Dayton Regional Stem Center – Science, Technology ...



Printable Resources

Next Generation Super Hero

Appendix A: Pre/Post-Test

Appendix B: Pre/Post-Test and Answer Key

Appendix C: Genes and Bitter Taste: Article and Sample Organizers

Appendix D: Take a Class Survey Lab Sheet

Appendix E: Probability and Heredity Lab Sheet

Appendix F: Argument Organizer and Essay Template

Appendix G: Malaria Resistant Mosquito Article

Appendix H: Perspectives on Probability and Exit Ticket

Appendix I: Coin Toss Activity

Appendix J: Nature’s Fury Phenotype Brainstorm Handout

Appendix K: Teacher Instructions :Next Generation Super Hero-Class

Appendix L: Teacher Instructions: Next Generation Super Hero-Team

Appendix M: Class Activity: Next Generation Super Hero

Appendix N: Team Activity: Next Generation Super Hero

Appendix O: Exponent Rules and Practice

Appendix P: Getting Started with “Scratch” – Instructions and Rubric

Appendix Q: Sample Allele Strips

Appendix A: Pre-Test/Post-Test

1. (2 points) A spinner and a number cube are used to play a game.

 

The sides of the cube are numbered 1 through 6. A player’s score is the sum of the numbers on which the spinner lands and the number rolled on the cube.

 

                  [pic]

 

Find the probability of getting a sum greater than 25. Show your work or provide an explanation to support your answer.

2. (2 points) Jennie calculated the probabilities of various events involving a coin.

What is the probability of a coin landing on heads all three times when the coin is flipped three times? Express your answer with a negative exponent instead of a fraction.

3. (2 points) A bag contains 2 Batman action figures, 4 Superman action figures and 6 Spiderman action figures. Mark draws a single action figure at random, records it and returns it to the bag. He does this 40 times. His results are shown:

Batman: 2

Superman: 18

Spiderman: 20

 

Based on theoretical probability, how does the number of Superman action figures Mark drew from the bag compare with the number he would be expected to draw?

4. (2 points) Simplify. Rewrite the term with a negative exponent as a part of showing your work.    [pic]

5. (2 points) What is the value of [pic] ? Rewrite the terms with negative exponents as a part of showing your work.    

6. (2 points) Determine whether the following statement is sometimes, always, or never true. Justify your answer.

A coin is used to model a simulation between two events.

7. (2 points – Simulation) A local comic book store is advertising a campaign to give prizes to 3 out of every 9 customers that come into the store. Describe a simulation that could be used to model this scenario.

What are the chances that two customers in a row win a price?

8. (2 points – Experimental & Theoretical) A class designed the spinner shown to award prizes at the school fair.

[pic] 

At the fair, 450 students spun the spinner. The class awarded 115 slices of pizza, 49 CDs and 78 DVDs. 

How do the actual results differ from the expected results?

Use the following information to answer questions 9-11

Gregor Mendel worked with pea plants to study genetics. He tracked seven traits from generation to generation. The traits are shown in the table below.

Pea Plant Traits Compared by Mendel

[pic]

(Aralin, 2010)

9. List one possible phenotype of the pea plants Mendel studied. Next the

phenotype, write its corresponding genotype. (2pts)

10. Complete the following Punnett Square to illustrate the probability of pea

plant offspring height. Next to the Punnett Square, write the fraction of tall

plants. (2 pts)

[pic]

11. Complete the Punnett square by filling in the parent genotypes so the

probability of short plants is 1/4.

[pic]

12. After you have complete the first cycle of the Engineering Design Process,

what is the next step?

Explain why this next step is a vital to the nature of Engineering Design.

13. The ability to roll your tongue is an inherited, dominant trait. A mother is able

to roll her tongue but her husband and son are not. What fraction of the

couple’s children would be expected to be able to roll their tongues? Explain

your reasoning with words or a Punnett square. (2pts)

14. If a pea plant has a genotype of Ggtt, what are the possible genetic

combinations that could be present in this plant?

A. Gt, gt

B. GT, gt

C. GT, Gt, gt

D. Gt, gt, gT, GT

15. In humans, the gene for brown eyes (B) is dominant and the gene for blue

eyes (b) is recessive. What are the possible combinations of genes in the

offspring of two brown-eyed heterozygous parents (Bb)?

A. BB only

B. bb only

C. Bb and bb only

D. BB, Bb, and bb only

16. Albinism in animals occurs when the animal has a lack of pigment in the skin,

fur or eyes. Albinism is caused by a recessive allele (a) of a particular gene,

and a normal pigment is caused by its dominant allele (A). What fraction of

the offspring of a normal heterozygous (Aa) animal and an albino animal (aa)

would be expected to have normal pigmentation?

A. 0/4

B. 1/4

C. 2/4

D. 4/4

17. Ear length in dogs is controlled by a gene that can occur in the dominant

form, (L), or the recessive form, (l). The color of a dog’s nose is controlled

by another gene which occurs in the dominant form, (B), or the recessive

form, (b). The table below shows the traits for these allele codes.

The following genotypes were found in a male dog and a female dog.

|Gene |Trait |

|L |long ears |

|l |short ears |

|B |black nose |

|b |non-black nose |

LlBb (male) llBB (female)

Which one of the following choices is true of the phenotype of offspring from

these parents?

A. All offspring will have long ears.

B. All offspring will have short ears.

C. All offspring will have black noses.

D. All offspring will have non-black noses.

Appendix B: Pre/Post-Test Answer Key

Pre/Post-Test Key

Answer Key

1. (2 points – Compound Probability)

Total outcomes: 4 (from the spinner) * 6 (from the cube) = 24

Desired outcomes: 6 options with 40 on spinner, 6 options with 30 on spinner, 1 option with 20 on spinner (6 on cube and 20 on spinner), so 6 + 6 + 1 = 13

The probability of getting a sum greater than 25 is 13/24.

2. (2 points – Compound Probability & Negative Exponents)

The probability of a coin landing on heads is ½.

