Why is my immune system killing
cells in my intestine?
I.
Name: Teacher Resource
Date: _______
Identify a Problem
When an organism has sensed that its immune system is
compromised—that is, when an injury occurs that triggers an immune
response, such as a serious burn or other wound—the body begins to
defend itself from infection. The lumen of the gut (the space within
the digestive system, running from the mouth to the anus) is an
external environment, exposed to countless species of bacteria (gut
flora). Thus the gut is essentially a danger zone, and the body begins
to enact protective mechanisms. For example, it will releasing
cellular signals to initiate programmed cell death (apoptosis) within
the most inner lining of the gut (epithelium), intending to limit the
quantity of bacteria that are able to infect the body by creating a
shield of dead cells. Normally these bacteria are useful in the gut,
since they aid in digestion, but their presence places an immune-compromised person at greater risk for
infection and more severe inflammation. These signals are intended to create a protective layer of dead
cells, reducing the chance the bacteria have for entering the body. However, they do so by injuring
surrounding tissue, such as the epithelium lining of the gut mucosa. But how does the body respond to
this? You are the scientist, so it’s up to you to figure it out!
1. Why is it significant that the cells are dying by apoptosis, rather than by necrosis?
(Remember: Apoptosis is cell death due to programmed cell signaling, and necrosis is cell
death due to external factors, like toxins.)
Students may respond that apoptosis is intentional, and that would be the correct line of thinking. It is
important to recognize that one difference between apoptosis and necrosis is that the cell death is
intentional with apoptosis, whereas necrosis is a casualty sustained by a tissue. Therefore, even though
an organism does sustain an injury to their skin when they are burned, any cell death in the intestine
would obviously not be a direct result of the burn. Cell death on the skin at the site of the injury would be
unintentional, but cell death in the intestine would have to be the body’s own mechanism.
2. Why would bacteria place an immune-compromised organism at risk?
The skin serves as a protective barrier for the body from the potentially dangerous bacteria in the
environment. The skin, which students may or may not realize, is contiguous with the lining of the gut;
this is what makes the lumen an external environment. Therefore a burn, which affects the skin on the
outside of the body, will create a wound site that leaves the internal environment of the body open to
receiving bacteria. The body of course does not want this to happen, so it makes an effort to establish a
barrier defending itself from these organisms. Students should respond in a way that shows they realize
that bacteria can cause infections, and that experiencing a wound elicits an immune response.
3. Why might it be dangerous to signal for apoptosis to begin within a tissue?
If cell death is allowed to accelerate in any given tissue, the tissue will eventually deteriorate, which
places an organism at risk. Thus homeostasis is needed to place these rates back into equilibrium.
4. Why might it be useful to initiate apoptosis in the gut lining when the immune system has
been compromised?
A layer of dead cells could make the intestine lining less pervious to the bacteria inhabiting the gut.
Page 1 of 9
Why is my immune system killing
cells in my intestine?
II.
Name: Teacher Resource
Date: _______
Design an Experiment
You have decided you want to see how the rate of apoptosis in the
gut epithelium changes over time after an organism’s immune
system has been compromised. You know that because an
organism requires homeostasis for survival, the rate of apoptosis
must eventually subside, but you must design an experiment to
determine precisely when or how this happens.
Homeostasis: an organism’s
ability to regulate its internal
environment to produce an
equilibrium between biological
forces.
You decide that your injury of choice will be a thermal burn (a burn from heat, rather than by chemicals
or electricity), since this type of injury will only directly damage skin and adipose tissue. Also, a thermal
burn will elicit an immune response without having to infect the body with a foreign organism (bacteria,
virus, fungus, etc.). In order to elicit the apoptosis in the gut lining, you know that the burn must be
severe—likely over 30% total body surface area (TBSA). Because thermal burns are injuries that humans
experience, but because humans cannot be used as test subjects in biology, your study may produce
results with clinical application. Because of possible clinical application in human burn patients, you
choose to use mice as the model organism for your experiment.
