Biotechnology Engineering - BT 2016 GATE Paper (Practice Test)


65 Questions MCQ Test GATE Past Year Papers for Practice (All Branches) | Biotechnology Engineering - BT 2016 GATE Paper (Practice Test)


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This mock test of Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) for GATE helps you for every GATE entrance exam. This contains 65 Multiple Choice Questions for GATE Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) (mcq) to study with solutions a complete question bank. The solved questions answers in this Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) quiz give you a good mix of easy questions and tough questions. GATE students definitely take this Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) exercise for a better result in the exam. You can find other Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) extra questions, long questions & short questions for GATE on EduRev as well by searching above.
QUESTION: 1

Q.No-1-5 Carry One Mark Each

The volume of a sphere of diameter 1 unit is than the volume of a cube of side 1 unit.

Solution:

Volume of a sphere of diameter 1 unit is  p = p and volume of a cube of side 1 unit is
13 = 1

QUESTION: 2

The unruly crowd demanded that the accused be without trial.

Solution:
QUESTION: 3

Choose the statement(s) where the underlined word is used correctly:

(i) A "prone" is a dried plum.
(ii) He was lying "prone" on the floor.
(iii) People who eat a lot of fat are "prone" to heart disease.

Solution:
QUESTION: 4

Fact: If it rains, then the field is wet.

Read the following statements:
(i) It rains
(ii) The field is not wet
(iii) The field is wet
(iv) It did not rain

Which one of the options given below is NOT logically possible, based on the given fact?

Solution:
QUESTION: 5

A window is made up of a square portion and an equilateral triangle portion above it. The base of the triangular portion coincides with the upper side of the square. If the perimeter of the window is 6 m, the area of the window in m2 is .

Solution:

QUESTION: 6

Q.No-6-10 Carry Two Marks Each

Students taking an exam are divided into two groups, P and Q such that each group has the same number of students. The performance of each of the students in a test was evaluated out of 200 marks. It was observed that the mean of group P was 105, while that of group Q was 85. The standard deviation of group P was 25, while that of group Q was 5. Assuming that the marks were distributed on a normal distribution, which of the following statements will have the highest probability of being TRUE?

Solution:
QUESTION: 7

A smart city integrates all modes of transport, uses clean energy and promotes sustainable use of resources. It also uses technology to ensure safety and security of the city, something which critics argue, will lead to a surveillance state.

Which of the following can be logically inferred from the above paragraph?
(i) All smart cities encourage the formation of surveillance states.
(ii) Surveillance is an integral part of a smart city.
(iii) Sustainability and surveillance go hand in hand in a smart city.
(iv) There is a perception that smart cities promote surveillance.

Solution:
QUESTION: 8

Find the missing sequence in the letter series. B, FH, LNP, _ _ _ _.

Solution:
QUESTION: 9

The binary operation is defined as a + b = ab+(a+b), where a and b are any two real numbers. The value of the identity element of this operation, defined as the number x such that a + x = a, for any a, is .

Solution:

QUESTION: 10

Which of the following curves represents the   Here x represents the abscissa and y represents the ordinate.

Solution:

 is a periodic function having period π and maximum value of y is 1 and y > 0.

QUESTION: 11

Q.No-11-35 Carry One Mark Each

Bacteria with two or more flagella at one or both ends are called

Solution:

Monotrichous bacteria have a single flagellum (e.g., Vibrio cholerae).
o Lophotrichous bacteria have multiple flagella located at the same spot on the bacteria's surfaces which act in concert to drive the bacteria in a single direction. In many cases, the bases of multiple flagella are surrounded by a specialized region of the cell membrane, the so-called polar organelle. lophotrichous (lo-fot? ri-kus) [lopho- + Gr. thrix hair] having two or more flagella at one or both ends; said of a bacterial cell
o Amphitrichous bacteria have a single flagellum on each of two opposite ends (only one flagellum operates at a time, allowing the bacteria to reverse course rapidly by switching which flagellum is active).
o Peritrichous bacteria have flagella projecting in all directions (e.g., E. coli).

QUESTION: 12

Which family of viruses has single stranded DNA?

Solution:

Although bacteriophages were first described in 1927, it was only in 1959 that Sinshemer
working with phage Phi X 174 showed that they could possess single-stranded DNA genomes. Despite this discovery until relatively recently it was believed that the majority of DNA viruses belonged to the double-stranded clade. Recent work suggests that this may not be the case with single-stranded viruses forming the majority of viruses found in sea water, fresh water, sediment, terrestrial, extreme, metazoan-associated and marine microbial mats. Many of these "environmental" viruses belong to the family Microviridae.[11] However, the vast majority has yet to be classified and assigned to genera and higher taxa. Because most of these viruses do not appear to be related or are only distantly related to known viruses additional taxa will be created for these. Families in this group have been assigned on the basis of the nature of the genome (circular or linear) and the host range. Eleven families are currently recognised.
o Family Anelloviridae
o Family Bacillariodnaviridae
o Family Bidnaviridae
o Family Circoviridae
o Family Geminiviridae
o Family Inoviridae
o Family Microviridae
o Family Mycodnaviridae
o Family Nanoviridae
o Family Parvoviridae
o Family Spiraviridae

QUESTION: 13

What will be the binding status of regulatory proteins in lac operon when concentrations of both lactose and glucose are very low in the culture medium?

