-the strawberry puree is a deep pink color.
-the detergent solution is a foggy white color
-after the detergent is added the strawberry puree turns a darker, deeper pink (is getting closer to a red-purple shade)
-chilled ethanol is a murky, white color
-once the ethanol is added the test tube divides into two sections, a smaller top section and a larger bottom section
-the top section is a whitish, clear foggy color that has small tiny bubbles in it, it appears kind of jelly like, has some small strawberry seeds in it, and has a white, foggy, thick blob sitting on the top near the surface of this section
-the bottom section is a deep red-pink color, with many strawberry seeds in it but appears to have a smooth texture
-the DNA takes awhile to attach to the stirring rod, and has a difficult time sticking
-once the DNA is removed and it has an applesauce like texture and appears to be slimy, it is a white foggy color and is about the size of a pea
- the top section is now completely clear and has small bubbles in it
-the bottom section remains the same
-it took three minutes for the DNA to separate itself from the strawberry and to appear in the beaker
2. Think about where DNA comes from in the cell. What role might the detergent solution have played in allowing you to see DNA?
DNA is stored inside the nucleus of the cell. The detergent solution helped to dissolve and “bust open”the membrane. This causes the DNA to be released.
3. Why did you not see a double helix?
A double helix was not seen because DNA is invisible to the naked eye, due to the fact that it is very thin/small. Many strands of DNA were grouped together into a thick bunch of many DNA groups which is what was seen. Similarly for example, a single strand of hair is often very hard or difficult to see alone, but when you look at a whole head of hair that’s all put together, all the little pieces add up to form a big picture.
4. If you compared the DNA found in the strawberry to that found in your cheek cells, how would you expect it to be similar and different?
If one compared the DNA found in the strawberry to that found in human cheek cells, they would find that strawberries are octoploid, meaning they have 8 copies of each chromosome. While the human cheeks are diploid, meaning that they have 2 copies of each chromosome. Another difference is that strawberries have 56 chromosomes total, have 7 types of chromosomes and are plant cells. Human cheeks cells have 46 chromosomes total, have 23 types of chromosomes and are animal cells. Also, they have different DNA coding.
Some similarities between human cheek cells and strawberry cells are that one cannot see a single strand of DNA. Also, they are structurally similar. They both have a double helix, both are made up of nucleotides, and they both have the same 4 nitrogen base pairs (adenine, guanine, thymine, and cytosine).
Mitosis Onion Root Tip Lab
1. What patterns exist in your data? In which stage of the cell cycle are most of the cells you examined? How does this data support what you know about the cell cycle?
According to my data, patterns in this lab took place during interphase, metaphase, and anaphase. Interphase always had the highest number of cells in its stage (sample 1 = 106 cells, sample 2 = 44 cells, making the average 75 cells compared to prophase which had an average of 10 [sample 1= 12, sample 2=8], metaphase which had an average of 3 [sample 1= 4, sample 2= 2], anaphase which had an average of 3 [sample 1= 3, sample 2=3], and telophase which had an average of 5.5 [sample 1= 5, sample 2= 6]). Metaphase had the same percentage of cells in both samples (3% in both samples), creating another pattern. Anaphase had the same number of cells in both samples (3), also creating another pattern. The cell cycle that had the most cells in it during our examination was interphase. This supports previously known knowledge about the cell cycle and interphase: cells spend the majority of their life in the interphase cycle, the time when cells grow and copy their DNA. The data from this experiment supports previously known information because of the fact that this stage had the most amount of cells in it.
2. Find the average percentage of cells in each stage of the cell cycle among the three samples. Assume that a cell takes 24 hours to complete one cell cycle. Calculate how much time is spent in each stage of the cell cycle. (Hint: Multiply the percentage of cells in each stage, as a decimal, by 24 hours).
Assuming that a cell takes 24 hours to complete one cell cycle, the average percentage of cells in each stage of the cell cycle in our samples are:
-Interphase: 18.17 hours (or 1089.94 minutes)
-Prophase: 2.63 hours (or 157.82 minutes)
-Metaphase: 0.75 hours (or 45.07 minutes)
-Anaphase: 0. 85 hours (or 50.83 minutes)
-Telophase: 1.71 hours (or 102.53 minutes)
To calculate the average percentage of cells in each stage first all of the times were converted into minutes. 24 hours = 1440 minutes. Next, the time in minutes was plugged into the following equation to get the results.
