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**To**:**BSE-L****Subject**:**Calculating infectivity of BSE****Subject**:**Re: BSE/22D****From**:**BSE-L; ral****From**:**BSE-L; ral**- Date: Thursday, April 24, 1997 3:53AM

At 09:47 AM 4/23/97 -0400, Antonio wrote: >I reached the 22D value on the following basis: > > "The contamination level of cow brain can be expected > to be about ten to eleven (10^11) prions / g [Brown et al., 1982]. > The nervous system is about 1/1000th of the body weight. > Then we may consider the maximum level of infected beef > contamination to be about ten to eight (10^8) prions/g, > irrespective of the tissue type. One hundred grams (a portion) > of beef will hold about 10^10 (ten to ten)prions. Then the > minimum number of D required to sanitize 100g beef would > be ten D (10D). The total number of D rises to 10+12=22D > to meet the usual basic requirement of a low survival > probability 10^ -12 (ten to minus twelve = the minimum > botulinum cook) and to assure an acceptable remote > probability of infection by BSE prions. > > Antonio This argument is flawed. Assuming, as you do, that the infectivity can be presumed to be a maximum of 10^11 prions/g in the worst-case (brain), and that the central nervous system is the only infectious tissue of consequence, then the "whole animal" infectivity could be expected to be 10^8 prions/g. Assuming, as you do, that ground beef exhibits the average carcass prion rate, then a 100 g portion would contain 10^10 prions. Also assuming, as [110 lines left ... full text available at <url:http://www.reference.com/cgi-bin/pn/go?choice=message&table=04_1997&mid=3819570&hilit=BRAIN+BRAINS+FEEDBACK> ] -------------------------------- Article-ID: 04_1997&3826312 Score: 78 Subject: Re: DVORAK typing

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