Solution:

Catabolite control of the lac operon. The operon is inducible by lactose to the maximal levels when CAMP and CAP form a complex. (a) Under conditions of high glucose, a glucose breakdown product inhibits the enzyme adenylate cyclase, preventing the conversion of ATP into CAMP. (b) Under conditions of low glucose, there is no breakdown product, and therefore adenylate cyclase is active and CAMP is formed. (c) When CAMP is present, it acts as anallosteric effector, complexing with CAP. (d) The CAMP–CAP complex acts as an activator of lac operon transcription by binding to a region within the lac promoter. (CAP = catabolite activator protein; cAMP = cyclic adenosine monophosphate.)

QUESTION: 14

Which of the following are TRUE for Treponema pallidum?
P. It is the causative agent of syphilis
Q. It is a spirochete
R. It is a non-motile bacterium
S. It is generally susceptible to penicillin

Choose the correct combination.

Solution:

Treponema pallidum is a spirochaete bacterium with subspecies that cause treponemal diseases such as syphilis, bejel, pinta, and yaws. The treponemes have a cytoplasmic and an outer membrane. Using light microscopy, treponemes are only visible using dark field illumination. They are Gram negative, but some regard them too thin to be Gram stained. T. p. pallidum is a motile spirochaete that is generally acquired by close sexual contact, entering the host via
breaches in squamous or columnar epithelium. The organism can also be transmitted to a fetus by transplacental passage during the later stages of pregnancy, giving rise to congenital syphilis. The helical structure of T. p. pallidum allows it to move in a corkscrew motion through a viscous medium such as mucus. It gains access to the host's blood and lymph systems through tissue and mucous membranes. Penicillin, the first antibiotic developed, was the first known effective antibiotic for T. pallidum and remains the treatment of choice today (12, 30, 94). The maximally treponemicidal serum concentration of penicillin is 0.36 μg/ml which can kill the organism in 6 to 9 hours (27). However, a concentration as low as 0.005 μg/ml can clear T. pallidum from chancres if it is maintained for 48 to 96 hours (27). In the rabbit model, viable treponemes may persist in the lymph nodes even after early lesions have been cleared. T. pallidum can regenerate if the serum penicillin concentration falls to sub-inhibitory levels for 24-30 hours.

QUESTION: 15

In a typical mitotic cell division cycle in eukaryotes, M phase occurs immediately after the

Solution:

Interphase
Interphase consists of 3 phases; G1 phase (growth), S phase (replication) and G2phase (preparation for division). During interphase, cells actively prepare for mitosis by gathering nutrients and conducting normal cellular functions i.e. they are metabolically active. Since cell division operates in a cycle, interphase is preceded by the previous cycle of mitosis.

When G2 is complete, the cell enters a relatively brief period of nuclear and cellular division, composed of mitosis and cytokinesis; together they define the mitotic phase (M phase) of the cell cycle involving the division of the mother cell into two daughter cells, genetically identical to each other and to their parent cell. This accounts for approximately 10% of the cell cycle.
The process of mitosis is complex and highly regulated. The M phase (nuclear division) has been characterised into five sequential phases (prophase, metaphase, anaphase and telophase) (PMAT) in which the pairs of chromosomes condense and attach to fibres that pull the sister chromatids to opposite sides of the cell. In a standard mammalian cell, mitosis can typically take around 1 hour to complete. The cell then divides in cytokinesis, to produce two identical daughter cells

QUESTION: 16

Which one of the following is NOT a therapeutic agent based on nucleic acid for the treatment of genetic disorders?

Solution:

Gene therapy is a technique for correcting defective genes responsible for disease development. Nucleic acid-based molecules (deoxyribonucleic acid, complementary deoxyribonucleic acid, complete genes, ribonucleic acid, and oligonucleotides) are utilized as research tools within the broad borders of gene therapy and the emerging field of molecular medicine. Deoxyribonucleic acidbased therapeutics includes plasmids, oligonucleotides for antisense and antigene applications, deoxyribonucleic acid aptamers, and deoxyribonucleic acidzymes, while ribonucleic nacid-based therapeutics includes ribonucleic acid aptamers, ribonucleic acid decoys, antisense ribonucleic acid, ribozymes, small interfering ribonucleic acid, and micro ribonucleic acid. To the evoiutionary unity or me

QUESTION: 17

ATP biosynthesis takes place utilizing the H+ gradient in mitochondria and chloroplasts. Identify the correct sites of H+ gradient formation.

Solution:

The mechanism of the two enzymes are similar, but their orientations differ. In chloroplasts, protons flow through the ATP synthase out of the thylakoid lumen into the stroma (where the ATP synthesized). In mitochondria the protons flow out of the intermembrane space into the mitochondrial matrix.

QUESTION: 18

Which one of the following is NOT an algorithm for building phylogenetic trees?

