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Felix Dyrek

                  Biological Investigation: The Variability of Organisms

Introduction:

This bean topic was done in order to show that a species differ in size and mass from even when it
developed in the same environment.

Research Question

Which differences does exist in an organism of the same group ?

Hypothesis

If a large number of Lima beans will be meassured in lengh and mass than there will be a small vary up to
no vary at all while some will differ much more from the average bean.

Hypothesis Explanation

An organism of a species and the genus wont look exactly the same as another one in nature.When we
clone it than but only than its possible.The organism will always differ,even if the differences are
small,from another one.

Compounds

160 Lima beans

Variables

length of beans (dependent)
mass of beans (dependent)

Apparatus

1 electronic scale
1 ruler
pencil & paper to record data

Method

Turn on the electronic scale.
Weigh each of the beans and write down the datas
Measure the length of the beans and note each value
Data Collection & Processing

  Length / cm   Length (sorted) / cm   Mass / g   Mass (sorted) / g

      2.4               1.2              1.91           1.07

      2.3               1.5              1.19            1.1

      2.1              1.72              1.79           1.13

      2.2               1.8              1.76           1.13

      2.4               1.8              1.86           1.14

      2.3               1.9              1.87           1.16

      2.6               1.9              1.52           1.19

      2.2               1.9              1.53            1.2

      2.2               1.9              1.24           1.21

      2.4               1.9              2.09           1.23

      2.1                2               1.96           1.23

      2.4                2               1.95           1.23

      2.2                2               1.68           1.24

      2.3                2               1.38           1.24

      2.4                2               1.51           1.24

      2.5                2               1.94           1.24

      2.2                2               2.14           1.25

      2.5                2               1.47           1.25

      2.2                2               1.57           1.26

      2.4                2               1.56           1.27

      2.7                2               1.39           1.27

      2.2                2               1.55           1.33

      2.6                2               1.59           1.35

      2.4                2               1.94           1.35

      2.3                2               1.57           1.37
2.2   2     1.84   1.38

2.6   2     1.86   1.38

2.2   2.1   2.12   1.39

1.9   2.1    2     1.39

2.3   2.1   1.48   1.4

2     2.1   1.79   1.42

2.1   2.1   1.74   1.43

2.3   2.1    2     1.44

2.4   2.1   1.55   1.45

2.1   2.1   2.17   1.46

2.5   2.1   1.39   1.46

2.2   2.1   1.23   1.46

2.3   2.1   2.16   1.46

2.3   2.1   1.54   1.47

2.3   2.1    2     1.47

2     2.1   1.68   1.48

2.4   2.1   1.65   1.48

2.1   2.1   1.48   1.48

2.4   2.1   1.95   1.49

2     2.1   1.24   1.5

2.4   2.1   1.49   1.5

2.1   2.1   1.7    1.5

2     2.1   1.27   1.5

2.1   2.1   1.25   1.51

2.4   2.1   1.8    1.51

2     2.1   1.25   1.51

2.2   2.2   1.5    1.52

2.1   2.2   1.95   1.52
2     2.2   2.25   1.52

2.4    2.2   1.61   1.53

2.2    2.2   1.24   1.53

2.5    2.2   1.65   1.53

2.9    2.2   1.23   1.54

2.2    2.2   2.04   1.55

2.4    2.2   1.07   1.55

2.2    2.2   1.45   1.55

2.4    2.2   1.6    1.56

2.5    2.2   1.65   1.57

2.7    2.2   1.71   1.57

 2     2.2   2.15   1.57

2.3    2.2   1.46   1.57

2.2    2.2   1.53   1.58