5. Why is it important that only skin and adipose tissue are directly damaged by the injury in
this experiment? (Hint: think about the need to establish controls.)
If other tissues were damaged, then that would establish more variables in the experiment. For example,
if the burn caused damage to the underlying muscles, could that be the cause of apoptosis in the gut?
6. Why is it important to elicit an immune response without the introduction of another
organism’s DNA?
Bacteria will work to take up residency within the host organism. Thus any immune response the body
has would be to kill bacteria and stop the infection already present, rather than to prevent an infection as
it is in this model.
7. Why is a mouse model be wise decision for a model organism? Why not use nematodes
(slender worms) or Drosophilia (fruit flies)?
Both mice and humans are mammals, and so they are evolutionarily and physiologically more similar
than humans are to nematodes or fruit flies. Because of their similarity, the data gained from experiments
with mice is more easily translatable into human studies than it would be with other organisms.
8. Why could humans not be used for this experiment? At what point can humans be used for
a study?
Experiments cannot be conducted on humans that would result in such serious injuries as are being
studied here. Because tissue samples are being taken from the intestine in this experiment, patients who
are already admitted cannot be used either because of their already-compromised immune system.
9. What do you think will happen to the rate of apoptosis over time after the burn? Why?
The body will need to return to homeostasis, and so the rate of apoptosis must eventually come back to
normal levels. For apoptosis to go unchecked would harm the tissue and the organism.
Page 2 of 9
Why is my immune system killing
cells in my intestine?
Name: Teacher Resource
Date: _______
Experimental Ethics:
In 2012, 450,000 individuals received burn injuries in the United States that required medical treatment,
about 40,000 of which required hospitalization. Once hospitalized, one of the greatest dangers for these
patients is the risk for infection. A burn injury drastically compromises the body’s immunity, so studies
such as this one are useful for doctors and nurses to know the best way to
care for these patients.
In order for a scientist to carry out an experiment, you must first gain the
proper approval. First, you must consult with the director of your
laboratory (that is, your Principle Investigator, or PI). Then, if you plans
to use animals in your experiment, you must—by law—submit a formal
proposal to the Institutional Animal Care and Use Committee (IACUC).
If you were using humans in an experiment that merited human study,
you would submit your proposal to an Institutional Review Board (IRB).
These groups ensure that the procedures for a proposed experiment avoid
animal mistreatment, and no experiment is allowed to proceed unless it
gets this approval. For example, according to the Animal Welfare Act of 1966, scientists must only
experiment with “the least degree of pain and suffering practicable to the animal.”
10. What sort of precautions and standards do you think would be appropriate to ensure that
this experiment would be completed with respect to the law and with respect to the animals
involved?
Students’ responses will vary. However, they should respond in some way that would treat the mice
nicely. Pain medication, decent housing, appropriate care, sufficient food and water, etc. are all
appropriate answers.
11. Experiments using animals have provided scientists and doctors the knowledge to treat
human conditions such as diabetes and cancer. The clinical applications of these
experiments have been a big reason to support animal studies. Do you think there is a
difference between a clinically-relevant experiment and a science-for-the-sake-of-science
experiment?
Students will likely respond that there is a difference, and that it is much easier to justify the use of animal
models in science if the experiments are geared toward helping solve a problem that could help save
human lives, such as improving the immune system of seriously injured patients.
12. Do you think all model organisms have the same needs for care in experiments? Other
examples of model organisms include E. coli (a bacterium), Drosophilia (fruit flies), Zebra
fish, guinea pigs, and the Rhesus monkey. How might you care for a Zebra fish differently
from a Rhesus monkey?
Students will not know what the exact protocols are for when using a fish model versus using a primate
model, but they do understand that a monkey is more like a human than a fish, in that it will have similar
behaviors, including responses to hunger, thirst, pain, etc. This level of development is something that
will change how scientists plan their experiments. For example, a primate will need ample space for
recreation and socialization, which is easy to accommodate because primates can easily be identified
from each other, but it may be less practical to keep fish in a single container for the purposes of
identification.
