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The Effect of Solute Concentration on the on the Change in Mass of a Cell

Info: 3255 words (13 pages) Example Research Project
Published: 1st Dec 2021

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Tagged: Biology

Background Knowledge


Cell membranes are semi-permeable by nature, and they exert control over substances passing through it, into and out of cells. During diffusion, substances will diffuse from an area of high concentration to an area of lower concentration, through a semi-permeable membrane, thus forming a concentration gradient. Solutes will continue diffusing until an equilibrium on both sides is reached.

Water will also diffuse through membranes during osmosis. The net movement of water is from an area of high solute concentration to an area of low concentration. This is what this experiment mostly covers.


Homeostasis is the maintenance of physical (for example, temperature or state of matter) and chemical (e.g., enzymes) conditions inside the body. One of the major factors that must be regulated is fluid balance. A lack of, or excess amount of fluid in the body can be fatal. Our body regulates fluid through two methods, osmoregulation and through our regular behavior.

Dialysis Tube

The dialysis tube acts as a model of a cell. It is an artificial semi-preamble membrane that is used when separating substances. In this experiment, the dialysis tube facilitates the flow of tiny molecules in the sugar concentrated solutions. It will be used in this experiment as it is a good teaching aid that demonstrates the principles of osmosis and movement of molecules across a semi-preamble membrane, as it will be extremely difficult to do the same experiment on an actual cell.

If the concentration of water is higher in the beaker than inside the tubing, the water will move into the tubing and it will increase in size. If the concentration of water is the same then there will be no movement and no change in size.

Aquaporin and Osmoregulation

Aquaporins, also known as ‘water channels’ are essential membrane proteins that forms pores in the membrane of cells. Aquaporin’s job is mainly transporting water between cells.

Osmoregulation is the active regulation of an organism’s body fluids through the regulation of osmotic pressure. Aquaporins are there to maintain the homeostasis of an organism’s fluid. Its job contains maintaining the fluid balance and concentration of electrolytes (salt concentration solution) to make sure that the organism’s body fluids are not too diluted or concentrations.

Osmotic pressure is the measure of the tendency of the movement of water into another solution by osmosis. The higher the osmotic pressure of a solution, the more water is attracted to it. The pressure must be exerted on the hypertonic side of a semi-permeable membrane in order to prevent diffusion of water from the side containing water.


In this experiment, the dialysis tubes are modelling what is happening in an organism’s cells. The objective of this experiment was to measure diffusion through dialysis tubes (artificial semi-permeable membrane). Certain smaller molecules, including water molecules, can move freely through a semi-permeable membrane, but larger molecules won’t pass as freely, and sometimes not at all. The size of the small pores in the dialysis tubing determines which substances can pass through the membrane, acting much like a filter.

Homeostasis is investigated in this experiment. When there is a high solute concentration in our cells, our cells use a protein called aquaporins to control water levels within the cell. This process is called osmoregulation.


If the concentration of the solution inside the cell differs from the concentration outside the cell, then there will be a net movement of water in or out of the cell.


Independent Variable

The IV is the concentration of sugar solution which the dialysis tube is placed in. It will change it by adding 5% of sugar concentration each time, starting from 0% concentration. The levels will be: 0%, 5%, 10%, 15%, and 20%. We will measure this by measuring by using same amount of water each time, and using that, each time we will calculate what percentage of sugar is needed, and then by using basic math figure out how much sugar is needed to make that percentage of sugar concentration with water.

We will do this experiment 3 times to ensure accuracy of the results. The variations of the IV will help answer my hypothesis as I would need to know how the sugar concentration solution affects the rate of osmosis.

Dependent Variable

The DV in this experiment is the change in mass of the dialysis tubes (which represent cells).

I would be measuring the change in mass of the dialysis bag by first measuring the mass before the experiment, then measuring the mass after the experiment. Using simple math, I would then deduct the final mass (after the experiment) from the initial mass (mass before the experiment).

Controlled Variables

Controlled Variables

How I will you ensure that the variable is controlled

Why will you ensure that the variable is controlled

The heat of the water (temperature)

I will ensure that this variable is controlled by using a water bath to control the temperature of the water to 37 degrees, which is the normal temperature of a cell. Furthermore, I will do this experiment in an environment where the temperature is controlled, for example, the biology lab where temperature is kept at a close constant.

A higher the temperature helps diffusion. As this specific experiment experiments on diffusion, not controlling the temperature of the water would make this experiment an unfair test. The temperature not being controlled will affect my experiment, thus making it hard to measure a direct relationship between my IV and DV.

Volume of the dialysis tube

I will control the volume of dialysis tube by using the same type of dialysis tube. On top of that, I would also measure with ruler the size and making sure that for each dialysis tube, the size is the same.

