Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1051.

Identify A, B ,C and D in the given diagram : A. `A-"ori", B-"amp"^(R), C-"tet"^(R),D-"Hind" III`B. `A-"Hind" III , B-"tet"^(R), C-"amp"^(R), D-"ori"`C. `A-"amp"^(R),B-"tet"^(R),C-"Hind" III,D-"ori"`D. `A-"tet"^(R),B-"Hind" III, C-"ori",D-"amp"^(R)`

Answer» Correct Answer - A
1052.

The complementary synthetic and random DNA are used as :A. TransposonsB. Passenger DNAC. Cloning vectorsD. Recombinant DNA

Answer» Correct Answer - B
Foreign`//`passenger DNA is a fragment of DNA molecule which is enzymatically isolated and cloned. The gene is identified on a genome and pulled out from it either before or after cloning . The cloned foreigh DNA fragment expresses normally as in parental cell.
1053.

A bacterial cell was transformed with a recombinant DNA that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reason could beA. human gene may have intron which bacteria cannot processB. amino acid condons for human and bacteria are differentC. human protein is formed, but degraded by bacteriaD. All of the above

Answer» Correct Answer - a
1054.

A bacterial cell was transformed with a re- combinant DNA that was generated using a human gene. However, the !ransform~d cells did not produce the desired protein. Reasons could beA. human gene may have intron which bacteria cannot processB. amino acid codons for humans and bac- teria are differentC. human protein is formed but degraded by bacteriaD. all of the above

Answer» Correct Answer - a
1055.

A bacterial cell was transformed with a recombinant DNA that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reason could beA. Human gane may have intron which bacteria cannot processB. Amino acid codons for humans and bacteria are differentC. Human protein is formed but degraded be bacteriaD. all of the above

Answer» Correct Answer - A
1056.

Who developed the immunological of selection of transformed cells ?A. Gunstein HognessB. BroomC. GilbertD. Both (b) and (c )

Answer» Correct Answer - a
1057.

Which one of these is not a tool of recombinant DNA technology ?A. VectorB. IntronsC. Restriction enzymeD. Polymerase enzyme

Answer» Correct Answer - B
1058.

A bacterial cell was transformed with a recombinant DNA that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reason could beA. Human gene may have intron which bacteria cannot process.B. Amino acid codons for humans and bacteria are differentC. Human protein is formed but degraded be bacteria.D. All of the above .

Answer» Correct Answer - A
1059.

Which of the following is a restricition endonuclease ?A. Hind IIB. RnaseC. Dnase ID. Protease

Answer» Correct Answer - A
1060.

Restriction endonucleases are enzymes which:A. make cuts at specific positions within the DNA moleculeB. recognise a specific nucleotide sequence for binding of DNA ligaseC. restrict the action of the enzyme DNA polymeraseD. remove nucleotides from the ends of the DNA molecule

Answer» Correct Answer - a
1061.

Restriction endonucleases are enzymes thatA. restricts the action of other enzymesB. break phosphodiester bond between specific nucleotides of DNAC. break nucleus into piecesD. break H-bond between two strands of DNA

Answer» Correct Answer - b
1062.

Consider the following statements. A. Endonucleases remove nucleotides from the ends of the DNA whereas, exonucleases make cuts a specific positions within the DNA. Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites. C. Asexual reproductions provides opportunities for variations and formulation of unique combinations of genetic setup. ltbrtgt D. DNA fragments can be separated by a technique known as gel electrophoresis. Of the above statements:A. A and B are correctB. B and D are correctC. A and C are correctD. A and C are correct

Answer» Correct Answer - B
1063.

One of the methods by which DNA connot be transformed to the host cell is by :A. gene gunB. microinjectionC. polymerase chain reactionD. disarmed pathogen vectors

Answer» Correct Answer - C
1064.

Restricition endonucleases :A. are used for in vitro DNA synthesis.B. are synthesized by bacteria as part of defence mechanism.C. are used in genetic engineering for ligation of two DNA molecules.D. are present in mammalian cells for degradation of DNA when the cell dies.

