How living in groups helps animals survive

Grade 9 Science Worksheet

Circle the best answer.

1. Read this first

   Group size and visits (model data)

   group size	visits a day
   4	1
   16	2
   64	4

2. 1.

   Luz reads the model table. Each visit takes one animal, picked at random. Which statement about one animal's daily chance of being lost is right?

   • The group of 64 loses 4 animals a day against 1, so its animals are worse off
   • A group 16 times larger meets 4 times the visits, so each animal's chance rises 4 times
   • Each animal's chance falls 16 times, since the group is 16 times larger
   • A group 16 times larger meets 4 times the visits, so each animal's chance falls 4 times

3. Ducks at day 10 (model data)

   group	ducks at start	ducks lost by day 10
   in flocks	30	3
   alone	60	24

4. 2.

   Milo reads the model data. Which statement compares how likely a duck was to be lost by day 10?

   • Ducks alone were 30 times as likely to be lost: 40% − 10%
   • Ducks alone were 8 times as likely to be lost: 24 ducks were lost alone against just 3 in flocks
   • Ducks alone were 4 times as likely to be lost: 40% (24 of 60) against 10% (3 of 30)
   • Ducks in flocks were 4 times as likely to be lost: 10% against 40%

5. Group size, risk and feeding (model data)

   group size	chance of being lost in a day (%)	share of the day spent feeding (%)
   2	8	70
   8	3	62
   32	2	40

6. 3.

   Noah reads the model data on one species. Which group size gives the best balance of risk and feeding?

   • Size 8, because the risk and the feeding time are equal there
   • Size 8: risk falls 5 points for 8 of feeding; size 32 gains 1 point of safety for 22 of feeding
   • Size 2: it has the most feeding time, so it is the pick
   • Size 32: it has the lowest chance of being lost, so more safety makes the better balance of the two

7. Two dances for one patch (the sun moves 15° clockwise each hour) (model data)

   dance	time	sun's compass bearing (°)	angle clockwise from vertical (°)
   dance A	12:00	180	40
   dance B	15:00	225	?

8. 4.

   Pia uses the model rule: bearing to the patch = sun's bearing + angle. Dance B is for the same patch. What angle should dance B show?

   • 40°, the same as dance A, since the patch has not moved
   • 5° counterclockwise from vertical, since 220 − 225 = −5
   • 45°, the amount the sun moved
   • 5° clockwise from vertical, since 225 − 220 = 5

9. Two waggle dances (model data)

   dance	waggle run (s)	angle clockwise from vertical (°)
   dance 1	2.0	40
   dance 2	2.0	130

10. 5.

    Ren reads two dances from one hive at the same time, using 1 s of run = 400 m. Which statement does the card support?

    • The bees found different resources, since the angles differ
    • Both patches are 800 m away, but in directions 90° apart
    • Patch 2 is farther, since its angle is larger
    • The two patches are the same place, since the runs match

11. Five dances for one patch (model data)

    dance	waggle run (s)
    1	2.3
    2	2.4
    3	2.5
    4	2.6
    5	2.7

12. 6.

    Zee reads five dances for one patch, using 1 s of run = 400 m. Which statement about the patch is best supported?

    • Likely about 1,000 m away, within about 80 m either side
    • Between 2.3 m and 2.7 m away
    • Exactly 1,000 m away, since the middle run is 2.5 s
    • About 920 m away, because dance 1 settles it

13. Two birds (model data)

    bird	air around it (°C)	gap from 38 °C (°C)
    alone	−17	?
    in a huddle pocket	20	?

14. 7.

    Ari uses the model rule: heat lost is proportional to the gap from a bird's 38 °C. How does a bird in a 20 °C pocket compare with one alone in −17 °C air?

    • About one third as much heat lost: a gap of 18 °C against 55 °C
    • About half as much heat lost: 20 °C against 38 °C
    • No heat lost, since 20 °C is above 0 °C
    • About 3 times as much heat lost: 55 °C against 18 °C

15. Use the table

    The huddling study (data: Gilbert et al. 2006)

    detail	entry
    birds fitted with data loggers	8
    what the loggers recorded	temperature around each bird
    colonies in the study	one

16. 8.

    Bo reads the study details. What can the study of 8 birds support?

    • It shows nothing, since any study of 8 birds is useless
    • It shows huddling raised the birds' body temperature to 37.5 °C
    • It shows every emperor penguin colony huddles for 1.6 h at a time, since the 8 birds are enough to speak for them
    • It shows what happened to these 8 birds; it cannot fix a value for every colony

17. Study card: minnows (model data)

    part of the study	entry
    fish in groups of 10	40 fish in a shaded tank
    fish alone	40 fish in a bright tank
    measured	fish surviving at day 10

18. 9.

    Esme reads the study card. Which flaw most weakens the claim that groups raise survival?

    • Light differed between the tanks too, so group living is not the only change
    • Fish alone act as the control group, which makes the whole result invalid for a comparison
    • Survival at day 10 cannot be used, so the study fails
    • Forty fish is always too few, so the study means nothing

19. Pond survey (model data)

    pool	group size	share surviving (%)	shelter plants (%)
    pool A	4	55	5
    pool B	12	75	40
    pool C	30	88	80

20. 10.

    Kira reads a pond survey. Does the table show that bigger groups cause better survival?

    • Yes: survival rises at each step in group size, so bigger groups cause better survival
    • No: a survey cannot be used as evidence
    • No: shelter rises with group size, so shelter may explain the survival
    • Yes: 88% in pool C is the highest, so groups of 30 work well