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Biotechnology Engineering - BT 2012 GATE Paper (Practice Test)


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65 Questions MCQ Test GATE Past Year Papers for Practice (All Branches) | Biotechnology Engineering - BT 2012 GATE Paper (Practice Test)

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Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 1

Q. No. 1 – 5 Carry One Mark Each
The cost function for a product in a firm is given by 2 5q , where q is the amount of production. The firm can sell the product at a market price of Rs.50 per unit. The number of units to be produced by the firm such that the profit is maximized is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 1

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 2

Choose the most appropriate alternative from the options given below to complete the following sentence:

Suresh’s dog is the one ________ was hurt in the stampede.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 3

Choose the grammatically INCORRECT sentence:

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 4

Which one of the following options is the closest in meaning to the word given below?

Mitigate

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 5

Choose the most appropriate alternative from the options given below to complete the following sentence:

Despite several __________ the mission succeeded in its attempt to resolve the conflict.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 6

Q. No. 6 – 10 Carry Two Marks Each

Wanted Temporary, Part-time persons for the post of Field Interviewer to conduct personal interviews to collect and collate economic data. Requirements: High School-pass, must be available for Day, Evening and Saturday work. Transportation paid, expenses reimbursed. Which one of the following is the best inference from the above advertisement?

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 6

Gender is not mentioned in the advertisement and (B) clearly eliminated

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 7

Given the sequence of terms, AD CG FK JP, the next term is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 7

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 8

Which of the following assertions are CORRECT?
P: Adding 7 to each entry in a list adds 7 to the mean of the list
Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list
R: Doubling each entry in a list doubles the mean of the list
S: Doubling each entry in a list leaves the standard deviation of the list unchanged

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 8

P and R always holds true Else consider a sample set {1, 2, 3, 4} and check accordingly

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 9

An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable Of X's shock absorbers, 96% are reliable. Of Y's shock absorbers, 72% are reliable. The probability that a randomly chosen shock absorber, which is found to be reliable, is made by Y is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 9

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 10

A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2x − 0.1x2 where y is the height of the arch in meters. The maximum possible height of the arch is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 10

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 11

Q. No. 11 – 35 Carry One Mark Each
In mismatch correction repair, the parental DNA strand is distinguished from the daughter strand by

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 11

The mismatch repair deals with correcting mismatch of the normal bases using BER or NER enzyme systems. The system assumes that the parental strands are methylated and the freshly synthesized daughter strands are non methylated.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 12

The basis for blue-white screening with pUC vectors is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 12

Both pUC and the bacterial own genome produce a faulty gene product of Lac Z gene. The lac Z fragment, whose synthesis can be induced by IPTG, is capable of intra-allelic complementation with a defective form of 0-galactosidase enzyme encoded by host chromosome (mutation lacZDM15)

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 13

Idiotypic determinants of an antibody are associated with the

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 13

Immunoglobulin idiotypes are serologically defined determinants associated with the variable (V) region of antibody molecules

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 14

Identification of blood groups involves

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 14

All the blood grouping reactions are agglutination reactions. In these the antibodies against A and B antigens when added, they bind to the cell surface and result in clumping of cells. These clumps or settling down of cells is referred to as agglutination.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 15

B-lymphocytes originate from the bone marrow whereas T-lymphocytes originate from

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 15

All T cells originate from haematopoietic stem cells in the bone marrow. B cells undergo their complete maturation in the thymus while, T cells during their early stages of maturation migrate to the thymus and then undergo a process of maturation. They are released as competent and mature T cells from Thymus.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 16

A humanized antibody is one in which the

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 16

A humanized chain is a chain in which the complementarity determining regions (CDR) of the variable domains are foreign (originating from one species other than human, or synthetic) whereas the remaining chain is of human origin.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 17

Dimethyl sulfoxide (DMSO) is used as a cryopreservant for mammalian cell cultures because

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 17

DMSO is used in cell freezing media to protect cells from ice crystal induced mechanical injury.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 18

Nude mice refers to

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 18

A nude mouse is a laboratory mouse from a strain with a genetic mutation that causes a deteriorated or absent thymus, resulting in an inhibited immune system due to a greatly reduced number of T cells.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 19

Heat inactivation of serum is done to inactivate

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 19

Heat-inactivation (heating to 56°C for 30 minutes) of serum is done to inactivate complement, a group of proteins present in sera that are part of the immune response.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 20

Choose the correct signal transduction pathway

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 20

In the G protein associated signal cascade system, binding of hormone to a cell surface receptor, like the 7TM (a 7 transmembrane alpha helices) which is a part of G protein. This will activate production of second messengers like cAMP by the enzyme adenylate cyclase. The cAMP in turn binds and activates down stream signal proteins line cAMP dependent protein kinase A (PKA).

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 21

A protein is phosphorylated at a serine residue. A phosphomimic mutant of the protein can be generated by substituting that serine with

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 21

To investigate the effect of serine 78 phosphorylation on p21 activity, replacement of serine 78 with aspartic acid is done, creating the phosphomimic p21S78D.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 22

A truncated polypeptide is synthesized due to a nonsense mutation. Where would you introduce another mutation to obtain a full-length polypeptide?