2.1    2.2   1.71   1.58

2.2    2.2   2.28   1.59

1.9    2.2   1.6    1.6

 2     2.2   1.85   1.6

2.9    2.2   1.98   1.6

1.5    2.2   1.52   1.6

1.2    2.2   1.27   1.6

1.72   2.2   2.32   1.6

2.5    2.2   1.5    1.61

2.1    2.2   1.26   1.62

2.1    2.2   2.15   1.62

2.2    2.2   2.1    1.63

 2     2.2   1.99   1.64

 2     2.2   2.02   1.64
2.3   2.2   1.87   1.64

2     2.2   1.55   1.65

1.9   2.2   1.46   1.65

2.2   2.2   1.78   1.65

1.8   2.2   1.46   1.68

2     2.2   1.62   1.68

2.4   2.2   1.38   1.7

2.2   2.2   1.51   1.7

2.2   2.2   1.8    1.71

1.9   2.3   1.64   1.71

2.2   2.3   1.4    1.71

2.2   2.3   1.51   1.74

2.1   2.3   1.84   1.74

2.3   2.3   1.6    1.75

2.2   2.3   1.98   1.75

2.2   2.3   1.35   1.76

2     2.3   1.6    1.76

2.1   2.3   2.24   1.76

2     2.3   1.13   1.78

2.1   2.3   1.21   1.78

2.1   2.3   1.7    1.79

2.1   2.3   1.57   1.79

2.4   2.3   1.64   1.8

2.1   2.3   1.8    1.8

2     2.3   2.3    1.8

2.3   2.3   1.35   1.83

2.1   2.3   1.9    1.84

2     2.3   1.71   1.84
2.1   2.3   1.52   1.84

2.4   2.3   1.74   1.85

1.9   2.3   2.04   1.85

2.1   2.3   1.2    1.86

2.5   2.3   1.6    1.86

2.4   2.4   1.86   1.86

1.8   2.4   1.46   1.86

2.3   2.4   2.25   1.87

2.3   2.4   2.29   1.87

2.2   2.4   1.16   1.9

2.3   2.4   1.24   1.9

2.5   2.4   1.75   1.91

2.5   2.4   2.06   1.92

2.2   2.4   1.98   1.92

2     2.4   1.62   1.94

2.2   2.4   1.44   1.94

2.2   2.4   1.84   1.95

2.2   2.4   1.5    1.95

2.1   2.4   1.92   1.95

2.2   2.4   1.47   1.95

2.1   2.4   1.5    1.96

2.2   2.4   1.23   1.98

2.4   2.4   1.9    1.98

2.3   2.4   1.43   1.98

2.6   2.4   2.01   1.99

2.2   2.4   1.92    2

2.3   2.4   1.78    2

2.3   2.4   1.33    2
2.4                 2.5                                     1.75        2.01

      2.2                 2.5                                     1.63        2.02

      2.2                 2.5                                     1.86        2.04

      2.5                 2.5                                     1.42        2.04

      2.1                 2.5                                     2.2         2.06

      2.3                 2.5                                     1.37        2.09

      2.5                 2.5                                     1.64         2.1

      2.2                 2.5                                     1.85        2.12

      2.3                 2.5                                     1.14        2.14

      2.3                 2.5                                     1.53        2.15

      2.5                 2.5                                     1.76        2.15

      2.2                 2.5                                     1.95        2.16

      2.2                 2.5                                     1.6         2.16

      2.1                 2.5                                     1.48        2.17

      2.5                 2.5                                     2.25         2.2

      2.5                 2.6                                     1.76        2.24

      2.4                 2.6                                     1.58        2.25

      2.4                 2.6                                     1.57        2.25

      2.5                 2.6                                     1.83        2.25

      2.3                 2.7                                     1.58        2.28

      2.2                 2.7                                     1.13        2.29

      2.3                 2.9                                     2.16         2.3

      2.2                 2.9                                     1.1         2.32


            Table 1: The values of the length and mass of each bean, unsorted and sorted.

The Classes for each variable:

Classes of length.

