Diffusion
and Osmosis through Dialysis Tubing
Purpose/Problem
The purpose of these experiments is
to investigate the influence of solute concentration on osmosis and diffusion.
Background Information
The membrane of the cell is semipermeable;
it must allow some products through such as waste, and materials for the
cells. However, it must also contain all
of the enzymes, DNA, and other materials necessary for cell function. In this lab, the focus will be on passive
transport through the semipermeable membrane, or the dialysis tubing. This has been made semi permeable through the
addition of microscopic holes in the sides of the tubing.
Hypothesis: Activity A
1.
If
the solution outside the bag tests positive for glucose, then glucose diffused
out of the dialysis bag and through the permeable membrane.
2.
If
the iodine diffuses into the bag, then it will turn black/blue.
3.
If
the starch diffuses out of the bag, the solution will turn black/blue.
Hypothesis: Activity B
1. If the bag is heavier after the experiment,
then the water was transferred in by osmosis.
Materials: Activity A
Dialysis
tubing, plastic cup, glucose/starch solution, distilled water, iodine-potassium
iodide solution, dropping pipet, glucose test strips, funnel
Materials: Activity B
Dialysis
tubing, plastic cups, distilled water, funnel, sucrose solutions, paper towels,
balance
Procedures: Activity A
1.
Pour 160-170 mL of distilled water into a
plastic cup. Add approx. 4mL of IKI
solution to the water and mix well. Record
solution color. Dip a glucose strip into
the solution and record results.
2.
Dip
a fresh glucose test strip into the glucose/starch solution. Record results.
3.
Obtain
a piece of dialysis tubing that has been soaked in water. Close one end by
knotting it.
4.
Using
a small funnel, pour 15mL of glucose starch solution in the dialysis bag. Expel the air. Tie off open end of bag, leaving enough room
to allow for expansion. Record
color. Immerse the dialysis in solution,
making sure it is completely covered.
5.
Wait
30 minutes.
6.
Remove
bag from cup, and cut a slit to insert the glucose test strip. Record results.
Procedures: Activity B
1.
Pour 160-170mL of distilled water into a
plastic cup. Label it with the sucrose
concentration.
2.
Form
a bag with the dialysis tubing by knotting off one end of it.
3.
Using
a small funnel, pour 15mL of sucrose solution into the dialysis bag. Expel the
air. Tie off the open end, leaving space for expansion.
4.
Dry
the bag and determine the mass. Record
result.
5.
Immerse
the bag in the water. Make sure it is
completely covered in water. Wait 30
minutes.
6.
Remove the bag and dry it on towels. Mass the bag and record results. Determine change in mass and record.
Results: Activity A
In the activity measuring the
diffusion through the semipermeable membrane, the results correlated with most
of our expectations. The IKI solution
was an orange brown color in the beginning, and ended the same, indicating that
starch did not diffuse through the dialysis tubing. The glucose starch solution inside the bag
started out clear and turned black/blue, indicating that the iodine diffused
into the bag. Our initial glucose
results tested positive for the glucose/starch solution, and negative for the
distilled water. Our final glucose
results were positive for both the water and the solution, indicating that the
glucose diffused out of the bag.
Results: Activity B
In
activity B, the experiment tested the change in mass when solutions of different
concentrations of sucrose were submerged in distilled water. The results show that the higher the
concentration, the more of a change in mass there was. For the 0.0M concentration, there was no
change. For the 0.2M concentration,
there was a 25.3% change in mass. For
the 0.4M concentration, there was a 46.1% change in mass. For the 0.6M concentration, there was a
53.76% change in mass. For the 0.8M
concentration, there was a 64.36% change in mass.
Conclusion: Activity A
Most of the above results for
activity A correlated with the hypotheses and predictions. However, the one result that was not seen was
the solution outside of the cup turning a black color. This indicates that the starch did not
diffuse out of the bag. This was
unexpected, but it allows us to draw conclusions about the size of the
molecules. We can determine that the
starch molecule is too large to fit through the microscopic holes in the
membrane, and this explains the results that we found here.
The one substance that was not
accounted for here was water. In order
to measure the water transfer, we could collect data about the mass of the bag
before and after the experiment. This
would mean that we had then accounted for all of the molecules in this
experiment.
