LE 7 -2024 Flashcards by

ECOR1 is a:
A. DNA ligase enzyme
B. A vector used for insulin synthesis
C. Restriction endonuclease
D. A plasmid used as a vector

C. Restriction endonuclease

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Construction of a recombinant DNA involves:
A. Cleaving DNA with restriction endonuclease and joining with ligase
B. Cleaving DNA with ligase and joining with endonuclease
C. Cleaving and joining DNA with restriction endonuclease
D. Cleaving DNA with restriction endonuclease and joining with polymerase

A. Cleaving DNA with restriction endonuclease
and joining with ligase

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“Thermal Cycler” is used in the reaction:
A. Enzyme linked immuno sorbant assay
B. Polymerase chain reaction
C. Ligation reaction
D. Immobilization reaction

B. Polymerase chain reaction

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The restriction Endonucleases are called so because they:
A. Have a very restrictive or site specific endonuclease activity
B. Cut DNA at a few restricted sites
C. Restrict the entry of foreign DNA into the cell by cleaving the DNA due to their endonuclease activity
D. Their distribution is restricted to only some bacterial cells

A. Have a very restrictive or site specific endonuclease activity

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A segment of DNA that reads from the same forward and backward is called:
A. Palindromic DNA
B. Plasmid DNA
C. Complementary DNA
D. Copy DNA

A. Palindromic DNA

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A part of nucleic acid used to find a gene by hybridization is called:
A. Vector
B. Clone
C. Probe
D. Cybrid

C. Probe

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cDNA, a term used in recombinant DNA technology means:
A. Competitive DNA
B. Complex DNA
C. Chemical DNA
D. Complementary DNA

D. Complementary DNA

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During the recent tsunami disaster a child was separated from its parents in Sri Lanka. Later with the help of technique the child was made to reunite with its true parents. The technique is:
A. DNA finger printing
B. Tissue culture
C. Gene therapy
D. Hybridoma technology

A. DNA finger printing

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Electrophoresis, a technique used in DNA fingerprinting helps to separate:
A. DNA segments
B. Tissues
C. Cells from DNA
D. RNA from DNA

A. DNA segments

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RFLP is:
A. Repeated fragment length polymorphism
B. Renewed fragment length polymorphism
C. Required fragment length polymorphism
D. Restriction fragment length polymorphism

D. Restriction fragment length polymorphism

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An example for autonomously replicating a mini chromosome is:
A. virus
B. Phage
C. Plasmid
D. Lichen

C. Plasmid

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In restriction endonuclease EcoR1, “E” stands for:
A. Extraction
B. Endonuclease
C. The first letter of the genus in which it is present
D. Endangered

C. The first letter of the genus in which it is present

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Which one of the following statement are attributes of plasmids, EXCEPT:
A. They are circular DNA molecule
B. They have antibiotic resistant genes
C. They have the ability of autonomous replication
D. They have DNA that is as long as chromosomal DNA

D. They have DNA that is as long as chromosomal DNA

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Which of the following statement about a vector is correct:
A. All vectors are plasmids only
B. Plasmids, phages can be used as vectors
C. Fungi can also be used as vectors
D. Cyanobacteria can also be used as vectors

B. Plasmids, phages can be used as vectors

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An ideal plasmid to be used for recombinant DNA technology must have:
A. Minimum amount of DNA
B. Relaxed replication control
C. One recognition site for one restriction endonuclease
D. All of these

D. All of these

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Prior to the production of recombinant insulin, insulin obtained from cows and pigs were given to patients. Some of the problems faced by this treatment was:
A. The insulin was not active
B. In some humans it induced antibody production
C. It reduces the weight of patients
D. Loss of memory power

A. The insulin was not active

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The first human protein produced through recombinant DNA technology is:
A. Insulin
B. Erythropoietin
C. Interferon
D. Somatostatin

A. Insulin

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A plasmid consisting of its own DNA with a foreign DNA inserted into it is called:
A. Recombinant DNA
B. Junk DNA
C. Non-coding DNA
D. None of the above

A. Recombinant DNA

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A recombinant DNA molecule is produced by joining together:
A. One mRNA with a DNA segment
B. One mRNA with a tRNA segment
C. Two mRNA molecules
D. Two DNA segments

