(i) (e) 5
Plant in tube 5 is kept in dark; hence, no photosynthesis can take place in this plant. Therefore, it will show least amount of starch.
(ii) (a) 1
Plant in tube 1 is kept in light; hence, it will perform photosynthesis and produce oxygen. No snail is present there to consume oxygen in this tube along with the plant. Thus, maximum oxygen will be present in it.
(iii) (a) 1
No snail is present in tube 1 to respire and give out carbon dioxide. Also, plant is in light; therefore, it will perform photosynthesis and utilise the available carbon dioxide. Hence, tube 1 will have least carbon dioxide.
(iv) (e) 5
Plant kept in tube 5 is in dark; so, no photosynthesis can take place. Thus, no food will be made and plant will die soon.