Biotechnology Principles and Processes Test

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Biotechnology Principles and Processes Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

To carry out a polymerase chain reaction (PCR), one must have the catalytic DNA polymerase and

A
a blueprint or gene map of the sequence one wish to copy

B
a number of primers from either side of the target DNA in order to get the polymerase replication process- going

C
a DNA synthesizer machine

D
a DNA probe

Solution

Polymerase Chain Reaction is a method widely used to make many copies of a specific DNA segment. Using PCR, a single (or more) copy of a DNA sequence is exponentially amplified to generate thousands to millions of more copies of that particular DNA segment. Components of PCR include DNA template, DNA polymerase, two DNA primers, nucleotides.
So, the correct option is 'a number of primers from either side of the target DNA in order to get the polymerase replication process-ongoing'.
Question 2
1 / -0

_________ are self-replicating, extrachromosomal segments of double-stranded circular and naked DNA, present in a bacterial cell.

A
Plasmids

B
Nucleoid

C
Mesosomes

D
Bacteriophages

Solution

Plasmid is a structure in a bacterial cell consisting of DNA that can exist and replicate independently of the chromosome. Plasmids provide genetic instructions for certain cell activities (e.g., antibiotic resistance). They can be transferred from cell to cell in a bacterial colony.
So, the correct answer is 'Plasmids'.
Question 3
1 / -0

Which of the molecule listed below is a product of fermentation of glucose by yeast?

A
$${\left({C}_{6}{H}_{10}{O}_{5}\right)}_{n}$$

B
$${C}_{2}{H}_{5}OH$$

C
$${C}_{6}{H}_{12}{O}_{6}$$

D
$${CH}_{3}OH$$

Solution

In yeast, the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions where pyruvic acid is converted to $${CO}_{2}$$ and $${C}_{2}{H}_{5}OH$$ (ethanol).
So, the correct answer is (B).
Question 4
1 / -0

Study the given figure carefully and select the incorrect statements regarding this.(i) It represents a typical agarose gel electrophoresis in which lane 1 contains undigested DNA(ii) Smallest DNA bands are formed at A and largest DNA bands are formed at B(iii) The separated DNA fragments can be visualized after staining in the visible light(iv) The separated DNA bands are cut out from the agarose gel and extracted from the gel piece. This step is known as elution.

A
(i) and (ii)

B
(ii) and (iii)

C
(ii) and (iv)

D
(i) and (iv)

Solution

The DNA fragments separate according to their size through the agarose gel, with smaller fragments moving farther away as compared to larger ones. The DNA fragments can be visualised by staining them with ethidium bromide followed by exposure to UV radiations.
Question 5
1 / -0

Gel electrophoresis is used for

A
Construction of recombinant DNA by joining with cloning vectors

B
Isolation of DNA molecules

C
Cutting of DNA into fragments

D
Separation of DNA fragments according to their size
Solution
Correct Option: D
Explanation:
- The fragments of DNA are separated by the method agarose gel electrophoresis.
- DNA fragments, being negatively charged, move towards the positive electrode or anode under the electric field.
- DNA fragments resolve according to their size.
- The separated DNA fragments can be seen on staining with fluorescent dye- ethidium bromide and followed by exposure to UV radiation.
- DNA fragments from orange-colored bands.
Hence, gel electrophoresis is used for the separation of DNA fragments according to their size.
Question 6
1 / -0

The sticky ends of a fragmented DNA molecule are made up of ________.

A
Calcium salts

B
Endonuclease enzyme

C
Unpaired bases

D
Methyl groups
Solution
- After digestion of DNA with certain restriction enzymes like Eco RI, there is one unpaired overhanging strand produced at each end.
- This overhang can easily re-attach to other ends that are complementary to it, thus known as "sticky ends".
So, the correct answer is option C.
Question 7
1 / -0

The restriction enzyme responsible for the cleavage of following sequence is ______________.

