9.E: Transposition of DNA (Exercises)
- Page ID
- 6912
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Question 9.5. Suppose you are studying a gene that is contained within a 5 kb EcoRI fragment for the wild type allele. When analyzing mutations in that gene, you found one that converted the 5 kb fragment to an 8 kb EcoRI fragment. Further analysis showed that the additional 3 kb of DNA was flanked by direct repeats of 6 bp, that the terminal 30 bp of the additional DNA was identical at each end but in an inverted orientation. Recombinant plasmids carrying the 8 kb EcoRI fragment conferred resistance to the antibiotic kanamycin in the host bacteria, whereas neither the parental cloning vector nor a recombinant plasmid carrying the 5 kb EcoRI fragment did. What do you conclude is the basis for this mutation? What other enzyme activities might you expect to be encoded in the additional DNA?
Use the following diagram to answer the next two questions. Transposase encoded by a transposable element (TE) has nicked on each side of the TE in the donor (black) replicon and made a staggered break in the recipient (gray) replicon, and the ends of the TE have been joined to the target (T) site in the recipient replicon. The strands of the replicons have been designated top (t) or bottom (b). The open triangles with 1 or 2 in them just refer to locations in the figure; they are not part of the structure.
Question 9.6. The action of DNA polymerase plus dNTPs, primed at positions 1, followed by ligase (with ATP or NAD) leads to what product or result? (In this scenario, nothing occurs at positions 2).
Question 9.7. The action of an endonuclease at the positions labeled 2 followed by DNA polymerase and dNTPs to fill in the gaps (from positions 1 to the next 5' ends of DNA fragments), and finally DNA ligase (with ATP or NAD) leads to what product or result?
Question 9.8. Refer to the model for a crossover intermediate in replicative transposition in Fig. 9.13. If the transposon moved to a second site on the same DNA molecule by replicative transposition (not to a different molecule as shown in the Figure), what are the consequences for the DNA between the donor and recipient sites?
Question 9.9.The technique of transposon tagging uses the integration of transposons to mutate a large numbers of genes while leaving a "tag" in the mutated gene to allow subsequent isolation of the gene using molecular probes (such as hybridization probes for the transposon). What is a good candidate for transposon tagging in mammalian cells?
Contributors and Attributions
Ross C. Hardison, T. Ming Chu Professor of Biochemistry and Molecular Biology (The Pennsylvania State University)