Answer3 - MCB 161 Problems for Lecture 3 Molecular Methods...

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MCB 161 S. Harmer Problems for Lecture 3 – Molecular Methods MBOC Problems book : 4-55 If nucleosomes were randomly positioned on the 225-bp segment of DNA, then the 146- bp fragments would be a collection of all possible 146-bp segments of the original DNA. Such a random collection would give a highly diverse set of fragments upon digestion with a restriction nuclease that cuts at a unique location. The generation of only two fragments after restriction digestion means that there is a strongly preferred location for a nucleosome on this piece of DNA, which gives rise to a unique 146-bp fragment. When this fragment is cut, it gives a 37-bp and a 109-bp fragment, which sum to 146. If the position of the restriction cut were given in the problem, you could have deduced where the nucleosome is situated on the 225-bp segment. (Some DNA sequences have a high affinity for the nucleosome because they bend easily due to high A:T content; see page 179 and Figure 7-38 in your text. Note that such sequences are not required for nucleosome formation.) 8-67 True. If each cycle doubles the amount of DNA, then 10 cycles equal a 2 10 -fold amplification (which is 1024), 20 cycles equal a 2 20 -fold amplification (which is 1.05 x 10 6 ), and 30 cycles equal a 2 30 -amplification (which is 1.07 x 10 9 ). (It is useful to remember that 2 10 is roughly equal to 10 3 or 1000. This simple relationship allows you to estimate the answer to this problem rapidly without resorting to your calculator. It comes in handy in a variety of contexts.) 8-80 ( Note there is a typo in the book; primer #1 is written incorrectly in your text but the correct primer sequences are below. ) The appropriate PCR primers are primer 1 (5’-GACCTGTGGAAGC) and primer 8 (5’- TCAATCCCGTATG). The first primer will hybridize to the bottom strand and prime synthesis in the rightward direction. The second primer will hybridize to the top strand and prime synthesis in the leftward direction. (Remember that strands pair antiparallel.) 8-87 A. The central four nucleotides of a BamHI site constitute a Sau3A site. Since Sau3A recognizes only these four nucleotides, all BamHI sites will be cut by Sau3A. The converse is not true because BamHI recognizes a 6-nucleotide site. Only that fraction of Sau3A sites with appropriate flanking nucleotides (specifically, a 5’ G and a 3’ C) will be
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This note was uploaded on 01/18/2010 for the course BIO SCI 104 taught by Professor Lin during the Spring '09 term at UC Davis.

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Answer3 - MCB 161 Problems for Lecture 3 Molecular Methods...

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