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 22

Non sense mutations causes premature stop to be introduced that would lead to truncated or incomplete protein synthesis. If the protein synthesis has to be continued, another mutation in t-RNA gene would continue be advised to get full length polypeptide

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 23

Protein-DNA interactions in vivo can be studied by

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 23

The strength of ChIP assays is their ability to capture a snapshot of specific protein: DNA interactions occurring in a system and to quantitate the interactions using quantitative polymerase chain reaction (qPCR).

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 24

The direction of shell coiling in the snail Limnaea peregra is a classic example of

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 24

The direction of shell coiling in snail is a classical example for extra chromosomal inheritance, as in this it is determined by maternal gene effects and not that of the offspring.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 25

During photorespiration under low CO2 and high O2 levels, O2 reacts with ribulose 1,5- bisphosphate to yield

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 25

During photo respiration under low CO2 and high O2 reacts with ribulose – 1, 5 bisphosphate to yield- one molecule each of 3-phosphoglycerate and 2- phosphoglycolate.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 26

Which one of the following is NOT a protoplast fusion inducing agent?

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 26

Colchicines is not a protoplast fusion inducing agent instead it functions as mitotic inhibitor

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 27

The activity of an enzyme is expressed in International Units (IU). However, the S.I. unit for enzyme activity is Katal. One Katal is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 27

One katak is 6 x 107 IU units

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 28

Identify the statement that is NOT applicable to an enzyme catalyzed reaction

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 28

Enzymes accelerate the rate of the reverse reaction as well as the forward reaction, it would be helpful to ignore any back reaction by which E1P might form ES. The velocity of this back reaction would be given by v = k-2[E][P].

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 29

An example of a derived protein structure database is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 29

Protein structural database: Primary database: PDB Secondary database: SCOP, CATH

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 30

An example of a program for constructing a phylogenetic tree is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 30

A program for constructing phylogenetic tree is PHYLIP. It is a Phylogeny Inference Package. It is a free computational phylogenetic package which has programs for inferring evolutionary tree constructions.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 31

Synteny refers to

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 31

Synteny refers to – gene duplication from a common ancestor. This is the condition in which two or more gene loci are present on the same chromosome. During evolution rearrangements may cause the loss of synteny else it will be retained. Translocations can cause gain of synteny.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 32

While searching a database for similar sequences, E value does NOT depend on the

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 32

E value does not depend on scoring system.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 33

In transmission electron microscopy, election opacity is greatly enhanced by treating the specimen with

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 33

In transmission electron microscope, electron opacity is greatly enhanced by treating the specimen with ferrous ammonium sulphate particles as coat on carbonaceous material and increases their opacity.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 34

The molarity of water in a water: ethanol mixture (15: 85, v/v) is approximately

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 34

18gm water in 100ml ethanol is 10M. Hence 15gm in 85ml is approximately- 8.5.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 35

The helix content of a protein can be determined using

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 35

The helix content of a protein can be determined using – circular dichromism spectroscopy. CD spectrum of unknown protein = A * [% alphahelical] + B * [% beta-sheet] + C * [% random coil]

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 36

Q. No. 36 – 65 Carry Two Marks Each

Which one of the following DNA sequences carries an invert repeat?

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 37

In zinc finger proteins, the amino acid residues that coordinate zinc are

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 37

Zinc fingers coordinate zinc ions with a combination of cysteine and histidine residues. They can be classified by the type and order of these zinc coordinating residues (e.g., Cys2His2, Cys4, and Cys6).

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 38

Match the entries in Group I with those in Group II.

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 38

MTT- binds to succinate dehydrogenase
Annexin V- specifically binds to phosphatidyl serine Methotrexate- allosterically inhibits dihydrofolate reductase Taxol- suppresses microtubule dynamics

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 39

In an exponentially growing batch culture of Saccharomyces cerevisiae, the cell density is 20gl−1 (DCW), the specific growth rate (μ )  is 0.41 and substrate uptake rate (v) is  16gl-1 h-1 . The cell yield coefficient Yx/s will be

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 39

Yield coefficient = mass of new cells formed/ mass of substrate consumed
New cells mass formed = cell density x specific growth rate = 20 x 0.4 = 8
Substrate consumed = 16 g/l/h
Yield coefficient = 8/16 = 0.5.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 40

A single base pair of DNA weighs 1.1 ×10-21 grams. How many picomoles of a plasmid vector of length 2750 bp are contained in 1μg of purified DNA?

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 40

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 41

Match the terms in Group I with the ploidy in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 42

What is the rank of the following matrix?

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 42

|A| = 5(0 + 40) − 3(0 + 56)

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 43

Match the products in Group I with the applications in Group II.