Page 3 of 9
Name: Teacher Resource
Date: _______
Why is my immune system killing
cells in my intestine?
III.
Analyze the Results ofan Experiment
You created two roughly equal groups of mice, one which will be burned and one which will be the sham
group. The sham group is the control group, receiving all the treatments of the burn group (the
experimental group) except for the actual burn, which includes being shaved and being given morphine in
their water. The two groups were housed in the same location and under the same conditions before and
after the burn. At several time points, as shown below, mice from each group were selected and
anaesthetized. Epithelial tissue was harvested from the small intestine in each mouse. The samples are
stained with a specific dye, such as trypan blue, which can only pass into cells with broken membranes
(dead cells). The tissue samples were then passed through a cellometer, which determined how viable the
tissue was. This instrument counts the number of stained cells in each sample, as well as the number of
unstained cells in each sample.
Burned mice
ID
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
B11
B12
Sham Mice
ID
S1
S2
S3
S4
S5
S6
S7
S8
S9
S10
S11
S12
S13
S14
Time
sacrificed
12 hours
12 hours
12 hours
24 hours
24 hours
24 hours
36 hours
36 hours
36 hours
48 hours
48 hours
48 hours
Total number of
Number of apoptotic
Percent of
cells in sample
cells in sample
apoptotic cells
24,500
2,800
11.43
24,200
3,100
12.81
22,600
2,500
11.06
21,800
1,500
6.88
25,100
1,300
5.18
24,300
1,600
6.58
25,000
900
3.60
22,500
800
3.56
21,700
600
2.76
24,500
190
0.78
24,300
140
0.58
22,300
120
0.54
Time
sacrificed
0 hours
0 hours
12 hours
12 hours
12 hours
24 hours
24 hours
24 hours
36 hours
36 hours
36 hours
48 hours
48 hours
48 hours
Total number of
Number of apoptotic
Percent of
cells in sample
cells in sample
apoptotic cells
0.40
25,300
100
0.53
22,600
120
2.65
23,800
630
1.98
24,700
490
1.61
24,900
400
0.66
22,800
150
0.53
24,500
130
0.52
23,300
120
0.42
23,600
100
0.63
25,300
160
0.55
21,900
120
0.46
26,200
120
0.59
22,000
130
0.48
21,000
100
Page 4 of 9
Name: Teacher Resource
Date: _______
Why is my immune system killing
cells in my intestine?
13. Calculate the percent of apoptotic cells in each tissue sample for each mouse. Record in the
charts above.
14. Calculate the average percent of apoptotic cells for each time point in both sham and
burned mice.
Burn
Sham
0 hours
12 hours
24 hour
36 hours
48 hours
0.46
11.77
2.08
6.21
0.57
3.31
0.53
0.63
0.51
15. Graph the average percentages in a double bar graph. Remember to use an appropriate
scale and labels!
Percent of apoptotic cells in gut epithelial tissue over time after burn
20
Percent of apoptotic cells
18
16
14
12
10
Burn
8
Sham
6
4
2
0
0
12
24
36
48
Time sacrificed after burn (hours)
16. What trends do you notice in the data based on your graph?
Students should notice that the sham group remains relatively constant and relatively low at each time
point. The burn group is much higher than the sham soon after the injury, but the levels of apoptosis in
the burn group eventually return to levels similar to the sham group after two days. Students may notice
that the sham group is higher at 12 hours than at other time points, and this will be important later in the
activity.
17. How do the data immediately compare to your expectations?
Students likely expect apoptosis to end at some point, since the concept of homeostasis would require the
body to return to equilibrium. However, they may be surprised at how much of a difference there is
between the burn group and the sham group, especially immediately after injury, or they may be
surprised at how long (or perhaps how quickly) it takes to return to an equilibrium.
Page 5 of 9
Name: Teacher Resource
Date: _______
Why is my immune system killing
cells in my intestine?
