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Chapter 14: Problem 19

Which enzyme initiates the splitting of the double DNA strand during replication? a. DNA gyrase b. helicase c. ligase d. telomerase

Short Answer

Expert verified
The enzyme that initiates the splitting of the double DNA strand during replication is helicase.

Step by step solution

01

Understanding DNA Replication

DNA replication is the process by which a double-stranded DNA molecule is copied to produce two identical DNA molecules.
02

Identify the Key Enzymes

Several enzymes are involved in DNA replication, each with a specific role. The main enzymes include DNA gyrase, helicase, ligase, and telomerase.
03

Role of DNA Gyrase

DNA gyrase is an enzyme that relieves strain while double-strand DNA is being unwound by helicase. It does not initiate the unwinding process.
04

Role of Helicase

Helicase is the enzyme responsible for unwinding the double-strand DNA, by breaking the hydrogen bonds between the nucleotide base pairs. This action initiates the splitting necessary for replication.
05

Role of Ligase

Ligase is an enzyme that joins DNA fragments together by forming a bond between the phosphate backbone. It does not initiate DNA strand splitting.
06

Role of Telomerase

Telomerase adds repetitive nucleotide sequences to the ends of chromosomes (telomeres) to prevent them from shortening during replication. It does not initiate DNA strand splitting.
07

Determine the Correct Enzyme

Given the roles of each enzyme, helicase is the enzyme that initiates the splitting of the double DNA strand during replication.
08

Final Answer

The enzyme that initiates the splitting of the double DNA strand during replication is helicase.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

helicase function
Let's delve into how helicase plays an essential role in DNA replication. Helicase is the enzyme that unwinds the double-stranded DNA. It starts the process by breaking the hydrogen bonds between the paired bases.
This separation is crucial for other enzymes to access the strands and begin copying the DNA.
Helicase works by moving along the DNA strand like a zipper, separating the two strands that form the double helix.
  • Breaks hydrogen bonds between nucleotides
  • Creates a single-stranded template for replication
  • Facilitates the work of other enzymes like DNA polymerase
    Understanding helicase is key to knowing how DNA replication begins and progresses.
DNA unwinding
DNA unwinding is a critical step in the DNA replication process. Before replication can begin, the DNA double helix must first be unwound. This job is performed by helicase breaking the hydrogen bonds that hold the complementary strands together.
As helicase unwinds the DNA, it creates what's called a replication fork, which looks like a Y-shaped structure. This fork is where the DNA strands separate and serve as templates for new DNA synthesis.
The process of unwinding also involves other enzymes like DNA gyrase, which relieves tension and prevents the DNA from becoming supercoiled.
  • Unwinding opens up the DNA strands
  • Formation of the Y-shaped replication fork
  • Involvement of other enzymes to stabilize the process
    DNA unwinding is a necessary precursor to the actual copying of DNA, laying the groundwork for accurate replication.
replication initiation
Replication initiation is the starting point of DNA replication. This phase is marked by the preparation and activation of the DNA sequences and the assembly of the necessary enzymes.
Helicase's role here is crucial as it begins to unwind the DNA helix, providing single-stranded templates. This allows other enzymes like primase to lay down RNA primers, which are short sequences that signal where DNA polymerase should start adding new nucleotides.
  • Initiation of DNA replication
  • Activation of origin sequences
  • Role of RNA primers by primase
    Proper initiation ensures that the DNA replication process starts accurately and continues efficiently.

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Most popular questions from this chapter

Compare and contrast the roles of DNA polymerase I and DNA ligase in DNA replication. a. DNA polymerase I removes the RNA primers from the developing copy of DNA. DNA ligase seals the ends of the new segment, especially the Okazaki fragments. b. DNA polymerase I adds the RNA primers to the already developing copy of DNA. DNA ligase separates the ends of the new segment, especially the Okazaki fragments. c. DNA polymerase I seals the ends of the new segment, especially the Okazaki fragments. DNA ligase removes the RNA primers from the developing copy of DNA. d. DNA polymerase I removes the enzyme primase from the developing copy of DNA. DNA ligase seals the ends of the old segment, especially the Okazaki fragments.

Discuss how mutations can increase variation within a population. a. Substitution mutations may cause a different amino acid to be placed at a specific location, causing small changes in the protein. Frame shift mutations usually cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. b. Substitution mutations may cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. Frame shift mutations may cause a different amino acid to be placed at a specific location, causing small changes in a protein. c. Substitution mutations may cause a different amino acid to be placed at a specific location, resulting in major changes to the protein and leading to radically different characteristics in the offspring. Frame shift mutations cause multiple amino acid differences in a protein, leading to small changes in the protein. d. Substitution mutations result in a different amino acid being placed at a specific position in a protein, causing small changes. Silent mutations could result in new characteristics possessed by an offspring when a stop codon is substituted for an amino acid.

Provide a brief summary of the Sanger sequencing method. a. Frederick Sanger鈥檚 sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. b. Frederick Sanger鈥檚 sequencing is a chain elongation method that is used to generate DNA fragments that elongate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. c. Frederick Sanger鈥檚 sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is joined together by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. d. Frederick Sanger鈥檚 sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a magnetic scanner.

Explain how the components of DNA fit together. a. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with thymine and guanine pairs with cytosine. Adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run anti parallel to each other. b. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form two hydrogen bonds and guanine and thymine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run anti parallel to each other. c. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. d. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run parallel to each other.

Discuss how the scientific community learned that DNA replication takes place in a semi conservative fashion. a. Meselson and Stahl experimented with E. coli. DNA grown in \(^{15} N\) was heavier than DNA grown in \(^{14} N.\) When DNA in \(^{15} N\) was switched to \(^{14} N\) media, DNA sedimented halfway between the \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating fifty percent presence of \(^{14} N\) This supports the semi-conservative replication model. b. Meselson and Stahl experimented with S. pneumonia. DNA grown in \(^{15} N\) was heavier than DNA grown in \(^{14} N\) When DNA in \(^{15} N\) was switched to \(^{14} N\) media, DNA sedimented halfway between the 1 \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating fifty percent presence of \(^{14} N\) This supports the semi-conservative replication model. c. Meselson and Stahl experimented with E. coli. DNA grown in \(^{14} N\) was heavier than DNA grown in \(^{15} N\) When DNA in \(^{15} N\) switched to \(^{14} N\) media, DNA sedimented halfway between the \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating fifty percent presence of \(^{14} N.\) This supports the semi-conservative replication model. d. Meselson and Stahl experimented with S. pneumonia. DNA grown in \(^{15} N\) was heavier than DNA grown in \(^{14} N.\) When DNA in \(^{15} N\) was switched to \(^{14} N\) media, DNA sedimented halfway between the \(^{15} N\) and \(^{14} N\) levels after one round of cell division, indicating complete presence of \(^{14} N.\) This supports the semi conservative replication model.
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