If you have a large group (35 or more) you may want to begin with two test tubes containing NaOH.

The final number of "infected" test tubes will vary depending on (1) the number of trades and (2) how many trades occur between two already infected tubes.

A possible method to find "patient zero" is to have each student write his or her name on the board and underneath it the names of students with whom he or she exchanged fluids in the order in which the exchanges occurred. Then, as a class, highlight the names of the currently "infected" people (shown in bold in sample).

This visual representation can help clarify which students may have infected one another, and in what order. Students who "test positive" and find that everyone with whom they traded also tested positive may be original carriers of the disease (Cal, Dee, Gib, and Hal in this example). It is likely that there will be several candidates for "patient zero." Cross-checking the history of each contact can narrow the field, but probably not to less than two candidates. (For example, here, Cal and Dee can be eliminated as "patient zero" because their first contacts, Bob and Ed, did not infect their own second contacts, Fran and Ilsa; either Gib or Hal, however, could be "patient zero.") If students are unable to reach a clear conclusion, the exercise will raise useful questions about the challenges facing real epidemiologists as they try to trace the sources of an infection.

Sample Chart Tracing Infection

	Ann	Bob	Cal	Dee	Ed	Fran	Gib	Hal	etc.
Exchange 1	Fran	Cal	Bob	Ed	Dee	Ann	Hal	Gib
Exchange 2	Jo	Fran	Gib	Hal	Ilsa	Bob	Cal	Dee
Exchange 3	Gib	Dee	Ed	Bob	Cal	Jo	Ann	Lin

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Students' first models may vary greatly. Students can test the 50:1 ratio of the volumes of Earth to Moon by comparing the diameters of their models to the actual planetary diameters. Earth's diameter = 12,756 km. The Moon's diameter = 3,476 km. Have students use these numbers to find that the Moon's diameter is about 1/4 of Earth's. They can then check to see that the diameters of their models show the same relationship.

Students' estimations of distance between Earth and the Moon again may vary widely. They can find the actual scale distance by first determining the scale of their models based on the ratio of their model Earth's diameter and planet Earth's actual diameter. For example, if a model of Earth measures 8.5 cm in diameter, then 1 cm represents 1,500 km and the distance between the Earth and Moon models should be approximately 2.6 m. At this scale, the space shuttle's orbit would be about 0.41 cm above model Earth's surface, and the Sun would be about 1,000 m away.

If students have trouble with the solar system map, suggest that they place the Sun at one end of the paper and Pluto at the other end. Then they can determine the scale based on the distance between them. A scale of about 1 cm = 50 million km works well, putting Pluto approximately 118 cm away from the Sun. The map's scale will not be detailed enough to show the distance of the Moon from Earth. For that, students would need to use a scale of about 1 cm = 1 million km. Then, a map showing both the Moon and Pluto would be about 59 m long.

To calculate positions of Alpha Centauri, the center of the Milky Way, and M31 on the solar system map, students need to determine the length of one light-year (in the scale they are using). For example, at a scale of 1 cm = 50 million km, one light-year (9,450,000,000,000 km) = 189,000 cm or 1.89 km. On that map, the distance from the Sun to Alpha Centauri would be about 8.13 km, from the Sun to the center of the Milky Way would be 56,700 km (about 4.5 times the distance around the earth), and from the sun to M31 would be 3.78 million km.

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1. they catch mice
2. they are cheaper
3. it wears well
4. what a man does . . .
5. I don't know him . . .
9. an automobile is more . . .
10. makes a wide, light surface
11. it avoids confusion . . .
12. it gives them control . . .
13. water expands . . .
picture 12. pig's tail
picture 13. crab's small left claw
picture 14. cat's shadow
picture 15. man's bowling ball
picture 16: net

Back to Testing Testing Activity

We are able to hold facts in short-term memory as long as we actively think about them. Once we stop, short-term memory fades after about 20 seconds. Short-term memory also has a finite capacity; it retains seven items, plus or minus two. Short-term memory is more accurately called "working" memory, suggesting a work space in which we actively organize and manipulate information, rather than a passive storage place.

Recall is one measure of memory. Students' graphs should show a decrease in short-term memory over time, usually represented by a classic, negatively accelerated curve. A week or two after the activity, ask the class to recall as many items as they can. Students may be surprised at what they remember -- that is, how flat the forgetting curve becomes after the initial steep decline.

Students may propose a number of valid hypotheses about the nature of memory. They may find that they are more likely to recall meaningful information -- names rather than other words, and words rather than nonsense syllables. They may also hypothesize that repetition aids memory: Repeated nonsense syllables are more likely to be recalled than those heard only once. In addition, items that stand out, such as the shouted syllable, are typically remembered more often than other nonsense syllables. Finally, students may observe that it is easiest to remember items near the beginning of the list; items near the end of the list are remembered more often than items just past the middle of the list.

