Gallus Egg Lab Report

Gallus Egg Lab Report
This study focuses on the means by which a Gallus gallus domesticus egg can demonstrate the
process of diffusion and osmosis through a course of distinct chemical changes along with a series
of scheduled measurements. Diffusion and osmosis are a critical aspect of human survival and an
essential concept for biology–students to understand. Diffusion can be characterized as the irregular
movement of atoms, particles, or molecules from a space that has a high concentration to an area
with low concentration. In other words, when there is an abundant amount of molecules in a
particular area, diffusion helps spread out the molecules until equilibrium is attained. Different
elements will affect the rate at which diffusion takes place. The larger distinction
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What Is The Change In Mass Of Sucrose Osmosis
Change in mass of potatoes submerged in sucrose solution
Rebekah Schmitz
Introduction: Diffusion explains the passive movement of materials into and out of the cell due to
the presence of a concentration gradient from a high to low concentration across a membrane
(Choinski and Karafit 2015).
In this experiment, we studied the effects of differing concentrations of sucrose solutions on samples
of potato material over time. This experiment focused on the effects of tonic solutions such as
hypertonic and hypotonic solutions. When a cell is placed in a hypertonic solution, the cell will
shrink due to the water moving from a high concentration to low concentration, but when a cell is
placed in a hypotonic solution, the cell will swell due to the ... Show more content on
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When a hospital prepares an IV saline solution, they have to be careful to use to correct
concentration of sodium chloride. The blood needs to remain isotonic, so the red blood cells do not
shrink or swell. If the solution is simply water, then a concentration gradient is formed and
hemolysis occurs (Osmosis in Red Blood Cells, 2015). Hemolysis is defined as the destruction of
red blood cells which leads to the release of hemoglobin from within the red blood cells into the
blood plasma (Hemolysis). To further test our experiment, one could let the potato samples sit
longer in the sucrose solution to see if time effects the amount of sucrose solution absorbed.
Alternatively, changing other environmental parameters, such as temperature or pH, may have
different effects on the rate of

Osmosis Lab
Lab Objective: The purpose of this experiment is to test the membrane dynamics of different solutes
when placed in unique solutions and the effect this has on diffusion rate, osmosis, and membrane
permeability.
Hypothesis: When observing molecule size in relation to diffusion rate, we expect smaller particles
to farther along the agar because the size of a particle and rate at which it diffuses correlates
negatively. In the osmosis and dialysis experiment, water should move down its concentration
gradient. The sucrose dialysis bag is hypertonic compared the DI water, so it will gain volume; and,
the dialysis bag containing DI water is hypotonic compared to the sucrose solution of the beaker, so
the bag will lose volume. Lastly, because Iodine ... Show more content on Helpwriting.net ...
Active transport can be broken down into two categories as well: primary active transport and
secondary active transport. Primary active transport begins with ligand binding to an integral protein
that can act as a transporter as well as an enzyme. As the ligand binds to the receptor, ATP
hydrolyzes, changing the configuration of the protein so the particle can transfer (Tandy 2016).
Secondary active transport pushes one substance against its concentration gradient in order to move
another substance as well. The energy created from the first substance moving provides energy to
move the second substance (Tandy 2016). When the two substances move in the same direction, this
is called cotransport, while countertransport refers to when they move in opposite directions
(Usenko

Osmosis Lab Report
The purpose of this investigation lab is to observe osmosis of water into a cell, and also how. The
goal of the experiment is to prove that a solution in an area of high concentration will move to an
area of low concentration. In this experiment, we used sheep blood, 0.9% NaCl, 10% NaCl, and
dH2O water to see what will cause when we add this to the cell. We will then record the rate of
diffusion by measuring how did the blood flow into each solution and recording them. First, we
placed with the wooden applicator stick, then dip the end into the blood and place tiny drop in the
slide to observe what it will do. In the first one we place 1 drop of 0.9% of NaCl with the drop of
sheep blood and absorb excess liquid, after that we observe the cells under the microscope with the
40x then 100x and then the high power of 400x, to see what it will look like in each lens. The
second one we put 1 drop of 10% NaCl solution into the slide with the drop of blood and use filter
paper to absorb excess liquid. After that we observe cells under the microscope with the 40x then
100x and then the high power of 400x, to see what this will do differently than the first one.

Investigating Correlation Between A Solution Salt...
The objective of this experiment is to determine the correlation between a solution's salt
concentration and its density. Also, it was to determine what equipment would be best to use to
measure volume precisely and accurately. Densities of a series of known salt solutions were trialed
three times and then averaged. By following the procedures, the unknown salt solution was
determined and graphed to calculate the percent salt to the sample.
Introduction
In chemistry there are often times an unknown solution's density must be determined by using the
known solutions. There are a variety of modern ways that chemist use to determine an unknown, but
the simplest method to find the identity is by using a common equation: p=M/V P is the density, M
is the mass, and V is the volume. Once the densities of the compounds are known, the density of the
unknown solution can be compared to the four known solutions. When being compared the
solutions can be graphed by plotting density against the percentages of NaCl. Once graphed there
will be an evident straight–line graph.
Experimental
General Methods and Materials.
Materials:
1. NaCl solutions of the following strengths: 6% 12% 18%, 24%
2. Unknown brine solution
3. Volumetric flasks, small beakers, droppers, 10 mL and 50 mL graduated cylinders
4. Scale that measures mass in grams
Procedure. For this experiment, the first task was to determine which equipment would be used to
measure volume accurately. The choices were a 10.0 mL

Transmittance Lab Report
The results of this experiment helped understand the concept of absorbance, transmittance and
concentration. Table 1 illustrates the percent transmittance, transmittance, and absorbance of the
made solutions. The percent transmittance was found by placing each test tube in the Spec20 and
recording its value. The transmittance is how much light is absorbed in the solution, its formula is
shown by Equation 1. The absorbance represents the amount of light absorbed by each solution, this
was found using Equation 2. Absorbance can also be found using Equation 4 as well, for this
specific formula you plug the percent transmittance directly in the equation instead of changing it to
transmittance. A calibration curve (Figure 1) was created with the data from Table 1. A calibration
curve for this experiment helps find the actual concentration of the made and unknown solutions.
The absorbance is on ... Show more content on Helpwriting.net ...
Since we were only given the percent transmittance, Equation 4 was used to find the absorbance.
Equation 2 could have been used as well by changing the percent transmittance to transmittance.
Using the equation from the line of best fit (Equation 5) to find the actual concentration of our made
solutions, I plugged in the absorbance for y and solved for the concentration (x). Our concentrations
were off due to the fact that we did not use the proper amount of milliliters from the previous
solution while making each solution. The purpose of this lab was to be able to use percent
transmittance, absorbance and concentration to find an unknown solutions concentration. Table 4
displays these three pieces of data, with them we can figure out what the unknown solutions
concentration is. Comparing the data from Table 1 and data from Table 4, we see that the absorbance
and concentration are similar to 0.2 M. We can conclude that the unknown concentration is close to
0.2