In order to establish classes for the data:

Take the lowest value and the greatest value, 1.2 cm and 2.9 cm
Find the difference of these values.
Divide this result by 8 (the amount of classes that are needed)




Now, with this value, you should add it starting with the lowest value of length, 1.2, and continue until you
have reached the highest value, 2.9. You will have 8 total classes.

    1st class: 1.2 + 0.2125 = 1.4125, 1.2 – 1.4125

    2nd class: 1.4125 + 0.2125 = 1.625, > 1.4125 – 1.625

    3rd class: 1.625 + 0.2125 = 1.8375, > 1.625 – 1.8375

    4th class: 1.8375 + 0.2125 = 2.05, > 1.8375 – 2.05

    5th class: 2.05 + 0.2125 = 2.2625, > 2.05 – 2.2625

    6th class: 2.2625 + 0.2125 = 2.475, > 2.2625 – 2.475

    7th class: 2.475 + 0.2125 = 2.6875, > 2.475 – 2.6875

    8th class: 2.6875 + 0.2125 = 2.9, > 2.6875 – 2.9




     Number of class                     Class                 Number of elements in         Median of values in each
                                                                      class                        class / cm

             1                       1.2 – 1.4125                         1                             1.2

             2                     > 1.4125 – 1.625                       1                             1.5

             3                     > 1.625 – 1.8375                       3                             1.8

             4                      > 1.8375 – 2.05                      22                              2

             5                      > 2.05 – 2.2625                      63                             2.2

             6                     > 2.2625 – 2.475                      47                             2.3

             7                     > 2.475 – 2.6875                      19                             2.5

             8                      > 2.6875 – 2.9                        4                             2.7
Table 2: The number of elements and median for each class of length.
Graph 1: representing the relation between the number of elements and each class of length.

Standard deviation of the values of length: 0.22

This was calculated by using Excel.

Classes of mass.

Use the same method that was used for classes of length:

Take the lowest value and the greatest value, 1.07g and 2.32g.
Find the difference of these values.




Divide this result by 8 (the amount of classes that are needed)




Now, with this value, you should add it starting with the lowest value of length, 1.07, and continue until
you have reached the highest value, 2.32. You will have 8 total classes.

        1st class: 1.07 + 0.15625 = 1.22625, 1.07 – 1.22625

        2nd class: 1.22625 + 0.15625 = 1.3825, >1.22625 – 1.3825

        3rd class: 1.3825 + 0.15625 = 1.53875, >1.3825 – 1.53875

        4th class: 1.53875 + 0.15625 = 1.695, >1.53875 – 1.695
5th class: 1.695 + 0.15625 = 1.85125, >1.695 – 1.85125

       6th class: 1.85125 + 0.15625 = 2.0075, >1.85125 – 2.0075

       7th class: 2.0075 + 0.15625 = 2.16375, >2.0075 – 2.16375

       8th class: 2.16375 + 0.15625 = 2.32, >2.16375 – 2.32


    Number of class                   Class                   Number of elements in   Median of values in each
                                                                  each class                 class / g

           1                     1.07 – 1.22625                        9                       1.14

           2                   >1.22625 – 1.3825                       18                      1.26

           3                   >1.3825 – 1.53875                       30                      1.48

           4                    >1.53875 – 1.695                       30                       1.6

           5                    >1.695 – 1.85125                       25                      1.78

           6                   >1.85125 – 2.0075                       25                      1.94

           7                   >2.0075 – 2.16375                       13                       2.1

           8                    >2.16375 – 2.32                        10                      2.25
Table 3: The number of elements and median for each class of mass.




Graph 2: representing the relation between the number of elements and each class of mass.
Standard deviation of the values of mass: 0.3

This was calculated by using Excel.

Conclusion and Evaluation

My hypothesis that organism of the same species will differ in length and mass but are more or less quite
so similar was correct.The measurements prrove it.Only 1.7cm and 1.25gr is the difference between the
smallest and largest bean during this experiment.The beans were very close to each other in length and
also in mass but still the differences exist.In general there were no obstructions during this
experiment.THe only thing which is worth to be mentioned is that we didnt know if these beans which
were brought up were the same and we also had no idea about the enviroment the beans were before.
The Investigation was succesfull.

