Germination Data Processing and Presentation, Lab Report Example

The aim of this experiment was to review the effect of the distinct salt concentrations of the solution which was used to irrigate the broad and mung beans. Different quantities of salt concentration were applied and enabled to form a solution in two liters of water. The broad beans and the mung beans were exposed to the different concentrations of the saline solution for seven days. In this investigation, the research question which will be made is the manner by which the concentration of salinity of the solution affects the broad bean and the mung bean’s ability to germinate (Hossain 325; Kerr 2; Mensah & Ihenyen 18). The goals of this experiment are to evaluate the broad bean and mung bean sprouts which grow the fastest in a collection of fifteen. The distinctions in the plant’s development will be measured.

Elevated levels of salinity in the area of the roots of germinating plants adversely influences the plant development, the decreased salinity in the root area of the plants enables faster plant development. The salinity has the capacity of influencing the comprehensive plant growth and the dimensions of the leaves (Hossain 325; Kerr 2; Mensah & Ihenyen 18). In this experiment the mung beans and the broad beans will be germinated. Notwithstanding , the elevated levels of salinity in the solution can cause the mung beans and the broad beans to develop at a less intensive rate, the elevated levels of salinity in the solution can also be the attribute of the broad bean and the mung bean sprout’s  early death. The aspect of possessing salinity in the root area of the plant implies that a decreased amount of water can be absorbed by the developing plant (Hossain 325; Kerr 2; Mensah & Ihenyen 18). The decreased amount of water absorption of the plant implies that the plant may dehydrate. The mung beans and broad bean can attain a height of 90cm and 30cm, respectively.  These dimensions are for fully developed plants.

The independent variables were the amount of sunlight received by each plant. The dependent variables were the quantity of seeds which germinated on a daily basis and the average time for germination of each of the plants. The rate of germination for each of the concentration of the Broad and mung beans was another dependent variable. The controlled variables where the amount of water given to each of the mung and broad bean sprouts and the salinity of the water which had been used in each of the petri dish groups.

Materials

In order to formulate the first saline solution, sixty two and a half milligrams of salt was dissolved into one liter of water. The salt was weighed on a digital scale which had an uncertainty of ± 0.1 mg. The water was held in a graduated cylinder which had an uncertainty of ± 1ml. The second solution applied 125mg of salt into a liter of water which had been held in a graduated cylinder. The graduated cylinder has an uncertainty of ± 1 ml. The scale which had been applied to measure the 125 mg of salt had an uncertainty of ± 0.1 mg. The following sample incorporated adding 250 mg of salt to one liter of water. The salt was once again weighed on the scale. The next solution was formed by aggregating five hundred mg of salt and addition it to a liter of water. . This is the manner by which the different solutions of 0.0625%, 0.1250%, 0.250 and 0.5% concentrations were formulated.  In total, five liter bottles were used for this experiment with saline solutions varying from zero to 0.5% concentration. Each of the liter saline solutions was labeled (Kerr 2).

• Five liter bottles of water , one with a salinity of 0%, the next with a salinity of 0.0625%, the third with a salinity of 0.125%, the fourth with a salinity of 0.250% and the last bottle with a salinity of 0.500%.
• Thirty petri dishes.
• Fifteen Mung beans
• Fifteen broad beans
• 25 ml graduated cylinder with an uncertainty of ± 0.1 ml.
• Ruler ( 30 cm with an uncertainty of ± 0.1).

Procedure

Each of the mung beans and the broad beans were placed into a petri dish. After all of the broad beans and mung beans had been labelled and placed individually into petri dishes, they were placed on a ledge outside of a window.  Three of the mung beans received 25 ml of water which had been taken from the water bottle which held a zero percent saline solution. Three of the mung beans received 25 ml of water which had been taken from the water bottle which held a 0.0625% saline solution. Three of the mung beans received 25 ml of water which had been taken from the water bottle which held a 0.125% saline solution. Three of the mung beans received 25 ml of water which had been taken from the water bottle which held a 0.250% saline solution. Three of the mung beans received 25 ml of water which had been taken from the water bottle which held a 0.5% saline solution (Kerr 3).

Three of the broad beans received 25 ml of water which had been taken from the water bottle which held a zero percent saline solution. Three of the broad beans received 25 ml of water which had been taken from the water bottle which held a 0.0625% saline solution. Three of the broad beans received 25 ml of water which had been taken from the water bottle which held a 0.125% saline solution. Three of the broad beans received 25 ml of water which had been taken from the water bottle which held a 0.250% saline solution. Three of the broad beans received 25 ml of water which had been taken from the water bottle which held a 0.5% saline solution.

