Biotechnology Eligibility Test Preparation

Biotechnology BET JRF Exam 2014 Original Solved Question Paper with Answer Key (Download PDF)


DBT BET JRF 2014 Question Paper PDF

Original (original) Previous Year (old) Solved Question Paper of DBT BET JRF 2014 Examination (Department of Biotechnology- Biotechnology Eligibility Test – Junior Research Fellowship) with Answer Key and Explanations as PDF. DBT BET JRF aspirants can download the question paper as single PDF file for your exam preparation. Please feel free to inform us for any mistakes in the answer key provided.

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Biotechnology Eligibility Test Preparation

DBT BET Previous Year Question Paper: 2012-2013 with Answer key: Part 4 (Question 101 – 150)


Telomer-structure

Three-dimensional structure of a telomere G-quadruplex (source wikipeida)

Biotechnology Eligibility Test (BET) for DBT JRF Award (2012-13)
Previous Year Question Papers with Answer Key, Explanations & References
Government of India, Ministry of Science & Technology
Department of Biotechnology, New Delhi
PART: B Set 4 (Questions 101– 150)


101. Telomeres consists of simple sequence repeats of

a. CATT rich strands that interact with protein
b. GCCT rich strands that interact with protein
c. CTTT rich strands that interact with protein
d. TTAGGG rich strands that interact with protein

Ans. (a)

Telomere is the sequence occurs at the end of chromosome. Telomere helps to maintain the integrity of chromosome by preventing the end to end fusion of different chromosome and deterioration of chromosomes by nucleases. In vertebrates telomere consists of TTAGGG sequence repeats (~2500 repeats).

102. True activators of transcription are transcription factors that binds to

a. Other proteins to enhance transcription
b. Promoters
c. Enhancers
d. Promoters and enhancers

Ans. (c)

Enhancers are cis acting transcription factor binding sites of DNA (50 – 1500 bp long) of specific genes

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Biotechnology Eligibility Test Preparation

DBT BET Previous Year Question Paper: 2012-2013 with Answer Key: Part 3 (Questions 51-100)


PLoSBiol3.5.Fig7ChromosomesAluFish

Alu sequence in Human Chromosomes (green fluorescence) Source Wikipedia

Biotechnology Eligibility Test (BET) for DBT JRF Award (2012-13)
Previous Year Question Papers with Answer Key, Explanations & References
Government of India, Ministry of Science & Technology
Department of Biotechnology, New Delhi

 PART: B Set 3 (Questions 51 – 100)


51. Suppressor tRNA mutations are those in which

a.       Transcription of tRNA genes is suppressed
b.      Translation form mRNA is suppressed due to absence of tRNA
c.       Amino acid is incorporated in place of a stop codon due to mutation in anticodon region of tRNA
d.      Charging of tRNA with cognate amino acids is suppressed due to mutation in amino acyl tRNA synthase enzyme

Ans. (c)

Suppressor tRNA: also called as nonsense suppressor, due to the suppressor tRNA mutations, an amino acid is incorporated into the stop codon due to mutation in the anticodons of tRNA so that the releasing factors of translation cannot binds to the complex and the translocation process continues.

52. Which of the following media is best suited for the selective growth of E. coli with genotype: Str+ His- leu- lys-?

a.       Minimal medium with thiamine, histidine, leucine and lysine
b.      Luria agar
c.       Minimal medium with thiamine and streptomycin
d.      Minimal medium with thiamine, histidine, leucine, lysine and streptomycin

Ans. (d)

Minimal medium: media that contains minimum nutrients possible for bacterial growth, usually without the presence of amino acids. They used by microbiologists and geneticists to grow “wild type” microorganisms and to select microbes against recombinants.

In the question, the given bacterium is mutated for Histidine, Leucine, and lysine and it has got an antibiotic resistant gene Str+.

53. Which of the following amino acid is coded by maximum number of codons?

a.       Leucine
b.      Tryptophan
c.       Valine
d.      Alanine

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Biotechnology Eligibility Test Preparation

DBT BET JRF Exam 2012-13 Previous Year Question Paper With Answer Key and Explanations Part 2


What are microsatellites?

