Molecular Basis of Inheritance NEET Test- master the concepts

Test your knowledge of the Molecular Basis of Inheritance with our NEET-specific Molecular Basis of Inheritance NEET Test Boost your biology preparation with chapter-wise questions and detailed explanations. Enhance your understanding of the Molecular Basis of Inheritance by taking our specialized NEET Test. This NEET-specific assessment is tailored to help you master the intricacies of this crucial biological concept. Elevate your biology exam readiness with comprehensive chapter-wise queries and thorough answer explanations.

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#1. The histone protein that attaches to DNA strands between nucleosomes is _________.

H2A

H2B

Answer: (1) H1
Explanation: Histone H1 is the linker histone protein that binds to the DNA between nucleosomes, stabilizing the chromatin structure.

#2. Calculate the number of nucleosomes that will attach to a small chromosome consisting of 500,000,000 nucleotides.

1000

50,00,00,000

12,50,000

25,00,00,000

Answer: (3) 12,50,000
Explanation: Each nucleosome contains about 200 base pairs of DNA. Therefore, 500,000,000200=12,50,000frac{500,000,000}{200} = 12,50,000200500,000,000=12,50,000 nucleosomes will attach.

#3. Which of the following is not a salient feature of the double-helix structure of DNA?

The two chains have anti-parallel polarity

The bases in two strands are paired through hydrogen bonds forming base pairs.

The plane of one base pair stacks over the other in the double helix.

The distance between a base pair in a helix is approximately equal to 3.4 nm.

Answer: (4)

The distance between a base pair in a helix is approximately equal to 3.4 nm.
Explanation: The actual distance between base pairs in the DNA double helix is 0.34 nm, not 3.4 nm.

#4. Calculate the base ratio for a DNA sample if 22% cytosine was found in it.

0.78

1.27

1.0

0.93

Answer: (3) 1.0
Explanation: According to Chargaff’s rule, cytosine pairs with guanine in equal amounts. Thus, the base ratio C:G=1:1text{C:G} = 1:1C:G=1:1.

#5. A diploid cell of an organism has 8.4×10108.4 times 10^{10}8.4×1010 base pairs. The length of the total DNA could be

28.56 m

28.56 Å

2.86 m

2.86 mm

Answer: (1) 28.56 m
Explanation: Each base pair contributes approximately 0.34 nm (or 3.4×10−103.4 times 10^{-10}3.4×10−10 m) to the DNA length. Therefore, 8.4×1010×3.4×10−10=28.568.4 times 10^{10} times 3.4 times 10^{-10} = 28.568.4×1010×3.4×10−10=28.56 m.

#6. Calculate the number of hydrogen bonds in DNA if out of 200 nitrogen bases 46 are cytosine molecules.

246

254

200

Answer: (2) 254
Explanation: Cytosine pairs with guanine via three hydrogen bonds. Thus, 46×3=13846 times 3 = 13846×3=138 H-bonds. The remaining 200−46=154200 – 46 = 154200−46=154 adenine-thymine pairs contribute 154×2=308154 times 2 = 308154×2=308. Total = 138+308=254138 + 308 = 254138+308=254

#7. Which of the following statements regarding R-strain of Streptococcus pneumoniae is incorrect?

R-strain bacteria have a mucous coat around it.

It produces rough colonies.

R-strain is a non-virulent strain

Griffith used R-II strain in his transforming experiment.

Answer: (1) R-strain bacteria have a mucous coat around it.
Explanation: The R-strain does not have a mucous coat, which makes its colonies appear rough and non-virulent.

#8. Fill in the blanks and select the option with all correct answers. (a) In eukaryotes, there is a set of (i) charged (ii) proteins called histones. (b) Histones are rich in the basic amino acid residues (iii) and (iv). (c) Histones are organized to form a unit of molecules called (v). (d) A typical nucleosome contains (vi) of DNA helix.

(i) – negatively; (iii) – glycine; (vi) – 800 bp

(ii) – basic; (iv) – arginine; (v) – octamer

(i) – positively; (ii) – acidic; (v) – solenoid

(iii) – Lysine; (iv) – Valine; (vi) – 200 bp

Answer: (2) (ii) – basic; (iv) – arginine; (v) – octamer
Explanation: Histones are positively charged, basic proteins rich in lysine and arginine, organized into octamer units, with ~200 bp of DNA wrapped around each nucleosome.

#9. Consider the following statements regarding the structure of B-DNA: (a) Backbone of the helix is made up of sugar-phosphate. (b) Parallel polarity of two chains. (c) Adenine binds to guanine by double hydrogen bonds. (d) The two chains are coiled in a right-handed fashion. (e) Pitch of the helix is 3.4 Å. (f) Roughly 10 base pairs exist in each turn. Which statements are correct?

(a), (c) and (d)

(a), (d) and (f)

(b), (c) and (f)

(b), (d) and (e)

Answer: (2) (a), (d) and (f)
Explanation: The backbone of B-DNA is sugar-phosphate, the chains are right-handed, and each turn contains approximately 10 base pairs. Adenine pairs with thymine, not guanine.

#10. A phosphate group is linked to (a) of a nucleoside through (b) linkage to form a corresponding nucleotide.

(a) – 5'-OH; (b) – N-glycosidic

(a) – 3'-OH; (b) – phosphodiester

(a) – 3'-OH; (b) – hydrogen bond

(a) – 5'-OH; (b) – phosphoester

Explanation: In a nucleotide, the phosphate group is attached to the 5′-OH group of the sugar via a phosphoester bond.

#11. Identify the mismatched pair.

Bacteriophage T4—48502 bp

Beads-on-string—Nucleosomes in chromatin structure

QB bacteriophage—RNA genetic material

ɸ×174 bacteriophage—5386 bp in DNA

Answer: (4) ɸ×174 bacteriophage—5386 bp in DNA
Explanation: The genetic material of the ɸ×174 bacteriophage is single-stranded DNA, but its length is 5386 bases, not base pairs.

#12. Which enzyme, if present in a medium, would affect transformation in bacteria?

DNase

RNase

Protease

More than one option is correct

Answer: (1) DNase
Explanation: DNase degrades DNA, which is the transforming principle. RNase and protease do not affect DNA transformation.

#13. What does “antiparallel strands” in DNA mean?

One strand turns clockwise.

One strand turns anti-clockwise.

The phosphate groups of two DNA strands, at their ends, share the same position.

The phosphate groups at the start of the two DNA strands are in opposite positions (poles).

Answer: (4) The phosphate groups at the start of the two DNA strands are in opposite positions (poles).
Explanation: DNA strands run in opposite directions (5′ to 3′ and 3′ to 5′), making them antiparallel.

#14. Which of the following does not follow the central dogma of molecular biology?

HIV

Pea

Mucor

Chlamydomonas

Answer: (1) HIV
Explanation: HIV is a retrovirus that reverses the central dogma using reverse transcriptase to synthesize DNA from RNA.

#15. Calculate the length of a DNA sample if it has 240 base pairs (bp).

340 Å

816 Å

1024 Å

120 Å

Answer: (2) 816 Å
Explanation: Each base pair contributes 0.34 nm (or 3.4 Å) to the DNA length. Thus, 240×3.4=816240 times 3.4 = 816240×3.4=816 Å.

#16. Which of the following statements is not correct?

The length of DNA is usually defined as the number of nucleotides present in it.

Escherichia coli has 3.3×1093.3 times 10^93.3×109 bp.

A nucleotide has three components—a nitrogenous base, a pentose sugar, and a phosphate group.

Cytosine is common for both DNA and RNA.

Answer: (2) Escherichia coli has 3.3×1093.3 times 10^93.3×109 bp.
Explanation: E. coli has 4.6×1064.6 times 10^64.6×106 base pairs, not 3.3×1093.3 times 10^93.3×109, which is the length of the human genome.

#17. Why do some strains of Streptococcus pneumoniae produce smooth shiny colonies (S)?

These strains have a coat made up of lipid and oils.

These strains have a mucous coat made up of polysaccharides.

These strains do not have any external coat.

These are non-virulent strains of bacteria.

Answer: (2) These strains have a mucous coat made up of polysaccharides.
Explanation: The polysaccharide mucous coat protects the bacteria and contributes to smooth shiny colonies.

#18. . A molecule that can act as a genetic material must fulfill all criteria except

It should be able to generate its replica.

It should chemically and structurally be stable

It should provide the scope for fast changes (mutations) required for evolution.

It should express itself in the form of Mendelian characters.

Answer: (4) It should express itself in the form of Mendelian characters.
Explanation: Genetic material does not directly express Mendelian characters; it codes for traits that are expressed through proteins.

#19. A DNA molecule in which both strands have radioactive thymidine is allowed to duplicate in an environment containing non-radioactive thymidine. What will be the exact number of DNA molecules that contain radioactive thymidine after three duplications?

One

Two

Four

Eight

Answer: (2) Two
Explanation: During the first duplication, the radioactive thymidine is incorporated into one strand of each of the two new DNA molecules. These radioactive strands persist as templates, so after three duplications, only two DNA molecules retain the radioactive thymidine in one of their strands.

#20. DNA polymerase enzyme is required for the synthesis of:

DNA from DNA

DNA from RNA

Both (1) and (2)

DNA from nucleosides

Answer: (1) DNA from DNA
Explanation: DNA polymerase catalyzes the replication of DNA using an existing DNA strand as a template.

