Osmosis Egg Lab

The Effect of Salt Solutions on Decalcified Eggs

Background Information:

     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.

Question:

How does the mass of an egg change after being submerged in distilled water, 5% salt solution & 15% salt solutions for 15 minutes?

 

Hypothesis:

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.

 

Procedure:

  1. 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.
  2. Record the weight of each egg on the data table (on Mac Airs)
  3. Get three 250mL beakers
  4. Fill one beaker with 150mL of 5% salt solution and, using a Sharpie, label it (indicating it’s the 5% salt solution)
  5. Fill a second beaker with 150mL of 15% salt solution and, using a Sharpie, label it (indicating it’s the 15% salt solution)
  6. Fill the 3rd beaker with 150mL of distilled water and, using a Sharpie, label it (indicating it’s the distilled water)
  7. 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.
  8. 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.
  9. Record the weight of each egg and any qualitative data in the appropriate column on the data table.
  10. Subtract the ‘before’ mass from the ‘after’ mass and record the results in the appropriate column on the data table.
  11. Calculate the percent change in mass and record it in the appropriate column on the data table.
  12. Analyze the information using a data table.

 

Variable:

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.

 

Control:

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.

 

Constants:

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.

 

Materials:

 

Data Table:

How Does the Mass of  Decalcified Eggs Change After Being Submerged in Distilled Water and 5%  &  15% Salt Solutions for 15 Minutes?

Graph:


Conclusion:

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.

 

Works Cited: 

“The Cell Membrane”. Web. 3 November  2011.

<http://people.usd.edu/~bgoodman/Membrane.htm>

“Biology 12 – Cell Membrane and Cell Wall Function: Chapter Notes”. Web. 3 November 2011.

<http://www.bio12.com/ch4/Raycroft%20Cell%20Membrane%20HO.pdf>

“Cell Organization and Functions”. Web. 3 November 2011.

<http://www.williamsclass.com/SeventhScienceWork/CellsOrganization.htm>

 

 

November 9, 2011.     Category: *Uncategorized, Labs, Unit 2- Diffusion and Transport.   2 Comments.

Elodea Lab

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.

October 31, 2011.     Category: *Uncategorized, Labs, Unit 2- Diffusion and Transport.   No Comments.

Transport Animations

How Diffusion Works. Answer questions below.

  1. Even in a lump of sugar, molecules are in constant what?

They are in constant motion.

  1. How is diffusion defined? (Use concentration gradient in your answer).

When molecules move from areas of their higher concentration to areas of their lower concentration.

 

How Facilitated Diffusion Works.  Answer questions below.

  1. What does the protein do in facilitated diffusion?

The protein acts as a selective corridor or passageway which helps molecules move across the membrane. The proteins change shape and form to do this.

  1. How much energy does facilitated diffusion require?(There must be a concentration gradient for particles, like amino acids, to move with diffusion).

No input of energy is required.

 

How Osmosis Works. Answer the following questions below.

  1. How is osmosis different than other forms of diffusion?

Osmosis just moves water across the cell membrane, not other molecules.

        2.  What particles move in what direction during osmosis?

Water moves in the direction where there is a high concentration of solute….and a lower concentration of water.

         3.  How does this compare to diffusion?

Diffusion is when any particles move from areas of high concentration to areas of low concentration.

  1. What happens to water molecules in an isotonic solution?

In an isotonic solution the concentration of solute is the same on both sides of the membrane. An isotonic solution neither gains or loses water.

 

How the Sodium-Potassium Pump Works. Answer questions below.

  1. What must be supplied to move sodium and potassium across the cell membrane from low concentration to high concentration?

ATP energy

  1. Is this example passive or active transport?

Active transport

  1. How is this different than diffusion or osmosis?

The molecules are moving against the concentration gradient and require ATP.

 

YouTube Endocytosis & Exocytosis Animation. Answer questions below.

  1. Is this video demonstrating passive transport or active transport?

Active transport

  1. Large molecules produced by the cell leave through what?

Exocytosis.

  1. Describe the difference between endocytosis and exocytosis.

Endocytosis is when particles/molecules are coming into the cell. Exocytosis is when particles/molecules are released from the cell.

October 26, 2011.     Category: *Uncategorized, Homework, Unit 2- Diffusion and Transport.   No Comments.

Practicing Scientific Methods and Graphing

 

 

Questions:
1. What questions is the farmer asking about clams and water temperature?

The farmer is asking: What is the optimum temperature for clam development?

OR

Do clams reproduce better  in warmer or cooler water?

 

2. What would be some of the constants for the farmer’s experiment?

Some of the constants for the farmer’s experiment would be the location, the type of ecosystem (water),  and the type of clam.

 

3. What is the farmer’s independent variable?

The farmer’s independent variable is the water temperature.

 

4. What is the farmer’s dependent variable?

The farmer’s dependent variable is the number of clams that are developing from fertilized eggs.

 

5.What would be an appropriate hypothesis for the farmer’s question (If, then)?

If the temperature of the water is warm, then more clams will develop from fertilized eggs.
6. What is the optimum temperature for clam development?

The optimum temperature for clam development is 30 degrees celcius.

October 25, 2011.     Category: *Uncategorized, Homework, Unit 2- Diffusion and Transport.   No Comments.

3.5 Review

1. How is active transport different than simple diffusion and facilitated diffusion?

Simple diffusion and facilitated diffusion are the flow of small molecules along the concentration gradient and do not require energy. Active transport is the flow of larger or charged molecules against their concentration gradient and does require the input of energy.

