DNA Structure & Types

Q1. DNA is composed of:
A. Deoxyribose, phosphate, nitrogenous base
B. Ribose, phosphate, nitrogenous base
C. Amino acids
D. Fatty acids

Q2. Chargaff’s rule states:
A. A = T, G = C
B. A + G = T + C
C. Both A and B
D. None

Q3. Watson and Crick proposed:
A. Triple helix DNA
B. Double helix DNA
C. Single-stranded DNA
D. Circular DNA only

Q4. DNA in prokaryotes is:
A. Linear
B. Circular
C. Helical RNA
D. Protein

Q5. Mitochondrial DNA is:
A. Linear
B. Circular
C. Double-stranded RNA
D. Single-stranded RNA

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RNA Structure & Types

Q6. tRNA carries:
A. Amino acids to ribosomes
B. Genetic information from DNA
C. rRNA synthesis
D. None

Q7. mRNA is:
A. Messenger RNA
B. Ribosomal RNA
C. Transfer RNA
D. None

Q8. rRNA forms:
A. Amino acids
B. Ribosome structure
C. Genetic code
D. tRNA anticodons

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DNA Replication

Q9. DNA replication is:
A. Conservative
B. Semi-conservative
C. Dispersive
D. None

Q10. Enzyme that unwinds DNA helix:
A. DNA polymerase
B. Helicase
C. Ligase
D. Primase

Q11. Leading strand is synthesized:
A. 3′ → 5′
B. 5′ → 3′
C. Both directions
D. Random

Q12. Okazaki fragments occur on:
A. Leading strand
B. Lagging strand
C. Both strands
D. None

Q13. DNA polymerase requires:
A. Primer
B. Template
C. dNTPs
D. All of the above

Q14. Replication fork is:
A. Site of transcription
B. Site of translation
C. Site where DNA unwinds for replication
D. None

Q15. Telomeres protect:
A. DNA ends from degradation
B. Genes from mutation
C. Proteins from misfolding
D. RNA from degradation

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Transcription

Q16. RNA polymerase synthesizes:
A. DNA
B. RNA
C. Proteins
D. Lipids

Q17. Template strand is:
A. 3′ → 5′ used for RNA synthesis
B. 5′ → 3′ used for RNA synthesis
C. Not used
D. Double-stranded

Q18. Transcription starts at:
A. Promoter
B. Terminator
C. Start codon
D. Stop codon

Q19. In eukaryotes, primary RNA transcript undergoes:
A. Capping
B. Polyadenylation
C. Splicing
D. All of the above

Q20. mRNA carries:
A. Genetic code for protein
B. Amino acids
C. Ribosome structure
D. tRNA anticodons

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Translation

Q21. Translation occurs at:
A. Nucleus
B. Ribosome
C. Mitochondria
D. Golgi apparatus

Q22. Start codon is:
A. UAA
B. AUG
C. UAG
D. UGA

Q23. Stop codons are:
A. AUG
B. UAA, UAG, UGA
C. AAA, GGG
D. None

Q24. tRNA contains:
A. Anticodon
B. Codon
C. Amino acid at 5′ end
D. Ribosome binding site

Q25. Peptide bond forms between:
A. Carboxyl and amino groups of adjacent amino acids
B. Ribose and phosphate
C. Base pairs
D. tRNA molecules

Q26. Translation ends when:
A. Ribosome reaches stop codon
B. DNA polymerase stops
C. RNA splicing occurs
D. None

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Genetic Code

Q27. Genetic code is:
A. Universal
B. Degenerate
C. Non-overlapping
D. All of the above

Q28. Each codon codes for:
A. One amino acid
B. One nucleotide
C. One tRNA
D. None

Q29. Number of codons:
A. 60
B. 64
C. 20
D. 3

Q30. AUG codes for:
A. Methionine
B. Tryptophan
C. Stop
D. Phenylalanine

Q31. Degeneracy of genetic code:
A. Single codon per amino acid
B. Multiple codons for same amino acid
C. All codons identical
D. None