#21. A DNA strand is directly involved in the synthesis of all the following except:

Another DNA

tRNA and mRNA

rRNA

Protein

Answer: (4) Protein
Explanation: DNA serves as a template for RNA synthesis (transcription). Proteins are synthesized using RNA as a template (translation), not directly from DNA.

#22. The Hershey and Chase experiment was precisely aimed to:

Use radioactive compounds to demonstrate DNA as the genetic material.

Prove that proteins, and not DNA, enter the bacterium as genetic material.

Determine which component of a virus enters the bacterial cell

Study the incorporation of 32P^{32}P32P and 35S^{35}S35S isotopes.

Answer: (3) Determine which component of a virus enters the bacterial cell.
Explanation: Hershey and Chase used radioactive isotopes to prove that DNA, and not protein, is the genetic material that enters bacteria during viral infection.

#23. Which of the following holds true for Escherichia coli?

It has a roughly 40-minute doubling time.

It is a parasite in the human digestive tract.

It is used for genetic engineering in plants.

It has 4.6×1064.6 times 10^64.6×106 base pairs of genetic material.

Answer: (4) It has 4.6×1064.6 times 10^64.6×106 base pairs of genetic material.
Explanation: E. coli has 4.6×1064.6 times 10^64.6×106 base pairs in its genome and is widely studied in genetics. It is not a parasite but a symbiotic organism in the human gut.

#24. What confers additional stability to the double-helix model of DNA apart from hydrogen bonds?

One base pair provides an inductive effect to the other.

One base pair is stacked over the other.

One base pair is perpendicular to another.

One base pair interacts with another base pair via steric hindrance.

Answer: (2) One base pair is stacked over the other.
Explanation: The stacking of base pairs provides additional stability to the DNA helix through hydrophobic interactions and van der Waals for

#25. The “beads-on-string” structure in chromatin is packaged to form ___________.

Nucleosomes

Solenoids

Chromatin fibers

Chromatids

Answer: (2) Solenoids
Explanation: The “beads-on-string” structure (nucleosomes) further folds into solenoids for efficient packaging within the nucleus.

#26. In the Hershey and Chase experiment, 35S^{35}S35S was used to label proteins. Which amino acids were actually labeled by it?

Tryptophan and Tyrosine

Proline and Histidine

Threonine and Serine

Methionine and Cysteine

Answer: (4) Methionine and Cysteine
Explanation: 35S^{35}S35S labels sulfur, which is present in the amino acids methionine and cysteine.

#27. DNA is a preferred genetic material over RNA due to one of the following reasons:

Bases are arranged linearly in a single strand.

Bases show less propensity for mutation

The 2′–OH group in RNA is more reactive and less stable.

None of these.

Answer: (3) The 2′–OH group in RNA is more reactive and less stable.
Explanation: DNA lacks the 2′–OH group, making it chemically more stable than RNA, which is more prone to hydrolysis.

#28. DNA (a) Acts as genetic material in all cellular organisms. (b) Was discovered by F. Miescher, who named it nuclein. (c) Is acidic in nature. (d) Cannot be digested by DNase.

(a) and (c)

(a), (b), (c) and (d)

(b), (c) and (d)

(a), (b) and (c)

Answer: (4) (a), (b) and (c)
Explanation: DNA is genetic material, was discovered as nuclein by F. Miescher, and is acidic in nature. However, it can be digested by DNase.

#29. In 1953, James Watson and F. Crick proposed the double-helix model of DNA and received the Nobel Prize. Their model of DNA was based on which two of the following findings? (a) X-ray diffraction studies by M. Wilkins and R. Franklin (b) Griffith’s experiment showing transformation (c) Hershey and Chase experiment demonstrating DNA as genetic material (d) Chargaff’s rule: (A + G) = (T + C)

(a) and (d)

(a), (b), (c), and (d)

(b), (c), and (d)

(a), (b), and (c)

Answer: (1) (a) and (d)
Explanation: The double-helix model of DNA was influenced by X-ray diffraction and Chargaff’s base-pairing rules.

#30. You analyzed DNA from a new virus. The base composition was 32% A, 17% C, 32% G, and 19% T. Which conclusion is most accurate?

The virus is extraterrestrial.

Double-stranded DNA has unconventional base pairing.

T was converted to C during isolation.

The genome is single-stranded

Answer: (4) The genome is single-stranded.
Explanation: Unequal proportions of bases (A ≠ T and G ≠ C) indicate a single-stranded DNA genome.

#31. X-ray crystallography provides information about the of DNA but is limited because of the of DNA. The technique relies on __ patterns.

structure; difficulty of purification; light absorption

dimensions; molecular weight; diffraction

molecular weight; shape; diffraction

dimensions; linearity; light absorption

Answer: (2) dimensions; molecular weight; diffraction
Explanation: X-ray crystallography reveals the 3D structure and dimensions of molecules by studying diffraction patterns.

#32. The nucleosome core is composed of which histones?

H1 and H2A

H2A and H2B

H3 and H4

Both (2) and (3)

Answer: (4) Both (2) and (3)
Explanation: The nucleosome core consists of two copies each of H2A, H2B, H3, and H4, while H1 stabilizes the structure externally.

#33. Which sequence correctly shows increasing levels of DNA condensation?

Nucleosomes → Solenoid → 30 nm fiber → Loops on scaffold → Heterochromatin

Euchromatin → Solenoid → 30 nm fiber → Nucleosomes

Nucleosomes → 30 nm fiber → Euchromatin → Heterochromatin

Nucleosomes → 30 nm fiber → Solenoid → Loops on scaffold

Answer: (1) Nucleosomes → Solenoid → 30 nm fiber → Loops on scaffold → Heterochromatin
Explanation: DNA condensation progresses from nucleosome organization to densely packed heterochromatin.

#34. The number of phosphodiester bonds in double-stranded DNA molecules is:

Two more than phosphoester bonds

One less than phosphoester bonds

One more than phosphoester bonds

Two less than phosphoester bonds

Answer: (3) One more than phosphoester bonds
Explanation: Phosphodiester bonds link nucleotides in the DNA backbone, which has one extra bond at each end.

#35. DNA packaging in eukaryotes:

Involves polyamines

Uses only NHC proteins

Requires acidic proteins in prokaryotes

Is more complex than in prokaryotes

Answer: (4) Is more complex than in prokaryotes
Explanation: Eukaryotic DNA is packaged with histones into chromatin, which is more intricate than the simpler prokaryotic nucleoid.

#36. Which of the following features applies to bacterial DNA?

Presence of intervening sequences

DNA does not show coiling

Linear single-stranded DNA

DNA can be chromosomal and extrachromosomal

Answer: (4) DNA can be chromosomal and extrachromosomal
Explanation: Bacteria have circular chromosomal DNA and may contain plasmids (extrachromosomal DNA).

#37. Identify the incorrect statement about nucleotides and nucleic acids.

N-glycosidic linkage forms a nucleotide

5-methyl uracil is present in DNA

RNA has an additional –OH group

A + T/G + C ratio is variable

Answer: (2) 5-methyl uracil is present in DNA
Explanation: DNA contains thymine, not 5-methyl uracil.

#38. Which bonds are involved in forming a nucleoside?

Only N-glycosidic linkage

Hydrogen bonds and N-glycosidic linkage

Only phosphodiester bond

All three bonds

Answer: (1) Only N-glycosidic linkage
Explanation: A nucleoside consists of a sugar linked to a nitrogenous base via an N-glycosidic bond.

#39. Which is not a function of RNA?

Helps synthesize DNA

Functions as an adapter

Acts as a catalyst

Functions mostly as a messenger

Answer: (1) Helps synthesize DNA
Explanation: RNA does not synthesize DNA, although some viruses use reverse transcription.

#40. RNA is , making it . DNA evolved with modifications for stability.

ribozyme; stable; physical; preferred

enzyme; unstable; natural; better

catalyst; unstable; chemical; stable

ribozyme; unstable; chemical; stable

Answer: (4) ribozyme; unstable; chemical; stable
Explanation: RNA’s instability as a ribozyme drove DNA evolution with chemical modifications like thymine

#41. Which statements about DNA and RNA are true? (a) DNA is better for genetic information transmission. (b) RNA codes proteins and expresses characters easily. (c) RNA mutates faster. (d) Uracil increases RNA stability. (e) Genetic material should allow mutations.

One

Two

Three

All

Answer: (3) Three
Explanation: (a), (b), and (c) are correct, but uracil does not increase RNA stability.

#42. Which statements are correct? (a) Both DNA and RNA are able to mutate. (b) RNA being unstable, mutates at a faster rate. (c) RNA is catalytic and reactive. (d) Thymine in DNA adds stability compared to uracil in RNA.

All are correct

All are incorrect

Only (a), (b), and (d) are correct

Only (d) is correct

Answer: (1) All are correct

Explanation: DNA and RNA both mutate, but RNA’s instability leads to faster mutations. RNA’s catalytic nature makes it reactive. Thymine’s methyl group stabilizes DNA.

#43. Identify the correct match. (a) Amount of mRNA – Number of ribosomes in a cell (b) RNA polymerase-III – Helps in synthesis of rRNA (c) Most stable RNA – mRNA (d) Catalytic RNA – Ribozyme

(a) and (b)

(b) and (d)

(a), (c), and (d)

All of these

Answer: 2. (b) and (d)

Explanation: RNA polymerase III is involved in tRNA synthesis and not rRNA. Ribozyme is RNA with catalytic activity. mRNA is not the most stable; rRNA is more stable.

