69 
Slide 5 
estihated probabilities 3 
Host 
Vector 
Survi va 1 b 
T ransmi ss i b i 1 
Ho selective 
‘ :y c disadvantage** 
Exhibit adverse 
consequences 41 
Robust 
E_. coli K* 12 
R100 
I0-I-1Q-2 
10*8 
10-2 
10-8 
Robust 
E_. col i K- 12 
pSCIOI 
o 
o 
IO-'9 
I0“ 2 
10*8 
E K2 l. 
:ol i K- 1 2 
pSCIOl 
<I0" 8 
<10-23 
IQ -2 
I0"8 
a Asscf*e shotgun' exper ->ent with -".anna li an c Per surviving escaped bacterium per 
DMA. day. 
b Per escaped bacteria per day. d Per cloned DNA segment. 
conservative guesses when no data are available. Estimates on probabili- 
ties of escape from various physical containment systems are left for Dr. 
Barkley to discuss this afternoon. 
In the first example, with a robust E^ coli K12 host, with a conjuga- 
tive R100 plasmid vector, one to ten percent of the escaped bacteria will 
survive each day, and one out of one hundred million of these surviving 
escaped bacteria should be able to transmit the recombinant DNA to some 
other bacterium in nature. These values assume that the escape occurs, 
either by accidental ingestion or by disposal down the drain. 
Most foreign DNA causes a microorganism that possesses it to be at a 
distinct selective disadvantage in competing with other bacteria. I thus 
consider that in only one percent of the cases will the foreign DNA be 
either beneficial to and/or no burden on the bacterial host. 
In estimating the probability for exhibiting adverse consequences, I 
have considered that most eukaryotic DNA sequences will not be expressed 
in bacteria, and that only one out of 10,000 of the one million mammalian 
genes present in the mammalian genome, if expressed, would be harmful to 
some organism in the biosphere. Thus, only one out of one million surviving 
cloned DNA segments might actually display detrimental properties. 
[210] 