Solution:

Methods for estimating phylogenies include neighbor-joining, maximum parsimony (also simply referred to as parsimony), UPGMA,Bayesian phylogenetic inference, maximum likelihood and distance matrix methods. Maximum parsimony (MP) is a method of identifying the potential phylogenetic tree that requires the smallest total number of evolutionary events to explain the observed sequence data. Some ways
of scoring trees also include a "cost" associated with particular types of evolutionary events and attempt to locate the tree with the smallest total cost. This is a useful approach in cases where not every possible type of event is equally likely - for example, when particularnucleotides or amino
acids are known to be more mutable than others...The maximum likelihood method uses standard statistical techniques for inferringprobability distributions to assign probabilities to particular possible phylogenetic trees.In bioinformatics, neighbor joining is a bottom up(agglomerative) clustering method for the creation of phylogenetic trees, created by Naruya Sait
ou and Masatoshi Nei in 1987. Usually used for trees based on DNAor protein sequence data, the algorithm requires knowledge of the distance between each pair of taxa (e.g., species or sequences) to form the tree..Bootstrap, jackknife, and permutation tests are common tests used in phylogenetics to estimate the significance of the branches of a tree. This process can be very time consuming because of the large number of samples that have to be taken in order to have an accurate confidence estimate’

QUESTION: 19

Cesium chloride density gradient centrifugation is commonly used for the separation of DNA molecules. The buoyant density, ρ, of a double stranded Cs+DNA is given by the equation ρ = 1.66 + 0.098 XG+C where XG+C denotes

Solution:

QUESTION: 20

Disaccharide molecules that contain β(1→ 4) glycosidic linkage are

Solution:

QUESTION: 21

Junctional diversity of antibody molecules results from

Solution:

Junctional diversity results from the imprecise joining of gene segments and from the addition of nucleotides to the DNA sequence at splice sites. TdT adds up to 15 nucleotides to the DNA sequence of human VH and JHregions. Junctional diversity affects predominantly CDR3; it is significant but difficult to quantify, since it also results in many nonproductive rearrangements.

QUESTION: 22

Which one of the following is NOT used for the measurement of cell viability in animal cell culture?

Solution:

Cell viability and cytotoxicity assays are used for drug screening and cytotoxicity tests of chemicals.Many have established methods such as Colony Formation method, Crystal Violet method, Tritium-Labeled Thymidine Uptake method, MTT, and WST methods, which are used for counting the number of live cells.Trypan Blue is a widely used assay for staining dead cells. In this method, cell viability must be determined by counting the unstained cells with a microscope or other instruments. However, Trypan Blue staining cannot be used to distinguish between the healthy cells and the cells that are alive but losing cell functions.Cellular enzymes such as lactate dehydrogenase, adenylate kinase, and glucose-6-phosphate dehydrogenase are also used as cell death markers, and there are several products available on the market. However, adenylate kinase and glucose-6-phosphate are not stable and only lactate dehydrogenase does not lose its activity during cell death assays. Therefore, cell death assays based on lactate dehydrogenase (LDH) activity are more reliable than other enzyme-based cell death assays. Enzyme-based methods using MTT and WST rely on a reductive coloring reagent and dehydrogenase in a viable cell to determine cell viability with a colorimetric method

QUESTION: 23

Which one of the following techniques relies on the spin angular momentum of a photon?

Solution:

For Raman spectra the molecules undergo transitions in which an incident photon is absorbed and another scattered photon is emitted. The general selection rule for such a transition to be allowed is that the molecular polarizability must be anisotropic, which means that it is not the same in all directions.[13] Polarizability is a 3-dimensionaltensor that can be represented as an ellipsoid. The polarizability ellipsoid of spherical top molecules is in fact spherical so those molecules show no rotational Raman spectrum. For all other molecules both Stokesand anti-Stokes lines[notes 5] can be observed and they have similar intensities due to the fact that many rotational states are thermally populated. The selection rule for linear molecules is ΔJ = 0, ± 2 . The reason for the values ±2 is that the polarizability returns to the same value twice during a rotation.[14] The value ΔJ = 0 does not correspond to a molecular transition but rather to Rayleigh scattering in which the incident photon merely changes direction. The selection rule for symmetric top molecules is ΔK = 0If K = 0, then ΔJ = ±2If K ≠ 0, then ΔJ = 0, ±1, ±2 Transitions with ΔJ = +1 are said to belong the an R series, whereas transitions with ΔJ = +2 belong to an S series. Since Raman transitions involve two photons, it is possible for the molecular angular momentum to change by two units.

QUESTION: 24

Which one of the following statements is NOT true?

Solution:

Transition state analogs (transition state analogues), are chemical compounds with a chemical structure that resembles the transition state of a substrate molecule in anenzyme-catalyzed chemical reaction. Enzymes interact with a substrate by means of strain or distortions, moving the substrate towards the transition state.[1] Theory suggests that enzyme inhibitors which resembled the transition state structure would bind more tightly to the enzyme than the actual substrate.[2] Transition state analogs can be used as inhibitors in enzyme catalyzed reactions by blocking the active site of the enzyme. Examples of drugs that are transition state analog inhibitors include flu medications such as the neuraminidase inhibitor oseltamivir and the HIV protease inhibitors saquinavir in the treatment of AIDS.
In non-competitive inhibition, the Km does not change. This is because Km is a measure of the affinity of the enzyme for its substrate and this can only be measured by active enzyme. The fixed amount of inactive enzyme in non-competitive inhibition does not affect the Km and the Km, therefore is unchanged

QUESTION: 25

Based on their function, find the ODD one out.