As stated before, in this experiment the dialysis tubes represent cells. If there is more volume in a dialysis tube, then it would have more space for water to flow through, thus creating another factor to complicate the experiment. As the rule is with all CV, if this variable wasn’t controlled, then it would make it hard for us to find a direct relationship between the IV and DV.

Time the dialysis tube is put inside water

By using a timer, I will make sure that the dialysis tube is put inside the sugar concentration the same amount of time.

This factor must be controlled, because just like with other CV, this is another variable that can affect the experiment. The more time the dialysis tube stays in water, the more mass it would lose. Not controlling this variable will make it hard for us to find a direct relationship between IV and DV.


5x 5000ml water

5x Graduated cylinder

5x beakers

Petri dish:1.27g

1x Scale

1x Stirring rod

5 x dialysis tubes (1 containing 0% of sugar concentration, and remaining 4 containing 10% of sugar concentration)

4x solutions of sugar (0%, 5%, 10%, 15%, 20%)

1x ruler (at least 30cm)

1x water bath

1x petri dish

1x timer

1x note taking device (laptop or notebook and pencil)


  1. Obtain the 5 beakers with their concentration of sugar of 0%, 5%, 10%, 15%, 20%. This will be your IV.
  2. Obtain 5 dialysis tubes. One will have 0% sugar concentration, while the rest (4) will have 10% sugar concentration.
  3. Place the petri dish on the scale, and make the weight 0g (tare the weight of petri dish)
  4. Place the 0% sugar concentration dialysis tube onto the petri dish and measure its mass, and record it on a note taking device. This will be the initial mass of the dialysis tube.
  5. Remove the dialysis tube and wait for the petri dish to go back to 0g. If applicable, clean the petri dish.
  6. Repeat step 4-5 four more times with the remaining dialysis tubes which all have 10% sugar concentration.
  7. Put the 0% sugar concentration dialysis tube into the 0% sugar concentrated water. The placement should look like this:

  1. Place the remaining 4 dialysis tubes, with their initial mass already recorded, into four beakers. The four beakers will have concentration of sugar as: 5%, 10%, 15%, 20%
  2. Start a timer and leave the dialysis tubes in the beakers for 24 hours.
  3. After a day has passed, tare the scale once again
  4. Remove the dialysis tube with 0% sugar concentration, and put it onto the petri dish. Record the mass of the dialysis tube. This will be the ‘final mass’
  5. Clean the petri dish, if applicable and tare the scale once again
  6. Repeat steps 11-12 four times with the other dialysis tubes in the other beakers (5% sugar concentration dialysis tubes extract from the 5% sugar concentrated water, and so forth until it reaches 20%)


Raw Data Table


% Concentration of Sugar Solution






Mass before (g)





Mass after (g)





Change in mass (g)





Processed Data Table

% Concentration of Sugar Solution

Change in mass (g)









Graph and Graph Analysis

Graph 1: How does osmosis affect the internal condition of the cell?

The graph above (Graph 1) shows a linear graph. This graph shows the relationship between the concentration of sugar solution to the change in mass. The graph is in a slow decline, as the slope is not steep. My graph indicates that as the concentration of sugar solution increases, the change in mass will decrease slowly.

In “% of concentration to change in mass” graphs, where the line crosses the horizontal axis is the isotonic point. By looking at Graph 1, we can see that the sugar concentration within the beaker was isotonic.


In this experiment, we expected that a net movement of water in or out of the dialysis tube would occur if the dialysis tube was placed in a different solute concentration then it’s internal environment.

The data supports my hypothesis. In this experiment, there is evidence that the difference in concentration between the inside and outside of the dialysis tube (which is representing cells) caused water to move in or out of it.

By looking at the results of the experiments, you can see that there is a decrease in the mass of dialysis tubes. This means that in a non-isotonic environment, there will be a net movement of water in or out of the cell to reach an equilibrium. Thus, this supports the validity of my hypothesis.

It is important for an organism, especially a mammal as mammals are warm blooded, to maintain a constant internal environment for its survival. The organism’s body maintains a constant internal environment through a process called homeostasis. An example of homeostasis is osmoregulation, as osmoregulation is the way osmosis is controlled by the animal to maintain a constant fluid balance.

If there is a higher solute concentration surrounding the cell, the water will then diffuse out of the cell and plasmolyze it. If there is a higher solute concentration inside the cell, the water will diffuse inside it and it would cause the cell to become lysed. Both conditions are not healthy for the cell. Thus, homeostasis and specifically osmoregulation is very important to make sure that the cell is isotonic with its external environment to prevent unhealthy situation from happening.

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