Since we need three heads in a row, then P(3 heads) = ½ * ½ * ½

Now 2^(-1) is equal to ½, so P(3 heads) = 2^(-3).

3. (2 points – Experimental & Theoretical)

The theoretical probability of picking a Superman is 4/12 because there are 4 Superman figures among (2+4+6=12) 12 total action figures. 4/12 simplifies to 1/3 and 1/3 of 40 chances is around 13 or 14. So, the theoretical chance of picking Superman in this sample space of 40 is 13 or 14. However, the results show that Superman was picked 18 times or 4 more times than expected. The experimental probability was higher than the theoretical by about 12%.

1 point – The correct theoretical probability of 4/12 is used in comparison statement.

1 point – There is a comparison between the 4/12 and 18/40 made in the explanation using differences, percentages, or fractions.

4. (2 points – Negative Exponents)

5² * 1/5² = 25 * 1/25 = 25 / 25 = 1

1 point – 5^(-2) rewritten as 1/25 or 1/(5²)

1 point – Simplified expression equaling 1

5. (2 points – Negative Exponents)

2^0 = 1

2^(-1) = 1/(2^1) = ½

2^(-2) = 1/(2^2) = ¼

So, 1 + ½ + ¼ = 1 and ¾ or 1.75

1 point – Rewrite the terms with negative exponents into fractions

1 point – Correctly simplify the expression to 1.75 or 1 and 3/4.

6. (2 points - Simulation)

A coin can sometimes be used to model a simulation between events. A coin would work perfectly for events with equal probabilities but would not work for events with unequal chances of occurring.

1 point – Correct identification of sometimes

1 point – Correct reasoning identifying the reason why a coin would not always work for two events

7. (2 points - Simulation)

A simulation that could be used for the comic book scenario is a spinner divided into three equal sections (with one identified as a winner and the others as not) or a number cube where two numbers are winners and the rest are not. Any simulation is accurate if a winner has a chance of winning 1/3 of the time.

The chances that two customers in a row win a prize is 1/3 * 1/3 = 1/9.

1 point – A simulation with a win occurring 1/3 of the time.

1 point – The correct answer of 1/9 for two people in a row winning.

8. The theoretical probabilities for the spinner are the following:

DVD = 1/8 CD = 1/8 No prize = 4/8 Pizza = 2/8

The experimental probabilities are:

DVD = 78/450 CD = 49/450 No prize = 208/450 Pizza = 115/450

So, the actual results differ from the expected results because the number of DVD winners should be closer to matching the number of CD winners since they have the same theoretical probabilities. The CD winners are under the expected amount while the DVD winners are over the expected amount winner.

1 point – Correct identification of theoretical probabilities for each category

1 point – Correct explanation of the inconsistency between CD and DVD winners (or argument for slight decrease in No Prize could also be mathematically proven)

Use the following information to answer questions 9-11.

Gregor Mendel worked with pea plants to study genetics. He tracked seven traits from generation to generation. The traits are shown in the table below.

Pea Plant Traits Compared by Mendel

[pic]

(Aralin, 2010)

9. List one possible phenotype of the pea plants Mendel studied. Next the

phenotype, write its corresponding genotype. (2pts)

10. Complete the following Punnett square to illustrate the probability of pea plant

offspring height. Next to the Punnett square, write the fraction of tall plants. (2

pts)

[pic]

11. Complete the Punnett square by filling in the parent genotypes so the

probability of short plants is 1/4.

[pic]

12. After you have complete the first cycle of the Engineering Design Process,

what is the next step?

After you have designed and tested during your first cycle the next step is

to redesign.

Explain why this next step is a vital to the nature of Engineering Design.

This step is vital because the Engineering Design Process is iterative. You may go through everything multiple times before you have a product an artifact that functions as needed.

13. The ability to roll your tongue is an inherited, dominant trait. A mother is able

to roll her tongue but her husband and son are not. What fraction of the

couple’s children would be expected to be able to roll their tongues? Explain

your reasoning with words or a Punnett square. (2pts)

2/4 of the children would be expected to roll their tongues. The mother

would have to be heterozygous dominant (Rr) for tongue rolling and the

husband would have to be homozygous (rr) recessive since he is not able

to roll his tongue. The possible combinations for all children would be Rr

and rr, meaning there is a 50% chance offspring would be able to roll their

tongues.

14. If a pea plant has a genotype of Ggtt, what are the possible genetic

combinations that could be present in this plant?

A. Gt,, gt

15. In humans, the gene for brown eyes (B) is dominant and the gene for blue

eyes (b) is recessive. What are the possible combinations of genes in the

offspring of two brown-eyed heterozygous parents (Bb)?

D. BB, Bb, and bb only

16. Albinism in animals occurs when the animal has a lack of pigment in the

skin, fur or eyes. Albinism is caused by a recessive allele (a) of a particular

gene, and a normal pigment is caused by its dominant allele (A). What

fraction of the offspring of a normal heterozygous (Aa) animal and an albino

animal (aa) would be expected to have normal pigmentation?

C. 2/4

17. Ear length in dogs is controlled by a gene that can occur in the dominant

form, (L), or the recessive form, (l). The color of a dog’s nose is controlled

by another gene which occurs in the dominant form, (B), or the recessive

form, (b). The table below shows the traits for these allele codes.

The following genotypes were found in a male dog and a female dog.

|Gene |Trait |

|L |long ears |

|l |short ears |

|B |black nose |

|b |non-black nose |

LlBb (male) llBB (female)

Which one of the following choices is true of the phenotype of offspring from

these parents?

C. All offspring will have black noses.

Appendix C: Genes and Bitter Taste Article and Sample Organizers

PTC: Genes and Bitter Taste

In 1931, a chemist named Arthur Fox was pouring some powdered PTC into a bottle. When some of the powder accidentally blew into the air, a colleague standing nearby complained that the dust tasted bitter. Fox tasted nothing at all. Curious how they could be tasting the chemical differently, they tasted it again. The results were the same. Fox had his friends and family try the chemical then describe how it tasted. Some people tasted nothing. Some found it intensely bitter, and still others thought it tasted only slightly bitter.