                                                                                                        1

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International baccalaureate biology sl investigation_beans by felix dyrek

  • 1. Felix Dyrek Biological Investigation: The Variability of Organisms Introduction: This bean topic was done in order to show that a species differ in size and mass from even when it developed in the same environment. Research Question Which differences does exist in an organism of the same group ? Hypothesis If a large number of Lima beans will be meassured in lengh and mass than there will be a small vary up to no vary at all while some will differ much more from the average bean. Hypothesis Explanation An organism of a species and the genus wont look exactly the same as another one in nature.When we clone it than but only than its possible.The organism will always differ,even if the differences are small,from another one. Compounds 160 Lima beans Variables length of beans (dependent) mass of beans (dependent) Apparatus 1 electronic scale 1 ruler pencil & paper to record data Method Turn on the electronic scale. Weigh each of the beans and write down the datas Measure the length of the beans and note each value
  • 2. Data Collection & Processing Length / cm Length (sorted) / cm Mass / g Mass (sorted) / g 2.4 1.2 1.91 1.07 2.3 1.5 1.19 1.1 2.1 1.72 1.79 1.13 2.2 1.8 1.76 1.13 2.4 1.8 1.86 1.14 2.3 1.9 1.87 1.16 2.6 1.9 1.52 1.19 2.2 1.9 1.53 1.2 2.2 1.9 1.24 1.21 2.4 1.9 2.09 1.23 2.1 2 1.96 1.23 2.4 2 1.95 1.23 2.2 2 1.68 1.24 2.3 2 1.38 1.24 2.4 2 1.51 1.24 2.5 2 1.94 1.24 2.2 2 2.14 1.25 2.5 2 1.47 1.25 2.2 2 1.57 1.26 2.4 2 1.56 1.27 2.7 2 1.39 1.27 2.2 2 1.55 1.33 2.6 2 1.59 1.35 2.4 2 1.94 1.35 2.3 2 1.57 1.37
  • 3. 2.2 2 1.84 1.38 2.6 2 1.86 1.38 2.2 2.1 2.12 1.39 1.9 2.1 2 1.39 2.3 2.1 1.48 1.4 2 2.1 1.79 1.42 2.1 2.1 1.74 1.43 2.3 2.1 2 1.44 2.4 2.1 1.55 1.45 2.1 2.1 2.17 1.46 2.5 2.1 1.39 1.46 2.2 2.1 1.23 1.46 2.3 2.1 2.16 1.46 2.3 2.1 1.54 1.47 2.3 2.1 2 1.47 2 2.1 1.68 1.48 2.4 2.1 1.65 1.48 2.1 2.1 1.48 1.48 2.4 2.1 1.95 1.49 2 2.1 1.24 1.5 2.4 2.1 1.49 1.5 2.1 2.1 1.7 1.5 2 2.1 1.27 1.5 2.1 2.1 1.25 1.51 2.4 2.1 1.8 1.51 2 2.1 1.25 1.51 2.2 2.2 1.5 1.52 2.1 2.2 1.95 1.52
  • 4. 2 2.2 2.25 1.52 2.4 2.2 1.61 1.53 2.2 2.2 1.24 1.53 2.5 2.2 1.65 1.53 2.9 2.2 1.23 1.54 2.2 2.2 2.04 1.55 2.4 2.2 1.07 1.55 2.2 2.2 1.45 1.55 2.4 2.2 1.6 1.56 2.5 2.2 1.65 1.57 2.7 2.2 1.71 1.57 2 2.2 2.15 1.57 2.3 2.2 1.46 1.57 2.2 2.2 1.53 1.58 2.1 2.2 1.71 1.58 2.2 2.2 2.28 1.59 1.9 2.2 1.6 1.6 2 2.2 1.85 1.6 2.9 2.2 1.98 1.6 1.5 2.2 1.52 1.6 1.2 2.2 1.27 1.6 1.72 2.2 2.32 1.6 2.5 2.2 1.5 1.61 2.1 2.2 1.26 1.62 2.1 2.2 2.15 1.62 2.2 2.2 2.1 1.63 2 2.2 1.99 1.64 2 2.2 2.02 1.64