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Gene Regulation (Prokaryotes & Eukaryotes)

Q32. Operon model proposed by:
A. Watson & Crick
B. Jacob & Monod
C. Mendel
D. Hershey & Chase

Q33. Lac operon is:
A. Repressible
B. Inducible
C. Constitutive
D. None

Q34. Trp operon is:
A. Repressible
B. Inducible
C. Constitutive
D. None

Q35. In prokaryotes, transcription and translation:
A. Occur simultaneously
B. Occur separately
C. Not related
D. None

Q36. Enhancers increase transcription in:
A. Prokaryotes
B. Eukaryotes
C. Both
D. None

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Mutation & Repair Mechanisms

Q37. Point mutation changes:
A. Entire gene
B. One nucleotide
C. Chromosome
D. Protein only

Q38. Sickle cell anemia is due to:
A. Frame shift mutation
B. Point mutation
C. Deletion
D. Inversion

Q39. UV light causes:
A. Thymine dimers
B. Frameshift mutation
C. RNA mutation
D. None

Q40. DNA repair mechanism excises incorrect bases:
A. Nucleotide excision repair
B. Base pairing
C. Replication
D. Translation

Q41. Mutagens include:
A. Chemicals
B. Radiation
C. Viruses
D. All of the above

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Recombinant DNA Technology

Q42. Enzyme that cuts DNA at specific sequences:
A. DNA polymerase
B. Restriction endonuclease
C. Ligase
D. Helicase

Q43. Vector used in cloning:
A. Plasmid
B. Ribosome
C. tRNA
D. mRNA

Q44. Gene amplification uses:
A. PCR
B. Transcription
C. Translation
D. Ligase only

Q45. DNA ligase function:
A. Cuts DNA
B. Joins DNA fragments
C. Synthesizes RNA
D. Synthesizes DNA

Q46. Recombinant DNA requires:
A. DNA fragment
B. Vector
C. Enzymes
D. All of the above

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Viruses as Genetic Material

Q47. Hershey-Chase experiment showed:
A. Protein is genetic material
B. DNA is genetic material
C. RNA is always genetic material
D. None

Q48. TMV (Tobacco Mosaic Virus) contains:
A. DNA
B. RNA
C. Both DNA & RNA
D. Protein only

Q49. Retrovirus contains:
A. DNA
B. RNA
C. Both
D. Protein

Q50. Reverse transcriptase synthesizes:
A. RNA from DNA
B. DNA from RNA
C. Protein from RNA
D. DNA from DNA

A. Standard / Conceptual MCQs (Remaining 20)

Advanced / Application-Based

Q51. Leading strand synthesized continuously because:
A. DNA polymerase moves 3′ → 5′
B. DNA polymerase moves 5′ → 3′
C. Primase synthesizes continuously
D. Ligase joins fragments

Q52. Lagging strand forms Okazaki fragments because:
A. DNA polymerase works 5′ → 3′ only
B. Helicase is slow
C. Primase synthesizes fragments
D. Ligase is inactive

Q53. In eukaryotes, DNA replication begins at:
A. Single origin
B. Multiple origins
C. Telomeres only
D. Promoter

Q54. Capping of mRNA occurs at:
A. 5′ end
B. 3′ end
C. Both ends
D. Internal nucleotides

Q55. Poly-A tail added to:
A. 5′ end
B. 3′ end
C. Both ends
D. tRNA

Q56. Spliceosome is composed of:
A. snRNA + proteins
B. DNA polymerase
C. Ribosome
D. tRNA

Q57. Genetic code is non-overlapping because:
A. Each nucleotide belongs to one codon only
B. Codons repeat
C. Amino acids overlap
D. None

Q58. Wobble hypothesis explains:
A. Redundancy of genetic code
B. One codon → multiple amino acids
C. Frameshift mutation
D. Transcription termination

Q59. Lac operon is turned on in presence of:
A. Glucose
B. Lactose
C. Tryptophan
D. Repressor

Q60. Trp operon is regulated by:
A. Inducer
B. Corepressor
C. Activator
D. Polymerase

