Name Class Date *PACKET NOTES & WORKSHEETS LAB GRADE

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1 Name Class Date *PACKET NOTES & WORKSHEETS LAB GRADE MEIOSIS is specialized cell division resulting in cells with the genetic material of the parents Sex cells called have exactly set of chromosomes, this state is called (1n) Regular cells have sets of chromosomes, this state is called (2n) and divide through Produce with different amount and type of These cells are (1n) : sex Examples of gametes are,, and plant pollen Meiosis is for reproduction Sexual reproduction is give DNA Gametes Are Haploid Gametes must have the genetic material of a normal cell If the genetic material in the gametes was not halved, when they combined the would have genetic material than the parents. THAT WOULD BE A! Stages Of Meiosis Meiosis resembles except that it is actually two divisions not one These divisions are called and Meiosis I 1. Interphase ( ) 2. Prophase 3. Metaphase 4. Anaphase 5. Telophase 6. Cytokinesis ( cells) Meiosis II *DNA does NOT replicate again 1. Prophase II 2. Metaphase II 3. Anaphase II 4. Telophase II 5. Cytokinesis II ( daughter cells)

2 Crossing Over of Chromosomes Crossing over- A process occurring during (Prophase I) where two pair up and segments of their. This is how we get which means that even though you and siblings have the same parents- you all look! (except for identical twins) In The Beginning Two sex cells, a and an, unite through the process of to form a, the single cell from which the organism develops. is the process of producing sperm and eggs (gametes) *Remember is to produce (diploid) cells n (Haploid) daughter cells

3 Name Class *GENETIC NOTES & WORKSHEETS LAB GRADE DAY 1: Mendelian Genetics Vocabulary A. Genetics- Study of B. Heredity- The passing on of characteristics (traits) from to C. Trait A particular that can vary from one individual to another. Ex: D. Fertilization- Joining of male and female (reproductive cells) during reproduction E. Genes- Section of a chromosome that determine the trait we will inherit Ex: Sally has the gene for brown hair F. Alleles- of one gene Ex: Brown, red, or blond hair are all different forms the hair color gene G. Dominant Gene- expressed if present for a trait Dominant allele = capital letter (B) H. Recessive Gene- Only expressed if are present for that trait Recessive allele = lowercase letter (b) PRACTICE! Use the term Dominant or Recessive to identify the type of trait shown. 1. TT 2. tt 3. Tt 4. SS 5. ss 6. Ss SPERM EGG I. Homozygous- Organisms have alleles for a trait Ex: RR (BIG/BIG) or rr (small/small) J. Heterozygous- Organisms have alleles for a trait EX: Rr (BIG/small) PRACTICE! Use the term Homozygous or Heterozygous to identify each of the following: 1. AA 2. Aa 3. Mm 4. gg 5. GG 6. Gg K. Genotype- the genetic makeup ( ) or combination of (one from mom and one from dad) Ex: QQ, Qq, qq L. Phenotype- characteristics of the trait Ex: Brown Hair (what you see) Gregor Mendel ( ) A. Austrian Monk B. The C. Cross Bread common to study the inheritance of traits through each generation. Mendel s Pea Plant Experiment A. P1 B. F1- C. F2- L

4 Mendel s Conclusions: A. Biological Inheritance is from one to the next. Ex: your traits were determined by your parents genes that were passed onto you. DAY 1 B. Law of During, the pairs of so that each gamete receives only one gene for each trait C. Law or Principal of Some alleles are and some are. Recessive traits will only show up if dominant is present Punnett Squares following Mendelian Rules of Genetics A. A Punnett Square is a used to the outcome of a particular cross or breeding experiment B. Used to determine the of offspring s genotypes and phenotypes C. This does determine how many offspring will be produced or exactly what the offspring s genotype and phenotype will be just the chances! D. There are five steps (you must always show your work) 1. Key- List the trait, Both alleles, and which phenotype corresponds with each allele 2. Parents- List the genotypes of each parent 3. Draw the Punnett square 4. Determine the genotype ratio 5. Determine the phenotype ratio Complete the following Punnett squares using all 5 steps SAMPLE PROBLEM: Wavy hair is dominant to straight hair. Cross two heterozygous parents. 1. Key: 2. Parents: X 3. Punnett Square 4. Genotype: 5. Phenotype: Worksheet Problems: YES, THIS IS FOR A GRADE. YOU need to complete it! For each phenotype below, list the POSSIBLE genotypes (remember to use the letter of the dominant trait) Brown eyes are dominant to blue eyes Homozygous Brown Heterozygous Brown blue Widow s peak is dominant to straight hair line Widow s peak Widow s peak straight hair line

