Molecular Basis of Inheritance

Q1. DNA is a long polymer of 
(A) Ribonucleotides (B) Deoxyribonucleotides (C) Amino acids (D) Monosaccharides
Ans: B

Q2. The length of DNA in an organism is usually expressed in terms of 
(A) Number of genes (B) Number of nucleotides or base pairs (C) Molecular weight only (D) Number of chromosomes
Ans: B

Q3. Which of the following is a purine base present in DNA? 
(A) Cytosine (B) Thymine (C) Adenine (D) Uracil
Ans: C

Q4. In DNA, thymine is present instead of 
(A) Adenine (B) Guanine (C) Cytosine (D) Uracil
Ans: D

Q5. A nucleoside is formed when a nitrogenous base is linked to the 
(A) 5′ carbon of sugar through phosphodiester bond (B) 1′ carbon of pentose sugar through N-glycosidic linkage (C) 3′ carbon through peptide bond (D) Phosphate group directly
Ans: B

Q6. Two nucleotides in a polynucleotide chain are joined by 
(A) N-glycosidic linkage (B) 3′-5′ phosphodiester linkage (C) Hydrogen bonds only (D) Peptide bonds
Ans: B

Q7. In a polynucleotide chain, the 5′-end has a free 
(A) Hydroxyl group at 3′ carbon (B) Phosphate moiety (C) Nitrogenous base (D) Ribose sugar
Ans: B

Q8. In RNA, an additional hydroxyl group is present at which position of the ribose sugar? 
(A) 1′ position (B) 2′ position (C) 3′ position (D) 5′ position
Ans: B

Q9. DNA was first identified as ‘Nuclein’ by 
(A) Watson and Crick (B) Friedrich Meischer (C) Griffith (D) Avery
Ans: B

Q10. The double helix model of DNA was proposed in 
(A) 1869 (B) 1928 (C) 1953 (D) 1958
Ans: C

Q11. The X-ray diffraction data for DNA structure was provided by 
(A) Watson and Crick (B) Maurice Wilkins and Rosalind Franklin (C) Meselson and Stahl (D) Hershey and Chase
Ans: B

Q12. According to Chargaff’s observation, in double stranded DNA the ratio of 
(A) A to G equals one (B) A to T equals one (C) G to C equals two (D) A + T to G + C equals one
Ans: B

Q13. The two strands of DNA in the double helix are 
(A) Parallel (B) Antiparallel (C) Identical in sequence (D) Held only by covalent bonds
Ans: B

Q14. Adenine pairs with thymine through 
(A) One hydrogen bond (B) Two hydrogen bonds (C) Three hydrogen bonds (D) Four hydrogen bonds
Ans: B

Q15. Guanine pairs with cytosine through 
(A) One hydrogen bond (B) Two hydrogen bonds (C) Three hydrogen bonds (D) No hydrogen bond
Ans: C

Q16. The pitch of the DNA double helix is 
(A) 0.34 nm (B) 3.4 nm (C) 34 nm (D) 0.034 nm
Ans: B

Q17. Approximately how many base pairs are present per turn in the DNA double helix? 
(A) 5 (B) 10 (C) 15 (D) 20
Ans: B

Q18. The distance between two consecutive base pairs in DNA helix is approximately 
(A) 0.034 nm (B) 0.34 nm (C) 3.4 nm (D) 34 nm
Ans: B

Q19. In DNA double helix, a purine always pairs with 
(A) Another purine (B) A pyrimidine (C) Any base randomly (D) The same purine
Ans: B

Q20. The backbone of the DNA polynucleotide chain is constituted by 
(A) Nitrogenous bases only (B) Sugar and phosphate (C) Only phosphate groups (D) Only nitrogenous bases
Ans: B

Q21. Stacking of base pairs one over the other in DNA provides 
(A) Flexibility to the helix (B) Additional stability to the helical structure (C) No structural role (D) Increased reactivity
Ans: B

Q22. The haploid content of human DNA is approximately 
(A) 4.6 × 10^6 bp (B) 3.3 × 10^9 bp (C) 48502 bp (D) 5386 bp
Ans: B

Q23. In prokaryotes like E. coli, DNA is organised in a region called 
(A) Nucleus (B) Nucleoid (C) Nucleosome (D) Chromosome
Ans: B

Q24. Histone proteins are rich in which positively charged amino acids? 
(A) Glutamic acid and aspartic acid (B) Lysine and arginine (C) Glycine and alanine (D) Valine and leucine
Ans: B

Q25. DNA wraps around a histone octamer to form a structure called 
(A) Nucleoid (B) Nucleosome (C) Chromatin fiber (D) Chromosome
Ans: B