[pic][pic]

The ratio of tasters to non-tasters varies between populations, but every group has some tasters and some non-tasters. On average, 75% of people can taste PTC, while 25% cannot.

The PTC Gene

Soon after its discovery, geneticists determined that there is an inherited component that influences how we taste PTC. Today we know that the ability to taste PTC (or not) is conveyed by a single gene that codes for a taste receptor on the tongue. The PTC gene, TAS2R38, was discovered in 2003.

There are two common forms (or alleles) of the PTC gene, and at least five rare forms. One of the common forms is a tasting allele, and the other is a non-tasting allele. Each allele codes for a bitter taste receptor protein with a slightly different shape. The shape of the receptor protein determines how strongly it can bind to PTC. Since all people have two copies of every gene, combinations of the bitter taste gene variants determine whether someone finds PTC intensely bitter, somewhat bitter, or without taste at all.

Natural Selection At Work

[pic]

PTC stands for phenylthiocarbamide. Also known as phenylthiourea, the chemical structure of PTC resembles toxic alkaloids found in some poisonous plants.

Although PTC is not found in nature, the ability to taste it correlates strongly with the ability to taste other bitter substances that do occur naturally, many of which are toxins.

Plants produce a variety of toxic compounds in order to protect themselves from being eaten. The ability to discern bitter tastes evolved as a mechanism to prevent early humans from eating poisonous plants. Humans have about 30 genes that code for bitter taste receptors. Each receptor can interact with several compounds, allowing people to taste a wide variety of bitter substances.

To Taste Or Not To Taste

If the ability to taste bitter compounds conveys a selective advantage, then shouldn't non-tasters have died off long ago? Why do so many people still carry the non-tasting PTC variant? Some scientists believe that non-tasters of PTC can taste another bitter compound. This scenario would give the greatest selective advantage to heterozygotes, or people who carry one tasting allele and one non-tasting allele.

The ability to taste PTC shows a dominant pattern of inheritance. A single copy of a tasting allele (T) conveys the ability to taste PTC. Non-tasters have two copies of a non-tasting allele (t).

Not So Simple After All

[pic]

Curiously, there are also tasting and non-tasting chimpanzees. Unlike non-tasting humans, chimps that cannot taste PTC appear to lack functional PTC receptors.

PTC sensitivity is often used as an example of a simple Mendelian trait with dominant inheritance. However, tasters vary greatly in their sensitivity to PTC. And while the PTC gene has about 85% of the total influence over whether someone is a taster or a non-taster, there are many other things that affect PTC tasting ability. Having a dry mouth may make it more difficult to taste PTC. What you ate or drank before sampling PTC paper may also affect your tasting ability. And an individual's sensitivity may change over time. Some people may find that they can taste PTC on some days, but not on others.

Potential Health Applications

Studies indicate that individuals with the "strong tasters" PTC gene variant were less likely to be smokers. This may indicate that people who find PTC bitter are more likely than non-tasters to find the taste of cigarettes bitter and may be less likely to smoke.

Other studies suggest that there may be correlations between the ability to taste PTC and preferences for certain types of foods. This may be why some of us think that broccoli is just too bitter to eat.

Personal Genetics Sample Chart:

|Taste PTC paper | |

|No Taste on PTC paper | |

Sample Punnett Square:

[pic]

Appendix D: Take a Class Survey Lab Sheet

Are Your Traits Unique?

1. Instruct students to place their finger on “ear lobes in the center of the circle of traits. Instruct students to select the small central circle that applies to them-either free ear lobes or attached ear lobes.

2. Instruct students to move their finger onto the next description that applies to them. Instruct students to continue using their finger to trace their traits until they reach a number on the outside rim of the circle.

3. Instruct students to record their number on lab sheet.

4. Instruct students to share their results with the class.

5. Instruct students to answer the following questions:

How many students ended up on the same number on the circle of traits? How many students were the only ones to have their number? What do the results suggest about each person’s combination of traits? Does your data support the hypothesis you proposed?

Take a Class Survey

| |Trait 1 |Number |Trait 2 |Number |

|A |Free ear lobes | |Attached ear lobes | |

|B |Hair on fingers | |No hair on fingers | |

|C |Widow’s peak | |No widow’s peak | |

|D |Curly hair | |Straight hair | |

|E |Cleft chin | |Smooth chin | |

|F |Smile dimples | |No smile dimples | |

1. Hypothesis:

2. Which traits controlled by dominant alleles were shown by a majority of students?

3. Which traits controlled by recessive alleles were shown by a majority of students?

4. My final number on outer rim ________

5. How many students ended up on the same number on the circle of traits? How many students were the only ones to have their number? What do the results suggest about each person’s combination of traits

6. Does your data support the hypothesis you proposed?

Appendix E: Probability and Heredity Lab Sheet

[pic]Probability and Heredity

Complete the two Punnett Squares below, and then answer the questions on a separate sheet of paper.

1. In the cross between two black guinea pigs shown in Punnett Square A, what is the probability that an offspring will be black?  White?

1. Is it possible that the cross between two black guinea pigs in Punnett Square A would no produce a white guinea pig?  Explain.

2. What color are the guinea pig parents in the cross shown in Punnett Square B?

3. Which guinea pig parent(s) in Punnett Square B is homozygous?  Which is heterozygous?  Explain how you know.

4. Calculate the probability that an offspring will be black in the cross in Punnett Square B.  What is the probability that an offspring will be white?