Q61. PCR amplifies DNA using:
A. DNA polymerase, primers, nucleotides
B. Ligase only
C. Restriction enzymes
D. Ribosome

Q62. Plasmid vector carries:
A. Antibiotic resistance gene
B. Gene of interest
C. Origin of replication
D. All of the above

Q63. Retrovirus uses:
A. DNA polymerase
B. RNA polymerase
C. Reverse transcriptase
D. Helicase

Q64. Telomerase adds:
A. Random nucleotides
B. Repeats at chromosome ends
C. RNA primers
D. Proteins

Q65. Mutation affecting start codon results in:
A. Normal protein
B. Translation initiation failure
C. Silent mutation
D. Termination

Q66. Thymine dimer caused by:
A. Chemicals
B. UV radiation
C. X-rays
D. RNA polymerase

Q67. Nucleotide excision repair corrects:
A. Thymine dimers
B. Frameshift mutation
C. Point mutation only
D. Deletions

Q68. Recombinant DNA is formed by:
A. Cutting DNA & joining with vector
B. Replication only
C. Transcription
D. Translation

Q69. DNA ligase is essential because:
A. Joins Okazaki fragments
B. Synthesizes RNA
C. Replicates DNA
D. Splices mRNA

Q70. Hershey-Chase experiment used:
A. Radioactive phosphorus & sulfur
B. UV radiation
C. Agarose gel
D. None

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B. Assertion–Reason MCQs (15)

Q1.
Assertion: DNA replication is semi-conservative.
Reason: Each daughter DNA has one parental and one new strand.

Q2.
Assertion: Okazaki fragments occur only on lagging strand.
Reason: DNA polymerase can synthesize DNA 5′ → 3′ only.

Q3.
Assertion: Genetic code is degenerate.
Reason: Multiple codons code for same amino acid.

Q4.
Assertion: Lac operon is inducible.
Reason: Lactose binds to repressor, allowing transcription.

Q5.
Assertion: Trp operon is repressible.
Reason: Tryptophan acts as corepressor.

Q6.
Assertion: mRNA undergoes capping and polyadenylation in eukaryotes.
Reason: These modifications protect mRNA and aid translation.

Q7.
Assertion: Retroviruses synthesize DNA from RNA.
Reason: Reverse transcriptase converts RNA → DNA.

Q8.
Assertion: tRNA carries amino acids.
Reason: Anticodon of tRNA pairs with mRNA codon.

Q9.
Assertion: DNA repair is essential for genome stability.
Reason: Mutations can cause diseases.

Q10.
Assertion: Genetic code is non-overlapping.
Reason: Each nucleotide belongs to only one codon.

Q11.
Assertion: Restriction enzymes are used in recombinant DNA.
Reason: They cut DNA at specific sequences.

Q12.
Assertion: PCR can amplify specific DNA sequences.
Reason: Primers and DNA polymerase allow selective amplification.

Q13.
Assertion: Spliceosome removes introns.
Reason: snRNA and proteins carry out splicing.

Q14.
Assertion: Okazaki fragments require ligase.
Reason: Ligase joins fragments to form continuous DNA.

Q15.
Assertion: Telomerase maintains chromosome ends.
Reason: Prevents loss of genetic material during replication.

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C. Difficult / Calculation / Diagram-Based MCQs (30)

Q1. Probability of all three codons coding for methionine in mRNA with 3 AUG codons:
A. 1/64
B. 1/27
C. 1/8
D. 1/16

Q2. tRNA with anticodon 3′–AUG–5′ carries:
A. Methionine
B. Phenylalanine
C. Serine
D. Valine

Q3. DNA polymerase synthesizes 1000 nucleotides in 5 minutes. Rate = ?
A. 200 nt/min
B. 100 nt/min
C. 50 nt/min
D. 500 nt/min

Q4. In eukaryotic mRNA, number of nucleotides removed as introns = 30% of transcript. If transcript = 1000 nt, introns = ?
A. 300 nt
B. 700 nt
C. 500 nt
D. 100 nt

Q5. Probability of 2 complementary bases pairing (A–T or G–C) randomly in 2 nt = ?
A. 0.25
B. 0.5
C. 0.75
D. 1

Q6. Diagram-based: Identify Okazaki fragments on lagging strand of replication fork.