Q26. A typical nucleosome contains approximately how many base pairs of DNA? 
(A) 100 bp (B) 200 bp (C) 500 bp (D) 1000 bp
Ans: B

Q27. The ‘beads-on-string’ structure of chromatin is visible under 
(A) Light microscope (B) Electron microscope (C) Compound microscope (D) Phase contrast microscope
Ans: B

Q28. Euchromatin is characterised as 
(A) Densely packed and transcriptionally inactive (B) Loosely packed and transcriptionally active (C) Highly condensed during interphase (D) Rich in heterochromatin proteins
Ans: B

Q29. Heterochromatin is 
(A) Transcriptionally active (B) Loosely packed and stains light (C) Densely packed and transcriptionally inactive (D) Absent in eukaryotic nuclei
Ans: C

Q30. Non-histone chromosomal proteins are required for 
(A) Formation of nucleosome (B) Higher level packaging of chromatin (C) DNA replication only (D) Transcription in prokaryotes
Ans: B

Q31. Frederick Griffith performed his transforming principle experiment on 
(A) Escherichia coli (B) Streptococcus pneumoniae (C) Bacillus subtilis (D) Staphylococcus aureus
Ans: B

Q32. In Griffith’s experiment, the virulent strain of pneumococcus was 
(A) R strain (rough colony) (B) S strain (smooth colony) (C) Both strains equally virulent (D) Neither strain
Ans: B

Q33. Griffith observed that heat-killed S strain mixed with live R strain 
(A) Did not affect mice (B) Killed the mice and live S bacteria were recovered (C) Produced only R bacteria (D) Had no transforming effect
Ans: B

Q34. The biochemical nature of the transforming principle was established by 
(A) Griffith alone (B) Avery, MacLeod and McCarty (C) Hershey and Chase (D) Watson and Crick
Ans: B

Q35. Avery and colleagues showed that transformation was inhibited by treatment with 
(A) Proteases (B) RNases (C) DNase (D) Lipases
Ans: C

Q36. Hershey and Chase used radioactive phosphorus to label 
(A) Viral protein (B) Viral DNA (C) Bacterial DNA (D) Both protein and DNA
Ans: B

Q37. Radioactive sulfur was used by Hershey and Chase to label 
(A) DNA (B) Protein (C) RNA (D) Polysaccharide coat
Ans: B

Q38. In Hershey-Chase experiment, the blender was used to 
(A) Lyse the bacteria (B) Remove viral coats from bacteria (C) Centrifuge the mixture (D) Label the phages
Ans: B

Q39. Bacteria infected with phages having radioactive DNA showed radioactivity in 
(A) Supernatant only (B) Bacterial cells (C) Both supernatant and pellet equally (D) Neither
Ans: B

Q40. The Hershey-Chase experiment conclusively proved that 
(A) Protein is the genetic material (B) DNA is the genetic material (C) RNA is the genetic material (D) Both DNA and protein enter the cell
Ans: B

Q41. Which property makes DNA a better genetic material than RNA? 
(A) RNA is more stable chemically (B) DNA is chemically less reactive and structurally more stable (C) RNA has thymine instead of uracil (D) DNA is single stranded
Ans: B

Q42. The presence of thymine instead of uracil in DNA provides 
(A) Increased reactivity (B) Additional stability (C) Faster mutation rate (D) Catalytic property
Ans: B

Q43. RNA mutates at a faster rate than DNA because 
(A) It is double stranded (B) It has reactive 2′-OH group and is less stable (C) It lacks uracil (D) It is found only in viruses
Ans: B

Q44. According to the RNA world hypothesis, the first genetic material was 
(A) DNA (B) RNA (C) Protein (D) Lipid
Ans: B

Q45. RNA can function as 
(A) Only messenger (B) Genetic material as well as catalyst (C) Only structural molecule (D) Only adapter
Ans: B

Q46. DNA replication is described as semiconservative because 
(A) Both strands are newly synthesised (B) Each daughter DNA has one parental and one new strand (C) Only one strand is conserved (D) The helix remains intact throughout
Ans: B

Q47. The semiconservative nature of DNA replication was experimentally proved by 
(A) Watson and Crick (B) Meselson and Stahl (C) Hershey and Chase (D) Griffith
Ans: B

Q48. In Meselson-Stahl experiment, E. coli was initially grown in medium containing 
(A) 14NH4Cl (B) 15NH4Cl (C) Radioactive phosphorus (D) Radioactive sulfur
Ans: B

Q49. After one generation of growth in 14N medium, Meselson-Stahl experiment showed 
(A) Only heavy DNA (B) Only light DNA (C) Hybrid DNA of intermediate density (D) Equal heavy and light DNA
Ans: C