Biology Eei Essay examples
| 2012 | | Toby Rheinberger
Lab Partner: Jeremy Saba |
[The Effect of surface area to volume ratio and solute concentration on osmosis.] | The aim of this
experiment was to test the effect of surface area on osmosis and the effect of solute concentration on
osmosis. To test this aim to hypotheses were devised. 1) If potato pieces are immersed in various salt
solutions, then the pieces with the greatest surface area to volume ratio will experience the greatest
weight change, because more water can move by osmosis across the potato cell's semi permeable
membrane. 2) If potato pieces are immersed in various salt solutions, then the pieces immersed in
the most concentrated solution will experience he greatest weight change, ... Show more content on
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A semi permeable membrane. (Biology at Aveson , 2007)
Figure 2. A semi permeable membrane. (Biology at Aveson , 2007) pass through the membrane and
into the cell.
Despite its importance osmosis may also damage cells by causing them to; a) shrink from water loss
or b) burst from too much water gain. Plant cells [fig 3] have adapted themselves to ensure that
these factors do not affect them, by forming a ridged wall, known as the cell wall, around their cells.
The cell wall maintains the shape of the cell, and prevents the cell from bursting in a hypotonic
medium by resisting water pressure. Plant cells have also adapted a larger vacuole, which occupies
80% or more of the cells cytoplasm (Davidson, 2004); allowing plants to store more water and
nutrients per cell. Vacuoles also play a structural role in plant cells; by swelling when liquids contact
them, plant vacuoles are able to control turgor pressure within the cell. This helps maintain the
structural integrity of the cell as well as providing the plant with suitable amounts of water and
nutrients; however the cell will never burst because the vacuole is contained within the cell wall. If
plant cells are deprived of water their vacuole will begin to shrink, yet due to the cell the wall, the
plant cell will be able to maintain its shape. [fig.4] Animal cells [fig 5] on the other hand do not have
this

Essay on Osmosis
Osmosis
Introduction
I obtained my scientific knowledge from the "AQA science: double award modular." This is an
experiment that will explore the osmosis theory
To make this experiment safe I will wear goggles to protect my eyes and also will make sure that
when using the borer to cut the potato I have a white tile underneath.
The experiment will be a fair test as I will keep the weight of the cut potato the same throughout the
water and sugar solutions. I will try to keep the potatoes surface area and volume the same. I will
make sure that both pieces of potato are placed in their solutions at the same time and pulled out
together. I will make the weight of the potatoes the same by ... Show more content on
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Example:
[IMAGE][IMAGE][IMAGE][IMAGE][IMAGE][IMAGE]In the root–hair cell of a plant the
surrounding soil will have a more concentrated water solution than inside the cell, thus leading to
the water solution in the soil osmosing into the root–hair cell. The cell which is next to the root hair
cell is separated by the partially permeable membrane. This leads to the more concentrated water
solution in the root hair cell osmosing into the cell next to it. This procedure of osmosis through the
cells carries on until the cells reach the xylem. The xylem finally carries the water solution upwards
to the areas that need it.
I have decided to vary the amount of sugar solution (%) in which the potato chip is placed in. I am
using this variable as I think it will give a good range of results because unlike changing the size of
the potato this should prove that the theory of higher concentration to a lower concentration is right.
To ensure accuracy I will use equipment such as the top pan balance to measure the weight of the
potatoes accurately. I will also use a pipette to measure the about of solution I put into the test tubes.
The drop sizes of the solution vary so I will try and keep them roughly the same size but there
possible could be some inaccuracy located. I will use a borer to cut parts of the potato and a white
tile

Osmosis And Its Effects On The Mass Of An Egg Lab Report
The effect of different environments on the mass of an egg
Daija brown
Abstract: Different environments have different effects on the mass of an egg. This was tested by
placing the egg in vinegar, corn syrup, and tap water. The vinegar increased the mass, the corn syrup
decreased the mass, and the tap water increased the mass. This is due to the different concentrations
of each substance and the process of osmosis. Particles of higher concentration move to regions of
lower concentration.
Introduction: Osmosis is the diffusion of water through a selectively permeable membrane. Due to
the difference in concentration gradient, particles that are dissolved can go from high concentration
to low concentration through a semipermeable membrane. The purpose of the investigation was to
observe the effects vinegar, corn syrup, and water has on the mass of an egg. For the vinegar my
hypothesis was, if the egg is placed in vinegar , then the shell will dissolve. For the corn syrup my
hypothesis was, if the egg is placed in corn syrup, then the egg will form a new shell. For the tap
water my hypothesis was, if the egg is placed in tap water, then the egg will stay the same.
Materials and Procedure: For the experiment the supplies we used were:
One egg
Vinegar
Corn syrup
Tap water
Masking tape
An empty cup
A balance
The procedure used in the investigation was:
On day 1: Used the masking tape to label the empty cup
Took the initial mass of the egg
Measured the initial

An experiment to investigate the effect of changing the...
An experiment to investigate the effect of changing the concentration of hydrochloric acid on the
rate of reaction with sodium thiosulpahte.
Aim
I am trying to find out how the concentration of hydrochloric acid affects the reaction time with
sodium thiosulpahte.
If a clear solution of sodium thiosulphate is mixed with hydrochloric acid then a reaction will occur
where a creamy–yellow precipitate of sulphur is produced.
H20
Text Box: + Text Box: + Text Box: + Na2S203 2HCL
Text Box: + 2NaCl SO2 S
Reactants Products
The sulphur takes a while to form and starts to turn the water cloudy after a short time. You cannot
tell when the reaction has stopped so a cross drawn on a piece of paper in pencil can be ... Show
more content on Helpwriting.net ...
They all increase the number of collisions.
Fair testing
* To make this a fair test I will only change one variable.
* Each experiment will be done three times so an average can be found.
* The same equipment will be used in each experiment.
Method
· Measure out 50ml of thiosulphate and add it to the conical flask.

· Draw a cross on a piece of paper and place the conical flask on it.
· Measure out 10ml of acid and add it to the conical flask and start timing.
· Look into the top of the conical flask and when the cross can no longer be seen stop the stop
watch.
· Do the experiment again but measure out 8ml of hydrochloric acid. Then add 2ml of distilled
water. The ratio for each experiment will therefore be 10:0, 8:2, 6:4, 4:6 and of acid to water
Conclusion
I conclude that whilst carrying out this investigation I found out that the more acid was in the
solution the cross on the paper disappeared quicker. This shows that the more concentrated the acid
is then the quicker the reaction time. My research about the collision theory was correct and the
more concentrated the solution is then the number of collisions will be increased. If the numbers of
collisions are increased then the reaction time will also be increased.
My prediction 'that the less concentrated the acid is then the longer it will take for the sodium
thiosulphate to react with the acid' is correct because experiment number 5