#44. Which of the following statements are correct? (a) Both DNA and RNA are able to mutate. (b) RNA being unstable, mutates at a faster rate. (c) RNA is also known to be catalytic, hence reactive. (d) The presence of thymine at the place of uracil confers additional stability to DNA.

All are correct

All are incorrect

Only (a), (b), and (d) are correct

Only (d) is correct

Answer: 3. Only (a), (b), and (d) are correct

Explanation: Statements (a), (b), and (d) are true as RNA is unstable and catalytic. The thymine in DNA adds stability compared to uracil in RNA.

#45. The differences between mRNA and tRNA include: (a) mRNA has a more elaborate three-dimensional structure due to extensive base pairing. (b) tRNA has a more elaborate three-dimensional structure due to extensive pairing. (c) tRNA is usually smaller than mRNA. (d) mRNA bears an anticodon, but tRNA has codons.

(a) and (c)

(b) and (c)

(b), (c), and (d)

All of these

Answer: 2. (b) and (c)

Explanation: Statement (b) is correct as tRNA has an elaborate 3D structure. Statement (c) is correct since tRNA is shorter than mRNA. Statement (d) is incorrect as tRNA has anticodons while mRNA contains codons.

#46. Which of the following statements is NOT correct with respect to the Meselson and Stahl experiment?

They proved that DNA replication is semi-conservative.

They used CsCl density gradient centrifugation for measuring the densities of DNA.

15N^{15}N15N is a radioactive isotope, while 14N^{14}N14N is the normal isotope of nitrogen.

They selected E. coli as their experimental model.

Answer: 3. 15N^{15}N15N is a radioactive isotope, while 14N^{14}N14N is the normal isotope of nitrogen.

Explanation: 15N^{15}N15N is not radioactive; it is a heavy stable isotope used for density labeling.

#47. Identify the incorrect statement about DNA replication.

The DNA polymerase on its own can initiate the process of replication.

Replication does not initiate randomly at any place in DNA.

In eukaryotes, the replication of DNA takes place at the S phase of the cell cycle.

A failure in cell division after DNA replication results in polyploidy.

Answer: 1. The DNA polymerase on its own can initiate the process of replication.

Explanation: DNA polymerase cannot initiate replication de novo; it requires an RNA primer synthesized by primase.

#48. If a DNA molecule is continuously replicated in 15N^{15}N15N medium, what is the percentage of lighter DNA in the fourth generation?

12.5%

25%

6.25%

Answer: 1. 12.5%

Explanation: In the fourth generation, 124=6.25frac{1}{2^4} = 6.25%241=6.25 of the DNA remains hybrid, and the rest is light.

#49. Which enzyme synthesizes DNA from an RNA template?

DNA-dependent RNA polymerase

DNA polymerase

Reverse transcriptase

RNA polymerase

Answer: 3. Reverse transcriptase

Explanation: Reverse transcriptase catalyzes the synthesis of complementary DNA (cDNA) from an RNA template.

#50. The Okazaki fragments in DNA chain growth:

Result in transcription.

Polymerize in the 3′→5′ direction and form the replication fork.

Prove the semi-conservative nature of DNA replication.

Polymerize in the 5′→3′ direction and explain 3′→5′ DNA replication.

Answer: 4. Polymerize in the 5′→3′ direction and explain 3′→5′ DNA replication.

Explanation: Okazaki fragments are short DNA sequences synthesized discontinuously on the lagging strand during DNA replication. They are polymerized in the 5′→3′ direction and joined later by DNA ligase.

#51. Which of the following statements are correct? (a) DNA replication is semi-discontinuous. (b) Growth of the lagging strand occurs in fragments. (c) DNA polymerase III can take part in proofreading. (d) Taylor fed dividing cells of Vicia faba leaves with radioactive thymine.

(a) and (d)

(a), (c), and (d)

(a), (b), and (c)

(a), (b), (c), and (d)

Answer: 4. (a), (b), (c), and (d)

Explanation: DNA replication is semi-discontinuous due to Okazaki fragments on the lagging strand. DNA polymerase III has proofreading capability. Taylor’s experiment proved semi-conservative replication using radioactive thymine.

#52. Which enzyme unwinds the DNA helix during replication?

Helicase

Gyrase

Primase

DNA polymerase III

Answer: 1. Helicase

Explanation: Helicase unwinds the DNA double helix by breaking hydrogen bonds between base pairs, creating the replication fork.

#53. Read the following statements and identify the correct ones: (a) Taylor used radioactive thymidine in Vicia faba to demonstrate semi-conservative replication. (b) DNA replication in eukaryotes occurs during the S phase. (c) Failure of cell division after DNA replication can lead to polyploidy. (d) Meselson and Stahl’s experiment demonstrated semi-conservative replication using NH4Cl in E. coli.

All are correct.

All are incorrect.

Only (d) is correct.

(a) and (d) are correct.

Answer: 4. (a) and (d) are correct.

Explanation: While (a) and (d) are correct, DNA replication occurs in the S phase (not proven by Taylor), and failure in division does result in polyploidy.

#54. Which of the following is false regarding DNA replication?

The ori site is where replication begins.

dNTPs serve as both substrates and energy sources.

Both strands are synthesized in the 3′→5′ direction.

DNA replication ensures the continuity of life

Answer: 3. Both strands are synthesized in the 3′→5′ direction.

Explanation: DNA synthesis occurs in the 5′→3′ direction for both leading and lagging strands. This is due to the structure of DNA polymerase.

#55. Why is DNA ligase essential during replication?

It joins fragments of the leading strand.

It joins fragments of the lagging strand.

It joins parental strands back together.

It converts 3′-dNTPs to 5′-dNTPs.

Answer: 2. It joins fragments of the lagging strand.

Explanation: DNA ligase connects Okazaki fragments on the lagging strand by forming phosphodiester bonds.

#56. Chargaff’s rule states that:

DNA must be replicated before cell division.

Cytosine equals guanine in amount.

Protein from infecting phages is detected in progeny phages.

Viruses enter cells without protein coats.

Answer: 2. Cytosine equals guanine in amount.

Explanation: Chargaff’s rule states that in DNA, the amount of adenine equals thymine, and cytosine equals guanine.

#57. Which statement about DNA replication is false?

Okazaki fragments initiate synthesis on the leading strand.

Replication forks are sites of active synthesis.

Error rates in replication are very low.

Ligases and polymerases work near replication forks.

Answer: 1. Okazaki fragments initiate synthesis on the leading strand.

Explanation: Okazaki fragments are formed only on the lagging strand, not the leading strand.

#58. Which is true regarding introns?

Introns are translated into proteins.

Introns are involved in exon shuffling.

Introns appear in mature mRNA.

Introns are only found in prokaryotic genes.

Answer: 2. Introns are involved in exon shuffling.

Explanation: Introns are non-coding sequences in pre-mRNA. They can promote genetic diversity through exon shuffling during evolution.

#59. RNA synthesis is controlled by:

Rho factor

Sigma factor

Endonuclease

RNA polymerase

Answer: 4. RNA polymerase

Explanation: RNA polymerase catalyzes RNA synthesis during transcription. Sigma factors aid in promoter recognition.

#60. The process of removing introns and joining exons in a defined order in a transcription unit is called

capping

splicing

tailing

transformation

Answer: (2) splicing

Explanation: Splicing is the process by which introns (non-coding regions) are removed from the pre-mRNA and exons (coding regions) are joined together. This occurs before the mRNA is translated into a protein.

#61. Both strands of DNA are not copied during transcription because

DNA segment will form two types of proteins.

Double-stranded RNA will be formed.

DNA will degenerate.

Both (1) and (2).

Answer: (1) DNA segment will form two types of proteins.

Explanation: During transcription, only one strand of the DNA, known as the template strand, is used to produce RNA. This ensures that only one type of protein is translated from the mRNA.

#62. Which is the initial step in the mRNA maturation process?

Polyadenylation

5′ capping

Splicing

Endonucleolytic cleavage

Answer: (2) 5′ capping

Explanation: The initial step in mRNA maturation is the addition of a 5′ cap, which is crucial for protecting the mRNA from degradation, helping ribosome binding during translation, and facilitating mRNA export from the nucleus.

#63. Methyl guanosine triphosphate is associated with

point mutation.

tautomerism.

capping.

Okazaki fragments.

Answer: (3) capping

Explanation: Methyl guanosine triphosphate is added to the 5′ end of the nascent mRNA during the capping process. This cap helps in stability, translation initiation, and protection of the RNA molecule.

#64. Answer: (3) capping Explanation: Methyl guanosine triphosphate is added to the 5′ end of the nascent mRNA during the capping process. This cap helps in stability, translation initiation, and protection of the RNA molecule. Shape

5′ – AUGAAUG – 3′

5′ – UACUUAC – 3′

5′ – CAUUCAU – 3′

5′ – GUAAGUA – 3′

Answer: (1) 5′ – AUGAAUG – 3′

Explanation: During transcription, the coding strand of DNA is used to synthesize mRNA. The sequence of the RNA is complementary to the template strand of the DNA. In this case, the mRNA sequence is 5′ – AUGAAUG – 3′, which corresponds to the coding strand’s sequence after replacing thymine (T) with uracil (U).

#65. In bacteria, in the context of mRNA, which of the following is not true?

Transcription and translation take place in different components.

Post-transcriptional modifications are required.

Translation occurs only after transcription.

None of these.

Answer: (2) Post-transcriptional modifications are required.