Q50. After two generations in Meselson-Stahl experiment, the DNA consisted of 
(A) Only hybrid DNA (B) Equal amounts of hybrid and light DNA (C) Only light DNA (D) Only heavy DNA
Ans: B

Q51. The main enzyme catalysing DNA polymerisation is 
(A) RNA polymerase (B) DNA-dependent DNA polymerase (C) DNA ligase (D) Reverse transcriptase
Ans: B

Q52. DNA polymerase can add nucleotides only in 
(A) 3′ to 5′ direction (B) 5′ to 3′ direction (C) Both directions simultaneously (D) Random direction
Ans: B

Q53. Replication on the lagging strand is 
(A) Continuous (B) Discontinuous forming Okazaki fragments (C) Absent (D) Bidirectional from both ends
Ans: B

Q54. Okazaki fragments are joined by the enzyme 
(A) DNA polymerase (B) DNA ligase (C) Helicase (D) Primase
Ans: B

Q55. Replication begins at a specific site called 
(A) Telomere (B) Origin of replication (C) Centromere (D) Promoter
Ans: B

Q56. Deoxyribonucleoside triphosphates serve as 
(A) Only substrates for polymerisation (B) Only source of energy (C) Both substrates and source of energy (D) Inhibitors of replication
Ans: C

Q57. In eukaryotes, DNA replication occurs during 
(A) G1 phase (B) S phase (C) G2 phase (D) M phase of cell cycle
Ans: B

Q58. Failure of cell division after DNA replication in eukaryotes may lead to 
(A) Haploidy (B) Polyploidy (C) No change in ploidy (D) Gene mutation only
Ans: B

Q59. The process of copying genetic information from DNA to RNA is called 
(A) Replication (B) Transcription (C) Translation (D) Transduction
Ans: B

Q60. During transcription, RNA is synthesised in 
(A) 3′ to 5′ direction (B) 5′ to 3′ direction (C) Both directions (D) Template independent manner
Ans: B

Q61. In a transcription unit, the strand that is transcribed is called 
(A) Coding strand (B) Template strand (C) Both strands equally (D) Non-template strand
Ans: B

Q62. The coding strand of DNA has the same sequence as 
(A) Template strand (B) mRNA except uracil replaces thymine (C) tRNA (D) rRNA
Ans: B

Q63. The promoter is located 
(A) Downstream of structural gene (B) Upstream of structural gene towards 5′ end (C) Within the structural gene (D) At the 3′ end of terminator
Ans: B

Q64. The terminator sequence in a transcription unit is located 
(A) Upstream of promoter (B) Towards 3′ end of coding strand (C) In the middle of structural gene (D) Before the start codon
Ans: B

Q65. In bacteria, there is 
(A) Only one type of RNA polymerase for all RNAs (B) Three different RNA polymerases (C) No RNA polymerase (D) RNA polymerase only for mRNA
Ans: A

Q66. In eukaryotes, RNA polymerase II transcribes 
(A) rRNA (B) tRNA and 5S rRNA (C) hnRNA (precursor of mRNA) (D) Only snRNA
Ans: C

Q67. RNA polymerase I in eukaryotes transcribes 
(A) mRNA (B) tRNA (C) rRNAs (28S, 18S, 5.8S) (D) snRNA
Ans: C

Q68. The process of removing introns and joining exons in hnRNA is called 
(A) Capping (B) Tailing (C) Splicing (D) Polyadenylation
Ans: C

Q69. In capping of hnRNA, which unusual nucleotide is added at 5′ end? 
(A) Adenine (B) Methyl guanosine triphosphate (C) Uracil (D) Cytosine
Ans: B

Q70. Tailing of hnRNA involves addition of 
(A) 50-100 adenylate residues at 5′ end (B) 200-300 adenylate residues at 3′ end (C) Guanylate residues (D) No nucleotides
Ans: B

Q71. The fully processed hnRNA that is transported out of nucleus is called 
(A) pre-mRNA (B) mRNA (C) tRNA (D) rRNA
Ans: B

Q72. The genetic code is a triplet code proposed by 
(A) Watson and Crick (B) George Gamow (C) Nirenberg (D) Khorana
Ans: B

Q73. How many codons code for amino acids out of the 64 possible triplets? 
(A) 60 (B) 61 (C) 62 (D) 64
Ans: B

Q74. The codon AUG codes for 
(A) Only methionine (B) Methionine and acts as initiator codon (C) Only isoleucine (D) Stop signal
Ans: B

Q75. UAA, UAG and UGA are 
(A) Initiator codons (B) Stop or terminator codons (C) Codes for tryptophan (D) Codes for methionine
Ans: B

Q76. The genetic code is said to be degenerate because 
(A) One codon codes for multiple amino acids (B) Some amino acids are coded by more than one codon (C) It is not universal (D) It has punctuations
Ans: B