A
EcoRI

B
Hindll

C
BamHI

D
EcoRII

Solution

HindII was the first discovered restriction endonuclease. It was isolated from Haemophilus influenzae. It produces blunt ends.
Question 8
1 / -0

If a plasmid vector is digested with EcoRI at a single site, then

A
One sticky end will be produced

B
Two sticky ends will be produced

C
Four sticky ends will be produced

D
Six sticky ends will be produced

Solution

A vector is a vehicle, usually an artificial DNA molecule, that carries a foreign genetic material into the cell. In this cell, it gets replicated. Restriction endonuclease enzymes create sticky ends that are then useful in producing a recombinant DNA molecule. There are 2 types of ends, Sticky ends and blunt ends. They cut the DNA strands 4 base base pairs from each other and create 5' overhang in one direction and complementary 5' overhang in another. When a 5' overhang remains it means that stranded base has ended in 5' phosphate region and similar is for 3' overhang.
So the correct answer "Two sticky ends will be produced".
Question 9
1 / -0

Which of the following statements is not correct regarding EcoRI restriction endonuclease enzyme?

A
It is isolated from Escherichia coli RY13

B
Its recognition sequence is 5'-GAATTC-3'

3'-CTTAAG-5'

C
It produces complementary blunt ends

D
None of these

Solution

The restriction endonuclease enzyme EcoRI was isolated from bacterium Escherichia coli RY13. It recognises the base sequence GAATTC in DNA duplex and cuts its strands between G and A as shown below:5'-G-A-A-T-T-C-3'
3'-C-T-T-A-A-G-5'
It results in complementary sticky ends.
Question 10
1 / -0

During insertional inactivation, the presence of a chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. The blue colour is produced by the enzyme

A
Glucosidase

B
Restriction endonuclease

C
Galactosidase

D
Taq polymerase

Solution

Alternative selectable markers have been developed which differentiate recombinants from the non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme, ( -galactosidase. This results into inactivation of the enzyme, which is referred i to as insertional inactivation. The presence of a chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. Presence of insert results into insertional inactivation of the -galactosidase and the colonies do not produce any colour, these are identified as recombinant colonies.
So, the correct answer is 'Galactosidase'.

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Re-Attempt Weekly Quiz Competition

Biotechnology Principles and Processes Test - 43

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Biotechnology Principles and Processes Test - 43

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SHARING IS CARING

Question 1

a blueprint or gene map of the sequence one wish to copy

a number of primers from either side of the target DNA in order to get the polymerase replication process- going

a DNA synthesizer machine

a DNA probe

Solution

Question 2

Plasmids

Nucleoid

Mesosomes

Bacteriophages

Solution

Question 3

$${\left({C}_{6}{H}_{10}{O}_{5}\right)}_{n}$$

$${C}_{2}{H}_{5}OH$$

$${C}_{6}{H}_{12}{O}_{6}$$

$${CH}_{3}OH$$

Solution

Question 4

(i) and (ii)

(ii) and (iii)

(ii) and (iv)

(i) and (iv)

Solution

Question 5

Construction of recombinant DNA by joining with cloning vectors

Isolation of DNA molecules

Cutting of DNA into fragments

Separation of DNA fragments according to their size

Solution

Question 6

Calcium salts

Endonuclease enzyme

Unpaired bases

Methyl groups

Solution

Question 7

EcoRI

Hindll

BamHI

EcoRII

Solution

Question 8

One sticky end will be produced

Two sticky ends will be produced

Four sticky ends will be produced

Six sticky ends will be produced

Solution

Question 9

It is isolated from Escherichia coli RY13

Its recognition sequence is 5'-GAATTC-3' 3'-CTTAAG-5'

It produces complementary blunt ends

None of these

Solution

Question 10

Glucosidase

Restriction endonuclease

Galactosidase

Taq polymerase

Solution

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Its recognition sequence is 5'-GAATTC-3'

3'-CTTAAG-5'