Conclusion: Activity B
This experiment tested the travel of
water into and out of the dialysis bag in relation to the concentration of
sucrose in the water in the bag. The
results were consistent with what we expected to see, so we can assume that all
of the variables were controlled other than the variable we were testing:
water. Our results do indeed support the
hypothesis that the water would transfer in by osmosis. The higher the concentration of sucrose in
the water in the bag, the higher the percent change of mass is. This shows that the more concentrated the
sucrose is, the more water is needed to dilute it, and the higher the change in
mass.
Presence
of Staphylococcus epidermidis,
Streptococcus epidermidis and Escherichia
coli in casual catch morning urine samples and casual catch afternoon
samples.
Purpose:
The
purpose of this lab is to discover the bacterial found on urogenital surfaces,
as well as how bacteria can enter the urinary tract and cause infections. This lab taught how to prepare a proper
bacterial culture as in a diagnostic laboratory. This lab was also intended to
draw conclusions on the sterility of morning urine and afternoon urine.
Background:
A
urinary tract infection is a bacterial infection. It is common among all people, but more so in
women. A urinary tract infection is
caused by bacteria from the urogenital area entering the urinary tract. They are diagnosed by performing a bacterial
culture and looking for abnormalities in species and levels. They are treated with antibiotics, but can
also be treated naturally with cranberry juice.
The lab deals specifically with the bacteria species E. coli and S. epidermidis. E. coli is a common fecal bacterium that
is being tested for in conjunction with urinary tract infections, and S. epidermidis is a common skin
bacterium. S. epidermidis is catalase-positive, so this lab will apply
hydrogen peroxide to the bacterium to ensure that it is not Streptococcus. This lab is also testing for bacteria
that are gram positive or negative by manipulating the agar chosen for growing
the samples. Gram positive bacteria will
not grow on the dyed agar, and gram negative bacteria will only grow on the
dyed agar because of the specific dietary requirements.
Hypotheses:
1.
If
E. coli is present in the urine
sample then green bacteria will grow on the EMB culture plate.
2.
If
S. epidermidis or Streptococcus are present in the urine
sample then the bacteria will grow on the phenylethanol culture plate.
3.
If
S. epidermidis is present in the
urine sample, then the hydrogen peroxide will react with the bacteria and break
down to water and gaseous oxygen.
4.
If
morning urine is less sterile than afternoon urine, then the morning urine
sample cultures will contain more species of bacteria than afternoon samples.
Materials:
Escherichia
coli, Staphylococcus epidermidis,
EMB agar, Phenylethanol agar, autoclave disposal bag, sterile petri dishes, 3%
hydrogen peroxide, sterile applicator sticks, sterile specimen containers,
antiseptic pads, 70% ethanol, cotton balls.
Methods:
1. Disinfect work surfaces and wash hands.
1. Disinfect work surfaces and wash hands.
2. Set out one petri dish of EMB agar and one
dish of phenylethanol agar.
3. Have one person from each group supply a casual
catch urine sample. Be sure that the container does not touch clothing or
body. The sample can be refrigerated
until ready to be used, but it should be taken as close to the laboratory
period as possible.
4. Remove a sterile applicator stick from the
package, and bei9ng careful not to touch the tip to any other surfaces, dip it
into the urine sample. Inoculate the
surface of the EMB agar, being sure to cover half of the surface. Be sure to inoculate heavily because the
bacteria will be diluted in the urine.
Label the dish.
5. Using a fresh stick, inoculate the
phenylethanol dish with the same urine sample using the same procedure as
before.
6. Incubate both the control and inoculated
plates at 37⁰ C
for 48 hours. Invert the plates to
prevent moisture from dripping down onto the agar and obscuring the results.
7. Observe the plates after 48 hours and note
the presence and appearance of the EMB and phenylethanol agars.
8. Test the S.
epidermidis colonies on the control phenylethanol plate with one drop of 3%
hydrogen peroxide. Repeat this step with
colonies on the urine-inoculated phenylethanol dishes. Observe whether bubbling occurs where the
hydrogen peroxide is in contact with the culture. Bubbling is a positive reaction for catalase;
no bubbling is a negative catalase reaction.
Record results.
Results:
On the
EMB culture plate, all samples were negative except for samples 2 and 8. On the phenylethanol, all samples were
positive except for sample 6 (Table 1). Samples 1,2,3,4,5, and 8 on the phenylethanol
culture plate tested positive for Staph., and samples 2,6, and 7 tested
negative. Sample 2 tested both positive
and negative for reaction to hydrogen peroxide.