Explanation: In bacteria, transcription and translation can occur simultaneously in the cytoplasm. Unlike eukaryotes, bacteria do not require post-transcriptional modifications like capping or splicing for mRNA maturation.

#66. Which of the following statements is true regarding introns?

Introns are the parts of mRNA that are translated.

Introns have no function.

In general, human genes have fewer introns than genes of other organisms.

Introns can be involved in exon shuffling.

Answer: (4) Introns can be involved in exon shuffling.

Explanation: Introns, although not coding for proteins, can play a role in exon shuffling, which can lead to genetic diversity through alternative splicing.

#67. What is correct for bacterial transcription?

mRNA requires processing to become active.

Translation can begin when mRNA is fully transcribed.

Transcription and translation take place in the same compartment.

Rho factor initiates the process.

Answer: (3) Transcription and translation take place in the same compartment.

Explanation: In bacteria, both transcription and translation occur in the cytoplasm, allowing for immediate translation of the newly synthesized mRNA.

#68. In eukaryotes, RNA polymerase III catalyses the synthesis of

all rRNA and tRNA.

mRNA, hnRNA, and snRNA.

5S rRNA, tRNA, and snRNA.

28S, 18S, and 5S rRNA.

Answer: (3) 5S rRNA, tRNA, and snRNA.

Explanation: RNA polymerase III in eukaryotes is responsible for synthesizing smaller RNAs, including 5S rRNA, tRNA, and small nuclear RNA (snRNA).

#69. Which of the following statements correctly describes the function of RNA polymerase during transcription?

RNA polymerase adds nucleotides in the 5′ to 3′ direction on the template strand.

RNA polymerase synthesizes mRNA in the 3′ to 5′ direction.

RNA polymerase moves along the DNA in the 3′ to 5′ direction on the coding strand.

RNA polymerase moves along the RNA strand to create a complementary copy of the DNA template.

Answer: (1) RNA polymerase adds nucleotides in the 5′ to 3′ direction on the template strand.

Explanation: RNA polymerase synthesizes the mRNA in the 5′ to 3′ direction, using the DNA template strand as a guide. The RNA strand is complementary to the DNA template strand, and transcription occurs in the 3′ to 5′ direction on the template strand.

#70. Of the ___ different possible codons, _ specify amino acids and ______ signal stop.

20; 17; 3

180; 20; 60

64; 61; 3

61; 60; 1

Answer: (3) 64; 61; 3

Explanation: There are 64 different possible codons in the genetic code, of which 61 codons specify amino acids and 3 codons signal the stop of protein synthesis.

#71. Given the base sequence 5′ AAC CAA GAC CCC 3′ in mRNA, how many codons will there be in the case of overlapping and non-overlapping codons, respectively?

4 and 10

10 and 4

12 and 10

10 and 8

Answer: (1) 4 and 10

Explanation: In non-overlapping codons, each codon is read separately, giving a total of 10 codons. In overlapping codons, some nucleotides are shared between adjacent codons, resulting in 4 distinct codons.

#72. Assume an organism has 6 types of nitrogenous bases on mRNA and 348 types of amino acids. How many nitrogenous bases are present in a codon?

Three

Four

Two

One

Answer: (1) Three

Explanation: Codons are made up of three nitrogenous bases, as three bases (a triplet) are required to specify one amino acid.

#73. What is the function of a nonsense codon?

Release the polypeptide chain from tRNA

Form an unspecified amino acid

Terminate the message for the gene-controlled protein synthesis

Convert a sense DNA into nonsense DNA

Answer: (3) Terminate the message for the gene-controlled protein synthesis

Explanation: Nonsense codons (or stop codons) signal the termination of translation, thereby ending the synthesis of the polypeptide chain.

#74. The degeneration of a genetic code is attributed to the:

First member of a codon

Second member of a codon

Entire codon

Third member of a codon

Answer: (4) Third member of a codon

Explanation: Degeneracy of the genetic code refers to the fact that different codons can code for the same amino acid, and this is primarily due to variability in the third position of the codon.

#75. Khorana synthesized two RNAs (a) with a repeat sequence of AB and (b) with a repeat sequence of ABC. The polypeptides coded by (a) and (b) are respectively:

Homopolypeptides in both (a) and (b)

Heteropolypeptides in both

Homopolypeptide in (a) and heteropolypeptide in (b)

Heteropolypeptide in (a) and homopolypeptide in (b)

Answer: (3) Homopolypeptide in (a) and heteropolypeptide in (b)

Explanation: A repeat sequence of AB would code for a homopolypeptide (repeating a single amino acid), while a repeat sequence of ABC would code for a heteropolypeptide (involving different amino acids).

#76. Select the incorrect statement(s): (a) Six codons do not code for any amino acid. (b) Codons are read in mRNA in a contiguous fashion. (c) Three codons function as stop codons. (d) The initiator codon AUG codes for methionine.

(a), (b), and (c)

(a), (b), and (d)

Only (b)

Only (a)

Answer: (4) Only (a)

Explanation: The statement that “six codons do not code for any amino acid” is incorrect, as only three codons are stop codons (not six).

#77. The genetic code translates the language of:

RNA into that of protein

RNA into that of DNA

Amino acids into that of RNA

Proteins into that of DNA

Answer: (1) RNA into that of protein

Explanation: The genetic code is responsible for translating the sequence of RNA into the sequence of amino acids in a protein.

#78. A codon consists of three bases, and there are four different kinds of bases in a nucleic acid altogether. How many different codons are possible?

Answer: (4) 64

Explanation: Since there are four different bases in nucleic acids (A, U, G, C), and codons are composed of three bases, the total number of possible codons is calculated as 43=644^3 = 6443=64.

#79. Identify the incorrect match among the following:

UUU – Phenylalanine

UAG – Sense codon

GUG – Valine

UGG – Tryptophan

Answer: (2) UAG – Sense codon

Explanation: UAG is a stop codon (nonsense codon), not a sense codon. Sense codons code for amino acids, while stop codons signal the termination of translation.

#80. Wobble helps to maintain the economy in the number of tRNA molecules because:

Genetic code is unambiguous and specific.

tRNA contains anticodons.

The first two bases in the codon are specific.

Genetic message is read in an open frame.

Answer: (3) The first two bases in the codon are specific.

Explanation: Wobble refers to the flexibility in base-pairing between the third base of the codon and the corresponding base in the tRNA anticodon, allowing multiple tRNAs to recognize a single codon. This reduces the number of tRNA molecules needed.

#81. Out of 64 codons, 61 codons code for 20 types of amino acids. This phenomenon is called:

Degeneracy of genetic code

Overlapping of genes

Universality of codons

Explanation: The degeneracy of the genetic code refers to the fact that multiple codons can code for the same amino acid. This redundancy allows for some variation without affecting protein synthesis.

#82. Which of the following is not true about the genetic code?

It is unambiguous.

A codon in mRNA is read in a non-contiguous fashion.

It is nearly universal.

it is degenerate.

Answer: (2) A codon in mRNA is read in a non-contiguous fashion.

Explanation: The genetic code is read in a contiguous fashion (one codon after another). The term “non-contiguous” would refer to a non-sequential reading, which does not apply to the genetic code.

#83. Which of the following codons are non-degenerate?

UUU and GUG

UGA and UAA

AUG and UGG

CAC and CAG

Answer: (3) AUG and UGG

Explanation: Non-degenerate codons are those that code for a single amino acid. AUG codes for methionine, and UGG codes for tryptophan, and each is specific to only one amino acid.

#84. A translational unit in mRNA includes the following: (a) Start codon (b) Stop codon (c) Untranslated regions (UTRs) (d) Coding sequence located between the start codon and termination codon Which of the following options correctly describes the components of the translational unit?

(a) and (c)

(a), (b), and (c)

(b), (c), and (d)

(a), (b), and (d)

Answer: (4) (a), (b), and (d)

Explanation: A translational unit includes the start codon, stop codon, and the coding sequence. The untranslated regions (UTRs) are part of the mRNA but are not considered part of the coding sequence.

#85. Given below are two statements (A and B) with blanks. Select the option that correctly fills the blanks in the statements. Statement (a): The ribosome consists of structural (i) and about 80 different (ii). In its inactive state, it exists as (iii) subunits. There are two sites in the large subunit for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a (iv) bond. The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme (v)_) for the formation of peptide bonds. Statement (b): A translational unit in (i)_ is the sequence of RNA that is flanked by the start codon (AUG) and the stop codon and codes for a polypeptide. An mRNA also has some additional sequences that are not translated and are referred to as untranslated regions (UTR). The UTRs are present at both the 5′ end (before the start codon) and the 3′ end (after the stop codon). They are required for efficient (ii)_ process.

(a) – (i) RNAs, (ii) Proteins, (iii) two, (iv) peptide, (v) ribozyme; (b) – (i) mRNA, (ii) translation

(a) – (i) Proteins, (ii) RNAs, (iii) two, (iv) peptide, (v) ribozyme; (b) – (i) mRNA, (ii) translation

(a) – (i) RNAs, (ii) Proteins, (iii) two, (iv) peptide, (v) ribozyme; (b) – (i) mRNA, (ii) transcription

(a) – (i) RNAs, (ii) Proteins, (iii) two, (iv) peptide, (v) cytozyme; (b) – (i) mRNA, (ii) translation

Answer: (1) (a) – (i) RNAs, (ii) Proteins, (iii) two, (iv) peptide, (v) ribozyme; (b) – (i) mRNA, (ii) translation

Explanation: In the ribosome, the structure consists of RNAs and proteins, and it operates in two subunits. It catalyzes peptide bond formation with the help of rRNA, acting as a ribozyme. The translational unit in mRNA includes both coding and untranslated regions essential for translation.