D. Two DNA segments

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A gene produced for recombinant DNA technology contains a gene from one organism joined to the regulatory sequence of another gene. Such a gene is called:
A. Oncogene
B. Junk gene
C. Chimeric gene
D. None of the above

C. Chimeric gene

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A group of genetically similar organisms obtained by a sexual reproduction is called:
A. Clone
B. Population
C. Assembly
D. None

A. Clone

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To be useful in the preparation of recombinant DNA, a plasmid must have:
A. No origin of replication
B. An origin of replication
C. The ability to alternate between the linear and circular forms
D. Restriction endonuclease activity

B. An origin of replication

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Restriction endonucleases have the ability of cutting:
A. DNA at random site
B. DNA at specific sites
C. DNA and RNA at random sites
D. BothAandB

B. DNA at specific sites

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Endonucleases, a group of enzymes cleave DNA:
A. Externally
B. Internally
C. Both Aand B
D. Neither A nor B

B. Internally

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The extra chromosomal, self-replicating, double stranded, closed, circular DNA molecules are called: A. Plasmids B. Phages C. Viruses D. Chloroplasts

A. Plasmids

Which of the following best describes semiconservative replication? A. The replication of DNA takes place at a defined period in the cell cycle B. A DNA molecule consists of one parental strand and one new strand C. The number of DNA molecules is doubled with every other replication D. The replication of DNA never takes place with 100% accuracy

B. A DNA molecule consists of one parental strand and one new strand

27. The final product of DNA replication is: A. mRNA, tRNA, and rRNA molecules B. A wide variety of proteins C. One DNA molecule D. 2 DNA molecules, each of which contains one new & one old DNA strand

D. 2 DNA molecules, each of which contains one new & one old DNA strand

28. Meselson and Stahl separated DNA from different generations using: A. Density gradient centrifugation B. Gel electrophoresis C. An electron microscope D. Differential radioisotope labelling

A. Density gradient centrifugation

29. Which of the following cause the unwinding of the DNA double helix? A. DNA polymerase B. DNA helicase C. RNA primer D. Primosome

B. DNA helicase

30. A replication fork is: A. Only seen in bacterial cells B. A Y-shaped structure where both DNA strands are replicated simultaneously C. A site where one DNA strand serves as a template, but the other strand is not replicated D. Is created by the action of the enzyme RNA polymerase

B. A Y-shaped structure where both DNA strands are replicated simultaneously

31. In replication, once the DNA strands have been separated, reformation of the double helix is prevented by: A. DNA helicase enzyme B. Single-stranded binding proteins C. DNA polymerases D. ATP

B. Single-stranded binding proteins

Enzymes called ____________ form breaks in the DNA molecules to prevent the formation of knots in the DNA helix during replication. A. Topoisomerases B. DNA polymerases C. RNA polymerases D. DNA ligases

A. Topoisomerases

33. Which of the following adds new nucleotides to a growing DNA chain? A. DNA polymerase B. DNA helicase C. Primase D. RNA polymerase

A. DNA polymerase

34. The reason why DNA synthesis only proceed in the 5 ́ to 3 ́ direction: A. DNA polymerases can only add nucleotides to the 3 ́ end of a polynucleotide strand B. The 3 ́ end of the polynucleotide molecule is more electronegative than the 5 ́ end C. That is the direction in which the two strands of DNA unzip D. That is the only direction that the polymerase can be oriented

A. DNA polymerases can only add nucleotides to the 3 ́ end of a polynucleotide strand

35. The 5 ́ end of each Okazaki fragment begins with: A. The same RNA primer that began synthesis on the leading strand B. A DNA primer binding to the template DNA C. DNA polymerase binding to the template DNA D. A separate RNA primer

D. A separate RNA primer

36. Primase is the enzyme responsible for: A. Introducing nicks into the DNA double strand in order to prevent the formation of knots B. Hydrolyzing ATP to facilitate DNA unwinding C. Making short strands of RNA at the site of replication initiation D. Forming a replication fork in the DNA double helix

C. Making short strands of RNA at the site of replication initiation

Okazaki fragments are joined together by: A. RNA polymerase B. DNA ligase C. DNA polymerase D. RNA ligase

