SCIENTIFIC INQUIRY. Ask a Question After learning background information, create questions that you hope to find the answers to.

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1 SCIENTIFIC INQUIRY Perform Experiments Test your hypothesis to see if it works. Formulate a Hypothesis Make a guess based on the background information as to what the answer might be. Ask a Question After learning background information, create questions that you hope to find the answers to. Draw Conclusions What evidence did you gather to support your hypothesis? Was your hypothesis correct? If not, why? Collect Data Gather observations pertaining to the question. Make sure to be honest with your observations, and record them even if they do not support your hypothesis.

2 ZEBRAFISH FACTS 1. Zebrafish are tropical, freshwater fish. 2. They are native to the Ganges River in East India, and to nearby countries. 3. They will eat small living organisms like insects and zooplankton. 4. They are eaten by larger fish, birds, and amphibians. 5. They grow to about 1 2 inches long and live two to five years. 6. Most have black stripes and black eyes. 7. They are kept by hobbyists and used in laboratories to learn about living things.

3 Name ZEBRAFISH WORD SEARCH (LEVEL 1) P Y D M O Y R B M E H B G E B J U E O Q P E X X O O Y E P U I H N Z M T R P D U M P N Y J J N D N I I E C G B O G O T T Y N D M E R N N M Q Z E T O E A J O I I T E A J T Y N Y N E A S J M A A T V N W G E P E L Y Z E T E S A S I T O T E H E E N C C G V X L Q S U S M P L T D I J V J D C A U S I A V L S I S E H T O P Y H A E V A A O Q N V Y V C L K H J R C G Q Y M S P N O I R O H C Z E E B E Z E B R A F I S H V F Q P R D Q O L A R V A P H D P Z U V ALLELE CHORION DNA DOMINANT EMBRYO EXPERIMENT GENE GENOTYPE HOMOZYGOUS HYPOTHESIS INDIA LARVA PHENOTYPE PUNNETT SQUARE RECESSIVE SOMITE WILDTYPE ZEBRAFISH

4 Name ZEBRAFISH WORD SEARCH SOLUTION (LEVEL 1) P Y D M O Y R B M E H B G E B J U E O Q P E X X O O Y E P U I H N Z M T R P D U M P N Y J J N D N I I E C G B O G O T T Y N D M E R N N M Q Z E T O E A J O I I T E A J T Y N Y N E A S J M A A T V N W G E P E L Y Z E T E S A S I T O T E H E E N C C G V X L Q S U S M P L T D I J V J D C A U S I A V L S I S E H T O P Y H A E V A A O Q N V Y V C L K H J R C G Q Y M S P N O I R O H C Z E E B E Z E B R A F I S H V F Q P R D Q O L A R V A P H D P Z U V ALLELE CHORION DNA DOMINANT EMBRYO EXPERIMENT GENE GENOTYPE HOMOZYGOUS HYPOTHESIS INDIA LARVA PHENOTYPE PUNNETT SQUARE RECESSIVE SOMITE WILDTYPE ZEBRAFISH

5 Name ZEBRAFISH WORD SEARCH (LEVEL 2) E E E O R M O X K M R H S I C X R X D Y D E L P E R Y U I H A F A P U R O N C G N P O J O H I W U E Y B E D E K O G T R L S D I Q R S M P E N T Y N I A E I N L S I Y E O L H Z A O R N M F I D T M I Y T E O N N V E R V A O T T E X J S R I E A G E U N R A Y E N S I E M L M S R E P V B J P N T S T O E T U G V R G J E J E N D E D L G E N E T I C S Z L S U H S L G Z S E P Y T O N E H P P X A H B E T T E P I P S O M I T E O H O M O Z Y G O U S D N A G ALLELE CHORION DNA DOMINANT EMBRYO EXPERIMENT GENE GENETICS GENOTYPE HETEROZYGOUS HOMOZYGOUS HYPOTHESIS INDIA LARVA MENDEL OBSERVATION PHENOTYPE PIPETTE PUNNETT SQUARE RECESSIVE SOMITE WILDTYPE YOLK ZEBRAFISH

6 Name ZEBRAFISH WORD SEARCH SOLUTION (LEVEL 2) E E E O R M O X K M R H S I C X R X D Y D E L P E R Y U I H A F A P U R O N C G N P O J O H I W U E Y B E D E K O G T R L S D I Q R S M P E N T Y N I A E I N L S I Y E O L H Z A O R N M F I D T M I Y T E O N N V E R V A O T T E X J S R I E A G E U N R A Y E N S I E M L M S R E P V B J P N T S T O E T U G V R G J E J E N D E D L G E N E T I C S Z L S U H S L G Z S E P Y T O N E H P P X A H B E T T E P I P S O M I T E O H O M O Z Y G O U S D N A G ALLELE CHORION DNA DOMINANT EMBRYO EXPERIMENT GENE GENETICS GENOTYPE HETEROZYGOUS HOMOZYGOUS HYPOTHESIS INDIA LARVA MENDEL OBSERVATION PHENOTYPE PIPETTE PUNNETT SQUARE RECESSIVE SOMITE WILDTYPE YOLK ZEBRAFISH