Biology Lab on Osmosis
Michelle Kim
2012.1.20
Biology
Iso–osmolar Concentration of Carrot Cells Lab
INTRODUCTION
For a more thorough understanding of this lab introduction, the concepts of, iso osmolar, membrane,
equilibrium, and concentration gradient evaluated. Iso osmolar can be known as the point in which
the substance of experimentation faces no change despite the amount of solute inside the solvent(
because the solvent and solute concentration is equal. This is where the line on a graph would cross
on the x–axis). Selectively permeable membrane can be defined as a microscopic double layer of
lipids and proteins that bounds cells and organelles and forms structures within cells and it controls
what comes in and out of the cell. Equilibrium is the ... Show more content on Helpwriting.net ...
The result showed the affects of osmosis, where particles moved to achieve equilibrium. The 0.2 M
sucrose had a +8.37% change in mass as a class average and our group and +9.52%. The initial mass
for these carrots were 2.1g and became 2.3g when the 2 days had ended. This happened because of
the same reason for the 0.0 M sucrose, where the concentration of particles were greater inside the
cell compared to the outside of the cell, causing the water molecules moving into the cell through
the semi permeable membrane. However, the solution the carrots were dunked into had a 0.2 M
sucrose, unlike the first solution. This caused the mass of the carrot to change less because there was
already some particles outside of the cell, allowing for equilibrium to become reached a lot faster
than 0.0 M sucrose. In 0.4 M sucrose, there was a class average of –2.34% and we had –4.76%. The
initial mass was 2.1g with a final mass of 2.0g. This was the first solution to have a negative
percentage change in mass from the other six. There was a decrease in mass of the carrot because
water had moved from the carrot sticks into the solution around is trying to achieve equilibrium. The
water molecules had moved from high concentration into lower concentration, moving our of the
carrot cells, causing them to

Taking a Look at RiboTALE Networks
RiboTALE networks studied were comprised of two modules and each module was based on a
separate plasmid. One of the modules, repressor module, expresses fusion of riboswitch (RS) from
an inducible PBAD promoter, second module, target module, contains a TALE binding site (TBS)
downstream of a regulated or constitutively expressed promoter. Ribosome binding site (RBS) and
gfp are present downstream to TBS (Fig. 1). gfp was used as proxy to quantitate the promoter
expression. Repressor modules were based on kanamycin resistance medium copy number BioBrick
plasmid, pSB3K3 (Shetty et al, 2008). Four different types of repressor modules, PR1T1, PR1T8,
PR2T1, and PR2T8, were constructed combinatorially by implementing two types of riboswitches
and two types of TALEs (supplementary table 1). Used riboswitches, Riboswitch–1 and 2, were
theophylline inducible (Lynch & Gallivan, 2009; Topp et al, 2010). TALES with either low
dissociation constant (Kd = 1.3±0.3, T8) or with high dissociation constant (Kd= 240±40, T1) were
selected (Meckler et al, 2013). Each of the TALEs recognizes a specific 19 bp target DNA
sequence.Two families of target modules were constructed. One of the modules was based on
regulated tetracycline promoter, PTet, and other was based on a family of constitutive promoter,
PConst, generating different level of outputs. Target modules contain two types of TALE binding
site, TBS1 or TBS8. TALE1 binds with TBS1 and TALE8 binds with TBS8 (Fig.1). All
measurements were

Β-Galactosidase Lab Report
With both the stock substrate and varying enzyme solutions prepared, the Spec20 spectrophotometer
was used to investigate the enzymatic activity of β–Galactosidase through an absorbance–based
assay. Using LoggerPro software on the computer to analyze the absorption data, the Spec20 was
calibrated before each run with 0.5 mL of the tested enzyme concentration at an absorbance of 420
nm. Data collection was then started, instantly followed by the addition of 0.5 mL of the stock 2.5
mM substrate solution, topping off the 1–mL cuvettes. Each of the nine varying enzyme
concentrations were split between the team and run for a total of 10 minutes. Upon completion, data
from each varying enzyme concentration was copied to a single Excel sheet and used ... Show more
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From the stock substrate solution of 2.5 mM, each group serially diluted at least one different
substrate concentration for a total of four different substrate concentrations to be investigated: 1.25
mM, 1.0 mM, 0.75 mM, 0.25 mM. The enzyme concentration was kept constant at 2.0 mM while
experimenting on the affect of varying enzyme concentration on the rate and product formation of
ONP. Enough 2.0 mM enzyme solution was prepared in the previous part of the project to supply
this assay. Using similar procedure to collect absorbance data as the first part, 0.5 mL of 2.0 mM
enzyme concentration was placed into the cuvette and used to calibrate the spectrometer at 420 nm.
Data was then started, with the immediate addition of 0.5 mL of varying substrate concentrations.
Each varying substrate concentration was split between the team and run for a total of 10 minutes,
with the exception of the 1.25 mM run. Upon completion, data from each varying substrate
concentration was copied to a single Excel sheet and used to produce an absorbance vs. time graph,
product formation vs. time graph, Michaelis Menten plot, and Lineweaver–Birk plot. This analysis
was used to calculate the V0,Vmax, and Km for β–Galactosidase

Egg Osmosis Lab
YEAR 11 SACE BIOLOGY THE EFFECT OF CONCENTRATION GRADIENT ON THE RATE
OF OSMOSIS Introduction: Water is the most abundant substance in any cell. All the chemical
processes of a cell involve water in some way. Osmosis is diffusion of water between a cell. The
following factors could affect the rate of osmosis: the solute concentration inside and out of the cell.
A raw chicken's egg with the shell removed can be used as a model of a cell to demonstrate the
process of osmosis.
Aim:
To determine the effect of changing the concentration gradient on the rate of osmosis. Hypothesis:
As the solute concentration of the water the egg is placed increases, the egg will shrink as the water
from inside it uses osmosis and moves towards the salt outside. ... Show more content on
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Pick up an egg with the spoon and carefully rinse it under tap water.
2. Place it on some paper towelling and blot it dry with another piece of paper towelling.
3. Place the egg on a watch glass and weigh it. Record the mass.
4. Place the egg in a beaker containing 10% salt solution ensuring there is enough liquid to cover the
egg.
5. Leave it in the solution for 10 minutes and then remove it with a spoon, pat it dry as before and
weigh it again. Record this mass.
6. Repeat steps 1 to 5 using the 5% salt solution, 2.5% salt solution and then distilled water.
7. Calculate the percentage mass change using the following formula: % mass change = Change in
mass x 100 Initial mass 1
Results:
1. Present the data in a table which includes the % mass change.
The effect of change in solute concentration on the mass of an egg Solute Concentration (%)
Initial Mass (grams)
Final Mass (grams)
Mass Change (grams)
Mass Change (%)
0
78.34
78.34
0
0

10
78.34
75.56
2.78
3.55
5
75.56

The Effect Of Osmosis On Potato Cells
BIOL 1208 Writing Assignment 2
Results and Discussion
These are to be written individually even if you worked in a group. This should be written in
paragraph format. Each section should include the answers to the following questions
Lab Topic: Effects of Osmosis with Potato Cells in Sucrose Solution
Introduction:
Small slices of potato are placed in six concentrations of sucrose: 0.0 M, 0.1 M, 0.2 M, 0.3 M, 0.4
M, and 0.5 M. The initial mass of potato is noted before being placed in the solution. The potato
remained in the solution for one hour then the potato was removed from the solution and dried off
before the final mass was noted. The null hypothesis is that the solutions of different molarities of
sucrose will not affect the ... Show more content on Helpwriting.net ...
Once the hour was over, the potato slices were removed from the solution, dried off in a paper
towel, and weighed once again. In the 0.0, 0.1, and 0.2 M of sucrose concentrations the final mass of
the potato increased. Therefore H2O enters the potato cell because the solution is hypotonic. In the
0.3, 0.4, and 0.5 M of sucrose concentration the final mass of potato decreases. The solution is
hypertonic which causes H2O to exit the potato cells through the semipermeable phospholipid
bilayer. The results prove that the different molarities of concentration affect the mass of the potato
in different ways; the mass of potato either increased or decreased depending on the molarity of
sucrose concentration.
Caption: The line of best–fit shows the pattern that as molarity increases in the sucrose
concentration then the percent change in mass of potato decreases.
Discussion:
Determining the Osmotic Concentration of Potato Tuber Cells When Placed in Sucrose Solutions
In this experiment, the osmotic concentration is found with potato slices placed in sucrose solutions.
Osmosis in this model is the net movement of water between the potato cell and the sucrose
solution. The movement of water is determined by the molarity of sucrose. As the molarity of
sucrose increased then the concentration in the solution also increased. H2O will move through the
cell membrane to areas of higher concentration in order to reach equilibrium. If cells are placed