B. DNA ligase

38. How is the chromosome of a bacterial cell replicated? A. Linear DNA molecule is replicated from multiple origins of replication bidirectionally B. Linear DNA molecule is replicated from one origin of replication bidirectionally C. Circular DNA molecule is replicated from one origin of replication bidirectionally D. Circular DNA molecule is replicated from one origin of replication unidirectionally

C. Circular DNA molecule is replicated from one origin of replication bidirectionally

39. The manner by which eukaryotic cells replicate their chromosomes: A. The linear DNA, from multiple origins of replication bidirectionally B. The linear DNA, from one origin of replication bidirectionally C. The circular DNA from multiple origins of replication bidirectionally D. The circular DNA from one origin of replication bidirectionally

A. The linear DNA, from multiple origins of replication bidirectionally

40. The correct designation for the DNA strand labelled “A” is: A. Leading strand template B. Okazaki fragment C. Lagging strand template D. RNA primer

A. Leading strand template

41. The segment labelled “F” is: A. Lagging strand B. Lagging strand template C. Okazaki fragment D. RNA primer

C. Okazaki fragment

The segment labelled “B” is: A. Leading strand B. Lagging strand C. Lagging strand template D. Okazaki fragment

A. Leading strand

43. The segment labelled “E” is: A. RNA primer B. Okazaki fragment C. Leading strand template D. Lagging strand template

A. RNA primer

43. The segment labelled “D” is: A. Replication fork B. Okazaki fragment C. Leading strand template D. Lagging strand template

A. Replication fork

E. coli DNA polymerase III: A. Can initiate replication without a primer B. Is efficient at nick translation C. Is the principal DNA polymerase in chromosomal DNA replication D. Represents over 90% of the DNA polymerase activity in E. coli cells

C. Is the principal DNA polymerase in chromosomal DNA replication

45. The 5' to 3' exonuclease activity of E. coli DNA polymerase I is involved in: A. Formation of a nick at the DNA replication origin B. Formation of Okazaki fragments C. Proofreading of the replication process D. Removal of RNA primers by nick translation

D. Removal of RNA primers by nick translation

46. What is the major polymerizing enzyme in E. coli? A. DNA pol I B. DNA pol II C. DNA pol III D. DNA pol IV

C. DNA pol III

47. What subunit of the major polymerizing enzyme in E. coli confers processivity? A. Gamma B. Delta C. Theta D. Beta

D. Beta

Among eukaryotes, what is the counterpart of the E. coli’s SSB protein? A. RPA B. RFC C. PCNA D. HD protein

A. RPA

What protein separates and unwinds the duplex DNA in eukaryotes? A. CMG complex B. DNA B C. MCM 2-7 D. CCD 45

C. MCM 2-7

What molecule once incorporated into the growing chain of DNA halts replication? A. Novobiocin B. 2’,3’-diseoxyadenosine triphosphate C. 2’-deoxycytidine triphosphate D. Actinomycin D

D. Actinomycin D

This type of mutation results in the premature termination of amino acid incorporation into a peptide chain and leads to a shorter polypeptide chain. A. Silent mutations B. Missense mutations C. Nonsense mutations D. Splice site mutations

C. Nonsense mutations

52. Which mutation involves a codon containing the changed base that may code for the same amino acid? A. Silent mutations B. Missense mutations C. Nonsense mutations D. Same sense mutations

A. Silent mutations

53. A type of DNA repair mechanism that involves activation of photolyase enzyme by UV light (320-370 nm) and splits abnormal base dimers apart. A. DNA polymerase B. Nucleotide excision repair (NER) C. Photoreactivation D. Demethylating DNA repair enzymes

C. Photoreactivation

54. This type of mutation may have no detectable effect because of the degeneracy of the genetic code. A. Silent mutations B. Missense mutations C. Nonsense mutations D. Frameshift mutations

A. Silent mutations

55. The type of mutation that alters gametes and are passed to the next generation. A. Somatic mutations B. Missense mutations C. Germline mutations D. Frameshift mutations

C. Germline mutations

56. This type of mutation will more likely occur if the changed base in the mRNA molecule were to be at the third nucleotide of a codon. A. Silent mutations B. Missense mutations C. Nonsense mutations D. Frameshift mutations

A. Silent mutations

57. Because of wobble, the translation of a codon is least sensitive to a change at which position? A. First B. Second C. Third D. Any position

