Name: Bio AP Mendelian Genetics & Chromosomal Inheritance

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1 Name: Bio AP Mendelian Genetics & Chromosomal Inheritance 1

2 ESSENTIAL KNOWLEDGE 3.A.3: The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring (Ex: Sickle cell anemia, Tay Sachs disease, Huntington s disease, X-linked color blindness, Trisomy 21/Downs syndrome, Klinefelter s syndrome, reproductive issues) 4.C.2: Environmental factors influence the expression of the genotype in an organism. (ex. Height and weight in humans, flower color based on soil ph, Density of plant hairs as a function of herbivory, Darker fur in cooler regions of the body in certain mammals species, Alterations in timing of flowering due to climate changes, presence of the opposite mating type on pheromone production in yeast and other fungi.) 4.C.4: The diversity of a species within an ecosystem may influence tha stability of the ecosystem, 3.A.4: The inheritance pattern of many traits cannot be explained by simple Mendelian genetics. (ex. Sex-linked genes, Mammal Y chromosome is small as does not carry many genes, Mammal females as XX and males are XY so sex linked recessive traits are always expressed in males. Some traits are sex limited and expression depends on the sex of the individual, such as milk production in female mammals and pattern baldness in males) 2

3 I. Mendelian Genetics A. Background History Genetics 1860 s: Mendel proposed that discrete inherited factors segregate and assort independently during gamete formation Cytology 1875: Cytologists worked out the process of mitosis 1890: Cytologists worked out the process of meiosis 1900: Three botanists independently rediscovered Mendel s principles of segregation and independent assortment (Correns, de Vries and von Seysenegg) 1902: Cytology and genetics converged as Walter Sutton and others noticed parallels between the behavior of Mendel s factors and the behavior of chromosomes. Chromosomes and genes are paired in diploid cells Homologous chromosomes separate and allele pairs segregate during restores the paired condition for both chromosomes and genes Gregor Mendel-Garden Experiments a. Worked with pea plants because they had a large number of different traits b. Control the mating because male and female reproductive structures are in the flower c. Developed true breeding plants through self-pollination d. Mated plants with different traits through cross-pollination (hybridization) 3

4 e. Examined the results of his experiments 4

5 Vocabulary a. Parental generation (P 1 ) - parents b. First filial generation (F 1 )- first generation offspring (filial - son) c. Second filial generation (F 2 )- second generation offspring (grandchildren) d. Character-heritable feature (ex. Pea shape, flower color) e. -physical appearance f. -genetic make-up (alleles determining a trait) g. Each individual has determining a trait h. (pure) -individual will have two of the same alleles for the trait i. (hybrid) -individual will have two different alleles for the trait B. Mendel s Laws Law of Dominance When one allele for a trait can the expression of another allele for the same trait allelestronger allele which is expressed alleleweaker allele which is masked Law of Segregation from each other in the formation of gametes Modern genetics-this is explained in the process of 5

6 3. Law of Independent Assortmenttwo or more characters in the formation of gametes Recall-modern genetics C. Mendel s Crosses & Punnett Square Basic Concepts a. Symbols/Terms Key letters are used to represent alleles. Dominant allele is represented by an Recessive allele is represented by the of the same letter Plant Trait Dominant Recessive Homozygous Trait Trait Plant (pure) Stem Length Tall Short Pure tall (TT) (T) (t) Pure short (tt) Seed Color Yellow Green Pure yellow (YY) (Y) (y) Pure green (yy) Seed Shape Round Wrinkled Pure round (RR) Flower Position (R) Axial (A) (r) Terminal (a) Pure Wrinkled (rr) Pure axial (AA) Pure terminal (aa) letter form Heterozygous Plant (hybrid) Hybrid tall (Tt) Hybrid yellow (Yy) Hybrid round (Rr) Hybrid axial (Aa) b. Genotype/Phenotype Genotype Phenotype Homozygous Dominant ( ) Hybrid ( ) Homozygous Recessive ( ) 6

7 c. Punnett Square-is a way of showing all the possible allele combinations in the offspring of a particular cross. Key: P=purple flower P=white flower Parental cross: Homozygous purple X White PP X pp P P p p Punnett Square F 1 Results: Phenotype: Genotype: F 1 cross: Hybrid purple X Hybrid purple Pp X Pp P p P p Punnett Square F 2 Results: Phenotype: Genotype: Steps for using Punnett Square 1. State the Key 2. Write the cross and show the gametes 3. Draw the Punnet square and place the egg alleles on one side of the box and the sperm allele on the other side of the box 4. Fill in the box representing all the alleles combinations possible in the offspring 5. Write the results, showing the phenotype and genotype ratios 7

8 2. Mendelian Inheritance (crosses) a. (monohybrid) Problems: 1) Mendel crossed pure pea plants that produced round seeds, with pure breeding pea plants that produced wrinkled seeds. In the F 1 all plants were round. He then crossed the F 1 plants. Determine the expected genotypic and phenotypic ratios in the F 2 generation. Key: Parental cross: Pure round X Pure wrinkled F 1 cross: Hybrid round X Hybrid round Punnett Square Punnett Square F 1 Results: Phenotype: Genotype: F 2 Results: Phenotype: Genotype: 2) When pea plants with axial flowers are crossed with pea plants having terminal flowers, all of the offspring have axial flowers. What were the genotypes of the parent plants? Key Cross: Axial X Terminal All offspring must have an allele to be axial. The unknown allele must be 8

