CASE STUDY: The Caruso Chronicles
Part I : Mosquitos: Ecosystem Service or Pest Control!
As the sun went down, Mike and Kelly were relaxing by the pool while Caruso, a two month old puppy
they recently found abandoned in the woods, had enough energy to power the east coast.
“Argh, these mosquitoes are so irritating!” exclaimed Mike as he sprayed enough bug repellant to kill an
elephant. Trying to sooth his mosquito bites, he said “A world without mosquitoes wouldn’t bother me
a bit.”
“Relax. For every pest there’s a predator.” replied Kelly.
“Well, I’m wondering where the nearest predator is then, as Mike swatted away one buzzing mosquito
after another. I just read an article in the newspaper that said because of the early warm spring and all
the wet weather we’ve been having, we’re probably going to see more mosquitoes. They need stagnant
or slow moving water to lay eggs.” Mike looked around the pool area and realized there might be places
where mosquitoes could easily make a comfy home for egg laying.”
Later that night Kelly thought about what would happen if there was a world without mosquitoes, but
also what if there was a world without mosquito control. Sure they’re annoying but she also knew they
spread diseases. Do they do any good she wondered? Kelly looked at a flyer reminding folks to practice
the “3D’s”:
Public Health Announcement
Take a Bite out of Disease
Mosquito season is upon us again. Dr. Public Health suggests the
public practice the 3D’s to protect themselves from mosquito borne
illnesses caused by the pathogens they carry like West Nile Virus
(WNV).
Drain: Keep stagnant water from forming by cleaning out drains and
containers on a weekly basis.
Dress: Wear long sleeves and pants to prevent mosquito bites spreading
pathogens in humans such as West Nile virus.
Defend: Wear insect repellent.
“Mosquito control isn’t just about human health”, says Dr. Health.
“Mosquitos are also responsible for transmitting pathogens to other
animals. They carry the pathogen that causes heart worm in dogs, and the
virus that causes Eastern Equine Encephalitis in horses.” This year dog and
cat owners need to be even more vigilant about protecting their animals.
“The mild winter and early spring provide the perfect environmental
conditions for mosquitos breeding”, said Dr. Kind Heart from the American
Heart Worm Society.
A healthy community is a community working to prevent the spread of
disease to people and pets alike.
For more information contact the American Mosquito Control Association.
Student Questions:
1. What if mosquitoes were eliminated altogether? Identify at least two consequences to an
ecosystem if mosquitoes were eliminated?
2. What evolutionary adaptations do mosquitos have that make them successful as insects?
3. In what way is adaptive evolution a function of natural selection?
Optional Debate Question
4. Does the public human health issue outweigh the impact of elimination of mosquitoes entirely
from an ecosystem?
Part II:
Caruso had not been to the veterinary clinic since they found him a year ago as a puppy. At that time he
had all his shots and a clean bill of health.
“Mike, something is wrong with Caruso. Caruso’s ears perked up when he heard his name and Kelly
placed her hands on each side of his face, reached her fingers behind his ears and gave him a comforting
scratch behind his ears. “He’s so tired acting. You know how energetic he usually is. He just seems like
he’s been getting more and more tired when he plays. Let’s take him to the clinic.”
At the Clinic
Dr. Veter: Hi Kelly, what’s going on with Caruso?
Kelly: Caruso just doesn’t seem to be himself. For the past couple of months he’s more tired than
usual” replied Kelly.
Dr. Veter: Let’s draw some blood and do a couple of tests. Has Caruso been on heartworm preventative
medication?
Mike: No.
Dr. Veter: I’d like to do a heartworm test. It won’t take very long, only 10 minutes or so.
Kelly: Okay, but how does the test work?
Dr. Veter: We’ll take some blood from Caruso. If Caruso is infected with the nematode that causes
heartworm, a protein that circulates in the blood from the nematode will be detected and the test will
turn up positive. The test is very specific and reliable.
The vet took blood samples and performed a heartworm test.
Dr. Veter: Good news, Caruso’s heartworm test was negative. We’ll wait to see what the blood work
tests tell us. We’ll have the results in just a couple of days. We’ll give you a call.