Q77. The codon is read in mRNA in a 
(A) Non-contiguous manner with punctuations (B) Contiguous fashion without punctuations (C) Random order (D) 3′ to 5′ direction only
Ans: B

Q78. The genetic code is nearly universal except in 
(A) All bacteria (B) Some mitochondrial codons and certain protozoans (C) All eukaryotes (D) Viruses only
Ans: B

Q79. Insertion or deletion of one or two bases in a gene causes 
(A) Silent mutation (B) Frameshift mutation (C) No change in reading frame (D) Point mutation only
Ans: B

Q80. Insertion or deletion of three bases or multiples of three 
(A) Always causes frameshift (B) Inserts or deletes one or more amino acids without shifting reading frame (C) Creates stop codon always (D) Has no effect on protein
Ans: B

Q81. tRNA is also called adapter molecule because it 
(A) Carries amino acids only (B) Has anticodon to read code and acceptor end to bind amino acid (C) Is part of ribosome (D) Acts as catalyst
Ans: B

Q82. The anticodon loop of tRNA has bases 
(A) Identical to codon (B) Complementary to the codon (C) Same as mRNA (D) Random sequence
Ans: B

Q83. There are no tRNAs for 
(A) Initiator codons (B) Stop codons (C) All amino acid codons (D) Methionine codon
Ans: B

Q84. Translation refers to 
(A) Synthesis of RNA from DNA (B) Polymerisation of amino acids to form polypeptide (C) Replication of DNA (D) Splicing of hnRNA
Ans: B

Q85. Charging of tRNA or aminoacylation requires 
(A) GTP only (B) ATP and specific amino acid (C) No energy (D) DNA template
Ans: B

Q86. Peptide bond formation is catalysed by 
(A) 23S rRNA (ribozyme) in bacteria (B) Protein enzyme only (C) DNA polymerase (D) mRNA
Ans: A

Q87. The ribosome has 
(A) Only one subunit (B) Large and small subunits (C) Three subunits (D) No protein component
Ans: B

Q88. Untranslated regions (UTRs) in mRNA are present 
(A) Only at 5′ end (B) Only at 3′ end (C) At both 5′ and 3′ ends (D) Within coding sequence
Ans: C

Q89. Initiation of translation begins when ribosome binds to 
(A) Any codon (B) Start codon AUG recognised by initiator tRNA (C) Stop codon (D) Promoter
Ans: B

Q90. A release factor binds to 
(A) Start codon (B) Stop codon to terminate translation (C) Any codon (D) Anticodon
Ans: B

Q91. In prokaryotes, control of gene expression occurs mainly at 
(A) Translational level (B) Rate of transcriptional initiation (C) Post-translational modification (D) mRNA transport
Ans: B

Q92. In lac operon, the regulatory gene (i gene) codes for 
(A) Beta-galactosidase (B) Repressor protein (C) Permease (D) Transacetylase
Ans: B

Q93. The z gene in lac operon codes for 
(A) Repressor (B) Beta-galactosidase (C) Permease (D) Operator
Ans: B

Q94. Lactose acts as inducer in lac operon because it 
(A) Binds to promoter (B) Inactivates the repressor by binding to it (C) Activates RNA polymerase (D) Binds to operator directly
Ans: B

Q95. In absence of inducer, the lac operon repressor 
(A) Allows transcription (B) Binds to operator and prevents transcription (C) Is degraded (D) Activates the operon
Ans: B

Q96. Regulation of lac operon by repressor is an example of 
(A) Positive regulation (B) Negative regulation (C) No regulation (D) Translational control
Ans: B

Q97. The lac operon is polycistronic because it has 
(A) One structural gene (B) Multiple structural genes under one promoter (C) Only regulatory gene (D) No operator
Ans: B

Q98. Human Genome Project was launched in 
(A) 1980 (B) 1990 (C) 2000 (D) 2003
Ans: B

Q99. The approximate number of base pairs in human genome is 
(A) 3 × 10^6 (B) 3 × 10^9 (C) 3 × 10^12 (D) 6.6 × 10^9
Ans: B

Q100. One of the goals of Human Genome Project was to 
(A) Sequence only coding regions (B) Identify all approximately 20000-25000 genes (C) Avoid ethical issues (D) Sequence only Y chromosome
Ans: B

Q101. The Human Genome Project was completed in 
(A) 1990 (B) 2003 (C) 2010 (D) 2020
Ans: B

Q102. Less than what percentage of human genome codes for proteins? 
(A) 10% (B) 2% (C) 50% (D) 90%
Ans: B

