statements is/are CORRECT?\nI. Cytosine is common to both DNA and RNA.\nII. Thymine is
present in DNA, while Uracil is present in RNA at the place of Thymine.\nIII. Thymine is also
chemically known as 5-methyl uracil.\nIV. Uracil has a methyl group at the 5th carbon position,
making it identical to Thymine.\n\nChoose the correct option:
(1)I, II, and III
(2)I and II only
(3)All of the above
(4)II, III, and IV
124.Assertion (A): The nitrogenous bases in a polynucleotide chain constitute the outer
structural backbone of the helix.\nReason (R): The backbone of a polynucleotide chain is
formed due to alternating pentose sugar and phosphate groups.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
125.Consider the genomic DNA sizes of different organisms mentioned in NCERT. Which of
the following relationships is/are CORRECT?\nI. The genome size of bacteriophage \lambda
(48502 bp) is approximately 9 times that of bacteriophage φ × 174 (5386 nucleotides).\nII. The
haploid genome of Escherichia coli (4.6 × 10^6 bp) is roughly 100 times smaller than the haploid
genome of humans (3.3 × 10^9 bp).\nIII. The total number of base pairs in a diploid human cell
is 6.6 × 10^9 bp.\nIV. The single-stranded DNA genome of bacteriophage φ × 174 contains base
pairs instead of single nucleotides.\n\nChoose the correct option:
(1)All of the above
(2)II and IV only
(3)I, II, and III
(4)I and III only
126.Assertion (A): The pitch of the DNA double helix is 3.4 nm, and the distance between
consecutive base pairs is 0.34 nm.\nReason (R): There are roughly 10 base pairs in each turn of
the DNA double helix, which means the helical turn angle is approximately 36^\circ per base
pair.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
127.Regarding the basic nature of eukaryotic histones, which of the following is/are
CORRECT?\nI. Histone proteins carry a positive charge because they are rich in basic amino
acid residues.\nII. The side chains of basic amino acids like Lysine and Arginine carry positive
charges at physiological pH.\nIII. The negatively charged phosphate backbone of DNA interacts
electrostatically with the basic residues on histones.\nIV. Histones are rich in acidic residues like
Aspartate and Glutamate, which stabilize the nucleosome core.\n\nChoose the correct option:
(1)I and II only
(2)II and IV only
(3)I, II, and III
(4)All of the above
128.Assertion (A): Chromatin is seen as a repeating 'beads-on-string' structure under an
electron microscope, where the 'beads' represent nucleosomes.\nReason (R): In a typical
mammalian cell, chromatin is organized into nucleosome repeating units containing
approximately 200 base pairs of DNA wrapped around a histone octamer.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
129.In Frederick Griffith's transformation experiments, which of the following statements are
CORRECT?\nI. Live S-strain bacteria injected into mice caused death.\nII. Live R-strain
bacteria injected into mice did not cause death.\nIII. Heat-killed S-strain bacteria injected into
mice caused death because the toxic polysaccharide coat was still intact.\nIV. Injecting a mixture
of heat-killed S-strain and live R-strain caused death, and Griffith recovered living S-strain
from the dead mice.\n\nChoose the correct option:
(1)All of the above
(2)I and II only
(3)II, III, and IV
(4)I, II, and IV
130.Assertion (A): Digestion of the purified transforming principle with DNase inhibits
transformation, whereas digestion with RNase or Protease has no effect.\nReason (R): DNase
digests DNA, which is the transforming principle, while RNase and Proteases degrade RNA and
proteins respectively, which do not cause transformation.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is true but (R) is false.
131.Regarding the radioisotopes used in the Hershey-Chase experiment, which of the
following statements is/are CORRECT?\nI. Bacteriophages grown in the presence of radioactive
phosphorus (32P) contain radioactive DNA because DNA has phosphorus.\nII. Bacteriophages
grown in the presence of radioactive sulfur (35S) contain radioactive protein because protein
contains sulfur.\nIII. DNA contains sulfur but protein does not, so radioactive sulfur labeled
DNA.\nIV. Protein contains phosphorus but DNA does not, so radioactive phosphorus labeled
protein.\n\nChoose the correct option:
(1)All of the above
(2)II and IV
(3)I, II, and III
(4)I and II
132.Assertion (A): In the Hershey-Chase experiment, centrifugation was used to separate the
radioactive viral capsids from the infected bacterial cells.\nReason (R): Centrifugation forces
the heavier bacterial cells to sediment at the bottom of the tube as a pellet, while the lighter viral
particles remain in the supernatant.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
133.Regarding the sugar carbons in DNA and RNA polynucleotides, which of the following
statements are CORRECT?\nI. The nitrogenous base is attached to the 1′-C position of the
pentose sugar.\nII. In DNA, the 2′-position has a hydrogen atom instead of an -OH group.\nIII.
In RNA, the 2′-position contains a free hydroxyl (-OH) group.\nIV. The phosphate group links
to the 3′-C of the nucleoside to form a nucleotide.\n\nChoose the correct option:
(1)I and III only
(2)II and IV only
(3)I, II, and III
(4)All of the above
134.Assertion (A): The distance between the two strands of the DNA double helix is
approximately uniform (2 nm) throughout its length.\nReason (R): In the double helix, a large
purine (double-ring structure) always pairs with a small pyrimidine (single-ring structure) on
the complementary strands.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
135.Regarding the discovery and structural elucidation of DNA, which of the following is/are
CORRECT?\nI. Friedrich Meischer identified DNA as an acidic substance in the nucleus in
1869 and named it 'Nuclein'.\nII. The elucidation of the double helix structure was delayed for
84 years due to technical limitations in isolating such a long polymer intact.\nIII. Watson and
Crick's proposal of the double helix in 1953 was partly based on X-ray diffraction data
produced by Maurice Wilkins and Rosalind Franklin.\nIV. Erwin Chargaff's rules of base
pairing were formulated before Watson and Crick proposed the double helix model.\n\nChoose
the correct option:
(1)II and IV only
(2)I and III only
(3)All of the above
(4)I, II, and IV
136.Assertion (A): Avery, MacLeod, and McCarty biochemically characterized the
transforming principle in Griffith's experiment, concluding that DNA was the genetic
material.\nReason (R): Prior to their work, the genetic material was widely thought to be a