The blood samples were sent to an outside lab for analysis. A Knott test was also requested. A Knott
test detects the microfilarial larval stage of D. immitis using microscopy and cell staining techniques.
Dr. Veter also knew that Wolbachia, an endosymbiotic group of bacteria, were essential to the
reproduction and survival of D. immitis. Wolbachia and D. immitis have a symbiotic relationship, but
Wolbachia have also been implicated in severe immune responses in animals. When dead worms
release Wolbachia into the blood stream the immune system goes into overdrive and have been
implicated in the formation of blood clots in the lungs specifically causing more problems and even
death in the animal (.
Student Questions:
1. Wolbachia are obligate endosymbionts. What does this mean?
2. What do you think would happen if an animal with a D. immitis infection was treated with an
antibiotic?
Part III: One week later…….
Dr. Veter received Caruso’s blood work and unexpectedly PCR results were also sent by the laboratory.
Patient: Caruso
TEST
Albumin (g/L)
Eosinophils
Total # x 103/uL
Globulin (Total Blood)
Hematocrit
Percent Packed Cell
Volume (%PCV)
Heartworm
Knott Test
Normal Range
2.7-4.4
3.2
At 2 Months
2.7
12 Months Later
0-1200
1.6-3.6
800
2.8
1300
3.8
36-60
Antigen Test for Adult
Female Nematode
Protein
52
Not performed
38
Negative
0 microfilaria/mL
Not performed
160 microfilaria/mL
Student Questions:
1. Why didn’t the veterinarian test Caruso for heartworm as a two month old puppy?
2. Hypothesize how Caruso could have a positive Knott test but have a negative heart worm test at
twelve months?
3. Dr. Veter specifically asked for eosinophil and globulin levels from Caruso’s blood sample. What
is the significance of this request?
Optional Extension Questions:
4. The heartworm test is an application of ELISA, Enzyme-Linked Immunosorbent Assay. ELISA is a
fundamental tool used in clinical immunology. How does this process work?
5. What other diseases has ELISA been used for in the detection of infection in humans? Are there
other non-disease ELISA applications?
PCR Test Results
La
dd
er
1
(-)
co
ntr
ol
H2
O
HH
2
H
(-)
contr
ol 2
Uninf
ected
Dog
3
(+)
co
ntr
ol
1
4
(+)
control
2
Dog
blood
Sample
5
Caruso
’s
Blood
Sampl
e
6
550bp
102bp
Figure 1. Detection of Wolbachia pipientis DNA by PCR
Using primers against ftsZ gene fragment (550 bp) and
D. immitis 18S ribosomal DNA gene fragment (102bp)
Lane 1, 100 base pair ladder, Lanes 2-3 negative controls;
Lane 4-5 positive control; Lane 6, Caruso’s Blood Sample.
Student Questions:
1. Lane 1 contains a 100bp ladder. What purpose does the ladder serve?
2. In Lanes 2 and 3, what is the purpose of each negative control?
3. In Lane 4, two gene fragment templates were used for the positive control. One gene fragment
is the 18S ribosomal DNA from D. immitis. The second is the ftsZ gene fragment from Wolbachia
(WO), which plays a role in controlling WO cell cycle.
a. Why was the 18S ribosomal DNA gene fragment from D. immitis used in the positive
control?
b. Why would a second positive control (Lane 5) be important?
4. Is Caruso, negative or positive for Wolbachia DNA? How do you know?
5. What diagnostic inference can be made about Caruso’s heartworm status? Why and is this
inference supported by the blood work results?
Part IV
Dr. Veter: Hi Kelly. The blood work is telling us that Caruso does have a heartworm infection. Let’s get
him in here and talk about a treatment plan. Fortunately for Caruso, we’ve caught it early.
In preparation for Caruso’s visit, Dr. Veter needed to decide the best course of treatment for the dog. D.
immitis is not easy to treat because the nematode also harbors the endosymbiotic bacterium
Wolbachia.
She reviewed a recent study about heartworm treatment (Bazzocchi et al., 2008). Doxycycline (DOXY) is
an antibiotic in the tetracycline family of antibiotics. Ivermectin (IVM) is an anti-parasite medication.
Here is an excerpt from the article*.