Q103. Repeated sequences in human genome 
(A) Code for most proteins (B) Have direct coding function (C) Shed light on chromosome structure and evolution (D) Are absent
Ans: C

Q104. Chromosome 1 in human genome has 
(A) Fewest genes (B) Most genes (C) No genes (D) Only regulatory sequences
Ans: B

Q105. SNPs stand for 
(A) Single nucleotide polymorphisms (B) Simple nucleotide proteins (C) Sequence nucleotide pairs (D) Satellite nucleotide polymorphisms
Ans: A

Q106. DNA fingerprinting is based on 
(A) Coding sequences (B) Polymorphism in repetitive DNA sequences like VNTR (C) Mitochondrial DNA only (D) Protein sequences
Ans: B

Q107. VNTR stands for 
(A) Variable number of tandem repeats (B) Very narrow tandem repeats (C) Variable nucleotide tandem repeats (D) Viral nucleotide tandem repeats
Ans: A

Q108. Satellite DNA in density gradient centrifugation forms 
(A) Major peak only (B) Minor peaks separate from bulk DNA (C) No visible peak (D) Same peak as bulk DNA
Ans: B

Q109. DNA fingerprinting technique was initially developed by 
(A) Watson (B) Alec Jeffreys (C) Nirenberg (D) Meselson
Ans: B

Q110. In DNA fingerprinting, the probe used is 
(A) Coding gene sequence (B) Radiolabelled VNTR (C) Ribosomal RNA (D) Mitochondrial DNA
Ans: B

Q111. DNA from which of the following can be used for DNA fingerprinting? 
(A) Only blood (B) Blood, hair follicle, skin, bone, saliva, sperm (C) Only bone marrow (D) Only muscle tissue
Ans: B

Q112. DNA fingerprinting is used in 
(A) Only criminal identification (B) Forensic science, paternity testing and genetic mapping (C) Only agricultural research (D) Only evolutionary studies
Ans: B

Q113. Monozygotic twins have 
(A) Completely different DNA fingerprints (B) Identical DNA fingerprints (C) Similar but not identical fingerprints (D) No DNA
Ans: B

Q114. The sensitivity of DNA fingerprinting has been increased by use of 
(A) Southern blotting only (B) Polymerase chain reaction (PCR) (C) Centrifugation (D) X-ray diffraction
Ans: B

Q115. Polymorphism in DNA arises due to 
(A) Only environmental factors (B) Mutations that are inheritable (C) Protein changes (D) RNA editing
Ans: B

Q116. If more than one variant (allele) at a locus occurs in population with frequency greater than 0.01, it is called 
(A) Mutation (B) DNA polymorphism (C) Silent variation (D) Epigenetic change
Ans: B

Q117. In the statement “RAM HAS RED CAP”, if one base (letter) is inserted between HAS and RED, it becomes 
(A) RAM HAS RED CAP (unchanged) (B) Reading frame shifts like frameshift mutation (C) No effect on meaning (D) Creates a stop signal
Ans: B

Q118. The Central Dogma of molecular biology states information flows from 
(A) Protein to DNA to RNA (B) DNA to RNA to Protein (C) RNA to DNA to Protein (D) Protein to RNA to DNA
Ans: B

Q119. In some viruses, the flow of information is reverse from 
(A) DNA to RNA (B) RNA to DNA (C) Protein to RNA (D) DNA to Protein
Ans: B

Q120. The process of making RNA from DNA is called transcription while making protein from RNA is called 
(A) Replication (B) Transcription (C) Translation (D) Transposition
Ans: C

Q121. Both DNA and RNA can act as genetic material but DNA is preferred for storage because it is 
(A) More reactive (B) Chemically less reactive and more stable (C) Single stranded (D) Catalytic
Ans: B

Q122. For transmission of genetic information, which is better? 
(A) DNA (B) RNA (C) Protein (D) Lipid
Ans: B

Q123. The structural gene in eukaryotes is mostly 
(A) Polycistronic (B) Monocistronic (C) Without introns (D) Without exons
Ans: B

Q124. Exons are 
(A) Intervening sequences that do not appear in mature RNA (B) Coding sequences that appear in mature RNA (C) Promoter sequences (D) Terminator sequences
Ans: B

Q125. Introns are 
(A) Expressed sequences in mature RNA (B) Intervening sequences removed during splicing (C) Always present in prokaryotes (D) Part of tRNA
Ans: B

Q126. The split gene arrangement with introns is more common in 
(A) Prokaryotes (B) Eukaryotes (C) Viruses (D) Bacteria
Ans: B

Q127. In bacteria, transcription and translation can be coupled because 
(A) There is nucleus (B) No separation of cytosol and nucleus (C) mRNA requires extensive processing (D) Ribosomes are absent
Ans: B

