1081091010 AngiospermsAmphibiansCartilaginous fishMammalsReptilesBony - PDF document

1081091010 AngiospermsAmphibiansCartilaginous fishMammalsReptilesBony FishBirdsCrustaceansInsects bp = � 50,000 izyajczkblqfreighttrainrunninsofastelizabethco

ttonqwftzxvbifyoudontbelieveimleavingyoujustcountthedaysimgonerxcvwpowentdowntothecrossroadstriedtocatchariderobertjohnsonpzvmwcomeonhomeintomykitchentrad.This is 100

times more complex (200 letters or 2000 bp).A solution of 1 mg DNA/ml is 0.0015 M (in terms of moles of bp per L) or 0.003 M (interms of nucleotides per L). We'll use

0.003 M = 3 mM, i.e. 3 mmoles nts per L. (nts =nucleotides).Consider a 1 mg/ml solution of each of the three DNAs. For DNA 1, this means that thesequence ab (20 nts

) is present at 0.15 mM or 150 (expressed as moles of to a known standard.If you have a standard of known genome size, you can calculate N from C0t1/2:NunknownNstand

ard=C0t12unknownC0t12standard(6)A known standard could be E. coliN = 4.639 x 106 bp orpBR322N C0t1/2 for the individual component is the C0t1/2(measured in the mixt

ure of components) ! f. For example the C0t1/2 for individual (pure)component 2 is 10-2 ! 0.2 = 2 ! 10-3 .Knowing the measured Cot1/2 for a DNA standard, one can calc

ulate the complexity of eachcomponent.complexityn = Nn = C0t1/2pure, n ! NstdC0t1/2std=C0t1/2pure,n!3!106bp10subscript n refers to the particular component, i.e. (1

, 2, 3, or 4)The repetition frequency of a given component is the total number of base pairs in that componentdivided by the complexity of the component. The total nu

mber of base pairs in that component isgiven by fn ! G.Rn = fn!!! 2.6 x 10713,0006-7Everything elsePreparation of normalized cDNA libraries for ESTsJust like the mRN

A populations used as the templates for reverse transcriptase, the cDNAsfrom a particular tissue or cell type will be composed of many copies of a very few, abundantmR

NAs, a fairly large number of copies of the moderately abundant mRNAs, and a small numberof copies of the rare mRNAs. Since most genes produce low abundance mRNA, a c

orrespondingsmall number of cDNAs will be made from most genes. In an effort to obtain cDNAs from mostgenes, investigators have normalized the cDNA libraries to remov

e the most abundant mRNAs.The cDNAs are hybridized to the template mRNA to a sufficiently high Rot (concentration of RNA http://www.ncbi.nlm.nih.gov/UniGene/index.htm

lcDNA clones and ESTs5Õ3ÕmRNAAAAA5Õ large scale sequencing1. Whole genomes can be sequenced both TotalEnzymesCategoryEubacteriaEscherichia coli4,639,2214,2891081,

254gram negativeHaemophilus influenzae1,830,1351,71774571gram negativeHelicobacter pylori1,667,8671,56643394gram negativeBacillus subtilis4,214,8144,100121819gram posi

tiveMycoplasma genitalium580,07346736202gram positiveMycoplasma pneumoniae816,39467733226gram positiveMycobacterium tuberculosis4,411,5293,91848-gram positiveAquifex a

eolicus1,551,3351,52250-hyperthermophilicbacteriumBorrelia burgdorferi1,230,6631,25623176lyme diseaseSpirocheteSynechocystis sp.3,573,4703,16649702cyanobacteriumArchae

bacteriaArchaeoglobus fulgidus2,178,400 NCBIhttp://www.ncbi.nlm.nih.govNucleic acid sequencesgenomic and mRNA, including ESTsProtein sequencesProtein structuresGenetic

and physical mapsOrganism-specific databasesMedLine (PubMed) , TBX1 104M. gen.H. infl.E. coliS. cer.C. ele.D. mel.Core proteomeGene numberNumber of genes, number in c

ore proteomeSpeciesFigure 4.22. Little change in core proteome size in eukaryotes Regions of human chromosomes homologous to regions of mouse chromosomes(indicated by

the colors). For example, virtually all of human chromosome 20 is homologous to aregion on mouse chromosome 2, and almost all of human chromosome 17 is homologous to a

region on mouse chromosome 11. More commonly, segments of a given human chromosomes arehomologous to different mouse chromosomes. Chromsosomes from mouse have morerear

rangements relative to humans than do chromosomes from many mammals, but thehomologous relationships are still readily apparent. chromosomes by removing most of the p

roteins, longloops of DNA are seen, emanating from a central scaffold epsgamgamdeltabetaLCRHPFH/ORGsHRASHBBPTHMYODp15Geimsa dark band D11S1cCK2-2, ZnFP104RRM1N159-H3AR