7 Name ZEBRAFISH WORD SEARCH (LEVEL 3) S W C Y N O I G V S G H N D T R N R U O O Y O E N I S E S P O N E O D L O R N N I S D T N I U M E C I K R B G O C E R E I E D N I M E T G M O T Y H L R O P A I N N I S A E E Y H T Z O U Y H N R E A R S T X P N O C Z O T S P C T T N E I U E J P O Y O D M H I J E T T P V M P Y Y G M L T E O O O P S S X E N H B O A I E N O K H I N Q O E T E O U K W Y O Y A N D C U U M Y T G S N O I T A V R E S B O A I T E Y T T G Y L A C I P O R T R T G P X C P P S L L E C M E T S E E R I O L E D N E M O M I C R O S C O P E V U Z E B R A F I S H A L L E L E E Z A X S C I T E N E G L A R V A ALLELE CHORION CONCLUSION DNA DOMINANT EMBRYO EXPERIMENT GENE GENETICS GENOME GENOTYPE HETEROZYGOUS HOMOZYGOUS HYPOTHESIS INDIA LARVA MENDEL MICROSCOPE MUTATION NOTOCHORD OBSERVATION OXYGEN PETRI DISH PHENOTYPE PIPETTE PUNNETT SQUARE RECESSIVE SOMITE STEM CELLS TROPICAL WILDTYPE YOLK ZEBRAFISH

8 Name ZEBRAFISH WORD SEARCH SOLUTION (LEVEL 3) S W C Y N O I G V S G H N D T R N R U O O Y O E N I S E S P O N E O D L O R N N I S D T N I U M E C I K R B G O C E R E I E D N I M E T G M O T Y H L R O P A I N N I S A E E Y H T Z O U Y H N R E A R S T X P N O C Z O T S P C T T N E I U E J P O Y O D M H I J E T T P V M P Y Y G M L T E O O O P S S X E N H B O A I E N O K H I N Q O E T E O U K W Y O Y A N D C U U M Y T G S N O I T A V R E S B O A I T E Y T T G Y L A C I P O R T R T G P X C P P S L L E C M E T S E E R I O L E D N E M O M I C R O S C O P E V U Z E B R A F I S H A L L E L E E Z A X S C I T E N E G L A R V A ALLELE CHORION CONCLUSION DNA DOMINANT EMBRYO EXPERIMENT GENE GENETICS GENOME GENOTYPE HETEROZYGOUS HOMOZYGOUS HYPOTHESIS INDIA LARVA MENDEL MICROSCOPE MUTATION NOTOCHORD OBSERVATION OXYGEN PETRI DISH PHENOTYPE PIPETTE PUNNETT SQUARE RECESSIVE SOMITE STEM CELLS TROPICAL WILDTYPE YOLK ZEBRAFISH

9 Name ZEBRAFISH CROSSWORD PUZZLE (LEVEL 1) Across 1 The defining feature of vertebrates 4 An individual with two identical copies of the same gene 6 A young organism that is physically very different from a juvenile or an adult 9 How we test our ideas in science Down 1 Unspecialized cells that can become any cell in the body (2 words) 2 The physical or detectable manner in which a gene is expressed. 3 These traits need only one copy of their gene present in order to be expressed 5 The study of biological inheritance of traits 7 A variant of a specific gene 8 A section of DNA that affects a specific trait

10 Name ZEBRAFISH CROSSWORD SOLUTION (LEVEL 1) 1S P I N E T 2P E 3D 4H O M O Z Y G O U S E C M N E I O L N T L 5G 6L A R V 7A Y S E N L 8G P N T L 9E X P E R I M E N T E N T L E I E C S Across 1 The defining feature of vertebrates 4 An individual with two identical copies of the same gene 6 A young organism that is physically very different from a juvenile or an adult 9 How we test our ideas in science Down 1 Unspecialized cells that can become any cell in the body (2 words) 2 The physical or detectable manner in which a gene is expressed. 3 These traits need only one copy of their gene present in order to be expressed 5 The study of biological inheritance of traits 7 A variant of a specific gene 8 A section of DNA that affects a specific trait

11 Name ZEBRAFISH CROSSWORD PUZZLE (LEVEL 2) Across 4 A developing organism that has not yet been born or hatched 6 Unspecialized cells that can become any cell in the body (2 words) 7 The country where wild zebrafish are most often found 9 A section of DNA that affects a specific trait 11 The physical or detectable manner in which a gene is expressed. 13 An individual with two identical copies of the same gene 15 The molecule containing all of an organism s genetic information Down 1 These traits need two copies of their gene present in order to be expressed 2 Surrounds and protects the fish embryo 3 The most common appearance of a certain organism in the wild 5 A change in a gene, sometimes resulting in a trait different from the parent 8 A variant of a specific gene 10 How we test our ideas in science 12 The genetic makeup for a certain trait 14 Provides nourishment to the fish embryo

12 Name ZEBRAFISH CROSSWORD SOLUTION (LEVEL 2) 1R 2C E H 3W 4E 5M B R Y O C O I U E R L 6S T E M C E L L S 7I N D I 8A A S O T L T 9G 10E N E I N Y L I X V 11P H E N O T Y P E E E L N E 12G E R E I N 13H O M O Z 14Y G O U S E O T 15D N A L Y T K P E Across 4 A developing organism that has not yet been born or hatched 6 Unspecialized cells that can become any cell in the body (2 words) 7 The country where wild zebrafish are most often found 9 A section of DNA that affects a specific trait 11 The physical or detectable manner in which a gene is expressed. 13 An individual with two identical copies of the same gene 15 The molecule containing all of an organism s genetic information Down 1 These traits need two copies of their gene present in order to be expressed 2 Surrounds and protects the fish embryo 3 The most common appearance of a certain organism in the wild 5 A change in a gene, sometimes resulting in a trait different from the parent 8 A variant of a specific gene 10 How we test our ideas in science 12 The genetic makeup for a certain trait 14 Provides nourishment to the fish embryo