Answer» Correct Answer - B
1065.

With respect to DNA fragmentation: Statement A : Gel electrophoresis and elution are two important processes. Statement B : After staining with ehidium bromide it has to be exposed to UV light.A. Only A is correctB. Both A and B are correct statementsC. Only B is correctD. Only A is correct and B is not correct

Answer» Correct Answer - B
1066.

In pBR 322, recognition segqences, that are present on the same type of selectable markerA. Pst I, Pvu IB. Bam HI, Sal IC. Both A and BD. None of the above

Answer» Correct Answer - C
In pBR 322, recognition sequences, Pst-I and Pvu-I are present on same type of selectable marker (`amp^(R)` gene) while BamH -I and Sal-I are present on `"tet"^(R)` gene
1067.

The genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline or kanamycin, etc. are considered useful selectable markers inA. E. coliB. Bacillus thuringiensisC. Meloidegyne incognitiaD. All of these

Answer» Correct Answer - a
1068.

Which of the following features are required to facilitate cloning into a vector ? a. Origin of replication (ori) b. Selectable marker c. Cloning siteA. a and bB. b and cC. a and cD. a, b and c

Answer» Correct Answer - D
Origin of replication (ori), selectable marker and cloning site are required to faciliate cloning into a vector
1069.

In the year 1963, the two enzymes responsible for restricting the growth of bacteriophage in E. coli were isolated. One of these added methyl groups to DNA, while the other cut DNA. The latter was calledA. restriction endonucleaseB. methylaseC. ligasesD. plasmid

Answer» Correct Answer - a
1070.

A mutant strain of `T_4` – Bacteriophage, R-II, fails to lyse the E-Coli but when two strains `R-II^X` and `R-II^Y` are mixed then they lyse the E.Coli. What may be the possible reason : -A. Bacteriophage transforms in wildB. It is not mutatedC. Both strains have similar cistronsD. Both strains have different cistons

Answer» Correct Answer - D
Two strains of `T_(4)` have different cistrons the codes for different proteins, which togther lyse the E.coli.
1071.

The function of selectable marker is :A. elimination of non-transformants and permiting transformantsB. eliminating transformants and permiting non-transformantsC. to destroy recognition sitesD. identify ori site

Answer» Correct Answer - A
1072.

In the year ....a... the two enzymes responsible for ...b... the growth of bacteriophage in E. coli were isolated. One of these added ..c... group to DNA, while other ...d... DNA. The later was called ...e...A. a - 1963, b - promoting, c -ethyl, d-join, e-DNA ligaseB. a-1963, b-restricting, c-methyl, d-join, e-DNA ligaseC. a-1972, b-restricting, c-methyl, d-cut, e-restriction endonucleasesD. a-1963, b-restricting, c-methyl, d-cut, e-restriction endonuclease

Answer» Correct Answer - D
In the year 1963 the two enzymes responsible for restricting the growth of bacteriophase in E. coli were isolated. One of these added methyl group of DNA, while other cut DNA. The later was called restriction endonucleases
1073.

Specify the role of  "cos " sites in bacteriophage lambda.

Answer»

“cos” sites are important for packaging DNA into phage head. 

1074.

How can we use LEU 2 gene as a selectable marker?

Answer»

The LEU2 gene codes for an enzyme which is needed for the synthesis of the amino acid leucine. Yeast cells having this plasmid can grow on a medium lacking leucine and hence can be selected over cells not containing the plasmid.

1075.

4 copies of ds DNA are subjected to polymerase chain reaction. How many copies would be obtained after 20 cycles?

Answer»

copies would be obtained after 20 cycles 4 x 220 .

1076.

If haemoglobin (Hb) of a normal individual and a sickle-cell anaemia patient are run in electrophoretic field, they will showA. same mobilitiesB. different mobilitiesC. Hb of patient will not move at allD. Hbs are immobile

Answer» Correct Answer - B
1077.

