Capture recapture method

CAPTURE, MARK, RELEASE AND RECAPTURE METHOD

Estimating the number of beans in an opaque bag

Material : Kidney beans, opaque bag, specimen tube, marker pen

Method

1.     About a beakerful of kidney beans is put in an opaque bag. (The kidney beans represent the population of animals.)

2.     An empty specimen tube is put into the bag filled with kidney beans and then taken out.  The kidney beans are placed in a petri dish and counted. (This represents the first capture).

3.     The beans are marked with a marker pen and the returned to the bag.

4.     The marked and the unmarked beans are mixed thoroughly by hand.

5.     An empty specimen tube is put into the bag again, filled with beans and then taken out.  One must not look into the bag when carrying out this step.  (This represent random recapturing).

6.     The kidney beans that have been scooped out are placed in a petri dish and counted. The number of beans that are found marked is noted. (This represent the number of animals found marked in the recapture).

                                                                                    

Calculation

The number of beans in the first scoop	X
The number of beans in the second scoop	Y
The number of beans found marked in the second scoop	Z
Therefore, an estimation of the number of beans in the bag	X x Y Z

                                                                                                       Z

Note: The accuracy of the method can be increased by

a)     taking out 2 scoops of beans each time instead of one

b)    repeating the experiment and finding the average of the results

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Discussion

1. The material used to mark the organism is important. Why?

    The substance used to mark the organism should not expose the

    organism to its predator, should not be poisonous, should not affect the

    activity of the organism, and yet not easily removed     

2. What is the estimated population size of beans in the bag?           

3. What is the actual population size of the beans?

4. Why is the estimated result different from the actual result?

     Estimation only whereas the second method is direct counting.

5. Why is direct counting difficult in real life situation?

     The organisms/animals move around, space/habitat is too big, you

     cannot capture them all, unlike the beans which are all in the bag.    

6. How can the accuracy of the data be improved?

·  Repeat the experiment and find the average of the results.

·   Mark a greater number of individuals.

7. State two assumptions made in this estimation method?

b.     The marked individuals can mix randomly in the population before the second sample is taken

c.      Marked and unmarked organisms in second sample are caught randomly

d.     The substance used to mark the organism should not be poisonous or affect the activity of the organism and yet not easily removed.

e.     The death rate and the birth rate are the same.

f.       The population to be estimated is stable/ the rate of the migration is equal to the rate of emigration of the organism. 

g.     There are no predators of the organism in the habitat to be studied.

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Question

1. A biologist originally marked 40 butterflies in Taman Bukit Cerakah, Shah Alam using tags. The marked butterflies were released back to where they were caught. After a week, the­ butterfly traps caught 200 butterflies. Of those 200, 80 were found to have tags. Based on this information, what is the estimated population size of the butterflies in Taman Bukit Cerakah?

2. A study of the garden snail population in a vegetable farm was carried out.  On the first trip, a total of 160 snail were caught, marked and released. A second trip was made after three days. There were 40 marked snails out of a total of 180 caught. Estimate the population of garden snails in the vegetable farm using the results obtained.

 

CAPTURE, MARK, RELEASE & RECAPTURE METHOD

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.