S. Albus Experiment
The aim of this experiment was to observe the effect different concentrations of Dettol had on the
growth of S. albus. The highest concentration of Dettol (100% – D) was predicted to be the most
effective solution to kill and stop the growth of this particular bacteria. This can be justified as S.
albus is gram–positive; easier bacteria to kill, developing resistance slower than a gram–negative.
Therefore, the highest concentration of antibiotic should be – and was – the most successful in
changing the pH levels, and killing S. albus. Through discussion and analysis of the results collected
through the experiment, it can be concluded that there is higher chance of killing and the growth of
this particular bacteria if 100% Dettol is used.
When ... Show more content on Helpwriting.net ...
It can be inferred that the more Dettol used, the larger the zone of inhibition will be. This was
predicted as Dettol is an antibacterial solution, therefore the higher the concentration is, the more
effective the substance is in killing bacteria; S. albus. Though the 100% Dettol concentration (D) is
clearly the most effective in stopping the growth of S. albus (average zone of inhibition diameter
being 20.12 mm), it may not be a healthy choice if using the product on skin. After looking closely
at the Agar plate, 'D' had burned away at parts of the disc, evidencing that using 100% Dettol on
your skin may lead to burns, justifying why it is not recommended on the company's bottle or
official website.
Through the analysis and understanding of the results table and concepts, it is evident that the
smallest diameter is solution 'A' (6mm) which can be compared to the largest diameter – that being
solution 'D' (23mm). This idea assists in justifying the hypothesis and research question as S. albus
was able to successfully grow in the incubator (hence why humans or warm–blooded animals are
generally the host of this bacteria), and the highest concentration of antibiotic was able to kill the
gram–positive

Osmosis To Determine The Mass Of A Potato
If I am to figure out the estimated concentration of sucrose found naturally inside the potato using
osmosis, then there will be a change in the mass of the potato because it will either be hypotonic or
hypertonic to the cell, as it will affect the size of the cell along with in. The 0 M, 0.2 M, and 0.4 M
solutions were are hypotonic to the cell. Hence, they contained less solute than the potato itself.
Therefore, the potato cell increased in mass because water was moving into the cell. It was more
concentrated had less water in comparison with outside the cell, so it moved in. However, the 0.6 M,
0.8 M, and 0.10 M solutions were hypertonic to the cell, causing the potato cell to decrease in mass
as water was moving out of the cell. There was ... Show more content on Helpwriting.net ...
This was because not all of them looked the same in shape. Also, for at least two of the slices we
used the curved ends of the potatoes, this meant parts that should have been there were already not
present for a cube shape. This was bad because much mass was eliminated in comparison with the
cube shaped ones actually coming from the middle area of the potato, where you get a sufficient
amount of the potato to actually use. This could remove more mass. Also, we only peeled off
enough skin as we could, although we did not have too. Some parts were too hard to peel, therefore,
a few potatoes had most of their skin removed, whereas, others mostly had at least half of their skin
remaining. This would also have a role in causing various wrong masses for the initial masses of the
potato, also affecting the final mass after it had been placed into the solution.
The osmosis potato lab is related to some aspect in the human body in that osmosis occurs in our red
blood cells. Osmosis helps us in containing a significant amount of water inside our

Sucrose Control Trial
The control trial for the fermentation rate experiment will be performed the same as outlined in the
procedure for 'Fermentation Investigation Week #1' in the Alverno College CH213 Laboratory
Manual, Fall 2016. In the second trial, the concentration of the sucrose in the reaction mixture will
be changed from 6% to 12%. Prior to conducting our modified experiment, we will first conduct the
original experiment with 6% sucrose that was done in week 1. To obtain 12% of sucrose we will use
10.00 mL of the prepared 6% (m/v) sucrose solution and we will then measure out 0.6 grams of a
solid sucrose. Once we measure out the 6.0 grams of sucrose we will dissolve that into the 10.00mL
of the prepared 6% (m/v) sucrose. This will allow us to obtain

Factors Affecting the Movement of Water Through Osmosis
Factors affecting the movement of water through osmosis
Introduction
In this I will be investigating what effects the movement of water through osmosis. Osmosis is the
diffusion of water. It is the process in which fluids pass through a partially–permeable membrane. It
is the movement of water from high water concentration to low water concentration. Plant cells react
to osmosis by hypertonic, isotonic and hypotonic.
Keywords
Hypertonic – is when the water outside of the cell is lower than that inside.
Isotonic – is when the net movement is the same in both directions.
Hypotonic – is when the water potential outside the cell is higher. Therefore the water has moved
into the cell and the cytoplasm is pushing against the cell ... Show more content on Helpwriting.net
...
I think that using potato cylinders to observe osmosis is the best method to use as it doesn't take too
long to do, it is simple to do and can give reliable results as the potato cylinders would be the same
length and width, and most likely the same weight as well, meaning it would be easier to tell which
solution had the greatest effect on the potato cylinders, by weighing the potato cylinders to see what
the difference in mass is.
Chosen Method 1) I will be taking several samples from the same potato by using a potato corer to
give me eighteen potato cylinders in equal size, a second potato is allowed if it is needed.
(Pictures from: http://eve.kean.edu/~breid/plantlab2/plant_2.html)
2) Then you measure all eighteen cylinders to see what their starting mass is, I will use a set of
scales to do this. 3) Once you have the starting mass you will need to add each potato cylinder to
different solutions, sucrose 0.1m/dm³, sucrose 0.2m/dm³, sucrose 0.4m/dm³, sucrose 0.5m/dm³ and
sucrose 1m/dm³, you will need to make sure there is enough solution to cover the potato cylinder.
4) Then you need to leave the potato cylinders in the solution for a long enough period of time, i.e.
an hour. 5) Once an hour is up you need to take the cylinders out of the solution and dry them off
with some paper towel. 6) After that weigh the potato cylinders again and calculate

Effect Of Sugar On Osmosis
The effect of sugar solutions (%)with different concentration levels on the rate of osmosis in
cylinders of potatoes(+/–0.01g)
Research question:What is the effect of sugar solutions(%) with different concentration levels on the
rate of osmosis in cylinders of potatoes(+/–0.01g)?
Hypothesis:The prediction that is stated is if the concentration level of the sugar solution is higher
then the mass of the potato cylinders will increase (hypertonic)because of a process known as
Osmosis.Osmosis is the chemical process of diffusion which is the passive movement of particles
from a region of high concentration to a low concentration.["Diffusion."And Osmosis]In other
words they allow certain substances to enter the cell and certain substances that ... Show more
Try 3–4 times to get as accurate results as possible.
The electronic mass kept on changing by saying error
This would affect the results as some of the results would be inaccurate
Wait one minute till the electronic mass turns to 0 then place the potato cylinder
Not enough experiments with different independent variables
Wouldn't of got accurate results therefore there would be an outlier in the graph
Repeat the same experiment 3–4 times
Work Cited:
"Plants For Kids."Osmosis in Potatoes Experiment. N.p., n.d. Web. 29 Oct. 2014.
"Osmosis Experiment For Kids: Potato, Water, and Salt."Udemy Blog. N.p., n.d. Web. 28 Oct. 2014.
"Simple Science Experiment: Osmosis with Potato Slices."Simple Science. N.p., n.d. Web. 28 Oct.
2014.
"Diffusion."And Osmosis. N.p., n.d. Web. 29 Oct. 2014.
"Welcome to the Living World.": Comparison between Diffusion and Osmosis. N.p., n.d. Web. 02
Nov.