C. Third

58. A mutation may appear that would then result in the premature termination of amino acid incorporation into a peptide chain and the production of only a fragment of the intended protein molecule. A. Silent mutations B. Missense mutations C. Nonsense mutations D. Frameshift mutations

C. Nonsense mutations

59. The type of mutations that only affects the individual in which the mutation arises. A. Somatic mutaions B. Missense mutations C. Germline mutations D. Frameshift mutations

B. Missense mutations

60. This amino acid is produced by just one coding triplet (codon). A. Arginine B. Leucine C. Methionine D. Serine

C. Methionine

61. The effect of this mutation will occur when a different amino acid is incorporated at the corresponding site in the protein molecule. A. Silent mutations B. Missense mutations C. Nonsense mutations D. Frameshift mutations

B. Missense mutations

62. This mutation result from deletion or insertion of nucleotides in DNA that generates altered mRNAs. A. Silent mutations B. Missense mutations C. Nonsense mutations D. Frameshift mutations

D. Frameshift mutations

63. The mRNA is read continuously from a start codon to a termination codon. Which among the following is a start codon? A. AUG B. UAA C. UAG D. UGA

A. AUG

64. Exposure of a cell to this agent can result in the covalent joining of two adjacent pyrimidines (usually thymines), producing a dimer. A. X-rays B. UV light C. Base analogs D. Alkylating agents

B. UV light

65. Exposure to this agent can cause double- strand breaks in DNA, which are potentially lethal to the cell. A. High-energy radiation B. Base analogs C. Alkylating agents D. Viruses

A. High-energy radiation

66. Which exemplifies the degeneracy of the genetic code? A. Methionine B. Arginine C. Tryptophan D. All of the above

B. Arginine

67. A type of base pair substitution that converts a purine- pyrimidine to the other purine-pyrimidine. A. Transition B. Base analogs C. Transversions D. Viruses

A. Transition

68. A type of base pair substitution that converts purine-pyrimidine to a pyrimidine-purine. A. Transition B. Base analogs C. Transversions D. . Viruses

C. Transversions

69. A type of base pair substitution that are more likely to result in nonsynonomous substitution. A. Transition B. Base analogs C. Transversions D. Viruses

A. Transition

70. A type of mutation wherein a base pair substitution results in substitution of an amino acid with similar chemical properties (protein function is not altered). A. Neutral nonsynonymous mutation B. Silent mutations C. Nonsense mutations D. Frameshift mutations

B. Silent mutations base pair substitution that results in the same amino acid.

71. A type of mutation which changes wild type (ancestral) to mutant (derived) gene. A. Neutral nonsynonymous mutation B. Silent mutations C. Reverse mutation D. Forward mutation

D. Forward mutation

72. A type of mutation which changes mutant (derived) to wild type (ancestral) gene. A. Neutral nonsynonymous mutation B. Silent mutations C. Reverse mutation D. Forward mutation

C. Reverse mutation

72. A type of mutation which a substation of an amino acid with similar chemical properties (protein functions is not altered). AA to AA A. Neutral nonsynonymous mutation B. Silent mutations C. Reverse mutation D. Forward mutation

A. Neutral nonsynonymous mutation

73. A type of spontaneous mutation wherein A or G are removed and replaced with a random base. A. Neutral nonsynonymous mutation B. Silent mutations C. Depurination D. Deamination

C. Depurination

A type of DNA repair mechanism where the 3’-5’ exonuclease activity corrects errors during the process of replication. A. DNA polymerase proofreading B. Photoreactivation C. Demethylating DNA repair enzymes D. Nucleotide excision repair

A. DNA polymerase proofreading

75. From the types of DNA repair mechanisms where damaged regions of DNA unwind and are removed by specialized proteins. A. DNA polymerase B. Nucleotide excision repair (NER) C. Methyl-directed mismatch repair D. DNA polymerase proofreading

B. Nucleotide excision repair (NER)

76. Which of the following antibiotics block the polypeptide exit tunnel on the 50s ribosome and prevent peptide chain elongation? A. Macrolides B. Clindamycin C. Streptogramins D. AOTA