Samples 1, 3, 4, 5, 6, and 7 had no bacteria on the EMB plates. Sample 2 had green colonies growing on the
EMB, and sample 8 had purple colonies growing.
All samples had white colonies growing on the phenylethanol except for
sample 6, which was negative. The
samples were taken from 3 males and 5 females.
All samples were casual catch samples taken in the afternoon except for
sample 7, which was a clean catch afternoon sample, and sample 8, which was a
casual catch morning sample. The control
plate for E. coli tested positive on
the EMB agar with green colonies, and negative on the phenylethanol agar. The control plate for Staph. tested negative on the EMB agar, and positive on the
phenylethanol agar, with white colonies showing.
Conclusion:
The data
suggest that the first hypothesis, which states that if E. coli is present in the urine sample, then bacteria will grow on
the EMB agar plate, can be accepted for sample 2, because green colonies were
present. All other samples, reject this
hypothesis suggesting no presence of E.
coli. The only urine samples that
tested positive for bacterial growth on
the EMB plate were samples 2 and 8.
The colonies on sample 8 were purple, not metallic green indicating that
the bacteria present was not E. coli. I conclude that the purple colonies must be a
vigorous fermenter in order to produce the purple color seen. Past this conclusion, it is unclear exactly
what bacterium was growing on that plate.
This leads to the conclusion that E.
coli can be present in urine, and will grow on the EMB agar plates. However, it also leads us to the conclusion
that bacterial growth on the EMB agar is not a definite proof of E. coli in the urine, because some
sample tested positive, but were clearly not E. coli because of their distinct purple color.
The
second hypothesis, which states that if S.
epidermidis or Streptococcus are
present in the urine sample then the bacteria will grow on the phenylethanol
culture plate, can be accepted. The data
show that many urine samples tested positive, and were determined to be either S. epidermidis or Streptococcus. Based on
these results, I conclude that when either bacteria is present in the urine
sample, it will show up on the phenylethanol agar.
The
third hypothesis, which states that if S.
epidermidis is present in the urine sample, then the hydrogen peroxide will
react with the bacteria and break down to water and gaseous oxygen, is accepted
in samples 1,2,3,4,5, and 8. This
suggests that that the samples are populated with S. epidermidis. Samples 2, 6
and 7 tested negative. Samples 6 and 7
showed no bacterial growth, so that can be deemed unimportant data in this
conclusion. Sample 2, however, did have
colonies growing on it, and the data suggest that bacteria of both types were
growing in the urine sample. Because of
the consistency of finding the S. epidermidis
bacteria in the urine samples, the conclusion that can be made is that it is a
common bacteria to be found in the urogenital area.
The
fourth hypothesis, which states that if morning urine is less sterile than
afternoon urine, then the morning urine sample cultures will contain more
species of bacteria than afternoon samples, can be accepted, but only on a
limited basis. The sample size for this
data set was alarmingly small, leading to the conclusion that a larger
conclusion cannot be determined from this data.
The data that is shown is also contradictory towards conclusions. The one morning sample that was taken did
have a significantly higher population of bacteria than sample 4 and 5, but was
similar in content to sample 2. The only
significant difference in the morning sample was the presence of the purple
colonies, but again, it is unknown as to what caused that. Therefore, on this hypothesis is can be
concluded that there was not enough information gathered, and much
contradictory results, that result in no conclusion being drawn.
The
interactions of Bacteriophage T4, antiserum T4 and E. coli in relation to bacterial infection prevention.
Purpose:
The
purpose of this lab is to explore the reaction between bacteriophage T4 and Escherichia coli, and the reactions
between the bacteriophage T4 and the homologous antibody. This lab presents the problem facing modern
medicine, namely, how to prevent the antibodies from attacking the
bacteriophage before it has destroyed the bacterium.
Background:
The
anticoliphage T4 serum used in this lab is produced through the autoimmune
response of a goat when the bacteriophage T4 is injected into it. The blood serum that is collected after this
process contains the antibody for bacteriophage T4. This antibody is capable of attacking free T4
virions. These antibodies are produced
by the B- cell lymphocytes in the goat’s immune system. These substances that stimulate the immune
system to create antibodies are called antigens. Specifically, an antibody will be produced
that is targeted against that specific antigen.
The trigger for the immune system in this situation has a “head” and
“tail”, where each end is a different antigen.
