Table 2.
Same legend as for Table 1, except that the experiment was made on two colonies of M. rubra and two colonies of M. sabuleti.
Days, food delivery times, and counting times (t1–t5) | M. rubra colonies | Mean | M. sabuleti colonies | Mean | ||
---|---|---|---|---|---|---|
A | B | A | B | |||
Control 1, no food | ||||||
t1 | 10 | 5 | 7.5 | 2 | 0 | 1.0 |
t2 = t1 + 20′ | 8 | 5 | 6.5 | 4 | 2 | 3.0 |
t3 = t1 + 40′ | 8 | 0 | 4.0 | 2 | 2 | 2.0 |
t4 = t1 + 60′ | 10 | 0 | 5.0 | 2 | 2 | 2.0 |
t5 = t1 + 80′ | 20 | 0 | 10.0 | 3 | 3 | 3.0 |
| ||||||
Control 2, food present | ||||||
t1 | 26 | 10 | 18.0 | 10 | 5 | 7.5 |
t2 = t1 + 20′ | 20 | 10 | 15.0 | 10 | 5 | 7.5 |
t3 = t1 + 40′ | 20 | 10 | 15.0 | 10 | 10 | 10.0 |
t4 = t1 + 60′ | 20 | 10 | 15.0 | 10 | 5 | 7.5 |
t5 = t1 + 80′ | 20 | 10 | 15.0 | 0 | 10 | 5.0 |
| ||||||
Day 1, food given at t1 and retrieved at the end of t1 | ||||||
t1 | 17 | 6 | 11.5 | 15 | 5 | 10.0 |
t2 = t1 + 20′ | 30 | 6 | 18.0 | 10 | 5 | 7.5 |
t3 = t1 + 40′ | 18 | 3 | 10.5 | 7 | 12 | 9.5 |
t4 = t1 + 60′ | 10 | 3 | 6.5 | 8 | 0 | 4.0 |
t5 = t1 + 80′ | 10 | 0 | 5.0 | 0 | 10 | 5.0 |
| ||||||
Day 2, food given at t2 and retrieved at the end of t2 | ||||||
t1 | 9 | 0 | 4.5 | 5 | 6 | 6.5 |
t2 = t1 + 20′ | 24 | 16 | 20.0 | 21 | 15 | 18.0 |
t3 = t1 + 40′ | 11 | 0 | 10.5 | 8 | 8 | 8.0 |
t4 = t1 + 60′ | 10 | 0 | 5.0 | 5 | 5 | 5.0 |
t5 = t1 + 80′ | 11 | 8 | 9.5 | 0 | 8 | 4.0 |
| ||||||
Day 3, food given at t3 and retrieved at the end of t3 | ||||||
t1 | 10 | 10 | 10.0 | 0 | 10 | 5.0 |
t2 = t1 + 20′ | 3 | 11 | 7.0 | 7 | 4 | 5.5 |
t3 = t1 + 40′ | 33 | 14 | 23.5 | 24 | 20 | 22.0 |
t4 = t1 + 60′ | 15 | 13 | 14.0 | 3 | 3 | 3.0 |
t5 = t1 + 80′ | 10 | 2 | 6.0 | 3 | 2 | 2.5 |
| ||||||
Day 4, food given at t4 and retrieved at the end of t4 | ||||||
t1 | 18 | 0 | 9.0 | 4 | 3 | 3.5 |
t2 = t1 + 20′ | 4 | 10 | 7.0 | 5 | 3 | 4.0 |
t3 = t1 + 40′ | 9 | 6 | 7.5 | 8 | 8 | 8.0 |
t4 = t1 + 60′ | 30 | 22 | 26.0 | 22 | 34 | 28.0 |
t5 = t1 + 80′ | 10 | 10 | 10.0 | 10 | 14 | 12.0 |
| ||||||
Day 5, food given at t5 and retrieved at the end of t5 | ||||||
t1 | 9 | 0 | 4.5 | 7 | 2 | 4.5 |
t2 = t1 + 20′ | 4 | 4 | 4.0 | 5 | 0 | 2.5 |
t3 = t1 + 40′ | 7 | 4 | 5.5 | 6 | 4 | 5.0 |
t4 = t1 + 60′ | 8 | 6 | 7.0 | 8 | 8 | 8.0 |
t5 = t1 + 80′ | 34 | 20 | 27.0 | 34 | 28 | 31.0 |