The broad beans and the mung beans which received the distinct concentrations of saline solution were labeled and allowed to sit on a window ledge in order to receive sunlight. The growth of the plants was measured on a daily basis for seven days. The data had been entered on the table which is demonstrated below. Two distinct tables were established for the mung beans and the broad beans.

Data Tables

Subsequent to entering the data on the tables, the mean was calculated. This was performed by taking the aggregate of all of the samples of each the concentrations and dividing that sum by the sample size. The formula which had been applied was: (day 1+ day2 + day 3 + day 4+ day 5 + day 6 + day 7) / 7 days. The mean for the mung beans which were placed in the zero percent saline solution was calculated by the following method (1 + 3 + 3 + 2 + 1 + 2 + 2/ 7). The mean of the sample which had been placed in the zero percent solution was 14/ 7 or   2.

The mean for the mung beans which were placed in the 0.0625% saline solution was calculated by the following method (2 + 1 + 3 + 2 + 2 + 1 + 2/ 7).  The mean of the sample which had been placed in the 0.0625% solution was 13/ 7 or 1.857.  The mean for the mung beans which were placed in the 0.125% saline solution was calculated by the following method (0 + 1 + 1 + 2 + 2 + 1 + 2 / 7). The mean of the sample which had been placed in the 0.125% solution was 10/7 or  1.429.

The mean for the mung beans which were placed in the 0.250% saline solution was calculated by the following method (0 + 0 + 0 + 2 + 0 + 1 + 0 / 7). The mean of the sample which had been placed in the 0.5% solution was 3/ 7 or 0.42857. The mean for the mung beans which were placed in the 0.5% saline solution was calculated by the following method (0 + 0+ 0+ 0+ 0+ 0 + 1/ 7). The mean of the sample which had been placed in the zero percent solution was 1/ 7 or   0.1429.

Subsequent to entering the data on the tables, the mean was calculated. This was performed by taking the aggregate of all of the samples of each the concentrations and dividing that sum by the sample size. The formula which had been applied was: (day 1+ day2 + day 3 + day 4+ day 5 + day 6 + day 7) / 7 days. The mean for the broad beans which were placed in the zero percent saline solution was calculated by the following method (0 + 2 + 2 + 1 + 3 + 2 + 3/ 7). The mean of the sample which had been placed in the zero percent solution was 13/ 7 or   1.857.

The mean for the broad beans which were placed in the 0.0625% saline solution was calculated by the following method (0 + 2 + 2 + 1 + 3 + 2 + 3/ 7).  The mean of the sample which had been placed in the 0.0625% solution was 11/ 7 or 1.571    .  The mean for the broad beans which were placed in the 0.125% saline solution was calculated by the following method (0 + 1 + 1 + 1 + 2 + 2 + 3 / 7). The mean of the sample which had been placed in the 0.125% solution was 10/7 or   1.42857.  The mean for the broad beans which were placed in the 0.250% saline solution was calculated by the following method (0 + 0 + 0 + 2 + 0 + 1 + 2 / 7). The mean of the sample which had been placed in the 0.5% solution was 5/ 7 or 0.714285. The mean for the broad beans which were placed in the 0.5% saline solution was calculated by the following method (0 + 0+ 0+ 0+ 0+ 0 + 1/ 7). The mean of the sample which had been placed in the zero percent solution was 1/ 7 or 0.1429.

The means were subsequently entered on a data table below:

Data Table for Mean Calculation

In order to calculate the standard deviation, the following formula will be applied:

Zero percent salinity group.

[(day 1 – mean_{mung zero percent})² + (day 2- mean_{mung zero percent})²+ (day 3 – mean_{mung zero percent})² + (day 4 – mean_mung) + ( day 5 – mean_{mung zero percent}) + (day 6 – mean_{mung zero percent}) + (day  7 – mean_{mung zero percent}) / total number of days – one day ]^{1/ 2} 

[(1- 2)² + (3- 2)² + (3- 2)² +( 2- 2)² +( 1- 2)² + ( 2- 2)² + ( 2- 2)² / 7- 1]² =(1 + 1+ 1+ 1+ 0 + 0 = 6 )^1/2 =  0.8145

[(day 1 – mean_{broad zero percent})² + (day 2- mean_{broad zero percent})²+ (day 3 – mean_{broad zero percent})² + (day 4 – mean_mung) + ( day 5 – mean_{broad zero percent}) + (day 6 – mean_{broad zero percent}) + (day  7 – mean_{broad zero percent}) / total number of days – one day ]^{1/ 2} 

[(0- 1.857)² + (2- 1.857)² + (2- 1.857)² + (1- 1.857)² + (3- 1.857)² + (2- 1.857)² + (2- 1.857)² + (0- 1.857)² / 7 – 1]² = (3.44845 + 0.020449 + 0.020449 + 0.73445 + 1.30645 + 0.020449 +0.020449  = 6.309/ 6)^1/2 = 0.9636

0.0625% salinity group.