Q128. Sigma factor in bacterial transcription is required for 
(A) Elongation (B) Initiation (C) Termination (D) Splicing
Ans: B

Q129. Rho factor in bacterial transcription is required for 
(A) Initiation (B) Elongation (C) Termination (D) Capping
Ans: C

Q130. In the given DNA sequence 3′-ATGCATGCATGCATGCATGCATGCATGC-5′ (template), the transcribed RNA would be 
(A) 5′-UACGUACGUACGUACGUACGUACGUACG-3′ (B) 5′-ATGCATGCATGCATGCATGCATGCATGC-3′ (C) 3′-UACGUACGUACGUACGUACGUACGUACG-5′ (D) Same as template
Ans: A

Q131. The largest known human gene is 
(A) Dystrophin (B) Insulin (C) Hemoglobin (D) Albumin
Ans: A

Q132. The average gene in human genome consists of approximately how many bases? 
(A) 300 (B) 3000 (C) 30000 (D) 300000
Ans: B

Q133. Almost what percentage of nucleotide bases are exactly the same in all humans? 
(A) 90% (B) 99.9% (C) 50% (D) 75%
Ans: B

Q134. Functions are unknown for over what percentage of discovered human genes? 
(A) 10% (B) 50% (C) 90% (D) 5%
Ans: B

Q135. BAC and YAC are used in Human Genome Project as 
(A) Sequencing machines (B) Vectors for cloning DNA fragments (C) Radioactive probes (D) Centrifuges
Ans: B

Q136. Frederick Sanger is credited with developing method for 
(A) DNA sequencing and amino acid sequencing in proteins (B) Only protein sequencing (C) Only RNA synthesis (D) PCR
Ans: A

Q137. Expressed Sequence Tags (ESTs) approach in HGP focused on 
(A) Sequencing whole genome blindly (B) Identifying genes expressed as RNA (C) Only non-coding DNA (D) Only mitochondrial DNA
Ans: B

Q138. Sequence Annotation in HGP refers to 
(A) Only sequencing (B) Assigning functions to different regions after sequencing (C) Removing introns (D) Adding poly A tail
Ans: B

Q139. The last human chromosome to be sequenced was 
(A) Chromosome 1 in May 2006 (B) Y chromosome (C) X chromosome (D) Chromosome 21
Ans: A

Q140. Microsatellites are 
(A) Coding sequences (B) Repetitive DNA sequences used in genetic mapping (C) Protein coding genes (D) tRNA genes
Ans: B

Q141. In DNA fingerprinting, the copy number of repeats in VNTR 
(A) Is same in all chromosomes of an individual (B) Varies from chromosome to chromosome (C) Is fixed for all humans (D) Does not show polymorphism
Ans: B

Q142. The size of VNTR after hybridisation varies from 
(A) 0.01 to 0.1 kb (B) 0.1 to 20 kb (C) 100 to 200 kb (D) 1 to 2 Mb
Ans: B

Q143. DNA fingerprinting bands are detected by 
(A) PCR only (B) Autoradiography after hybridisation (C) Light microscopy (D) Centrifugation
Ans: B

Q144. The technique of DNA fingerprinting involves all except 
(A) Isolation of DNA (B) Digestion by restriction endonucleases (C) Electrophoresis and Southern blotting (D) Protein sequencing
Ans: D

Q145. If E. coli DNA length is 1.36 mm and distance between bp is 0.34 nm, number of base pairs is approximately 
(A) 4 × 10^6 (B) 4.6 × 10^6 (C) 3.3 × 10^9 (D) 6.6 × 10^9
Ans: A

Q146. Theoretically, how many nucleosomes (beads) are present in a mammalian cell with 6.6 × 10^9 bp? 
(A) About 3.3 × 10^7 (B) About 6.6 × 10^7 (C) About 200 (D) About 10
Ans: A

Q147. The Central Dogma was proposed by 
(A) Watson and Crick (B) Francis Crick (C) George Gamow (D) Nirenberg
Ans: B

Q148. In lac operon, permease is coded by 
(A) z gene (B) y gene (C) a gene (D) i gene
Ans: B

Q149. Transacetylase in lac operon is coded by 
(A) z gene (B) y gene (C) a gene (D) i gene
Ans: C

Q150. Very low level of lac operon expression is always present so that 
(A) Repressor is made (B) Lactose can enter the cell when available (C) Glucose is metabolised (D) No transcription occurs
Ans: B

Q151. Glucose or galactose cannot act as inducers for lac operon because 
(A) They are not substrates (B) They do not bind and inactivate the repressor (C) They repress the operon (D) They are toxic
Ans: B

Q152. The repressor of lac operon is synthesised 
(A) Only when lactose is present (B) Constitutively (all the time) (C) Only during starvation (D) Never
Ans: B