Solution:

Two types of small ribonucleic acid (RNA) molecules – microRNA (miRNA) and small interfering RNA (siRNA) – are central to RNA interference. RNAs are the direct products of genes, and these small RNAs can bind to other specific messenger RNA (mRNA) molecules and either increase or decrease their activity, for example by preventing an mRNA from producing a protein. RNA interference has an important role in defending cells against parasitic nucleotide sequences – viruses and transposons. It also influences development.A small hairpin RNA or short hai rpin RNA(shRNA) is an artificial RNA molecule with a tight hairpin turn that can be used to silence target gene expression via RNA interference (RNAi).Expression of shRNA in cells is typically accomplished by delivery ofplasmids or through viral or bacterial vectors

QUESTION: 26

Prandtl number is the ratio of

Solution:

The Prandtl number (Pr) or Prandtl group is a dimensionless number, named after the German physicist LudwigPrandtl, defined as the ratio of momentum diffusivityto thermal diffusivity

QUESTION: 27

Fed batch cultivation is suitable for which of the following?

P. Processes with substrate inhibition processes with product inhibition R. High cell density cultivation

Solution:

The types of bioprocesses for which fed-batch culture is effective can be summarized as follows: 1. Substrate inhibition Nutrients such as methanol, ethanol, acetic acid, and aromatic compounds inhibit the growth of microorganisms even at relatively low concentrations. By adding suchsubstrates properly lag-time can be shortened and the inhibition of the cell growth markedly reduced.
2. High cell density (High cell concentration)
To achieve very high cell concentrations, e.g.50-100 g of dry cells/L, in a batch culture a high initial concentrations of the nutrients in the medium are needed. At such high concentrations of the nutrients become inhibitory, even though they have no such effect at the normal concentrations used in batch cultures

QUESTION: 28

A biological process is involved in the _________ treatment of industrial effluent.

Solution:

Secondary treatment removes dissolved and suspended biological matter. Secondary treatment is typically performed by indigenous, water-borne micro-organisms in a managed habitat. Secondary treatment may require a separation process to remove the micro-organisms from the treated water prior to discharge or tertiary treatment

QUESTION: 29

In dead-end filtration, rate of filtration is

Solution:

The main modeling equation for the dead-end filtration at constant pressure dropis represented by Darcy’s law: where Vp and Q are the volume of the permeate and its
volumetric flow raterespectively (proportional to same characteristics of the feed flow), μ
is dynamic viscosity of permeating fluid, A is membrane area, Rm and R are the respective resistances of membrane and growing deposit of the foulants. Rm can be interpreted as a membrane resistance to the solvent (water) permeation. This resistance is a membraneintrinsic property and is expected to be fairly constant and independent of the driving force, Δp

*Answer can only contain numeric values
QUESTION: 30

The power required for agitation of non-aerated medium in fermentation is kW.
Operating conditions are as follows:
Fermentor diameter = 3 m
Number of impellers = 1
Mixing speed = 300 rpm
Diameter of the Rushton turbine = 1 m
Viscosity of the broth = 0.001 Pa.s
Density of the broth = 1000 kg.m-3
Power number = 5


Solution:

QUESTION: 31

Which one of the following is the most suitable type of impeller for mixing high viscosity (viscosity > 105 cP) fluids?

Solution:

*Answer can only contain numeric values
QUESTION: 32

Runs scored by a batsman in five one-day matches are 55, 75, 67, 88 and 15. The standard deviation is________


Solution:

*Answer can only contain numeric values
QUESTION: 33

The positive Eigen value of the following matrix is ___________ .


Solution:

QUESTION: 34

The Laplace transform F(s) of the function f(t) = cos (at), where a is constant, is ___________.

Solution:

*Answer can only contain numeric values
QUESTION: 35

The value of the integral 


Solution:

QUESTION: 36

Q.No-36-65 Carry Two Marks Each

Which combination of the following statements is CORRECT for cyanobacteria?
P. They can perform oxygenic photosynthesis
Q. Usually filamentous forms are involved in nitrogen fixation
R. Nitrogen fixation occurs in heterocysts
S. They cannot grow in a mineral medium exposed to light and air

Solution:

Because they are photoautotrophs, cyanobacteria can be grown in simple mineral media. Vitamin B12 is the only growth factor that is known to be required by some species. Media must be supplemented with the essential nutrients needed to support cell growth, including sources of nitrogen, phosphorus, trace elements, etc. Toxigenic strains of cyanobacteria are deposited in international-type culture collections. Clonal cultures are distributed for research, taxonomic work and teaching purposes.

QUESTION: 37

Which set of the following events occurs during the elongation step of translation?

P. Attachment of mRNA with the smaller subunit of ribosome
Q. Loading of correct aminoacyl-tRNA into the A site
R. Formation of a peptide bond between the amino acyl-tRNA in the A site and the peptide chain that is attached to the peptidyl-tRNA in the P site
S. Dissociation of the ribosomal subunits
T. Translocation of peptidyl-tRNA from the A site to the P site of the ribosome

Solution:

During chain elongation, each additional amino acid is added to the nascent polypeptide chain in a three-step microcycle. The steps in this microcycle are (1) positioning the correct aminoacyltRNA in the A site of the ribosome, (2) forming the peptide bond and (3) shifting the mRNA by one codon relative to the ribosome On the other hand, non-competitive inhibitors do NOT bind to the active site of the enzyme and do not resemble the substrate. Therefore, addition of more substrate cannot eliminate the effect of the inhibitor. As a result, there is always a fixed amount of enzyme inactive in non-competitive inhibition. As you recall, when you change the amount of enzyme, you change the Vmax (from last lecture), so in the presence of a non-competitive inhibitor, the Vmax decreases.