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For best results with this activity, students should understand polymer structure, osmosis/diffusion, and hypertonic, hypotonic, and isotonic environments.

Superabsorbents are made of a polymer network and can absorb large amounts of water due to their ionic nature and interconnected structure. This network creates a membrane that allows water diffusion. The polymer absorbs water by osmosis: The large concentration of sodium inside the polymer network makes water flow into the polymer until equilibrium is reached, or the concentration of ions in the polymer equals the concentration of ions outside the polymer. (From a biological point of view, the polymer is hypertonic relative to the outside environment, which is hypotonic. Therefore, water diffuses into the polymer until equilibrium is reached.) Water is caught and held inside the polymer network by hydrogen bonds that it forms with the sodium acrylate monomers, forming a gel.

Sodium polyacrylate can absorb about 800 times its weight in distilled water (800 g water : 1 g polymer) and 300 times its weight in tap water. Absorbency decreases as the water's ion concentration increases. In diapers, the polymer's absorbency is only 30 times its weight due to urine's salt concentration (0.9% NaCl [aq]) plus the mechanical pressure from a baby's weight.

Students' results may differ from these absorbencies. Students should hypothesize why their results differ based on the method they used. For the suggested method, the amount of time allotted for absorption, variations in initial weights of dry polymer, and human error in determining when saturation is reached can lead to slightly different absorbency ratios. To vary this activity, students might keep the amount of water constant and determine the amount of polymer needed to absorb it. Or, they might vary salt concentration and determine the amount of polymer that can be gelled.

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An autoradiograph is a photographic image that lets us visualize segments of DNA. A process called gel electrophoresis separates DNA fragments by length. Each piece of DNA is labeled with a radioactive marker and recorded on x-ray film. The resulting series of bands show the different fragment sizes and can be used to compare DNA from different individuals.

The autoradiographs show that the letters in the child's genotype match those of its mother and the alleged father; therefore the alleged father is most likely biologically related to the child.

DNA-matching results for autoradiographs 1-4
	#1	#2	#3	#4
child	A/C	E/G	I/K	M/O
mother	A/D	G/G	I/J	M/N
alleged father	B/C	E/F	H/K	L/O

According to the University of Utah, this genetic test based on four markers indicates that there is more than a 99 percent probability that the alleged father is the child's biological parent. Because statistical calculations prevent a 100 percent probability of paternity, most courts accept a probability of 99.5 percent or greater as evidence of paternity. An "exclusion" result, where the patterns do not match between a child and the alleged father, is unequivocal evidence of non-paternity. Remind students that genetic tests can only absolutely disprove, not prove, relationship.

Back to DNA "Fingerprinting" Activity

Each of the isotopes in the activity -- represented by pennies, dice, or sugar cubes -- has a different half-life based on rate of decay. The rate of decay for each is determined by the likelihood that one unit of the isotope will be removed after a trial. The probability that a penny lies heads up is 1:2; that a sugar cube shows a dot is 1:6; and that a die shows a 1 or 2 is 1:3. Based on these probabilities, students will remove about half the pennies, about one-sixth of the sugar cubes, and about one-third of the dice after each trial. Thus a half-life for pennies will be about 1 trial; for sugar cubes, 3.8 trials; and for dice, 1.7 trials. The slopes of the graphs should show the number of radioactive, or undecayed, atoms declining exponentially and gradually approaching zero.

Just as it cannot be predicted exactly when an individual sugar cube will show a dot side, it cannot be predicted exactly when an unstable nucleus will decay. However, if a large enough number of sugar cubes are tossed, the probability of a dot side up occurs at a predictable rate. Similarly, a large group of identical nuclei will decay at a predictable rate. Scientists can use this predictable decay rate to estimate the age of materials that contain radioactive substances. One way is by comparing the ratio of radioactive atoms to stable daughter atoms. The greater the percentage of daughter atoms, the older the specimen. A formula based on the rate of decay helps pinpoint the actual age.

Procedure 2 of the activity helps assess students' understanding of half-life. If students have difficulty, suggest that they choose an age for their rock that is a multiple of the half-life. For example, suppose a rock has a half-life of 3 trials and 1 trial equals 1,000 years. Its half-life (or the time it takes for half of its atoms to decay) would be 3,000 years. If this rock were 6,000 years old, it would have gone through two half-lives. Therefore, given 80 original atoms, 40 would have decayed after the first half-life and the other 40 would not have decayed. Over the second half-life, half of the remaining 40 radioactive atoms would decay, leaving 20 undecayed atoms and 60 daughter atoms. A box representing this 6,000-year-old rock would have 20 of the original atoms and 60 daughter atoms.

Back to The Dating Game Activity