protein, and their conclusion was instantly and universally accepted by all biologists in 1944.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
137.Read the following statements regarding the Central Dogma and identify the CORRECT
ones:\nI. It represents a unidirectional flow of genetic information from DNA → RNA →
Protein.\nII. In some viruses, the flow of information is bidirectional, meaning it can flow from
Protein back to RNA.\nIII. The process of reverse transcription copies viral RNA into DNA
inside host cells.\nIV. The Central Dogma was proposed by Francis Crick alone.\n\nChoose the
correct option:
(1)II and IV only
(2)I and III only
(3)I, III, and IV
(4)All of the above
138.Assertion (A): Chromatin fibers coil and condense to form chromosomes during the
Metaphase stage of cell division.\nReason (R): Non-histone Chromosomal (NHC) proteins are
required for the higher-level packaging of chromatin into chromosomes.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
139.Watson and Crick's DNA double helix model is characterized by specific base-pairing
rules. Which of the following statements is/are CORRECT?\nI. Adenine pairs with Thymine
through two hydrogen bonds.\nII. Guanine pairs with Cytosine through three hydrogen
bonds.\nIII. The hydrogen bonds are formed between the nitrogenous bases projecting inwards
from the sugar-phosphate backbone.\nIV. The hydrogen bonds are covalent interactions that
hold the two strands together tightly.\n\nChoose the correct option:
(1)I, II, and III
(2)I and II only
(3)II, III, and IV
(4)All of the above
140.Assertion (A): If one strand of a DNA double helix has 5′ → 3′ polarity, the
complementary strand must align in the same 5′ → 3′ direction.\nReason (R): The two strands
of DNA run parallel to each other to ensure that complementary base pairing can occur without
twisting the sugar-phosphate backbone.
(1)Both (A) and (R) are false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is true but (R) is false.
141.Regarding the structural stability of the DNA double helix, which of the following
features is/are CORRECT according to NCERT?\nI. The hydrogen bonds between the
complementary nitrogenous bases hold the two strands together.\nII. The plane of one base pair
stacks over the other in the double helix.\nIII. Stacking of base pairs confers stability to the
helical structure in addition to hydrogen bonds.\nIV. The covalent phosphodiester bonds
between the base pairs on opposite strands stabilize the helix horizontally.\n\nChoose the correct
option:
(1)All of the above
(2)I and II only
(3)I, II, and III
(4)II and IV only
142.Assertion (A): Friedrich Meischer named the substance he isolated from the cell nucleus
'Nuclein' because it was basic in nature.\nReason (R): DNA contains an abundance of basic
amino acid residues carrying positive charges in their side chains.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are false.
143.Regarding the polarity and chemical groups at the ends of a polynucleotide chain, which
of the following is/are CORRECT?\nI. The 5′-end of the chain contains a free phosphate group
attached to the 5′-carbon of the pentose sugar.\nII. The 3′-end of the chain contains a free
hydroxyl (-OH) group attached to the 3′-carbon of the pentose sugar.\nIII. The backbone of the
chain is composed of alternating sugars and bases.\nIV. The nitrogenous bases project outwards
from the sugar-phosphate backbone.\n\nChoose the correct option:
(1)I and II
(2)I, II, and III
(3)II and IV
(4)All of the above
144.Assertion (A): Digestion of the transforming principle with Protease or RNase does not
inhibit the transformation of rough Streptococcus pneumoniae cells into smooth cells.\nReason
(R): The transforming principle is DNA, which is selectively degraded by Protease and RNase,
proving that protein and RNA are genetic materials.
(1)Both (A) and (R) are false.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
145.Given that the distance between consecutive base pairs is 0.34 × 10-9 m and a haploid
human genome has 3.3 × 10^9 bp. Which of the following is/are CORRECT?\nI. The total
length of DNA in a haploid human cell is approximately 1.1 meters.\nII. The total length of DNA
in a diploid human cell is approximately 2.2 meters.\nIII. The number of nucleosomes in a
diploid human cell is approximately 3.3 × 10^7.\nIV. The total length of DNA in an E. coli cell
with 4.6 × 10^6 bp is approximately 1.36 mm.\n\nChoose the correct option:
(1)I, II, and IV only
(2)II and III only
(3)All of the above
(4)I and III only
146.Assertion (A): In prokaryotes like Escherichia coli, the DNA is scattered throughout the
cell because they do not have a defined nucleus.\nReason (R): Prokaryotic DNA is negatively
charged and is held with some basic, positively charged proteins in a region termed 'nucleoid'.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
147.Identify the CORRECT statements regarding the recovery of bacteria in Frederick
Griffith's experiments:\nI. Griffith injected a mixture of heat-killed S-strain and live R-strain
into mice.\nII. The injected mice died of pneumonia.\nIII. From the dead mice, Griffith
recovered living S-strain bacteria.\nIV. From the dead mice, Griffith recovered living R-strain
bacteria that had acquired a shiny coat.\n\nChoose the correct option:
(1)All of the above
(2)I and II only
(3)I, II, and III
(4)II and IV only
148.Assertion (A): In RNA, every nucleotide residue has a free hydroxyl (-OH) group present
at the 2′-position of the ribose sugar.\nReason (R): The presence of this additional reactive
group makes RNA chemically more stable and less reactive than DNA.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are false.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
149.Regarding the application of Erwin Chargaff's rules, which of the following is/are
CORRECT?\nI. The rule states that the ratio of Adenine to Thymine, and Guanine to Cytosine,
is constant and equals one.\nII. Chargaff's rules apply strictly to double-stranded DNA
molecules.\nIII. In a single-stranded RNA virus, the ratio of A/U is always equal to one.\nIV. In
a single-stranded DNA bacteriophage like φ × 174, the ratio of A/T does not necessarily equal
one.\n\nChoose the correct option:
(1)I and II only
(2)II, III, and IV
(3)I, II, and IV
(4)All of the above
150.Assertion (A): The covalent linkage that links a nitrogenous base to a pentose sugar is
called an N-glycosidic linkage.\nReason (R): The linkage connects the nitrogenous base to the 5′-
carbon of the pentose sugar to form a nucleoside.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
151.Regarding the structural stability differences between DNA and RNA, which of the