QUESTION: 38

A DNA sequence, 5’-ATGGACGTGCTTCCCAAAGCATCGGGC-3’, is mutated to obtain
P. 5’-ATGGACGTGCTTCaCAAAGCATCGGGC-3’
Q. 5’-ATGGACGTGCTTCCCgAAAGCATCGGGC-3’
R. 5’-ATGGACGTGCTTCC-AAAGCATCGGGC-3’
S. 5’-ATGGACGTGCTTCCCAAtGCATCGGGC-3’
T. 5’-ATGGACGaGCTTCCCAAAGCATCGGGC-3’
[Point mutations are shown in the lower case or ‘-’ within the sequences]
Which of the above mutant sequences DO NOT have frame-shift?

Solution:
QUESTION: 39

Which of the following events occur during the stationary phase of bacterial growth?
P. Rise in cell number stops
Q. Spore formation in some Gram-positive bacteria such as Bacillus subtilis
R. Cell size increases in some Gram-negative bacteria such as Escherichia coli

S. Growth rate of bacterial cells nearly equals their death rate
T. Decrease in peptidoglycan crosslinking

Solution:

The AbrB protein of the spore-forming bacterium Bacillus sabtilis is a repressor of numerous genes that are switched on during the transition from the exponential to the stationary phase of growth.The stationary phase is often due to a growth-limiting factor such as the depletion of an essential nutrient, and/or the formation of an inhibitory product such as an organic acid. Stationary phase results from a situation in which growth rate and death rate are equal. The number of new cells created is limited by the growth factor and as a result the rate of cell growth matches the rate of cell death. tely digested by lysozyme due to de-N-acetylation of glucosamine, which occurs on 17.3% of muropeptides. The cross-linking index of the polymer changes with the growth phase. It is highest in late stationary phase, with a value of 33.2 or 44% per muramic acid residue, as determined by reverse-phase high-pressure liquid chromatography or gel filtration Division should occur subsequent to carbon source exhaustion. For example, an approximately two fold increase in cell number has been reported during the first part of stationary phase for both Beneckea natriegens and Escherichia coli (Wade, 1952; Nazly et al.., 1980). The increase in cell number in the stationary phase consequently leads to a reduction in the cell size (Henrici. Growth rate of bacterial cells nearly equals their death rate

QUESTION: 40

Select the CORRECT combination of genetic components that are essential for the transfer of TDNA segment from Agrobacterium tumefaciens to plant cells.

Solution:

Ti Plasmid: a: T-DNA, b: Vir genes , c: Replication origin , d: Opines catabolism genes

QUESTION: 41

Match the secondary metabolites (Column-I) with the corresponding plant species (Column-II).

Solution:

Norman strain of P. Somniferum, also developed in Tasmania, produces down to 0.04% morphine but with much higher amounts of thebaine andoripavine, which can be used to synthesise semi-synthetic opioids as well as other drugs like stimulants, emetics, opioid antagonists, anticholinergics, and smooth-muscle agents.The callus tissue of Tagetes erecta maintained on revised Murashige and Skoog's medium (RT) as static cultures showed the presence of insecticidal pyrethrins. The percentage of pyrethrins further increased by feeding the tissue with various concentrations of ascorbic acid. All Datura plants contain tropane alkaloids such as scopolamine, hyoscyamine, and atropine, primarily in their seeds and flowers. Because of the presence of these substances, Datura has been used for centuries in some cultures as a poison.[6][7] There can be a 5:1 toxin variation between plants, and a given plant's toxicity depends on its age, where it is growing, and the local weather conditions. These variations makes Datura exceptionally hazardous as a drug. Vinblastine and vincristine are excellent anticancer drugs but their current production using plants is non-abundant and expensive. In order to make these drugs readily available to the patients at affordable prices, we isolated the endophytic fungi from Catharanthus roseus plant and found a fungus AA-CRL-6 which produces vinblastine
and vincristine in appreciable amounts

QUESTION: 42

A variety of genetic elements are used in the transgenic plant research. Match the genetic elements Column-I) with their corresponding source (Column-II).

Solution:

Enhanced expression of the CAT marker gene suggests increased expression of selectable marker genes, thus improving the selection procedure in cereal transformation. A combination of maize ubiquitin 1 promoter with the bar gene, conferring resistance against the herbicide phosphinotricin, has already been constructed by Christensen and Quail (pAHC25, unpublished)
The promoter region of the Agrobacterium tumefaciens T-cyt gene was linked in a translational fusion to the coding DNA of the reporter gene uidA (for beta-glucuronidase or GUS protein; EC 3.2.1.31) and to nos 3’ flanking DNA [1].
They are flanked by the nopaline synthase (nos) or the cauliflower mosaic virus (CaMV) 355 promoters on one side by the nos polyadenylation signal on the other (2) The bar gene was originally cloned from Streptomyces hygroscopicus, an organism which produces the tripeptide bialaphos as a secondary metabolite.

