English
Karnataka Board PUCPUC Science 2nd PUC Class 12

While isolating DNA from bacteria, which of the following enzymes is not required? - Biology

Advertisements
Advertisements

Question

While isolating DNA from bacteria, which of the following enzymes is not required?

Options

  • Lysozyme

  • Ribonuclease

  • Deoxyribonuclease

  • Protease

MCQ
Advertisements

Solution

Deoxyribonuclease

Explanation:

While isolating DNA from bacteria, deoxyribonuclease is not required because it degrades DNA itself. Ribonuclease, protease, and lysozyme are necessary to remove RNA, proteins, and to break cell walls, respectively. 

shaalaa.com
Tools of Recombinant DNA Technology - Restriction Enzymes
  Is there an error in this question or solution?
Chapter 11: Biotechnology : Principles and Processes - MULTIPLE CHOICE QUESTIONS [Page 76]

APPEARS IN

NCERT Exemplar Biology [English] Class 12
Chapter 11 Biotechnology : Principles and Processes
MULTIPLE CHOICE QUESTIONS | Q 9. | Page 76
Nootan Biology [English] Class 12 ISC
Chapter 13 Principles and Processes of Biotechnology
Test Your Progress | Q 1. 43. | Page 531

RELATED QUESTIONS

Mention the difference in the mode of action of exonuclease and endonuclease.


How does a restriction nuclease function? Explain


Name and describe the technique that helps in separating the DNA fragments formed by the use of restriction endonuclease


Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.


Explain briefly:

Restriction enzymes and DNA


How does restriction endonuclease function?


Explain the roles of the following with the help of an example each in recombinant DNA technology :

Restriction Enzymes


Answer the following question.
Explain the significance of palindromic nucleotide sequence in the formation of recombinant DNA.


Answer the following question.
Write the use of restriction endonuclease in the formation of recombinant DNA.


The total number of nucleotide sequences of DNA that code for a hormone is 1530. The proportion of different bases in the sequence is found to be Adenine = 34%, Guanine = 19%, Cytosine = 23%, Thymine = 19%.

Applying Chargaff’s rule, what conclusion can be drawn?


The DNA fragment separated on an agarose gel can be visualized by staining with ______.


Restriction enzymes ______.


Molecular scissors, which cut DNA at specific site is ______.


DNA strands on a gel stained with ethidium bromide when viewed under UV radiation, appear as ______


A specific recognition sequence identified by endonucleases to make cuts at specific positions within the DNA is ______


Which of the given statements is correct in the context of visualizing DNA molecules separated by agarose gel electrophoresis?


In agarose gel electrophoresis, DNA molecules are separated on the basis of their ______.


The role of DNA ligase in the construction of a recombinant DNA molecule is ______.


Which of the following bacteria is not a source of restriction endonuclease?


Which of the following statements does not hold true for restriction enzyme?


Restriction enzymes that are used in the construction of recombinant DNA are endonucleases which cut the DNA at ‘specific-recognition sequence’. What would be the disadvantage if they do not cut the DNA at specific-recognition sequence?


A plasmid DNA and a linear DNA (both are of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.


How does one visualise DNA on an agarose gel?


A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gel with ethidium bromide, no DNA bands were observed. What could be the reason?


CTTAAG
GAATTC

  1. What are such sequences called? Name the enzyme used that recognizes such nucleotide sequences.
  2. What is their significance in biotechnology?

Carefully observe the given picture. A mixture of DNA with fragments ranging from 200 base pairs to 2500 base pairs was electrophoresed on agarose gel with the following arrangement.

(a) What result will be obtained on staining with ethidium bromide? Explain with reason.

(b) The above setup was modified and a band with 250 base pairs was obtained at X.

What change(s) were made to the previous design to obtain a band at X? Why did the band appear at position X?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×