Answers to Pg 87-89
8.1 Intraspecific competition and interspecific competition
Aim : To sudy the intraspecific competition and interspecific competition in plants
Hypothesis : The greater the intraspecific and interspecific competition, the shorter the plant.
a) MV : Type of seedlings
b) RV : Heights of seedlings
c) FV : Amount and type of soil//amount of water//light intensity//amount of seedling//distance between seedlings
Results:
Seedling tray
|
Height of plants (cm)
| ||||||||||
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
Mean
| |
A (Paddy)
|
25.1
|
24.2
|
26.3
|
24.0
|
23.8
|
25.4
|
25.8
|
26.1
|
25.7
|
26.9
|
25.33
|
B (Maize)
|
30.8
|
32.3
|
34.1
|
33.4
|
34.0
|
32.8
|
32.6
|
33.0
|
34.1
|
32.1
|
32.92
|
C (Paddy)
|
20.2
|
20.3
|
19.8
|
19.1
|
18.2
|
16.1
|
18.4
|
16.8
|
17.0
|
19.4
|
18.52
|
C (Maize)
|
38.4
|
40.1
|
40.5
|
38.8
|
38.8
|
39.4
|
41.2
|
40.8
|
41.1
|
39.7
|
39.88
|
**Please draw bar chart**
Discussion
1. A : Intraspecific competition
B : Intraspecific competition
C : Interspecific competition
2. The paddy plants in A are taller than those in C because interspecific competition in C causes the paddy plants to lose to maize plants. The shorter paddy plants do not absorb enough sunlight and their root systems are not extensive enough to absorb enough water.
3. Dry mass
Conclusion
Hypothesis is accepted. The greater the intraspecific and interspecific competition, the shorter the plant.
Answers to Pg 89-91
8.2 : Quadrat sampling Technique
Aim : To determine the percentage cover, frequency and density of Mimosa pudica using the Quadrat sampling technique
Results
Quadrat
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
Covered area (m2)
|
0.2
|
0.3
|
0
|
0
|
0.4
|
0.3
|
0.3
|
0.4
|
0.2
|
0.3
|
Number if individuals
|
5
|
7
|
0
|
0
|
8
|
6
|
5
|
10
|
5
|
4
|
Percentage cover = 2.4/10 x 100 = 24%
Frequency = 8/10 x 100 = 80 %
Density = 50/10 = 5 inviduals per m2
Discussion
1. Easy to use and suitable for studying the distribution of plants.
2. Increase the number of quadrats.
3. It is difficult to count the number of plants as some plants are creepers and they tend to grow overlapping each other.
Conclusion
The estimated percentage cover of Mimosa pudica is 24%. The estimated frequency and density is 80% and 5 plants per m2 respectively.
Amswers to Pg 91-93
8.3 Capture, mark, release and recapture method
Aim : To estimate the population size of woodlice using the capture, mark, release and recapture method
Results
Number of woodlice in first capture
|
30
|
Number of woodlice in second capture
|
25
|
Number of marked woodlice in second capture
|
10
|
Estimated population size of woodlice
|
30 x 25 = 75
10
|
Discussion
1. To prevent the ink from harming the woodlice and from being washed off.
2. a) To allow the marked woodlice to mix freely with the rest of the woodlice.
b) Because the population may change due to emigration, death or birth of woodlice over a long period of time.
3.
- The marked woodlice can mix randomly in the population before the second sample is taken.
- Marked and unmarked woodlice in second sample are caught randomly
- The substance used to mark the woodlice should not be poisonous or affect the activity of the woodlice and yet not easily removed.
- The death rate and the birth rate are the same.
- The population to be estimated is stable/ the rate of the migration is equal to the rate of emigration of the woodlice.
- There are no predators of woodlice in the habitat to be studied. (any 2)
Conclusion
The estimated population size of woodlice is 75.
Answers to Pg 95-100
8.4 : Effect of temperature, pH, light intensity and nutrients on the activity of yeast.
Results
Boiling tube
|
Temperature (oC)
|
Condition of lime water
|
A
|
0
|
Clear
|
B
|
20
|
Slightly cloudy
|
C
|
35
|
cloudy
|
D
|
45
|
Slightly cloudy
|
E
|
65
|
clear
|
Discussion
1. There is no activity in A as the temperature is too low for enzymes in the yeast to function whereas enzymes in the yeast in E are denatured as the temperature is too high.
2. When the yeast cells are more active, more carbon dioxide is produced. This causes the lime water to be more cloudy.
3. A lot of foam is produced.
Conclusion
Hypothesis is accepted. The activity of yeast is highest at an optimum temperature of 35oC.
Pg 97
Boiling tube
|
pH
|
Condition of lime water
|
A
|
5
|
cloudy
|
B
|
7
|
Slightly cloudy
|
C
|
9
|
Clear
|
Discussion
1. The activity of yeast is very high in A because yeast works best in an acidic condition wheres the yeast cells were not active in C as it is alkaline.
2. The lime water turns cloudy faster than before because the medium for the reaction of yeast is now acidic.
Conclusion
Hypothesis is accepted. The activity of yeast is highest in an acidic condition of pH of 5.
Pg 98-99
Boiling tube
|
Distance of light source (cm)
|
Condition of lime water
|
A
|
20
|
Clear
|
B
|
30
|
Slightly cloudy
|
C
|
40
|
cloudy
|
D
|
50
|
Very cloudy
|
Discussion
1. When the distance of the light source from the boiling tube increases, the lime water is more cloudy.
2. When the distance of the light source increases, the light intensity decreases.
Conclusion
Hypothesis is accepted. The lower the light intensity, the higher the activity of yeast.
Answers to pg Pg 99-100
Boiling tube
|
Distance of light source (cm)
|
Condition of lime water
|
A
|
20
|
Slightly cloudy
|
B
|
30
|
cloudy
|
C
|
40
|
Very cloudy
|
D
|
0 (distilled water)
|
Clear
|
Discussion
1. In C, there is a lot of foam whereas there is no foaming in D.
2. There is no activity in D as there is no nutrient. Activity of yeast increases when the concentration of glucose solution increases from A to C.
Conclusion
Hypothesis is accepted. The higher the concentration of nutrient, the higher the activity of yeast.
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