QUESTION: 43

Match the type of chromosomal inheritance (Column-I) with the corresponding genetic disease or trait (Column-II).

Solution:

Examples of diseases with autosomal dominant inheritance include myotonic muscular dystrophy and Huntington disease.examples of diseases with autosomal recessive inheritance include sickle cell anemia and cystic fibrosis.Examples of diseases with X-linked dominant inheritance are hypophosphatemic ricketsm, oral-facial-digital syndrome type I, and Fragile X syndrome.Examples of diseases with X-linked recessive inheritance include Ducheme muscular dystrophy, hemophilia A and hypohidrotic or anhidrotic ectoder maldysplasia

*Answer can only contain numeric values
QUESTION: 44

A crossing was performed between the genotypes DdEeFfgg and ddEeFfGg. Assuming that the allelic pairs of all genes assort independently, the proportion of progeny having the genotype ddeeffgg is expected to be %.


Solution:
*Answer can only contain numeric values
QUESTION: 45

The equilibrium potential of a biological membrane for Na+ is 55 mV at 37 °C. Concentration of Na+ inside the cell is 20 mM. Assuming the membrane is permeable to Na+ only, the Na+ concentration outside the membrane will be mM.
(Faraday constant: 23062 cal.V-1.mol-1, Gas constant: 1.98 cal.mol-1.K-1)


Solution:
QUESTION: 46

A 1.2 kb DNA fragment was cloned into BamHI and EcoRI sites located on a 2.8 kb cloning vector.The BamHI and EcoRI sites are adjacent to each other on the vector backbone. The vector contains an XhoI site located 300 bp upstream of the BamHI site. An internal XhoI site is present in the gene sequence as shown in the figure. The resultant recombinant plasmid is digested with EcoRI and XhoI and analyzed through 1% agarose gel electrophoresis. Assuming complete digestion with EcoRI and XhoI, the DNA fragments (in base pairs) visible on the agarose gel will correspond to:

Solution:
QUESTION: 47

Find the INCORRECT combination.

Solution:

B cells derive from the bone-marrow and as mature B cells recirculate regularly through secondary lymphoid organs in search of signs of infection. On their surface they carry cellsurface immunoglobulin, the B cell receptor (BCR), which is specific for antigen. After antigen encounter, B cells initially congregate at the boundary between B cell follicles and T cell areas in search of T cell help.Immunoglobulin class switching, also known as isotype switching, isotypic commutationor class-switch recombination (CSR), is a biological mechanism that changes a B cell's production of immunoglobulin (antibodies) from one type to another, such as from theisotype IgM to the isotype IgG. Class switching occurs after activation of a mature B cell via its membrane-bound antibody molecule (or B cell receptor) to generate the different classes of antibody, all with the same variable domains as the original antibody generated in the immature B cell during the process of V(D)J recombination, but possessing distinct constant domains in their heavy chains. The fragment crystallizable region (Fc region) is the tail region of an antibody that interacts with cell surface receptors called Fc receptors and some proteins of thecomplement system. Secondary lymphoid tissue provides the environment for the foreign or altered native molecules (antigens) to interact with the lymphocytes. It is exemplified by the lymph nodes, and the lymphoid follicles in tonsils,Peyer's patches, spleen, adenoids, skin, etc. that are associated with the mucosa-associated lymphoid tissue

QUESTION: 48

Which of the following statement(s) is/are CORRECT for antigen activated effector T cells?
P. CD4+ cells make contact with macrophages and stimulate their microbicidal activity
Q. CD4+ cells make contact with B cells and stimulate them to differentiate into plasma cells
R. CD8+ cells make contact with B cells and stimulate them to differentiate into plasma cells
S. CD8+ cells make contact with virus infected cells and kill them

Solution:

In contrast with CD8 T cells,CD4 T cells differentiate into several subsets of effector T cells with a variety of different functions. The main functional classes are TH1,TH2, TH17, and the regulatory T cells. A recently recognized class specialized for providing help to B cells in the lymphoid follicles is called the T follicular helper cell, or THF . The subsets, particularly TH1,TH2, and TH17, are defined on the basis of the different combinations of cytokines that they secrete (Fig.9.28). The first to be distinguished were the TH1 and TH2 sub-sets, hence their names. TH1 cells help control bacteria that can set up intravesicular infections in macrophages, such as the mycobacteria that cause tuberculosis and leprosy. These bacteria are taken up by macrophages in the usual way but can evade the killing mechanisms described in Chapter 3. If a TH1 cell recognizes bacterial antigens displayed on the surface of an infected macrophage, it will interact with the infected cell to activate it further, stimulating the macrophage’s microbicidal activity to enable it to kill its resident bacteria. The final step of B-cell activation occurs when a CD4 helper T cell recognizes and binds with an antigen-MHC II complex on the B cell. This binding causes the CD4 cell to secrete cytokines, which then stimulate the B cell to proliferate and differentiate into two types of cells plasma cells and memory B cells. The plasma cells are the cells that make antibodies. CD8’ T lymphocytes can kill virus-infected cells by at least two mechanism which, although acting on different signal pathways, both will end in the induction of apoptotic killing machinery of the affected target cells.

