Disease-Detectives-C-Key.pdf (Nationals 2011)

7/27/2019 Disease-Detectives-C-Key.pdf (Nationals 2011)

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____________ Total Score

____________ Rank

School Name: _________________________________

Team Number: ________________________________

Student Name(s) (1): ______________________________

(2): ______________________________

ANSWER KEY

DIVISION CDISEASE DETECTIVES

National Science OlympiadUniversity of Wisconsin

Madison, WisconsinMay 21, 2011

Developed by theCareer Paths to Public Health Program

Centers for Disease Control and Prevention (CDC)

Page 2 Page 8 Page 13 Page 19 Page 25

Page 3 Page 9 Page 14 Page 21 Page 26Page 4 Page 10 Page 15 Page 22 Page 27Page 6 Page 11 Page 17 Page 23 Page 28

Page 7 Page 12 Page 18 Page 24


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Part A: Food Illness/Safety

1. (2 pts.) (The ongoing, systematic collection, analysis, interpretation, and dissemination of dataregarding a health-related event for use in public health action to reduce morbidity and mortality

and to improve health.)

2. (2 pts.) (Persons with a severe illness are more likely to be tested than those with a milder one.).

3. (2 pts.) (Norovirus = 14663/5461731 = 0.002685; Salmonella= 19336/1027561 = 0.01882;Campylobacter= 8463/845024 = 0.010015.)

4. (1 pt.) (Toxoplasma gondii.)

5. (1 pt.) (Norovirus.)

6. (1 pt.) (STEC O157.)

7. (1 pt.) (Neither.)

8. (2 pts.)(E. coli enterotoxigenic. 142/2 = 71)

9. (3 pts.) (Elderly, young children, pregnant women, immunocompromised.)

10.(2 pts.) (Need a control or comparison group.)

11.(1 pt.) b. 20 seconds

12.(1 pt.) (Chilling slows growth.)

13.(1 pt.) (Cooking kills bacteria.)

14.(1 pt.) (Freezing slows growth and can kill bacteria. Give half credit for each answer.)

15.(1 pt.) b. 40-140 degrees F

16.(1 pt.) 2 (Half credit for any answer of less than or equal to 4 hours that is not 2.)

17.(1 pt.) a. USDA (U.S. Department of Agriculture)

18.(1 pt.) (Farm__ _processing__ _transportation__ __retail__ _home table)(Students should have farm and home table. 1 point for 3 middle steps as 1/3 credit)

19.(1 pt.) binary fission

20.(2 pts.)(4 hours.)


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Part B: Botulism1

21.(1 pt.) (A bacterium or bacteria.)

22.(1 pt.) (Indirectvehicle borne; half credit is allowed for only one of the two answers)

23.(2 pts.) (The father and son could have shared other meals or other exposures that could haveresulted in these symptoms. The additional 4 cases strengthened the association between thesymptoms and the restaurant. Outbreak is defined as common source with more than 2 unrelated

people.) (1 pt. for each bolded pt.)

24.(2 pts.) (Age or mental status of respondents, interval between eating and interview, severity ofillness, alcohol consumption.)

25.(3 pts.) (A person who ate at the Greek restaurant on April 8 or April 9 and who had C. botulinumdetected in their stool.)

26.(2 pts.) (18/30 = 60%) (2 pts. for showing work and having the correct answer; 1 pt. if failed toshow the work, but has correct answer; no points if % is not shown.)

27.(2 pts.) (Double vision is more severe than blurred vision or more severe cases were more likely tobe diagnosed. The suspect cases may not have had botulism.)

28.(1 pt.) (Epi curve or histogram.)

29.(1 pt.) (The time between when someone is exposed to a pathogen and onset of symptoms.)

30.(1 pt.) (April 9-13.)

31.(1 pt.) (Median: 2 days.)

32.(1 pt.) (A point source.)

33.(2 pts.) (An outbreak is point source if it has a common source [e.g. contaminated food or water]and the exposure occurs over a finite period of time. 2 pts. 1 for common source and 1 for finite

period.)

34.(1 pt.) (Dinner on 04/09.)

Table 5.

Ate the Food Did not Eat the Food

Ill Total Ill Total

Black olives 19 93 11 105

Greek salad 12 94 18 104

Gyros 12 90 18 108

Spaghetti 5 27 25 171

1While the outbreak presented here occurred many years ago (1994), these are examples of classic investigations used for thiscompetition.