Q153. In presence of inducer, the repressor becomes 
(A) More active (B) Inactive (C) Degraded completely (D) Converted to activator
Ans: B

Q154. Human genome contains 
(A) 3164.7 million bp (B) 3.3 billion bp exactly (C) 4.6 million bp (D) 5386 bp
Ans: A

Q155. The Y chromosome has 
(A) Most genes (B) Fewest genes (C) No genes (D) Same number as X
Ans: B

Q156. Scientists have identified about how many locations of single-base DNA differences (SNPs) in humans? 
(A) 1.4 million (B) 14 million (C) 140 million (D) 1.4 billion
Ans: A

Q157. Bioinformatics developed rapidly due to 
(A) Need for data storage, retrieval and analysis in HGP (B) Only protein research (C) Agricultural needs only (D) No relation to HGP
Ans: A

Q158. Transfer of technologies to industries was one of the 
(A) Accidental outcomes (B) Goals of HGP (C) Failures of HGP (D) Unrelated aspects
Ans: B

Q159. Ethical, legal and social issues (ELSI) were addressed in HGP because 
(A) They were ignored (B) Project could raise such issues (C) No issues arose (D) Only scientists were involved
Ans: B

Q160. Non-human model organisms sequenced in HGP context include 
(A) Only humans (B) Bacteria, yeast, C. elegans, Drosophila, rice, Arabidopsis (C) Only viruses (D) Only mammals
Ans: B

Q161. The two major approaches in HGP methodologies were 
(A) ESTs and Sequence Annotation (B) Only whole genome sequencing (C) Only protein sequencing (D) Only RNA sequencing
Ans: A

Q162. In DNA fingerprinting, polymorphism is detected because 
(A) All DNA sequences are identical (B) Repetitive sequences show high degree of variation in copy number (C) Coding genes vary (D) No variation exists
Ans: B

Q163. Alec Jeffreys used which as probe for DNA fingerprinting? 
(A) A coding gene (B) A satellite DNA showing high polymorphism (VNTR) (C) Ribosomal DNA (D) Mitochondrial DNA
Ans: B

Q164. After hybridisation with VNTR probe, the autoradiogram shows 
(A) Single band for all individuals (B) Many bands of differing sizes giving characteristic pattern (C) No bands (D) Only one band per person
Ans: B

Q165. DNA fingerprinting pattern is 
(A) Same for all humans (B) Unique for every individual except identical twins (C) Same for siblings always (D) Dependent on age
Ans: B

Q166. The process opposite to transcription (RNA to DNA) in some viruses is called 
(A) Translation (B) Reverse transcription (C) Replication (D) Splicing
Ans: B

Q167. If both strands of DNA are transcribed, it would 
(A) Produce identical proteins (B) Code for two different proteins and form double stranded RNA (C) Have no effect (D) Stop all cellular activity
Ans: B

Q168. The reason both strands are not copied during transcription is 
(A) Energy saving only (B) It would complicate genetic information transfer and form dsRNA (C) DNA is too long (D) RNA polymerase cannot transcribe both
Ans: B

Q169. The reference point for defining promoter and terminator is made with respect to 
(A) Template strand (B) Coding strand (C) Both equally (D) mRNA directly
Ans: B

Q170. Switching position of promoter with terminator would 
(A) Have no effect (B) Reverse definition of coding and template strands (C) Stop transcription (D) Create mutation
Ans: B

Q171. In the lac operon diagram, when inducer is absent, transcription 
(A) Proceeds normally (B) Is prevented by repressor bound to operator (C) Occurs at high rate (D) Produces only repressor
Ans: B

Q172. In presence of inducer in lac operon, 
(A) Repressor binds tightly (B) Repressor is inactivated allowing transcription (C) Operator is deleted (D) No mRNA is made
Ans: B

Q173. All three gene products (z, y, a) in lac operon are required for 
(A) Glucose metabolism (B) Metabolism of lactose (C) DNA replication (D) Cell division
Ans: B

Q174. The i gene in lac operon is called regulatory gene because it 
(A) Codes for beta-galactosidase (B) Codes for repressor that regulates the operon (C) Is part of structural genes (D) Has no function
Ans: B

Q175. Operon is a common arrangement in 
(A) Eukaryotes (B) Bacteria and prokaryotes (C) Only viruses (D) Only plants
Ans: B

Q176. The ribosome acts as catalyst for peptide bond formation through 
(A) Its protein component (B) 23S rRNA (ribozyme) (C) mRNA (D) tRNA only
Ans: B

Q177. A translational unit in mRNA is flanked by 
(A) Promoter and terminator (B) Start codon AUG and stop codon (C) Only UTRs (D) Exons only
Ans: B