[(day 1 – mean_{mung 0.0625%})² + (day 2- mean_{mung 0.0625%})²+ (day 3 – mean_{mung 0.0625%})² + (day 4 – mean_mung0.0625%) + ( day 5 – mean_{mung 0.0625%}) + (day 6 – mean_{mung 0.0625%}) + (day  7 – mean_{mung 0.0625%}) / total number of days – one day ]^{1/ 2} 

[(2-  1.857)² + (1-  1.857)² + (3-  1.857)²+ (2-  1.857)² + (2-  1.857)² + (1-  1.857)² +(2-  1.857)²/ 7 – 1]^1/2= (0.020449 + 0.7345 + 1.30645 + 0.020449 + 0.020449 +  0.73445 + 0.020449/ 6 ]^1/2 = 0.68759

[(day 1 – mean_{broad 0.0635%})² + (day 2- mean_{broad 0.0635%})²+ (day 3 – mean_{broad 0.0635%})² + (day 4 – mean _{broad 0.0635%}) + ( day 5 – mean_{broad 0.0635%}) + (day 6 – mean_{broad 0.0635%}) + (day  7 – mean_{broad 0.0635%}) / total number of days – one day ]^{1/ 2} 

[(0-0.1571)² + (2 -0.1571)² +  (3 -0.1571)² +  (1 -0.1571)² +  (1 -0.1571)² + (1 -0.1571)² +(1 -0.1571)² / 7- 1]² = (0.02468 +  3.396 + 0.7105 + 0.7105 + 0.7105 + 0.7105 / 6)^1/2 = 1.02164912764

0.125% salinity group.

[(day 1 – mean_{mung 0.125%})² + (day 2- mean_{mung 0.125%})²+ (day 3 – mean_{mung 0.125%})² + (day 4 – mean_mung) + ( day 5 – mean_{mung 0.125%}) + (day 6 – mean_{mung 0.125%}) + (day  7 – mean_{mung 0.125%}) / total number of days – one day ]^{1/ 2} 

[(0 – 1.429)² + (1 – 1.429)² + (1 – 1.429)²+ (2 – 1.429)²+ (2 – 1.429)²+ (1 – 1.429)² + (2 – 1.429)²/ 7- 1]^1/2= (2.042041 + 0.18041 +0.18041 + 0.571 + 0.571 + 0.18041 + 0.571= 0.8461945

[(day 1 – mean_{broad 0.125%})² + (day 2- mean_{broad 0.125%})²+ (day 3 – mean_{broad 0.125%})² + (day 4 – mean_mung) + ( day 5 – mean_{broad 0.125%}) + (day 6 – mean_{broad 0.125%}) + (day  7 – mean_{broad 0.125%}) / total number of days – one day ]^{1/ 2} 

(0 – 1.429)² + (1 – 1.429)²+ (1 – 1.429)²+ (1 – 1.429)²+ (2 – 1.429)²+ (2 – 1.429)²+ (3 – 1.429)²/ 7- 1]²=

2.042041 + 0.18041 + 0.18041 +0.18041 + 0.571 + 0.571 + 1.571/ 6 )^1/2= 0.9396161502

0.250% salinity group.

[(day 1 – mean_{mung 0.250%%})² + (day 2- mean_{mung 0.250%%})²+ (day 3 – mean_{mung 0.250%%})² + (day 4 – mean_mung) + ( day 5 – mean_{mung 0.250%%}) + (day 6 – mean_{mung 0.250%%}) + (day  7 – mean_{mung 0.250%%}) / total number of days – one day ]^{1/ 2} 

[(0 – 0.4257)² + (0 – 0.4257)² + (2 – 0.4257)² + (0 – 0.4257)² + (1 – 0.4257)² + (0 – 0.4257)² + (0 – 0.4257)²/ 6] ^½  =  (0 .18122 + 2.380 + 0.3298 + 0.18122 + 0.3298 + 0.18122 + 0.18122/6) ^½   =0 .694814

[(day 1 – mean_{broad 0.250%})² + (day 2- mean_{broad 0.250%})²+ (day 3 – mean_{broad 0.250%})² + (day 4 – mean_mung) + ( day 5 – mean_{broad 0.250%}) + (day 6 – mean_{broad 0.250%}) + (day  7 – mean_{broad 0.250%}) / total number of days – one day ]^{1/ 2} 

[(0 – 0.714285)²  +( 0 – 0.714285)² +( 0 – 0.714285)² +( 2 – 0.714285)² + ( 0 – 0.714285)² +( 1 – 0.714285)² + ( 0 – 0.714285)²/ 7- 1]² = (0.510203 + 0.510203 + 0.510203 + 1.653 + 0.08163 + 0.510203 + 0.510203)/6]^1/2  = 0.84514

0.5% salinity group.