5 DAY 1 1. In rabbits, black hair color is dominant to white. Cross homozygous black rabbit with a white rabbit. 1. Key: 2. Parents: X 4. Genotype: 5. Phenotype: 2. In roses, red is dominant to yellow. Cross a homozygous red rose with a heterozygous red rose 1. Key: 2. Parents: X 4. Genotype: 5. Phenotype: 3. In cats, long tails are dominant to short tails. Cross two heterozygous long tail cats. 1. Key: 2. Parents: X 4. Genotype: 5: Phenotype: 4. In lizards, green skin color is dominant to purple. Cross a heterozygous green lizard with a purple lizard. 1. Key 2. Parents: 4: Genotype Ratio 5: Phenotype Ratio Match the terms with the definitions. 1. Genetics A. the weaker of a pair of genes 2. Dominant B. When the pair of genes are identical 3. Genes C. the study of heredity 4. Homozygous D. The passing of traits from parents to their young 5. Gamete E. the stronger of a pair of genes 6. Recessive F. sex cells, wither male or female 7. Heterozygous G. segment of DNA that codes for a particular protein 8. Heredity H. the physical appearance of an organism 9. Phenotype I. the actual genetic make-up of an organism 10. Genotype J. when the pair of genes are different

6 DAY In humans, a widow s peak is dominant to a straight hair line. Cross homozygous widow s peak with a straight hair line. 1. Key: 2. Parents: X 4. Genotype: 5. Phenotype: 12. In humans, a widow s peak is dominant to a straight hair line. Cross a straight hair line with a straight hair line. 1. Key: 2. Parents: X 4. Genotype: 5. Phenotype: 13. Mendel discovered that yellow pea color is dominant to green pea color. Cross a green pea plant with a heterozygous yellow pea plant. 1. Key: 2. Parents: X 4. Genotype: 5. Phenotype: 14. Using Mendel s information about pea plants, (yellow is dominant to green) cross a homozygous yellow pea plant with a heterozygous yellow pea plant. 1. Key: 2. Parents: X 4. Genotype: 5. Phenotype: Get STAMPED!

7 Day 2 Mendelian Genetics- Dihybrid Crosses (Crossing 2 Traits/Factors) Review of Monohybrid Crosses Remember, monohybrid crosses involve only trait Example: In fruit flies, red eyes are dominant over white eyes. RR x rr In this example you are only examining the trait. Dihybrid Cross: the study of pairs of contrasting traits at the same time Example: Fur color WITH Coat Texture Fur Color: Coat Texture: B: Black R: Rough b: White r: Smooth Mother is black fur AND rough coat BbRr Father is black fur AND rough coat BbRr Notice that each parent has so that results in (1 trait= 2 alleles) The Law of Independent Assortment During formation, segregating pairs of unit (alleles) assort of each other. The two traits are inherited totally independently of each other. Example: Fur color is inherited independently of coat texture. Sample problem 1: We will cross a heterozygous individual with another heterozygous individual. Their genotypes will be BbRr x BbRr Step 1: Find ALL possible gametes that can be made from each parent. Remember, each gamete must have one B and one R. USE THE FOIL METHOD B b R r x B b R r Step 2: Arrange all possible gametes for one parent on the top of your Punnett Square and the other parent on the side Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth Step 3: Fill in the Punnett Square (find the possible genotypes of the offspring) ACA Level