13 ZEBRAFISH CROSSWORD PUZZLE (LEVEL 3) (MAY ALSO BE DONE ON JOURNAL PAGE 20) `8 19 Across 4 Surrounds and protects the fish embryo 6 The genetic makeup for a certain trait 8 Element in air and dissolved in water that both humans and fish need 11 How we test our ideas in science 14 A variant of a specific gene 16 The defining feature of vertebrates 18 An embryonic body segment that will turn into skin, muscle, and bone 19 A young organism that is physically very different from a juvenile or an adult Down 1 An individual with two different versions of the same gene 2 These traits need only one copy of their gene present in order to be expressed 3 The molecule containing all of an organism s genetic information 5 Unspecialized cells that can become any cell in the body (2 words) 6 The study of biological inheritance of traits 7 The type of plant that Mendel experimented with 9 Mendel s experiments laid the groundwork for modern genetics 10 The most common appearance of a certain organism in the wild 12 The physical or detectable manner in which a gene is expressed. 13 A section of DNA that affects a specific trait 15 Where a fish embryo gets its nourishment from 17 The country where wild zebrafish are most often found

14 ZEBRAFISH CROSSWORD SOLUTION (LEVEL 3) (FROM JOURNAL PAGE 20) 1H 2D 3D E 4C H O R I O N T M A E I 5S R 6G E N O T Y 7P E 8O X Y 9G E N E A E E Z R 10W N N M A Y 11E X 12P E R I M E N T C G G H L T E 13G O O E D I 14A L L E L E U R N T C L N S 15Y O Y S 16S P 17I N E 18S O M I T E P N L Y E D K P I E 19L A R V A Across 4 Surrounds and protects the fish embryo 6 The genetic makeup for a certain trait 8 Element in air and dissolved in water that both humans and fish need 11 How we test our ideas in science 14 A variant of a specific gene 16 The defining feature of vertebrates 18 An embryonic body segment that will turn into skin, muscle, and bone 19 A young organism that is physically very different from a juvenile or an adult Down 1 An individual with two different versions of the same gene 2 These traits need only one copy of their gene present in order to be expressed 3 The molecule containing all of an organism s genetic information 5 Unspecialized cells that can become any cell in the body (2 words) 6 The study of biological inheritance of traits 7 The type of plant that Mendel experimented with 9 Mendel s experiments laid the groundwork for modern genetics 10 The most common appearance of a certain organism in the wild 12 The physical or detectable manner in which a gene is expressed. 13 A section of DNA that affects a specific trait 15 Where a fish embryo gets its nourishment from 17 The country where wild zebrafish are most often found

15 VOCABULARY ANSWER KEY (FROM JOURNAL PAGE 15) 1. A RECESSIVE trait can be carried in a person s genes and not be apparent or expressed in the individual. 2. A SOMITE is a segment of a vertebrate embryo that will eventually develop into skin, muscles, and vertebrae. 3. A branch of biology that studies inheritance throughout families. GENETICS 4. The developmental stage in which the embryo spreads out and encases the yolk. EPIBOLY 5. Coloration of living things, including eye and skin color, is produced by proteins called PIGMENTS. 6. A tool used to predict probability in offspring. PUNNETT SQUARE 7. A DOMINANT trait is expressed from a person s genes even when there is only one copy of that parental DNA present. 8. Unspecialized cells that can multiply repeatedly and can potentially develop into many types of specialized cells. STEM CELLS 9. The typical appearance of an organism in a natural population. WILDTYPE 10. The protective outer membrane surrounding the developing zebrafish embryo. CHORION 11. The full DNA sequence of an organism. GENOME 12. The NOTOCHORD is a flexible rod-shaped structure that is an early developmental stage of the vertebral column. 13. The expression of genes in an individual, such as hair color or eye shape, is one s PHENOTYPE. 14. Describes identical alleles for a given trait. HOMOZYGOUS 15. The genetic makeup for a given trait, not necessarily seen by the eye. GENOTYPE 16. Describes an individual that has two different alleles for a given trait. HETEROZYGOUS

16 DETERMINE THE ZEBRAFISH MORTALITY RATE (ALSO FOUND ON JOURNAL PAGE 11) Directions: Using the formulae below and the data recorded in the charts on the daily observation pages in your agent handbook, calculate the mortality rate of your zebrafish. Formulae: Step 1: Subtract Day 5 total (including embryos AND larvae) from Day 2 total to get the total number of expired embryos. SHOW YOUR WORK: - = Step 2: Total number of expired embryos Day 2 total = X SHOW YOUR WORK: / = Step 3: X multiplied by 100 equals the mortality rate of your zebrafish. SHOW YOUR WORK: * 100 = % What percentage of your embryos did not survive? How does this percentage compare to those of the other groups? What variables might have contributed to this mortality rate?

