Molecular 50 questions
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The document discusses various aspects of the human diploid genome, including its composition and gene count, as well as details about the Human Genome Project. It highlights key milestones, vector usage, and genetic similarities among humans, while also covering DNA replication, transcription factors, and molecular mechanisms of gene expression. Additionally, the text covers other genetic concepts such as restriction enzymes, splicing, and mutations.
![In the entire human genome, coding
DNA constitute: [AIIMS May 2014]
a. 2%
b. 1%
c. 0.1%
d. 45](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-17-320.jpg)
![Mitochondrial DNA is [AIPGMEE 2006]
a) Open circular
b) Closed circular
c) Nicked circular
d) Linear](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-29-320.jpg)
![Which one of the following is the complementary
sequence of 5′ TTAAGCGTAC 3′? [AIPGMEE 2006]
a) 5’ GTACGCTTAA 3′
b) 5’ AATTCGCATG 3’
c) 5’ CATGCGAATT 3’
d) 5’ TTAAGCGTAC 3’](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-31-320.jpg)
![DNA estimation can be done by
[AIIMS May 2012]
a. Spirometer
b. Spectrophotometer
c. pH meter
d. Sphygmomanometer](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-43-320.jpg)
![After digestion by restriction endonucleases DNA
strands can be joined again by: [AIIMS May 2011]
a. DNA polymerase
b. DNA ligase
c. DNA topoisomerase
d. DNA gyrase](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-51-320.jpg)
![If a bacteria acquires the gene for only restriction
endonuclease by horizontal transfer [AIPG 10]
a) Bacteria will be able to destroy host DNA
more efficiently
b) Bacteria will die due to methylase deficiency
c) Bacteria will multiply faster
d) Proof reading improved](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-83-320.jpg)
![CAP in Lac operon is an example of
[AIPGMEE 07, AIIMS Nov 11]
a) Positive regulator
b) Negative regulator
c) Constitutive expression
d) Attenuation](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-132-320.jpg)
![Splicing activity is a function of [AIIMS May 2010]
a. mRNA
b. snRNA
c. tRNA
d. rRNA](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-151-320.jpg)
![apoB48 & apoB100 production is
due to [AIIMS May 11]
a) alternative splicing
b) mRNA editing
c) alternative polyadenylation
d) mRNA silencing](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-167-320.jpg)
![Apo B-48 and Apo B-100 is synthesized from
the same mRNA; the difference between them
is due to [AIIMS May 2011]
a. RNA spicing
b. Allelic exclusion
c. Deamination of cytidine to uracil
d. Upstream repression](https://image.slidesharecdn.com/molecular50questions-160918184653/85/Molecular-50-questions-169-320.jpg)
Molecular 50 questions
- 2. Human diploid genome consists of a) 2 billion base pairs b) 3 billion base pairs c) 5 billion base pairs d) 6 billion base pairs
- 4. Human diploid genome consists of a) 2 billion base pairs b) 3 billion base pairs c) 5 billion base pairs d) 6 billion base pairs Haploid Diploid
- 5. Human diploid genome consists of a) b) c) d) 6 billion base pairs
- 6. Estimated number of human genes a) 20 to 25 thousand b) 50 thousand c) 60 thousand d) 1 lakh
- 9. Human Genome Project • The Human Genome Project was initiated by _____________ • The prime objective of HGP was ________________________________ • The private company involved in Human genome sequencing in parallel with HGP was _____________________ • The first draft of HGP was published in 2001 in the journal _____________ • ____________are the vectors widely used in HGP. • According to HGP, genetic similarity between all humans is _______________
- 10. Human Genome Project (1990- 2003) • The Human Genome Project was initiated by NIH and DOE. • The prime objective of HGP was to sequence the entire base pairs that makes up the 24 chromosomes. • The private company involved in Human genome sequencing in parallel with HGP was Celera Genomics. • The first draft of HGP was published in 2001 in the journal “Nature” • BAC and YAC are the vectors widely used in HGP. • According to HGP, genetic similarity between all humans is 99.9%.
- 12. Human genome sequencing is complete but annotation is yet to be done.