# of cells in each cycle / total number of cells in that cycle x 100 = % of cells in that cycle
3.The cells in the root of an onion are actively dividing. How might the numbers you count here be different than if you had examined cells from a different part of the plant?
The cells in the root of an onion contain the largest amount of cell growth, the highest percentage of cellular activity, and the highest percentage of cells going through mitosis. Because of this, if cells from a different part of the plant were examined, the cellular activity would be slower.
4. Using the data from Table 2, create a graph in Google Spreadsheets using the % of cells in each stage of Mitosis. Add a title and label the graph. Upload a picture of the graph to your post.
- 4 plants (obtained from the same offspring) in 4″ pots
- Potting soil
- Compound light microscope
- 250 mL beaker
- Mystery solution
- Distilled water
- 75 watt plant light
- Stirring rod
- Sharp blade
- 12 microscope slides
- Collect 4 potted plants and label each pot accordingly – ’0 drops’, ’5 drops’, ’10 drops’, and ’15 drops’
- Measure the height of each plant and record data
- Place on a shelf 2 feet under plant light
- Pour 150 mL of distilled water into beaker
- Water potted plant labeled ’0 drops’ with the distilled water (this is the control)
- Pour 150 mL of distilled water into beaker
- Add 5 drops if mystery solution, stir to mix using stirring rod
- Pour solution into potted plant labeled ’5 drops’
- Rinse beaker
- Repeat steps 6 – 9, putting the appropriate number of drops into the beaker for the appropriate potted plant
- Water each potted plant with its appropriate solution every 2 days
- After 7 days, measure the height of each plant and record qualitative and quantitive data
- Repeat for a period of one month
- After one month, remove each plant from its pot and collect three – 1 mm root tip samples from each plant using the sharp blade
- Place each sample on its own microscope slide
- Label each slide, noting which pot they came from and numbering them 1,2, 3, & 4.
- View each slide under the microscope and count the number of cells in each of the following cell stages: interphase, prophase, metaphase, anaphase, and telophase
- Record data and analyze results
- Create a data table and graph indicating the number of cells and percentage of cells in each part of the cell cycle for each sample
The Effect of Salt Solutions on Decalcified Eggs
The cell membrane forms a boundary between a cell and the outside environment, it surrounds the cytoplasm of a cell. It is made up of a double layer of phospholipids embedded with proteins. The arrangement of the molecules that make up a cell membrane are described as a fluid mosaic, because it is flexible and floats, and the variety of molecules stud the membrane similar to the arrangement of tiles in a mosaic. The cell membrane is selectively permeable, which means that the cell membrane controls what substances pass in and out through the membrane. This characteristic of cell membranes plays a big role in passive transport; the movement of substances across the cell membrane without any input of energy by the cell. The energy for passive transport comes entirely from kinetic energy that the molecules have. Because of kinetic energy, molecules are in constant motion. The simplest type of passive transport is diffusion, which is the movement of small molecules from an area of high concentration to an area of lower concentration. Diffusion always moves down a concentration gradient, the difference between the concentration (or the strength of) of a solute in one place and its concentration in an adjacent area.
Osmosis is a type of diffusion in which water molecules diffuse across a semi-permeable membrane from a region of high concentration to one of low concentration. The net direction of osmosis depends on the concentration of solutes (the particles which are dissolved in a solvent) on the two sides of the cell membrane. In a hypertonic solution, the concentration of solutes in the solution that the cell is in is higher than the solute concentration of a cell’s cytoplasm. This causes the cell to shrink or shrivel due to water leaving the cell, and turgor pressure is lost. In plants, when this occurs, plasmolysis occurs, which is the reason plants wilt. A hypotonic solution is a solution where the solute concentration of the solution that the cell is in is lower than the solute concentration of a cell’s cytoplasm.The net movement of water is from the outside to the inside of the cell. The effect on the cell is that the cell swells and may burst, and eventually die. In an isotonic solution, the concentration of solutes on both sides of the membrane is the same and so the net movement of water is zero. When the concentration of dissolved substances is equal inside the cell to outside the cell this is known as dynamic equilibrium. Cells in an isotonic environment have no difficulty keeping the movement of water across the cell membrane in balance.
