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The Chemistry of DNA Replication All DNA polymerases synthesize new DNA strands in the \(5^{\prime} \rightarrow 3^{\prime}\) direction. In some respects, replication of the antiparallel strands of duplex DNA would be simpler if there were also a second type of polymerase, one that synthesized DNA in the \(3^{\prime} \rightarrow 5^{\prime}\) direction. The two types of polymerase could, in principle, coordinate DNA synthesis without the complicated mechanics required for lagging strand replication. However, no such \(3^{\prime} \rightarrow 5^{\prime}\) -synthesizing enzyme has been found. Suggest two possible mechanisms for \(3^{\prime} \rightarrow 5^{\prime}\) DNA synthesis. Pyrophosphate should be one product of both proposed reactions. Could one or both mechanisms be supported in a cell? Why or why not? (Hint: You may suggest the use of DNA precursors not actually present in extant cells.)

Short Answer

Expert verified
Both mechanisms are unlikely to be supported due to thermodynamic and evolutionary constraints.

Step by step solution

01

Understanding the Replication Directionality

DNA polymerases synthesize DNA in the 5′ to 3′ direction by adding nucleotides to the 3′-end of the growing DNA strand through a condensation reaction. This involves the attack by the 3′-hydroxyl group on the new base's α-phosphate, releasing pyrophosphate.
02

Considering Alternative Chemical Reaction

For a hypothetical 3′ to 5′ synthesis, one suggestion is that the polymerase could use a different precursor, such as a nucleotide triphosphate with the 3′ position activated instead of the 5′. The release of pyrophosphate could still occur if the precursor had high-energy bonds cleaved.
03

Proposed Mechanism with Reversing the Chemistry

Another mechanism could involve reversing the standard condensation reaction to incorporate the nucleotide through a nucleophilic attack by a 5′-phosphate from an incoming nucleotide on the growing chain's 3′-end phosphate, still resulting in pyrophosphate formation.
04

Evaluating the Viability of Mechanisms

Neither proposed mechanisms are likely to be supported in cells. The reason being the thermodynamic instability and high-energy requirement for such atypical reactions which are not natural, as well as the absence of evolutionary advantage for alternative chemistry.

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Key Concepts

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

DNA polymerases
DNA polymerases are essential enzymes that play a crucial role in the process of DNA replication. They synthesize new strands of DNA by reading an existing template strand and creating a complementary strand. DNA polymerases work primarily in the cell's nucleus during DNA replication and repair processes.
Key functions of DNA polymerases include:
  • Adding nucleotides to a growing DNA strand
  • Ensuring accuracy through proofreading
  • Repairing damaged DNA
Their activity is vital for cell division, as they ensure that DNA is accurately duplicated for new cells. Without DNA polymerases, cells would not be able to replicate DNA properly, leading to genetic errors.
5′ to 3′ synthesis
The synthesis of new DNA strands by DNA polymerases occurs in a specific direction: from the 5′ (5 prime) end to the 3′ (3 prime) end. This directionality is vital because of the structure of DNA and the nature of the chemical reactions involved.
During DNA replication, nucleotides are added to the 3′ end of the growing strand. The process involves a nucleophilic attack by the 3′-hydroxyl group of the last nucleotide on the DNA strand onto the α-phosphate group of the incoming dNTP (deoxynucleoside triphosphate).
This reaction releases a molecule of pyrophosphate, providing energy to drive the reaction forward. Thus, 5′ to 3′ synthesis ensures that DNA strands are extended in an efficient and energy-favorable manner, aligning with the inherent chemical structure of nucleotides.
lagging strand replication
In DNA replication, the polymerase synthesizes the leading strand continuously but faces challenges with the lagging strand. The antiparallel nature of DNA means the lagging strand is oriented in a 3′ to 5′ direction against the replication fork movement.
To resolve this, the lagging strand is synthesized in short fragments, known as Okazaki fragments. These fragments are formed in the 5′ to 3′ direction as the fork unwinds further, and later, they are joined to form a complete strand.
The process requires:
  • RNA primers to initiate the synthesis of each Okazaki fragment
  • Additional enzymes, like DNA ligase, to join fragments
This process is efficient yet more complex, highlighting the intricacies of cellular machinery to maintain accurate DNA replication.
pyrophosphate
In the context of DNA replication, pyrophosphate plays a key role as a byproduct of the polymerization process. During the synthesis of new DNA strands, each nucleotide addition involves forming a phosphodiester bond.
The reaction starts with the 3′-hydroxyl of the existing strand attacking the phosphate group of an incoming nucleotide triphosphate. This results in the breaking of a high-energy bond and the release of pyrophosphate ( PP_i ), which comprises two linked phosphate groups.
The breakdown of pyrophosphate into two inorganic phosphates by pyrophosphatase is crucial. This hydrolysis reaction is thermodynamically favorable and drives the DNA polymerization process, ensuring that the synthesis reaction proceeds in the forward direction. Thus, the generation and subsequent hydrolysis of pyrophosphate is a central aspect of the energy dynamics involved in DNA replication.