following statements are CORRECT?\nI. The 2′-OH group in RNA ribose sugar makes it
chemically reactive and easily degradable.\nII. The presence of Thymine in DNA instead of
Uracil in RNA confers additional stability to DNA.\nIII. Double-stranded RNA is chemically
more stable than double-stranded DNA because of extra hydrogen bonds.\nIV. DNA is
chemically less reactive and structurally more stable than RNA, making it a better genetic
storage molecule.\n\nChoose the correct option:
(1)I, II, and IV
(2)II, III, and IV
(3)All of the above
(4)I and II only
152.Assertion (A): RNA was the first genetic material and can act as a catalyst in metabolic
reactions.\nReason (R): Catalytic RNA molecules are highly stable because they can fold into
complex 3D structures that resist chemical degradation.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
153.Regarding the expression of genetic information, which of the following is/are CORRECT
according to NCERT?\nI. DNA can directly code for the synthesis of proteins without requiring
an RNA intermediate.\nII. RNA can directly code for the synthesis of proteins, which is why it
can easily express characters.\nIII. DNA is dependent on RNA for the synthesis of proteins.\nIV.
The protein-synthesizing machinery has evolved around RNA.\n\nChoose the correct option:
(1)II, III, and IV
(2)II and III only
(3)All of the above
(4)I and II only
154.Assertion (A): DNA has evolved from RNA with chemical modifications that make it more
stable.\nReason (R): Double-stranded DNA has complementary strands and resists changes by
evolving a process of repair.
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
155.In Matthew Meselson and Franklin Stahl's experiment (1958), which of the following
statements is/are CORRECT?\nI. The E. coli cells were grown for many generations in a
medium containing 15NH_4Cl.\nII. 15N is a radioactive isotope of nitrogen that was detected by
Geiger counters.\nIII. 15N is a heavy, stable isotope of nitrogen that can be separated from 14N
based on density.\nIV. CsCl density gradient centrifugation was used to separate the heavy and
light DNA strands.\n\nChoose the correct option:
(1)I, III, and IV
(2)I and III only
(3)II and IV only
(4)All of the above
156.Assertion (A): After transferring E. coli from 15N to 14N medium, the DNA extracted
after 20 minutes consists entirely of hybrid density DNA.\nReason (R): E. coli divides every 20
minutes, meaning that after 20 minutes (one generation), each DNA molecule contains one
parental 15N strand and one newly synthesized 14N strand.
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
157.Regarding the experiments conducted by Herbert Taylor and colleagues (1958), which of
the following is/are CORRECT?\nI. They proved that DNA in chromosomes replicates
semiconservatively.\nII. They used radioactive thymidine to trace the newly synthesized
DNA.\nIII. The experiments were conducted on E. coli bacteria.\nIV. The experiments were
performed on Vicia faba (faba beans).\n\nChoose the correct option:
(1)II and III only
(2)I and II only
(3)I, II, and IV
(4)All of the above
158.Assertion (A): Failure in cell division after DNA replication results in polyploidy, a
chromosomal anomaly.\nReason (R): DNA replication occurs during the M-phase of the cell
cycle, which is followed immediately by cytokinesis.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
159.Consider the kinetic features of DNA replication in E. coli. Which of the following
statements is/are CORRECT according to NCERT?\nI. The main polymerizing enzyme is DNA-
dependent DNA polymerase.\nII. E. coli has 4.6 × 10^6 bp in its genome and completes
replication in 18 minutes.\nIII. The average rate of polymerization is approximately 2000 base
pairs per second.\nIV. DNA-dependent DNA polymerase catalyzes the reaction with high speed
but low accuracy, leading to frequent mutations.\n\nChoose the correct option:
(1)All of the above
(2)I and III only
(3)I, II, and III
(4)II and IV only
160.Assertion (A): Deoxyribonucleoside triphosphates (dNTPs) act as substrates and also
provide energy for the DNA polymerization reaction.\nReason (R): The two terminal
phosphates in a deoxyribonucleoside triphosphate are high-energy phosphates, similar to those
in ATP.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
161.Regarding the expression of genetic information as Mendelian characters, which of the
following is/are CORRECT?\nI. RNA can directly code for the synthesis of proteins, which
allows it to express characters easily.\nII. DNA is completely independent of RNA for protein
synthesis.\nIII. DNA requires an RNA intermediate to translate its code into a protein
sequence.\nIV. The entire protein-synthesizing machinery of a cell has evolved around
RNA.\n\nChoose the correct option:
(1)I and II only
(2)II, III, and IV
(3)All of the above
(4)I, III, and IV
162.Assertion (A): Viruses containing RNA genomes show high rates of mutation and evolve
much faster than DNA-based systems.\nReason (R): RNA is chemically highly stable, allowing
mutations to remain intact without being corrected by repair systems.\n
(1)Both (A) and (R) are false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
163.Regarding the RNA World hypothesis in NCERT, which of the following statements are
CORRECT?\nI. Splicing, translation, and metabolism are essential life processes that evolved
around RNA.\nII. RNA can act as a genetic repository as well as a biocatalyst.\nIII. DNA
evolved from RNA via chemical modifications that made it more stable.\nIV. DNA is a single-
stranded nucleic acid that is chemically more stable than double-stranded RNA.\n\nChoose the
correct option:
(1)All of the above
(2)I, II, and III
(3)I and II only
(4)II and IV only
164.Assertion (A): DNA is preferred for the storage of genetic information, whereas RNA is
better for the transmission of genetic information.\nReason (R): DNA is chemically less reactive
and structurally more stable, whereas RNA can directly translate its code into proteins.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
165.In Meselson and Stahl's experiments, E. coli cells initially grown in 15N were transferred
to 14N medium. Which of the following describes the DNA density proportions correctly?\nI.
After 20 minutes (Generation 1), all DNA molecules have a hybrid density.\nII. After 40 minutes
(Generation 2), the DNA consists of equal amounts of light and hybrid density molecules.\nIII.
After 60 minutes (Generation 3), the DNA consists of 75\% light and 25\% hybrid density
molecules.\nIV. Centrifugation in CsCl density gradients separates these molecules based on
radioactivity.\n\nChoose the correct option:
(1)I and II only