Q7. Number of possible codons = 64. Probability a codon codes for leucine (6 codons) = ?
A. 3/32
B. 1/6
C. 1/4
D. 3/16

Q8. PCR with 3 cycles produces how many DNA molecules from 1 template?
A. 8
B. 6
C. 4
D. 3

Q12. In a PCR experiment, starting with 1 DNA molecule, how many molecules will be present after 5 cycles?
A. 16
B. 32
C. 64
D. 128

Q13. EcoRI recognizes the sequence GAATTC. A 1 kb DNA fragment with 3 EcoRI sites is digested. How many fragments result?
A. 2
B. 3
C. 4
D. 5

Q14. A tRNA has anticodon 3′–GAC–5′. The mRNA codon it binds to is:
A. 5′–CUG–3′
B. 5′–GAC–3′
C. 5′–CAG–3′
D. 5′–GUC–3′

Q15. AUG codon is the start codon. A mutation changes AUG → AUA. Consequence:
A. Protein synthesis initiates normally
B. Translation does not initiate
C. Frameshift mutation occurs
D. Silent mutation

Q16. In a DNA sequence, 25% of bases are adenine. What percentage of cytosine is present?
A. 25%
B. 50%
C. 10%
D. 20%

Q17. If 40% of nucleotides in DNA are thymine, then percentage of guanine = ?
A. 10%
B. 20%
C. 30%
D. 40%

Q18. In eukaryotic transcription, 1200 nt of pre-mRNA has 30% introns. Mature mRNA length = ?
A. 840 nt
B. 360 nt
C. 1200 nt
D. 400 nt

Q19. Probability that a random codon codes for a stop signal (UAA, UAG, UGA) = ?
A. 1/16
B. 1/20
C. 3/64
D. 3/61

Q20. In a replication fork diagram, identify:
A. Leading strand
B. Lagging strand
C. Okazaki fragments
D. All of the above

Q21. A plasmid vector of 5 kb carries a 1 kb gene insert. After restriction digestion with a single cutter, how many fragments will appear?
A. 1
B. 2
C. 3
D. 4

Q22. Reverse transcriptase synthesizes DNA from:
A. RNA template
B. DNA template
C. Protein template
D. Lipid template

Q23. Probability that 2 consecutive codons code for leucine (6 codons for leucine, total 64 codons) = ?
A. 36/4096
B. 1/64
C. 1/32
D. 1/128

Q24. Telomerase adds 6 nt repeats to chromosome ends. If 50 repeats are added, total nucleotides added = ?
A. 50
B. 100
C. 300
D. 600

Q25. A 900 bp DNA has 20% thymine. Total A + T pairs = ?
A. 180
B. 360
C. 400
D. 450

Q26. A mutation changes a codon from UUU → UUC. Effect:
A. Missense mutation
B. Nonsense mutation
C. Silent mutation
D. Frameshift mutation

Q27. A gene of 300 codons undergoes deletion of 1 nucleotide. Effect:
A. Silent mutation
B. Nonsense mutation
C. Frameshift mutation
D. Missense mutation

Q28. In a ribosome diagram, identify:
A. Large subunit
B. Small subunit
C. tRNA binding sites (A, P, E)
D. All of the above

Q29. Probability of a random codon NOT coding for methionine (AUG) = ?
A. 63/64
B. 1/64
C. 1/3
D. 60/64

Q30. A recombinant plasmid is made by ligating a 1 kb gene into a 3 kb plasmid. Total plasmid size = ?
A. 1 kb
B. 3 kb
C. 4 kb
D. 5 kb