In this lab we are exploring the transport of water across the membrane of an egg by osmosis. The membrane of the egg is permeable to water and depending on the extracellular environment, water will move into or out of the egg. The egg resembles how cells in the body regulate the flow of molecules across the cell membrane. The purpose of this experiment is to understand osmosis and water transport using eggs.
How does the mass of an egg change after being submerged in distilled water, 5% salt solution & 15% salt solutions for 15 minutes?
If an egg is placed in the 5% salt solution then its mass will decrease slightly. If an egg is placed in the 15% salt solution then its mass will decrease and it will become even smaller than the egg placed in the 5% salt solution. If an egg is place in distilled water then it will swell and its mass will increase.
- Collect three decalcified eggs (eggs without their outer shell) and determine the dry mass of each egg by individually putting the eggs into a weighing boat and placing them on a balance.
- Record the weight of each egg on the data table (on Mac Airs)
- Get three 250mL beakers
- Fill one beaker with 150mL of 5% salt solution and, using a Sharpie, label it (indicating it’s the 5% salt solution)
- Fill a second beaker with 150mL of 15% salt solution and, using a Sharpie, label it (indicating it’s the 15% salt solution)
- Fill the 3rd beaker with 150mL of distilled water and, using a Sharpie, label it (indicating it’s the distilled water)
- Making sure to record which egg is being placed into what beaker, carefully place the 3 individual eggs into their own beaker using the spoon.
- After the eggs have been in the beakers 15 minutes, remove them from the beakers using the spoon, and place them back into the weighing boat and weigh them on the balance.
- Record the weight of each egg and any qualitative data in the appropriate column on the data table.
- Subtract the ‘before’ mass from the ‘after’ mass and record the results in the appropriate column on the data table.
- Calculate the percent change in mass and record it in the appropriate column on the data table.
- Analyze the information using a data table.
The independent variable is the salt concentration of the solutions, the thing added to an experiment that causes the dependent variable to change.
The dependent variable is the eggs, the variable being observed.
The control is the egg’s mass without any solution, what is used to show what would happen if nothing was done to the egg.
The constants are the 250mL beakers, the temperature of the solutions, the amount of solution (150mL), the type of eggs, the time in the solutions, and the balance.
- Three decalcified eggs
- 150mL distilled water
- 150mL 5% salt water
- 150mL 15% salt water
- 3 – 250mL beakers
- Masking tape
- Weighing boat
- Mac Airs
How Does the Mass of Decalcified Eggs Change After Being Submerged in Distilled Water and 5% & 15% Salt Solutions for 15 Minutes?
This lab studied how the percentage of salt solute effects the mass of an egg. The hypothesis stated that the egg in the 5% salt solution would have a slight decrease in mass size, that the egg in the 15% salt solution would have the largest amount of change (and would have the largest decrease in mass), and that the egg in the 0% salt solution/distilled water would be the only egg to increase its mass size. The data from the experiment contradicted the hypothesis.