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35.(2 pts.) Fill in the gaps in the above table. (2 pts., pt for each correct answer.)

36.(1 pt.) (The number of individuals who ate spaghetti.)

37.(2 pts.) Relative risk compares exposed and diseased among 2 groups; have an exposed (sick) groupand an unexposed (not sick) group; compare risk (attack rates) among exposed and unexposed (1 ptfor relative risk; 1 point for reason)

38.(6 pts.) (Eggplant dip or potato dip. 1 each for eggplant and potato dip, 2 points for each set ofcalculations, only 1 point each if used odds ratio.)

EGGPLANT: 6/9 = 0.6667

24/189 = .126984RR = 5.25

POTATO DIP : 19/22 = .8636364

11/176 = .0625RR = 13.818181

39.(2 pts.) (The potato dip; it has the highest relative risk or RR.) (2 pts.1 for each answer.)

40.(2 pts.) (The risk of becoming ill was 13.8 times higher for restaurant patrons who are potato dipcompared to patrons who did not eat potato dip.) (2 pts. 1 for correct interpretation of RR and onefor specifying potato dip in the answer.)

41.(2 pts.) (Next to one another in the fridge; sharing serving utensils; put in the same dish to serve;both foods taken from serving dish and put on personal plate; contaminated ingredient used in bothfoods; both were contaminated, confounding [always eaten together]; next to each other on same

plate or eaten together simultaneously; accept all reasonable answers.)

42.(1 pt.) (The CI provides information on the precision, reliability (significance) of the estimate of the

relative risk.)

43.(2 pts.) (Not significant, because the CI crosses 1.0.) (2 pts. 1 for no and 1 for CI crossing 1).

44.(1 pt.) (I would expect electromyography to have a lower PPV than a stool test.)

45.(2 pts.) (PPV is the proportion of positive test results that accurately help diagnose a case. Theoutbreak investigators felt confident enough to define a confirmed case according to a positive stool

test, but only used a positive electromyography test to indicate that a case was probable. Therefore,we can assume that a positive stool test more reliably predicts a case of botulism and therefore has ahigher PPV than electromyography.) (ACCEPT ANY REASONABLE ANSWER.)


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46. (2 pts.) Fill in the 2 x 2 table below. (2 pts. for each correctly-filled box.)

47.(4 pts.) (Sensitivity = 20 / (20 + 5) = 20/25 = 80%; Specificity = 5 / (5 + 10) = 5/15 = 33% (4 pts. 1 each for calculations; 1 each for correct answers.)

48.(4 pts.) (Yes. The high sensitivity of the test (80%) indicates that persons with botulism will likelybe detected by the test. However, the low specificity (33%) suggests that persons who do not havebotulism are also likely to have a positive electromyography test. Therefore, because a positive testis inclusive of casesbut not exclusive of non-casesinvestigators were wise to not consider all

positive electromyography tests as confirmed cases; but rather to use it as an indication of aprobable case.) (ACCEPT ANY ANSWER THAT CORRECTLY INTERPRETS SENSITIVITY,

SPECIFICITY, AND LINKS IT BACK TO CASE DEFINITIONS.) (4 pts. 1 for yes, 1 forrelating high sensitivity to the answer; 1 for relating low specificity to the answer; and 1 for the link

between sensitivity and specificity.)

Test (+) Test (-)

Botulism (+)20 5

Botulism (-)10 5


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Part C: Lassa Fever2

49.(1 pt.) Figure 3b. (correct answer)

50.(1 pt.) (10/26 = 38.5%.)

51.(1 pt.) (7/26 = 26.9%.)

52.(3 pts.) Airborne transmission, Direct contact, Vehicle borne transmission

53.(1 pt.) zoonosis.

54.(1 pt.) (Reservoirif they only have host; minus 0.5 point for Div. C, 1 pt. for Div. B.)

55.(1 pt.) fomites

56.(6 pts.)Case Classification

2 (Not a case)

3 (Suspect)

6 Confirmed

8 (Probable)

9 (Probable)

10 (Suspect)

57. (1 pt.) (3/12 = 25% (or 250 per 1000, other units are acceptable.)

2 Some of the studies and data presented in this event have been fabricated or pooled in an effort to develop a cohesive event.However, we believe the underlying concepts and descriptions of the epidemiology of Lassa fever to be scientifically sound.