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Most popular questions from this chapter

DNA Repair Mechanisms Vertebrate and plant cells often methylate cytosine in DNA to form 5 -methylcytosine (see Fig. \(8-5\) a). In these same cells, a specialized repair system recognizes G-T mismatches and repairs them to \(\mathrm{G}=\mathrm{C}\) base pairs. How might this repair system be advantageous to the cell? (Explain in terms of the presence of 5-methylcytosine in the DNA.)

DNA Repair in People with Xeroderma Pigmentosum The condition known as xeroderma pigmentosum (XP) arises from mutations in at least seven different human genes (see Box \(25-1\) ). The deficiencies are generally in genes encoding enzymes involved in some part of the pathway for human nucleotide- excision repair. The various types of XP are denoted A through G (XPA, XPB, etc.), with a few additional variants lumped under the label \(\mathrm{XP}-\mathrm{V}\) Cultures of fibroblasts from healthy individuals and from patients with XPG are irradiated with ultraviolet light. The DNA is isolated and denatured, and the resulting single-stranded DNA is characterized by analytical ultracentrifugation. (a) Samples from the normal fibroblasts show a significant reduction in the average molecular weight of the single-stranded DNA after irradiation, but samples from the XPG fibroblasts show no such reduction. Why might this be? (b) If you assume that a nucleotide-excision repair system is operative in fibroblasts, which step might be defective in the cells from the patients with XPG? Explain.

Importance of DNA Topoisomerases in DNA Replication DNA unwinding, such as that occurring in replication, affects the superhelical density of DNA. In the absence of topoisomerases, the DNA would become overwound ahead of a replication fork as the DNA is unwound behind it. A bacterial replication fork will stall when the superhelical density \((\sigma)\) of the DNA ahead of the fork reaches +0.14 (see Chapter 24 ). Bidirectional replication is initiated at the origin of a 6,000 bp plasmid in vitro, in the absence of topoisomerases. The plasmid initially has a \(\sigma\) of \(-0.06 .\) How many base pairs will be unwound and replicated by each replication fork before the forks stall? Assume that both forks travel at the same rate and that each includes all components necessary for elongation except topoisomerase.

Fidelity of Replication of DNA What factors promote the fidelity of replication during synthesis of the leading strand of DNA? Would you expect the lagging strand to be made with the same fidelity? Give reasons for your answers.

A Connection between Replication and Site-Specific Recombination Most wild strains of Saccharomyces cerevisiae have multiple copies of the circular plasmid \(2 \mu\) (named for its contour length of about \(2 \mu \mathrm{m}\) ), which has \(\sim 6,300\) bp of DNA. For its replication the plasmid uses the host replication system, under the same strict control as the host cell chromo- somes, replicating only once per cell cycle. Replication of the plasmid is bidirectional, with both replication forks initiating at a single, well- defined origin. However, one replication cycle of a \(2 \mu\) plasmid can result in more than two copies of the plasmid, allowing amplification of the plasmid copy number (number of plasmid copies per cell) whenever plasmid segregation at cell division leaves one daughter cell with fewer than the normal complement of plasmid copies. Amplification requires a sitespecific recombination system encoded by the plasmid, which serves to invert one part of the plasmid relative to the other. Explain how a site-specific inversion event could result in amplification of the plasmid copy number. (Hint: Consider the situation when replication forks have duplicated one recombination site but not the other.)

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