“A total of 20 young adult beagles were used. The animals were born and raised in a mosquito-proof
environment and did not have any prior exposure to natural infection with D. immitis. Seven male and
nine female adult heartworms harvested from dogs approximately 8 months post infection were
introduced by intravenous transplantation (via a jugular vein) into each dog (McCall et al., 2008).
Approximately 6 weeks later (Day 0), the dogs were randomly allocated to four groups of five dogs each.
Beginning on Day 0, the dogs in one group were given weekly prophylactic doses of IVM (6ug/kg) orally
for 34 weeks. Weekly treatment was chosen to evaluate if previously reported adulticide effect following
monthly treatment could be achieved in a shorter period of time. The dogs in another group were given
DOXY (10mg/kg/day) orally from Weeks 0-6, 10-12, 16-18, 22-26, and 28-34. The dogs in the third group
were given a combination of IVM and DOXY at the same dosages and treatment schedules as used for
the first and second groups, respectively. The remaining group served as the non-treated (or infection )
group.”
Student Questions:
1. Describe the experimental design of this study. Is there anything missing or more information
you need to know about?
2. Based on the above information, is this study a controlled experiment? Explain why or why not?
3. Formulate a research question based on the information.
The table below shows some of the data collected from the study.
4. Why are the researchers testing the effectiveness of an antiobiotic?
5. In Table 1, the researchers present range values. What is range statistically and why have the
data been incorrectly reported?
6. What does (+) SD mean?
7.
8.
9.
10.
What does 95% Confidence limits represent?
What does percent efficacy mean?
Formulate a null hypothesis for this experiment.
What is the significance of a P value and what does P< 0.001 represent at the bottom left of the
table?
11. Using information provided in the table, are there differences between the treatment groups?
Explain using the statistical information presented and percent efficacy. Is your null hypothesis
supported or rejected?
The graph below represents data collected over 36 weeks of treatment.
Figure 1. Mean microfilarial counts (+ SDs) obtained over a 36 week period.
Student Questions:
12. Some of the data points have bars on them. According to the authors, what do these bars
represent and what does it mean?
Synthesis Question:
13. Based on the graph and Table 1, which treatment would be most effective against D. immitis?
Explain.
*Information presented based on the research study from Bazzocchi et al., 2008 has been modified to
meet the objectives of the case study. To review the research study in full and without changes refer to
literature citation section below.
Answer Key
Part I:
Answers to Questions:
Q1: Mosquitoes are part of ecosystems and serve as food for others in a food chain. Larvae are food
for fish and other aquatic animals; adult mosquitoes are food for birds, bats, dragonflies, and spiders.
Mosquitoes get their nourishment from plant nectar and serve as pollinators. Mosquito larvae even
help plants like the pitcher plant grow, by producing nitrogen as a by-product of the waste from other
insects. Small fish feed on the larvae too.
Q2: Structural and Functional Components: Mosquitoes have wings, are light bodied, and have a
proboscis that enables them to feed. Male and female mosquitoes get their nutrients in the form of
nectar. Both male and females eat flower nectar for nourishment but only females need other protein
nourishment to promote the development of eggs. The mosquito uses its sensory palps to find a meal!
Male and female stylets are morphologically different as a result. Females have a piercing stylet that is
slender and needlelike with ridges while the male stylet is less neddlelike and does not have tiny ridges
to pierce skin. Females have chemicals in their saliva that help them draw blood out of the animal
without it clotting.
Habitat Adaptations: Legs to walk on water and used for weight balance; aquatic life cycle adaptations.
Q3: As favorable traits increase (due to changes in allelic frequencies that enhance survival and
reproduction), so too does the match between a species and its environment.
Q4: Debate Question. Does the public human health issue outweigh the impact of elimination of
mosquitoes entirely from an ecosystem?
Part II:
Answers to Questions
Q1: Wolbachia are obligate endosymbionts. What does this mean? Endosymbionts are organisms that
live in the body or cells of another organism. An obligate endosymbiont cannot survive without the
host. Wolbachia pipientis cannot survive without the host, D. immitis, in this case.
Q2: D. immitis is a eukaryotic organism so antibiotics will not work on a eukaryotic organism.