17 MONOHYBRID PUNNETT SQUARE PRACTICE Background: A Punnett Square is a visual tool used by scientists to determine the possible combinations of genetic alleles in a cross. Since genes are inherited randomly and independently, Punnett Squares are useful for looking at just one gene combination (monohybrid) or a whole series of combinations (dihybrid for two traits, etc.) To make a Punnett Square, draw a box and then divide it into four smaller squares. On the top of the box write the letters that correspond with the alleles for one parent. Write the alleles for the other parent along the left side of the box. Use an upper case letter for a dominant allele and a lower case letter for a recessive allele. Fill in the smaller squares with the alleles from the corresponding rows and columns. The resulting alleles combinations are the possible genotypes for the offspring from that cross. Example: A green pea plant (GG) is being crossed with a green pea plant (Gg). G G G GG GG g Gg Gg Genotypes: 50% GG, 50% Gg Phenotypes: 100% green Activity: Use the provided Punnett Squares to determine the genotypes and phenotypes of each cross. 1. A green pea plant (Gg) is crossed with a yellow pea plant (gg). 2. A tall plant (TT) is crossed with a tall plant (Tt). 3. A tall plant (Tt) is crossed with a short plant (tt). 4. A red flower (Rr) is crossed with a white flower (rr).

18 5. A white flower (rr) is crossed with a white flower (rr). 6. A black chicken (Bb) is crossed with a black chicken (Bb). For the following problems, list the parent genotypes, draw and fill in a Punnett square, and then list the offspring genotypes and phenotypes. 7. A homozygous dominant brown mouse is crossed with a heterozygous brown mouse (tan is the recessive color). 8. Two heterozygous white (brown fur is recessive) rabbits are crossed. 9. Two heterozygous red flowers (white flowers are recessive) are crossed. 10. A homozygous tall plant is crossed with a heterozygous tall plant (short is the recessive size). 11. A heterozygous white rabbit is crossed with a homozygous brown rabbit.

19 MONOHYBRID PUNNETT SQUARE SOLUTIONS 1. A green pea plant (Gg) is crossed with a yellow pea plant (gg). G g g Gg gg g Gg gg Genotypes: 50% Gg, 50% gg Phenotypes: 50% green, 50% yellow 2. A tall plant (TT) is crossed with a tall plant (Tt). T T T TT TT t Tt Tt Genotypes: 50% TT, 50% TT Phenotypes: 100% 3. A tall plant (Tt) is crossed with tall a short plant (tt). Genotypes: 50% Gg, 50% gg Phenotypes: 50% green, 50% yellow T t t Tt tt t Tt tt Genotypes: 50% Tt, 50% tt Phenotypes: 50% tall, 50% short 4. A red flower (Rr) is crossed with a white flower (rr). R r Genotypes: 50% Rr, 50% rr r Rr rr Phenotypes: 50% red, 50% white r Rr rr 5. A white flower (rr) is crossed with a white flower (rr). r r r rr rr r rr rr Genotypes: 100% rr Phenotypes: 100% white 6. A black chicken (Bb) is crossed with a black chicken (Bb). The recessive color is white. B b B BB Bb b Bb bb Genotypes: 25% BB, 50% Bb, 25% bb Phenotypes: 75% black, 25% white 7. A homozygous dominant brown mouse is crossed with a heterozygous brown mouse. The recessive color is tan. B B B BB BB b Bb Bb The parents genotypes are BB and Bb. The offspring s genotypes are 50% BB, 50% Bb, which means their phenotypes are 100% brown. 8. Two heterozygous white (brown fur is recessive) rabbits are crossed. W w W WW Ww w Ww ww The parents genotypes are Ww and Ww. The offspring s genotypes are 25% WW, 50% Ww, and 25% ww, which means their phenotypes are 75% white and 25% brown.

20 9. Two heterozygous red flowers (white flowers are recessive) are crossed. R r R RR Rr r Rr rr The parents genotypes are Rr and Rr. The offspring s genotypes are 25% RR, 50% Rr, and 25% rr, which means their phenotypes are 75% red and 25% white. 10. A homozygous tall plant is crossed with a heterozygous tall plant (short is the recessive size). T T T TT TT t Tt Tt The parents genotypes are TT and Tt. The offspring s genotypes are 50% TT and 50% Tt, which means their phenotype is 100% tall. 11. A heterozygous white rabbit is crossed with a homozygous brown rabbit. W w w Ww ww w Ww ww The parents genotypes are Ww and ww. The offspring s genotypes are 50% Ww and 50% ww, which means their phenotypes are 50% white and 50% brown.

21 DIHYBRID PUNNETT SQUARE PRACTICE Directions: In rabbits, gray hair (G) is dominant to white hair (g), and black eyes (B) are dominant to red eyes (b). These two traits are independent of each other. In other words, a female rabbit with the genotype GgBb may produce eggs with the alleles GB, Gb, gb, or gb. To predict the probability of these sorts of crosses, we will make a dihybrid Punnett Square. Activity: 1. What are the phenotypes (descriptions) of rabbits that have the following genotypes: Ggbb ggbb ggbb GgBb 2. A male rabbit with the genotype GGbb is crossed with a female rabbit with the genotype ggbb The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. Gb Gb Gb Gb gb How many out of 16 are: gb gb gb Grey, red-eyed Grey, black-eyed White, red-eyed White, black-eyed 3. A male rabbit has the genotype GgBb. Determine the gametes produced by this rabbit. Hint: There are 4 possible combinations. 4. A female rabbit has the genotype ggbb. Determine the gametes (eggs) produced by this rabbit. 5. Use the gametes from #3 and #4 to set up a Punnett Square below. Put the male's gametes on the top and the female's gametes down the side. Then fill out the square and determine what kind of offspring would be produced from this cross and in what proportion. 6. An aquatic arthropod called a Cyclops has antennae that are either smooth or barbed. The allele for barbs is dominant. In the same organism, resistance to pesticides is a recessive trait. Make a key to show all the possible genotypes (and phenotypes) of this organism.

