BOTANY: OSTERHOUT AND HAAS 
89 
The agreement between the observed and the calculated values is very sat- 
isfactory except at the start. In this connection it may be pointed out 
that at the beginning of a reaction disturbances are to be expected. 
It is therefore evident that the assumption justifies itself by giving an ade- 
quate quantitative explanation of the observed results. The question then 
arises whether it is a natural one. It would seem very probable that the 
light produces a substance which accelerates the reaction and unless this 
substance is produced in unlimited amount there must come a time when the 
rate will become steady (or fall off). The assumption therefore seems to be 
reasonable. 
It is attractive to form a hypothesis as to the nature of the catalyst. One 
might be tempted to suppose that it is chlorophyll but for the fact that some 
plants which are deep green may not photosynthesize as rapidly as those 
which possess less chlorophyll. 11 It is of course possible that the less active 
plants are deficient in some essential factor other than chlorophyll. On the 
other hand it may be necessary for chlorophyll to be transformed by the 
light from an inactive into an active form, 12 so that the rate of photosynthesis 
depends on the amount of 'active chlorophyll' present This would be anal- 
ogous to the well known activation of enzymes by various means. 
An equally satisfactory quantitative explanation is obtained if we suppose 
the amount of photosynthesis to correspond to the amount of a substance 
P, produced (under the influence of light) by the reaction 
S^M-^P, 
in which S represents a constant source, (i.e., a substance which does not 
appreciably diminish during the experiment). 
We may suppose that in the morning, before the frond is exposed to the 
light, S alone is present. On exposure to light the formation of M and P 
occurs. The amount of M will then increase until it reaches a constant value 
(when its rate of formation is equal to its rate of decomposition) but the 
value of P will continually increase, since it does not undergo decomposition. 
When M has reached a constant value we find (putting K as the velocity con- 
stant of the reaction M —*P) that the amount of M decomposed in 1 minute 
(unit time) is KM; this is also the amount of P which is formed in 1 minute 
and since the reaction S— >M produces just enough of M to balance the 
loss of M (by transformation into P) the amount of M produced each minute 
is KM. Hence if we start in the morning with S alone there will be pro- 
duced each minute KM and all of this will be transformed into P except what 
is present at any moment as M. Hence the amount of P produced in the 
time T is KMT - M. 
We may, for convenience, put M — 1 when it has attained its constant 
value; the rate of increase of P is then constant and we find that it takes 20.4 
minutes to produce 1 unit of photosynthesis. Hence KMT = 1. Substi- 
