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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.
4. Enzymes catalyze reaction by:
a. Binding regulatory proteins
b. Covalently modifying active site residues
c. Binding substrates with great affinity
d. Selectively binding the transition state of a reaction with high affinity
Ans. (d). Selectively binding the transition state of a reaction with high affinity
Enzymes speed up the rate of reaction by reducing the activation energy of reactants. Activation energy is the energy required to start a chemical reaction. Enzymes adopt several methods to reduce the activation energy of reactants so that the reaction can proceed. Among different methods, the selective but high affinity binding of substrate molecules in transition state is the most common method to reduce activation energy.
Transition state: Short-lived molecular moment in which events such as bond breakage, bond formation, and charge development have proceeded to the precise point at which decay to either substrate or product is equally likely. Transition state of a molecule is not a chemical species and it should not be confused with a reaction intermediate such as enzyme substrate complex (ES) or enzyme product complex (EP).
The difference between the energy levels of the ground state and the transition state is known as the activation energy. Activation energy in enzyme kinetics is represented as ΔG‡. The rate of a reaction reflects on activation energy; the higher is the activation energy, the slower will be the reaction and vice versa.
For more details, refer: Lehninger’s Principles of Biochemistry, Ed. 5, by Lehninger et. al., Chapter 6, Enzymes, Page 186-188
Formation of PE from PS (source wikipedia)
5. Which of the following is a common reaction used for the formation of phosphatidyl ethanolamine in bacteria?
a. Decarboxylation of phosphatidyl serine
b. Demethylation of phospatidyl choline
c. Reaction of ethanolamine with CDP-diacylglycerol
d. Reaction of CDP-ethanolamine with CDP-diacylglycerol
Ans. (a). Decarboxylation of phosphatidyl serine
Phosphatidyl ethanolamine, phosphatidyl choline and phosphatidyl serine are the membrane lipids. They belong to the category glycerol-phospho-lipids. In bacteria, the phospatidylethanolamine is synthesized from phosphatidylserine by decarboxylation with the enzyme Phosphatidylserine decarboxylase.
Please see: Membrane Lipids: Properties, Structure and Classification
For details, refer: Lehninger’s Principles of Biochemistry, Ed. 5, by Lehninger et. al., Chapter 21, Lipid Biosynthesis, 21.3 Synthesis of membrane lipids, Page 824 – 831
6. Holliday junction is observed during:
a. Mitosis
b. Interphase
c. Recombination
d. DNA Repair
Ans. (c). Recombination
Holliday Junction (source wikipedia)
Genetic recombination during meiosis involves the physical breakage of individual DNA molecules and the ligation of the split ends from one DNA duplex with the split ends of the duplex from the homologous chromosome. The process of recombination should be highly precise and should occur without any addition or deletion of even a single base pair. In order to ensure the accuracy, the recombination process depends on the complementary base sequences that exist between a single strand from one chromosome and the homologous strand of another chromosome. The process of recombination is also assisted molecular machinery of cellular DNA repair.
Holliday junction and Holiday Model: The Holiday model explain the molecular mechanism of recombination between the non-sister chromatids of homologous chromosomes during meiotic crossing over. It is named after Robin Holliday, who proposed the existence of such a structure in 1964
For more details, refer: Cell and Molecular Biology by Gerald Karp, Ed.6, Chapter- 14: Cellular reproduction, Page 597-598
7. In humans, XX males and XY females are rare, such rare sexes are due to:
a. Deletion of XY chromosome
b. Deletion of Y chromosome
c. XY translocation
d. Duplication of X chromosome
Ans. (c). XY Translocation
Translocation is structural lesion (aberration) in the chromosomes happens through the exchange of segments of chromosomes between non homologous pairs.