After collecting three decalcified eggs and weighing each egg separately; three test groups were created. Each individual egg was placed in one of three beakers: #1 with distilled water had an egg initially weighing 90.50 grams, beaker #2 with 5% salt solution had an egg initially weighing 83.87 grams, and beaker #3 with 15% salt solution had an egg initially weighing 86.25 grams at the start of the experiment. Each beaker contained 150mL of solution. After sitting in the solutions for 15 minutes the eggs were removed and weighed again. The egg in beaker #1 then weighed 92.98 grams, resulting in a +2.48 gram change in mass. The egg in beaker #2 then weighed 85.57 grams, resulting in a +1.7 gram change in mass. Lastly, the egg in beaker #3 then weighed 87.47 grams, resulting in a +1.22 gram change in mass. The comparison of the percentage change of the three eggs in this experiment resulted in the following overall percent change in mass: egg #1= +2.74%, egg#2= +2.02%, and egg #3= +1.41% change. To recap, osmosis is the flow of water through a semi-permeable membrane from an area of higher water concentration to an area of lower water concentration. The experiment results showed that all the eggs increased in mass size, slightly swelling, showing the effects that osmosis has on a solution. Osmosis has taken place in the eggs in the 5% salt solution and 15% salt solution. The water has moved from an area of higher concentration (the beaker) to an area of lower concentration (egg). The mass of the eggs increased because the water concentration of the salt water solution in the beakers is higher than the water concentration inside the eggs. In beaker #1, the beaker with distilled water, the concentration of water outside the egg was greater than the concentration of water inside the egg. Therefore, the water moved from the beaker to the egg.
The main limitation in this experiment was the short amount of time the eggs sat in the solutions. This may have caused some inaccurate data since there wasn’t a lot of time for osmosis to occur dramatically. It would be interesting to do this experiment again, allowing more time to see how much time would have been needed to reach a state of equilibrium, an isotonic environment. Also, the difference in the initial size/mass of the eggs varied somewhat which may have contributed to the rate at which osmosis occured. If doing this experiment again, including a broader sampling of the independent variable (the salt) and more dependent variables (the eggs) would allow a larger population to be tested and then averaged. Also, measuring the remaining volume of the solutions after the eggs have been removed would help to chart the changes in mass. It would also be interesting to see the results if other independent variables (such as sugar or corn syrup or vinegar) were included. Errors may have occurred due to incorrect measurements of the weights of the eggs. Recording wrong measurements could lead to a false interpretation of the results.
One example of osmosis in the human body involves water and nutrient absorption in the stomach, small intestine, and colon. Osmosis gradients lead water to flow into the body through tiny gaps between the cells in the intestines and absorb nutrients and food into the blood to keep humans alive. Another example of osmosis in the human body is the instinct of thirst. If the water volume in the body falls below a certain threshold, or if the salt concentration becomes too high, the brain signals thirst. Also, the kidneys use osmotic gradient to make urine more concentrated or more diluted, regulating the bodies fluids. The kidneys expel water and salt to maintain optimal levels for the cells in the body to function. Overall, this was a fun and beneficial lab which demonstrated how osmosis occurs in the human body.
“The Cell Membrane”. Web. 3 November 2011.
“Biology 12 – Cell Membrane and Cell Wall Function: Chapter Notes”. Web. 3 November 2011.
“Cell Organization and Functions”. Web. 3 November 2011.
Elodea Lab Questions:
1. What affect did each solution have on the Elodea samples? Be sure to describe the affect for each solution sample.
-Solution A: caused the cell to shrink
-Solution B: caused the cell to expand
-Solution C: caused the cell to remain the same.
2. Which solutions where hypertonic, hypotonic, and isotonic?
-Solution A was hypertonic, Solution B was hypotonic, and Solution C was isotonic.
3. What solution’s concentration was closest or most similar to the cell’s concentration? Explain your answer.
-Solution C was closest to the cell’s concentration because the size and shape of the cells stayed the same and the concentration of the water was the same inside and outside of the cell (isotonic). In Solution C, the molecules balance out and are in dynamic equilibrium.
4. How does a plant’s cell wall affect the movement of water in and out of a plant cell?
The cell wall is a structure that surrounds the cell membrane and provides strength and rigidity to cells. The cell wall is not selectively permeable; things can easily pass through the wall.The survival of plant cells depends on their ability to balance water uptake and loss. In plants, the rigid cell wall prevents the membrane from expanding too much. The net uptake or loss of water by a plant cell occurs by osmosis, the passive transport of water across the membrane, in an effort to remain isotonic. The direction of water movement depends on solute concentration and physical pressure. When the cell begins to swell (hypotonic solution), it pushes against the cell wall, producing turgor pressure. When the cell shrinks, as in a hypertonic solution, the plant cell loses water and turgor pressure making the cell flaccid; plasmolysis.