#86. Listed below are the stages in the cellular synthesis of a protein. Arrange these stages in the correct sequence: (a) Movement of mRNA from the nucleus to the cytoplasm (b) Linking of adjacent amino acid molecules (c) Transcription of mRNA from a DNA template (d) Formation of the polypeptide chain (e) Attachment of the mRNA strand to a ribosome

(a) → (c) → (b) → (e) → (d)

(a) → (e) → (c) → (d) → (b)

Answer: (3) (c) → (a) → (e) → (b) → (d)

Explanation: First, mRNA is transcribed from the DNA (c). Then it moves from the nucleus to the cytoplasm (a) and attaches to a ribosome (e). Next, amino acids are linked together (b) to form the polypeptide chain (d).

#87. Transcription and translation of a gene composed of 30 nucleotides would form a protein containing not more than ________ amino acids.

Answer: (1) 10

Explanation: Each codon is made up of 3 nucleotides and codes for one amino acid. So, 30 nucleotides will form 10 codons, which can produce a protein with a maximum of 10 amino acids.

#88. Protein synthesis in an animal cell occurs:

On ribosomes present in cytoplasm as well as in mitochondria.

On ribosomes present in the nucleolus as well as in cytoplasm.

Only on ribosomes attached to the nuclear envelope and endoplasmic reticulum.

Only on the ribosomes present in cytosol.

Answer: (1) On ribosomes present in cytoplasm as well as in mitochondria.

Explanation: Protein synthesis in animal cells occurs on ribosomes in the cytoplasm and also in mitochondria, as mitochondria have their own ribosomes to synthesize some of their own proteins.

#89. Which one of the following statements is true for protein synthesis?

Amino acids are directly recognized by mRNA.

The third base of the codon is less specific.

Only one codon codes for an amino acid.

Every tRNA molecule has more than one amino acid attachment site.

Answer: (2) The third base of the codon is less specific.

Explanation: This refers to the “wobble hypothesis,” where the third base of a codon is less critical in determining which amino acid is added, allowing some flexibility in pairing.

#90. The gene responsible for the synthesis of a polypeptide chain is called:

Promoter gene

Structural gene

Regulator gene

Operator gene

Answer: (2) Structural gene

Explanation: Structural genes encode the information needed to build proteins, including the polypeptide chains. The promoter gene is involved in initiating transcription, while regulator genes control gene expression.

#91. Choose the wrong statement in the process of protein synthesis.

After uncoiling of the DNA molecule, one strand acts as a template for the formation of mRNA.

In the presence of DNA polymerase enzyme, the mRNA is formed based on the triplet codes.

The mRNA that leaves the nucleus reaches the cytoplasm and gets attached to the ribosomal subunit.

The amino acids are transferred from the intracellular amino acid pool to the active ribosomes by the tRNA.

Answer: (2) In the presence of DNA polymerase enzyme, the mRNA is formed based on the triplet codes.

Explanation: mRNA formation occurs during transcription, which is catalyzed by RNA polymerase, not DNA polymerase. DNA polymerase is involved in DNA replication.

#92. Select the correct statement regarding protein synthesis.

When the small subunit of the ribosome encounters an mRNA, the process of translation begins.

Peptidase catalyses the formation of peptide bonds.

UTRs are present between the start codon and stop codon.

At the end of translation, the release factor binds to the initiation codon.

Answer: (1) When the small subunit of the ribosome encounters an mRNA, the process of translation begins.

Explanation: The small ribosomal subunit binds to mRNA and initiates translation at the start codon (AUG). Peptidyl transferase, not peptidase, catalyzes peptide bond formation.

#93. The fifth charged tRNA attaches itself to __________ of the ribosome in translation.

Shine-Dalgarno (SD) sequence

P site

A site

Anti SD sequence

Answer: (3) A site

Explanation: During translation, the incoming tRNA with an amino acid binds to the A site of the ribosome, where codon-anticodon pairing occurs.

#94. The UTRs are located:

Between the start and stop codons of a translational unit

Before the start codon and after the stop codon of a translational unit

Between the promoter and terminator of a transcriptional unit

Before the promoter and after the terminator of a transcriptional unit

Answer: (2) Before the start codon and after the stop codon of a translational unit

Explanation: The untranslated regions (UTRs) are located at the 5′ end (before the start codon) and the 3′ end (after the stop codon) of mRNA. These regions are important for the regulation of translation.

#95. Choose the correct statements about protein synthesis: (a) The binding of mRNA to the ribosome is loose and reversible. (b) The ribosome, composed of tRNA and proteins, consists of a small and large subunit. (c) Each of the 20 amino acids has at least one specific tRNA molecule. (d) tRNA-Met binds to the first codon (AUG); this is called the START codon. (e) The second tRNA binds to the second codon so that their attached amino acids are positioned next to each other. (f) The newly formed polypeptides may undergo post-transcriptional modifications.

(a), (c), and (d)

(b), (d), and (f)

(b), (e), and (f)

(a), (e), and (f)

Answer: (2) (b), (d), and (f)

Explanation:

Statement (b) is correct as the ribosome consists of two subunits.

Statement (d) is correct as the first codon is always AUG (START codon) which binds tRNA-Met.

Statement (f) is true because newly synthesized polypeptides may undergo modifications such as phosphorylation or glycosylation.

#96. Which of the following statements about mutations is incorrect?

Deletion and insertion of base pairs cause frame-shift mutations.

Cancer cells commonly show chromosomal aberrations.

UV and gamma rays are mutagens.

A change in a single base pair of DNA does not cause mutation.

Answer: (4) A change in a single base pair of DNA does not cause mutation.

Explanation:
A change in a single base pair can lead to a mutation, such as in point mutations or missense mutations. Therefore, statement (4) is incorrect.

#97. In the given mRNA strand, 5′ – AUG GUG CUA ACA UUC GUC UAG – 3′ If the 15th nitrogenous base gets mutated to G, which type of mutation could be observed?

Point mutation

Missense mutation

Transversion

More than one option is correct

Answer: (4) More than one option is correct

Explanation:
A mutation in a single nitrogenous base is a point mutation, which could result in a missense mutation if the codon now codes for a different amino acid. If the mutation involves changing a purine to a pyrimidine or vice versa, it would be a transversion.

#98. Imagine an error during DNA replication where an A is replaced by a C in a gene. What effect will this likely have on the cell?

The amino acid sequence will be completely changed.

An amino acid will be missing.

An incorrect amino acid will be incorporated.

An additional amino acid will appear.

Answer: (3) An incorrect amino acid will be incorporated.

Explanation:
A substitution of one base for another could lead to an incorrect amino acid being incorporated into the polypeptide, depending on the codon that is changed.

#99. If the 25th codon (UAU) in a gene encoding a polypeptide of 50 amino acids is mutated to UAA, what will happen?

A polypeptide of 24 amino acids will be formed.

wo polypeptides of 24 and 25 amino acids will be formed.

A polypeptide of 49 amino acids will be formed.

A polypeptide of 25 amino acids will be formed.

Answer: (1) A polypeptide of 24 amino acids will be formed.

Explanation:
UAA is a stop codon, so the polypeptide synthesis will terminate prematurely at the 25th position, leading to a polypeptide of 24 amino acids.

#100. In the given DNA segment, ATG ACC AGG ACC CCA ACA, if the first base is mutated, how will it affect the coding of this DNA segment?

Complete change in the type and sequence of amino acids.

Change in the first amino acid only.

No change in the sequence.

One amino acid less in the protein.

Answer: (1) Complete change in the type and sequence of amino acids.

Explanation:
A mutation in the first base of a codon could change the entire sequence of amino acids, as the codon specifies which amino acid is added to the growing polypeptide chain.

#101. Imagine an error during DNA replication where a T is replaced by a G in a gene. What effect will this likely have on the cell?

The amino acid sequence of one of its proteins will be completely changed.

An amino acid will be missing from one of its proteins.

One of its proteins might contain an incorrect amino acid.

An amino acid will be missing from one of its proteins.

Answer: (3) One of its proteins might contain an incorrect amino acid.

Explanation:
A substitution of one base for another can lead to a change in the codon, potentially resulting in an incorrect amino acid being incorporated into the protein.

#102. What are mutations?

Heritable changes in the sequence of DNA bases that produce an observable phenotype.

Heritable changes in the sequence of DNA bases.

Mistakes in the incorporation of amino acids into proteins.

Heritable changes in the mRNA of an organism.

Answer: (1) Heritable changes in the sequence of DNA bases that produce an observable phenotype.

Explanation:
Mutations are changes in the DNA sequence that can affect an organism’s phenotype. Some mutations can be inherited, while others may not be expressed phenotypically.

#103. Which of the following mutations are required for evolution to occur?

Frame-shift mutations

Missense mutations

Non-sense mutations

Silent mutations

Answer: (2) Missense mutations

Explanation:
Missense mutations, which result in a change to a single amino acid, can introduce variation in protein function and structure, which may contribute to evolutionary processes.

#104. E. coli cells with a mutated z gene of the lac operon cannot grow in a medium containing only lactose as the energy source because:

They cannot synthesize functional β-galactosidase.

They cannot transport lactose from the medium into the cell.

The lac operon is constitutively active in these cells.

In the presence of glucose, E. coli cells do not utilize lactose.

Answer: (2) They cannot transport lactose from the medium into the cell.