Osmosis Lab Report
The Rate of Osmosis changing due to the Internal and External concentrations of NaCl reaches and
surpasses the Equilibrium of the egg.
Osmosis is the movement of solutes through a semipermeable membrane from a low concentration
to a higher concentration until the equilibrium is met ("Osmosis – Biology–Online Dictionary",
2017). Through the practical we were measuring the change in mass on the egg as it was placed in
increasing NaCl Solutions starting from Distilled water and ending at 10% NaCl. The change in
mass that is being measured is solely reliant on the the percentage of NaCl that we will be using.
The change in mass is reliant due to the rate of osmosis that the egg will have as it changes from one
solution to the next.
The Diagram above is showing how the movement of a low concentration is moving through the
semipermeable membrane towards the higher concentrated area. Water molecules are able to move
freely around which allows for the process of osmosis without requiring energy for the movement
due to being able to move through the membrane. For this experiment this process will be the
solutions that the egg is placed in, Distilled water and NaCl, to gain the results to predict the rate of
osmosis through the change in mass.
Therefore, the aim of this experiment is to calculate and discuss how the rate of change in mass with
an increase of concentration of sodium chloride will show how the rate osmosis will differ on an egg
due to its internal concentration.

How Does Diffusion And Osmosis Affect The Rate Of Diffusion
Abstract
Cells and molecules in the environment are constantly moving and changing, for cells to function
properly there is a need for equilibrium to be met. The size of the cell and the solution outside of the
cell affects the rate of diffusion and osmosis in the cell. Cells are constantly trying to reach an
equilibrium with the molecules and substances around it, which is why there are such terms as:
hypertonic, hypotonic and isotonic. The procedures allowed testing of whether or not surface area or
volume increased diffusion and how different substance control diffusion. Cells are constantly
moving to reach equilibrium through diffusion and osmosis.
Purpose
The size of a cell and the solution that surrounds it affects the ability

Osmosis and Cell
Name: ______________________________________ Date: ________________________
Student Exploration: Osmosis
Vocabulary: cell membrane, concentration, diffusion, dynamic equilibrium, osmosis, semipermeable
membrane, solute, solvent
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
1. Suppose you were trapped on a desert island with no sources of fresh water. Should you drink
water from the ocean? Explain why or why not.
_________________________________________________________________________
_________________________________________________________________________
2. What do you think would happen if you watered your houseplants with salt water? ... Show more
Observe: Wait until the numbers are not changing very much. What do you notice about the solute
concentrations inside and outside of the cell? _______________________________ This situation
is called dynamic equilibrium.
5. Experiment: Click Reset. Check that the Solute outside is 10 and the Initial cell volume is 40%.
To calculate the solvent concentration, divide the number of solvent particles by the total number of
particles, and then multiply by 100. (Note: The Gizmo only displays the solute concentrations.)
A. What is the solvent concentration inside the cell? ____________________________
B. What is the solvent concentration outside the cell? ___________________________
C. Where is there a higher solvent concentration? _____________________________
D. Click Play. Do most of the solvent particles move into or out of the cell? (Hint: Does the cell
expand or shrink?) ______________________________________________

6. Experiment: Click Reset, and set the Solute outside to 1.
A. What is the solvent concentration inside the cell? ____________________________
B. What is the solvent concentration outside the cell? ___________________________
C. Where is there a higher solvent concentration? _____________________________
D. Do you think the cell will get larger or smaller?

Correlation Between Accuracy And Precision
i. The Objectives of Part A of the experiment include (1) to determine the accuracy and precision of
the variation in mass across packets of sweetener (2) to determine the accuracy and precision
variation in mass across balances (3) to gain a relative understanding of the accuracy and precision
of such values by both comparing the values to each other, as well as to the Nominal mass of the
sweetener (4) to therefore determine the difference by finding the value of the mean and the
standard deviation for each (5) To use such information to understand the relationship between
accuracy and precision ii. The Objectives for Part B of the experiment are as follows: (1) To
determine the mean values and standard deviation for the added volume of (a) Graduated Cylinder
(b) Serological Pipet (c) Volumetric Pipet. (2) To determine the accuracy and precision of (a)
Graduated Cylinder (b) Serological Pipet (c) Volumetric Pipet by comparing the values of the mean
and standard deviation for each volume of 10 mL. (3) To therefore compare such values and further
relay the difference in the use of the equipment. (4) To use such information to understand the
relationship between accuracy and precision iii. The Objectives for Part C of the experiment include
(1) To calculate the dilution necessary for various molarities of a specific solution (2) To use these
calculations to prepare such solutions (3) To use such solutions to understand the relationship
between the concentration of a dilution

Essay on Determination of an Equilibrium Constant
Determination of an Equilibrium Constant
Abstract: In this experiment, two reactions were run to determine the molar absorptivity and the
equilibrium constant of FeSCN2+. The main principles used in this lab are equilibrium, LeChatlier's
Principle, Beer's Law and Spectrocopy. The first reaction was run to completion using LeChatier's
Principle and the second reaction was run to equilibrium. A spectrophotometer was used to measure
absorbances. Using a graph of absorbance versus concentration of FeSCN2+ was used to determine
that the molar absorptivity constant was 3670. Beer's Law was used to determine that the average
equilibrium constant was 33.1793.
Introduction:
The purpose of this experiment is to determine the value of the ... Show more content on
Helpwriting.net ...
LabQuest was not used during this experiment.
Results:
Table 1. Standard Solutions
Solutions 1 2 3 4 5 mL Fe(NO3)3 25.00 25.00 25.00 25.00 25.00 mL HSCN 5.00 10.00 15.00 20.00
25.00 mL HNO3 70.00 65.00 60.00 55.00 50.00
Total mL 100.00 100.00 100.00 100.00 100.00
Intial [HSCN] 3.0*10–5 6.0*10–5 9.0*10–5 1.2*10–4 1.5*10–4
Table 2. Equilibrium Solutions
Solutions 1 2 3 4 5 mL Fe(NO3)3 5.00 5.00 5.00 5.00 5.00 mL HSCN 1.00 2.00 3.00 4.00 5.00 mL
HNO3 4.00 3.00 2.00 1.00 0.00
Total mL 10.00 10.00 10.00 10.00 10.00
Initial [Fe3+] 1.0*10–3 1.0*10–3 1.0*10–3 1.0*10–3 1.0*10–3
Intial [HSCN] 2.0*10–4 4.0*10–4 6.0*10–4 8.0*10–4 1.0*10–4
Table 3. Absorbance of Standard and Equilibrium Solutions
Solutions 1 2 3 4 5
Standard Solutions
Abs 0.135 0.223 0.345 0.456 0.509
Equilibrium Solutions
Abs 0.046 0.089 0.136 0.166 0.217

