CASE STUDY: The Caruso Chronicles
Part I:
Caruso had not been to the veterinary clinic since Kelly and Mike found him a year ago as a puppy. The
veterinarian estimated that Caruso was about two-months old. At that time he had all his shots, blood
drawn and sent away to a laboratory. Caruso received 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 has 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 II: One week later…….
Dr. Veter received Caruso’s blood work and to his surprise 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 (ELISA)
Knott Test
Normal Range
2.7-4.4
At 2 Months
12 Months Later
3.2
2.7
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 technique work?
5. What other diseases has ELISA been used for in the detection of infection in humans? Are there
other non-disease ELISA applications?
Part III:
The PCR test results for Caruso were provided and analyzed by Dr. Veter. It had been a long time
since he had to think about this type of test.
PCR Test Results
1
2
3
4
5
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 antibiotic?
5. In Table 1, the researchers present range values. What is range statistically? Have the data been
reported correctly?
6. What does (+) SD mean?
7. In the graph below, some of the data points have bars on them. According to the authors, what
do these bars represent and what do they mean?
8. What does 95% Confidence limits represent?
9. What does percent efficacy mean?
10. Formulate a null hypothesis for this experiment.
11. What is the significance of a P value and what does P< 0.001 represent at the bottom left of the
table?
12. Using information provided in the table and the graph below, 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.
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 of the case study.
Answer Key
Part I
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 II
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
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
Part III
Answers to Questions
Q1: Lane 1 contains a 100bp ladder. What purpose does the ladder serve? A DNA ladder is used for
fragment size determination.
Q2: 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.
Q3: 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.
Q4: 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.
Q5: 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? Have the data been
reported correctly?
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: 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.
Q8: 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.
Q9: What does percent efficacy mean? How effective the particular treatment was at reducing adult
worms.
Q10: 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.
Q11: 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.
Q12: 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.
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 Resources
PART I:
Concepts for Focus:
Life cycles, 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.
Background Resources
Microbial Life
http://serc.carleton.edu/microbelife/index.html
Heartworm background
http://www.metapathogen.com/heartworm/
http://www.merckvetmanual.com/mvm/index.jsp?cfile=htm/bc/11300.htm
Heartworm testing info
http://www.marvistavet.com/html/body_diagnosis_of_heartworm_disease.html
Video of HW infection
http://www.heartwormsociety.org/pet-owner-resources/canine.html
Heartworm map
http://www.veterinarypracticenews.com/vet-breaking-news/2011/05/13/heartworm-diagnosed-inevery-state-in-10-survey-finds.aspx
Microbe wiki
http://microbewiki.kenyon.edu/index.php/Dirofilaria_immitis
Wolbachia Web site
https://www.wolbachiawebsite.org/about.cfm
Wolbachia Project
http://discover.mbl.edu/index.html
Werren, J. H. (1997). Biology of Wolbachia. Annual Review of Entomology, 42, 587-609.
Part II:
Concepts for Focus
Nematode life cycle, immunology methods
Student Learning Objectives
Students will explain the application of clinical immunological methods to pathogen detection.
Students will distinguish the difference between prokaryotic and eukaryotic cells.
Background Resources
Heartworm micofilaria measurement and blood testing
http://www.heartwormsociety.org/veterinary-resources/canine-guidelines.html#3
http://www.drkaslow.com/html/proteins_-_albumin__globulins_.html
ELISA Tutorials
http://www.sumanasinc.com/webcontent/animations/content/ELISA.html
http://www.biology.arizona.edu/immunology/activities/elisa/main.html
Examples of other ELISA kits and uses
http://www.rapidtest.com/index.php?product=Parasitology-ELISA-kits&cat=17
Bazzocchi, C., Genchi, C., Paltrinieri, S., Lecchi, C., Mortarino, M., & Bandi, C. (2003). Immunological role
of the endosymbionts of Dirofilaria immitis: the Wolbachia surface protein activates canine neutrophils
with production of IL-8. Veterinary Parasitology, 117(1–2), 73-83.
Part III:
Concepts for Focus
Polymerase chain reaction, biotechnology, data interpretation
Student Learning Objectives
Students will explain the function of controls in a PCR test.
Students will interpret and make inferences from PCR results.
Background Resources
PCR Virtual Lab
http://learn.genetics.utah.edu/content/labs/pcr/
Biology Animation: PCR
http://www.dnalc.org/resources/animations/pcr.html
NCBI PCR
http://www.ncbi.nlm.nih.gov/projects/genome/probe/doc/TechPCR.shtml
PCR Video Lecture
http://www.youtube.com/watch?v=_YgXcJ4n-kQ
http://www.youtube.com/watch?v=OK7_ReXhVaQ
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.
Background Resources
Basics of Standard Deviation
http://www.mathsisfun.com/data/standard-deviation.html
Mean, Median, and Mode Review
http://www.mathsisfun.com/mean.html
Scientific Process
http://undsci.berkeley.edu/index.php
Research Methods Knowledge Base
http://www.socialresearchmethods.net/kb/contents.php
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.