(2)All of the above
(3)I, II, and III
(4)II and IV only
166.Assertion (A): Taylor and colleagues in 1958 proved that chromosomal DNA replication in
Vicia faba is semiconservative.\nReason (R): They used radioactive thymidine to selectively
label the newly synthesized DNA strands in chromosomes.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
167.Regarding directionality at the replication fork, which of the following is/are CORRECT
according to NCERT?\nI. DNA-dependent DNA polymerase catalyzes polymerization only in
the 5′ → 3′ direction.\nII. Synthesis is continuous on the template strand with 3′ → 5′
polarity.\nIII. Synthesis is discontinuous on the template strand with 5′ → 3′ polarity.\nIV. The
discontinuous fragments are joined together by the enzyme DNA polymerase III.\n\nChoose the
correct option:
(1)I, II, and III
(2)I and II only
(3)All of the above
(4)II and IV only
168.Assertion (A): In genetic engineering, any random foreign DNA piece can replicate and
propagate in host cells without being linked to vector DNA.\nReason (R): DNA polymerase
requires a specific origin sequence (ori) to initiate replication, which is provided by the host
genome or vector DNA.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
169.A molecule must fulfill which of the following criteria to act as a genetic material?\nI. It
should be able to direct its own duplication (Replication).\nII. It should chemically and
structurally be stable.\nIII. It should provide scope for slow changes (mutations) that are
required for evolution.\nIV. It should be able to express itself in the form of Mendelian
Characters.\n\nChoose the correct option:
(1)I and III only
(2)All of the above
(3)I, II, and IV only
(4)II and III only
170.Assertion (A): DNA polymerase must catalyze the addition of nucleotides with a high
degree of speed during E. coli replication.\nReason (R): Replication of the 4.6 × 10^6 base pairs
in E. coli takes approximately 18 minutes, requiring an average polymerization rate of 2000 bp
per second.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
171.Regarding the final measurements in the Hershey-Chase experiment, identify the
CORRECT statements:\nI. Radioactivity (32P) was detected in the bacterial cells, which
sedimented in the pellet.\nII. Radioactivity (35S) was detected in the liquid supernatant, which
contained viral ghosts.\nIII. The pellet in the 35S setup was highly radioactive, showing viral
proteins entered the cells.\nIV. The supernatant in the 32P setup was highly radioactive,
indicating DNA did not enter the cells.\n\nChoose the correct option:
(1)All of the above
(2)II and IV
(3)I and II
(4)I, II, and III
172.Assertion (A): RNA can act as genetic material, but RNA genomes are highly unstable
compared to DNA genomes.\nReason (R): RNA is single-stranded and lacks complementary
repair templates, which, combined with the reactive 2′-OH group, results in high mutation rates.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
173.Regarding the division time of E. coli and the Meselson-Stahl experiment details, which of
the following are CORRECT?\nI. E. coli divides every 20 minutes under standard growth
conditions.\nII. E. coli cells replicate their entire genome of 4.6 × 10^6 bp in 18 minutes.\nIII.
The division time of E. coli is identical to the genome replication time of 18 minutes.\nIV. The
division time includes cell growth and cytokinesis, which takes slightly longer than the DNA
synthesis window.\n\nChoose the correct option:
(1)I, II, and IV
(2)I and II only
(3)II and III only
(4)All of the above
174.Assertion (A): DNA is structurally and chemically more stable than RNA because it
contains Thymine instead of Uracil.\nReason (R): Thymine is also known as 5-methyl uracil,
and the presence of this methyl group at the 5th carbon position provides resistance to chemical
mutations.\n
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
175.Which of the following descriptions of a eukaryotic nucleosome are CORRECT?\nI. A
histone octamer is composed of two molecules each of H2A, H2B, H3, and H4.\nII. The histone
H1 is not part of the core histone octamer but binds to linker DNA.\nIII. The DNA is wrapped
around the histone octamer core, making approximately 1.65 turns.\nIV. The negatively charged
histones bind to the positively charged phosphate groups of DNA.\n\nChoose the correct option:
(1)I, II, and III
(2)II and IV only
(3)I and II only
(4)All of the above
176.Assertion (A): Proteins cannot act as genetic material because they lack the ability to
direct their own duplication.\nReason (R): The duplication of genetic material is based on the
rules of base pairing and complementarity, which are absent in proteins.\n
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)(A) is true but (R) is false.
177.If E. coli replicates its genome of 4.6 × 10^6 bp in 18 minutes, what will be the average
replication rate?\nI. The total time in seconds is 1080 seconds.\nII. The replication rate is
approximately 4259 bp per second if replication is unidirectional.\nIII. The replication rate is
approximately 2130 bp per second if replication is bidirectional.\nIV. NCERT states the average
rate of polymerization has to be approximately 2000 bp per second.\n\nChoose the correct
option:
(1)All of the above
(2)I, II, and IV only
(3)I and IV only
(4)II and III only
178.Assertion (A): RNA genomes mutate at a faster rate, which allows RNA viruses to evolve
faster than DNA-based viruses.\nReason (R): RNA is chemically highly stable and resists
deamination, allowing mutations to accumulate without breaking the genome.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are false.
179.Regarding the biochemical characterization of the transforming principle (1933-44),
which of the following is/are CORRECT according to NCERT?\nI. Avery, MacLeod, and
McCarty purified proteins, DNA, and RNA from heat-killed S cells.\nII. They tested these
purified chemicals for their ability to transform live R cells into S cells.\nIII. They discovered
that only the purified DNA fraction was able to cause transformation.\nIV. They concluded that
proteins were the actual transforming substance, but DNA was a necessary cofactor.\n\nChoose
the correct option:
(1)I and II only
(2)I, II, and III
(3)All of the above
(4)II and IV only
180.Assertion (A): DNA is preferred over RNA for the storage of genetic information in most
organisms.\nReason (R): DNA is chemically less reactive and structurally more stable than RNA
because it lacks the 2′-OH group and has Thymine.
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
181.Regarding the transcription unit in DNA, which of the following statements are
CORRECT?