Wolbachia is a prokaryotic organism. Antibiotics specifically target bacterial cells. Because D. immitis
has a symbiotic relationship with WO and requires WO for reproduction and survival, killing the
symbiont WO would also likely impact D. immitis and interrupt its life cycle.
Part III
Answers to Questions
Q1: Students will need to apply their understanding of the nematode life cycle and knowledge of what
the heartworm test tests for (female protein antigen) to arrive at the conclusion that there would be no
adult worm protein made at that young age to have an antigen reaction.
Q2: Low HW burden one compared to four females for example and enough antigen has not been
accumulated. Students might also capitalized on sensitivity of the test. The level of antigen is directly
related to the number of females present; the microfilaria were passed from mother to puppy
transplacentally but did not have adult D. immitis at that time.
Q3: Caruso had high levels of eosinophils, which are specialized white blood cells of the immune system
and indicate a parasitic infection. High globulin levels indicate an immune system response to infection
and production of antibodies. The host immune system will attempt to eliminate the parasite producing
antibody globulins and at the same time increasing other immune system white blood cells such as
eosinophils that target parasitic infectious agents.
Q4: There are two forms of ELISA to detect antigens: direct and indirect ELISA assays.
ELISA is used in research and applied settings. Probably some of the most familiar tests include the
human HIV and hepatitis tests, but ELISA kits are used for a variety of pathogen infection testing along
with autoimmune, cancer, allergy and food toxin testing.
Q5: What other diseases has ELISA been used for in the detection of infection in humans? Are there
other non-disease ELISA applications? See this web site for many different tests currently in use and
approved by the FDA http://www.rapidtest.com/index.php?product=Parasitology-ELISA-kits&cat=17
Q6: Lane 1 contains a 100bp ladder. What purpose does the ladder serve? A DNA ladder is used for
fragment size determination.
Q7: In Lanes 2 and 3, what is the purpose of the negative control? Show a result with no WO present.
The water control is essential for detecting contamination or non-specific amplification of your reaction.
The second negative control, confirms a negative result in a dog not infected by D.immitis and therefore,
no WO should be detected.
Q8: In Lane 4, two gene fragment templates were used for the positive control. One gene fragment is
the mitochondrial cytochrome oxidase 1 (CO1) gene from D. immitis. The second is the ftsZ gene
fragment from Wolbachia (WO) which plays a role in regulating WO cell division.
a.
Why was the 18S ribosomal gene fragment from D. immitis used in the positive
control? If DNA isolation is successful they will get a band for 18S ribosomal DNA even if there is
no WO gene fragment. This important to know that the DNA extraction was successful and that
WO is actually being amplified. Perhaps the WO ftsZ gene amplicon was not detected. This
would alert the researcher that there is something wrong with the WO ftsZ gene amplification
process (i.e. primer specificity) because if D. immitis is present, then WO should also be
detectable. Based on the paper from Rossi et. al (2010) PCR sensitivity is extremely high in the
context of WO DNA ftsZ gene detection.
b. Why would a second positive control (Lane 5) be important? Shows conformation of WO DNA
and therefore D. immitis infection in an infected dog.
Q9: Is Caruso, negative or positive for Wolbachia DNA? How do you know? Positive. Caruso’s blood
DNA sample showed a 550bp fragment identifying the WO ftsZ gene fragment.
Q10: What diagnostic inference can be made about Caruso’s heartworm status from the PCR results?
Why and can this inference supported by the blood work results? Caruso has heartworm because WO is
symbiotic with D. immitis. The results from the blood work, the Knott’s test specifically through
identification of D. immitis microfilariae, also support D. immitis infection.
Part IV
Answers to Questions
Q1: Describe the experimental design of this study. Is there anything missing or more information you
need to know about? The experimental design does not identify the gender of the animals, nor does it
identify if there were differences in nematode counts before group assignment.
Q2: Based on the above information, is this study a controlled experiment? Explain why or why not?
Yes, this is a controlled study. The researchers included a group not receiving any treatment. The
control group offers a comparison group for the researchers.
