Answers to Hands on 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 35^oC.

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.

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.

Answers to Hands on 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.

1. The marked woodlice can mix randomly in the population before the second sample is taken.
2. Marked and unmarked woodlice in second sample are caught randomly
3. The substance used to mark the woodlice should not be poisonous or affect the activity of the woodlice and yet not easily removed.
4. The death rate and the birth rate are the same.
5. The population to be estimated is stable/ the rate of the migration is equal to the rate of emigration of the woodlice.
6. There are no predators of woodlice in the habitat to be studied. (any 2)

4. Increasing the number of catches//repeat the experiment
Conclusion
The estimated population size of woodlice is 75.

Answers to Hands on 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 m<sup>2

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.</sup>

Answers to Hands on 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 Hands on Pg 76-79

7.1 : Anaerobic respiration in yeast

Aim : To investigate anaerobic respiration in yeast
Hypothesis : Anaerobic respiration by yeast produces carbon dioxide and ethanol
a) MV : Presence or absence of yeast
b) RV : Change in colour of bicarbonate indicator and smell of ethanol
c) FV : Temperature, volume of solution
Discussion :

1. To remove oxygen
2. To pevent oxygen from dissolving in the solution
3. Carbon dioxide is produced and it will turn limewater cloudy
4. Yes. Ethanol is produced and can be detected through the smell of alcohol.

Conclusion : The hypothesis is accepted. Anaerobic respiration by yeast produces carbon dioxide and ethanol.

7.2 : Effects of vigorous exercising on the rates of respiration and heartbeat

Aim : To study the effects of vigorous exercising on the rates of respiration and heartbeat
Hypothesis : The more vigorous the exercise, the higher the rates of respiration and heartbeat.
a) MV : At rest or vigorous activity
b) RV : Rates of respiration and heartbeat
c) FV : Temperature, student

Condition	Rate of respiration (counts per min)				Rate of respiration (counts per min)
	1	2	3	Mean	1	2	3	Mean
At rest	14	15	14	14	72	71	72	72
After vigorous exercise	27	27	28	27	88	86	88	87

Discussion

1. The rate of respiration increases when the exercise is more vigorous.
2. Vigorous exercise actually increases the rate of heartbeat.
3. An increase in the rate of respiration and the rate of heartbeat will increase the oxygen content in the body. Cell respiration increases due to an increase in oxygen content. This results in an increase in carbon dioxide content in the body as more carbon dioxide is released through cell respiration.

Conclusion

The hypothesis is accepted. The more vigorous the exercise, the higher the rates of respiration and heartbeat