EXAMINATION OF WATER FOR MICRO-ORGANISMS 73 



the third day six, on the fourth day four, and on the 

 fifth day two. Taking a concrete example with wort r 

 suppose No. 1 flask (10 c.c. wort -4- 1 c.c. water) be- 

 comes turbid on the second day, the product is (1 x 8); 

 No. 2 flask (10 c.c. wort + '75 c.c. water) becomes- 

 turbid on the third day, the product is (2 x 6) ; No. 3 

 flask (10 c.c. wort + '5 c.c. water) also becomes turbid 

 on the third day, the product is (3 x 6) ; No. 4 flask 

 (10 c.c. wort + '25 c.c. water) becomes turbid on the 

 fourth day, the product is (4 x 4). Then adding these 

 products together, we obtain 



(1 x 8) + (2 x 6) + (3 x 6) + (4 x 4) 

 8 + 12 + 18 + 16 = 54 



as the numerical expression for the energy with which 

 the particular water is capable of producing growths in 

 wort. 



In the case of beer the above factors are multiplied 

 by f , so that they are respectively 



10 x f = 17 ; 8 x f = 13-3 ; 6 x f = 10 ; 4 x f = 6-7 ; 

 2 x |= 3-3; 



and in the concrete example given above, if the results 

 had been obtained with beer instead of wort, the nume- 

 rical expression for the energy of growth possessed by 

 the water in respect of beer would be 



(1 x 13-3) + (2 x 10) + (3 x 10) -f (4 x 6-7) 



13-3 + 20 + 30 + 26-8 = 90-1. 



NUMERICAL DETERMINATION OF BACTERIA ix WATER 

 (MIQUEL'S METHOD) 



In a previous chapter (see p. 28) an account is- 

 given of the method of isolating particular micro- 

 organisms from any given material by means of the 

 dilution method. In Miquel's process of water exami- 