QUESTION: 49

Which one of the following statements regarding G proteins is INCORRECT?

Solution:

In the resting state, G protein forms a heterotrimer, consisting of GDP-bound form of the G protein α subunit (Gα(GDP)) and G protein βγ subunit (Gβγ).GTP binding changes the conformation of switch regions within the alpha subunit, which allows the bound trimeric G protein (inactive) to be released from the receptor, and to dissociate into active alpha subunit (GTP-bound) and beta/gamma dimer. The alpha subunit and the beta/gamma dimer go on to activate distinct downstream effectors, such as adenylyl cyclase, phosphodiesterases, phospholipase C, and ion channels. Many different mammalian cell-surface receptors are coupled to a trimeric signal-transducing G
protein. Ligand binding to these receptors activates their associated G protein, which then activates an effector enzyme to generate an intracellular second messenger GDP to dissociate and to be replaced with GTP (GDP-GTP exchange), which is turn causes dissociation of the G protein trimer, releasing a −GTP and bg subunits; these are the ‘active’ forms of the G protein, which diffuse in the membrane and can associate with various enzymes and ion channels, causing activation of the target (Fig. 3.9). It was originally through that only the a subunit had a signaling function, the bg complex serving merely as a chaperone to keep the flighty a subunits out of range of the various effector proteins that they might otherwise excite. However, the bg complexes actually make assignations of their own, and control effectors in much the same way as the a subunits. Association of a or bg subunits with target enzymes or channels can cause either activation or inhibition, depending on which G protein is involved (see Table 3.3).

QUESTION: 50

In animal cell culture, a CO2 enriched atmosphere in the incubator chamber is used to maintain the culture pH between 6.9 and 7.4. Which one of the following statements is CORRECT?

Solution:

The optimal pH range of 7.2 to 7.4 can be maintained by supplementing the medium with sodium bicarbonate and regulating the level of CO2 in the atmosphere above the medium as shown by the reaction below: H2O + CO2 + NaHCO3 H+ + Na+ + 2HCO3 – In tissue culture, cells are grown either in open systems (where there is free exchange of the atmosphere immediately above the medium with the atmosphere of the incubator) or in closed systems (where the two atmospheres are kept separate). The buffering system employed in the medium needs to be matched to the culture system. Otherwise the cells may be subject to metabolic stress which will impair their performance. In closed systems the level of CO2 is regulated by the metabolism of the cells. The culture vessel must be sealed (flasks tightly capped) to retain any CO2 generated by the cells.
Consequently, closed systems provide additional protection against contamination and have simpler incubator requirements than open systems. Closed systems usually require media with buffers based on Hanks’ balanced salt solution having relatively low levels of sodium bicarbonate. In open systems, humidity (to reduce evaporation) and a means of regulating CO2 levels (if the culture medium contains sodium bicarbonate) are required during incubation to maintain the pH of the culture medium. Open systems usually require the higher levels of sodium
bicarbonate found in Earle’s salt solution combined with a 5 to 10% CO2 atmosphere supplied by the incubator. In general, 1.2 g/L to 2.2 g/L of sodium bicarbonate is used with 5% CO2 whereas 3.7 g/L sodium bicarbonate is used with 10% CO2 . The exact amount will depend upon the medium formulation.

QUESTION: 51

Choose the CORRECT combination of True (T) and False (F) statements about microcarrier used in animal cell culture.
P. Higher cell densities can be achieved using microcarriers
Q. Microcarriers increase the surface area for cell growth
R. Microcarriers are used for both anchorage- and nonanchorage-dependent cells
S. Absence of surface charge on microcarriers enhances attachment of cells

Solution:

Microcarrier beads in bioreactor systems are being applied as a large scale production method. SoloHill is developing a family of unique microcarrier products which are superior to those currently available. The overall aim is to develop a novel small microcarrier bead (38-63 micrometers) which will foster the aggregation of cells. This will encourage the use of microcarriers for conditions where higher cell densities have been shown to be advantageous. For anchorage-dependent cells, microcarrier culture provides the advantage of greatly increased surface area for cell growth and increasing the growth area per unit volume in the culture.

*Answer can only contain numeric values
QUESTION: 52

In an assay of the type II dehydroquinase of molecular mass 18 kDa, it is found that the Vmax of the enzyme is 0.0134 μmol.min-1 when 1.8 μg enzyme is added to the assay mixture. If the Km for the substrate is 25 μM, the kcat/Km ratio will be ×104 M-1.S-1.


Solution:
*Answer can only contain numeric values
QUESTION: 53

The molar extinction coefficients of Trp and Tyr at 280 nm are 5690 and 1280 M-1.cm-1, respectively. The polypeptide chain of yeast alcohol dehydrogenase (37 kDa) contains 5 Trp and 14 Tyr residues. The absorbance at 280 nm of a 0.32 mg.mL-1 solution of yeast alcohol dehydrogenase measured in a cuvette of 1 cm pathlength will be . (Assume that the molar extinction coefficient values for Trp and Tyr apply to these amino acids in the yeast alcohol dehydrogenase).