18. At which time points are the differences between the two groups statistically significant?
Use the Unpaired T-test. Remember that n is your number of categories, that x (without a
bar) is the individual datum for each category, and that α = 0.05.
Your Null Hypothesis: H0 = No difference between the burn and the sham groups
Your Alternative Hypothesis: H1 = Burn group > Sham group
12 hour:
Average for Burn: 𝑥̅𝑏 = 11.77
# of categories for burn: 𝑛𝑏 = 3
Average for Sham: 𝑥̅𝑠 = 2.08
# of categories for sham: 𝑛𝑠 = 3
∑(𝑥𝑏 − 𝑥̅𝑏 )2 = (11.43 − 11.77)2 + (12.81 − 11.77)2 + (11.06 − 11.77)2 = 1.3209
∑(𝑥𝑠 − 𝑥̅𝑠 )2 = (2.65 − 2.08)2 + (1.98 − 2.08)2 + (1.61 − 2.08)2 = 0.558
Variance: 𝑠 2 =
T-Statistic: 𝑡 =
∑(𝑥𝑠 −𝑥̅𝑠 )2 +∑(𝑥𝑏 −𝑥̅𝑏 )2
𝑛𝑠 +𝑛𝑏 −2
𝑥̅𝑏 −𝑥̅𝑠
1
1
2
√𝑠 (𝑛 +𝑛 )
𝑠
𝑏
=
1.3209+0.558
3+3−2
11.77−2.08
=
1 1
3 3
= 0.470
= 17.31
√0.470( + )
Degrees of freedom: 𝑑𝑓 = 𝑛𝑠 + 𝑛𝑏 − 2 = 3 + 3 – 2 = 4
P-Value:
24 hour:
_______ < p < 0.001
P-Value Comparison:
Can the null hypothesis be rejected?
Yes
Average for Burn: 𝑥̅𝑏 = 6.21
# of categories for burn: 𝑛𝑏 = 3
Average for Sham: 𝑥̅𝑠 = 0.57
# of categories for sham: 𝑛𝑠 = 3
p < a
∑(𝑥𝑏 − 𝑥̅𝑏 )2 = (6.88 − 6.21)2 + (5.18 − 6.21)2 + (6.58 − 6.21)2 = 1.6467
∑(𝑥𝑠 − 𝑥̅𝑠 )2 = (0.66 − 0.57)2 + (0.53 − 0.57)2 + (0.52 − 0.57)2 = 0.0122
Variance: 𝑠 2 =
T-Statistic: 𝑡 =
∑(𝑥𝑠 −𝑥̅𝑠 )2 +∑(𝑥𝑏 −𝑥̅𝑏 )2
𝑛𝑠 +𝑛𝑏 −2
𝑥̅𝑏 −𝑥̅𝑠
1
1
√𝑠2 (𝑛 +𝑛 )
𝑠
𝑏
=
=
0.0122+1.6467
3+3−2
6.21−0.57
1 1
3 3
= 0.4147
= 10.73
√0.4147( + )
Degrees of freedom: 𝑑𝑓 = 𝑛𝑠 + 𝑛𝑏 − 2 = 3 + 3 – 2 = 4
P-Value:
_______ < p < 0.001
Can the null hypothesis be rejected?
P-Value Comparison:
p < a
Yes
Page 6 of 9
Name: Teacher Resource
Date: _______
Why is my immune system killing
cells in my intestine?
