Amount of Food Protein and the Accumulation of Body Protein in Fish:IV. Influence of Some Feeding Conditions on the Growth of Fish
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1. The rate of growth of fish can be taken for constant till increase in body weight amounts to 60% of the initial, even when the fish is kept daily on the same ration. 2. Divide the whole feeding period during which fish grows with constant velocity in n equal intervals, and let the increase of body protein for unit initial body weight in the first interval be p1 and that in the whole period be p and the increase of body protein, in an arbitrary i-th interval for unit body weight at the beginning of this interval be p1. Then we have the following relations. 1/Pi=1/p1+(i-1)k (1) in which k is a constant which denotes the ratio of body protein to body weight, and p1 is expressed as already proved*, in the following formula: p1=e-a1-b1x(x-x0) or p1=A1e-b1x(x-x0), where x is the feeding amount of protein in the first interval and x0 is that equivalent to maintenance protein. p=np1=nA1e-b1x(x-x0) (2) If during the whole period fish grows with constant velocity even on the same daily ration, we can rewrite the formula (2) into "the equation of accumulation of body protein" which represents the relation between the amount of food protein and the accumulation of body protein. p=A1e-b1/n(nx)(nx-nx0) (3) 3. Let the equation of accumulation of body protein when a given amount of food protein is divided equally to each day and administered in a certain unit period be p1=Ae-blx(x-x0). Now suppose that the same amount of protein is given in a period of n units, and that the rate of growth of fish remains still unchanged, then the accumulation of body protein pn will be pn=A1e-b1/nx, /suo(x-nx0) (4) The value of pn is maximum when n=1/2x0(√b12x02+4b1x0-b1x0)x. Denote this value of n by n0, then n0 is proportional to the total amount of food protein x. Therefore, under the above conditions the daily amount of food protein which gives the maximum growth is independent of x. Putting the value of n0 into (4), we have pn0=A1e-b1x0/B(1-B)x=Kx, where B=1/2(√b12x02+4b1x0-b1x0). It is seen that if food protein, of which total amount is given, is administered in the optimum period, accumulation of body protein is proportional to the given amount of food protein. 4. Suppose that under the conditions that the rate of growth of fish is constant even with the same daily ration, both the total amount of food protein, 2α, and the feeding period are given. If the given period is divided equally and the feeding amounts of protein in both intervals are a-ε and a+ε respectively, the total increase of body protein p will be, p=Ae-b(a-α)(a-α-x0)+Ae-b(a+α)(a+α-x0), where x0 is the feeding amount of, protein equivalent to the maintenance protein in the half period.
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