Human shows XY sex determination system. Individuals with XX (homogametic) will be female and individuals with XY will be males (heterogametic). These two chromosomes (X and Y) are called sex chromosomes. The actual sex determination process is done by the Y chromosome and more than that X chromosomes do not have any role in sex determination in human. The Y chromosome in human is very small and contain only very few genes. The most important part of Y chromosome is the SRY region (sex determining region of Y). The SRY region encodes a protein called TDF (testis determining factor) which induce testis development in the embryo and such embryo will develop into male individuals. In the absence of TDF, the gonadial initials in the embryo develop female reproductive structures and that embryo will develop into a female individual. If an individual with sex chromosomal constitution XX and one of the X chromosomes carries the SRY region of Y chromosome will be a male. Similarly, an individual with XY chromosome in which, if the Y chromosome lacks the SRY region, (lost due to deletion) will be a female.
For more details on sex determination, refer: Principles of Genetics, Ed. 7 by Tamarin:, Chapter 5, Sex determination, Page 82 – 90
8. Induction of β-galactosidase activity by IPTG is due to
a. Stimulation of lac repressor function
b. IPTG binding to lac operon and inducing transcription
c. IPTG binding to lac I gene product and inhibiting its activity
d. Inhibition of β-galactosidase degradation
Ans. (c). IPTG binding to lac I gene product and inhibiting its activity
IPTG: Isopropyl β-D-1-thiogalactopyranoside is a structural analogue of allolactose. It can activate the lac operon just as allolactose of lactose and thus it acts as an inducer of any gene which is cloned under the lac promoter. The superiority of IPTG over allolactose is that, IPTG cannot be hydrolyzed by the cellular β-galactosidase enzyme and hence the concentration of IPTG in the cell can be maintained constantly for long time. Similar to allolactose, IPTG can binds to the lac repressor allosterically and this releases the lac repressor from the lac operator and allowing RNA polymerase to transcribe the genes.
lac repressor is the product of lac I gene.
For more details: Principles of Gene Manipulation and Genomics, Ed. 7 by Primrose, Chapter 15: Advanced transgenic technology, Page 301 – 305
9. Which of the following enzymes doesn’t require a primer?
a. RNA dependent DNA polymerase
b. DNA dependent DNA polymerase
c. DNA dependent RNA polymerase
d. Taq DNA polymerase
Ans. (c). DNA dependent RNA polymerase
DNA dependent RNA polymerization is the usual translation processes in which RNA polymerase synthesize different types RNAs (rRNA, mRNA, tRNA etc.) complementary to the DNA strand. RNA polymerase enzyme do not require a primer to initiate the synthesize RNA.
RNA dependent DNA polymerization is the process of reverse transcription; here DNA is synthesized using an RNA template by a special enzyme called reverse transcriptase.
DNA Dependent DNA Polymerization is the usual DNA replication process by the DNA polymerase enzyme, DNA polymerase always requires a primer to initiate DNA synthesis.
10. Which one of the following antibiotics attaches to 50S ribosome and inhibits its peptidyl-transferase activity?
a. Penicillin
b. Chloramphenicol
c. Trimethoprim
d. Amphotericin
Ans. (b). Chloramhenicol
Chloramphenicol is a broad spectrum antibiotic which inhibits the bacterial growth by specifically inhibiting the bacterial protein synthesis. Chloramphenicol binds to the 50S subunit of ribosome and there it specifically binds to the 23S rRNA. A part of 23S rRNA is the Peptidyl transferase enzyme which catalyzes the formation of peptide bind during protein synthesis. It specifically binds to A2451 and A2452 residues.
Penicillin: antibiotic obtained from Penicillium notatum, inhibit bacterial growth by preventing the peptidoglycan cell wall formation by inhibiting the enzyme transpeptidase. Transpeptidase enzyme add shot protein cross links in the mucopolysaccharide chains (of NAM and NAG).
Trimethoprim: another antibiotic, trimethoprim binds to dihydrofolate reductase and inhibits the reduction of dihydrofolic acid (DHF) to tetrahydrofolic acid (THF). THF is an essential precursor in the thymidine synthesis pathway and interference with this pathway inhibits bacterial DNA synthesis.
Amphotericin B: an antifungal drug originally extracted from Streptomyces nodosus. Amphotericin B interfere the membrane permeability by binding with ergosterol, a component of fungal cell membranes. This amphotericin ergosterol complex forms trans-membrane channel that leads to K+, Na+, H+ and Cl ions leakage, which ultimately leads to cell death.