5. Animal cells don’t have cell walls. How does this change an animal cell’s ability to survive in a hypertonic or hypotonic environment?
In animal cells there is no cell wall. If the water potential of the solution around the cell is too high (hypotonic environment), the cell will swell and burst; if it is too low (hypertonic environment), the cell shrinks. In order to survive it’s important to keep a constant water potential inside animal bodies otherwise they cannot survive.
I am studying the process of diffusion and how iodine is affected by a semi-permeable membrane filled with starch. I am studying this lab to watch and see how diffusion works. This topic is important because in order to survive our cells need to be able to transport materials into and out of our cells. One way of bringing materials into and out of the cell is through the process of diffusion. I already know that diffusion is the movement of molecules in a fluid or gas from a region of higher concentration to a region of lower concentration. I also know that diffusion is a type of passive transport- a type of movement of molecules across a cell membrane without energy input from the cell. I have observed food coloring diffusing through various types of water.
Sugars provide living things with energy. All carbohydrates are sugars. Starch is a complex carbohydrate made up of simple sugars. Starch is the polymer of a monosaccharide (one sugar) , glucose (the simplest sugar). This means that the complex carbohydrates are units of sugar bonded together in large chains. Some examples of starch are: corn, peas, potatoes and winter squash, and breads.
Does starch or iodine travel through a semi-permeable membrane (dialysis tubing)?
If iodine travels through a semi-permeable membrane, then diffusion will occur and the starch will turn into the tea-color.
- Soak dialysis tubing in water for approximately 2 minutes.
- Open dialysis tubing by placing between fingers and moving fingers back-and-forth.
- Tie dialysis tube by twisting, folding and using piece of string approximately 2 cm from edge.
- Add 8mL of starch solution to dialysis tube.
- Seal second end of dialysis tube by repeating Step 3.
- Rinse dialysis tube well with water and dry gently.
- Fill 250mL beaker with 200ml of water.
- Add drops of idoine until the soluiton is “tea” color, approximately 20 drops.
- Place dialysis tube in the beaker and let sit for 20 minutes.
- Record observations every 5 minutes.
I predict that the iodine will diffuse through the dialysis tube and react with the starch, causing the starch to become the “tea” color.
According to my hypothesis, iodine would travel through a semi-permeable membrane and it would turn the liquid inside the dialysis tube a tea-color. My hypothesis was incorrect. Iodine travels through a semi-permeable membrane because iodine molecules are smaller than starch molecules. Since we were unable to achieve results from our experiment, we used data from another class to complete this lab. Since iodine molecules are small, they are able to diffuse across the semi-permeable membrane from an area of high concentration to an area of low concentration turning the liquid (starch) dark purple. The starch molecule was too big to pass across the semi-permeable membrane so the liquid outside the membrane remained clear. Errors occurred because we were unable to tie the dialysis tube tight enough, which caused it to leak. In addition, there could have been a measurement error putting in too little or too much iodine or starch. If I were to repeat this experiment I would conduct multiple trials, and I would allow more time setting up the experiment (sealing the tubing) to get more accurate results.
If food coloring is added to sugar, salt, and plain water (all at room temperature), then the food coloring will diffuse the fastest in the plain room-temperature water.
1. Fill a beaker with 100mL of room temperature water.
2. Grab both of the pre-arranged jugs, one containing room temperature salt water, the other containing room temperature sugar water, and pour 100mL of each liquid into their own beakers.