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58.(2 pts.) Possible answers include:

Ensure every case is diagnosed consistently

Using objective criteria

Sensitive and specific case definition captures only true cases of the disease inquestion

So that we can compare cases across time and place with some certainty that theyre

really the same condition To detect differences in disease occurrence

Accept all reasonable answers

59.(2 pts.) (Because of its non-specific symptoms, Lassa fever can easily be confused with

other illnesses common in the tropics, like malaria, typhoid, and other viral hemorrhagic

fevers. In low-income countries, laboratory facilities may not be available to perform

diagnostic tests.)

60. (2 pts.) (You would want to use a loose case definition because it is important to identify

every single case so that you can begin planning what action to take. It would be better to

include some patients who do not in fact have Lassa to make sure that you dont miss any

true [and potentially infectious] cases. 1 point for broad/narrow; 1 point for explanation)

61.(4 pts.)(1 pt for proper labels, 1 point for correct numbers, 2 points for calculations; minus2 points if not odds ratio)

Households with Lassa

Fever

Households without

Lassa FeverStoring open food 29 5

Not storing open food 55 117

(12.34 or = 29*117/55*5 = 3393/275 = 12.34)

62.(2 pts.) (Households that store open food have 12.34 times the odds (more likely) to have

Lassa fever than people who live in households that do not store open food.)


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63.(4 pts.)(1 pt for proper labels, 1 point for correct numbers, 2 points for calculations; minus

2 points if not odds ratio)

Households with Lassa

Fever

Households without

Lassa FeverGood Housing Quality 34 73

Poor Housing Quality 50 49

(0.456 OR = 34*49/50*73 = 1666/3650 = 0.456)

64.(2 pts.) (Households that have good housing quality have 0.456 times the odds (54.4% less

likely) to have Lassa fever than people who live in households that have poor housing

quality. Also can be 1/0.456 or 2.193 times the odds (119.3 % more likely).)

65.(1 pt.) protective

66.(1 pt.) (Case control.)

67.(2 pts.)

X= hospitalized patients

Y= total casesX/Y = proportion of total cases that are hospitalized

.01Y=.15X

.01 = .15X/Y

.01/.15 = X/Y

(About 6.7% of cases are hospitalized, because of the 1520% range.) (Accept anyanswer from 56.7%.)

68.(1 pt.)

.01Y.15X

.01 .15X/Y

.01/.15 X/Y(Underestimate.)

69.(1 pt.) (Nosocomial, healthcare-associated infections, or HAIs)

70.(1 pt.) (19)

71.(3 pts.) (Wear gloves, masks, protective clothing, patient isolation, and dont reusesyringes.)


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72.(2 pts.) (Hospitals have more severe cases and have later stages of the disease.)

73.(4 pts.)

A: Possible advantages include:

Serious illness, so many cases go to the hospital Because medical professionals can provide a correct diagnosis

Because medical records are available from hospitals

Hospitalized patients are a captive audience and will answer questions aboutrisk factors

Poor community-based surveillance

Easier to collect data

B: Possible disadvantages include:

No information on less severe cases that dont go to the hospital or moresevere cases that die before they get to the hospital

Selection bias for people who can get to the hospital Might miss very rural or very poor people who cant get to the hospital

Eighty percent of patient-cases are asymptomatic

74.(5 pts.) (Use rodent-proof containers, dispose of garbage away from the home, keep a cleanhouse, have cats, keep food off the ground, use mouse traps, health education, and do not

eat rats.)

75.(3 pts.) (Stay away from infected people, bandage sores and cuts, do not exchange bodilyfluids.)

Table 8.

Viral IsolationAntibody Test Positive Negative

Positive 22 7

Negative 3 68

Total 25 75

76.(2 pts.) a = 22, c = 30.88*25 = 22;

2522 = 3

77.(1 pt.) (7/75 = 9.33333% = 9.33%.)

78.(1 pt.) (4/25 = 16%.)

79.(1 pt.) (51/55 = 92.7272% = 92.7%)

80.(1 pt.) (92.7% of samples that test negative on the DNA Amplification [PCR] test arenot actually infected with Lassa fever virus. Use the number calculated in the previous

question.)81.

Disease-Detectives-C-Key.pdf (Nationals 2011)

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