I. The template strand has 3′ → 5′ polarity, and transcription proceeds in the 5′ → 3′ direction.

II. The coding strand has 5′ → 3′ polarity, and its sequence is identical to the synthesized RNA
(except Thymine is replaced by Uracil).

III. The promoter and terminator define the boundaries of the transcription unit.

IV. All reference points for describing the location of regulatory sequences are made with
respect to the template strand.
Choose the correct option:
(1)I, II, and III
(2)II, III, and IV
(3)I and II only
(4)All of the above
182.Assertion (A): The promoter of a transcription unit is located at the 5′-end (upstream) of
the structural gene.

Reason (R): The location of the promoter is defined relative to the polarity of the template
strand of the transcription unit.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is true but (R) is false.
183.Regarding the division of labor among eukaryotic RNA polymerases, which of the
following pairings is/are CORRECT?

I. RNA Polymerase I transcribes rRNAs (28S, 18S, and 5.8S).

II. RNA Polymerase II transcribes hnRNA (precursor of mRNA).

III. RNA Polymerase III transcribes tRNA, 5S rRNA, and snRNAs.

IV. Eukaryotic organelles (mitochondria/plastids) contain their own RNA polymerases.
Choose the correct option:
(1)I, II, and III only
(2)II and IV only
(3)All of the above
(4)I and III only
184.Assertion (A): Eukaryotic hnRNA must undergo post-transcriptional processing before it
can function as mature mRNA.

Reason (R): Eukaryotic genes are split genes containing non-coding intervening sequences
(introns) that must be removed via splicing.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
185.Read the following statements regarding post-transcriptional modifications of hnRNA
and identify the CORRECT ones:

I. In capping, an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′-end of
hnRNA.

II. In tailing, adenylate residues (200-300) are added at the 3′-end in a template-independent
manner.

III. Splicing involves the removal of exons and joining of introns in a defined order.

IV. Once capping, tailing, and splicing are complete, hnRNA is now called mature mRNA.
Choose the correct option:
(1)All of the above
(2)II, III, and IV
(3)I, II, and IV
(4)I and II only
186.Assertion (A): George Gamow, a physicist, argued that because there are only 4 bases, the
genetic code must be composed of triplets to code for 20 amino acids.

Reason (R): A doublet code (combinations of 2 bases) would generate only 16 codons (4^2),
which is insufficient to code for 20 amino acids, whereas a triplet code generates 64 codons
(4^3).
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
187.Regarding the scientists who helped decipher the genetic code, which of the following
is/are CORRECT?

I. Har Gobind Khorana developed a chemical method for synthesizing RNA molecules with
defined combinations of bases (homopolymers and copolymers).

II. Marshall Nirenberg developed a cell-free system for protein synthesis, which was essential
for deciphering the code.

III. Severo Ochoa discovered that DNA polymerase can synthesize RNA in a template-
dependent manner.

IV. Polynucleotide phosphorylase (Severo Ochoa enzyme) was used to polymerize RNA with
defined sequences in a template-independent manner.
Choose the correct option:
(1)All of the above
(2)II and III only
(3)I and II only
(4)I, II, and IV
188.Assertion (A): The codon AUG has dual functions in translation: it codes for methionine
and acts as the initiator codon.

Reason (R): Some amino acids are coded by more than one codon, which is referred to as the
degeneracy of the genetic code.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
189.Which of the following is/are CORRECT regarding the salient features of the genetic
code?

I. 61 codons code for amino acids, while 3 codons (UAA, UAG, UGA) do not code for any amino
acids and function as stop codons.

II. The code is unambiguous, meaning one codon codes for only one amino acid.

III. The code is nearly universal, with exceptions found in mitochondrial codons and some
protozoans.

IV. Codons are read in mRNA in a punctuated fashion, with spacer bases between triplets.
Choose the correct option:
(1)I and II only
(2)All of the above
(3)I, II, and III
(4)II and IV only
190.Assertion (A): Insertion or deletion of one or two bases in a gene alters the reading frame
from the point of mutation downstream.

Reason (R): Insertion or deletion of three bases (or multiples of three) inserts or deletes one or
more codons, hence one or more amino acids, without altering the reading frame downstream.
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
191.Assertion (A): RNA polymerase is capable of catalyzing all three steps of transcription
(initiation, elongation, and termination) on its own in bacteria.

Reason (R): The association of RNA polymerase with initiation factor (σ) and termination factor
(ρ) alters its specificity to initiate or terminate transcription.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)(A) is true but (R) is false.
192.Which of the following statements regarding transcription in bacteria is/are CORRECT?

I. A single DNA-dependent RNA polymerase transcribes all types of RNA (mRNA, tRNA,
rRNA) in bacteria.

II. The initiation factor (σ) remains permanently bound to the RNA polymerase core enzyme
throughout the elongation process.

III. The termination factor (ρ) is required to release the nascent RNA and RNA polymerase
from the DNA template.

IV. Translation can begin much before the mRNA is fully transcribed in bacteria.
Choose the correct option:
(1)I, III, and IV
(2)All of the above
(3)I and III only
(4)II, III, and IV
193.Assertion (A): In bacteria, translation can begin much before the mRNA is fully
transcribed, and transcription and translation can be coupled.

Reason (R): Bacteria lack a nuclear membrane to separate transcription (in nucleus) and
translation (in cytoplasm), allowing both processes to occur in the same compartment.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
194.Regarding the complexities of transcription in eukaryotes, which of the following
statements is/are CORRECT?

I. There are at least three RNA polymerases in the nucleus, in addition to the RNA polymerases
found in organelles.

II. RNA Polymerase I is responsible for transcribing precursor of mRNA (hnRNA).

III. RNA Polymerase II transcribes 28S, 18S, and 5.8S rRNAs.

IV. RNA Polymerase III transcribes tRNA, 5S rRNA, and snRNAs (small nuclear RNAs).
Choose the correct option:
(1)I and IV
(2)II and III only
(3)I, II, and IV
(4)All of the above
195.Assertion (A): The split-gene arrangement in eukaryotes is interpreted as an ancient
feature of the genome.

Reason (R): The presence of introns is reminiscent of antiquity, and the process of splicing
represents the dominance of the RNA world.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)(A) is true but (R) is false.
196.Which of the following descriptions of post-transcriptional processing in eukaryotes are
CORRECT?

I. Primary transcripts (hnRNA) are non-functional and contain both exons and introns.

II. Splicing removes introns and joins exons in a defined order.

III. In capping, methyl guanosine triphosphate is added to the 3′-end of hnRNA.

IV. In tailing, adenylate residues (200-300) are added at the 5′-end in a template-independent
manner.
Choose the correct option:
(1)I, II, and III
(2)II and IV only
(3)I and II only
(4)All of the above
197.Assertion (A): A doublet code of four nitrogenous bases would generate 16 codons, which
is mathematically insufficient to code for 20 amino acids.

Reason (R): In a doublet code, the number of combinations is calculated as 4^2 = 16 because
there are 4 base options and 2 positions in a codon.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
198.Regarding the biochemical deciphering of the genetic code, which of the following is/are
CORRECT?

I. Har Gobind Khorana synthesized RNA molecules with defined combinations of bases
(homopolymers and copolymers) using chemical methods.

II. Severo Ochoa's enzyme is polynucleotide phosphorylase, which polymerizes RNA with
defined sequences in a template-dependent manner.