8 Step 4: Figure out the genotypic and phenotypic ratios: How many of the offspring would have a black, rough coat? How many of the offspring would have a black, smooth coat? How many of the offspring would have a white, rough coat? How many of the offspring would have a white, smooth coat? Phenotypic ratio : : : Sample Problem 2: What are the genotypic and phenotypic ratios in the offspring resulting from a cross between two people heterozygous for eye color and eyelash length? What is the phenotype of the parents in this cross? B Brown b Blue L - Long l short Parents are heterozygous for both traits. The cross is B b L l x B b L l. STEP 1: Determine the possible gametes that the parents can produce. USE FOIL for independent assortment STEP 2: Enter the possible gametes at the top and side of the Punnett square. Always keep the same letter first and capitals of each first. STEP 3: Complete the Punnett square by writing the alleles from the gametes in the appropriate boxes. The alleles from the gamete above the box and the alleles from the gamete to the side of the box are combined inside each of the boxes. The letters inside each box represent the probably genotypes of the offspring resulting from the cross. STEP 4: Determine the genotypes of the offspring. Count each row from left to right (as you read). Mark each one as it is used. Determine the phenotypes of the offspring. Count each row vertically (top to bottom). Genotypes Phenotypes ACA Level

9 Practice Problems: In mice, the ability to run normally is a dominant trait. Mice with this trait are called running mice (R). The recessive trait causes mice to run in circles only. Mice with the trait are called waltzing mice (r). Hair color is also inherited in mice. Black hair (B) is dominant over brown hair (b). For each of the following problems, draw a Punnett Square in the space provided and fill in the information on the indicated lines. CREATE THE KEY: 1. Cross a heterozygous running, heterozygous black mouse with a homozygous running, homozygous black mouse. Parents: X 2. Cross a homozygous running, homozygous black mouse with a heterozygous running, brown mouse. Parents: X 3. Cross a waltzing, brown mouse with a waltzing, brown mouse. Parents: X Get Stamped!

10 Day 3 : Non-Mendelian Incomplete Dominance *Exceptions to the Mendel Rule* Notes: Incomplete Dominance A. A cross between organisms with phenotypes Ex: Red x White B. Produce offspring with a phenotype that is a of the parental traits Ex: Pink is a blend of red & white C. EXAMPLE: If a homozygous flowered snap dragon plant (RR) is crossed with a homozygous the F1 offspring will have flowered snap dragon plant (WW), all of flowers. 1. RED flower x WHITE flower PINK flower *be careful- notice that the alleles are ALL capitalized D. The of individuals is (in the middle) between those of two homozygotes Sample Problem: In another flower, if red (RR) and blue (BB) flowers are crossed, they produce a 3 rd purple (RB) flower What would be the genotype ratio and phenotype ratio if you crossed two purple flowers? Cross of two purple flowers (parents) X What are gamete possibilities? genotype ratio phenotype ratio Can you have a heterozygous red or hybrid blue flower? Practice Incomplete dominance - When a flowering plant homozygous for red flowers is crossed with one homozygous for white flowers, all of the offspring bear pink flowers. Cross a homozygous red flowered plant with a homozygous white flower plant. A. Key: B. Parents: x C. Genotype: D. Phenotype: What is the genotype of the F 2 generation from the problem above? (*you take any 2 offspring/babies from the Punnett square above and cross them) A. Key: B. Parents: x C. Genotype: D. Phenotype: ACA

11 Inheritance of color in rats The same principle of inheritance of coat color is also shown in some rodents. When a rat homozygous for white fur is crossed with one homozygous for black fur, the offspring are all gray in color. Show the result in crossing two gray rats. Sho w your work. a. Key: b. Parents: x c. Genotype: d. Phenotype: ***Explain how people who raise rats could maintain only rats with white fur: Practice Incomplete Dominance show all work. 1. In noses, dry nostrils are incompletely dominant to runny nostrils. A cross will result in a slimy nostril. A. Cross a slimy nostril with a dry nostril. a) Key: DD = dry nose RR = runny nose DR=Slimy nose b) Parents: x c) Genotype: d) Phenotype: B. Cross a dry nostril with a runny nostril. a) Key: b) Parents: x c) Genotype: d) Phenotype: C. Cross two slimy nostrils. a) Key: b) Parents: x c) Genotype: d) Phenotype: Get STAMPED!