11. DNA from a host sample can be amplified by a process known as the polymerase chain reaction (PCR). Which of the following is required for PCR?
a. Knowledge of the genetic sequence to be amplified
b. A single nucleotide primer
c. A universal probe to detect the amplified product
d. A heat sensitive DNA polymerase enzyme
Ans. (a). Knowledge of genetic sequence to be amplified
All the components in the PCR reaction such as DNA polymerase enzyme, dNTPs, buffer containing Mg2+ ions, primer etc. are equally important for the proper amplification of target DNA. DNA polymerase enzyme requires a shot RNA primer to start the DNA synthesis. The pair of primers, (forward and reverse primer), binds to the denatured template stand by sequence complementarity. Primers are synthesized complementary to the template strand and thus for the PCR amplification of a gene of interest, we have to know the sequence (complete or at least some portion) of the template strand that we are going to amplify.
12. Gluconeogenesis is not capable of making glucose from:
a. Adenine
b. Lactate
c. Acetyl CoA
d. Palmitate
Ans. (d). Palmitate
Gluconeogenesis: formation of glucose for non-carbohydrate sources such as pyruvate, lactate, glycerol, and glucogenic amino acids.
Palmitate: a common fatty acid present in plants and animals. Palmitate cannot be converted to glucose directly; however the beta oxidation of palmitate can release acetyl CoA, which can be converted to glucose.
For more details on gluconeogenesis, refer: Lehninger’s Principles of Biochemistry, Ed. 5, Chapter: 14.4 Gluconeogenesis, Page 551-558
13. Glycosylation of proteins occurs in the
a. Peroxisome
b. Mitochondrion
c. Lysosome
d. Endoplasmic reticulum
Ans. (d). Endoplasmic reticulum
Glycosylation: attachment of carbohydrate moieties to the proteins, for the production of glyco-conjugates such as glycoproteins and proteoglycans. There are two types of glycosylation, N linked glycosylation and O linked glycosylation. The process of glycosylation occurs inside the lumen of endoplasmic reticulum
For details of glycosylation process, refer: Lehninger’s Principles of Biochemistry, Ed. 5, Chapter: Protein Synthesis, Protein Metabolism, Page 1101 – 1106
14. The 20 different amino acids found in proteins are normally coded by:
a. 59 codons
b. 60 codons
c. 61 codons
d. 63 codons
Ans. (c). 61 codons
The alphabets of DNA contains four letters (nucleotides), they are A, T, G and C. As we know the genetic code is triplet code (contain three nucleotides) there will be 43 possibilities. There are a total of 64 codons (4 X 4 X 4 = 64), among these 64 codons; three are stop codons which do not code for any amino acids. Thus there will be 61 codons available for coding the 20 amino acids. The stop codons are UAA, UAG and UGA in the mRNA.
For more details on genetic code, refer: Genetics: A Conceptual Approach, Ed. 2 by Benjamin A. Pierce, Chapter 15. Genetic Code and Translation, Page 410-414
15. How many microliters of 0.1 M solution of sodium chloride will make 10 ml of 5 mM sodium chloride?
a. 200
b. 100
c. 500
d. 10
Ans. (c). 500
Use the N1V1 = N2V2 equation
N1 = 0.1
V1 = ?
N2 = 5
V2 = 10
? X 0.1 = 5 X 10
? X 0.1 = 50
? = 50/0.1 = 500
16. Which of the following is NOT found inside the eukaryotic nucleus?
a. Nucleolus
b. Cajal bodies
c. PML bodies
d. Centrosomes
Ans. (d). Centrosome
Centrosome is found outside the nucleus. Centrosome composed of centrioles and associated spindle firbers and it forms the spindle poles during cell division.
Cajal bodies: they are non-membrane bound special small sub-organelles present in the nucleus of actively dividing cells and they mainly consist of proteins and scaRNA. First described as nucleolar accessory bodies by Santiago Ramon Cajal. Functions of Cajal body (CBs) may be (1) SnRNPs biogenesis, (2). Maturation and recycling of histone mRNA and (3). Telomere maintenance. Addition of nucleotide to telomere with the help of RNA requires the helps of Cajal bodies.