22 7. A Cyclops that is resistant to pesticides and has smooth antennae is crossed with one that is heterozygous for both traits. Show the genotypes of the parents. x 8. Set up a Punnett Square for the cross. How many are smooth, resistant? How many are smooth, not resistant? How many are barbed, resistant? How many are barbed, not resistant? 9. A Cyclops that is homozygous dominant for the barbed gene and is resistant to pesticides is crossed with one that is resistant to pesticides but not barbed. What proportion of the offspring will be barbed and resistant? 10. If two Cyclops that are heterozygous for both traits are crossed, what are the resulting phenotypes and in what proportion?

23 DIHYBRID PUNNETT SQUARE SOLUTIONS 1. What are the phenotypes (descriptions) of rabbits that have the following genotypes: Ggbb Gray fur, red eyes ggbb White fur, red eyes ggbb White fur, black eyes GgBb Gray fur, black eyes 2. A male rabbit with the genotype GGbb is crossed with a female rabbit with the genotype ggbb The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. Gb Gb Gb Gb gb GgBb GgBb GgBb GgBb gb GgBb GgBb GgBb GgBb gb Ggbb Ggbb Ggbb Ggbb gb Ggbb Ggbb Ggbb Ggbb How many out of 16 are: Grey, red-eyed 8 Grey, black-eyed 8 White, red-eyed 0 White, black-eyed 0 3. A male rabbit has the genotype GgBb. Determine the gametes produced by this rabbit. Hint: There are 4 possible combinations. GB, Gb, gb, and gb. 4. A female rabbit has the genotype ggbb. Determine the gametes (eggs) produced by this rabbit. gb, gb, gb, and gb 5. Use the gametes from #3 and #4 to set up a Punnett Square below. Put the male's gametes on the top and the female's gametes down the side. Then fill out the square and determine what kind of offspring would be produced from this cross and in what proportion. GB Gb gb gb gb GgBB GgBb ggbb ggbb gb GgBB GgBb ggbb ggbb gb GgBb Ggbb ggbb ggbb gb GgBb Ggbb ggbb ggbb Grey, red-eyed 2 (12.5%) Grey, black-eyed 6 (37.5%) White, red-eyed 2 (12.5%) White, black-eyed 6 (37.5%)

24 6. An aquatic arthropod called a Cyclops has antennae that are either smooth or barbed. The allele for barbs is dominant. In the same organism, resistance to pesticides is a recessive trait. Make a key to show all the possible genotypes (and phenotypes) of this organism. BBPP = barbed, not resistant BBPp = barbed, not resistant BBpp = barbed, resistant bbpp = smooth, not resistant bbpp = smooth, resistant BbPP = barbed, not resistant BbPp = barbed, not resistant Bbpp = barbed, resistant bbpp = smooth, not resistant 7. A Cyclops that is resistant to pesticides and has smooth antennae is crossed with one that is heterozygous for both traits. Show the genotypes of the parents. bbpp x BbPp 8. Set up a Punnett Square for the cross. bp bp bp bp BP BbPp BbPp BbPp BbPp Bp Bbpp Bbpp Bbpp Bbpp bp bbpp bbpp bbpp bbpp bp bbpp bbpp bbpp bbpp How many are smooth, resistant? 4 (25%) How many are smooth, not resistant? 4 (25%) How many are barbed, resistant? 4 (25%) How many are barbed, not resistant? 4 (25%) 9. A Cyclops that is homozygous dominant for the barbed gene and is resistant to pesticides is crossed with one that is resistant to pesticides but not barbed. What proportion of the offspring will be barbed and resistant? Bp Bp Bp Bp bp Bbpp Bbpp Bbpp Bbpp bp Bbpp Bbpp Bbpp Bbpp bp Bbpp Bbpp Bbpp Bbpp bp Bbpp Bbpp Bbpp Bbpp All (100%) of the offspring will be barbed and resistant to pesticides. 10. If two Cyclops that are heterozygous for both traits are crossed, what are the resulting phenotypes and in what proportion? BP Bp bp bp BP BBPP BBPp BbPP BbPp Bp BBPp BBpp BbPp Bbpp bp BbPP BbPp bbpp bbpp bp BbPp Bbpp bbpp bbpp Smooth and resistant: 1 (6.25%) Smooth and not resistant: 3 (18.75%) Barbed and resistant: 3 (18.75%) Barbed and not resistant: 9 (56.25%)

25 DIHYBRID CROSS Name A cross (or mating) between two organisms where two genes are studied is called a DIHYBRID cross. The genes are located on separate chromosomes, so the traits themselves are unrelated. BB = black Bb = black bb = white LL = short hair Ll = short hair ll = long hair Fill out the genotypes of each of the offspring to determine how many of each type of offspring are produced. Phenotypic ratios - How many, out of 16 are: Black, Short Black, Long White, Short White, Long

26 How many of the offspring are: Black, Short Black, Long White, Short White, Long How many of the offspring are: Black, Short Black, Long White, Short White, Long