III. Marshall Nirenberg's cell-free system for protein synthesis was instrumental in translating
the genetic code.

IV. The triplet codon dictionary was completely established using only computational modeling
without any in vitro protein synthesis.
Choose the correct option:
(1)All of the above
(2)II and IV only
(3)I, II, and III
(4)I and III only
199.Assertion (A): Out of 64 codons, only 61 codons code for amino acids, while the remaining
3 codons do not code for any amino acids.

Reason (R): The codons UAA, UAG, and UGA are stop codons (terminator codons) and do not
have any corresponding tRNAs in the cell.
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
200.Which of the following features of the genetic code is/are CORRECT?

I. The code is degenerate, meaning some amino acids are coded by more than one codon.

II. The code is unambiguous and specific, meaning one codon codes for only one amino acid.

III. The codon is read in mRNA in a contiguous fashion without any punctuations.

IV. The code is universal, meaning from bacteria to human, a specific codon always codes for the
exact same amino acid without any exceptions.
Choose the correct option:
(1)I, II, and IV
(2)I, II, and III
(3)All of the above
(4)II and III only
201.Assertion (A): A classical example of a point mutation is the change of a single base pair
in the gene for beta-globin chain that results in the change of amino acid residue glutamate to
valine.

Reason (R): Point mutations only occur when a purine is replaced by another purine, which is
called a transition mutation.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
202.Read the statements regarding insertion and deletion mutations in DNA and identify the
CORRECT ones:

I. Insertion or deletion of one or two bases changes the reading frame from the point of
mutation onwards.

II. Insertion or deletion of three bases or its multiple inserts or deletes one or multiple codons,
without altering the reading frame from that point onwards.

III. Frameshift mutations provide the solid genetic proof that the codon is a triplet and is read in
a contiguous manner.

IV. Deletion of a single base pair in the middle of a gene will alter the amino acid sequence both
upstream and downstream of the deletion point.
Choose the correct option:
(1)All of the above
(2)I and II only
(3)I, II, and III
(4)II, III, and IV
203.Assertion (A): Francis Crick postulated the presence of an adapter molecule that would
on one hand read the code and on the other hand bind to specific amino acids.

Reason (R): Amino acids have no structural specialties to read the genetic code on mRNA
directly.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
204.Which of the following features of tRNA is/are CORRECT according to NCERT?

I. tRNA was known as sRNA (soluble RNA) before its genetic adapter role was postulated.

II. The tRNA has an anticodon loop that has bases complementary to the codon on mRNA.

III. The tRNA has an amino acid acceptor end at its 3′-end, which binds to amino acids.

IV. For initiation, there is a specific tRNA referred to as initiator tRNA, and there is also a
specific tRNA for termination.
Choose the correct option:
(1)I, II, and III
(2)II, III, and IV
(3)All of the above
(4)I and II only
205.Assertion (A): Translation begins when the small subunit of the ribosome encounters the
mRNA and binds to it.

Reason (R): The ribosome binds mRNA at the 3′-untranslated region (UTR) because translation
proceeds in the 3′ → 5′ direction on the mRNA.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is false but (R) is true.
206.Regarding the elongation phase of translation, which of the following is/are CORRECT?

I. Charged tRNAs sequentially bind to the appropriate codon in the ribosome by
complementary base pairing.

II. The ribosome moves from codon to codon along the mRNA in the 5′ → 3′ direction.

III. Peptide bond formation between amino acids is catalyzed by a ribozyme in bacteria.

IV. Amino acids are added one by one, and the sequence of amino acids is determined by the
sequence of bases in the tRNA.
Choose the correct option:
(1)I, II, and III
(2)II, III, and IV
(3)All of the above
(4)I and II only
207.Assertion (A): In eukaryotes, gene expression can be regulated at the transcriptional level
during primary transcript synthesis, but not during splicing or export.

Reason (R): Splicing and export of mRNA occur in the cytoplasm where no transcriptional
control proteins can bind.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)Both (A) and (R) are false.
208.Identify the CORRECT statements regarding the regulation of gene expression in E. coli:

I. The genes in an operon are expressed only when the substrate lactose is present and glucose is
absent.

II. The regulation of gene expression is driven by metabolic, physiological, or environmental
conditions.

III. The development and differentiation of embryo into an adult is a result of coordinated
regulation of gene sets.

IV. Beta-galactosidase enzyme levels in E. coli increase million-fold within seconds of adding
lactose, regardless of glucose levels.
Choose the correct option:
(1)All of the above
(2)I, II, and III
(3)II and III only
(4)II and IV only
209.Assertion (A): The operator region of the lac operon is located adjacent to the structural
genes and binds to the repressor protein.

Reason (R): Each operon has its own specific operator and specific repressor; for example, the
lac operator is specifically recognized by the lac repressor only.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)(A) is false but (R) is true.
210.Regarding the basis of DNA fingerprinting, which of the following statements is/are
CORRECT?

I. DNA fingerprinting involves identifying differences in specific regions of DNA called
repetitive DNA.

II. Repetitive DNA is separated from bulk genomic DNA as a major peak during density
gradient centrifugation.

III. The classification of satellite DNA into micro-satellites, mini-satellites, etc., is based on base
composition, length of segment, and number of repetitive units.

IV. Satellite DNA sequences code for structural proteins that are highly polymorphic.
Choose the correct option:
(1)All of the above
(2)I and III only
(3)I, II, and III
(4)II and IV only
211.Regarding the structure of tRNA (the adapter molecule), which of the following
statements is/are CORRECT according to NCERT?

I. tRNA has an anticodon loop that has bases complementary to the codon on mRNA.

II. tRNA has an amino acid acceptor end at its 5′-end, which binds to amino acids.

III. The secondary structure of tRNA is depicted as a clover-leaf shape.

IV. The actual 3D structure of tRNA is an inverted L-shape.
Choose the correct option:
(1)II and IV only
(2)I, III, and IV
(3)All of the above
(4)I and III only
212.Assertion (A): Untranslated regions (UTRs) are present in mRNA at both the 5′-end
before the start codon and the 3′-end after the stop codon.

Reason (R): UTRs are sequences of mRNA that are not translated into protein, but are required
for the efficient translation process.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
213.Regarding the bacterial translation machinery, which of the following is/are CORRECT
according to NCERT?