DNA samples from specimens of Littorina plena amplified using PCR with primers targeting a variable simple sequence repeat (SSR, a.k.a. microsatellite) locus, run on a 5% polyacrylamide gel, visualized by silver staining (source wikipedia)

Biotechnology Eligibility Test (BET) for DBT JRF Award (2012-13)
Previous Year Question Papers with Answer Key, Explanations & References
Government of India, Ministry of Science & Technology
Department of Biotechnology, New Delhi

PART: A Set 2 (Questions 26 – 50)

26.  Which one of the following is the natural host for pseudo-rabies virus?

a.      Dog
b.      Man
c.       Swine
d.      Horse

Ans. (c). Swine

Pseudorabies (Aujeszky’s disease) is a viral disease of swine caused by Suid herpesvirus 1 (SuHV1) belongs to the family Herpesviridae. Other species such as sheep, cattle, and human are also susceptible to SuHV1. This disease is nothing to do with rabies virus. The notorious Rabies virus belongs to the family Rhabdoviridae which cause rabies in human and animals.

27.  Which one of the following is the causative agent of fowl cholera?

a.       V. cholera
b.       P. multocida
c.       E. coli
d.       S. pullorum

Ans. (b). P. multicida

Fowl cholera or avian cholera is caused by Pasteurella multocida. Fowl cholera is considered as Zoonoses

Zoonoses: an infectious disease of animals that can be transferred to human, Example: Rabies

S. pullorum is a poultry adapted species of Salmonella

28.  The wavelengths of light that penetrate the least into the ocean are

a.       Red and violet
b.      Red and yellow
c.       Blue and brown
d.      Green and blue

Ans. (b). Red and Yellow

Yellow, orange and red waves have the least energy in the visible spectrum, thus they cannot penetrated deep into the sea. This is the reason why most of the deep sea animals (fishes) are red in colour. In the deep sea, due to absence of enough light the red colour appears as black so that these animals can be escaped from their predators.

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Biotechnology Eligibility Test Preparation

DBT BET JRF Exam: Previous Year Question Paper With Answer Key and Explanations: BET 2012 – 2013 Part 1


What is immunoglobulin domain

(Image Source Wikipedia)

Biotechnology Eligibility Test (BET) for DBT JRF Award (2012-13)
Previous Year Question Papers with Answer Key, Explanations & References
Government of India, Ministry of Science & Technology
Department of Biotechnology, New Delhi

PART: A Set 1 (Questions 001 – 025)


1. The immunoglobulin fold is made up of:

a.       Seven alpha helical segments
b.      A beta barrel
c.       A sandwich of two parallel beta sheets
d.      A sandwich of two antiparallel beta sheets

Ans. (d). A sandwich of two antiparallel beta sheets

Immunoglobulin fold consists of a pair of β-sheets each built of antiparallel β-strands connected by a disulfide bridge. This β strands surrounds a central hydrophobic core. Immunoglobulin fold is one of the most predominant domains encoded by the human genome.

For more details refer: Biochemistry by Stryer, Ed.6, Chapter: 33, Immune systems, Page 951


2. RNA is analyzed for the location of hairpin folds. Which of the following sequence could form a mini-hairpin?

a.       AGGUUUCCU
b.      AAAAAAAAA
c.       AGGUUUGGA
d.      AGGUUUAGG

Ans. (a). AGGUUUCCU

This RNA molecule can loop around itself and can form double stranded structure due to the sequence complementarity. First A is complementary to last U, second and third nucleotide (UU) is complementary to CC. The middle UUU forms the junction of loop and they do not form the complementary hydrogen bonds.


3. Increasing the concentration of which of the following would most effectively anatomized the inhibition of protein synthesis by puromycin?

a.       ATP
b.      eIF2, GTP
c.       Aminoacyl-tRNAs
d.      Peptidyl-tRNAs

Ans. (c). Aminoacyl-tRNA

Puromycin is an antibiotic obtained from a bacterium Streptomyces alboniger, which inhibit protein synthesis by specifically inhibiting the translation process. Puromycin inhibits protein translation by premature chain termination process. A part of the puromycin structurally mimics the 3’ end of Aminoacyl-tRNA molecule and hence they can enter to the ‘A’ site of ribosome and form a nascent peptide-puromycin adduct. This adducts formation results in the release of peptide from the ribosome. Puromycin compete with aminoacyl-tRNA for the A site of ribosome. Thus the inhibitory effect of puromycin can be theoretically overcome by increasing the concentration of the competitor, i.e., aminoacyl-tRNA.

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