Therefore, in the blood serum sample collected, there are two distinct
clones of the antibody. Despite this,
studies have shown that only the antibody that attacks the tail of the virus
has an effect on the virus infectivity.
These antibodies neutralize unadsorbed virions, but cannot have an
effect on a virion that has already attached to a bacterium. The free floating antibodies produced by the
goat’s immune system mirror the workings of the human system as well, so the
results of this lab can be used to tackle the ideas and problems presented by
the concept of battling bacterial infections with virulent phages. This is becoming increasingly relevant in
today’s world, because there are increasingly more antibiotic-resistant
bacterial infections coming into existence.
A problem yet to be solved is the elimination of the bacteriophages by
the antibodies before they have performed their job of eliminating the
bacteria. The removal of the phage by
the body’s natural processes does not happen instantaneously, but rather
proceeds at a measurable rate. This is
the process being addressed in this lab.
Question:
Might a
virulent phage be administered to destroy a particular bacterial infection?
Materials:
Antiserum
and coliphage T4 set, 18 tubes dilation and culture broth, 6 tubes soft agar, 6
base-layer agar plates, 21 pipets, 1 mL, marking pen, 5 procedure flowsheets.
Results:
On each
plate, the measurements were determined subjectively, by estimating the percent
coverage of both phage and E. coli on each culture plate. This lab was measuring the concentrations of E. coli and bacteriophage T4 on soft
agar plates. The first plate was P-5,
meaning that it consisted of phage at a dilution of 10-5. The results on this agar show E. coli at a concentration of 5% on the
plate, and the bacteriophage at a concentration of 95% (Table 1). This plate presented with a dense film of
bacteriophage colonies with sparse scatted milky white E. coli colonies. The P-6
plate, or phage at a dilution of 10-6, showed 5% E. coli and 70% bacteriophage. This plate had sparsely scattered milky
white E. coli colonies, and densely
scattered tiny white bacteriophage colonies.
The 2A/P-5 plate consisted of phage at a 10-5 dilution
allowed to interact with the antibody for 2 minutes. The results show 30% E. coli and 30% bacteriophage, an equal ratio. This plate held medium sized milky white E. coli colonies, and scattered tiny
white phage colonies. The 4A/P-5 plate
consisted of phage at 10-5 dilution and the phage was allowed to
interact with the antibody for 4 minutes.
The results show 25% E. coli
concentration and a 15% bacteriophage concentration. The plate held small milky white scattered E. coli colonies, and scattered tiny
white bacteriophage colonies. The 8A/P-5
consisted of phage at 10-5 dilution allowed to interact with the
antibody for 8 minutes. The plate showed
a 20% E. coli concentration and a 5%
bacteriophage concentration. The plate
held several large white E. coli
colonies and scattered tiny white bacteriophage colonies. The 8A/P-5X plate consisted of phage at a 10-5
dilution allowed to interact with the antibodies for 8 minutes. This plate showed no signs of life.
Table 1:
|
Plate Name
|
E. coli
|
Bacteriophage
|
Description
|
|
P-5
|
5%
|
95%
|
filmy layer, scattered
small E. coli colonies
|
|
P-6
|
5%
|
70%
|
small white milky
colonies, tons of tiny white dots
|
|
2A/P-5
|
30%
|
30%
|
medium white milky
colonies, scattered lots of white dots
|
|
4A/P-5
|
25%
|
15%
|
small white milky
colonies, densely scattered white dots
|
|
8A/P-5
|
20%
|
5%
|
Several large white
milky colonies, sparsely scattered white dots
|
|
8A/P-5X
|
0%
|
0%
|
devoid of all life on
the plate
|
Figure
1:
Conclusion:
The
results of this lab can be applied to the question of bacterial infection
treatment through bacteriophages. The
effects of the different factors in this lab could lead to more ideas on how to
oppress the natural antibody response in human systems. Our results on the P-5 and P-6 plates showed
varying levels of phage growth, which was clearly due to the dilution of the
phage added to that plate. There were no
antibodies to inhibit the growth on these plates, so therefore we see an
explosion of phage growth, and a very slight presence of e. coli. The explosive growth seemed to be less on the
plate labeled P-6, which correlates because this is a higher dilution of the
phage. On plate 2A/P-5, there appeared
to be a dramatic decrease in amount of phage, and almost equal parts phage and
e. coli. This makes sense, because this
plate allowed interaction between the antibodies and the phage for 2
minutes. A lesser amount of phage would
be expected here, due to the fact that it is being incapacitated by the antibody. On plate 4A/P-5, there is recorded a drop in
phage on the plate, most likely due to the fact that this plate allowed
interaction between the phage and the antibody for 4 minutes instead of 2. On plate 8A/P-5, there was a drop in numbers
of phage colonies, again likely due to the fact that the interaction time
between the antibody and the phage was increased to 8 minutes. Here, it would be safe to conclude that the
more time the bacteriophage is exposed to the antibodies, the less survives on
the plate. Plate 8A/P-5X showed no signs
of life, which could be predicted, given the fact that this plate contains no
e. coli, and the phage was allowed to interact for 8 minutes with the
antibodies. The combination of being
attacked, and having no host cells to reside in clearly resulted in complete
inability for the phage to take hold.
Out of all of this, it can be concluded that the phage does increasingly
worse the more time it is exposed to an antibody, and when it has no host
cells, it stands not a chance. One
result that was found that has difficulty being described is the decrease in
the numbers of e. coli colonies as the numbers of phage decreased. The expected result would be that as the
number of phages preying on it decreased, the number of e. coli colonies seen
would increase, and flourish in the absence of their predator. This leaves the unanswered question, what is
causing the decline in E. coli
colonies? It is possible that if this
could be identified, it could lead to yet another treatment option for these
kinds of bacterial infections. Given
that some of the possible concerns of bacterial therapy may include difficulty
containing the virus, this data suggests that the phage cannot survive without
cells to invade, and so could be contained.
One of the questions still to be answered here is how the body’s natural
immunological response could be repressed so that the phage could do its job,
because clearly the phage does not function nearly as well where it is
necessary for it to be attacked as well as attack. Is there a possibility for a phage that is
not recognized by the body as an invader, which can seek out and destroy these
bacterial invaders? This would be a huge
advance in modern medicine, if this form of infection management could be
achieved.
However,
the conclusions drawn in this lab are not necessarily to be trusted, because
the sample size here was atrociously small.
Because of technical difficulties during this lab, the number of plates
where the results were uncontaminated was much smaller than expected. The nature of this lab means that if one
step in the series was disrupted or performed incorrectly, all of the results
would need to be scrapped. Also, the number
of labs performed independently would need to be increased, so that all results
were not dependent on the prior person’s accuracy. Therefore, for the conclusions made here to
be trusted, the sample size would need to increase significantly.
Allele Shuffling of Single Gene Traits in Drosophila
Introduction
The purpose of the experiment conducted was to determine if the shuffling of alleles of single gene traits is indeed random. The fruit fly, or Drosophila, is an ideal choice for the experiment because of the fast turn-over, or reproductive, rate, require little space, and are easy to feed and handle. They also only have four pairs of chromosomes, making it easy to identify and observe genetic shuffles. They have been used for many genetic studies, meaning that there is a wealth of information, as well as many genetically pure strains for additional research. In the experiment conducted with the Drosophila, if the p-value is greater than 5%, then allele shuffling can be said to be random.
Materials
Sorting brushes, sorting cards, fly morgue, microscope, wild-type flies, mutant A- Se, mutant B- Vg, mutant C- W-, petri dishes, culture vials, potatoes, foam plug, fly nap kit
Methods
Warning: Don’t Inhale fly nap. Also, the fruit flies may not be native to your area, so do not release them!
1. Anesthetize all flies in separate vials.
2. Examine each type of fly; be sure to note eye color and wing shape for each type.
3. Make sure you are able to differentiate between the males and females; practice on the flies you have out.
4. Put the flies in the morgue.
5. Examine the previously created homozygous crosses and score the F1 generation.
6. Set up your F1 cross and the culture vial.
7. Put excess F1 flies in the morgue.
8. Wait for the results of the F1 cross to be born, and remove the F1 adults to the morgue.
9. Once the F2 generation has hatched, score them.
10. Put scored flies in morgue.
Results and Analysis
For cross 1(Sese X Sese), there were 871 red-eyed and 326 sepia-eyed flies counted, with a total of 1197. In order to discover the number of each type of fly there would be expected from this number, multiply the total found by the decimal forms of the numbers expected. Sample Equation: 1997*.75=897.75. The expected numbers were 897.75 and 299.25, respectively.