12 Day 4: Non-Mendelian Genetics: Codominant Traits with Blood Typing Codominance A. A cross between organisms with _ phenotypes 1. Ex: Red x White B. Produce offspring with a phenotype that displays traits at the same time 1. Ex: Red and White striped C. EXAMPLE: If a homozygous flowered snap dragon plant (RR) is crossed with a homozygous flowered snap dragon plant (WW), all of the F1 offspring will have flowers. 1. RED flower x WHITE flower RED AND WHITE flower Human Codominance A. Sickle-Cell Anemia 1. An individual who is heterozygous for sickle-cell alleles will express BOTH and blood cells. The oxygen carrying protein hemoglobin differs by one amino acid than the regular cause the shape to change: Normal RBC are and abnormal are or sickle shaped. 2. Abnormally shaped blood cells,, and result in tissue damage and. 3. Heterozygous individuals are said to have the sickle-cell trait because they show signs of sickle-cell related disorders if the availability of oxygen is reduced. B. Blood types 1. Blood type is determined by alleles. This means there are more than two types of alleles possible that can make up a pairing. : red blood cells produce A antigens (protein) on the outside of the cell Allele is expressed I A because it is : red blood cells produce B antigens (protein) on the outside of the cell Allele is expressed I B because it is : red blood cells will produce an antigen (protein) Allele is expressed i because it is 2. Determining Blood types is necessary before you receive a blood transfusion because incompatible red blood cells together or clot causing death. 3. Your immune system or antibodies recognizes the red blood cells belonging to you. If cells with a of antigen enter your body your immune system will attack them. ACA Level

13 Complete the following chart: Day 4 Phenotype or blood type Type A All genotypes possible Type B Type AB (Codominant) Type O (recessive) ***The true universal blood donor are O- ***The true universal recipient is AB+ SAMPLE PROBLEMS: Cross a parent with Type AB blood with a parent that has heterozygous type A blood. Parents: X Genotype: Phenotype: Cross a heterozygous type A blood father with a heterozygous type B blood mother. Parents: X Genotype: Phenotype: Practice Problems: Include all 5 steps where necessary. 1. Type A blood can be expressed as AA or AO. A. Complete the Punnett squares to determine if two parents with type A blood can produce a baby with type O blood. AA x AA AA x AO AO x AO B. What is the genotype of a child with type O blood? C. Is it possible for parents with type A blood to have a child with type O blood? If yes, what must be the genotypes of the parents? X ACA Level

14 2. A man with homozygous A blood has children with a woman with heterozygous B blood. What are the possible blood types for their children? Parents: X Day 4 Genotype: Phenotype: 3. Pick one child from problem #2 and cross that person with a person with heterozygous A blood. Parents: X Genotype: Phenotype: 4. Suppose that a man with type A blood marries a young woman who has type AB blood. What blood types would you expect to find among their children. Remember to take all possibilities into account. **What are ALL the possible blood types for the offspring? 5. Mr. Jones was taken to court and charged with being the father of a woman s child. He is homozygous type A blood. She is heterozygous type B blood. The child is type O blood. Is Mr. Jones the father? Show your work. Parents: X Genotype: Phenotype: **Is Mr. Jones the father of the baby? Explain. 6. In some cows, the genes for brown hair (B) and for white hair (W) are co-dominant. Cows with alleles for both brown and white hair, express both brown and white hairs. This condition gives the cattle a reddish color, and is referred to as Roan (BW). Cross a brown bull with a roan cow. Parents: X Genotype: Phenotype: Get STAMPED!