Vocabulary Assessment

Match each term with its definition by writing the letter of the correct definition on the line beside the term.

|_____ 1.  Heterozygous |a.  a chart that shows all the possible combinations of alleles that can result from a |

| |genetic cross |

|_____ 2.  Punnett Square |b.  a number that describes how likely it is that an event will occur |

|_____ 3.  Genotype |c.  an organism that has two identical alleles for a trait |

|_____ 4.  Codominance |d.  an organism's physical appearance |

|_____ 5.  Probability |e.  an organism's genetic makeup, or allele combinations |

|_____ 6.  Homozygous |f.  an organism that has two different alleles for a trait |

|_____ 7.  Phenotype |g. inheritance pattern in which the alleles are neither dominant nor recessive |

[pic]Probability and Heredity (Answer Key)

Complete the two Punnett Squares below, and then answer the questions on a separate sheet of paper.

[pic]

2. In the cross between two black guinea pigs shown in Punnett Square A, what is the probability that an offspring will be black?  White?

¾ black ¼ white

5. Is it possible that the cross between two black guinea pigs in Punnett Square A would no produce a white guinea pig?  Explain.

Yes if both parents were homozygous dominant for black color

6. What color are the guinea pig parents in the cross shown in Punnett Square B?

One white one black

7. Which guinea pig parent(s) in Punnett Square B is homozygous?  Which is heterozygous?  Explain how you know.

White (two recessive) Black one dominant one recessive

8. Calculate the probability that an offspring will be black in the cross in Punnett Square B.  What is the probability that an offspring will be white?

½ 1/2

|__f___ 1.  Heterozygous |a.  a chart that shows all the possible combinations of alleles that can result from a |

| |genetic cross |

|__a___ 2.  Punnett Square |b.  a number that describes how likely it is that an event will occur |

|__e___ 3.  Genotype |c.  an organism that has two identical alleles for a trait |

|__g___ 4.  Codominance |d.  an organism's physical appearance |

|__b___ 5.  Probability |e.  an organism's genetic makeup, or allele combinations |

|__c___ 6.  Homozygous |f.  an organism that has two different alleles for a trait |

|__d__ 7.  Phenotype |g. inheritance pattern in which the alleles are neither dominant nor recessive |

Appendix F: Argument Organizer and Essay Template

Malaria Resistant Mosquitoes

Your position on the issue: __________________________

Choose supporting facts from the reading selection and explain why each scaffolds your position.

Support: Explanation:

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

| | |

Malaria Resistant Mosquitoes

Defense

Do you believe that malaria resistant mosquitoes should be bred?

1. State the position of your group regarding the question above.

2. Use specific support from the article to scaffold your position.

3. Explain each supporting fact from the article and make connections to your

position.

4. Form a conclusion that leaves the reader with a point that they may take with

them.

| |

| |

| |

| |

| |

| |

| |

| |

| |

| |

| |

| |

| |

| |

|Essay Rubric |

| |4 |3 |2 |1 |

| Position |Writer has clearly stated |Writer has stated their |The writer is missing |The position or intent of|

| |their position and has made|position and their intent |either their position or |support is missing from |

| |it clear their intent of |of support, but one is not |intent of support. |the essay and the other |

| |support. |clear. | |is not clear. |

|Support |There are four pieces of |There are three pieces of |There are two pieces of |There is one piece of |

| |evidence from the article |evidence from the article |evidence from the article |evidence from the article|

| |listed in support of the |listed in support of the |listed in support of the |listed in support of the |

| |writer’s position. |writer’s position. |writer’s position. |writer’s position. |

|Connections |All the pieces of evidence |Three out of four pieces of|Two out of four pieces of |One out of four pieces of|

| |make clear connections to |evidence make clear |evidence make clear |evidence makes clear |

| |the writer’s position. |connections to the writer’s|connections to the |connections to the |

| | |position. |writer’s position. |writer’s position. |

|Conclusion |The writer has formed a |The writer has formed a |The writer has formed a |The writer has formed a |

| |strong conclusion and it |conclusion and it leaves |strong conclusion with no |weak conclusion with no |

| |leaves the reader with a |the reader with a weak |clear stance on their |clear stance on their |

| |strong point about their |point about their position.|position. |position. |

| |position. | | | |

|Mechanics |All parts of the essay have|There are fewer than two |There are fewer than three|There are fewer than four|

| |correct spelling and |mistakes in either spelling|mistakes in either |mistakes in either |

| |grammar. |or grammar. |spelling or grammar. |spelling or grammar. |

Score: ____________ /20

Appendix G: Malaria Resistant Mosquito Article

Malaria-Resistant Mosquitoes Bred in Lab for First Time

Releasing such mosquitoes in strategic locations could dramatically reduce the spread of malaria

Jun 18, 2012 |By Lacey Johnson and ClimateWire

|

[pic]

Wikimedia Commons/Centers for Disease Control and Prevention

Advertisement

Scientists may have developed a new tool for combating malaria, according to a recent study in the Proceedings of the National Academy of Sciences.

After more than 20 years of genetic experimentation, researchers have discovered how to breed malaria-resistant mosquitoes that are unable to infect humans with their bites.

"We see a complete deletion of the infectious version of the malaria parasite," said Anthony James, a microbiology and molecular genetics professor at the University of California, Irvine, and the lead author of the report. This can help to "significantly reduce human sickness and death," he added.

With the help of fellow researchers from the Pasteur Institute in Paris, James and his colleagues were able to alter the DNA of Anopheles stephensi mosquitoes, which are major transmitters of the most deadly strain of malaria -- Plasmodium falciparum.

By microinjecting a specially engineered gene into the mosquitoes' eggs, the scientists produced insects that were unable to transmit the disease when they reached adulthood.

More importantly, the gene that James' team created was dominant. In other words, introducing it into a wild population of mosquitoes would achieve the same result as placing a group of brown-eyed humans into a blue-eyed population: gradually, fewer children would be born with the recessive, blue-eyed gene.

This means that releasing the mosquitoes in strategic locations could dramatically reduce the spread of malaria, James said.

A disease poised to spread

According to the World Health Organization, more than 650,000 people died from malaria in 2010, most of whom were African children. Some researchers believe this number could climb even higher due to climate change, which is expected to increase rainfall in many regions. More puddles and swampland would provide additional breeding grounds for mosquitoes -- translating into more cases of malaria (ClimateWire, Nov. 21, 2011).