4. Add one drop of food coloring to each of the beakers simultaneously.
5. Start your stopwatch as the food coloring enters each beaker.
6. Record qualitative and quantitative observations.
7. Record the time that the food coloring has diffused for each beaker.
8. Repeat steps 1-7
9. Clean up your lab area.
Use the following table to record qualitative and quantitative data for your experiment:
|Liquid Type:||Qualitative Data||Quantitative Data|
|Control Trial 1: (Room-Temperature Water)||Drops to the bottom and disperses everywhere from the bottom of the beaker.||Finished Diffusing: 13:06|
|Control Trial 2: (Room-Temperature Water)||Sank to the bottom quickly and dispersed.||Finished Diffusing: 8:49|
|Variable 1 Trial 1: (Sugar Water)||Expands downward in a circle-like form.||Finished Diffusing: 8:17|
|Variable 1 Trial 2: (Sugar Water)||Spiraled out toward the side of the beaker.||Finished Diffusing: 5:11|
|Variable 2 Trial 1: (Salt Water)||Looks like strands slowly going down from the surface.||Finished Diffusing: 9:49|
|Variable 1 Trial 2: (Salt Water)||Swirly, spread from middle out.||Finished Diffusing: 9:58|
|Control||Variable 1||Variable 2|
|Average Rate of Diffusion in Trials||11:17||7:04||9:54|
The average times of the liquids are: 11:17 (for the control), 7:04 (for the sugar water), and 9:54 (for the salt water). My hypothesis that the control would diffuse the fastest was incorrect. It was actually the sugar water that diffused the fastest, and the control water diffused the slowest. The extra chemicals in the sugar and salt water may have helped speed up the process. Human error affected our data collection. Inaccurate timing and observation errors could’ve occurred in our experiment which could have affected the results. This experiment correlates how we use diffusion in everyday tasks like adding detergent to a load of laundry and washing our hair, where detergent and shampoo mix in with water. If I were to repeat this experiment I would only have one beaker going at a time. This would allow better attention to be made when making observations and measuring the diffusion time, and reduce the likelihood of errors.
2000 mL e flask
5 mL of water
3 drops of ammonia
3 drops of phenolphtalein
2 pieces of 7 cm string
1. Soak the dialysis tube in water for 2 minutes
2. Open the dialysis tube by pinching the dialysis tube between your fingers and moving it around
3. Seal one end of the dialysis tube by twisting, folding over and tying the end with string
4. Add 4 mL of water to the dialysis tube
5. Add 3 drops of phenolphthalein to the dialysis tube
6. Seal/ tie the other end of the dialysis tube by twisting, folding over, and tying the end with string
7. Rinse the dialysis tube in water and dry it off
8. Add 3 drops of ammonia to the E Flask. Be sure not to let any ammonia get on the sides or opening of th flask.
9. Tape the dialysis tube to the inside of the flask
I think that it will change colors.
-it turned pink (it started light pink then gradually faded to a deep pink)
- it turned pink within 5 seconds
The Plasma Membrane is the thin outer boundary in all cells that regulates the traffic of chemicals between the cell and its surroundings.
1.What do you notice about the way the “membrane” moves?
—The membrane moves easily and is flexible. I noticed that the liquid in the bubble kind of gives and takes. As you move the bubble up and down, or tilt it in a certain direction, the liquid inside the bubble shifts. And as you expand the bubble, the liquid “tightens” and pops. As you bring the sides closer together the liquid relaxes and up to a point, it releases the liquid.
2.Plasma membranes can also reform because they act like a liquid. How could this help the cell bring materials in and out?
—The plasma membrane creates a passage for materials to pass through and move in and out. It is also able to reseal itself.
3.What occurs when you try sticking a pencil through the membrane?
—The bubble pops and the membrane breaks when you try sticking a pencil through it.
4.Describe how the second outcome is similar to how materials get into and out of the cell by vesicles (look up in text if unfamiliar with this term).
—The second time, when we dipped the pencil in the bubble solution, it went through the membrane smoothly instead of resisting the membrane and popping the bubble. Since the membrane did not break, it acted as a semi-permeable membrane which allows only certain materials to go into and out of the cell.
5. How would a prokaryotic cell membrane be different then the simulated membranes you formed?
—The prokaryotic cell membrane would just pop and not allow other materials inside it.
Analyse & Conclude Questions:
1. What characteristics do all of the cells have in common? List as many as you can.
All cells have: plasma membrane, ribosomes (make proteins), DNA (genetic information, chromosomes), and cytoplasm (semi-fluid substance)
2. Identify the unique characteristics of each cell type (plant vs animal).
Plant cells have a cell wall and chloroplasts; animal cells do not.
3. What type of cells did you examine, eukaryotic or prokaryotic, for each cell? Explain how you know.
Eukaryotic, because they have a nucleus; and prokaryotic cells don’t have a true nucleus.