A. Standard / Conceptual MCQs – Answers + Brief Explanations (Q1–Q70)

Q.No	Answer	Explanation
1	A	DNA is composed of deoxyribose, phosphate, and nitrogenous bases.
2	C	Chargaff’s rule: A = T, G = C; also A+G = T+C.
3	B	Watson & Crick proposed double helix DNA.
4	B	Prokaryotic DNA is circular.
5	B	Mitochondrial DNA is circular.
6	A	tRNA carries amino acids to ribosomes.
7	A	mRNA is messenger RNA.
8	B	rRNA forms structural & functional ribosome component.
9	B	DNA replication is semi-conservative (each daughter has one parental strand).
10	B	Helicase unwinds DNA helix.
11	B	Leading strand is synthesized 5′ → 3′ continuously.
12	B	Okazaki fragments occur on lagging strand.
13	D	DNA polymerase requires primer, template, dNTPs.
14	C	Replication fork = site where DNA unwinds for replication.
15	A	Telomeres protect chromosome ends from degradation.
16	B	RNA polymerase synthesizes RNA from DNA template.
17	A	Template strand is 3′ → 5′; complementary RNA is 5′ → 3′.
18	A	Transcription starts at promoter.
19	D	Eukaryotic pre-mRNA undergoes capping, polyadenylation, splicing.
20	A	mRNA carries genetic code for protein synthesis.
21	B	Translation occurs on ribosomes.
22	B	AUG is start codon (Methionine).
23	B	Stop codons: UAA, UAG, UGA.
24	A	tRNA contains anticodon complementary to mRNA codon.
25	A	Peptide bond forms between amino acids’ carboxyl and amino groups.
26	A	Translation ends at stop codon.
27	D	Genetic code is universal, degenerate, non-overlapping.
28	A	Each codon codes for one amino acid.
29	B	64 codons exist in total.
30	A	AUG codes for Methionine.
31	B	Degeneracy = multiple codons for same amino acid.
32	B	Operon model proposed by Jacob & Monod.
33	B	Lac operon is inducible.
34	A	Trp operon is repressible.
35	A	Transcription & translation occur simultaneously in prokaryotes.
36	B	Enhancers increase transcription in eukaryotes.
37	B	Point mutation affects single nucleotide.
38	B	Sickle cell anemia caused by point mutation in β-globin.
39	A	UV light causes thymine dimers.
40	A	Nucleotide excision repair removes incorrect bases.
41	D	Mutagens include chemicals, radiation, viruses.
42	B	DNA polymerase synthesizes 5′ → 3′, requires template & primer.
43	D	Ribosome subunits synthesize protein.
44	A	Epigenetic regulation affects gene expression.
45	D	Recombinant DNA requires vector, gene fragment, enzymes.
46	B	Restriction enzymes cut DNA at specific sequences.
47	B	Hershey-Chase experiment proved DNA is genetic material.
48	B	TMV contains RNA.
49	B	Retrovirus contains RNA.
50	B	Reverse transcriptase synthesizes DNA from RNA.
51	B	Leading strand synthesized continuously 5′ → 3′.
52	A	Lagging strand forms Okazaki fragments due to 5′ → 3′ polymerase limitation.
53	B	Eukaryotic replication starts at multiple origins.
54	A	mRNA capping occurs at 5′ end.
55	B	Poly-A tail added at 3′ end.
56	A	Spliceosome = snRNA + proteins.
57	A	Non-overlapping code = each nucleotide belongs to one codon.
58	A	Wobble hypothesis explains redundancy of code.
59	B	Lactose induces lac operon.
60	B	Tryptophan acts as corepressor in trp operon.
61	A	PCR amplifies DNA using primers, DNA polymerase, nucleotides.
62	D	Plasmid vector carries ORI, gene, and antibiotic marker.
63	C	Retrovirus uses reverse transcriptase.
64	B	Telomerase adds repeats at chromosome ends.
65	B	Mutation in start codon prevents translation initiation.
66	B	UV light forms thymine dimers.
67	A	Nucleotide excision repair corrects thymine dimers.
68	A	Recombinant DNA formed by cutting & joining DNA with vector.
69	A	DNA ligase joins Okazaki fragments.
70	A	Hershey-Chase used radioactive phosphorus & sulfur to track DNA.

