Biology Problem Set 5.5

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Biology Problem Set 5.5
2011-03-31 01:28:51
biology exam problem set

Missing #8, #12, #13, #14
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  1. Although DNA polymerases require both a template and a primer, the following DNA was found to serve as a substrate for DNA polymerase in vitro in the absence of any additional DNA or RNA. Explain how this could happen and give the sequence of the product of this reaction. Note: dNTPs and DNA polymerase are supplied but no primase is around in this in vitro DNA synthesis reaction.

    The blue region and the yellow region of this DNA are complementary to each other, thus the blue region can fold back to hybridize with the yellow region. The 3’-end of the blue region can be used as a primer (hence no need for primase) and be extended following a standard DNA synthesis process.
  2. Compare and contrast E. coli polymerase I and III.
    • Function:
    • DNA pol I - maturation of Okazaki fragments
    • DNA pol III - leading and lagging strand synthesis

    • 5' to 3' polymerase:
    • Both

    • 3' to 5' (proofreading) exonuclease:
    • Both
    • 5' to 3' exonuclease:
    • Only DNA pol I
    • Speed:
    • DNA pol I works at a significantly slower rate than DNA pol III (10 nucleotides/sec as opposed to 1000)
    • Processivity:
    • DNA pol I has a much lower processivity than DNA pol III
    • Composition:
    • DNA pol I is composed of a single protein, while DNA pol III is a multi-protein complex
  3. Explain why the two replicating DNA strands are labeled “leading” and “lagging”.
    Because DNA synthesis has to follow the 5’ to 3’ direction, new DNA synthesis using one of the templates (ssDNA) can start right away (leading strand) while the DNA synthesis on the complementary template has to wait until enough room on the template is generated (lagging strand). (See diagram from problem set)
  4. What are Okazaki fragments? Why are they needed? How are they synthesized?
    • An Okazaki fragment is a relatively short fragment of DNA (with an RNA primer at the 5' terminus) created on the lagging strand during DNA replication.
    • They are needed because the lagging strand has to be synthesized in a delayed fashion and in pulses.
    • They are synthesized in the same way as the leading strand, i.e. a primase generates an RNA primer that is used by DNA polymerase III to synthesize new DNA.
  5. What is the role of helicase? Why is its activity very much dependent on ATP?
    • The role of helicase is to unwind the double-stranded DNA, making them single standed templates for the synthesis of new DNA.
    • Helicase is dependent on ATP because it needs the energy provided by ATP hydrolysis to break apart the two DNA stranded held together by H-bonds and base-stacking held together by hydrophobic interactions.
  6. Describe the chemical bond that is formed by the action of DNA ligase. Your answer should include the nature of the bond and the entities linked by this bond.
    It is a phosphodiester linkage between the OH on the 3' end of one DNA molecule and the phosphate on the 5’ end of another DNA molecule.
  7. What is a telomerase? How does it work? Why is a telomerase inhibitor a potential treatment for cancer?
    • A telomerase is an enzyme responsible for extending the repeated telomere sequences usually existing at the end of linear chromosome DNA molecules in eukaryotic cells.
    • It is actually an enzyme carrying an RNA template that is used to extend the 3’end of a DNA molecule.
    • Cancer cells need to have high level of telomerase activity to sustain fast cell division, otherwise the linear DNA molecules become shorter each time the cell divides, and eventually cell division will stop due to shortening of DNA at its ends.Thus, a telomerase inhibitor should slow down cell division in cancer cells.
  8. Explain the key differences between PCR and “cycle” dideoxy sequencing.
    • While both PCR and cycle sequencing use a thermostable Taq DNA polymerase and reiterative DNA synthesis controlled by temperature cycling, there are several important differences:
    • PCR uses two different primers for exponential amplification of a target region, while cycle sequencing only uses one primer per linear amplification. Remember that when we say “one primer,” we mean one particular primer sequence – but we will be adding a huge number of molecules, each with the same nucleotide sequence.
    • Another important difference is that the cycle sequencing reaction must include fluorescently labeled dideoxynucleotide triphosphate “dye terminators,” but the nucleotides added to the PCR reaction will be comprised of only dNTP's.
  9. How does “next generation” sequencing differ from the Sanger sequencing technology (describe in a few sentences). Name two current areas of investigation made possible by next-generation sequencing.
    • The Titanium 454 machine (this one we discussed in lecture) and the Solexa/Illumina machine, both use two key changes in approach. First, there is no cloning. Fragments of DNA are amplified and then used for sequencing while linked to a surface (either a flow cell or a bead), saving all the work of cloning and subcloning the DNA in the sample. Second, CCD cameras are used to observe synthesis as it occurs, recording either a color-coded flash of light for incorporation of a specific base (all four present), or a flash of light following incorporation of a specific base (only one at a time present). This eliminates the need for size-separation of sequencing products – no gel electrophoresis is needed – and allows sequencing of thousands of DNA fragments at the same time.
    • Several kinds of experiments are made possible by next-generation sequencing. Once we have a good version of a genome sequence, we can “resequence” DNA from individuals of that species, allowing analysis of population genetics, analysis of the genomes of cancer cells, etc. It also allows “metagenomics” – sequencing and comparing pools of DNA in a sample taken from the environment. Another big use is “ChIP” experiments – chromatin immunoprecipitation. Here we cross-link the chromosomal proteins to DNA, shear the cross-linked chromatin to generate small fragments, and selectively precipitate that fragments that have a particular protein (such as RNA polymerase) bound by using specific antibodies; sequencing the DNA fragments from the bound material gives us a map of the sites in the genome where RNA polymerase is bound.
  10. Conventional Taq DNA polymerase lacks a 3’ to 5’ exonuclease activity. What is the consequence of the absence of this activity for PCR?
    The error frequency in DNA polymerization in a PCR reaction is high because of the lack of a “proof-reading” activity.