An important objective of biotechnology in agricuture section is to:A. increase plant weightB. decrease seed numberC. increase nitrogen contentD. produce pest resistant varieties of plant

Answer» Correct Answer - D
1078.

How can bacterial DNA be released from the bacterial cell for biotechnology experiments?

Answer»

The bacterial cell wall is digested by the enzyme lysozyme to release DNA from the cell.

1079.

Indicate the inheritance pattern, genomic location and mutation in any two diseases caused by single gene mutations which follow mendelian inheritance. Also, specify the genomic location in any two diseases resulting from gene polymorphisms with complex inheritance. 

Answer»
Single-gene mutations which follow mendelian inheritanceGene polymorphisms which has complex inheritance
Cystic Fibrosis (Cystic Fibrosis Transmembrane Conductance Regulator CFTR gene)
1. Inheritance: autosomal recessive disease
2. Genomic location: Chromosome 7 (7q31.2)
3. Mutation: The most common mutation is a deletion of 3 bps resulting in the loss of codon no. 508, which codes for phenylalanine
Common late-onset Alzheimer's disease
1. Inheritance: Major cause is epsilon4 allele of the gene coding for apolipoproteinE (APOE)
2. Genomic location: Chromosome 19 (19q13) and recently Chromosome 10 (10q21).
Huntington disease (Huntingtin gene HTT)
1. Inheritance: autosomal dominant
2. Location: Chrosome 4 (4p16.3)
3. Mutation: increased number of CAG repeats more than 35 times
Migraine
1. Susceptibility locus:
Chromosome 6p12.
2 - 6p21.1 and Chromosome 1q31
1080.

The autonomously independent self replicating extranuclear DNA imparting cartain factors to some bacterium is called:A. cosmidB. plastidC. plasmidD. phagemid

Answer» Correct Answer - C
1081.

Autonomously replicating circular extrachromosomal DNA is calledA. ChromatinB. PlasmidC. Palindromic nucleotide sequenceD. Nucleoid

Answer» Correct Answer - B
Autonomously replicating circular extrachromosomal DNA is called plasmid
1082.

A part from DNA in the bacterial nucleoid there is a circular extrachromosomal DNA called:A. plasmidB. mesosomeC. chromosomeD. none of these

Answer» Correct Answer - A
1083.

If a plasmid vector is digested with EcoRl at a single site, thenA. one sticky end will be producedB. two sticky ends will be producedC. four sticky ends will be producedD. six sticky ends will be produced

Answer» Correct Answer - B
Plasmid is a circular DNA, if it is digested at a single site, one fragment will be produced with two sticky ends.
1084.

Explain palindromic nucleotide sequence with the help of a suitable example.

Answer»

Palindrome in DNA is a sequence of base pairs that reads the same on two strands when orientation of reading is the same.

Example: 5' -------- GAATTC ----------3'

                3' -------- CTTAAG ----------5'

1085.

Explain palindromic nucleotide sequence with the help of a suitable example.

Answer»

The palindrome in DNA is a sequence of base pairs that reads same on the two strands when orientation of reading is kept the same. For example, the following sequences reads the same on the two strands in 5′ → 3′ direction. This is also true if it is read in the 3′ → 5′ direction.

5′ — — GAATTC — — 3′

3′ — — CTTAAG — — 5′

1086.

Identify the palindromic sequence in the followingA. `(GA AT T C)/(CT TU UG)`B. `(GGATC C)/(C CTAGG)`C. `(C CTG G)/(GGA C C)`D. `(C DATA)/(GCT A A)`

Answer» Correct Answer - B
The palindromes in DNA are base pair sequences that are the same when read forward (left or right) or backward (right to left) from a central axis of symmetry. Thus, `(GGATC C)/(C CTAGG)` is a palindromic sequences.
1087.

What is the utility of the bacterium, Thermus aquaticus?A. It is used in RELP mappingB. It is used to create recombinant plasmidsC. It is used in automated DNA sequencingD. It facilitates the polymerase chain reaction

Answer» Correct Answer - D
1088.

