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 saidbecause of the early warm springand 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”:

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

  1. 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. immitisinfection was treated with an antibiotic?

Part III: One week later…….

Dr. Veter received Caruso’s blood work and unexpectedly PCR resultswere also sent by the laboratory.

Patient: Caruso

TEST / Normal Range / At 2 Months / 12 Months Later
Albumin (g/L) / 2.7-4.4 / 3.2 / 2.7
Eosinophils
Total # x 103/uL / 0-1200 / 800 / 1300
Globulin (Total Blood) / 1.6-3.6 / 2.8 / 3.8
Hematocrit
Percent Packed Cell
Volume (%PCV) / 36-60 / 52 / 38
Heartworm / Antigen Test for Adult Female Nematode Protein / Not performed / Negative
Knott Test / 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:

  1. 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?
  2. 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

Figure 1. Detection of Wolbachia pipientis DNA by PCR

Using primers against ftsZ gene fragment (550 bp) and

D. immitis18S 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.
  1. Why was the 18S ribosomal DNA gene fragment from D. immitis used in thepositive control?
  2. Why would a second positive control (Lane 5) be important?
  1. Is Caruso, negative or positive for Wolbachia DNA? How do you know?
  2. 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. Theanimals 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.

  1. Why are the researchers testing the effectiveness of an antiobiotic?
  2. In Table 1, the researchers present range values. What is range statistically and why have the data been incorrectly reported?
  3. What does (+) SD mean?
  4. What does 95% Confidence limits represent?
  5. What does percent efficacy mean?
  6. Formulate a null hypothesis for this experiment.
  7. What is the significance of a P value and what doesP 0.001represent at the bottom left of the table?
  8. 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:

  1. 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:

  1. 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

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.

  1. 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.
  1. Why would a second positive control (Lane 5) be important? Shows conformation of WO DNA and therefore D. immitisinfection 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 specificallythrough 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 discussionabout 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 largerthe standard deviation the greater the difference from the mean, the smaller the SD the more tightly grouped the data is to the mean.