Haemol. 
100% 
90% 
80% 
70% 
60% 
50% 
40% 
30% 
20% 
10% 
286 
041% 0.43% 045% 047% 049% 051% 053% 055% 057% sn 
so ' 
mmm Ist Bleeding 26-I||-1912, ------- 2nd Bleeding 1-IV-1912, m.n. 3rd Bleeding 3-IV-1912, 
Fig. 2. graphical representation of Table B. 
TABLE B, denoting the haemolysis caused by each NaCl-concentration in the 
normal and the anaemic animal. 
eee ereen eene neee eere ae en 
| Ist Venesection | 2nd Venesection | 
3rd Venesection 
At 0.57 pCt. NaCl 
” 
0.55 
0.53 
0.51 
0.49 
0.47 
0.45 
0.43 
0.41 
” 
” 
12 pCt. 
” 
12 pCt. 
” 
„ 
4. New erythrocytes are built up from old ones. 
uoIjnjos 
ayy Ul punoy sem uogojowoeH 
The values and curves of fig. 2 may also be viewed in another 
way. On the first day, for instance, a solution of 0.57 °/, NaCl causes 
12 °/, haemolysis, whilst at 0.55 °/, NaCl 18°/, of the erythrocytes 
had disappeared. Hence there were then 18 °/, — 12 °/, = 6 °/, blood- 
corpuscles which were just unable to withstand a solution of 0.55 °/, 
NaCl. For 6°/, of the blood corpuscles 0.55 °/, NaCl is just the 
minimum concentration they can bear. 
