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by: Marcella Lebsack


Marcella Lebsack
GPA 3.81


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Class Notes
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This 13 page Class Notes was uploaded by Marcella Lebsack on Thursday October 22, 2015. The Class Notes belongs to BIOL 4320 at Texas Tech University taught by Densmore in Fall. Since its upload, it has received 27 views. For similar materials see /class/226494/biol-4320-texas-tech-university in Biology at Texas Tech University.




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Date Created: 10/22/15
Directions for accessing website for Powerpoint presentations for BIOL 43205320 httpwwwf cutvbiottueduldensmorelMolecular Biol 091 2007htm Then look for the pertinent lecture Quick review from last Thursday regarding Cot analysis If you think about it eukaryotic genomes are made up of various classes of different sequences of varying size and copy number From the curves we have seen so far we note that short sequences that are highly repetitive in nature reassociate faster than longer sequences that are in low copy number Cot analysis can be used to estimate the total length of a genome and to estimate just how many size classes of sequences exist how large on average each class is as well as the average number of copies that each sequence class has Fraction dissociated Cot Analysis 1 10 I 20 of total curve C 08 L 27 of curve B 06 I I 04 l I I l 53 of curve A I I I I I I I I I I I I 1 l J l 4 I 10 4 10 3 10 2 10 1 10 101 Io2 103 C 1 Observed Cot1 2 C0112 C 0 C0112 B C0112 A values I l l I l I I 102 103 104 105 106 107 10quot Empirically Mo39ecu39ar weight in base pairs determined Fraction dissociated Cot Analysis 2 Real cot 2 I l J 10 I 3 10 2 1o I 10 101 10 values 0 t 160 bp 23 x 1o 5 bp quot 42 07 bP Avg size of II I I I I I I fragments in 102 103 104 105 106 107 108 each fraction Molecular weight in base pairs Copy number for each fraction is inversely proportional 10 I to the observed t12 value of each fractIon and therefore to the Cot1 2 value of each fraction 08 39C 9 VS 6 06 O U 2 I U I E 04 I 5 I S I LL 0 2 39 I I I I I I I I I I I I 4 1 3 I I Observed Cot 2 2 1 0 1 2 3 10 10 10 10 10 10 10 10values C t Cot12 C 0 Cot12 B Cot12 A l l l l I l 102 103 104 105 106 107 108 Molecular weight The observed Cot 2 for the largest fraction is A 102 and we assume that it represents the largest fraction of fragments thus we also assume that it includes all unique sequences To estimate the copy number for other fractions we need to get a baseline value to use as a controlthat baseline copy needs to be 1 It turns out that we can use the observed Cot 2 for fraction A as the denominator in the following equation 1observed Cot 2 A observed Cot 2 A since the observed Cot 2 A 102 then 1 102102 1 copy which is what it should be Use 102 as the denominator to estimate the copy number of the other fractions as well For Fraction B the observed Cot 2 value is 10 then the copy number for Fraction B will be 110 102 100 copies And for Fraction C observed Cot 2 is 103 the copy number will therefore be 1 103 102 105 copies Copy number for each fraction is inversely proportional 10 to the observed t12 value of each fraction and therefore to the Cot1 2 value of each fraction 08 39C 9 3 06 O 8 C E 04 6 U I 02 I I I Cole 105 Copy 102 Copy I I I 4 1 3 I 2 1 1 IO 41 I2 30bserved Cot 2 10 10 10 10 10 10 10 10values C t Cot12 C 0 Cot12 B Cot12 A l l l l I l 102 103 104 105 106 107 108 Molecular weight We can now estimate the entire genome length For Fraction A Average fragment size 42 x 107 bp and copy number 1 For Fraction B Average fragment size 23 x 105 bp and copy number 100 For Fraction C Average fragment size 160 bp and copy number 105 Entire haploid genome length product of Fraction A size and copy number product of Fraction B size and copy number product of Fraction C size and copy number 42 x 107 23 x 107 16 x 107 81 107 bp Since there are about 660 daltons per base pair this equals 54 x 1010 daltons for a haploid genome and 2 x 54 x 101 daltons for a diploid genome which 107 x 1011 daltons Reannealing and hybridization a second look If one was going homology between species which fraction of sequences would probably be best to use Why Fig 614 mm WWW mm in mm W m i u i u W 55w mm 2 mm mm X cuiXiz39ni and mm mm M W to raw 115d mvr Now we are ready to try our hands at estimating several paramaters from some Cot curves in a problem For problem 1 that has been handed out you will be expected to 1 Estimate the number of different size classes and the average size of each class 2 Estimate the copy number for each of the size classes that you have recognized 3 Estimate the total haploid genome size in base pairs and daltons DNA Topology Most prokaryotic DNA molecules are circular at least part of the time including genomes of phages bacterial chromosomes and plasmids Some eukaryotic DNAs are also circular organellar DNAs mitochondrial and plastid DNAs and indeed there are regions of nuclear chromosomal DNA that actually may be functionally circular how might that be The ends of a linear DNA molecule are flexible and free to rotate if they are covalently attached together or to something else there is only so far that the chains can twist about one another without breaking the bonds they are said to be topologically constrained An illustration of the effects of constraining or twisting the DNA follows There are several topologic properties of covalently closed circular ccc DNA that are important to know Linking number Lk number of times we would have to pass one strand through the other to totally separate two strands Lk Tw Wr Tw is the twist the number of times one strand wraps around the other Tw twist Wr is the number of times the long axis of the Wr writhe double helix crosses over itself Writhe can either be interwound long axis twisted around itself or toroidal long axis wound in a cylindrical manner Lk bpl105 for DNA B Lkquot is the linking number of a fully relaxed cchNA under physiological conditions a Relaxed Lk Lkquot b c Apparently relaxed 1 1 actually overwound 1 3 caused by negative J supertWIsting fer or DNAbindings proteins proteins mpmsammse a Interwound writhe Copyright 2004 Pearson Education Inc publishing as Benjamin Cummings Topological states of covalently closed circular DNA Fig 617


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