27 PUNNETT SQUARE SCENARIOS Activity: Below are sample scenarios that can be used to practice making Punnett Squares and interpreting the results. 1. Tom and Tina decided they wanted to start a family. Tom knew his grandfather had sickle-cell disease. Sickle-cell disease is a recessive disorder that causes blood cells to stiffen and take on a crescent shape. These blood cells have a difficult time moving through the blood vessels and can cause health problems. Tina didn t know of any family members that had it. Still, before they got pregnant they both went to the doctor for genetic tests. The doctor found that even though Tom and Tina were healthy, they both carried the gene for sickle-cell disease. Make a Punnett Square to determine the probability of Tom and Tina s child having sickle-cell disease. 2. Some cattle are naturally hornless, a condition called polling. Polling is dominant over the more common horned trait. A polled bull is mated with three cows with the following offspring: What are the genotypes of the four parents? Cow A (horned) = polled calf Cow B (horned) = horned calf Cow C (polled) = horned calf

28 3. Short hair is dominant over long hair in guinea pigs. A long-haired male guinea pig and a heterozygous short-haired female are kept in the same cage and mate numerous times, ultimately producing 32 total offspring. What is the most likely distribution of short hair and long hair among their offspring? 4. Tay-Sachs disease is due to a homozygous recessive genotype (nn). About 4% of the alleles in a given population are n. This means that about 96% of the alleles are N. Use the Punnett Square below to determine the approximate percentage of each genotype within the population. N = n = N = n =

29 PUNNETT SQUARE SCENARIO SOLUTIONS 1. Tom and Tina decided they wanted to start a family. Tom knew his grandfather had sickle-cell disease. Sickle-cell disease is a recessive disorder that causes blood cells to stiffen and take on a crescent shape. These blood cells have a difficult time moving through the blood vessels and can cause health problems. Tina didn t know of any family members that had it. Still, before they got pregnant they both went to the doctor for genetic tests. The doctor found that even though Tom and Tina were healthy, they both carried the gene for sickle-cell disease. Make a Punnett Square to determine the probability of Tom and Tina s child having sickle-cell disease. S s S SS Ss s Ss ss The child has a 25% chance of having sickle-cell disease. 2. Some cattle are naturally hornless, a condition called polling. Polling is dominant over the more common horned trait. A polled bull is mated with three cows with the following offspring: What are the genotypes of the four parents? Cow A (horned) = polled calf Cow B (horned) = horned calf Cow C (polled) = horned calf Since being horned is recessive, Cow A and Cow B must both be homozygous recessive (pp). Since the polled bull has horned (homozygous recessive) offspring via Cow B, the bull must be heterozygous (Pp), as per this Punnett Square: P p p Pp pp p Pp pp If he were homozygous dominant (PP), all of his offspring via Cow B would be heterozygous polled, as per this Punnett Square: P P p Pp Pp p Pp Pp Similarly, Cow C must also be heterozygous (Pp), as she has horned offspring with the heterozygous bull, as per this Punnett Square: P p P PP Pp p Pp pp

30 3. Short hair is dominant over long hair in guinea pigs. A long-haired male guinea pig and a heterozygous short-haired female are kept in the same cage and mate numerous times, ultimately producing 32 total offspring. What is the most likely distribution of short hair and long hair among their offspring? The male must be homozygous recessive (ll), as he has long hair. Mating him with the heterozygous (Ll) female results in the following Punnett Square: L l l Ll ll l Ll ll Thus, the most likely distribution is about 50% short-haired (all heterozygous) and 50% long-haired, or 16 of each. In practice, it is unlikely to be an exact split, but it is the most likely single outcome, and the actual result will probably be close to it. 4. Tay-Sachs disease is due to a homozygous recessive genotype (nn). About 4% of the alleles in a given population are n. This means that about 96% of the alleles are N. Use the Punnett Square below to determine the approximate percentage of each genotype within the population. N=0.96 n=0.04 N=0.96 ~0.922 ~0.038 n=0.04 ~0.038 ~0.002 Those without the allele (NN) make up about 92.2% of the population. Heterozygous carriers (Nn) make up about 3.8% + 3.8% = 7.6%. Those with the disease (nn) make up about 0.2%.

31 FURTHER GENETICS ACTIVITIES ONLINE The GEEE! in Genome Home page Lots of good genetics information and games, both to do online and to download and do in the classroom nature.ca/genome/index_e.cfm The Biology Corner Worksheets on a variety of biological topics Genetics Practice Problems Online activities covering monohybrids, dihybrids, trihybrids, incomplete dominance, codominance, linked genes, and more biology.clc.uc.edu/courses/bio105/geneprob.htm Working out Punnett Square Examples More online activities covering monohybrids, dihybrids, and incomplete dominance Punnett square on Wikipedia Simple explanations and examples en.wikipedia.org/wiki/punnett_square Practice Punnett Square Problems Simple problems with step-by-step solutions frank.mtsu.edu/~rseipelt/web212b/ho/212practpun.htm Punnett Square Calculator Quickly create your own Punnett square problems online, complete with solutions and percentages Virtual Lab: Punnett Squares Tutorial/activity showing how to predict inheritance in fruit flies These are, of course, only a few of the websites that have genetic problems, worksheets, and activities. Search for Punnett square problems, genetic worksheets, heredity problem worksheet, etc., and see what else you can find!