- 13. Harper 30 p.102
- 16. Components of Human genome
- 17. In the entire human genome, coding DNA constitute: [AIIMS May 2014] a. 2% b. 1% c. 0.1% d. 45
- 21. 30% of human genome is made of repetitive DNA
- 24. All are true about microsatellite repeats EXCEPT a) Consist of 2 to 6 bp repeated up to 50 times b) Trinucleotide repeat is the most common microsatellite DNA c) Exists as both dispersed and grouped tandem arrays d) Number of repeats may vary on the two chromosomes
- 26. Our genome is not our own! Don’t forget Human genome contains pro-virus! e.g. ERVW-1
- 28. Number of genes encoded by mitochondrial genome a) 37 b) 47 c) 57 d) 67
- 29. Mitochondrial DNA is [AIPGMEE 2006] a) Open circular b) Closed circular c) Nicked circular d) Linear
- 30. Mitochondrial DNA is a) Closed circular, diploid b) Closed circular, haploid c) Closed circular, polyploid d) None
- 31. Which one of the following is the complementary sequence of 5′ TTAAGCGTAC 3′? [AIPGMEE 2006] a) 5’ GTACGCTTAA 3′ b) 5’ AATTCGCATG 3’ c) 5’ CATGCGAATT 3’ d) 5’ TTAAGCGTAC 3’
- 32. Why purine pairs with pyrimidine? Why purine-purine & pyrimidine- pyrimidine base pairing is not possible?
- 34. 5′ TTAAGCGTAC 3′ G 5’
- 35. 5′ TTAAGCGTAC 3′ TG 5’
- 36. 5′ TTAAGCGTAC 3′ ATG 5’
- 37. 5′ TTAAGCGTAC 3′ 3’ AATTCGCATG 5’
- 38. 5′ TTAAGCGTAC 3′ 3’ AATTCGCATG 5’ 5’ GTACGCTTAA 3′
- 39. Which one of the following is the complementary sequence of 5′ TTAAGCGTAC 3′? a)5’ GTACGCTTAA 3′ b) 5’ AATTCGCATG 3’ c) 5’ CATGCGAATT 3’ d) 5’ TTAAGCGTAC 3’
- 40. Double stranded RNA exist in a) A - DNA like conformation b) B - DNA like conformation c) Z - DNA like conformation d) None
- 43. DNA estimation can be done by [AIIMS May 2012] a. Spirometer b. Spectrophotometer c. pH meter d. Sphygmomanometer
- 45. DNA gyrase is a) Eukaryotic DNA topoisomerase I b) Eukaryotic DNA topoisomerase II c) Prokaryotic DNA topoisomerase I d) Prokaryotic DNA topoisomerase II
- 46. Type I Type II Makes cut on single strand Makes cut on both strands ATP is not needed. ATP is needed. Relaxes negative supercoils in E. coli; Both negative and positive in Eukaryotes. Relaxes both negative & positive supercoils. +ve supercoil is relieved by introducing -ve supercoils. Ciprofloxacin & Nalidixic acid are antimicrobial agents as they inhibit bacterial topoisomerase II (DNA gyrase). Campothecin & Etoposide are anticancer drugs as they inhibit human topoisomerase I & II respectively.