Explanation:
The z gene in the lac operon codes for the enzyme permease, which is involved in the transport of lactose into the cell. Without this protein, E. coli cannot utilize lactose as an energy source.

#105. In the regulation of gene expression in prokaryotes, which of the following statements is correct? (a) Lactose acts as the suppressor for gene expression. (b) Tryptophan acts as the inducer for gene expression. (c) The regulator gene produces the repressor molecule.

.Only (a) is correct

Only (b) is correct

Only (c) is correct

Both (a) and (b) are correct

Answer: (3) Only (c) is correct

Explanation:
In prokaryotes, the regulator gene produces a repressor molecule that controls the expression of operons, such as the lac and trp operons. Lactose is an inducer, and tryptophan is a corepressor, not an inducer.

#106. In eukaryotes, which of the following statements is correct?

Different structural genes controlling the same metabolic process require one regulator.

Different structural genes controlling the same metabolic process may be present together on the same chromosome or on different chromosomes.

Mostly one metabolic process is controlled by one operon.

All of these.

Answer: (2) Different structural genes controlling the same metabolic process may be present together on the same chromosome or on different chromosomes.

Explanation:
In eukaryotes, multiple structural genes involved in the same metabolic process may be located on the same chromosome or across different chromosomes, unlike in prokaryotes where operons regulate related genes together.

#107. β-galactosidase in E. coli is synthesized to catalyze the hydrolysis of _ into __ and glucose.

Galactose; lactose

Galactose; glucose

Lactose; galactose

Maltose; galactose

Answer: (3) Lactose; galactose

Explanation:
β-galactosidase breaks down lactose into two monosaccharides: galactose and glucose, as part of the metabolic process in E. coli.

#108. In bacteria, the structural genes can be turned off when:

The end product of a reaction combines with the repressor protein and activates it.

The substrate combines with the repressor protein and inactivates it.

The substrate combines with the promoter.

The substrate combines with its enzyme and produces a repressor molecule.

Answer: (1) The end product of a reaction combines with the repressor protein and activates it.

Explanation:
In negative gene regulation, when the end product (e.g., tryptophan) accumulates, it binds to the repressor protein, activating it, which then binds to the operator and blocks transcription.

#109. Which enzyme/s will be produced in a cell with a nonsense mutation in the lac Y gene?

Lactose permease

Transacetylase

Lactose permease and transacetylase

β-galactosidase

Answer: (1) Lactose permease

Explanation:
A nonsense mutation in the lacY gene would produce a truncated protein, preventing the synthesis of functional lactose permease. However, transacetylase and β-galactosidase would still be produced as they are encoded by other genes in the operon.

#110. Select the correct pair of statements about the tryptophan operon: (a) In the presence of tryptophan, it is switched on. (b) Five structural genes are present in it. (c) Its regulatory gene is represented by i. (d) It is a repressible operon.

(b) and (d)

(a) and (b)

(b) and (c)

(a) and (c)

Answer: (1) (b) and (d)

Explanation:
The tryptophan operon is a repressible operon with five structural genes (b). It is repressed when tryptophan is present, making it a repressible operon (d).

#111. Which of the following is a wrong statement?

VNTR belongs to a class of mini-satellite DNA.

DNA sequencers work on the principle developed by Frederick Sanger.

HGP was coordinated by the US Department of Health and the National Institute of Energy.

Satellite DNA normally does not code for proteins.

Answer: (3) HGP was coordinated by the US Department of Health and the National Institute of Energy.

Explanation:
The Human Genome Project (HGP) was coordinated by the National Institutes of Health (NIH) and the Department of Energy (DOE) but not specifically by the National Institute of Energy as stated in (3).

#112. Sequencing the whole genome of an organism into coding and non-coding sequences is termed as:

Sequence determination

Sequence detection

Sequence annotation

Sequence completion

Answer: (3) Sequence annotation

Explanation:
Sequence annotation refers to the process of identifying and describing the functional elements within the genomic sequence, including coding and non-coding regions.

#113. Which of the following is not a salient feature of the Human Genome Project?

The human genome contains 3614.7 million nucleotide bases.

Less than 2% of the genome codes for proteins.

Over 50% of the genes discovered are yet to be investigated regarding their functions.

Chromosome Y has the fewest genes.

Answer: (4) Chromosome Y has the fewest genes.

Explanation:
Chromosome Y has fewer genes compared to other chromosomes, but it is not the one with the fewest. Chromosome 21 has the fewest genes in the human genome.

#114. In humans, which of the following statements is correct? (a) Non-coding DNA is the most abundant. (b) The function of more than 50% of discovered genes is unknown. (c) Less than 2% of the genome codes for protein. (d) The total number of genes is 30,000.

(a) and (c)

(a), (b), (c), and (d)

(b), (c), and (d)

(a), (b), and (c)

Answer: (4) (a), (b), and (c)

Explanation:
In the human genome, non-coding DNA is the most abundant, and over 50% of the discovered genes have unknown functions. Less than 2% of the genome codes for proteins.

#115. Which of the following statements is correct about the aim of the Human Genome Project?

To develop ways of mapping the human genome with increasing fine-level precision.

To store this information in databases and develop tools for data analysis.

To address the ethical, legal, and social issues that may arise from this project.

All of these.

Answer: (4) All of these.

Explanation:
The Human Genome Project aimed to map the entire human genome, store this data, and analyze it, while also addressing ethical, legal, and social implications.

#116. The last human chromosome that was completely sequenced in May 2006 is:

Chromosome 22

Chromosome 14

Chromosome 1

Chromosomes X and Y

Answer: (3) Chromosome 1

Explanation:
Chromosome 1, the largest human chromosome, was the last to be fully sequenced as part of the Human Genome Project in May 2006.

#117. In sequence annotation, which of the following is involved?

All expressed genes are identified.

Sequences that occur only once in the genome are tagged.

All coding and non-coding sequences are involved.

Sequencing of genes with exact locations and the order of known bases.

Answer: (3) All coding and non-coding sequences are involved.

Explanation:
Sequence annotation involves identifying both coding and non-coding regions, including genes, regulatory elements, and repetitive sequences.

#118. Lily has 18 times more DNA than the human genome, yet it produces fewer proteins than humans because its DNA has:

Introns only.

More introns and fewer exons.

Exons only.

More exons and fewer introns.

Answer: (2) More introns and fewer exons.

Explanation:
Lily’s genome contains a higher proportion of introns (non-coding regions) compared to humans, leading to fewer proteins being produced despite the larger DNA content.

#119. Satellite DNA Which of the following statements are correct about Satellite DNA? (a) It is classified into categories like micro satellites, mini satellites, etc., based on base composition, length of segments, and the number of repetitive units. (b) It generally does not code for any protein. (c) It shows polymorphism. (d) It forms the basis of DNA fingerprinting. Select the correct option:

(a) and (c)

(a), (b), (c), and (d)

(b), (c), and (d)

(a), (b), and (c)

Answer: (2) (a), (b), (c), and (d)
Explanation: Satellite DNA is non-coding, shows variability, and forms the basis of DNA fingerprinting due to the repetitive nature of the DNA segments.

#120. An RFLP is a

DNA probe used for hybridization.

Variation of a DNA segment cut by restriction enzyme(s).

Recessive form of a deleterious allele.

Restriction enzyme used to cut DNA.

Answer: (2) Variation of a DNA segment cut by restriction enzyme(s).
Explanation: RFLP (Restriction Fragment Length Polymorphism) refers to differences in DNA fragment sizes after being cut by restriction enzymes.

#121. DNA fingerprinting works because

Genes containing the same alleles make it simple to compare different individuals.

PCR allows amplification of proteins from single cells.

There are multiple alleles for some DNA sequences, making it possible to obtain unique patterns for each individual.

DNA in the skin cells is very diverse.

Answer: (3) There are multiple alleles for some DNA sequences, making it possible to obtain unique patterns for each individual.
Explanation: DNA fingerprinting relies on the variability of satellite DNA sequences between individuals.

#122. When bands of RNA are transferred to a nitrocellulose membrane for identification, the blotting is called

Southern Blotting.

Northern Blotting.

Western Blotting.

Eastern Blotting.

Answer: (2) Northern Blotting.
Explanation: Northern blotting is used for detecting RNA, while Southern blotting is for DNA.

#123. In PCR, it is _____ that creates single-stranded template molecules.

Heat

High salt concentration.

DNA polymerase.

Exonuclease.

Answer: (1) Heat.
Explanation: Heat is used in PCR to denature the DNA, creating single-stranded template molecules.

#124. In gel electrophoresis, all the DNA fragments move towards the base of the gel with variable velocities because

DNA is positively charged.

DNA is slightly negatively charged.

DNA fragments have different molecular weights.

Both (2) and (3)

Answer: (4) Both (2) and (3).
Explanation: DNA is negatively charged and moves towards the positive electrode, with smaller fragments traveling faster than larger ones.

#125. Western Blot hybridization is used for

DNA analysis.

RNA analysis.

Protein analysis.

Polysaccharide analysis

Answer: (3) Protein analysis.
Explanation: Western blotting is used to detect specific proteins, typically after electrophoresis.

#126. The polymerase chain reaction (PCR)

) Is a method for sequencing DNA.

Is used to transcribe specific genes.

Amplifies specific DNA sequences.

Does not require DNA replication primers.

Answer: (3) Amplifies specific DNA sequences.
Explanation: PCR is a technique used to amplify specific DNA sequences for analysis.