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QUESTION: 54

The activity of lactate dehydrogenase can be measured by monitoring the following reaction: Pyruvate + NADH → Lactate + NADThe molar extinction coefficient of NADH at 340 nm is 6220 M-1.cm-1. NAD+ does not absorb at this wavelength. In an assay, 25 μL of a sample of enzyme (containing 5 μg protein per mL) was added to a mixture of pyruvate and NADH to give a total volume of 3 mL in a cuvette of 1 cm pathlength. The rate of decrease in absorbance at 340 nm was 0.14 min-1. The specific activity of the enzyme will be ___________μmol.min-1.mg-1.


Solution:
QUESTION: 55

Analysis of a hexapeptide using enzymatic cleavage reveals the following result:
. Amino acid composition of the peptide is: 2R, A,V, S, Y
. Trypsin digestion yields two fragments and the compositions are: (R, A, V) and (R, S, Y) PDF

. Chymotrypsin digestion yields two fragments and the compositions are: (A, R, V, Y) and (R,S)
. Digestion with carboxypeptidase A yields no cleavage product.
Given: Trypsin cleaves at carboxyl side of R.
Chymotrypsin cleaves at carboxyl side of Y.
Carboxypeptidase A cleaves at amino side of the C-terminal amino acid (except R and K) of the peptide.
The correct amino acid sequence of the peptide is:

Solution:
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QUESTION: 56

The empirical formula for biomass of an unknown organism is CH1.8O0.5N0.2. To grow this organism, ethanol (C2H5OH) and ammonia are used as carbon and nitrogen sources, respectively. Assume no product formation other than biomass. To produce 1 mole of biomass from 1 mole of ethanol, the number of moles of oxygen required will be __________ .


Solution:
*Answer can only contain numeric values
QUESTION: 57

Saccharomyces cerevisiae is cultured in a chemostat (continuous fermentation) at a dilution rate of 0.5 h-1. The feed substrate concentration is 10 g.L-1. The biomass concentration in the chemostat at steady state will be __________. gL- 1
Assumptions: Feed is sterile, maintenance is negligible and maximum biomass yield with respect to substrate is 0.4 (g biomass per g ethanol). Microbial growth kinetics is given by 
where μ is specific growth rate (h-1), μm = 0.7 h-1, Ks = 0.3 g.L-1 and s is substrate concentration (g.L-1).


Solution:
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QUESTION: 58

Decimal reduction time of bacterial spores is 23 min at 121°C and the death kinetics follow first order. One liter medium containing 105 spores per mL was sterilized for 10 min at 121°C in a batch sterilizer. The number of spores in the medium after sterilization (assuming destruction of spores in heating and cooling period is negligible) will be ×107.


Solution:
QUESTION: 59

A bioreactor is scaled up based on equal impeller tip speed. Consider the following parameters for small and large bioreactors:

Assuming geometrical similarity and the bioreactors are operated in turbulent regime, what will beP2/P1?

Solution:

Physical properties of the broth in geometrically similar conditions and fully baffled fermentors are assumed to be same. Items relevant to liquid behavior in agitated fermentor vessel are: Power requirement of the agitation, P, or power requirements of agitation in gassed systems, Pg, rotation speed of the impeller, n, and pumping rate of impeller, F. For turbulent liquid motion,

*Answer can only contain numeric values
QUESTION: 60

An enzyme converts substrate A to product B. At a given liquid feed stream of flow rate 25 L.min-1 and feed substrate concentration of 2 mol.L-1, the volume of continuous stirred tank reactor needed for 95% conversion will be L. Given the rate equation 
Where -rA is the rate of reaction in mol.L-1.min-1 and CA is the substrate concentration in mol.L-Assumptions: Enzyme concentration is contant and does not undergo any deactivation during the
reaction.


Solution:
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QUESTION: 61

A protein is to be purified using ion-exchange column chromatography. The relationship between HETP (Height Equivalent to Theoretical Plate) and the linear liquid velocity of mobile phase is given by:

where H is HETP (m) and u is linear liquid velocity of mobile phase (m.s-1). The values of A, B and C are 3×10 8 m2.s-1, 3 s and 6×10-5 m, respectively. The number of theoretical plates based on minimum HETP for a column of 66 cm length will be .


Solution:
*Answer can only contain numeric values
QUESTION: 62

An enzyme is immobilized on the surface of a non-porous spherical particle of 2 mm diameter. The immobilized enzyme is suspended in a solution having bulk substrate concentration of 10 mM. The enzyme follows first order kinetics with rate constant 10 s-1 and the external mass transfer coefficient is 1 cm.s-1. Assume steady state condition wherein rate of enzyme reaction (mmol.L-1.s-1) at the surface is equal to mass transfer rate (mmol.L-1.s-1). The substrate concentration at the surface of the immobilized particle will be mM ______________.


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QUESTION: 63

The initial conditions for this second order homogeneous differential equation are y(0) = 1 and dy/dx =3 at x=0 The value of y when x = 2 is .


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*Answer can only contain numeric values
QUESTION: 64

The value of determinant A given below is .


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*Answer can only contain numeric values
QUESTION: 65

Consider the equation  Given a = 4, b = 1 and c = 9, the positive value of S at which V is maximum, will be _________.


Solution: