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The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.
During DNA replication, the breaking of H-bonds is performed by :
- a)Topoisomerase
- b)Gyrase
- c)Helicases
- d)None
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The DNA replication is semi-conservative is proved by an experiment ...
DNA replication starts with unwinding of DNA duplexes which are held together by hydrogen bond. Helicases move along the double stranded DNA and separate the strands by breaking hydrogen bonds between base pairs.
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Most Upvoted Answer
The DNA replication is semi-conservative is proved by an experiment ...
The breaking of hydrogen bonds during DNA replication is performed by helicases.
Helicases:
- Helicases are enzymes that play a crucial role in DNA replication by unwinding the double-stranded DNA molecule.
- These enzymes use energy from ATP hydrolysis to break the hydrogen bonds between the base pairs and separate the DNA strands.
- The separation of the DNA strands creates a replication fork, which allows the DNA replication machinery to access the DNA template strands.
Role in DNA Replication:
- During DNA replication, the first step is the unwinding of the double helix structure of DNA.
- Helicases bind to the origin of replication, which is a specific sequence on the DNA molecule.
- Once bound, helicases start moving along the DNA molecule, unwinding the double helix as they progress.
- As the helicases move along the DNA, they break the hydrogen bonds between the base pairs, separating the two DNA strands.
- This process creates two single-stranded DNA templates that can be used as templates for DNA synthesis.
Directionality:
- Helicases move in a 5' to 3' direction along the template DNA strand.
- The 5' to 3' directionality refers to the orientation of the DNA molecule, with the 5' end having a phosphate group and the 3' end having a hydroxyl group.
- The helicases break the hydrogen bonds between the base pairs starting from the 5' end and moving towards the 3' end of the DNA molecule.
- This directionality is important as it allows DNA replication to proceed in the 5' to 3' direction, which is necessary for the proper synthesis of the new DNA strand.
Conclusion:
In conclusion, helicases are enzymes that play a critical role in DNA replication by breaking the hydrogen bonds between the base pairs and unwinding the DNA molecule. They move in a 5' to 3' direction along the DNA template strand, creating a replication fork and allowing the DNA replication machinery to access the DNA template strands for synthesis. Hence, the correct answer is option 'C' - Helicases.
Helicases:
- Helicases are enzymes that play a crucial role in DNA replication by unwinding the double-stranded DNA molecule.
- These enzymes use energy from ATP hydrolysis to break the hydrogen bonds between the base pairs and separate the DNA strands.
- The separation of the DNA strands creates a replication fork, which allows the DNA replication machinery to access the DNA template strands.
Role in DNA Replication:
- During DNA replication, the first step is the unwinding of the double helix structure of DNA.
- Helicases bind to the origin of replication, which is a specific sequence on the DNA molecule.
- Once bound, helicases start moving along the DNA molecule, unwinding the double helix as they progress.
- As the helicases move along the DNA, they break the hydrogen bonds between the base pairs, separating the two DNA strands.
- This process creates two single-stranded DNA templates that can be used as templates for DNA synthesis.
Directionality:
- Helicases move in a 5' to 3' direction along the template DNA strand.
- The 5' to 3' directionality refers to the orientation of the DNA molecule, with the 5' end having a phosphate group and the 3' end having a hydroxyl group.
- The helicases break the hydrogen bonds between the base pairs starting from the 5' end and moving towards the 3' end of the DNA molecule.
- This directionality is important as it allows DNA replication to proceed in the 5' to 3' direction, which is necessary for the proper synthesis of the new DNA strand.
Conclusion:
In conclusion, helicases are enzymes that play a critical role in DNA replication by breaking the hydrogen bonds between the base pairs and unwinding the DNA molecule. They move in a 5' to 3' direction along the DNA template strand, creating a replication fork and allowing the DNA replication machinery to access the DNA template strands for synthesis. Hence, the correct answer is option 'C' - Helicases.
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Community Answer
The DNA replication is semi-conservative is proved by an experiment ...
During replication of DNA ,the unwinding of DNA strand with the help of Enzymes DNA helicase. The breaking of DNA is due to hydrogen bond .the separation of each strand now act as template strand.
Thus, option C is correct.
Thus, option C is correct.
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The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer?
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The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer?.
The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The DNA replication is semi-conservative is proved by an experiment conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.During DNA replication, the breaking of H-bonds is performed by :a)Topoisomeraseb)Gyrasec)Helicasesd)NoneCorrect answer is option 'C'. Can you explain this answer?.
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