Table 4. Equilibrium Concentration of FeSCN2+
Solutions 1 2 3 4 5
[FeSCN2+] 1.2534*10–5 2.425*10–5 3.7057*10–5 4.5230*10–5 5.9130*10–5
Table 5. Equilibrium Constants
Solution Equilibrium Constant
1 33.855
2 33.072
3 34.1827
4 31.385
5 33.4018
Average 33.1793
Graph 1. Absorbance vs. Concentration Table 6. Molar Absorptivity of FeSCN2+
Molar Absorptivity 3670
Equilibrium concentration of FeSCN2+ (Solution 1)
Abs = εl[X]
0.046 = 3670 × 1 ×

estimating osmolarity Essay
Estimating Osmolarity by change in weight
Abstract
Water flows in and out of cells in an attempt to attain a state of equilibrium. The concentration of
solutes to solvent in the cells environment is the cause of the water flow. Plant and animal cells can
be negatively affected or positively affected due to the concentration balance in their environment.
Potato cells were used to see the affects of sucrose in different concentrations. In some
concentrations a weight change was seen in the potato.
Introduction
In the osmotic concentration lab potatoes were used to see the affects of different concentrations of
sucrose on the weight of the potatoes. Solanum tubersummore commonly ... Show more content on
Helpwriting.net ...
Materials & Methods
In the osmotic lab seven cups , a marker, metric ruler, digital balance, paper towels, a knife and
seven cups containing solutions 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6m of sucrose were used in the
experiment.
The marker was used to label the seven cups with the seven different concentrations of sucrose . A
cork borer was used to cork the potato and then the metric ruler was used to measure seven 5cm
long samples cut with scalpel . The samples were weighed on the digital balance to the nearest
tenths of a gram. The weight of each potato cube was recorded on table under initial weight. As each
mass was taken the potato core were placed in the solutions of sucrose and stirred every 15 min
where the initial mass was recorded. After letting the potatoes soak in their designated sucrose
solutions from 9:28–10:36 they were removed and blotted lightly with the paper towels. The
potatoes where then measured on the digital balance and the changes in weight were recorded on
table under final weight. The change in weights if any was then calculated between each individual
potato's initial and final weight.
Results
After we finished recording the data of the experiment it became evident that the solution of 0.3 was
the osmolarity of the potato. It showed a 0 percent change in this solution.
Sucrose Molarity

Time in solution 9:28am–10:36am
0.0
0.1
0.2
0.3
0.4
0.5
0.6
Final weight
3.4g
3.2g
3.2g
3.1g
3.0g

Research Report On Orchard Inoculation Studies
Results
Orchard inoculation studies
In the orchard studies of 2014 season, significant variations (P<0.05) on disease severity (Table 4).
Disease severity increased with increasing conidial concentration. The conidial concentration of
5000/ml was minimum for 1% disease severity in mature detached fruits in lab. The interaction
between inoculation date and inoculum concentrations indicated the dependency of disease
development on growth stages of fruits besides conidial concentrations (Fig. 5).
Assessment of airborne inoculum in the orchard
DNA was extracted from the aerial samples collected in weekly basis from May 19 to July 14 in
2014 and from May 11 to July 5 in 2015. Regular PCR conducted with P. prunicola mildew specific
primer and ... Show more content on Helpwriting.net ...
Again, in the July increase in Cp value was observed in both years.
Discussion
This study showed that powdery mildew disease infection on sweet cherry fruits is effected by P.
prunicola inoculum or conidial concentration. At least 500 conidia / ml was needed to develop
significant fruit infection in cherry. Generally, disease incidence and disease severity increased in
cherry fruits with increasing inoculum concentration. A positive relationship between inoculum
concentration and disease development has also been demonstrated in the other powdery mildew
species (Carisse, 2009; Van der Heyden et al., 2013). However, very high concentration may cause
auto–inhibition of conidial germination as reported in other fungal pathogens (Leandro et al., 2003).
Additional studies should be carried out to find out highest conidial concentration threshold for
maximum cherry fruits disease incidence and severity. Orchard experiments of this study suggested
that infection of cherry fruits by P. prunicola also depends on the development stage of fruits.
Increased resistance to powdery mildew infection with maturing fruit has been reported in many
powdery mildew – host fruit systems such as grapes (Gadoury et al., 2003), hop (Wolfenbarger et
al., 2012), pears (Spotts, 1984), and apples (Russ and Blatter 1990). However, results from this
study showed that fruit became increasingly susceptible to P.

What Is The Concentration Of Salt Concentration In Osmosis
Based on the data, the 1% salt solution had the least change in mass. Its mass only went down by 0.1
grams. The 5% and 10% solutions caused the mass to decrease, while 0% salt solution was the only
one that lead to an increase in the mass. From these results, it can be concluded that as the salt
concentration get higher, the potato's mass starts to decrease.When there is barely any salt in the
solution the mass will increase. When comparing the turgidity of the cells a similar type of pattern
emerged. As the salt concentration got higher, the potato became flaccid. However, at lower salt
concentrations the potato remained crisp. An increase in the salt concentrations ended up changing
the turgidity of the potatoes and making them flaccid.
These results arose because of the different solutions the cells were put. When the potato mass,
decreased the cells of the potatoes were in an hypertonic solution because the solute concentration
outside the cells is greater than in the cell so water will flow out of the cells to maintain equilibrium.
Since there was osmosis of water out of the cell it lost weight. This also explains why cells appeared
flaccid. The loss of water can cause it to loose its shape and shrink A source mentions, in
hypertonice soltions, " water rushes out of the cell, and the cell shrinks away from the rigid [cell]
wall, resulting in plasmolysis"
(http://www.phschool.com/science/biology_place/biocoach/biomembrane1/isotonic.html). The
opposite happened in the 0% salt solution. The potato gained mass because the solute concentrations
was greater inside the cell than the outside which caused water to move into the cell. A microscopic
view reveals "the vacuoles of plant cells appear noticeably larger" (https://sciencing.com/happens–
hypertonic–hypotonic–isotonic–environments–8624599.html ) The water intake kept it study and
crisp. .However, too much water could have caused many potato cells to burst. Lastly, the 1%
solution was isotonic to cells because its mass remained about the same. In an isotonic solution the
water movement in and out of the cells is balanced to there is" no net gain, making it stay the same.
We had a 0.1 gram decrease but, that could be due measuring errors. In fact, one source