Equation for finding Chi-squared:
In order to determine whether the numbers fall between reasonable limits, you perform the chi-square equation. Sample Equation:X2=(871-897.75)2/897.75=0.7971. This value is then compared to a chart of values, and depending on your degrees of freedom (=options-1) you determine the p-value. If it is less than the 5% bracket, you can consider it to be significantly off. The total Chi-square value was 3.19. For cross 2 (VgvgSese X VgvgSese) there were 197 vestigial winged red eyed flies, 231 vestigial winged sepia eyed flies, 708 normal winged red eyed flies, and 240 normal winged sepia eyed flies. Some of these numbers deviated significantly from the expected numbers; we expected 258, 86, 774, and 258 respectively. The total chi-square value for this cross was 265.78. For cross 3 (W+w- X W+w-) there were observed 228 white eyed males, 184 white eyed females, 240 wild eyed males, and 214 wild eyed females. The expected number for each category was 216.5. The total chi-square value for this cross was 8.07.
The purpose of this experiment was to determine if the allele shuffling of single gene traits in Drosophila is indeed random. The hypothesis was that the shuffling of the single gene traits is indeed random. Had the hypothesis been confirmed, we would see that the chi-square values would all correlate with a p-value greater than 5%. Cross 1 reinforces and supports the hypotheses. However, not all other data correlated with the hypothesis. There were several numbers that contradicted the hypothesis, specifically the values for crosses two and three (VgvgSese X VgvgSese and W+w- X W+w-). The values for vestigial winged red-eyed flies had a chi value of 14.42, which is much higher than it should be, and this shows that something is amiss. The values for vestigial winged sepia-eyed flies were also off, and the chi square value was 244.48. This is very far off of what it should be, and this show that there was something severely amiss in this cross. There are several options for what could have caused this. It is possible that the room was not kept at an even temperature, and this could have affected the results of the cross. The most plausible possibility, however, is that we had a mislabeled vial, or people who counted inaccurately, for whatever reason. Both numbers that were off were ones that had vestigial wings. This leads me to believe that the problem was in the misidentification of sepia and red eyed vestigial winged flies. The fact that there were so many people doing crosses could be helpful in terms of more numbers making more accurate data, except for the fact that the numbers are not easily tracked back to the original counter. This means that there are not repercussions for people doing a lazy job, or even genuinely being unable to correctly identify the flies. This dilutes the accuracy of our experiment, and basically negates the results. However, the huge number of mislabeled flies (approx. 119) suggests that there was something bigger going on. In order to confirm the hypothesis for this dihybrid cross, I would suggest that the experiment be run again, and this time have a large number of flies being bred by one person, to see if the results are again off. The other option is to run each cross through the ratios again, but using individual data instead of that from over forty people. The anomalies can then be detected, and a more accurate picture can be formed. The results gained from this experiment are not ones that are watertight; therefore to achieve a more accurate result, the experiment should be run again. Conclusions about the accuracy of the hypothesis cannot be determined from this data, due to the uncertainty surrounding the results.
| Ss X Ss | |||||
| Phenotype | Observed | Expected | Chi-square value | p-Value | Significant (Y/N) |
| Red-eyed | 871 | 897.75 | 0.7971 | >5% | N |
| Sepia-eyed | 326 | 299.25 | 2.3911 | >5% | N |
| Total | 1197 | 1197 | 3.1882 | >5% | N |
| VvSs X VvSs | |||||
| Phenotype | observed | expected | Chi-square Value | p-Value | Significant(Y/N) |
| Vestigial red | 197 | 258 | 14.4224 | <5% | Y |
| vestigial sepia | 231 | 86 | 244.4767 | <5% | Y |
| Normal Red | 708 | 774 | 5.6279 | >5% | N |
| Normal Sepia | 240 | 258 | 1.2558 | >5% | N |
| Total | 1376 | 1376 | 265.7828 | <5% | Y |
| Ww X Ww | |||||
| Phenotype | Observed | Expected | Chi-square Value | p-Value | Significant (Y/N) |
| White-Eyed Males | 228 | 216.5 | 0.6109 | >5% | N |
| White-Eyed Females | 184 | 216.5 | 4.8788 | >5% | N |
| Wild-Eyed Males | 240 | 216.5 | 2.5508 | >5% | N |
| Wild-Eyed Females | 214 | 216.5 | 0.0289 | >5% | N |
| Total | 866 | 866 | 8.0694 | <5% | Y |