People who have never been exposed to the disease also run the highest risk of infection, so mosquitoes may spread malaria to countless new victims as they follow the rains into fresh territories, say experts.

To make James' malaria-fighting research a reality, millions of mosquitoes would need to be bred in a lab and released into the wild at key intervals.

"We have to figure out how these things are going to scale up," explained James, who says the process of caring for tropical mosquitoes can be very labor-intensive. "This is not something that people are going to be doing in their garage."

Aside from malaria, he believes, the research could ultimately be tailored to prevent other mosquito-borne diseases, such as the West Nile virus and dengue fever.

"I'm pretty enthusiastic that in five years, we'll have tools we'll be able to use," although the ethical, social and legal aspects will likely slow progress, he said.

Reprinted from Climatewire with permission from Environment & Energy Publishing, LLC. , 202-628-6500

Appendix H: Perspectives on Probability and Exit Ticket

Perspectives on Probability

[pic]

Exit Ticket

1. (4 points) Compare and contrast the difference between experimental and theoretical probability. Include how simulations impact both in your answer.

2. (4 points) Create and describe a scenario requiring a simulation. Be specific as you create your setting. What is the theoretical probability of your event?

3. (4 points) Justify the math model chosen for your description above. Complete the simulation using technology and analyze the results.

Perspectives on Probability Answer Key

Exit Ticket

1. (4 points) Compare and contrast the difference between experimental and theoretical probability. Include how simulations impact both in your answer.

Sample Answer: Theoretical probability is used before an event as an attempt to make a prediction about the likelihood of an event occurring. This is mathematically calculated. On the other hand, experimental probability is based on what actually happens and is calculated after an event. Both probabilities calculate chances from the same set of circumstances in the same way, dividing the desired outcomes by the total number of outcomes.

1 point – describe how theoretical is unique

1 point – describe how experimental is unique

1 point – describe how the probabilities are similar

1 point – clarity and conventions of writing are specific and correct, respectively

2. (4 points) Create and describe a scenario requiring a simulation. Be specific as you create your setting. What is the theoretical probability of your event?

Answers will vary

1 point – describe a scenario with clarity

1 point – describe a scenario with specificity

1 point – correctly create a situation that needs a simulation to predict an event

1 point – correctly determine the theoretical probability of the event created

3. (4 points) Justify the math model chosen for your description above. Complete the simulation using technology and analyze the results.

Answers will vary

1 point – identify an accurate tool to create the simulation

1 point – justify why the tool accurately reflects the simulation

1 point – correctly create a situation that needs a simulation to predict an event

1 point – analyze results by comparing simulation with theoretical probabilities

Appendix I: Coin Toss Activity

Coin Toss Activity to Demonstrate 50-50 Probability

1. Place a piece of tape on both sides of a penny and a nickel.

2. Use a permanent marker to label one side of each coin with an “E” to represent the dominant gene. Label the other side of each coin with an “e” to represent the recessive gene.

3. Flip the two coins together 50 times.

4. Organize your results in a data table using tally marks to record the gene combinations that result from each flip of the coins (“EE’”, “ee,” and “Ee”).

5. What percentage of the flips represents each of the gene combinations? Record percentages in the data table.

6. Record class results in the data table and determine the total number of “EE,”, “ee,” and “Ee” results.

7. What percentage of the flips represent each of the gene combinations? Record your percentages in the data table.

8. Compare your percentages for each gene combination with the class percentages. What do you notice about the percentages?

|Gene Combination | |% Flips Represented by Gene | |% Flips Represented by Gene |

| |My Results |Combination |Total Class Results |Combination |

| | | | | |

|“EE” | | | | |

| | | | | |

|“ee” | | | | |

| | | | | |

|“Ee” | | | | |

Appendix J: Nature’s Fury Phenotype Brainstorm Handout

Name: ____________________________ Date: ______

Nature’s Fury: Eyewitness Reports of Natural Disasters

Phenotype Brainstorm

Directions: Read an eyewitness’s report of their experience during a natural disaster. Complete the graphic organizer below to determine what phenotypes may have made it easier for that person to survive.

Natural Disaster: ______________________________________________

Location: ________________________________________________

Date: ________________________________________

|Adversity |Phenotype | |

|A brief description of a specific adversity being|A physical feature that may help a person |Explanation |

|faced in the natural disaster. |overcome the listed adversity. |An explanation of how the phenotype may help the |

| | |person overcome the adversity. |

| | | |

| | | |

| | | |

| | | |

| | | |

Appendix K: Teacher Instructions: Next Generation Super Hero - Class

Gene Pool

Begin with a set of alleles:

|V – normal vision |v – super vision |

|F – normal feet |f – fast feet |

|S – normal strength |s – super strength |

|E – normal ears |e – super ears |

|N – normal nose |n – super nose |

We can write the probability of a certain event as Prob(event). And the sum of the probabilities of all cases of an event add up to 1.

Population Statistics

Considering feet, what are the possible parent genotypes: FF, Ff, fF, ff

From this population of genotypes, let us compute

Prob(F allele) = 4/8 = 1/2

Prob(f allele) = 4/8 = 1/2

We can combine probabilities in two ways. If we want to know the probability that one event OR another event, maybe both, occurs, and those events are independent, we add them up:

Prob(F allele OR f allele) = 1/2 + 1/2 = 1

If we want to know the probability one event AND another event occurring together (we call this the joint probability), we multiply them

Prob(F allele AND f allele) = 1/2 x 1/2 = 1/4

Review Punnett Squares

Say we have two parents, FF and Ff, let’s review the probabilities of different genotypes in the offspring.

From FF parent, what is Prob(F allele)? Prob(F) = 1, Prob(f) = 0

From Ff parent, what are Prob(F allele) and Prob(f allele)? Prob(F) = 1/2, Prob(f) = 1/2

| |F |f |

|F |FF |Ff |

|F |FF |Ff |

From the table, what is the probability of the offspring having genotype FF? Ff? ff?