15 Day 5: Non- Mendelian Genetics: SEX- LINKED TRAITS A. Determining Sex 1. Humans have a number of chromosomes or of the pairs are autosomes, they are the for males and females 3. The 23 rd pair of chromosomes in males and females These are the sex chromosomes and are indicated by and determine the baby s sex. B. Inheriting Sex linked traits 1. Genes located on the are called sex- linked traits 2. Because the Y chromosome is small it carries, including the male determinant sex gene. 3. Males who receive a on the X chromosomes will express the because he cannot inherit on the Y chromosome. There is dominant allele to the recessive gene 4. have a higher percent chance of expressing the recessive trait C. Examples in Humans 1. Hemophilia Causes a problem with Caused by a allele on the chromosomes Males: 1/10,000 Need recessive allele from carrier mom Females: 1/10,000,000 Need recessive alleles; one from mom and one from dad Sex- Linked Traits 2. Color Blindness People who have red-green color blindness cannot differentiate between the two colors. Caused by a allele on the X chromosomes ACA worksheet

16 Sample problem 1: A colorblind man marries a woman who is not colorblind. Assuming there is no colorblindness in the woman s ancestr what will most likely be the genotype and phenotypes of their sons and daughters? Show your work a. Key: e. Punnett Square b. Parents: Day 5 c. Genotype Ratio: d. Phenotype Ratio: USE THIS CHART TO HELP YOU ANSWER THE PRACTICE PROBLEMS THAT FOLLOW!! Colorblindness- C= normal c = colorblindness Hemophilia- H=normal h=hemophilia PHENOTYPE GENOTYPE PHENOTYPE GENOTYPE Female normal X X Female normal, Carrier Female, color-blind Male normal X Y Male, color-blind Female normal X X Female normal, Carrier Female, hemophiliac Male normal X Y Male hemophiliac *Males either have colorblindness or they don t. They only have one X chromosome that was received from the mothe ** More males have hemophilia and color blindness because males need one gene for hemophilia or Color-blindness to occur. Sample problem2: WORK TOGETHER WITH TEACHER A normal woman marries a normal man producing two colorblind sons, two normal sons and three normal daughters. *HINT You should work backwards- put in the offspring first to figure out parents. 1. Key: 3. Punnett Square 2. Parents: 3. Geno Ratio: 4. Pheno Ratio: ACA worksheet

17 PRACTICE PROBLEMS: SHOW ALL 5 STEPS! Day 5 A. A female carrier of hemophilia marries a normal male. What are the genotypes and phenotypes of the possible offspring? Show your work. 1. Key: 3. Punnett Square 2. Parents: 3. Geno Ratio: 4. Pheno Ratio: B. Normal man X Hemophiliac woman 1. Key: 3. Punnett Square 2. Parents: 3. Geno Ratio: 4. Pheno Ratio: C. Hemophiliac man X Heterozygous normal woman 1. Key: 3. Punnett Square 2. Parents: 3. Geno Ratio: 4. Pheno Ratio: D. A normal woman marries a normal man. Two sons that are hemophiliacs and one son is normal. What is the mother s genotype? (LOOK AT SAMPLE #2 for help) 1. Key: 3. Punnett Square 2. Parents: 3. Geno Ratio: 4. Pheno Ratio: Get STAMPED!

18 Day 6: Genetic Pedigree Charts A pedigree is a which shows the within a family. Pedigrees are used to: how the traits are passed from one generation to the next. inferring (carriers) Pedigrees are based on the individual s. So if someone has the trait being studied they are phenotypically. How to read a pedigree Square = or Circle = or Shaded means Phenotypically Affected or Un-shaded means Phenotypically Unaffected or A line connecting a male and a female represents a marriage. A line coming off of a male and a female represents their. Sometimes the pedigree will show half-filled shapes. This indicates a carrier. What is a carrier? Practice: 1. How many children did this first couple have? 2. How many grandchildren did this first couple have? 3. How many people in this family are affected? 4. How many people can you be certain are carrying the trait? 5. If the allele for this trait is the letter e, assign a genotype to all of the individuals you are sure of. 6. So which individual is the only one we cannot be sure of? TURN IN PACKET FOR LAB GRADE!!