2. How is active transport similar to facilitated diffusion?

They both require a transport protein to move molecules across the cell membranes.

3. List two characteristics that almost all transport proteins share.

1) All transport proteins span the membrane

2) Most transport proteins change shape when they bind to a target molecule or         molecules.

4. List the key distinguishing feature of active transport.

Active transport requires chemical energy to move substances against its concentration gradient.

5. Most active transports proteins use energy from the breakdown of _______.

ATP

6.  A cell may transport a substance in________________ if the substance is too large to cross the membrane

vesicles

7. During endocytosis the vesicle membrane fuses with a lysosome and the membrane and its contents are broken down by_______________.

lysosomal enzymes

8. What term means “cell eating” and describes a type of endocytosis?

Phagocytosis is a type of endocytosis in which the cell membrane engulfs large particles.

9. The prefix exo- means “out of” and the prefix endo- means “taking in” how do these meanings relate to the meaning of exocytosis and endocytosis?

Exocytosis is the release of  (out of) substances out of a cell by the fusion of a vesicle with the membrane. Endocytosis is the prices of taking liquids or fairly large molecules into a cell by engulfing them into a membrane.

10. What process drives molecules across a membrane against a concentration gradient?

Active transport

Complete the following table:

Endocytosis Characteristics Characteristics of Both Exocytosis Characteristics
Brings molecules in the cellLeads to the creation of vesiclesPrimary function is getting nutrients

Phagocytosis (a type of endocytosis) engulfs pathogens in the body and destroys them

Both use vesicles for their molecular transport  Releases/expels moleculesLeads to the destruction of vesiclesPrimary function is expelling waste

October 25, 2011. Tags: , , .    Category: *Uncategorized, Homework, Unit 2- Diffusion and Transport.   No Comments.

Starch and Iodine Lab

Background:

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.


Question:

Does starch or iodine travel through a semi-permeable membrane (dialysis tubing)?

 

Hypothesis:

If iodine travels through a semi-permeable membrane, then diffusion will occur and the starch will turn into the tea-color.

 

Procedure:

  1. Soak dialysis tubing in water for approximately 2 minutes.
  2. Open dialysis tubing by placing between fingers and moving fingers back-and-forth.
  3. Tie dialysis tube by twisting, folding and using piece of string approximately 2 cm from edge.
  4. Add 8mL of starch solution to dialysis tube.
  5. Seal second end of dialysis tube by repeating Step 3.
  6. Rinse dialysis tube well with water and dry gently.
  7. Fill 250mL beaker with 200ml of water.
  8. Add drops of idoine until the soluiton is “tea” color, approximately 20 drops.
  9. Place dialysis tube in the beaker and let sit for 20 minutes.
  10. Record observations every 5 minutes.

 

Prediction:

I predict that the  iodine will diffuse through the dialysis tube and react with the starch, causing the starch to become the “tea” color.

 

Observations:

 

Conclusion:

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.

October 23, 2011.     Category: *Uncategorized, Labs, Unit 2- Diffusion and Transport.   2 Comments.

Graphing Practice

October 21, 2011.     Category: *Uncategorized, Homework, Unit 2- Diffusion and Transport.   No Comments.

Passive Transport Warm-Up

1. Particles have kinetic energy.

2. Particles move from areas of high concentration to areas of low concentration.

3. During diffusion particles move from areas of high concentration to areas of low concentration.

4. Diffusion does not require energy to occur.

5. Osmosis is the movement or diffusion of water.

6. The cool-aid example.

October 18, 2011.     Category: *Uncategorized, Unit 2- Diffusion and Transport, Warm-ups.   No Comments.

Diffusion Rates Lab

Hypothesis:

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.

 

Procedure:

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

 

 

Conclusion:

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.

October 18, 2011.     Category: *Uncategorized, Labs, Unit 2- Diffusion and Transport.   1 Comment.

3.4 Review

Directions: Describe how particles will move in the following image through simple diffusion.

        

Through simple diffusion, the particles will move through the membrane from the outside to the inside. They will one from a higher concentration to a lower concentration.

 

 

Directions: Define the following terms:

 

  1. Passive Transport:      the movement of molecules across  a cell membrane from areas of higher concentration to areas of lower concentration without energy input from the cell.
  2. Diffusion:      the movement of molecules in a fluid or gas from a region of higher concentration to a region of lower concentration. It results from the natural motion of particles which causes molecules to collide and scatter.
  3. Osmosis:     the diffusion of water across a semipermeable membrane from an area of higher water concentration to an area of lower water concentration.

 

Directions: Describe the type of solution for each red blood cell in the image below

Solution 1: A hypertonic solution has more solutes than a cell. More water exits a cell in hypertonic solution causing the cell to shrivel or even die.

Solution 2: A solution is isotonic to a cell if it has the same concentration of dissolved particles as the cell. Water molecules move into and out of the cell at an equal rate so the cell size stays constant. particles

Solution 3: A hypotonic solution has fewer solutes than a cell. More water enters a cell causing the cell to expand or even burst.

 

 

Define Facilitated Diffusion:

Facilitated Diffusion is the diffusion of molecules across a membrane through transport proteins.

 

Create an image like that above to demonstrate facilitated diffusion.

 

October 18, 2011.     Category: *Uncategorized, Homework, Unit 2- Diffusion and Transport.   No Comments.

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