Now, from the probability rules, what is the Prob(F from Father and F from Mother)?

Prob(F from Father and F from Mother)?= 1 x 1/2 = 1/2

What is Prob(F.father and f.mother) = 1 x 1/2 = 1/2

What is Prob(f.father and f.mother) = 0 x 1/2 = 0

We will use these tools to predict the likelihood of producing a Next Generation Super Hero.

Set up the First Generation Example

1. Select a random parent from the Gene Pool (will be a heterozygous parent drawn from the pool).

a. The randomly drawn parent will have a genotype defined by two traits.

2. Assume our father (second parent) is a homozygous recessive parent for the two genes in the random parent– a superhero on both the features of the random parent.

a.

3. Set up the Punnett Square

a. Place the random parent (mother) on the top row

i. In the Mother’s genotype, there are two alleles, each with two possible values, so we use the notation:

Mother1.1 = first gene, first allele value

Mother1.2 = first gene, second allele value

Mother2.1 = second gene, first allele value

Mother2.2 = second gene, second allele value

ii. For example, if the Mother is SsEe, then

Mother1.1 = S

Mother1.2 = s

Mother2.1 = E

Mother2.2 = e

b. Place the homozygous recessive parent (father) on the left side

i. We label these alleles rec1 and rec2 for the two recessive values

| |Mother1.1-Mother2.1 |Mother1.1–Mother2.2 |Mother1.2–Mother2.1 |Mother1.2– Mother2.2 |

|rec1-rec2 |Mother1.1-rec1 - Mother2.1-rec2 |Mother1.1-rec1 - |Mother1.2-rec1 - |Mother1.2-rec1 - |

| | |Mother2.2-rec2 |Mother2.1-rec2 |Mother2.2-rec2 |

|rec1-rec2 |Mother1.1-rec1 - Mother2.1-rec2 |Mother1.1-rec1 - |Mother1.2-rec1 - |Mother1.2-rec1 - |

| | |Mother2.2-rec2 |Mother2.1-rec2 |Mother2.2-rec2 |

|rec1-rec2 |Mother1.1-rec1 - Mother2.1-rec2 |Mother1.1-rec1 - |Mother1.2-rec1 - |Mother1.2-rec1 - |

| | |Mother2.2-rec2 |Mother2.1-rec2 |Mother2.2-rec2 |

|rec1-rec2 |Mother1.1-rec1 - Mother2.1-rec2 |Mother1.1-rec1 - |Mother1.2-rec1 - |Mother1.2-rec1 - |

| | |Mother2.2-rec2 |Mother2.1-rec2 |Mother2.2-rec2 |

From the columns of the table, how likely is the offspring to have each possible from Parent 1? (similar questions about the other alleles and from Parent 2 on the rows)

Using the probability rules, from the random parent, find

Prob( Mother1.1) = 1/2

Prob( Mother1.2) = 1/2

Prob( Mother2.1 ) = 1/2

Prob( Mother2.2) = 1/2

Using the probability rules, from the homozygous recessive parent (father), find

Prob(Dom1 allele) = 0

Prob(rec1 allele) = 1

Prob(Dom2 allele) = 0

Prob(rec2 allele) = 1

In the offspring, find probabilities of various allele combinations

Prob(Mother1.1 and rec1) = 1/2 x 1 = 1/2 Prob(Mother2.1 and rec2) = 1/2 x 1 = 1/2

Prob(Mother1.1 and Dom1) = 1/2 x 0 = 0 Prob(Mother2.1 and Dom2) = 1/2 x 0 = 0

Prob(Mother1.2 and rec1) = 1/2 x 1 = 1/2 Prob(Mother2.2 and rec2) = 1/2 x 1 = 1/2

Prob(Mother1.1-rec1 and Mother2.1-rec2) = 1/2 x 1/2 = 1/4

Prob(Mother1.1-rec1 and Mother2.1-Dom1) = 1/2 x 0 = 0

Prob(Mother1.1-rec1 and Mother2.2-rec2) = 1/2 x 1/2 = 1/4

Second Generation

From this first generation, what are the unique genotypes available for parents?

Genotype 1: Mother1.1-rec1 & Mother2.1-rec2

Genotype 2: Mother1.1-rec1 & Mother2.2-rec2

Genotype 3: Mother1.2-rec1 & Mother2.1-rec2

Genotype 4: Mother1.2-rec1 & Mother2.2-rec2

Let’s look at the Punnett Squares for the possible offspring from each of these parents with the homozygous recessive super hero parent

Genotype 1:Mother1.1-rec1 & Mother2.1-rec2 Genotype 2: Mother1.1-rec1 & Mother2.2-rec2:

| |M1.1-M2.1 |M1.1-rec2 |rec1-M2.1 |rec1-rec2 |

|se |SsEe |Ssee |ssEe |ssee |

|se |SsEe |Ssee |ssEe |ssee |

|se |SsEe |Ssee |ssEe |ssee |

|se |SsEe |Ssee |ssEe |ssee |

From the columns of the table, how likely is the offspring to have e allele from Parent 1? (similar questions about the other alleles and from Parent 2 on the rows)

Using the probability rules, from parent SsEe, find

Prob(E allele) = 1/2

Prob(e allele) = 1/2

Prob(S allele) = 1/2

Prob(s allele) = 1/2

Using the probability rules, from parent ssee, find

Prob(e allele) = 1

Prob(E allele) = 0

Prob(S allele) = 0

Prob(s allele) = 1

In the offspring, find

Prob(S and s) = 1/2 x 1 = 1/2 Prob(E and e) = 1/2 x 1 = 1/2

Prob(S and S) = 1/2 x 0 = 0 Prob(E and E) = 1/2 x 0 = 0

Prob(s and s) = 1/2 x 1 = 1/2 Prob(e and e) = 1/2 x 1 = 1/2

Prob(Ss and Ee) = 1/2 x 1/2 = 1/4

Prob(Ss and EE) = 1/2 x 0 = 0

Prob(Ss and ee) = 1/2 x 1/2 = 1/4

Prob(SS and Ee) = 0 = Prob(SS and EE) = Prob(SS and ee)

Prob(ss and Ee) = 1/2 x 1/2 = 1/4

Prob(ss and EE) = 1/2 x 0 = 0

Prob(ss and ee) = 1/2 x 1/2 = 1/4

Do these numbers match the proportions in the tables?

