Problem 7
Question
A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture? \begin{equation}\begin{array}{l}{\text { (A) DNA polymerase }} \\ {\text { (B) DNA ligase }} \\ {\text { (C) Okazaki fragments }} \\ {\text { (D) primase }}\end{array}\end{equation}
Step-by-Step Solution
Verified Answer
B - DNA ligase
1Step 1 - Understand the symptoms
The DNA replication process occurs, but results in DNA strands paired with numerous short segments.
2Step 2 - Identify possible missing components
Review the provided options and their roles in DNA replication:(A) DNA polymerase - an enzyme that synthesizes new DNA strands.(B) DNA ligase - an enzyme that connects DNA fragments.(C) Okazaki fragments - short DNA fragments synthesized on the lagging strand.(D) Primase - an enzyme that synthesizes RNA primers needed for DNA polymerase to start synthesis.
3Step 3 - Analyze the role of each component
DNA polymerase synthesizes DNA and cannot be missing if replication occurs. Primase helps start the replication process and its absence would stop replication. Okazaki fragments are part of the process and cannot be missing. DNA ligase connects Okazaki fragments into a continuous strand.
4Step 4 - Identify the missing component
Since replication occurs but results in fragmented DNA, DNA ligase is likely missing because it is responsible for linking the short DNA segments, Okazaki fragments, into a complete strand.
Key Concepts
DNA polymeraseDNA ligaseOkazaki fragmentsprimase
DNA polymerase
DNA polymerase is an essential enzyme in the DNA replication process. It has the critical role of synthesizing new DNA strands by adding nucleotides one by one to the growing DNA chain. This enzyme works by reading the existing DNA strand, known as the template strand, and creating a complementary strand. Without DNA polymerase, the replication process cannot proceed, as no new DNA molecules would be formed. Different types of DNA polymerase exist in eukaryotic and prokaryotic organisms, each with specific functions in replication and repair.
DNA ligase
DNA ligase is another crucial enzyme in DNA replication. Its main job is to join the short DNA fragments, known as Okazaki fragments, which are created intermittently on the lagging strand during replication. Enzymes like DNA polymerase can synthesize DNA only in one direction, so the lagging strand forms in short bursts. These fragments need to be pieced together to form a continuous strand of DNA. DNA ligase performs this task by creating phosphodiester bonds between the fragments. In its absence, the DNA would remain fragmented and incomplete, which is exactly the issue described in the exercise.
Okazaki fragments
During replication, the DNA double helix unwinds, creating a leading and a lagging strand. Okazaki fragments are short sequences of DNA nucleotides which are synthesized discontinuously and later linked together. They form on the lagging strand due to the antiparallel nature of DNA. The leading strand is synthesized continuously, but the lagging strand must be copied in small segments to match the 5’ to 3’ direction required by DNA polymerase. After they are synthesized, DNA ligase joins these fragments to form a complete DNA strand.
primase
Primase is an enzyme that plays a key role in the initiation phase of DNA replication. It synthesizes a short RNA primer that provides a starting point for DNA polymerase to begin synthesis. Without primase, DNA polymerase would not know where to start making the new DNA strand. The primer is later removed and replaced with DNA nucleotides. It's important to note that the absence of primase would halt the replication process entirely. This is because DNA polymerase cannot initiate replication on its own; it relies on the RNA primer provided by primase to begin.
Other exercises in this chapter
Problem 5
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