The average metabolic rate O2 consumption in Lkg times h for
Solution
the Q10 is given by the formula - Q10 = (R2/R1) ^ (10/T2-T1)
we have to calculate Q10 for fishes that were kept at 10 and 28 degrees that are normally mid range - 10 to 20
therefore T2 becomes 10 and 28
T1 becomes 10 and 20
selecting the average rate values from the table, we have 4 values :
answer a : Q10 = 2.2 ^ (10 / 10- 10) = 2.2^ (10/0) = it means result is undefined because anything divided by zero is undefined.
Q10 = 3.1 ^ ( 10 / 10 - 20) = 3.1 ^(10/10) = 3.1
Q10 = 1.5 ^ ( 10 / 28 - 10) = 1.5 ^ (10/18) = 1.25
Q10 = 1.6 ^ ( 10/ 28 - 20) = 1.6 ^ ( 10/8) = 1.80
the group of fishes that shows a higher Q10 means that their rate increases by a higher factore for every 10 degrees temperature raised,
since q10 for fishes at 20 degrees was higher, therefore they were more sensitive to the change in temperature.
answer b : the animals that were kept at 28 degrees from 10, 20 and 30 degrees respectively, have their average metabolic rates as 1.5, 1.6 and 2.3. therefore the highest metabolic rate was seen in the 30 degree organism, which means that they show complete acclimation.