I. The ribosome consists of structural RNAs and about 80 different proteins.

II. In bacteria, 23S rRNA acts as a ribozyme and catalyzes peptide bond formation.

III. The enzyme peptidyl transferase is a purely proteinaceous enzyme in bacteria.

IV. The ribosome exists as two subunits: a large subunit and a small subunit.
Choose the correct option:
(1)All of the above
(2)I, II, and IV
(3)I and II only
(4)II and III only
214.Assertion (A): In the lac operon, the structural genes z, y, and a are transcribed together
as a polycistronic mRNA.

Reason (R): The products of these genes are: z codes for beta-galactosidase, y codes for
transacetylase, and a codes for permease.
(1)Both (A) and (R) are false.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
215.Regarding the induction of the lac operon, which of the following statements is/are
CORRECT according to NCERT?

I. Lactose or allolactose acts as the inducer that inactivates the repressor.

II. A very low level of expression of lac operon must be present in the cell all the time, otherwise
lactose cannot enter the cell.

III. Lactose enters the cell through the action of permease enzyme.

IV. In the presence of glucose, lactose is rapidly transported into E. coli cells, inducing the
operon.
Choose the correct option:
(1)All of the above
(2)II and IV only
(3)I, II, and III
(4)I and II only
216.Assertion (A): The methodology of Human Genome Project called Expressed Sequence
Tags (ESTs) focused on sequencing the entire genome, including coding and non-coding regions.

Reason (R): Sequence Annotation is the methodology where functions are assigned to different
regions after sequencing the whole genome.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
217.In the sequencing phase of the Human Genome Project, which of the following is/are
CORRECT according to NCERT?

I. The total DNA from a cell is isolated and converted into random fragments of relatively
smaller sizes.

II. These fragments are cloned in suitable hosts using specialized vectors called BAC (Bacterial
Artificial Chromosomes) and YAC (Yeast Artificial Chromosomes).

III. The hosts used for cloning were bacteria and yeast.

IV. The fragments were sequenced using automated DNA sequencers based on a method
developed by Frederick Sanger.
Choose the correct option:
(1)I and IV only
(2)I, II, and III only
(3)All of the above
(4)II and IV only
218.Assertion (A): Chromosome 1 of the human genome was the last chromosome to be
completely sequenced, with its sequencing completed in May 2006.

Reason (R): Chromosome 1 is the smallest human chromosome, containing the fewest genes
(231 genes) compared to other chromosomes.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
219.Regarding Variable Number of Tandem Repeats (VNTR) used in DNA fingerprinting,
which of the following is/are CORRECT according to NCERT?

I. VNTR belongs to a class of satellite DNA referred to as mini-satellite.

II. A small DNA sequence is arranged tandemly in many copy numbers.

III. The size of VNTR in a genome varies from 0.1 to 20 kb.

IV. The copy number of VNTR is identical in all individuals of a population, which allows
standard comparison.
Choose the correct option:
(1)II, III, and IV
(2)I, II, and III
(3)I and II only
(4)All of the above
220.Assertion (A): During DNA fingerprinting, the separated DNA fragments are transferred
(blotted) to synthetic membranes, such as nitrocellulose or nylon.

Reason (R): The transfer of fragments is performed before the digestion of DNA by restriction
endonucleases to stabilize the long genomic DNA strands.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
221.Regarding the hybridization step in DNA fingerprinting, which of the following is/are
CORRECT according to NCERT?

I. Hybridization is performed using labeled VNTR probes.

II. The probe binds complementary sequences on the blotted membrane.

III. The hybridized fragments are detected by autoradiography, which yields dark bands on X-
ray film.

IV. The probe used for hybridization is an unlabeled double-stranded bulk DNA segment.
Choose the correct option:
(1)I and II only
(2)All of the above
(3)I, II, and III
(4)II and IV only
222.Assertion (A): During density gradient centrifugation of human DNA, bulk genomic DNA
forms a major peak, while repetitive DNA forms smaller peaks called satellite DNA.

Reason (R): Repetitive DNA sequences have a different base composition (A-T rich or G-C rich)
compared to bulk genomic DNA, altering their density.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
223.Regarding the classification of satellite DNA, which of the following is/are CORRECT
according to NCERT?

I. Satellite DNA is classified into categories like micro-satellites and mini-satellites.

II. The classification is based on the base composition (A-T rich or G-C rich).

III. The classification is based on the length of the segment and the number of repetitive units.

IV. Satellite DNA codes for specialized proteins that assist in chromosome condensation.
Choose the correct option:
(1)All of the above
(2)I, II, and III
(3)II, III, and IV
(4)I and II only
224.Assertion (A): DNA fingerprinting from a single hair follicle or blood stain is completely
identical to the DNA fingerprinting of a saliva or bone sample from the same individual.

Reason (R): DNA from every tissue (such as blood, hair, semen, skin, bone, saliva etc.) from an
individual shows the same degree of polymorphism, which makes it a useful identification tool
in forensic applications.
(1)(A) is false but (R) is true.
(2)(A) is true but (R) is false.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
225.Regarding the negative regulation of the lac operon, which of the following is/are
CORRECT according to NCERT?

I. The regulatory i gene codes for the repressor protein.

II. In the absence of inducer, the active repressor binds to the operator region, preventing RNA
polymerase from transcribing the operon.

III. The i gene repressor is synthesized constitutively (all the time) in E. coli cells.

IV. In the presence of inducer, the repressor binds to the promoter, blocking transcription
initiation.
Choose the correct option:
(1)All of the above
(2)I and II only
(3)II, III, and IV
(4)I, II, and III
226.Assertion (A): In eukaryotes, gene expression can be regulated at the transcriptional
level, processing level, transport level, and translational level.

Reason (R): Eukaryotes have a compartmentalized cell structure where the transcription occurs
in the nucleus and translation occurs in the cytoplasm.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
227.Regarding the goals and details of the Human Genome Project (HGP) as per NCERT,
which of the following is/are CORRECT?

I. The HGP was a 13-year project coordinated by the US Department of Energy and the
National Institutes of Health.

II. The project aimed to sequence approximately 3 billion chemical base pairs of the human
genome.

III. If the sequence data were stored in books containing 1000 pages each, with 1000 letters per
page, it would require 3300 books to store the sequence of a single human cell.

IV. The project was completed in May 2006 with the sequencing of Chromosome 21.
Choose the correct option:
(1)I and II only
(2)II, III, and IV
(3)All of the above
(4)I, II, and III
228.Assertion (A): Besides humans, the genomes of many non-human model organisms have
also been completely sequenced as part of genomic studies.