32 EXTENDED RESEARCH SOLUTIONS (FROM JOURNAL PAGES 18 AND 19) A family tree is shown below. The genotypes of the grandparents regarding a recessive genetic mutation (a) that gives rise to a genetic disorder are given. Grandfather 1 (Aa) Grandmother 1 (AA) Grandfather 2 (Aa) Grandmother 2 (Aa) Father Mother Child 1. What percentage chance does the child have of inheriting the disorder? Of being an asymptomatic carrier? Of not carrying the mutation at all? Show your work below. Grandparents 1: Grandparents 2: A a A AA Aa A AA Aa A a A AA Aa a Aa aa The father has a 50% chance of being AA, and 50% chance of being Aa. The mother has a 25% chance of being AA, 50% chance of being Aa, and 25% chance of being aa. This creates six possible crosses resulting in the child. Carrying the probabilities through each gives these results: Possibility 1 (50% x 25% = 12.5%): Possibility 2 (50% x 50% = 25%): A A A AA AA A AA AA A A A AA AA a Aa Aa 100% x 12.5% = 12.5% AA 50% x 25% = 12.5% AA 50% x 25% = 12.5% Aa Possibility 3 (50% x 25% = 12.5%): Possibility 4 (50% x 25% = 12.5%): A A A a a Aa Aa A AA Aa a Aa Aa A AA Aa 100% x 12.5% = 12.5% Aa 50% x 12.5% = 6.25% AA 50% x 12.5% = 6.25% Aa

33 Possibility 4 (50% x 50% = 25%): Possibility 6 (50% x 25% = 12.5%): A a A AA Aa a Aa aa A a a Aa aa a Aa aa 25% x 25% = 6.25% AA 12.5% x 50% = 6.25% Aa 25% x 50% = 12.5% Aa 12.5% x 50% = 6.25% aa 25% x 25% = 6.25% aa We now compile the probabilities for each result from each cross: 12.5% % % % = 37.5% AA 12.5% % % % % = 50% Aa 6.25% % = 12.5% aa In other words, the child has a 12.5% chance of inheriting the disorder, a 50% chance of being an asymptomatic carrier, and a 37.5% chance of not carrying the mutation at all. Please note that simply counting the three genotypes and dividing by the total number of genotypes is not a valid method, as it assumes that all six possible crosses are weighted equally, which is not the case. Doing so on this question will coincidentally yield the correct answer, but would lead to false results on the next question.

34 2. You discover that those with the disorder rarely live beyond the age of 5 years. How does this change the above percentages? Show your work below. Any aa results in the parents can be ignored, as they would not live long enough to reproduce. Thus, the mother now has ~33.3% (1 in 3) chance of being AA and ~66.7% (2 in 3) chance of being Aa. The father s percentages remain the same. This leads to four possible crosses: Possibility 1 (50% x ~33.3% = ~16.7%): Possibility 2 (50% x ~66.7% = ~33.3%): A A A A A AA AA A AA AA A AA AA a Aa Aa 100% x ~16.7% = ~16.7% AA 50% x ~33.3% = ~16.7% AA 50% x ~33.3% = ~16.7% Aa Possibility 3 (50% x ~33.3% = ~16.7%): Possibility 4 (50% x ~66.7% = ~33.3%): A a A a A AA Aa A AA Aa A AA Aa a Aa aa 50% x ~16.7% = ~8.3% AA 25% x ~33.3% = ~8.3% AA 50% x ~16.7% = ~8.3% Aa 50% x ~33.3% = ~16.7% Aa 25% x ~33.3% = ~8.3% aa Compiling the probabilities as before: ~16.7% + ~16.7% + ~8.3% + ~8.3% = 50% AA ~16.7% + ~8.3% % = ~41.7% Aa ~8.3% aa This means that the child now has about an 8.3% (1 in 12) chance of inheriting the disorder (and thus likely dying young), about a 41.7% (5 in 12) chance of being an asymptomatic carrier, and a 50% chance of not carrying the mutation at all. Notice at this point that counting each possible genotype and dividing by the total number of genotypes would yield incorrect results, as the four possible crosses are not weighted equally.

35 SCENARIO ASSESSMENT 1 Students tested three different temperatures of water to determine if temperature had an effect on the development and hatching of zebrafish embryos from their chorions. Three separate groups of zebrafish embryos were allowed to develop for three days. Each group consisted of 100 embryos and was placed in a constant temperature for the three-day development. The students recorded the number of hatched larvae on Day 3. The students had a null hypothesis: Different temperatures will not affect the hatching of zebrafish embryos. Temperature Number of Embryos Hatched What conclusion(s) can best be drawn from the data? What variables could be changed to improve (or better support) the conclusion of this experiment?

36 SCENARIO ASSESSMENT 2 (MAY ALSO BE DONE ON JOURNAL PAGE 16) Students created a simulated shoreline in their mating tanks by tipping half of their tanks at a 25 incline, allowing the water level to go from about three inches deep to zero inches deep in the same tank. The students had a null hypothesis: Tilting the mating tanks to produce a simulated shoreline in each tank will have no effect on the number of eggs laid. Results: 30 out of 50 untilted tanks produced eggs. The producing tanks had an average of 62 eggs per tank. 28 out of 50 tilted tanks produced eggs. The producing tanks had an average of 112 eggs per tank. What conclusion(s) can be drawn from the data? What variables could be changed to improve (or better support) the conclusion of this experiment?