  • 5. 2.3 2.2 1.87 1.64 2 2.2 1.55 1.65 1.9 2.2 1.46 1.65 2.2 2.2 1.78 1.65 1.8 2.2 1.46 1.68 2 2.2 1.62 1.68 2.4 2.2 1.38 1.7 2.2 2.2 1.51 1.7 2.2 2.2 1.8 1.71 1.9 2.3 1.64 1.71 2.2 2.3 1.4 1.71 2.2 2.3 1.51 1.74 2.1 2.3 1.84 1.74 2.3 2.3 1.6 1.75 2.2 2.3 1.98 1.75 2.2 2.3 1.35 1.76 2 2.3 1.6 1.76 2.1 2.3 2.24 1.76 2 2.3 1.13 1.78 2.1 2.3 1.21 1.78 2.1 2.3 1.7 1.79 2.1 2.3 1.57 1.79 2.4 2.3 1.64 1.8 2.1 2.3 1.8 1.8 2 2.3 2.3 1.8 2.3 2.3 1.35 1.83 2.1 2.3 1.9 1.84 2 2.3 1.71 1.84
  • 6. 2.1 2.3 1.52 1.84 2.4 2.3 1.74 1.85 1.9 2.3 2.04 1.85 2.1 2.3 1.2 1.86 2.5 2.3 1.6 1.86 2.4 2.4 1.86 1.86 1.8 2.4 1.46 1.86 2.3 2.4 2.25 1.87 2.3 2.4 2.29 1.87 2.2 2.4 1.16 1.9 2.3 2.4 1.24 1.9 2.5 2.4 1.75 1.91 2.5 2.4 2.06 1.92 2.2 2.4 1.98 1.92 2 2.4 1.62 1.94 2.2 2.4 1.44 1.94 2.2 2.4 1.84 1.95 2.2 2.4 1.5 1.95 2.1 2.4 1.92 1.95 2.2 2.4 1.47 1.95 2.1 2.4 1.5 1.96 2.2 2.4 1.23 1.98 2.4 2.4 1.9 1.98 2.3 2.4 1.43 1.98 2.6 2.4 2.01 1.99 2.2 2.4 1.92 2 2.3 2.4 1.78 2 2.3 2.4 1.33 2
  • 7. 2.4 2.5 1.75 2.01 2.2 2.5 1.63 2.02 2.2 2.5 1.86 2.04 2.5 2.5 1.42 2.04 2.1 2.5 2.2 2.06 2.3 2.5 1.37 2.09 2.5 2.5 1.64 2.1 2.2 2.5 1.85 2.12 2.3 2.5 1.14 2.14 2.3 2.5 1.53 2.15 2.5 2.5 1.76 2.15 2.2 2.5 1.95 2.16 2.2 2.5 1.6 2.16 2.1 2.5 1.48 2.17 2.5 2.5 2.25 2.2 2.5 2.6 1.76 2.24 2.4 2.6 1.58 2.25 2.4 2.6 1.57 2.25 2.5 2.6 1.83 2.25 2.3 2.7 1.58 2.28 2.2 2.7 1.13 2.29 2.3 2.9 2.16 2.3 2.2 2.9 1.1 2.32 Table 1: The values of the length and mass of each bean, unsorted and sorted. The Classes for each variable: Classes of length. In order to establish classes for the data: Take the lowest value and the greatest value, 1.2 cm and 2.9 cm Find the difference of these values.
  • 8. Divide this result by 8 (the amount of classes that are needed) Now, with this value, you should add it starting with the lowest value of length, 1.2, and continue until you have reached the highest value, 2.9. You will have 8 total classes. 1st class: 1.2 + 0.2125 = 1.4125, 1.2 – 1.4125 2nd class: 1.4125 + 0.2125 = 1.625, > 1.4125 – 1.625 3rd class: 1.625 + 0.2125 = 1.8375, > 1.625 – 1.8375 4th class: 1.8375 + 0.2125 = 2.05, > 1.8375 – 2.05 5th class: 2.05 + 0.2125 = 2.2625, > 2.05 – 2.2625 6th class: 2.2625 + 0.2125 = 2.475, > 2.2625 – 2.475 7th class: 2.475 + 0.2125 = 2.6875, > 2.475 – 2.6875 8th class: 2.6875 + 0.2125 = 2.9, > 2.6875 – 2.9 Number of class Class Number of elements in Median of values in each class class / cm 1 1.2 – 1.4125 1 1.2 2 > 1.4125 – 1.625 1 1.5 3 > 1.625 – 1.8375 3 1.8 4 > 1.8375 – 2.05 22 2 5 > 2.05 – 2.2625 63 2.2 6 > 2.2625 – 2.475 47 2.3 7 > 2.475 – 2.6875 19 2.5 8 > 2.6875 – 2.9 4 2.7 Table 2: The number of elements and median for each class of length.