36 hour:
Average for Burn: 𝑥̅𝑏 = 3.31
# of categories for burn: 𝑛𝑏 = 3
Average for Sham: 𝑥̅𝑠 = 0.53
# of categories for sham: 𝑛𝑠 = 3
∑(𝑥𝑏 − 𝑥̅𝑏 )2 = (3.60 − 3.31)2 + (3.56 − 3.31)2 + (2.76 − 3.31)2 = 0.4491
∑(𝑥𝑠 − 𝑥̅𝑠 )2 = (0.42 − 0.53)2 + (0.63 − 0.53)2 + (0.55 − 0.53)2 = 0.0225
Variance: 𝑠 2 =
T-Statistic: 𝑡 =
∑(𝑥𝑠 −𝑥̅𝑠 )2 +∑(𝑥𝑏 −𝑥̅𝑏 )2
𝑛𝑠 +𝑛𝑏 −2
𝑥̅𝑏 −𝑥̅𝑠
1
1
√𝑠2 (𝑛 +𝑛 )
𝑠
𝑏
=
0.4491+0.0225
=
3+3−2
3.31−0.53
=
1 1
3 3
0.1179
= 9.916
√0.1179( + )
Degrees of freedom: 𝑑𝑓 = 𝑛𝑠 + 𝑛𝑏 − 2 = 3 + 3 – 2 = 4
P-Value:
_______ < p < 0.001
P-Value Comparison:
Can the null hypothesis be rejected?
48 hour:
p<a
Yes
Average for Burn: 𝑥̅𝑏 = 0.63
# of categories for burn: 𝑛𝑏 = 3
Average for Sham: 𝑥̅𝑠 = 0.51
# of categories for sham: 𝑛𝑠 = 3
∑(𝑥𝑏 − 𝑥̅𝑏 )2 = (0.78 − 0.63)2 + (0.58 − 0.63)2 + (0.54 − 0.63)2 = 0.0331
∑(𝑥𝑠 − 𝑥̅𝑠 )2 = (0.46 − 0.51)2 + (0.59 − 0.51)2 + (0.48 − 0.51)2 = 0.0098
Variance: 𝑠 2 =
T-Statistic: 𝑡 =
∑(𝑥𝑠 −𝑥̅𝑠 )2 +∑(𝑥𝑏 −𝑥̅𝑏 )2
𝑛𝑠 +𝑛𝑏 −2
𝑥̅𝑏 −𝑥̅𝑠
1
1
2
√𝑠 (𝑛 +𝑛 )
𝑠
𝑏
=
=
0.0331+0.0098
3+3−2
0.63−0.51
1 1
3 3
= 0.01073
= 1.41
√0.01073( + )
Degrees of freedom: 𝑑𝑓 = 𝑛𝑠 + 𝑛𝑏 − 2 = 3 + 3 – 2 = 4
P-Value:
0.10 < p < 0.15
Can the null hypothesis be rejected?
P-Value Comparison:
p>a
No
19. Why is an unpaired t-test appropriate for this data set?
The data is continuous, since the percents of apoptosis could have a range of anywhere between 0 and
100. There are two populations, and the populations are unpaired, because rather than examine the rates
pre-treatment and post-treatment in the same group, we are measuring treatment versus no treatment in
two groups. Thus we must use an unpaired t-test.
Page 7 of 9
Why is my immune system killing
cells in my intestine?
Name: Teacher Resource
Date: _______
20. Based on the p-values, which time points show a statistical difference between the two
groups of mice? Why is this statistic important?
The time points at 12, 24, and 36 hours all show a statistical difference between the burn group and the
sham group, but the time point at 48 hours shows no statistically significant difference. Thus, by 48 hours,
the burn group has returned to homeostasis.
21. What sort of questions could these data answer?
These questions will likely depend on what trends the students notice in the data, but possible questions
could be: How fast do rates of apoptosis in the gut subside after an injury? How elevated are the rates of
apoptosis after an injury? How much apoptosis happens in the gut without an injury?
22. What sort of questions do you have about this phenomenon that these data not support?
Students may have any number of questions, some of which may include: How do cells undergo
apoptosis? What signals are the cells receiving to undergo apoptosis? Where are the signals coming
from? Is this only happening in the gut, or is this happening on the skin as well? Unfortunately any given
data set can only answer so many questions, but students should be instructed that it is natural to end an
experiment with more questions than when the experiment began, for this spark of inquiry is the true
nature of science.
23. What could be an alternative explanation for these data? (That is, what if the data set is
valid, but your hypothesis is still wrong?)