Q3: Formulate a research question based on the information. This may become an important
discussion point with students because the teacher can determine if students are distinguishing the
difference between a question and a hypothesis. It can also be a springboard for discussion about the
importance of having a question that can be tested scientifically.
Q4: Why are the researchers testing the effectiveness of an antibiotic? D. immitis shares a symbiotic
relationship with the bacterium Wolbachia pipientis. Killing the bacterium may have an adverse effect
on D. immitis. There is a secondary implication for the health of the dog that addresses the immune
response to the release of WO as noted above.
Q5: In Table 1, the researchers present range values. What is range statistically and why have the data
been incorrectly reported? Range represents how far apart the two extremes are in a data set. Range
represents the difference between the minimum and maximum values and is a single number not a
‘range’. The range has been incorrectly reported
Q6: What does SD (+) mean? SD stands for standard deviation. Standard deviation is another way to
represent spread of data and specifically represents how far away data values are from the mean. SD
therefore represents variability. The larger the standard deviation the greater the difference from the
mean, the smaller the SD the more tightly grouped the data is to the mean.
Q7: What does 95% Confidence limits represent? This tells us how close our estimate of any sample will
be to the mean. 95% of any given sample population will be covered within 2 standard deviations and
would fall between the upper and lower limits identified in the paper.
Q8: What does percent efficacy mean? How effective the particular treatment was at reducing adult
worms.
Q9: Formulate a null hypothesis for this experiment. Ho There will be no difference in reduction of
adult worms between the control group and treatment type.
Q10: What is the significance of a P value and what does P< 0.001 represent at the bottom left of the
table? P value tells you how much statistical evidence there is to reject the notion that the differences
between groups is due to chance alone. A typical level of less than 0.05 says that there is less than a 5%
chance that the difference between groups is due to chance alone. Another way to look at it is that
there is a 95% chance that the difference between groups is real. A P value of less than 0.001 means the
statistical chance that the difference between groups is random, or due to chance, is 0.1%. Stated
another way there is a 99.9% chance the difference between groups is real and not because of chance.
Q11: Using information provided in the table, are there differences between the treatment groups?
Explain using the statistical information presented and percent efficacy. There was not statistical
difference between the IVM or DOXY treatments alone with regard to the mean number of worms
recovered compared to the control group. The IVM/DOXY combination resulted in the greatest
adulticidal effect at 78.26%. Based on the P value of 0.001, this is a statistically significant difference in
worm reduction in the IVM/DOXY treatment group compared to the control group.
Q12: Some of the data points have bars on them. What do these bars represent? In this study, as
reported by the authors they represent standard deviation. SD represents how far from the mean the
data point deviates. This information may be of limited use because there can be overlap of the bars
and yet there may still be significant differences between the data points. SD bars wouldn’t necessarily
reflect that but does show how the data varies within the group.
Synthesis Question:
Q13: Based on the Graph and Table 1, which treatment would be most effective against D. immitis?
Explain. What other statistical information would you like to have to strengthen your claim particularly
with respect to the graph?
IVM/DOXY combination was most effective at reducing the number of adult worms and microfilaria over
the 36 weeks of treatment. The control group had the highest microfilarial/mL throughout the 36 weeks
compared to the IVM/DOXY and even the other two treatment groups. A P value for the data presented
would help support this claim with respect to the decreases in microfilaria (Figure 1). The P value of less
than 0.001 shows that there is statistical difference between the number of adult worms after the
IVM/DOXY treatment compared to the control and rejects the null hypothesis.
(It should be noted that in the original article the authors give a P value of less than 0.05 for control
versus IVM/DOXY data presented in the graph.)
Teacher Notes and Background Information
Case Overview and Background
Purpose of case
Where the case has been used
Kinds of students
Classroom Management
How do you run the case in the classroom? DETAILS
Time devoted to each phase of the discussion? Mention pitfalls to avoid in the discussion.
Preclass Assignments?
Questions to facilitate discussion? Identify any useful discussions that might come from the questions.
Ways the question can be used, ways it might be used differently if used in more than one type of class
etc. Identify the questions and briefly describe their use.
Introductory Questions?
Challenge Questions?
Decision Forcing Questions?
Summary Questions?
PART I:
Concepts for Focus:
Food chains, trophic levels, adaptive evolution, natural selection.