- 48. Which is NOT true about eukaryotic DNA ligase? a) Catalyses the formation of a phosphodiester bond b) NAD+ is energy source c) ATP is the energy source d) Can act only on dsDNA
- 50. DNA ligase
- 51. After digestion by restriction endonucleases DNA strands can be joined again by: [AIIMS May 2011] a. DNA polymerase b. DNA ligase c. DNA topoisomerase d. DNA gyrase
- 52. Deamination of methylated cytosine will produce a) Uracil b) Thymine c) Hypoxanthine d) Adenine
- 57. Deamination of cytosine (Non- enzymatic)
- 61. In humans, unmethylated CpG islands are found in a) Operator b) Promoter c) Introns d) Exons
- 65. Which is the most processive DNA polymerase? a) DNA Polymerase I b) DNA Polymerase II c) DNA Polymerase III d) None
- 68. Which is a template independent DNA polymerase? a) Poly A polymerase b) CCA adding enzyme c) Terminal deoxynucleotidyl transferase d) Polynucleotide kinase
- 69. Which is a template independent DNA polymerase? a) Poly A polymerase b) CCA adding enzyme c) Terminal deoxynucleotidyl transferase d) Polynucleotide kinase
- 70. In Eukaryotes, mRNA is produced by a) RNA Polymerase I b) RNA Polymerase II c) RNA Polymerase III d) None
- 73. TRUE about cap0 of mRNA is a) 7 methyl guanosine attached by 5’5’ thioether bond b) 7 methyl guanosine attached by 3’5’ phosphodiester bond c) 7 methyl guanosine attached by 5’5’ phosphodiester bond d) 7 methyl guanosine attached by 5’5’ triphosphate bond
- 76. Which is NOT true about telomerase? a) Ribonucleoprotein b) Reverse transcriptase c) RNA containing ribozyme d) Active in cancer cells
- 78. Restriction enzymes a) are produced by viruses b) cut only single stranded DNA c) can cut the DNA anywhere d) restrict the expression of viral DNA
- 83. If a bacteria acquires the gene for only restriction endonuclease by horizontal transfer [AIPG 10] a) Bacteria will be able to destroy host DNA more efficiently b) Bacteria will die due to methylase deficiency c) Bacteria will multiply faster d) Proof reading improved
- 84. Isoschizomers & Neoschizomers • Isoschizomers are pairs of restriction enzymes specific to the same recognition sequence. • Neoschizomers are a subset of isoschizomers that recognize the same sequence, but cleave at different positions from the prototype.
- 85. DNA glycosylase is involved in a) Mismatch repair b) Base excision repair c) Nucleotide excision repair d) Direct repair
- 88. Equivalent of pribnow box in eukaryotes is a) Hogness box b) GC box c) CAAT box d) None
- 92. What strategy in transcription factor research allows for the simultaneous identification of all of the genomic sites bound by a given transcription factor under a given set of physiological conditions? a) Microarray a) Dnase I sensitivity b)Chromatin immuno precipitation sequencing (ChIPseq) c) FISH
- 93. 466
- 94. What reactions among transcription proteins greatly expand the diversity of regulatory factors that can be generated from a small number of polypeptides? a) Recombination b)Homodimerization c) Heterozygosity & Heterodimerization d)Trimerization
- 96. Most eukaryotic transcription factors contain at least two domains, each of which mediate different aspects of transcription factor function; these domains are a) RNA binding domain and repression domain b) Activation domain and repression domain c) DNA binding domain and activation domain d) DNA binding domain and ligand binding domain e) RNA binding domain and the activation domain
- 99. The DNA segment from which the primary mRNA transcript is copied or transcribed is called a) Coding strand b) Initiator methionine domain c) Translation unit d) Template strand
- 101. Which of the forces or interactions listed below play the predominant role in driving RNA secondary and tertiary structure formation? a) Hydrophilic repulsion b) Formation of complementary base pair regions c) Hydrophobic interaction d) Van der Waals interactions e) Salt bridge formation
- 103. What is the approximate number of base pairs associated with a single nucleosome? a) 146 b) 292 c) 73 d) 1460 e) 900
- 105. What is on the 3′ end of all functional, mature tRNAs? a) The cloverleaf loop b) The anticodon c) The sequence CCA d) The codon
- 109. Which of the following cis acting elements typically resides adjacent to or overlaps with many prokaryotic promoters? a) Regulatory gene b) Structural gene(s) c) Repressor d) Operator e) Terminator
- 111. HindIII is a restriction endonuclease. Which of the following is most likely to be the recognition sequence for this enzyme? a) AAGAAG b) AAGAGA c) AAGCTT d) AAGGAA e) AAGTTC