Q178. UTRs in mRNA are required for 
(A) Coding amino acids (B) Efficient translation process (C) Splicing (D) Replication
Ans: B

Q179. During elongation in translation, the ribosome moves 
(A) From stop to start codon (B) Codon to codon along mRNA adding one amino acid at a time (C) Randomly (D) Only once
Ans: B

Q180. Amino acids are activated and linked to tRNA in presence of 
(A) GTP (B) ATP (C) CTP (D) No energy source
Ans: B

Q181. If the mRNA sequence is AUG UUU UUC UUC UUU UUU UUC, the amino acid sequence would be 
(A) Met-Phe-Phe-Phe-Phe-Phe-Phe (B) Met-Leu-Leu-Leu-Leu-Leu-Leu (C) Stop-Phe-Phe-Phe-Phe-Phe (D) Met-Ser-Ser-Ser-Ser-Ser
Ans: A

Q182. Predicting nucleotide sequence from amino acid sequence Met-Phe-Phe-Phe-Phe-Phe faces difficulty because 
(A) Code is unambiguous (B) Code is degenerate (multiple codons for Phe) (C) No tRNA exists (D) Translation does not occur
Ans: B

Q183. The two properties of genetic code illustrated by degeneracy and the above difficulty are 
(A) Universality and triplet nature (B) Degeneracy and lack of punctuation (C) Ambiguity and overlapping (D) Non-degeneracy
Ans: B

Q184. Sickle cell anemia is caused by 
(A) Frameshift mutation (B) Point mutation changing glutamate to valine in beta globin (C) Deletion of whole gene (D) Viral infection
Ans: B

Q185. Effect of point mutation in beta globin gene resulting in sickle cell anemia shows 
(A) RNA is genetic material (B) Change in nucleic acid causes change in amino acid sequence of protein (C) No relation between gene and protein (D) Only DNA mutates
Ans: B

Q186. The development and differentiation of embryo into adult is result of 
(A) Only replication (B) Coordinated regulation of expression of several sets of genes (C) Random gene expression (D) Only translation
Ans: B

Q187. In prokaryotes, the operator region is 
(A) Far from promoter (B) Adjacent to promoter elements in most operons (C) Part of structural gene (D) Absent
Ans: B

Q188. Each operon has 
(A) Same operator and repressor for all (B) Its specific operator and specific repressor (C) No repressor (D) Only activator
Ans: B

Q189. Lac operator is present 
(A) In all operons (B) Only in lac operon and interacts specifically with lac repressor (C) In trp operon also (D) Only in eukaryotes
Ans: B

Q190. Jacob and Monod elucidated 
(A) DNA structure (B) Transcriptionally regulated system of lac operon (C) Genetic code (D) Replication mechanism
Ans: B

Q191. In lac operon, all three structural genes are needed together because they function in 
(A) Different unrelated pathways (B) Same or related metabolic pathway of lactose metabolism (C) DNA repair (D) Cell wall synthesis
Ans: B

Q192. The repressor protein in lac operon binds to 
(A) Promoter (B) Operator region (C) Structural genes (D) Ribosome
Ans: B

Q193. When repressor binds to operator in lac operon, it 
(A) Activates transcription (B) Prevents RNA polymerase from transcribing the operon (C) Helps ribosome bind (D) Adds poly A tail
Ans: B

Q194. The lac operon can be visualised as regulation of enzyme synthesis by 
(A) Its product (B) Its substrate (lactose) (C) Glucose always (D) No regulator
Ans: B

Q195. Positive regulation of lac operon is 
(A) The only regulation (B) Also present but beyond scope at this level (C) Absent (D) Same as negative regulation
Ans: B

Q196. The total estimated cost of HGP at beginning was approximately 
(A) 1 billion dollars (B) 9 billion US dollars (C) 100 billion dollars (D) 500 million dollars
Ans: B

Q197. If human genome sequences were stored in books (1000 letters/page, 1000 pages/book), approximately how many books would be needed? 
(A) 330 (B) 3300 (C) 33000 (D) 3.3 million
Ans: B

Q198. HGP was closely associated with rapid development of 
(A) Only microscopy (B) Bioinformatics (C) Only agriculture (D) Only medicine
Ans: B

Q199. Knowledge from HGP can lead to 
(A) Only understanding evolution (B) New ways to diagnose, treat and prevent disorders (C) No practical application (D) Only plant breeding
Ans: B

Q200. DNA fingerprinting has much wider application than forensics including 
(A) Only criminal cases (B) Paternity testing, genetic mapping and evolutionary studies (C) Only agriculture (D) No other use
Ans: B

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Molecular Basis of Inheritance

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