Osmosis Egg Lab
The purpose of this experiment was to determine the sugar (glucose) content in the egg and how
different solutions affected the mass of the eggs, my hypotheses is the higher level of glucose the
eggs sit in the solution will continue to rise and become hypertonic. In this experiment it called for 4
eggs and placed them in beakers. One was filled with ¾ of water, one with 0.5 glucose, 1.5 M
glucose and the last was 2.0 M glucose. The eggs were weighed in the beginning and then every 15
minutes for an hour, recording the mass of the egg. With the numbers recorded the percent change in
mass was calculated. Once this was completed and looked over I found my hypothesis to be wrong.
The egg was tested in 2.0 M glucose started out with a mass of 93.5M ... Show more content on
Helpwriting.net ...
This explains passive transport by movement of molecules without adding energy. Osmosis is the
result of diffusion. A process of water or any fluid moves from the area of less concentration of
dissolved particles to a region of high concentration of dissolved particles across a semi–permeable
membrane. Osmosis is important to the cell because the cell gains water along with important
molecules and particle for growth. In the body there are three solute concentrations. First is isotonic
solute which is two solutions that have the same osmotic pressure across a semi permeable
membrane. In this solute it allows water to move freely across the membrane without changing the
concentration of solutes on either side. The concentration of solutes is equal inside and outside the
cell so water moves across the membrane in both directions maintaining cell size. The second is
Hypotonic solution in this solution is it any solution that has a lower osmotic pressure than another
solution. The solution has less solute and more water than another solution. The third and final is
hypertonic solution and it has a greater concentration of solutes on the outside of a cell when
compared with inside of a cell. The purpose of this experiment was to find the concentrations of
solutes in the eggs by observing the process of osmosis. When the eggs are placed in different
"tonic" solutions they increase in weight or decrease in weight. To properly measure the eggs they
are removed from the beaker, dried with a paper towel and weighed on a scale in a weight

Osmosis Lab Report
The Effects of Solute Concentration on Osmosis
Introduction:
Diffusion is the movement of certain molecules toward the direction of equilibrium, and is an
important process in eukaryotic and prokaryotic cells. Osmosis is a specific type of diffusion that
refers to the diffusion of water, which is an important factor in many types of cells without cell walls
because the water potential of a cell must be similar or equal to its environment in order for the cell
to function properly. In cells, a semi permeable membrane regulates the diffusion of certain
molecules, selectively allowing substances to travel inside and outside of the cell and its organelles.
Dialysis bags are used to imitate the structure of cell membranes. They are also semi–permeable,
allowing water molecules to flow through while blocking larger sucrose molecules. Potato cell
membranes act similarly, allowing water to enter and exit but not facilitating the movement of
sucrose molecules. Because of this, the concentration of sucrose in the environment of a cell
membrane or in a dialysis bag will affect the direction of net water movement by changing the
amount of free water available for osmosis. Water potential is determined by the concentration of
sucrose within the membrane and its aqueous environment. The dialysis tubing experiment will test
the effect that different concentrations of sucrose within the semi–permeable membrane will have on
the net direction of water diffusion. The solutions within the

Reaction Rates of Sodium Thiosulphate and Hydrochloric...
Reaction Rates of Sodium Thiosulphate and Hydrochloric Acid at Various Concentrations
Introduction/ plan
––––––––––––––––––
I am experimenting into the reaction rates of Sodium Thiosulphate and
Hydrochloric acid when under different concentrations.
Both Sodium Thiosulphate and Hydrochloric acid are clear solutions but when they react together
they produce a cloudy solution, the solution become so cloudy you cannot see through it. To
measure the rate of reaction I am going to print off a cross from the computer onto a piece of paper,
place it under the beaker with the solution in it and record the time taken for the cross not to be
visible.
The Symbol equation for my experiment is:
Na S O (aq) + 2HCl (aq) 2NaCl (aq) ... Show more content on Helpwriting.net ...
By doing this it is changing the concentration. I am not going to carry out an experiment using 0ml
Sodium Thiosulphate, 10mls Hydrochloric acid and 30mls water, as there would be no Sodium
Thiosulphate for the hydrochloric acid to react with. I will do each experiment 3 times and then
work out averages for each one, as this will give me a more accurate set of results. Altogether
including averages I will have 40 separate results. From repeating each experiment 3times I will be
able to discover any drastic anomalies if there are any.
Preliminary Work
––––––––––––––––
Sodium Thiosulphate (mls) 30
Hydrochloric Acid (mls) 10
Water (mls) 0
Time Taken for Cross to Disappear (mins) 01.10
Prediction
––––––––––

I predict that as the volume of Sodium Thiosulphate decreases the reaction time will increase, as
there will be a lower concentration of
Sodium Thiosulphate for the Hydrochloric

Osmosis Lab Report
INTRODUCTION: If a cell is to perform its functions it must maintain a steady state in the midst of
an ever changing environment. This steadiness is maintained by the regulation of movement of
materials into and out of the cell. To achieve this control, cells are bounded by a delicate membrane
that differentiates between different substances, slowing down the movement of some while
allowing others to pass through. The membrane is said to be differentially permeable since not all
substances penetrate equally well. Selective permeability is a characteristic of healthy, intact cells.
When a cell is seriously damaged, the membrane becomes permeable to virtually to everything and
substances can move freely in and out of the cell. (Marieb 2013) ... Show more content on
Helpwriting.net ...
(Marieb 2013) Osmosis, is a special kind of diffusion, is defined as the diffusion of water through a
semi–permeable membrane from a higher concentration to that of a lower concentration. The
concentration of water is inversely related to the concentration of solutes. If the water can diffuse
across the membrane, both water and solutes will move down their concentration gradients through
the membrane. (Marieb 2012) There are three terms used to describe the concentrations of solute
particles of different solutions: Hypotonic– When compared to another solution of different
concentration, the solution contains the lower concentration of solute particles. Water moves across
a semi–permeable membrane out of a hypotonic solution. Hypertonic– When compared to another
solution of different concentration, the solution that contains the higher concentration of solute
particles. Water moves across a semi–permeable membrane into a hypertonic solution. Isotonic–
Having the same concentration of solute particles as another solution. There is no net flow of water
across the membrane. In animal cells, the movement of water is affected by the relative solute
concentration of the plasma membrane. As water moves out of the cell, the cell shrink and if water
moves into the cell, the cell swells and may even burst.
PURPOSE: The objective of this

Graph 6a Lab Report
Graph 6a (Substrate) – Represents a xy scatterplot with linear regression, which show the change in
product concentration over the change in time at different substrate The data in the graph also gives
the value of V0 at corresponding substrate concentration.
Graph 6b (Substrate) – Represents a xy scatterplot depicting velocity of enzyme–catalyzed at
multiple substrate concentrations.
To find the concentration of the different absorbance in this experiment, a modified version of the
Beer's Law equation was used (C=A/k). The k which represents the slope in the equation was
determine by using the date from Table 1 and points plotted in Graph 1. k=6.8339 the A in the
equation is the measured absorbance which was determine by using a spectrophotometer. ... Show
more content on Helpwriting.net ...
Cellobiase produced more product when placed at the right temperature and pH. These experiment
also showed how the simplest change in any environmental factor could cause a change in function
of the enzyme. This experiment was a good source to show the effects of environment on enzyme
activity. The amount of product produce could be monitored by the amount of p–nitrophenol
present, which couldn't be done with using Cellobiose. Factor that could have been improved would
have been testing the rate of reaction at a longer time then 2 minutes and at more enzyme/substrate
concentrations. The data produce from the enzyme and substrate concentration parts never showed
where the reaction reached Vmax. The most important factor of this experiment was the ability to
use the artificial substrate, it allowed the experimenter to visual see before measuring the
absorbance if product was made from substrate. It correlated that absorbance may be a direct
measurement of enzyme activity because in every experiment as the concentration of product (p–
nitrophenol and glucose) increase so did the absorbance. Since these were the only experiments
done that was the only conclusion that could be constructed. Since absorbance can't be measured
with colorless enzymes and substrates one can't really know if absorbance is a direct measurement
of enzyme