- 113. A physician would like to determine the global patterns of gene expression in two different types of tumor cells in order to develop the most appropriate form of chemotherapy for each patient. Which of the following techniques would be most appropriate for this purpose? a) Enzyme linked immunosorbent assay b) Microarray c) Northern blot d) Southern blot e) Western blot
- 117. A mutation to this sequence in eukaryotic mRNA will affect the process by which the 3ʹend polyA tail is added to the mRNA. a) AAUAAA b) CAAT c) CCA d) GU… A … AG e) TATAAA
- 119. Which of the following eukaryotic transcription factor recognises the promoter? a) Rho b) Pribnow box c) Sigma d) TFIID
- 120. Which of the following prokaryotic transcription factor recognises the promoter? a) Rho b) Pribnow box c) Sigma d) TFIID
- 122. Poly A tail is translated into a) Polylysine b) Polyglutamate c) Polyproline d) None
- 124. Poly A is translated into a) Polylysine b) Polyglutamate c) Polyproline d) None
- 125. A 20-year-old man with a microcytic anemia is found to have an abnormal form of β globin (Hemoglobin Constant Spring) that is 172 amino acids long, rather than the 141 found in the normal protein. Which of the following point mutations is consistent with this abnormality? a) CGA → UGA b) GAU → GAC c) GCA → GAA d) UAA → CAA
- 128. A transfer RNA (tRNA) molecule that is supposed to carry cysteine (tRNAcys) is mischarged, so that it actually carries alanine (alatRNAcys). Assuming no correction occurs, what will be the fate of this alanine residue during protein synthesis? a) It will be incorporated into a protein in response to a codon for alanine. b) It will be incorporated into a protein in response to a codon for cysteine. c) It will be incorporated randomly at any codon. d) It will remain attached to the tRNA because it cannot be used for protein synthesis. e) It will be chemically converted to cysteine by cellular enzymes
- 129. The ZYA region of the lac operon will be maximally expressed if: a) cyclic AMP levels are low b) glucose and lactose are both available c) The attenuation stem–loop is able to form d) the CAP site is occupied
- 132. CAP in Lac operon is an example of [AIPGMEE 07, AIIMS Nov 11] a) Positive regulator b) Negative regulator c) Constitutive expression d) Attenuation
- 133. Klenow fragment lacks a) 5'→3' polymerase b) 3'→5' exonuclease c) 5'→3' exonuclease activity d) None
- 135. Similarity between DNA polymerase & Glycogen synthase a) Both need a primer b) Both enzymes are processive in adding substrates c) Either of the above d) None
- 136. Basic protein that binds to DNA in spermatid a) Histone b) Protamine c) Polyamine d) None
- 138. Signal recognition particle contains a) 28S RNA b) 60S RNA c) 40S RNA d) 7S RNA
- 144. Carboxy terminal domain (CTD) of RNA polymerase is phosphorylated by a) TFIID b) TFIIA c) TFIIB d) TFIIH
- 146. The frequency of transcription is determined by a) TATA box b) Inr element c) DPE d) CAAT box
- 147. Which is NOT a DNA binding domain? a) Zinc finger b) Helix turn Helix c) Helix loop helix d) None
- 149. Sequence of an intron always starts and ends with a) GU, AG b) AG, GU c) GA, GU d) UG, AG
- 151. Splicing activity is a function of [AIIMS May 2010] a. mRNA b. snRNA c. tRNA d. rRNA
- 157. Group I Self-Splicing is mediated by a) Adenosine b) Guanosine c) Cytosine d) None
- 159. 5S RNA is transcribed by a) RNA polymerase I b) RNA polymerase II c) RNA polymerase III d) All of the above
- 162. In mitochondrial mRNA, UGA codes for a) Methionine b) N formyl-methionine c) Tryptophan d) None
- 165. Which is NOT a synonymous codon pair? a) CAU & CAC b) AUU & AUC c) AUG & AUA d) AUU & AUC
- 167. apoB48 & apoB100 production is due to [AIIMS May 11] a) alternative splicing b) mRNA editing c) alternative polyadenylation d) mRNA silencing
- 169. Apo B-48 and Apo B-100 is synthesized from the same mRNA; the difference between them is due to [AIIMS May 2011] a. RNA spicing b. Allelic exclusion c. Deamination of cytidine to uracil d. Upstream repression
- 170. During translation, initiator tRNA binds to a) A site b) P site c) E site d) None
- 172. P – bodies are a) present in the nucleus b) the site of protein synthesis c) the sites of translation repression d) exclusively involved in the destruction of RNA
- 175. The molecular mechanism behind the alternative production of membrane bound and secretory IgM is a) Allelic exclusion b) Somatic hypermutation c) Class switching d) Alternate polyadenylation
- 177. Molecular mechanism responsible for affinity maturation of antibodies is a) Somatic hypermutation b) RNA editing c) RNA splicing d) Allelic exclusion
- 179. The molecular mechanism behind T cells expressing only one type of T cell receptor is a) Allelic exclusion b) Somatic hypermutation c) Class switching d) Alternate polyadenylation