#127. Choose the correct statements with respect to satellite DNA.

Shows high degree of polymorphism.

Forms the basis of DNA fingerprinting.

Part of both coding and non-coding sequences.

More than one option is correct.

Answer: (4) More than one option is correct.
Explanation: Satellite DNA is highly polymorphic, non-coding, and is key in DNA fingerprinting.

#128. Repetitive sequences are stretches of DNA with repeated bases in a genome, but (a) These sequences have no transcriptional function. (b) These are associated with the euchromatin region. (c) These help to identify a person based on DNA specificity. Select the correct option:

All are correct.

Only (b) is incorrect.

Both (a) and (b) are correct.

Both (b) and (c) are incorrect.

Answer: (3) Both (a) and (b) are correct.
Explanation: Repetitive sequences are often non-transcribing, found in heterochromatin regions, and assist in DNA profiling.

#129. The microsatellites have simple sequences of repeated

11–6 bp

1–6 bp

10 bp

50 bp

Answer: (2) 1–6 bp
Explanation: Microsatellites are short, repetitive sequences typically 1–6 base pairs long.

#130. What forms the basis of DNA fingerprinting?

The relative proportions of purines and pyrimidines in DNA.

The relative difference in DNA occurrence in blood, skin, and saliva.

The relative amount of DNA in the ridges and grooves of fingerprints.

Satellite DNA occurring as highly repeated short DNA segments.

Answer: (4) Satellite DNA occurring as highly repeated short DNA segments.
Explanation: DNA fingerprinting relies on highly repetitive satellite DNA sequences.

#131. The VNTR belongs to a class of satellite DNA referred to as

Mini-satellite.

NOR.

Selectable marker.

Polymorphic DNA.

Answer: (1) Mini-satellite.
Explanation: VNTRs (Variable Number Tandem Repeats) are a class of mini-satellite DNA.

#132. Identify the wrong statement from the following.

Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acids found in living systems.

RNA, though it also acts as a genetic material in some viruses, mostly functions as a messenger.

DNA functions as adapter, structural, and in some cases as a catalytic molecule.

The determination of the complete nucleotide sequence of the human genome during the last decade has set in a new era of genomics.

Answer: (3) DNA functions as adapter, structural, and in some cases as a catalytic molecule.
Explanation: DNA is primarily genetic material, not typically involved in catalytic or structural roles like RNA.

#133. Find the incorrect statement.

N-glycosidic linkage between nitrogenous base and sugar forms a nucleotide.

5-methyl uracil is present in DNA.

In RNA, every nucleotide residue has an additional –OH group.

The ratio of (A+T)/(G+C) is variable.

Answer: (2) 5-methyl uracil is present in DNA.
Explanation:

5-methyl uracil is present in RNA as thymine, but in DNA, thymine has a methyl group attached to the uracil structure.
Other options are correct as per molecular biology basics.

#134. In recent years, DNA sequences (nucleotide sequences) of mtDNA and Y chromosomes were considered for the study of human evolution, because

they are small, and therefore, easy to study.

they are uniparental in origin and do not take part in recombination.

their structure is known in great detail.

they can be studied from the samples of fossil remains.

Answer: (2) they are uniparental in origin and do not take part in recombination.
Explanation:

mtDNA and Y chromosomes are inherited uniparentally, which makes them ideal for tracing lineage without the complications of recombination.

#135. Consider the following statements: (a) rRNA provides the template for the synthesis of proteins. (b) tRNA brings amino acids and reads the genetic code. (c) RNA polymerase binds to the promoter and initiates transcription. (d) A segment of DNA coding for a polypeptide is called an intron.

(a) and (b) are correct

(b) and (d) are correct

(a), (b), and (c) are correct

(b) and (c) are correct

Answer: (4) (b) and (c) are correct

Explanation:

Statements (b) and (c) are correct because tRNA is involved in bringing amino acids and decoding the mRNA, and RNA polymerase initiates transcription by binding to the promoter.
Statements (a) and (d) are incorrect, as rRNA does not serve as a template for protein synthesis, and an intron is not involved in coding for polypeptides.

#136. Which is the true statement? (a) DNA is chemically less reactive and structurally more stable than RNA. (b) Viruses having RNA genomes evolve faster. (c) A molecule that can act as genetic material should be stable chemically and structurally. (d) For the transmission of genetic information, RNA is better.

(a) and (b)

(a) and (c)

(b) and (d)

(a), (b), and (c)

Answer: (1) (a) and (b)

Explanation:

Statements (a) and (b) are correct because DNA is chemically less reactive and structurally more stable than RNA, and viruses with RNA genomes evolve faster due to higher mutation rates.
Statement (c) is true but is not relevant to the context of the question, which focuses on evolution and transmission.
Statement (d) is incorrect because DNA, not RNA, is better suited for transmitting genetic information due to its stability.

#137. Read the following statements and choose the correct option. (a) RNA polymerase associates transiently with the ‘Rho’ factor to initiate transcription. (b) In bacteria, transcription and translation take place in the same compartment. (c) Nirenberg developed a method for protein synthesis in cells. (d) When hnRNA undergoes the capping process, adenylate residues are added at the 3′ end in a template-independent manner. (e) hnRNA is the precursor of mRNA.

(a), (b), and (d)

(b), (c), and (e)

(a), (c), and (e)

(b), (d), and (e)

Answer: (2) (b), (c), and (e)

Explanation:

Statements (b), (c), and (e) are correct:

In bacteria, transcription and translation occur in the same compartment due to the lack of a nucleus.
Nirenberg developed methods to study protein synthesis in cells.
hnRNA is the precursor to mRNA, processed through capping, splicing, and polyadenylation.

Statement (a) is incorrect because RNA polymerase associates transiently with the ‘Rho’ factor for termination of transcription, not initiation.
Statement (d) is incorrect because during capping, a guanine cap is added at the 5′ end, not adenylate residues at the 3′ end.

#138. Find the wrongly matched pair

HarGobind Khorana – Synthesized RNA molecules

George Gamow – Codon is triplet

Maximum number of genes – Chromosome no. 21

Alec Jeffreys – DNA fingerprinting

Answer: (3) Maximum number of genes – Chromosome no. 21
Explanation:

Chromosome 21 has about 200-300 genes, not the maximum number. Chromosome 1 has the most genes.

#139. The enzyme that is responsible for the transcription of RNA from a DNA template is:

DNA polymerase

RNA polymerase

Reverse transcriptase

Ligase

Answer: (2) RNA polymerase
Explanation:

RNA polymerase is responsible for synthesizing RNA from a DNA template during the process of transcription.

#140. Which of the following is not a feature of the genetic code?

It is universal.

It is degenerate.

It is overlapping.

It is non-ambiguous.

Answer: (3) It is overlapping.
Explanation:

The genetic code is not overlapping; each nucleotide is part of only one codon. The genetic code is universal, degenerate, and non-ambiguous, meaning a specific codon corresponds to a specific amino acid.

#141. In a double-stranded DNA, if the percentage of adenine is 30%, what will be the percentage of guanine?

30%

40%

20%

50%

Answer: (3) 20%
Explanation:

According to Chargaff’s rules, adenine pairs with thymine, and guanine pairs with cytosine. If adenine is 30%, thymine will also be 30%, leaving 40% for guanine and cytosine. Since guanine pairs with cytosine in equal amounts, guanine will be 20%.

#142. The central dogma of molecular biology describes the flow of genetic information. Which of the following processes is involved in this flow?

Transcription

Translation

Replication

All of the above

Answer: (4) All of the above
Explanation:

The central dogma involves three key processes: DNA replication (for copying genetic material), transcription (for synthesizing RNA from DNA), and translation (for synthesizing proteins from RNA).

#143. Which of the following is true about transcription in eukaryotes?

Transcription occurs in the cytoplasm.

The newly formed RNA is a copy of the DNA coding strand.

RNA polymerase is the only enzyme involved in the process.

The mRNA produced undergoes modifications such as capping, polyadenylation, and splicing before leaving the nucleus.

Answer: (4) The mRNA produced undergoes modifications such as capping, polyadenylation, and splicing before leaving the nucleus.
Explanation:

Transcription in eukaryotes occurs in the nucleus. The mRNA is modified before leaving the nucleus to ensure stability and translation efficiency.

#144. The process of transcription is initiated when RNA polymerase binds to the:

Ribosome

Promoter region of DNA

Coding region of DNA

Terminator region of DNA

Answer: (2) Promoter region of DNA
Explanation:

RNA polymerase binds to the promoter region of DNA to initiate transcription.

#145. Which of the following processes directly results in the production of a protein?

Replication

Transcription

Translation

Splicing

Answer: (3) Translation
Explanation:

Translation is the process by which mRNA is decoded to produce a specific protein.

#146. Which of the following is a difference between prokaryotic and eukaryotic transcription?

In eukaryotes, RNA is synthesized in the nucleus, while in prokaryotes it occurs in the cytoplasm.

Eukaryotes require a primer to start transcription, while prokaryotes do not.

Eukaryotes have more types of RNA polymerases than prokaryotes.

All of the above.

Answer: (4) All of the above
Explanation:

All statements are correct: Eukaryotic transcription takes place in the nucleus, prokaryotes synthesize RNA in the cytoplasm, eukaryotes require multiple RNA polymerases, and eukaryotic transcription involves additional processes such as capping and splicing.

#147. Which of the following is/are true about RNA processing in eukaryotes?