HGD11S30, D11S191HBBSMPD1HPXST5WEE1RBTN1HPXST5RBTN1D11S189D11S569CALCAPTHMYODfrom Higgins et al.BreakpointsJ1-series84b5010179TEL chromatin are histones.Composition of

chromatinVarious biochemical methods are avialable to isolated chromatin from nuclei. Chemical analysis ofchromatin reveals proteins and DNA, with the most abundant

proteins being the histones. Acomplex set of less abundant histones are referred to as the nonhistone chromosomal proteins.The histones and DNA present in equal masse

s.Mass RatioDNA: histones: nonhistone proteins: RNA= 1: 1: 1: 0.1Histones are small, basic (positively charged), highly conserved proteins. They bind toeach

other to form specific complexes, around which DNA wraps to form nucleosomes. Thenucleosomes are the fundamental repeating unit of chromatin.There are 5 histones, 4

in the core of L2CN nucleosome core showing solenoid in the 30 ) are as follows:Component fCot 1/2(measured)fast0.210-4medium0.410-1slow0.4104a)Use the information

provided to calculate the Cot 1/2(pure), the complexity (N), andthe repetition frequency (R) for each component. Assume that the slowly renaturing component issingle

copy.b)Calculate the genome size (G) of the armadillo under the assumption that the slowlyrenaturing component is single copy.c) globin gene is induced how much in MEL

cells treated with DMSO?DNA probemRNA uninducedinducedMEL cellsT cells}Source of RNAb) The previous assay gives the relative amounts of the mRNA under the two condit

ions,and this is an extremely powerful and widely used assay. But what does this mean in terms ofmRNA molecules per cell, i.e. how does the abundance change upon indu

ction? One can alter thisassay somewhat to get a measure of abundance, similar in principle to the calculations in SectionVIIF. First, one needs a measure of the num

ber of mRNA molecules per cell. Let's say that youharvested 107 MEL cells and isolated 3 2525ovalbumin00d) In general, what is the distribution of mRNAs in a partic

ular type of differentiated cell,i.e. how abundant are the different complexity classes of mRNA?Use of databases of sequences, mutations, gene on the E. coli chromo

some? Go back to the Entrez server (where youclicked on Nucleotides before). Click on Genomes, and then select Escherichia coli. Enter "argI"in the Search window (d

on't enter the quotes, and that is the letter I "eye" not a "one").4.7Is the E. coli OTC protein related to any other proteins in the sequence databases? Youneed to

get the protein sequence, which you can do by clicking on argI while you are at the genome - humanpair, we will have to use a different tool to see the alignment. At

the Blast server top page (whereyou selected Basic Blast search before), select Blast 2 sequences. This utility allows you to enterany two sequences and generate a pa

irwise alignment by the program Blast2. You should use thehuman and E. coli OTC protein sequences or their accession numbers, and be sure to chooseblastp as the prog

ram. When doing this in July of 1998, I ran into a problem with the utilitymaking a duplicate of each sequence I entered (I don't know if that was a problem at my end

ortheirs); this is likely a temporary condition. If you encounter a problem, try a different Server, suchas the Sequence Analysis Server at http://genome.cs.mtu.edu/

sas.html. Choose Pairwise SequenceAlignment, enter your sequences and run GAP or SIM on protein sequences.Chromatin4.10One of the important early pieces of evidence t

hat helped define the structure of thenucleosome was the pattern of nuclease cleavage in chromatin. In this experiment, chromatin wastreated briefly with an enzyme, m

icrococcal nuclease, that degrades DNA, then all protein wasremoved and the and the purified DNA resolved by electrophoresis. A regular pattern of broadbands was seen

; the average sizes of the DNA fragments were multiples of 200 bp, i.e. 200, 400,600, 800 bp, etc. What does this result tell you about chromatin structure? The ba

nds of DNAbands were thick and spread out rather than sharp; what does this tell you about the positions ofcleavage by micrococcal nuclease?4.11Which histones are in t

he core of the nucleosome? What are the protein-proteininteractions in the core? What protein domains mediate these interactions?4.12The mammalian virus SV40 has min

ichromosomes in which the circular duplex DNA ispackaged into nucleosomes. When histones are removed from the minichromosomes, the resultingDNA is found to be negativ

ely supercoiled. What does this tell you about the state of the DNA inthe minichrosomes and the path of the DNA around the nucleosome?4.13Are the following statements