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Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - GATE MCQ


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30 Questions MCQ Test - Biotechnology Engineering - BT 2016 GATE Paper (Practice Test)

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) for GATE 2024 is part of GATE preparation. The Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) questions and answers have been prepared according to the GATE exam syllabus.The Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) MCQs are made for GATE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) below.
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Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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.

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 1

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 2

The unruly crowd demanded that the accused be without trial.

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Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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.

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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?

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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 .

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 5

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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?

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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.

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 8

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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 .

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 9

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 10

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 10

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 11

Q.No-11-35 Carry One Mark Each

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 11

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).

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 12

Which family of viruses has single stranded DNA?

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 12

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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?

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 13

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.)

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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.

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 14

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.

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 15

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 15

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 16

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 16

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 17

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 17

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.

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 18

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 18

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’

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 19

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 20

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 20

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 21

Junctional diversity of antibody molecules results from

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 21

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.

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 22

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 22

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 23

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 23

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.

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 24

Which one of the following statements is NOT true?

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 24

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 25

Based on their function, find the ODD one out.

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 25

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 26

Prandtl number is the ratio of

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 26

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 27

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 28

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

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 28

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

Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 29

In dead-end filtration, rate of filtration is

Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 29

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
Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - 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


Detailed Solution for Biotechnology Engineering - BT 2016 GATE Paper (Practice Test) - Question 30

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