D. AOTA

77. What is the role of the poly-A tail during translation initiation? A. It stimulates recruitment of the 40s ribosomal subunit to mRNA B. It recognizes the stop codons during the termination phase of translation C. It binds to transfer RNA and facilitates peptide bond formation D. It attaches to the Methionine bound to the methylguanosine cap structure and initiates the transcription of the upstream nucleotide sequences

A. It stimulates recruitment of the 40s ribosomal subunit to mRNA

78. The attachment of phenylalanine to a tRNA with an anticodon that reads “AAA” is facilitated by which enzyme? A. Aminoacyl tRNA synthetase B. RNA polymerase C. Peptidyltransferase D. Eukaryotic initiation factor

A. Aminoacyl tRNA synthetase

Which of the following is the first ANTICODON read during translation? A. UAB B. UAC C. GAG D. AAG

B. UAC

The genetic code is said to be degenerate. Therefore, which of the following is TRUE to the definition of degeneracy? A. UUU always codes for Phenylalanine only B. All throughout the process of evolution, the genetic code has undergone several revisions C. The change of the codon CUU to UUU alters the amino acid incorporated into the protein D. Leucine has as many as six codons

A. UUU always codes for Phenylalanine only

Which inhibitor blocks protein synthesis in eukaryotes but not in prokaryotes? A. Puromycin B. Cycloheximide C. Chloramphenicol D. Tetracycline

A. Puromycin

Which inhibits peptidyl transferase in the 60S ribo- somal subunit in eukaryotes, presumably by binding to an rRNA component.? A. Puromycin B. Cycloheximide C. Chloramphenicol D. Tetracycline

B. Cycloheximide

82. What is the sequence of the nucleotide that would give the amino acid sequence Met-Leu-Cys-Lys-Ala? A. 5’AUGCUGUGUAAGGCU3’ B. 3’AUGCUCAGUAAGGCA5’ C. 5’AUGUUGCGGAAUGUU3’ D. 3’AUGUUACGGAAUGCG5’

A. 5’AUGCUGUGUAAGGCU3’

83. If a nucleotide change from A→G is transition, a change from C→T is: A. Framshift mutation B. Transposition C. Transversion D. Transition

D. Transition

Which statement is true regarding eukaryotic protein synthesis? A. Many ribosomes can translate the same mRNA molecule simultaneously. B. Even before transcription is completed, translation of the mRNA can begin. C. The 18s rRNA of the 50s ribosomal subunit is responsible for the peptidyltransferase activity that joins the amino acids. D. Diptheria toxin inhibits prokaryotic but not eukaryotic protein synthesis.

C. The 18s rRNA of the 50s ribosomal subunit is responsible for the peptidyltransferase activity that joins the amino acids.

85. What is the role of the A site in translation? A. Binds to the tRNA holding the growing polypeptide chain of amino acids. B. Binds to incoming to aminoacyl-tRNA which has the anti-codon for the corresponding codon in the mRNA C. Binds to the amino acid that corresponds to the amino terminal end of the growing peptide chain D. Binds to the t-RNA on the acceptor arm of the translation exit site

B. Binds to incoming to aminoacyl-tRNA which has the anti-codon for the corresponding codon in the mRNA

86. A mutation in the Shine-Dalgarno sequence will result in: A. Termination of translation in bacteria B. Termination of transcription in eukaryote C. Inhibition or activation of translation in bacteria D. Inhibition of translation in eukaryotes

C. Inhibition or activation of translation in bacteria

87. The wobble hypothesis allows for non-standard base pairing between: A. DNA template and RNA transcript B. Codon and anticodon C. Exons and introns D. Template and coding strands

B. Codon and anticodon

88. Mutation of a gene that produces a nonsense effect results in: A. No change in the nucleotide sequence of the mRNA encoded by the gene B. No change in the amino acid sequence of the protein encoded by the gene C. No translation of the gene D. Premature termination of amino acid incorporation

C. No translation of the gene

89. Insulin enhances protein synthesis by: A. Increasing the activity of peptidyltransferase B. Promoting the binding of eIF-4E to the 5’cap of the mRNA C. Increasing the synthesis of the rRNA that makes up the 40s and 60s subunits of ribosomes D. Promoting the synthesis of essential amino acids from amphibolic intermediates

B. Promoting the binding of eIF-4E to the 5’cap of the mRNA

90. The sequence of a segment of a gene is given below. 5’TGTCCCTCTAGGGCA 3’ If this DNA segment is transcribed and translated, what will be the sequence of the amino acid of the polypeptide? A. Ser-Gly-Arg-Pro-Arg B. Cys-Pro-Arg-Gly-Thr C. Arg-Pro-Arg-Gly-Ser D. Thr-Gly-Arg-Pro-Cys

B. Cys-Pro-Arg-Gly-Thr

91. If point mutation occurred such that a G was changed to a T as shown below, what will be the most likely effect? 5’AGTCCCTCTAGG[T]CA 3’. A. Incorporation of a different amino acid B. Premature termination of amino acid incorporation C. Reading through of a normal termination codon D. No detectable effect

B. Premature termination of amino acid incorporation

92. The set of DNA and RNA sequences that determine the amino acid sequences used in the synthesis of an organism's proteins is the same. A. Degenerate B. Unambiguous C. Non-overlapping D. No punctuation E. Universal F. Nonsense

E. Universal Universality: The genetic code is virtually universal, that is, its specificity has been conserved from very early stages of evolution, with only slight differences in the manner in which the code is translated.

93. Given a specific codon, only a single amino acid is indicated. A. Degenerate B. Unambiguous C. Non-overlapping D. No punctuation E. Universal F. Nonsense

B. Unambiguous

94. Multiple codons decode the same amino acid. A. Degenerate B. Unambiguous C. Non-overlapping D. No punctuation E. Universal F. Nonsense

A. Degenerate

95. Three of the 64 possible codons do not code for specific amino acids. A. Degenerate B. Unambiguous C. Non-overlapping D. No punctuation E. Universal F. Nonsense

F. Nonsense

96. Isolated from castor bean, inactivates 28s ribosome by catalyzing the N-glycolytic cleavage of adenine. A. Puromycin B. Diphtheria C. Ricin D. Chloramphenicol E. Tetracycline

C. Ricin

97. Has an exotoxin that catalyzes the ADP ribosylation of EF-2. A. Puromycin B. Diphtheria C. Ricin D. Chloramphenicol E. Tetracycline

B. Diphtheria

98. Inhibits protein synthesis in prokaryotes and eukaryotes. A. Puromycin B. Diphtheria C. Ricin D. Chloramphenicol E. Tetracycline

A. Puromycin

99. Binds to 23s rRNA, inhibiting peptidyl transferase activity. A. Puromycin B. Diphtheria C. Ricin D. Chloramphenicol E. Tetracycline

D. Chloramphenicol

100. Prevents the binding of aminoacyl t-RNA to the bacterial ribosome A site. A. Puromycin B. Diphtheria C. Ricin D. Chloramphenicol E. Tetracycline

E. Tetracycline

C. No translation of the gene

24. A plasmid consisting of its own DNA with a foreign DNA inserted into it is called 1. recombinant DNA 2. non-coding DNA 3. junk DNA 4. none of the above

1. recombinant DNA

39. A gene for insulin has been inserted into a vector for the purpose of obtaining its protein product only. Such a vector is called 1. expression vector 2. suppression vector 3. storage vector for genomic library 4. none of the above

1. expression vector

40. Expression vectors are those 1. produce protein products 2. used for genomic libraries 3. used for chromosome synthesis 4. used for finger printing

1. produce protein products

41. E. coli is generally used for gene cloning because 1. it supports the replication of recombinant DNA 2. it is easy to transform 3. it is free from elements that interferes with replication and recombination of DNA 4. all of these

4. all of these

42. An ideal plasmid to be used for recombinant DNA technology must have 1. minimum amount of DNA 2. relaxed replication control 3. one recognition site for one restriction endonuclease 4. all of these

4. all of these

43. Transgenic organisms are 1. produced by gene transfer technology 2. extinct organisms 3. naturally occurring and endemic 4. produced by traditional plant breeding technique

1. produced by gene transfer technology

44. Transfer of recombinant plasmid into E. Coli cells needs 1. heat treatment 2. UV rays treatment 3. Cacl2 treatment 4. lysis

1. heat treatment

67. A small, 15-30 bases long nucleotide sequences used to detect the presence of complementary sequences in DNA sample during DNA finger printing is called 1. RFLP 2. Probe 3. VNTR 4. reporter gene

2. Probe

LE 7 -2024 Flashcards

(112 cards)