What is the probability of a Next Generation Super Hero (homozygous recessive on all traits) in the first generation? 1/4

Second Generation

From this first generation, what are the unique genotypes available for parents?

Heterozygous dominant for both traits – SsEe

Heterozygous dominant for one trait, homozygous recessive for the other – ssEe, Ssee

Homozygous recessive for both traits – ssee

Second Generation

From this first generation, what are the unique genotypes available for parents?

Heterozygous dominant for both traits – SsEe

Heterozygous dominant for one trait, homozygous recessive for the other – ssEe, Ssee

Homozygous recessive for both traits – ssee

Let’s look at the Punnett Squares for the possible offspring from each of these parents with the ssee super hero parent

| |SE |

|F – regular feet |f – Fast feet |

|S – normal strength |s – Super strength |

|E – normal ears |e – Enormous ears |

|N – normal nose |n – Super nose |

A parent in the Gene Pool is defined by two of the above traits, containing 4 alleles.

Preparation:

1. Base on the size of the desired student group (call it Group Size), set up the gene pool with the Group Size number of parents for each trait pair

a. With 5 traits, there are 10 possible trait pair groups

b. For a class of 30, set up to groups of 3 students.

Select Groups:

1. Each student selects a parent from the Gene Pool

a. Each parent will have two genes, a pair of the above, and is heterozygous

2. Students group together according to the parents randomly drawn from the gene pool. Their pedigree and super hero will have the two traits for the specific group.

Set up the statistical model

1. Recall population statistics (probabilities of alleles), and rules for combining probabilities.

2. What is the desired allele combination for each group’s super hero

a. This will be homozygous dominant for the two traits in the parents

Working in the small groups, students will use the statistical model strategy learned in the class activity to try to solve the following statistical model design challenge.

The challenge:

Given a set of parents from the gene pool, your challenge is to use your knowledge of statistics and genetics to determine a model that will produce the highest likelihood of a Next Generation Super Hero after 3 generations.

Constraints:

a) One of your first generation parents is homozygous dominant (e.g. AABB).

b) Each student in the team will have one possible parent drawn from the gene pool.

c) Must determine which order to use those randomly selected parents

3. Using the worksheet, begin to set up a plan. Students should use the process of setting up the probability equations learned in the whole class exercise.

4. Generation 1:

a. Determine one homozygous dominant parent, place on the left side of the Punnett Square,

b. Students select one of their random parents, place on top of Punnett Square

i. It is up to the students to select which of their randomly drawn parents should be the used in the first generation

c. Based on the alleles in the two parents, determine the number of offspring and the probability of each type, following the probability equations

d. What is the probability of each type of offspring for the Generation 1 parents?

e. What is the probability of the Next Generation Super Hero in Generation 1?

5. Generation 2:

a. Students select a second random parent from their gene pool (not the one selected for Generation 1); if still using Punnett Squares, place on the left side of all squares

i. Students will need the number of squares equal to the number of possible unique genotypes in Generation 1.

ii. This should be a maximum of 4 unique Generation 1 offspring genotypes.

b. For each unique Generation 1 genotype, compute probabilities of various offspring with the Generation 2 random parent.

i. If still using Punnett Squares, the unique genotypes will be on the top side

ii. Determine the possible offspring for each parent combination (there will be up to 4 sets of offspring)

iii. Find the probabilities of each offspring within each parent combination

iv. Find the overall probabilities of each offspring

c. What is the probability of each type of offspring for the Generation 2 parents?

d. What is the probability of the Next Generation Super Hero in Generation 2?

6. Generation 3:

a. Students select a third random parent from their gene pool (not the ones selected for Generation 1 or 2)

i. Students should not need to rely on Punnett Squares here; they should use equations.

ii. There should be a maximum of 7 unique Generation 2 offspring genotypes.

b. For each unique Generation 2 genotype, compute probabilities of various offspring with the Generation 3 random parent.

i. Determine the possible offspring for each parent combination (there will be up to 7 sets of offspring)

ii. Find the probabilities of each offspring within each parent combination

iii. Find the overall probabilities of each offspring

c. What is the probability of each type of offspring for the Generation 3 parents?

d. What is the probability of the Next Generation Super Hero in Generation 3?

7. What is the probability of Next Generation Super Hero for each generation?

8. REDESIGN: Could you get a higher probability if you use the randomly drawn parents in a different order?

9. Students should use a pedigree to illustrate their proposed model, labeling types of offspring and their likelihood at each generation.

Alternative orderings:

1. For generation 1: if we choose 1 homozygous parent and one heterogygous parent, then the largest number of possible types of offspring is 4

a. If homozygous dominant + homozygous dominant, only offspring is homozygous dominant. Same for homozygous recessive pair

b. Homozygous dominant + homozygous recessive parent also only 1 type of offspring

c. Homozygous dominant/recessive + heterozygous produces up to 4 types of offspring

d. Two heterozygous parents can produce up to 9 types of offspring in the first generation

2. Options for selecting the parents:

a. They use 1 homozygous dominant and 1 random for generation 1; successive generations are random parent with the possible types of offspring from Gen 1

b. They use 1 random parent and a specific heterozygous parent.

c. They use 2 randomly drawn parents and have to determine which 2 to start with from the pool they have chosen.

d. They use 1 random parent and any second parent of their choice.

Options b, c, and maybe d could produce 9 types of offspring in the first generation. Would be a little more challenging than the 4 worked out in the class activity, especially if they want to write out all the Punnett squares, but that emphasizes why the math is useful.

For Reference:

The following show the types of offspring (and so the total number of possible offspring) for the different possible parent combinations. Use this as a guide to help students make sure they have the right numbers of genotypes for each generation of their model.