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B. Assertion–Reason MCQs – Answers + Explanations (Q1–Q15)

Q.No	Answer	Explanation
1	A	Both true; semi-conservative replication gives 1 old + 1 new strand.
2	A	Both true; lagging strand synthesized in Okazaki fragments due to 5′ → 3′ limitation.
3	A	Both true; genetic code is degenerate (multiple codons for same amino acid).
4	A	Both true; lactose binds repressor → lac operon transcription.
5	A	Both true; tryptophan acts as corepressor for trp operon.
6	A	Both true; capping & poly-A tail protect mRNA and aid translation.
7	A	Both true; retroviruses reverse-transcribe RNA → DNA.
8	A	Both true; tRNA carries amino acids, anticodon pairs with mRNA codon.
9	A	Both true; DNA repair is essential to prevent mutations/disease.
10	A	Both true; genetic code is non-overlapping.
11	A	Both true; restriction enzymes cut DNA at specific sequences for recombinant DNA.
12	A	Both true; PCR selectively amplifies DNA using primers and DNA polymerase.
13	A	Both true; spliceosome removes introns in eukaryotic pre-mRNA.
14	A	Both true; ligase joins Okazaki fragments to form continuous strand.
15	A	Both true; telomerase prevents chromosome shortening during replication.

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C. Difficult / Calculation / Diagram-Based MCQs – Answers + Explanations (Q1–Q30)

Q.No	Answer	Explanation
1	C	PCR: 1 molecule → 2^5 = 32 molecules (5 cycles).
2	A	tRNA with anticodon 3′–GAC–5′ pairs with codon 5′–CUG–3′; carries corresponding amino acid.
3	A	DNA polymerase rate = 1000 nt / 5 min = 200 nt/min.
4	A	30% introns in 1000 nt → 300 nt removed.
5	B	Two consecutive nucleotides: A–T or G–C probability = 0.5 per base pair.
6	D	Diagram shows leading, lagging, Okazaki fragments at replication fork.
7	A	Leucine 6 codons / total 64 → probability one codon = 6/64; two codons = (6/64)^2 = 36/4096.
8	A	PCR amplification: n cycles → 2^n molecules; 3 cycles → 8.
9	C	1 kb DNA with 3 sites → 4 fragments after digestion.
10	B	Start codon mutation prevents translation initiation.
11	A	Mismatch repair corrects ~99% replication errors.
12	A	PCR requires primers, DNA polymerase, nucleotides for amplification.
13	C	EcoRI cuts at GAATTC; 3 sites → 4 fragments.
14	A	Anticodon 3′–GAC–5′ pairs with 5′–CUG–3′ codon.
15	B	AUG → AUA mutation blocks translation initiation.
16	D	DNA: 25% A → 25% T → remaining 50% = 50% C+G → C = 25%.
17	B	40% T → 40% A → remaining 20% C+G → G = 10%.
18	A	1200 nt pre-mRNA, 30% introns → 1200 – 360 = 840 nt mature mRNA.
19	C	Stop codons: 3/64 probability.
20	D	Diagram includes leading, lagging strands & Okazaki fragments.
21	B	Single cutter splits plasmid+insert into 2 fragments.
22	A	Reverse transcriptase synthesizes DNA from RNA template.
23	A	Two leucine codons: (6/64)^2 = 36/4096.
24	C	50 repeats × 6 nt/repeat = 300 nt added.
25	B	20% T → 20% A → total A+T = 40% of 900 bp = 360 nt.
26	C	UUU → UUC mutation does not change amino acid → silent mutation.
27	C	Single nucleotide deletion → frameshift mutation.
28	D	Diagram shows ribosome subunits + tRNA binding sites (A, P, E).
29	A	Probability NOT methionine = 63/64.
30	C	Plasmid 3 kb + insert 1 kb → total 4 kb.

Disclaimer:

All MCQs and answers are for educational purposes only. They are based on NCERT and NEET syllabus and do not represent official NEET questions.