Addition of a 5' cap

Splicing of introns

Addition of a poly-A tail at the 3′ end

All of the above

Answer: (4) All of the above
Explanation:

All the processes mentioned are part of RNA processing in eukaryotes, which includes capping, splicing, and polyadenylation to produce a mature mRNA molecule.

#148. Which of the following types of RNA is involved in protein synthesis?

mRNA

rRNA

tRNA

All of the above

Answer: (4) All of the above
Explanation:

mRNA carries the genetic code, rRNA forms the structural component of ribosomes, and tRNA brings amino acids to the ribosome for protein synthesis.

#149. he first codon of mRNA that codes for an amino acid is:

UGA

UAA

AUG

UGC

Answer: (3) AUG
Explanation:

The codon AUG serves as the start codon in mRNA, coding for the amino acid methionine in eukaryotes and formylmethionine in prokaryotes.

#150. Which of the following represents a sequence of bases in DNA that would code for the following mRNA sequence: 5′-AUG-GUA-CGC-UGG-3′?

3'-TAC-CAT-GCG-ACC-5'

3'-ATG-GTA-CGC-ACG-5'

5'-ATG-GTA-CGC-ACC-3'

3'-TAC-GTA-CGC-ACG-5'

Answer: (1) 3′-TAC-CAT-GCG-ACC-5′
Explanation:

The DNA sequence that would transcribe to the given mRNA sequence is the reverse complement of the mRNA.

#151. The process by which the mRNA code is used to assemble a protein is known as:

Replication

Transcription

Translation

Mutation

Answer: (3) Translation
Explanation:

Translation is the process by which the sequence of codons in mRNA is used to assemble amino acids into a polypeptide chain, forming a protein.

Finish

Results

#1. The histone protein that attaches to DNA strands between nucleosomes is _________.

#2. Calculate the number of nucleosomes that will attach to a small chromosome consisting of 500,000,000 nucleotides.

#3. Which of the following is not a salient feature of the double-helix structure of DNA?

#4. Calculate the base ratio for a DNA sample if 22% cytosine was found in it.

#5. A diploid cell of an organism has 8.4×10108.4 times 10^{10}8.4×1010 base pairs. The length of the total DNA could be

#6. Calculate the number of hydrogen bonds in DNA if out of 200 nitrogen bases 46 are cytosine molecules.

#7. Which of the following statements regarding R-strain of Streptococcus pneumoniae is incorrect?

#10. A phosphate group is linked to ____ (a) ____ of a nucleoside through ____ (b) ____ linkage to form a corresponding nucleotide.

#11. Identify the mismatched pair.

#12. Which enzyme, if present in a medium, would affect transformation in bacteria?

#13. What does “antiparallel strands” in DNA mean?

#14. Which of the following does not follow the central dogma of molecular biology?

#15. Calculate the length of a DNA sample if it has 240 base pairs (bp).

#16. Which of the following statements is not correct?

#17. Why do some strains of Streptococcus pneumoniae produce smooth shiny colonies (S)?

#18. . A molecule that can act as a genetic material must fulfill all criteria except

#19. A DNA molecule in which both strands have radioactive thymidine is allowed to duplicate in an environment containing non-radioactive thymidine. What will be the exact number of DNA molecules that contain radioactive thymidine after three duplications?

#20. DNA polymerase enzyme is required for the synthesis of:

#21. A DNA strand is directly involved in the synthesis of all the following except:

#22. The Hershey and Chase experiment was precisely aimed to:

#23. Which of the following holds true for Escherichia coli?

#24. What confers additional stability to the double-helix model of DNA apart from hydrogen bonds?

#25. The “beads-on-string” structure in chromatin is packaged to form ___________.

#26. In the Hershey and Chase experiment, 35S^{35}S35S was used to label proteins. Which amino acids were actually labeled by it?

#27. DNA is a preferred genetic material over RNA due to one of the following reasons:

#28. DNA (a) Acts as genetic material in all cellular organisms. (b) Was discovered by F. Miescher, who named it nuclein. (c) Is acidic in nature. (d) Cannot be digested by DNase.

#30. You analyzed DNA from a new virus. The base composition was 32% A, 17% C, 32% G, and 19% T. Which conclusion is most accurate?

#31. X-ray crystallography provides information about the ______ of DNA but is limited because of the ______ of DNA. The technique relies on ______ patterns.

#32. The nucleosome core is composed of which histones?

#33. Which sequence correctly shows increasing levels of DNA condensation?

#34. The number of phosphodiester bonds in double-stranded DNA molecules is:

#35. DNA packaging in eukaryotes:

#36. Which of the following features applies to bacterial DNA?

#37. Identify the incorrect statement about nucleotides and nucleic acids.

#38. Which bonds are involved in forming a nucleoside?

#39. Which is not a function of RNA?

#40. RNA is ______, making it ______. DNA evolved with ______ modifications for ______ stability.

#41. Which statements about DNA and RNA are true? (a) DNA is better for genetic information transmission. (b) RNA codes proteins and expresses characters easily. (c) RNA mutates faster. (d) Uracil increases RNA stability. (e) Genetic material should allow mutations.

#42. Which statements are correct? (a) Both DNA and RNA are able to mutate. (b) RNA being unstable, mutates at a faster rate. (c) RNA is catalytic and reactive. (d) Thymine in DNA adds stability compared to uracil in RNA.

#43. Identify the correct match. (a) Amount of mRNA – Number of ribosomes in a cell (b) RNA polymerase-III – Helps in synthesis of rRNA (c) Most stable RNA – mRNA (d) Catalytic RNA – Ribozyme

#44. Which of the following statements are correct? (a) Both DNA and RNA are able to mutate. (b) RNA being unstable, mutates at a faster rate. (c) RNA is also known to be catalytic, hence reactive. (d) The presence of thymine at the place of uracil confers additional stability to DNA.

#46. Which of the following statements is NOT correct with respect to the Meselson and Stahl experiment?

#47. Identify the incorrect statement about DNA replication.

#48. If a DNA molecule is continuously replicated in 15N^{15}N15N medium, what is the percentage of lighter DNA in the fourth generation?

#49. Which enzyme synthesizes DNA from an RNA template?

#50. The Okazaki fragments in DNA chain growth:

#51. Which of the following statements are correct? (a) DNA replication is semi-discontinuous. (b) Growth of the lagging strand occurs in fragments. (c) DNA polymerase III can take part in proofreading. (d) Taylor fed dividing cells of Vicia faba leaves with radioactive thymine.

#52. Which enzyme unwinds the DNA helix during replication?

#54. Which of the following is false regarding DNA replication?

#55. Why is DNA ligase essential during replication?

#56. Chargaff’s rule states that:

#57. Which statement about DNA replication is false?

#58. Which is true regarding introns?

#59. RNA synthesis is controlled by:

#60. The process of removing introns and joining exons in a defined order in a transcription unit is called

#61. Both strands of DNA are not copied during transcription because

#62. Which is the initial step in the mRNA maturation process?

#63. Methyl guanosine triphosphate is associated with

#64. Answer: (3) capping Explanation: Methyl guanosine triphosphate is added to the 5′ end of the nascent mRNA during the capping process. This cap helps in stability, translation initiation, and protection of the RNA molecule. Shape

#65. In bacteria, in the context of mRNA, which of the following is not true?

#66. Which of the following statements is true regarding introns?

#67. What is correct for bacterial transcription?

#68. In eukaryotes, RNA polymerase III catalyses the synthesis of

#69. Which of the following statements correctly describes the function of RNA polymerase during transcription?

#70. Of the _______ different possible codons, _______ specify amino acids and ________ signal stop.

#71. Given the base sequence 5′ AAC CAA GAC CCC 3′ in mRNA, how many codons will there be in the case of overlapping and non-overlapping codons, respectively?

#72. Assume an organism has 6 types of nitrogenous bases on mRNA and 348 types of amino acids. How many nitrogenous bases are present in a codon?

#73. What is the function of a nonsense codon?

#74. The degeneration of a genetic code is attributed to the:

#75. Khorana synthesized two RNAs (a) with a repeat sequence of AB and (b) with a repeat sequence of ABC. The polypeptides coded by (a) and (b) are respectively:

#76. Select the incorrect statement(s): (a) Six codons do not code for any amino acid. (b) Codons are read in mRNA in a contiguous fashion. (c) Three codons function as stop codons. (d) The initiator codon AUG codes for methionine.

#77. The genetic code translates the language of:

#78. A codon consists of three bases, and there are four different kinds of bases in a nucleic acid altogether. How many different codons are possible?

#79. Identify the incorrect match among the following:

#80. Wobble helps to maintain the economy in the number of tRNA molecules because:

#81. Out of 64 codons, 61 codons code for 20 types of amino acids. This phenomenon is called:

#82. Which of the following is not true about the genetic code?

#83. Which of the following codons are non-degenerate?

#84. A translational unit in mRNA includes the following: (a) Start codon (b) Stop codon (c) Untranslated regions (UTRs) (d) Coding sequence located between the start codon and termination codon Which of the following options correctly describes the components of the translational unit?

#10. A phosphate group is linked to (a) of a nucleoside through (b) linkage to form a corresponding nucleotide.

#31. X-ray crystallography provides information about the of DNA but is limited because of the of DNA. The technique relies on __ patterns.

#40. RNA is , making it . DNA evolved with modifications for stability.

#70. Of the ___ different possible codons, _ specify amino acids and ______ signal stop.

#107. β-galactosidase in E. coli is synthesized to catalyze the hydrolysis of _ into __ and glucose.