4. Why do you think it was necessary to add methylene blue and iodine to the slides?
To enhance detail and to clearly show the nuclei and cell walls.
5. What type of cell (plant or animal) is the amoeba? Use evidence to justify and explain your reasoning.
The amoeba is an animal cell. My reasoning comes from the fact that they have more than one nuclei. Also, amoebas eat algae, bacteria and plant cells.
Enrichment – Human Blood Cells
4. Describe how the function of red and white blood cells affects their structure.
Red blood cells deliver oxygen and remove waste throughout your body. They are shaped like discs which increases their surface area, and they are flexible. White blood cells defend your body against infection. They destroy invading organisms and assist the removal of dead or damaged tissue cells. They contain nuclei and are larger than red blood cells. White blood cells can change shape so they can gulf bacteria.
Title: The Effect of Bacterial Growth on Band Instruments Placed in an Incubator
How does bacteria grow on various band instruments/surfaces? And which band instrument contains/will grow the most bacteria?
If bacteria grows on the mouth piece of all musical instruments, then the tuba is the instrument that will have the most bacteria grow on it because it gets a lot of moisture on it.
The bacteria from the band instruments/surfaces that are put in the incubator and agar plates to grow
The growth of the bacteria
1. Show the title, question, hypothesis, variables, and procedure to Mr. Rott for approval
2. Get the petri dish and cotton swabs from Mr. Rott
3. Without opening the petra dish, use a marking pen to divide the dish into four quadrants and label them on the bottom of the petra dish (lids are not labeled because they can easily fall off or be mixed up.)
4. Identify three surfaces your group would like to test and also one control: A, B, C, D.
5. Label quadrant 1: piano
6. Label quadrant 2: tuba
7. Label quadrant 3: mallet stick
8. Label quadrant 4: control
9. Walk down to the band room
10. Using sterile cotton swabs, rub a cotton swab over the surface of each of the items (a tuba’s mouthpiece, a mallet stick, and the piano (in the hall).
11. Rub the cotton swab from each source of data in the designated area of the petri dish.
12. Rub an unused/sterile cotton swab in the designated quadrant/control quadrant in the petri dish).
13. Tape the petri dish shut around the edges of the container.
14. Label a piece of tape “Group 3″ and place across the lid of the petri dish.
15. Give the petri dish to Mr. Rott.
16. Let bacteria sit and grow in the incubator for 5 days.
17. Wash hands thoroughly.
18. Disinfect counter tops and other areas using bleach.
|Objects Tested||Bacterial Colonies|
|Quadrant 1||Piano Keys||None (0)|
|Quadrant 2||Tuba Mouthpiece||None (0)|
|Quadrant 3||Chimes Mallet||None (0)|
|Quadrant 4||Control||None (0)|
A. Restating The Hypothesis:
If bacteria grows on the mouth piece of all musical instruments, then the tuba is the instrument that will have the most bacteria grow on it because it gets a lot of moisture on it.
B. State The Answer To Your Question:
The answer to our question is unknown. Due to an error in our data collection we were not able to conclude which band instrument grew/contained the most bacteria.
C. Explain Your Results Using Data
Our results are unknown due to an error in our data collection.
D. Describe Any Observed Errors
We had an error in our data collection, instead of rubbing the cotton swab from each source on the bottom of the petri dish, we rubbed the cotton swab on the lid of the petri dish.
E. Describe Any Modifications to Improve the Experiment
Some modifications that could be done to improve the experiment would be to correctly collect the data by rubbing the cotton swab from each source and swab the bottom of the petri dish instead of the lid. Another modification that could be done to improve the experiment would be to measure the bacteria’s growth daily in the incubator to see/measure how bacteria grows over a period of time.
F. Describe What Could Be Done Next to Further Study the Topic
To further study this topic, we could measure the growth of the bacteria daily, we could conduct this experiment on other band instruments/surfaces, and lastly, we could test/swab the instruments after a holiday or summer break to see if the amount of bacterial growth is affected by a gap in the use of the various instruments.