[(day 1 – mean_{mung 0.5%})² + (day 2- mean_{mung 0.5%})²+ (day 3 – mean_{mung 0.5%})² + (day 4 – mean_mung) + ( day 5 – mean_{mung 0.5%}) + (day 6 – mean_{mung 0.5%}) + (day  7 – mean_{mung 0.5%}) / total number of days – one day ]^{1/ 2} 

[(0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (1 – 0.1429)²/ 7- 1]² = ( 02042041 +  02042041 + 02042041 + 02042041 + 02042041 + 02042041 + 18041) 6]^1/2  = 0.6910283

[(day 1 – mean_broad0.5)² + (day 2- mean_broad0.5)²+ (day 3 – mean_broad0.5)² + (day 4 – mean_mung) + ( day 5 – mean_broad0.5) + (day 6 – mean_broad0.5) + (day  7 – mean_broad0.5) / total number of days – one day ]^{1/ 2} 

[(0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (0 – 0.1429)² + (1 – 0.1429)²/ 7- 1]² = 0.6910283

In order to convert the standard deviation of plants into standard deviation as a percentage, the standard deviation of plants will be taken  and divided  by each mean.  This result will be multiplied by one hundred. This will provide the standard deviation in percent.

Discussion

The mung and broad bean sprouts were subsequently measured with a ruler which had an uncertainty of ±0.1mm. The final outcomes of the experiment on the seventh day were: the group which possessed zero percent salinity for the mung beans demonstrated a growth of 100 mm. The second group which had been grown in the petri dishes which had a solution with a salinity of 0.0625% grew 10.5 mm. The group of mung bean sprouts which were grown i9n the petri dish which had a solution of 0.125 % salinity grew 100.5 mm. The fourth group of mung bean sprouts which were grown in a petri dish which had a salinity of 0.255 grew 51 mm. The final group of mung beans which were placed in the Petri dish which possessed a salinity of 0.5% grew to 46 mm.

The group which possessed zero percent salinity for the broad beans demonstrated a growth of 110 mm. The second group which had been grown in the petri dishes which had a solution with a salinity of 0.0625% grew 105 mm. The group of broad bean sprouts which were grown in the petri dish which had a solution of 0.125 % salinity grew 100. mm. The fourth group of broad bean sprouts which were grown in a petri dish which had a salinity of 0.255 grew 50 mm. The final group of broad beans which were placed in the Petri dish which possessed a salinity of 0.5% grew to 40 mm. The mean outcome demonstrated that by possessing additional salt in the water increased the potential of the sprouts’ dehydration. The plants do not always die as a result of extra salinity in the water .  Research has demonstrated that elevated salinity only facilitates the plants growth in a small measure (Hossain 325; Kerr 2; Mensah & Ihenyen 18).

The groups which possessed salinity concentrations of 0.250% and 0.5% demonstrated minimal growth. The elevated indexes of salinity demonstrate that the plants will decrease in dimension . The errors in the experiment could have been caused by the initial salt measurements, the amount of sunlight which had been received by the plants and the aspect of evaporation which would have caused the samples to experience a higher salinity.  The experiment could have been conducted more effectively if the external conditions had been more controlled. The results of the experiment confirmed the research hypothesis .

Works Cited

Hossain, M. M. M. N. A. Miah, M. A. Rahman, M. A. Islam & M. T. Islam. : effect of salt stress on growth and yield attributed of mungbean.” Bangladesh Research Publications Journal 1.4( 2008): 324-326 <.http://www.bdresearchpublicatioins.com/admin/journal/upload/08035>

Kerr, Stephen. “Lab effects of soil salinity on seed germination.” AP Environmental  Science, February 2005. Web. 3 March 2014 <http:/www.csun.3edu/scince/kerr/Stephen/salinity.pdf>

Mensah, J. K. & J. Ihenyen. “ Effects of salinity of germination, seedling establishment and yield of three genotypes of mung vean (Vigne mungo L. Hepper) in Edo State m Nigeria.” Nigerian Annals of Natural Science 8.2(2009): 17-24.<http://nansjounal.org/contents/nans2009v9n1.017-024.pdf?Herr