Describe the roles of heat, primers and the bacterium Thermus aquaticus in the process of PCR.

Answer»

• Heat denatures or helps in separation of DNA into two strands.

• Primer–Enzyme DNA Polymerase extend the primers using the nucleotides provided in the reaction and the genomic DNA as template.

Thermus aquaticus: It is the source of thermostable DNA polymerase or Taq polymerase.

1089.

The extraction of DNA fragment from agarose gel is called_______ .

Answer» Correct Answer - Elution
1090.

Which of the following step is not involved in basic steps in genetically modifying an organismA. Identification of DNA with desirable geneB. Introduction of the identified DNA into the hostC. Amplification of DNA by using PCRD. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny

Answer» Correct Answer - C
1091.

The DNA probe CTTCAAT will hybridize DNA containing:A. GAAGTTAB. GUUGAAUC. CTTCAATD. GAAGAAT

Answer» Correct Answer - A
1092.

Which one of the following technique is not used for early molecular diagnosis ?A. polymerase chain reactionB. polyacrylamide gel electrophoresisC. Recombinant DNA technologyD. Enzyme linked immunosorbent assay

Answer» Correct Answer - C
1093.

There are three basic steps in genetically modifying an organism. Arrange these steps in correct sequence (a) Introduction of the identified DNA inot the host. (b) Maintenance of introduced DNA in the host and transfer of the DNA to its progeny. (c) Identification of DNA with desirable genesA. `a rarr b rarr c`B. `b rarr c rarr a`C. `c rarr b rarr a`D. `c rarr a rarr b`

Answer» Correct Answer - D
There are three basic steps in genetically modify organism. I these steps in correct sequence are
c. Identification of DNA with desirable genes.
b. Maintenance of introudced DNA in the host and transfer of the DNA of its progeny.
a. Introduction of the identified DNA into the host
1094.

In DNA segment the probe binds is identified by its size by using a technique called:A. DNA probeB. DNA denaturationC. DNA polymorphismD. None of the above

Answer» Correct Answer - D
1095.

In blue-white screening the recombinants can be identified asA. colourless coloniesB. blue coloured coloniesC. red coloured coloniesD. None of the above

Answer» Correct Answer - a
1096.

Give reasons for the following:(i) Kappa casein is involved in micelle stabilization of milk proteins.(ii) Whey protein detoxifies xenobiotics.(iii) Curd is used as pro-biotic.

Answer»

(i) kappa casein contains a lipid molecule. 2/3 of the protein is hydrophobic

(ii) Whey protein results in elevation of a tripeptide glutathione in cells which detoxifies xenobiotics.

(iii) Curd is a good source of beneficial bacteria which can colonise the intestinal tract. 

1097.

The publication of ‘Atlas of Protein Sequences and Structure’ under the editorship of Margaret O’ Dayhoff was a pioneering effort. Why? 

Answer»

Macromolecular Sequences first compiled in this Atlas

Development of computer methods for comparison of protein sequences. Detection of various features from sequences like duplications, evolutionary histories, alignments etc. 

1098.

A specific DNA sequence is identified from a heterogeneous population of DNA molecules on the basis of DNA-DNA hybridization. (a) Identify the technique. (b) Schematically depict the steps of DNA separation technique. 

Answer»

(a) Southern hybridization

(b) (i) Digestion of genomic DNA with restriction enzymes

(ii)  Electrophoresis on agarose gel

(iii) Transfer to a nitrocellulose membrane (Southern Blotting)

(iv) Hybridization with labeled probe

(v) Detection of labeled probe. 

1099.

Based on Genomic studies, why do people say that different species and organisms had a common ancestor 100 million years ago. 

Answer»

It is due to SNP that is single nucleotide polymorphism

It is estimated that differences between human and chimpanzee is only 1-3% while human and mouse share about 97.5% of their working DNA 

1100.

RNA interfeence is essential for theA. cell defenceB. cell proliferationC. micropropagationD. cell differentiation

Answer» Correct Answer - A