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Stuff to Do. memory +. analysis. 0 pts. 5 pts. memory only. integration of information. Midterm I questions due 1/31. Email me your question (with answers), if you have the capability, mail complete questions, figures, etc. and all,

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midterm i questions due 1 31

memory +


0 pts

5 pts

memory only

integration of information

Midterm Iquestions due 1/31
  • Email me your question (with answers),
    • if you have the capability, mail complete questions, figures, etc. and all,
    • if not, write questions, with instructions…i.e. in Figure 2 of x paper, blah, blah, blah,
  • Friday afternoon, I’ll post the questions on the WEB page, on Monday, you’ll have time to work on them together, in class.
cycle sequencing chain termination a dna polymerase application




Taq DNA Polymerase w/ Buffer


Polymerization until Taq hits ddNTP.

Cycle SequencingChain Termination…a DNA polymerase application.





dNTPs and ddNTPS



Linked on Course WEB Page.

Cycle Sequence Tutor

…and an animation,


Whole Genome Assembly

Hierarchical Clone-by-Clone

Map First: then sequence

Sequence First: then map

genome sequencing strategy 1
Genome Sequencing Strategy#1

Clone-by Clone Approach

  • Order clones along the genome, then sequence,
    • not dependent on acceleration of sequencing capacity,
    • not dependent on advanced computer analysis,
    • not dependent on ‘as-of-yet’ sequencing technologies,
    • “repeats” not as big a problem?
    • heavy up-front demand for human labor.

Genomic Libraries

…how many clones to cover a genome?

vectors carry insert dna

Bacterial Artificial Chromosome

Yeast Artificial Chromosome

Vectors(carry insert DNA)




Plasmid E. coli up to 15 kb,

Phage E. coli up to 25 kb,

Cosmid E. coli up to 45 kb,

BAC E. coli 100-500 kb,

YAC Yeast 250-1000 kb.

plasmid/phage hybrid

genomic sequences and coverage

p finding clone



Genomic Sequences and Coverage

N = ln(1 - .9999)

ln(1 - v/2,900,000,000)

v = average vector insert size

plasmid (5 kb) = 5.3 x 106

phage (20 kb) = 1.3 x 106

BAC (125 kb) = 2.2 x 105

YAC (500 kb) = 27,000 clones

contigs contig uos s equences

Clone 2

Contigs(Contiguos Sequences)

Find overlapping ends…

Clone 1


…Restriction Fragment Length Polymorphisms (RFLPs).



Restriction enzymes cut

specific DNA…


Fragment lengths

provide clone

identification data.

contigs contig uos s equences1
Contigs(Contiguos Sequences)

Find overlapping ends…

Merge good pairs of reads into longer contigs…

Find the minimal Tilling Path,

- minimum set of overlapping clones that cover the genome.

minimal tilling path

Fig. 2

Identify minimal overlapping clones.

Minimal Tilling Path

Shotgun Sequence Each Clone

b acterial a rtificial c hromosomes bacs
Bacterial Artificial ChromosomesBACs
  • Universal Priming Sites,
    • On the vector, flanking the genomic insert.
shotgun self quiz

~ 8x - 10x coverage:

To shotgun sequence 10,000 bp, you’d need 80k - 100k bp of sequence, or ~160 - ~180 sequencing reactions.

But, 10,000 bp, at 500 bp per sequencing reaction could be done in as few as 20 sequencing reactions.

Why Shotgun?

structural genomic strategies 2
Structural Genomic Strategies#2

Whole Genome Assembly Approach:

  • Sequence first, then order,
    • dependent on advances in computer analysis and sequencing technologies,
    • dependent on automated labor.
read pairs mate end pairs



Read Pairs = Mate End Pairs
  • Paired End Sequencing,
    • sequence both ends of the vector insert, using vector derived primers,
      • Maintain mate pair data.





example sequence output example 5 kb insert
Example Sequence Output(example: 5 kb insert)

5’ read(543 bp)-atatgtatattgaattacatacatattattaatgcacatttttatccggagttgtggaccatagaaagacatattgactcctcaaagtaaattctgcatgttacattgaaatcataggctaaatttgagatgcactatttttagaaagtgtagagaaaaggacaggaagaaataagcgaaagctttggtaagccaccaaacctgattactggaagaaaagaaaaaagttccgagaatagagttagatcgctggtgagggttttaaatggaacacaacaatggttgttttagagtgtgttattcttttgtatttataccttctcataggtttcttgtaatacacgcttcttcctctctctccctctctcttatggcctcgtcttgaaagcgtcttgcatgctaagagaaggctttagagcaaggagagaagggagaagttgatttatacgtccatcggatatatcttctttttatatctgtctctcttttaaggaagaaaaatggcgactgaattctcgtgggatgaaatcaagaaagaaaatg...

- rest of insert (unsequenced, ~3.9 kb) -

...ggcttgaaatatttggggcaaacaagcttgaagagaaatcagagaacaagtttttgaaattcttggggttcatgtggaatcctctctcatgggttatggagtctgctgcaatcatggctattgttttagctaatggaggaggaaaggcgccggattggcaagattttatcggtattatggtgttgcttatcatcaactccaccataagtttcatcgaggagaacaatgctggcaatgccgctgctgctctcatggcaaatcttgcaccaaagactaaggtatgcaaatttctcaatacatatatataggtatgtattttctaaaaaggagagttatataacctatgtgtgaatgtaggtgttgagagatggtaaatggggggagcaagaggcttcaatcttggttccgggtgatttgataagcatcaaattgggtgacattgttcctgctgatgctcgtctcctcgaaggagatcctttaaaaattgaccaatctgctcttactggtgaatcccttccaaccaccaaacacccaggagat - 3’ read(540 bp)

…plus trace data files associated with these sequence runs.

project comparisons nyt 10 3 2002
Project Comparisons(NYT: 10/3/2002)
  • Decoding the genome of Plasmodium falciparum, the most dangerous of the four single-cell parasites that cause malaria, took six years and cost about $20 million, paid for by the Wellcome Trust of London, the National Institutes of Health in Bethesda, Md., and other sources. Dr. Malcolm J. Gardner of the Institute for Genomic Research in Rockville, Md., led a large team of scientists there and at the Sanger Centre near Cambridge in England. Completion of the falciparum genome was first announced at a conference in Las Vegas in February.
  • The genome of Anopheles gambiae, the primary carrier of the parasite, was begun more recently and took a mere 15 months even though its genome is far larger — some 278 million units of DNA encoding 14,000 genes compared with the parasite's 23 million units of DNA and 5,268 genes. The mosquito team was led by Dr. Robert A. Holt of Celera Genomics in Rockville. The $14 millioncost was born by the National Institutes of Health, by Genoscope in France and other sources.



  • WGA,
  • Shotgun Sequencing,
  • Hybrid Approach.




  • Please read…

Science 291: 1304-1315