37 BIOETHICS SCENARIOS New biomedical technologies have brought about new ways to improve the lives and health of many people. With these new innovations also come new questions of how they should be properly used. Such things as stem cell research, artificial fertilization, transgenics, and gene manipulation to enhance characteristics or eradicate diseases have caused many researcher, government leaders, and ordinary people to think hard about what life is and what it means. Below are a few examples of situations with possible ethical implications. These situations are adapted from the Designing Babies curriculum by Nancy Lapotin, Maggie Raczek, and Sue Ann Jones Dobbyn. You can use these to stimulate discussion among your students about what they would do and why if they were presented with these dilemmas. You and your students can also discuss examples from the news, such as cloning, human euthanasia, and stem cell research. Down syndrome Tracy is 11 weeks pregnant with her third child. She and her husband have two healthy children and are looking forward to another. She was not planning to have the amniocentesis procedure to test for fetal health problems because it carried with it the possibility of miscarriage. When her doctor told her about a new and safer testing method, a simple blood test and special sonogram, she gladly agreed. The test results showed that the fetus has a high chance of having Down syndrome. An extra 21 st chromosome in the fetus results in Down syndrome, although the severity of the disorder cannot be determined by pre-natal testing. The severity of the disorder may range from no mental deficit to sever mental retardation. They show a greater incidence of hearing defects and congenital heart failure and have a life expectancy of 50+ years. They are also known to be affectionate, happy people with the potential for a fulfilling life. The doctor has brought up the possibility of aborting the fetus. Should Tracy abort the fetus? Athletic enhancement The year is Mary has worked hard her whole life to bring her family up out of poverty, but she still struggles. She has one child and is planning a second when she is given a unique opportunity to participate in a program allowing her to enhance her unborn son. Scientists can alter the gene that makes HGH (human growth hormone), significantly increasing the height of the child with low probability of side effects. Combined with a good chance of inheriting Mary s athletic ability, the child could become a professional athlete. Should Mary enhance her child? Achondroplasia Allen and Elaine have been happily married for several years and recently found that they are expecting a child. Elaine has achondroplasia, an autosomal dominant genetic disorder that causes dwarfism. Because the disorder is dominant any child of theirs has a fifty percent chance of being a dwarf. Elaine has decided to do genetic testing to determine whether the fetus has the gene for achondroplasia. This decision has infuriated many of Allen and Elaine s achondroplagic friends who value their unique culture. Elaine cares very much about her friends opinion. What should Allen and Elaine do if the gene tests show that the fetus is carrying the gene for achondroplasia?

38 Huntington s Disease Janice and James have one child already, and they are planning on having another baby. Janice s father, Moe, is 45 years old and was just diagnosed with Huntington s Disease, a genetic disorder that doesn t show up until later in life (35-50 years old). A person with the disorder has no symptoms until the onset of the disease, which starts out with clumsiness and motor skill decline, and progresses to full neurologic decline and death. There is no cure for this disease and it is autosomal dominant. This means that if one of your parents has the disease you will have a 50% chance of having it also. Janice and James s doctor has just informed them of 2 different types of tests that are available to determine whether a person is carrying the gene for Huntington s. The first test could be performed to tell if Janice or her son, Maurice, have the gene. The second test could be performed to tell whether or not an embryo that was formed by in vitro fertilization (IVF) has the gene. The prospective parents can then decide on which embryos to have implanted in the mother s uterus. It is assumed that they would choose an embryo that was free of the Huntington s gene. This procedure is very expensive, and there is an increased risk of miscarriage compared to a natural conception. What should Janice and James do? Gender selection Marina and Jason have three daughters. They only want to have four children and they would really like to have a son. They love all their daughters; they just really want to have a son also. Until recently, these desires would have to be left to chance. Physicians now have available a technique called PGD (Pre-Implantation Diagnosis). This allows physicians to screen embryos for a wide range of diseases, as well as for gender. The future parents can then decide which embryos they want to implant and which ones they don t want implanted into the mother s uterus. Any leftover embryos will be destroyed if they are not healthy or given to infertile couples so they can be implanted and hopefully brought to term. What should Marina and Jason do? IQ enhancement Greg smoked a lot of marijuana in high school and consequently has a very low sperm count. Julie and Greg will have to use artificial insemination if they want a child. Since a sperm donation will have to be placed in Julie anyway whether it comes from Greg or not, they are considering another option. They can afford to pay for sperm from the Mensa sperm bank. (Currently no such sperm bank exists, but sperm donors may identify themselves as Mensa members.) Mensa is an organization made up of people with IQs in the top 2% of the population. While the Mensa sperm donor increases the chances of having a child with a high IQ, it will not be Greg s biological child. Which sperm should they choose?

39 DESIGN AN EXPERIMENT (MAY ALSO BE DONE ON JOURNAL PAGE 17) Directions: After working with zebrafish, you have become a zebrafish researcher. Design an experiment you would like to do using the zebrafish model. List a testable hypothesis, the steps of your experiment, the materials you ll need, and the types of data you would collect to support or negate your hypothesis.

40 ZEBRAFISH DEVELOPMENT FLIPBOOK Materials: You ll need 5 sheets of stiff 8.5 x 11 paper (cardstock or heavy photo paper work well), a copier that will accept the heavy paper, a medium-sized binder clip, and scissors. What to do: Copy each of the following pages, 1-5 onto a sheet of heavy paper (the pages are in color, but you can copy them in black-and-white, too). Cut out the flipbook pages on the dotted lines. Stack the pages in order, including the four blank pages at the end. Tap the stack of pages on a tabletop to make sure the right hand sides are lined up. Use the binder clip on the left-hand side of the stack to hold your book together. Flip! This series of images depicts the first 48 hours of zebrafish development, starting out as a single cell, and ending as a hatched larva. If you look carefully at individual pages, you will see cell division (pages 1-10), development of eyes and other organs (starting at page 21), and development of muscles and pigment (starting at page 28). Thanks to Exploratorium.edu for making this, and to Mount View Middle School 2009 for sharing it with us!

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