Students’ answers will vary widely, but perhaps examining the rates of mitosis in the gut epithelium may
negate the hypothesis. If cells are continuing to divide faster than they die, even at elevated rates, then a
layer of dead cells would not be accumulating, but rather the living cells would be multiplying. The data
showing that rates of apoptosis increase and subside, however, would still be valid.
24. Why would there not be a time point for burn mice at 0 hours after burn?
In reality, if this were data from human burn victims, it would take time to transport a patient to
a hospital where tests could be performed, especially because burn clinics are few and far
between in many parts of the country. Students may offer any number of answers.
25. Notice that the percent of apoptotic cells for sham mice is higher at the 12-hours time point
than at the other time points. Why do you think this is so?
Students’ answers may vary, but this could be due to a reaction the sham mice have to the treatments
associated with the sham group. Recall that the sham group receives all treatments that the burn group
receives except for the actual burn. This could be a reaction to morphine, which could have effects on the
body, or being shaved, which could change thermoregulation, etc.
Page 8 of 9
Name: Teacher Resource
Date: _______
Why is my immune system killing
cells in my intestine?
26. Of course, this situation is only one of the many where apoptosis is a useful mechanism for
an organism. For example, scientists often measure the rates of apoptosis within a tumor to
determine how well or how poorly a particular therapy works for a cancer patient. The
three patients below are each undergoing a different cancer therapy. Make an inference
about the relative size of their tumor in each of the following situations. Is the tumor
growing, shrinking, or remaining the same size?
Patient A:
The rate of apoptosis > The rate of mitosis
shrinking
Patient B:
The rate of apoptosis = The rate of mitosis
remaining the same
Patient C
The rate of apoptosis < The rate of mitosis
growing
27. Use another unpaired t-test to determine if this 12-hours time point for sham mice is
significantly different from the other time points.
Null Hypothesis: H0 = 12-hour sham group is not different from the other sham groups
Alternative Hypothesis: H1 = 12-hour sham group > other sham groups
Average for 12 hour Sham: 𝑥̅1 = 2.08
# of categories for 12hr Sham: 𝑛1 = 3
Average for Other Sham: 𝑥̅2 = 0.52
# of categories for sham: 𝑛2 = 11
∑(𝑥1 − 𝑥̅1 )2 = (2.65 − 2.08)2 + (1.98 − 2.08)2 + (1.61 − 2.08)2 = 0.558
∑(𝑥2 − 𝑥̅2 )2 = (.4 − .52)2 + (.53 − .52)2 + (.66 − .52)2 + (.53 − .52)2 +
(.52 − .52)2 + (.42 − .52)2 + (.63 − .52)2 + (.55 − .52)2 +
(.46 − .52)2 + (.59 − .52)2 + (.48 − .52)2 = 0.0673
Variance: 𝑠 2 =
T-Statistic: 𝑡 =
∑(𝑥2 −𝑥̅ 2 )2 +∑(𝑥1 −𝑥̅1 )2
𝑛2 +𝑛1−2
𝑥̅1 −𝑥̅2
1
1
√𝑠2 (𝑛 +𝑛 )
1
2
=
0.558+0.0673
3+11−2
2.08−0.52
=
1 1
11 3
= 0.0521
= 10.493
√(0.0520)( + )
Degrees of freedom: 𝑑𝑓 = 𝑛2 + 𝑛1 − 2 = 3 + 11 – 2 = 12
P-Value: _______ < p < 0.001
P-Value Comparison:
Can the null hypothesis be rejected?
Yes
p<a
28. Are the sham mice at the 12-hour time point significantly different from the other sham
mice? Why or why not might they be different? Does this at all affect your original
conclusions from the data?
The sham mice at the 12 hour time point are significantly different from the other sham. However, this
has no effect on our conclusions because the original hypothesis of the experiment was to determine
whether (or when) the burn groups are significantly different from the sham groups, and at 12 hours,
burn is significantly greater than sham.
Page 9 of 9