  • 9. Graph 1: representing the relation between the number of elements and each class of length. Standard deviation of the values of length: 0.22 This was calculated by using Excel. Classes of mass. Use the same method that was used for classes of length: Take the lowest value and the greatest value, 1.07g and 2.32g. Find the difference of these values. Divide this result by 8 (the amount of classes that are needed) Now, with this value, you should add it starting with the lowest value of length, 1.07, and continue until you have reached the highest value, 2.32. You will have 8 total classes. 1st class: 1.07 + 0.15625 = 1.22625, 1.07 – 1.22625 2nd class: 1.22625 + 0.15625 = 1.3825, >1.22625 – 1.3825 3rd class: 1.3825 + 0.15625 = 1.53875, >1.3825 – 1.53875 4th class: 1.53875 + 0.15625 = 1.695, >1.53875 – 1.695
  • 10. 5th class: 1.695 + 0.15625 = 1.85125, >1.695 – 1.85125 6th class: 1.85125 + 0.15625 = 2.0075, >1.85125 – 2.0075 7th class: 2.0075 + 0.15625 = 2.16375, >2.0075 – 2.16375 8th class: 2.16375 + 0.15625 = 2.32, >2.16375 – 2.32 Number of class Class Number of elements in Median of values in each each class class / g 1 1.07 – 1.22625 9 1.14 2 >1.22625 – 1.3825 18 1.26 3 >1.3825 – 1.53875 30 1.48 4 >1.53875 – 1.695 30 1.6 5 >1.695 – 1.85125 25 1.78 6 >1.85125 – 2.0075 25 1.94 7 >2.0075 – 2.16375 13 2.1 8 >2.16375 – 2.32 10 2.25 Table 3: The number of elements and median for each class of mass. Graph 2: representing the relation between the number of elements and each class of mass.
  • 11. Standard deviation of the values of mass: 0.3 This was calculated by using Excel. Conclusion and Evaluation My hypothesis that organism of the same species will differ in length and mass but are more or less quite so similar was correct.The measurements prrove it.Only 1.7cm and 1.25gr is the difference between the smallest and largest bean during this experiment.The beans were very close to each other in length and also in mass but still the differences exist.In general there were no obstructions during this experiment.THe only thing which is worth to be mentioned is that we didnt know if these beans which were brought up were the same and we also had no idea about the enviroment the beans were before. The Investigation was succesfull. 1