Number of genome copies in a gram of soil

prokaryotic cells per gram of soil: <= 1e10
- bionumbers
genome copies per prokaryotic cell:

Number of genome copies in X ug of soil DNA

mean DNA conc. in a cell (Fg): 3-18
- bionumbers



In [16]:

    
gradientDNA_low = 5e-6 
gradientDNA_high = 8e-6 
cellDNA_low = 3e-15 
cellDNA_high = 18e-15 

print "Number of genome copies in the gradient: "
print '  High: {0:.2e}'.format(gradientDNA_high / cellDNA_low)
print '  Low: {0:.2e}'.format(gradientDNA_low / cellDNA_high)









    



Number of genome copies: 
  High: 2.67e+09
  Low: 2.78e+08

Number of bacterial 16S rRNA copies in X ug of soil DNA

16S rRNA copies per genome: 1-15
- mean copy #: 4.2
- http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0057923



In [17]:

    
mean_16S_copy = 4.2

print "Number of 16S rRNA copies in the gradient: "
print '  High: {0:.2e}'.format(gradientDNA_high / cellDNA_low * mean_16S_copy)
print '  Low: {0:.2e}'.format(gradientDNA_low / cellDNA_high * mean_16S_copy)









    



Number of 16S rRNA copies in the gradient: 
  High: 1.12e+10
  Low: 1.17e+09

Notes:

Will need to simulate >= 1 billion DNA fragments containing a 16S rRNA gene
>=10 billion DNA fragments containing a 16S rRNA gene could be present in one gradient

Abundance cutoff for a large BD distribution due to diffusion

Based on E. coli simulations, an abs abundance of 1e8 will produce a fragment BD distribution spanning most of the gradient



In [20]:

    
total_abund_high = 1.12e10
total_abund_low = 1.17e9

print 'Relative abundance where taxon should be observed across most of a gradient: '
print '  High: {0:.2f}%'.format(1e8 / total_abund_high * 100)
print '  Low: {0:.2f}%'.format(1e8 / total_abund_low * 100)









    



Relative abundance where taxon should be observed across most of a gradient: 
  High: 0.89%
  Low: 8.55%

Notes:

Numbers seem reasonable based on Ashley's data.
- Ashley's data indicates high end is more likely the case for her dataset.



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