PS1 key - MCB110 Spring 2011 Problem Set 1 1....

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Unformatted text preview: MCB110 Spring 2011 Problem Set 1 1. Why are polynucleotides that are GC ­rich more stable than those that are AT ­rich? Does this mean they have a higher or lower Tm? G base pairs with C through 3 hydrogen bonds, comparing to 2 in an AT base pair, thus making GC ­rich polynucleotides more stable. As a result, they require higher temperature to be separated, thus higher Tm. 2. What kind of information can proteins recognize in the major grooves of dsDNA that facilitates binding? Where is this information coming from, i.e. the backbone, bases, or sugars? Proteins recognize different functional groups such as methyl groups, hydrogen bonds, hydrogen donors or hydrogen acceptors on the bases. Thus minor groove is a poor place to look for information because it cannot distinguish differences in a GC and a CG pair, or AT from a TA pair. 3. Write an 8bp palindromic sequence. Up to you! 4. You found a protein with unknown functions. It exists as a homodimer and each monomer has a protruding alpha ­helix that has a positively charged surface. What is a possible function for this protein? What might it bind to and where? Either transcription factor or endonuclease would be accepted, or just DNA binding protein in general that can bind to the negatively charged backbone of a nucleic acid duplex. There are other possibilities of course but you don’t have to know those 5. Is dsRNA more accessible by proteins in its major or minor groove? Why? Explain in terms of structure. It’s more accessible in the minor groove because its major grooves are buried deeply within the duplex since dsRNA takes on the more “squished” A ­form, with larger diameters of duplex cross section and shorter vertical distance between base pairs with respect to the 5’ ­3’ axis. 6. Recently, scientists at NASA ­Ames reported in a controversial article that they have found a strain of bacteria that uses arsenic in its genetic information (Wolf ­ Simon et al. Science. 2010) a) What could arsenic replace if it is in the nucleic acids of this bacterium? Hint: look at a periodic table. Phosphorus b) Considering As is very unstable, this bacterium must have developed ways to stabilize its nucleic acids. List THREE different types of energetic motivation that it could improve for DNA or RNA duplexes to form spontaneously, and their directions of action (with respect to the 5’  ­ ­> 3’ axis). H ­bonds—horizontal between bases in a pair Base stacking—vertical between base pairs along the axis (rungs on a ladder) Electrostatic repulsions to keep backbone negative charges as far away from each other as possible—twist…not really a direction c) Can you think of a simple initial experiment to test this claim with techniques that we’ve learned in class so far? Describe how you would set it up and include the controls. (Note normally this strain grows in an environment that can have arsenic concentration as high as 200uM). Grow the bacteria on media containing either 200uM As or equal concentration of phosphorus. If this bacterium incorporates As instead of P into its genomic content, then when you purify plasmid DNA from bacteria in the two conditions and run a gel, you should be able to see DNA bands in samples from bacteria grown on 200uM As, but no bands from phosphorus ­containing media. However, if this bacterium is not an obligate “arsenic ­phile”, i.e. it doesn’t absolutely require As to survive, then it could produce DNA even when grown on phosphorus, which was what the authors have found. Furthermore, they actually saw an increased growth when phosphorus is supplemented into arsenic media. So “are aliens among us”? Not quite… 7. Based on the structure below, what might this molecule bind and how? Because it’s a planar structure with aromatic rings, this compound probably acts as an intercalator in DNA in the same way as ethidium bromide. This is in fact SYBR green, a common fluorophore used to detect dsDNA. When these molecules are not intercalated into the duplex, they are not very fluorescent, but once they intercalate, their Pi electrons have more space to move around and interact with the Pi orbitals found in the bases, which increases fluorescence. 8. You are studying a wild ­type strain of bacteria that express a restriction endonuclease that recognizes the sequence: a) What type of overhang will this restriction enzyme leave? Blunt, 3’ overhang or 5’ overhang? 3’ ­overhang b) You want to introduce a plasmid from another bacteria into this species. Will this be successful if the DNA you want to introduce is left as is? Why and why not? No, because this host WT bacterium would methylate its own DNA at the sequence GGGCCC to keep its restriction endonuclease from cleaving its own DNA. But other species of bacteria will produce different endonucleases that recognize different sites so their GGGCCC sequences will not likely to be methylated, or protected. Therefore, if you introduce a foreign DNA without any modifications, the foreign DNA will be digested by the enonucleases in the host bacterial species. 9. If a DNA polymerase lacks a 3’ ­5’ exonuclease activity, do you expect it to be more or less accurate than E. coli DNA polymerase I? Explain. Less accurate. The 3’ ­5’ exonuclease is the proofreading activity that increases fidelity. 10. Is the 3’ ­5’ exonuclease activity of DNA PolI simply the reverse reaction of the 5’ ­ 3’ polymerase activity? Explain. No. The polymerase starts with dNTP, adds dNMP to the chain and in the process, releases PPi. The exonuclease simply removes dNMP without synthesizing dNTPs. 11. Name the 4 requirements for most DNA polymerase to function and the 2 traits they must possess to be able to replicate an entire genome. Template, primer, dNTP and magnesium. Fast rate of replication and high processivity (the ability to add many nucleotides before falling off) 12. How can DNA polymerase I distinguish and control its two activities, polymerase or 3’ ­5’ exonuclease, that’s favored at a particular moment? Explain in terms of structure and kinetics. What happens to the 3’OH of the growing chain during this decision? DNA PolI can recognize a mismatch by the change mismatches can induce in the active site, and thus can no longer bind as well to the 3’OH of the growing chain. Think of it as loosening its grip by shifting the palm (active site). This loosening effect slows down polymerase, so while the 3’OH of the growing chain can flip to the active site of the exonuclease, its 3’ ­5’ cutting activity has faster kinetics, thereby favoring correction over adding new nucleotides. ...
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This document was uploaded on 09/12/2011.

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