PML bodies: are punctate structures found in the nucleus of certain cells. They are also called as nuclear bodies and nuclear dots.
17. Haemolytic disease of the newborn due to Rhesus incompatibility depends up on the:
a. Mother possessing Rh antigen not present on the baby’s red cells
b. Transplacental passage of IgM anti-Rh antibodies
c. Transplacental passage of IgG anti-Rh antibodies
d. Production of cytotoxic antibodies in the body
Ans. (c). Transplacental passage of IgG anti-Rh antibodies
(18). Hemoglobin shows sigmoidal curve for oxygen saturation. What is the shape of curve for myoglobin oxygen binding?
a. Linear
b. Hyperbolic
c. Sigmoidal
d. Bell shaped
Ans. (b). Hyperbolic
Myoglobin is the oxygen storage protein of muscles, the oxygenated myoglobin it is responsible for the deep red colour of muscles.
For details of the oxygen saturation curve of myoglobin and hemoglobin, refer: Biochemistry, Ed. 4 by Garrett and Grisham, Chapter 14, Enzyme regulation, Page 468 – 471.
19. For transcription to occur in the lactose operon, an induced must be present so that
a. The repressor can binds to the operator
b. The repressor does not binds to the operator
c. The induced can binds to the operator
d. The inducer does not binds to the operator
Ans. (b). The repressor does not bind to the operator
The inducer (lactose or allolactose) binds to the repressor (the gene product of lac I) and thereby making it unable to binds to the operator. Once the operator is free, RNA polymerase can synthesize the structural genes.
For more details of lac operon, refer: Cell and Molecular Biology by Gerald Karp, Ed.6, Chapter- 12. Cell Nucleus and Control of Gene Expression, Biological Operons, Page 501 – 503
20. Which of the following is NOT a feature of mutagenic action of 5-Bromo-deoxyuridine?
a. It acts as growing cells
b. It forms base pair with A in its rare form
c. It induces transitions
d. It affects only one strand of DNA
Ans. (b). It forms base pair with A in its rare form
Boromodeoxyuridine: is a synthetic analogue of thymidine.
During replication Bromodeoxyuridine is incorporated to DNA in place of Thymine
21. The action potential results from
a. Decrease in negative charge inside the nerve fibre
b. Increase in positive charge outside the nerve fibre
c. Opening of voltage gated sodium channels
d. Activation of sodium potassium pump
Ans. (c). Opening of voltage gated sodium channels
22. A recombinant vaccine is available for which one of the following cancers?
a. Adult T cell leukemia
b. Colon carcinoma
c. Glioblastoma
d. Cervical carcinoma
Ans. (d). Cervical carcinoma
The only cancer which can be prevented by vaccination is cervical cancer. Cervical cancer is caused by a virus called Human Papiloma Virus (HPV). HPV infection can be controlled by vaccination and thereby it can be used as a vaccine against cervical cancers. Similarly human Hepatitis B virus which causes hepatocellular carcinoma can also be controlled by vaccination and thus it can also be considered as a vaccine against cancer. However, in the case of hepatocellular carcinoma, HBV infection is not the only reason, other agents both living and nonliving can also cause hepatocellular carcinoma.
23. A patch-clamp device is used to
a. Measure the strength of an electrochemical gradient
b. Study the properties of individual neurotransmitters
c. Infuse different kinds of ions into an axon
d. Study the properties of individual membrane channels
Ans. (d). Study the properties of individual membrane channels
24. Which type of neurons among the following are predominantly lost in Alzheimer’s disease?
a. Cholinergic
b. Serotonergic
c. Noradrenergic
d. Histaminergic
Ans. (a). Cholinergic
25. Circadian rhythm is regulated by the
a. Hypothalamus
b. Suprachiasmatic nucleus
c. Amygdala
d. Basal ganglia
Ans. (b). Suprachaismatic nucleus
Suprachiasmatic nucleus (SCN) is a tiny region of the hypothalamus of brain, situated directly above the optic chiasma. It is responsible for controlling circadian rhythms
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