If parents AABB + AABB ( AABB

If parents aabb + aabb ( aabb

If parents AABB + aabb ( AaBb

If parents AABB + AaBb ( AABB, AABb, AaBB, AaBb

If parents aabb + AaBb ( AaBb, aaBb, Aabb, aabb

If parents AABB + aaBb ( AaBB, AaBb

If parents AABB + Aabb ( AABb, AaBb

If parents aabb + aaBb ( aaBb, aabb

If parents aabb + Aabb ( Aabb, aabb

If parents AaBb+AaBb ( AABB, AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, aabb

If parents AaBb+aaBb ( AaBB, AaBb, Aabb, aaBB, aaBb, aabb

If parents AaBb+AaBB ( AABB, AaBB, aaBB, AABb, AaBb, aaBb

If parents AaBb+Aabb ( AABb, AAbb, aaBb, aabb

If parents aaBb+aaBb ( aaBB, aaBb, aabb

If parents aaBb+Aabb ( AaBb, aaBb, Aabb, aabb

If parents aaBb+AaBB ( AaBB, AaBb, aaBB, aaBb

Appendix M: Class Activity: Next Generation Super Hero

Appendix N

Appendix O

Appendix N: Team Activity: Next Generation Super Hero

Appendix O: Exponent Rules and Practice

Answer Keys

Appendix P: Getting Started with Scratch – Instructions & Rubric

Instructions: How to set up and/or use “Scratch”

1. Open Scratch from its location on your computer.

2. Click on the scissors icon above the provided cat “sprite” then click on the cat to clear

it.

3. In the lower right hand corner click on the tab that is labeled “paint a new sprite”

[pic]

4. Create a sprite that fits the characteristics of your first generation super creature then

click “ok”. (see rubric for requirements)

5. In the middle window click on the “costumes” tab. The sprite you created will show

up here. Click “copy” the Click “edit”. Give the sprite the characteristics of the

second generation super creature then click “ok”.

[pic][pic]

6. Click “copy” – then Click “edit”. Give the creature the characteristics of the third

generation super creature.

7. On the left choose “variables” tab. At the bottom of the choices choose “make a list”.

Create the list according to prompts. Lable it “Generations”. Click “add thing” to

make a list of the new characteristics each “Super Hero” gains in successive

generations.

[pic]

8. In the middle window choose the scripts tab. Go to the left and choose controls

[pic]

8. Now follow the process below:

a. Drag over the following control: “when sprite is clicked”

b. Click “looks” and drag over: “next costume”

[pic]

c. click on the sprite in order to see if it changes costumes

[pic]

e. If all work as desired check your “costumes to make sure that they match the way you wish your super hero to look.

Super Hero Presentation Rubric

| |4 |3 |2 |1 |

|Sprite Properties |Each sprite has humanoid |Each sprite has only three | Each sprite has only two | Each sprite has only one |

| |characteristics including: |humanoid characteristics |humanoid characteristics. |humanoid characteristic. |

| |Feet, arms, head, and | | | |

| |facial features. | | | |

|List Properties |The generations list is |The generations list is |The generations list is |There is either no |

| |labeled and clearly states |not labeled or clearly |not labeled or clearly |properties or there is one |

| |three different generation |states only two different |states only one |property but no label |

| |properties. |generation properties. |generation property. |indicating what it |

| | | | |represents. |

|Animations |When clicked, the sprite |When clicked, the sprite |When clicked, the sprite |When clicked, the sprite |

| |moves through four |moves through three |moves through two |does not move through any |

| |different representations. |different representations. |different representations.|other representations. |

|Colors |Each Sprite costume |Each Sprite costume |Each Sprite costume |Each Sprite costume |

| |includes four different |includes three different |includes two different |includes one color. |

| |colors. |colors. |colors. | |

|Mechanics |All parts of list are |Title or one of the |Title and/or two |Title and/or three |

| |spelled correctly. |generations are misspelled.|generations are |generations are misspelled.|

| | | |misspelled. | |

Team Name:___________________________

Score:______ out of 20

Appendix Q: Sample Allele Strips

|VVFf |VVee |

|VVff |VvEE |

|VvFF |vvEE |

|vvFF |VvEe |

|VvFf |vvEe |

|vvFf |Vvee |

|Vvff |VVNn |

|VVSs |VVnn |

|VVss |VvNN |

|VvSS |vvNN |

|vvSS |VvNn |

|VvSs |vvNn |

|vvSs |Vvnn |

|Vvss |ffEE |

|VVEe |FfEe |

|FFss |ffEe |

|FFSs |Ffee |

|FfSS |FFNn |

|ffSS |FFnn |

|FfSs |FfNN |

|ffSs |ffNN |

|Ffss |FfNN |

|FFEe |FfNn |

|FFee |ffNn |

|FfEE |Ffnn |

|SSEe |SSNn |

|SSee |SSnn |

|SsEE |SsNN |

|ssEE |ssNN |

|SsEe |SsNn |

|ssEe |ssNn |

|Ssee |Ssnn |

|EENn |EeNn |

|EEnn |eeNn |

|EeNN |Eenn |

|eeNN | |

-----------------------

(T)

(g)

(a)

(g)

(i)

(y)

(G)

(A)

(I)

(G)

(Y)

(r)

(R)

(t)

(g)

(a)

(g)

(i)

(y)

(G)

(A)

(I)

(G)

(Y)

(r)

(R)

(t)

(T)

[pic]

2. Punnett Square B:

Bb

Bb

bb[pic]b

bb[pic]b

B

b

B

b

1. Punnett Square A:

Experimental

Simulation

Theoretical

Experimental

Simulation

Tool used in place of an experiment to verify predictions made from theoretical probabilities.

Calculate chance from same set of circumstances.

Action carried out to demonstrate nature of experiment.

Desired total outcome

Used to predict actual Events.

Theoretical

Experimental

Results calculated from the events of an activity.

Mathematically calculated to make predictions of events.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download