QUANTITATIVE MICROSCOPIC DETERMINATIONS 143 



which might be confused with bacteria; such as milks, frozen eggs, 

 sausage meats, pastes of all kinds, etc. Proceed as follows: Deliver 

 J-f o cc - f the properly sampled, mixed, and diluted (if necessary) por- 

 tion of the substance to be examined upon a well cleaned (by means of 

 alcohol) slide and spread this volume (namely J/fo cc -) evenly and 

 uniformly over an area of 10 sq. cm. (2 cm. by 5 cm. = 10 sq. cm.), 

 let it air dry upon a leveling table (drying may be hastened somewhat 

 by using moderate heat), add a drop or two of alcohol and again dry; 

 stain for from one to three minutes in a few drops of a 10 per cent, 

 solution of methyl-blue (or Hoffman's violet), wash gently in water, 

 and if desired decolorize somewhat by means of acidulated alcohol, 

 and again wash in water. Set aside until air dry and make the counts 

 by means of the oil immersion objective. The counting should be 

 done carefully, selecting ten fields each from five or more different 

 areas of the mount, and determine the average of the entire count. 

 The computation is simple. We will suppose that the substance under 

 examination was a highly contaminated gelatine. The field of view 

 of the oil immersion was HQ sq. mm., the dilution used was 1 : 100, 

 and the average number of bacteria per field was found to be 30, then 

 the total number of bacteria per gram of the gelatine would be 1,500,- 

 000,000 (50 X 30 X 100 X 1,000 X 10 = 1,500,000,000). The prob- 

 lem may be stated more in detail as follows : If there are 30 bacteria 

 in y$Q of 1 sq. mm., then there would be 50 times thirty bacteria in 

 1 sq. mm., or 1,500 bacteria. Since the dilution employed in this 

 case was 1-100, two ciphers must be added, making 150,000 bacteria 

 per 1 sq. mm. of the original substance. In the entire slide mount 

 there are 1,000 sq. mm., hence representing 150,000,000 bacteria 

 when figured back to the original undiluted state. Since the entire 

 amount of the substance upon the slide was Jfo cc. (or 3/fo gram), we 

 must multiply by ten in order to reduce to 1 cc. or 1 gram. The 

 special advantage of the staining method lies in the fact that bacteria, 

 as a rule, stain differently from other minute organic particles. Small 

 crystals which might be mistaken for bacteria do not take the stain 

 at all. The instructor in charge must assist the student in the use of 

 the hemacytometer. 



The area upon which the material is to be spread (on the slide) 

 is indicated as follows: Upon a piece of white card board mark the 

 area above indicated by means of ruler, pen and India ink, and lay the 

 slide over this ruled card when spreading the material. 



Substances deficient in albuminous matter for fixing purposes (on 

 the thoroughly cleaned slide) should have added to them a small 

 quantity of egg albumin solution. The pipette should be clean when 