Osmosis in Potato Tubers
Osmosis in Potato Tubers
Andrew Dickson Background
When a plant cell is bathed in a solution of the same concentration (isotonic) as its intracellular
environment, its mass and volume remain the same. This is because water enters and leaves the cells
at the same rate. There is no net loss or gain of water by osmosis.
Samples of cells can be placed in a range of solutions of different concentration. The cells will gain
water by osmosis when placed in solutions which are more dilute (hypotonic) than the intracellular
environment. They will therefore gain mass. The cells will lose water in those solutions that are
more concentrated (hypertonic) than the intracellular environment and so lose mass. The
concentration of the ... Show more content on Helpwriting.net ...
Some of the potatoes expanded and increased in mass whilst others shrunk and decreased in mass.
This was caused by the diffusion of water from the cells to the sucrose solution or vice–versa
through the process of osmosis. During osmosis water will diffuse across a semi–permeable
membrane from a region of low solute concentration to a region of high solute concentration.
The potato cylinders in both test tubes A and B with sucrose concentrations of 0% and 5%
respectively, had a noticeable increase in mass, whereas the potato in test tube C even though having
an increase in mass, was only marginal. This suggests that the sucrose concentration in test tube C,
which was 10%, is almost the same concentration (isotonic) as the potato's cells intracellular
environment because the mass remains relatively the same. This is because the movement of water
into and out of the cells occurs at the same rate and results in no net loss or gain of water by
osmosis. This also implies that the potato cylinders in test tubes A and B where bathed in a sucrose
solution that was more dilute (hypotonic) than the intracellular environment of the potato's cells.
The cells of the potato in test tubes A and B gained water by osmosis and therefore gained mass.
However, the results show that in test tubes D and E with sucrose concentrations of 15% and 20%
respectively, that there was a decrease in mass of the potato cylinders. This was

Osmosis Egg Lab Report
Sharifah Leben
Anatomy & Physiology 1
Dr. Pasley
17 February 2016
Osmosis Egg Lab
Central Question
The purpose of this lab is to determine what is the concentration of the unknown solution along with
what concentration would be more isotonic to the chicken egg?
Hypothesis
If we placed our chicken egg into a higher percentage of sucrose solution for a set period of 60
minutes then the egg will gain weight. If the egg gained weight then the sucrose solution it was
placed in had a low concentration that would make it hypotonic. If the egg loses weight then sucrose
solution it was placed in had a high concentration that made it hypertonic. If there is no change in
weight in the chicken egg when placed in the concentration the solution will happen to be isotonic.
Method ... Show more content on Helpwriting.net ...
The solutions are 0% sucrose, 10% sucrose, 20% sucrose, 30% sucrose, 40% sucrose and an
unknown sucrose concentration. We then weighed each egg separately to the nearest gram in order
to have an initial starting weight to compare to the results throughout our experiment being
conducted. The eggs were then placed in each beaker for 12 intervals at a time. After every 12
minutes the eggs were taken out and weighed to see if the weight of the egg changed. With a total of
five intervals (12, 24, 36, 48, 60) the steps were repeated till the egg had reached the total time of 60
minutes. The changes in weight of the eggs were then added into a data table showing the weight of
the chicken eggs in grams vs. the time in minutes. In a second data table the weight changes (g) vs.
time (min.) between the eggs were taken and used the difference from each time and subtracted it
from the initial

Effect of Sucrose Solution on Osmosis Essays
Effect of Sucrose Solution on Osmosis
Aim: The aim of the experiment is to show how varying the concentration of sucrose solution
affects osmosis by changing different molar solutions of sucrose and water and how it affects the
potato. Introduction: In this investigation I will be exploring the effect of varying concentration of
sucrose sugar solution on the amount of activity between the solution and the potatoes.
Osmosis is the movement of water molecules across a partially permeable membrane from a high
water concentration to a low water concentration. Osmosis can be effected different things such as:
– Temperature – The warmer the solution, the quicker the osmosis will go. This is because the water
... Show more content on Helpwriting.net ...
The potato cells, took in, or gave out the water depending on the concentration of the solution it is
surrounded in.
The results were fine and by looking at the mass measured before the experiment, you can see that
there is no reading which seems to be out of the line. As the weights before the experiment range
between 2.31g and 2.46g, this tells us that the potato pieces were cut well, and I believe accurate
enough.
The results show that:
– Osmosis actually took place in the experiment.
– As the sucrose solution increased, the mass decreased, and as the sucrose solution decreased, the
mass increased
Evaluation: I followed the plan correctly; I believe I gained accurate and sufficient enough results to
conclude the experiment, and to prove my prediction.
To make this experiment better, I believe that I could have done more than one experiments to see if
anything has gone wrong, but since

Resistance Of Movement Of The Cell
If a cell is to perform its functions it must maintain a steady state in the midst of an ever changing
environment. This steadiness is maintained by the regulation of movement of materials into and out
of the cell. To achieve this control, cells are bounded by a delicate membrane that differentiates
between different substances, slowing down the movement of some while allowing others to pass
through. The membrane is said to be differentially permeable since not all substances penetrate
equally well. Selective permeability is a characteristic of healthy, intact cells. When a cell is
seriously damaged, the membrane becomes permeable to virtually to everything and substances can
move freely in and out of the cell. (Marieb 2013) Movement of ... Show more content on
Helpwriting.net ...
(Marieb 2013) Osmosis, is a special kind of diffusion, is defined as the diffusion of water through a
semi–permeable membrane from a higher concentration to that of a lower concentration. The
concentration of water is inversely related to the concentration of solutes. If the water can diffuse
across the membrane, both water and solutes will move down their concentration gradients through
the membrane. (Marieb 2012) There are three terms used to describe the concentrations of solute
particles of different solutions: Hypotonic– When compared to another solution of different
concentration, the solution contains the lower concentration of solute particles. Water moves across
a semi–permeable membrane out of a hypotonic solution. Hypertonic– When compared to another
solution of different concentration, the solution that contains the higher concentration of solute
particles. Water moves across a semi–permeable membrane into a hypertonic solution. Isotonic–
Having the same concentration of solute particles as another solution. There is no net flow of water
across the membrane. In animal cells, the movement of water is affected by the relative solute
concentration of the plasma membrane. As water moves out of the cell, the cell shrink and if water
moves into the cell, the cell swells and may even burst.
PURPOSE: The objective of this experiment is to

Osmosis in Potatoes
How is Osmosis Affected By the Concentration
PLANNING
Independent Variable
The independent variable in this experiment is the thing that I will be changing in the experiment.
This is the range of concentrations that have been obtained by diluting the original 1M solution of
Sucrose. This is therefore the only thing that is going to be altered as I am investigating how
different percentage of water molecules affects the weight of the the potato thus the rate and
direction of osmosis.
Dependant Variable
The dependant variable will be the mass of the potato as this is the variable I will be measuring.
From this I can see the change in mass(g) and so work out the amount of water that has moved and
the rate of the osmosis in a ... Show more content on Helpwriting.net ...
To makes sure that this does happen and that the controlled variables are controlled I will be doing a
second test to ensure that the results are reliable.
[pic]
| |(Sample of Potato) Change in Length/mm |
CONCENTRATION / M |1 |2 |3 |mean | |0.0 |50 |52 |54 |52 | |0.2 |50

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