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 43

Digoxin- cardiovascular disorder

Stevioside- sweetener

Atropine- Muscle relaxant

Vinblastine- anticancer agent

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 44

Determine the correctness or otherwise of the following Assertion (a) and Reason (r).
Assertion : The production of secondary metabolites in plant cell cultures is enhanced by the addition of elicitors
Reason : Elicitors induce the expression of enzymes responsible for the biosynthesis of secondary metabolites

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 45

Determine the correctness or otherwise of the following Assertion (a) and Reason (r).
Assertion : Plants convert fatty acids into glucose
Reason : Plants have peroxisomes

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 45

Plants can make glucose from fatty acids, but this is only because they are able to use the glyoxylate cycle instead of the Krebs cycle.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 46

Determine the correctness or otherwise of the following Assertion (a) and Reason (r).
Assertion : In direct somatic embryogenesis, embryos are developed without going through callus formation
Reason : This is possible due to the presence of pre- embryonically determined cells

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 46

Somatic embryogenesis can be initiated either directly without going through the callus phase from predetermined embryonic cells or indirectly through callus proliferation and differentiation into embryonic cells within the
callus tissue

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 47

Match the entries in Group I with the process parameters in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 48

Match the downstream processes in Group I with the products in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 49

Determine the correctness or otherwise of the following Assertion (a) and Reason (r).
Assertion : Cell mass yield of methylotrophic yeast is more on methanol compared to glucose
Reason : Methanol has a greater degree of reductance compared to glucose.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 50

A disease is inherited by a child with a probability of 1/4. In a family with two children, the probability that exactly one sibling is affected by this disease is

Detailed Solution for Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 50

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 51

Match the organisms in Group I with the entries in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 52

Match the entries in Group I with the methods of sterilization in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 53

Match the high energy compounds in Group I with the biosynthetic pathways for the molecules in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 54

Match the vitamins in Group I with the processes/reactions in Group II.

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 55

Consider the data set 14, 18, 14, 14, 10, 29, 33, 31, 25. If you add 20 to each of the values, then

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 56

An enzymatic reaction is described by the following rate expression.

Which one of the following curves represents this expression?

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 57

A bacterial culture (200 μl containing 1.8 ×109) cells was treated with an antibiotic Z (50μg per ml) for 4 h at 37oC. After this treatment, the culture was divided into two equal aliquots.
Set A: 100μl was plated on Luria agar.
Set B: 100μl was centrifuged, the cell pellet washed and plated on Luria agar.
After incubating these two plates for 24 h at 37°C, Set A plate showed no colonies, whereas the Set B plate showed 0.9 ×109 cells. This experiment showed that the antibiotic Z is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 58

Common Data Questions: 58 & 59
In a muscle, the extracellular and intracellular concentrations of Na+ are 150 mM and 12 mM and those of K+ are 2.7 mM and 140 mM, respectively. Assume that the temperature is 25oC and that the membrane potential is –60mV, with the interior more negatively charged than the exterior.
( R= 8.314 J mol-1 K-1 ; F= 96.45kJ mol-1 V-1)

The free energy change for the transport of three Na+ out of the cell is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 59

In a muscle, the extracellular and intracellular concentrations of Na+ are 150 mM and 12 mM and those of K+ are 2.7 mM and 140 mM, respectively. Assume that the temperature is 25oC and that the membrane potential is –60mV, with the interior more negatively charged than the exterior.
( R= 8.314 J mol-1 K-1 ; F= 96.45kJ mol-1 V-1)

The free energy change for the transport of two K+ into the cell is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 60

Common Data Questions: 60 & 61

The purification data for an enzyme is given below:

The fold purification for each step is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 61

The purification data for an enzyme is given below:

The yield (%) for each step is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 62

Statement for Linked Answer Questions: 62 & 63
An E. coli cell of volume 10 -12cm3  contains 60 molecules of lac-repressor. The repressor has a binding affinity (Kd) of 10-8M and 10-9M  with and without lactose respectively, in the medium

The molar concentration of the repressor in the cell is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 63

An E. coli cell of volume 10 -12cm3  contains 60 molecules of lac-repressor. The repressor has a binding affinity (Kd) of 10-8M and 10-9M  with and without lactose respectively, in the medium

Therefore the lac-operon is

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 64

Statement for Linked Answer Questions: 64 & 65
β- Galactosidase bound to DEAE-cellulose is used to hydrolyze lactose to glucose and galactose in a plug flow bioreactor with a packed bed of volume 100 liters and a voidage (ε) of 0.55. The Km and Vmax for the immobilized enzyme are 0.72 gl-1 and 18gl-1 h-1 ,respectively. The lactose concentration in the field stream is 20gl-1 , and a fractional conversion of 0.90 is desired. Diffusional limitations may be ignored.

The residence time required for the steady state reactor operation will be

Biotechnology Engineering - BT 2012 GATE Paper (Practice Test) - Question 65

β- Galactosidase bound to DEAE-cellulose is used to hydrolyze lactose to glucose and galactose in a plug flow bioreactor with a packed bed of volume 100 liters and a voidage (ε) of 0.55. The Km and Vmax for the immobilized enzyme are 0.72 gl-1 and 18gl-1 h-1 ,respectively. The lactose concentration in the field stream is 20gl-1 , and a fractional conversion of 0.90 is desired. Diffusional limitations may be ignored.

The feed flow rate required for the above bioconversion will be

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