Student Learning Objectives:
Students will apply their understanding of organismal relationships in food chains and trophic levels.
Students will apply their understanding of adaptive evolution to the characteristics of an organism that
define structure and function.
Students will explain the relationship between natural selection and adaptive evolution.
Some background information on food chains, trophic levels, and ecosystem basics would be useful.
Knowledge of the mosquito life cycle would be helpful, but not necessary.
Mosquito as Vector and Nematode life cycle
Two critical environmental factors are necessary to promote a mosquito population and the spread of D.
immitis: temperature and water availability.
The mosquito is the vector for the transmission of heartworm. Mosquitoes become infected with the
first stage larva (L1) of the nematode Dirofilaria immitis while taking a blood meal from an infected
animal. Female mosquitoes bite. Males do not. Female mosquitoes of the Culix pipiens species require
a blood meal to promote egg development. The L1 stage of developing nematodes are called
microfilaria. Once the female mosquito is infected, the microfilaria incubate and develop to the L3 stage
within the Malpighian tubules of the mosquito. The L3 microfilaria migrate to the labium of the
mosquito. In the process of the bite, an infected female mosquito will deposit microfilaria near the bite
opening. The dog is the primary host for the nematode. The microfilaria wriggle their way into the
channel created by the mosquito bite and incubate under the skin where the L3 stage microfilaria
develop into L4 stage larvae. The L4 stage larvae migrate to other tissues including adipose and muscle
and complete its L5 stage. L5 then pass into the bloodstream where they migrate to the pulmonary
arteries.
The nematode could technically be called a vascular pathogen, with worm infections reaching into the
chambers of the heart by virtue of their length and overcrowding in the pulmonary vessels. Worms can
be found in the pulmonary arteries and right ventricle as a result.
Seventy to ninety days later they will develop into adults. Mating occurs approximately 6-9 months
after the adults become sexually mature and females release microfilaria into the bloodstream.
[INSERT mosquito/heartworm life cycle?]
PART II:
Concepts for Focus:
Life cycles, immunology, prokaryotic and eukaryotic cells,
Student Learning Objectives:
Students will apply their understanding of nematode life cycle to parasitic infections.
Students will formulate hypotheses using their understanding of nematode life cycle and symbiotic
relationship with Wolbachia.
Students will explain the application of clinical immunological methods to pathogen detection.
Students will distinguish the difference between prokaryotic and eukaryotic cells.
[INSERT BACKGROUND INFO]
Part III:
Concepts for Focus
Nematode life cycle, PCR,
Student Learning Objectives
Students will explain the function of controls in a PCR test.
Students will interpret and make inferences from PCR results.
[INSERT BACKGROUND INFO]
Part IV:
Concepts for Focus
Data interpretation, hypothesis formation, descriptive statistics
Student Learning Objectives
Student will apply their understanding of experimental design.
Students will use their understanding of experimental design to explain a controlled study.
Students will use statistics to analyze data to arrive at conclusions.
Students will formulate hypotheses and use statistics to reject or accept a hypothesis.
[INSERT BACKGROUND INFO]
References
Information on Culex pipiens
Information on D. immitis
Information on Wolbachia pipientis
Information on Statistics
Literature Citations
Bazzocchi, C., Mortarino, M., Grandi, G., Kramer, L. H., Genchi, C., Bandi, C., …McCall, J. W. (2008).
Combined ivermectin and doxycycline treatment has microfilaricidal and adulticidal activity against
Dirofilaria immitis in experimentally infected dogs. International Journal for Parasitology, 38(12), 14011410.
McCall, J. W., Genchi, C., Kramer, L., Guerrero, J., Dzimianski, M. T., Supakorndej, P., . . . Carson, B.
(2008). Heartworm and Wolbachia: therapeutic implications. Veterinary Parasitology, 158, 204-214.
Rossi, M. I. D., Aguiar-Alves, F., Santos, S., Paiva, J., Bendas, A., Fernandes, O., & Labarthe, N. (2010).
Detection of Wolbachia DNA in blood from dogs infected with Dirofilaria immitis. Experimental
Parasitology, 126(2), 270-272.