Reason (R): Organisms like E. coli, Saccharomyces cerevisiae (yeast), Caenorhabditis elegans (a
free-living pathogenic nematode), Drosophila (fruit fly), and plants like rice and Arabidopsis
have been sequenced.
(1)(A) is true but (R) is false.
(2)(A) is false but (R) is true.
(3)Both (A) and (R) are true and (R) is the correct explanation of (A).
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
229.Regarding the salient features of the human genome discovered by HGP, which of the
following are CORRECT according to NCERT?

I. The human genome contains 3164.7 million base pairs.

II. The average gene consists of 3000 bases, but sizes vary greatly; the largest known human
gene is dystrophin with 2.4 million bases.

III. The total number of genes is estimated at 30000, and almost all (99.9\%) nucleotide bases
are exactly the same in all people.

IV. The functions are unknown for over 50\% of the discovered genes.
Choose the correct option:
(1)I, II, and III only
(2)All of the above
(3)I and III only
(4)II and IV only
230.Assertion (A): Scientists have identified about 1.4 million locations where single-base
DNA differences (SNPs) occur in humans.

Reason (R): SNPs stand for Single Nucleotide Polymorphisms, and mapping them helps trace
human history and find disease-associated sequences.
(1)(A) is true but (R) is false.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
231.Regarding a polynucleotide chain, which of the following descriptions is/are CORRECT?

I. The 5′-end of the polymer refers to the free phosphate group on the 5′-carbon of the pentose
sugar.

II. The 3′-end of the polymer refers to the free hydroxyl group on the 3′-carbon of the pentose
sugar.

III. The nitrogenous bases project outwards from the sugar-phosphate backbone.

IV. Two nucleotides are linked via a 3′-5′ phosphodiester linkage.
Choose the correct option:
(1)I and II only
(2)II, III, and IV
(3)I, II, and IV
(4)All of the above
232.Assertion (A): If E. coli is grown in a 14N medium for three generations (60 minutes) after
transfer from a 15N medium, the ratio of hybrid to light DNA molecules is 1:3.

Reason (R): Semiconservative replication preserves the two parental 15N template strands as
hybrid DNA molecules in all subsequent generations, while all new strands are synthesized using
14N.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)(A) is false but (R) is true.
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true and (R) is the correct explanation of (A).
233.Regarding the split-gene arrangement in eukaryotes, which of the following descriptions
is/are CORRECT according to NCERT?

I. The structural gene in a transcription unit is monocistronic in eukaryotes and polycistronic in
prokaryotes.

II. Exons are the coding or expressed sequences that appear in mature or processed RNA.

III. Introns are the intervening sequences that do not appear in mature or processed RNA.

IV. The split-gene arrangement is a characteristic feature of prokaryotic structural genes.
Choose the correct option:
(1)I, II, and III
(2)I and II only
(3)II, III, and IV
(4)All of the above
234.Assertion (A): The genetic code is completely and absolutely universal, with no exceptions
across any living organisms or organelles.

Reason (R): A codon codes for the exact same amino acid in all species, from bacteria to
humans, because the translation machinery is conserved.
(1)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is false but (R) is true.
(4)Both (A) and (R) are false.
235.Regarding the tRNAs present in a cell, which of the following is/are CORRECT
according to NCERT?

I. For initiation, there is a specific tRNA referred to as initiator tRNA.

II. There are no tRNAs for stop codons (UAA, UAG, UGA).

III. tRNA is specific, meaning there is a separate tRNA for each amino acid.

IV. tRNA has an anticodon loop that contains bases identical to the codon on mRNA.
Choose the correct option:
(1)I and II only
(2)II and IV only
(3)I, II, and III
(4)All of the above
236.Assertion (A): In the first phase of translation, amino acids are activated in the presence
of ATP and linked to their cognate tRNA.

Reason (R): The charging of tRNA (aminoacylation) is required because peptide bond formation
between two amino acids is an energetically unfavorable reaction.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)(A) is false but (R) is true.
237.Regarding the functions of the lac operon structural gene products, which of the
following descriptions is/are CORRECT according to NCERT?

I. Beta-galactosidase (coded by z) hydrolyzes lactose into glucose and galactose.

II. Permease (coded by y) increases the cell's permeability to beta-galactosides.

III. Transacetylase (coded by a) transfers an acetyl group to beta-galactosidase to activate it.

IV. All three gene products are required for the metabolism of lactose in E. coli.
Choose the correct option:
(1)I, II, and IV
(2)All of the above
(3)II, III, and IV
(4)I and II only
238.Assertion (A): The lac operon is described in NCERT as being under negative regulation,
but it can also be under positive regulation.

Reason (R): Positive regulation of the lac operon is mediated by a repressor protein that
activates transcription initiation when bound to the promoter.
(1)Both (A) and (R) are true and (R) is the correct explanation of (A).
(2)Both (A) and (R) are false.
(3)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(4)(A) is true but (R) is false.
239.Regarding the gene distribution on human chromosomes discovered by HGP, which of
the following is/are CORRECT according to NCERT?

I. Chromosome 1 has the most genes (2968 genes).

II. Chromosome Y has the fewest genes (231 genes).

III. The average size of a human chromosome is 3000 base pairs.

IV. The genes are distributed evenly across all 24 human chromosomes.
Choose the correct option:
(1)All of the above
(2)I, II, and III
(3)I and II
(4)II and IV
240.Assertion (A): The Variable Number of Tandem Repeats (VNTR) probe shows a high
degree of specificity and hybridization to genomic DNA fragments.

Reason (R): The size of the VNTR repeat sequence in the human genome ranges from 0.1 to 20
kb, and the copy number varies between individuals.
(1)(A) is false but (R) is true.
(2)Both (A) and (R) are true and (R) is the correct explanation of (A).
(3)(A) is true but (R) is false.
(4)Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
NEET ]240[ Contd...

Answer Key

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22
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64
72
82
94
102
111
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583
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682
693
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714
721
731
743
751
764
773
784
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803
811
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833
844
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873
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891
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923
934
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951
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971
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1032
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1123
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1303
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1344
1353
1364
1373
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1391
1401
1413
1424
1431
1442
1453
1463
1473
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1493
1503
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1932
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2013
2023
2034
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2092
2102
2112
2121
2132
2144
2153
2162
2173
2183
2192
2202
2213
2222
2232
2243
2254
2262
2274
2281
2292
2304
2313
2324
2331
2344
2353
2361
2371
2384
2393
2404