9 b. How can you determine if a purple flowered plant is pure or hybrid? Used to determine the genotype of an unknown individual Ex. Organisms exhibiting the dominant phenotype can be Cross the unknown organism with a and examine the offspring The appearance of individual trait will determine that the unknown was Problems: 1) How can a breeder determine whether a tall pea plant is pure or hybrid? Unknown Tall plant X plant Key T = tall t= short If the unknown is pure (TT) then If the unknown is hybrid (Tt) Then t t t t T Tt Tt T Tt Tt T Results: Phenotype: Tt Tt t Results: Phenotype: tt tt 9

10 Let s Practice: 1. In watermelons, solid green color is dominant over the stripped pattern and short shape is dominant over long shape. Using the Punnett square method, show the offspring of (1)a cross of a pure green watermelon and a pure stripped watermelon and (2) a cross between a hybrid short melon and a long melon. (a) (b) 2. In fruit flies, gray body is dominant over ebony body. Two gray fruit flies are mated and produce 186 gray bodied and 58 ebony bodied offspring. Determine the parental genotype. 10

11 3. In humans, albinism is recessive to normal skin pigmentation. If an albino female marries a heterozygous male, what percentage of their offspring would most likely be normal? 4. In guinea pigs, black fur is dominant over white fur. How can a breeder determine whether a black guinea pig is homozygous or heterozygous black? 5. In fruit flies, dumpy wings are shorter and broader than normal wings. The allele for normal wings (D) is dominant to the allele for dumpy wings (d). Two normal-winged flies were mated and produced 300 normal-winged flies and 100 dumpy-winged flies. The parents were probably (1) DD and DD (2) DD and Dd (3) Dd and Dd (4) Dd and dd (5) dd and dd 6. Suppose that in sheep, a dominant allele (B) produces black hair and a recessive allele produced white hair. If you saw a black sheep, you would be able to identify (1) its phenotype for hair color (2) its genotype for hair color (3) the genotypes for only one of its parents (4) the genotypes for both of its parents (5) the phenotypes for both of its parents. 11

12 c. Inheritance of genes on chromosomes (dihybrid cross) Genes on separate chromosomes will and follow the law of 12

13 Punnett Square needs to be expanded to account for the formation of more gamete combinations and more combinations in the offspring Key Y = Yellow y = green R=round r=wrinkled Cross: Parent Generation Yellow Round X Green Wrinkled F 1 Generation 100% hybrid Yellow Round F 2 Generation Results: 13

14 Lets Practice: 1) Determine the results of a cross between two hybrid tall-yellow pea plants. Key Cross: Results: 2) In humans brown eyes is determined by a dominant allele B and blue eyes is determined by a recessive allele b; free earlobes are determined by a dominant allele F and attached earlobes by a recessive allele f. A browneyed man with attached earlobes (whose mother was blue-eyed) marries a blue-eyed woman with free ear-lobes (whose father had attached earlobes. What phenotypes may be expected among their children? 14

15 3) In cocker spaniels, black color is dominant over red color, and solid color is dominant over white spotting. A solid red male was mated to a black and white spotted female. They had five puppies: one black, one red, one black and white, and two red and white. What were the genotypes of the parents? d. Rule of Probability The Law of Segregation and the Law of Independent Assortment are chance events and will therefore follow rules of probability; The chance that alleles from an egg and alleles from a sperm will join together is a zygote is the product of their individual probabilities Ex. The chance that an egg from a Pp mother will receive a p is The chance that a sperm from a Pp father will receive a p is The chance that the ova and sperm will fertilize is (Rule of Multiplication) Problem: 1) What is the probability that a dihybrid cross between two plants with the genotypes TtYy and TtYy will produce offspring with the genotype Ttyy? Since the traits are on separate chromosomes and are inherited independently, set up two crosses and multiply the probabilities. 15

16 Cross 1 Cross 2 Tt X Tt Yy X Yy T t Y y T t TT Tt Y YY Yy Tt tt y Yy yy Chance of Tt is Chance of yy is Chance of the offspring having both combinations is 2) What is the probability that a trihybrid cross between two organisms with the genotypes AaBbCc and AaBbCc will produce an offspring with the genotype aabbcc? 3) What fraction of the offspring of parents each with the genotypes Kk Ll Mm will be kk LL Mm? 16

17 Summary of Mendelian Genetics: Cross Ratio Monohybrid 3 expressing dominant phenotype:1 expressing recessive phenotype Dihybrid 9 expressing BOTH dominant phenotypes 3 expressing one dominant and one recessive phenotype 3 expressing one recessive and one dominant phenotype 1 expressing BOTH recessive phenotypes Test Cross Used to determine an unknown genotype by mating with the PURE recessive II. Extending Mendelian Genetics A. Incomplete Dominance Cross: Japanese Four O clock Flowers Key: =red =white P 1 Red Flower X White Flower x F 1 100% Pink Flowers ( ) Pink Flowers X Pink Flower C R C W x C R C W C W C W B. Codominance Complete Dominance (A is dominant) Incomplete Dominance (A is incompletely dominant) Codominance (no dominance) AA and Aa have the phenotype Aa =Intermediate phenotype (Phenotype is the two homozygotes) Aa = Both alleles are in the phenotype 17

18 Codominance- the inheritance is characterized by the full in the heterozygote MN blood Group the locus codes for the production of surface glycoprotein on red blood Genotype Blood Type cells. MM M (M glycoprotein) The MN blood type NN N (N glycoprotein) produces both surface MN MN (M & N glycoprotein) glycoproteins. alleles are fully expressed in the heterozygote C. Multiple Alleles alleles determine a phenotype Any one individual will possess only of the multiple alleles Key will use a letter and the allele will be expressed as a superscript Example: Human Blood Type ( A blood, B blood, AB blood, O blood) There are THREE alleles for Human blood type: Key: =Dominant allele = Recessive allele = A antigen (protein) = B antigen (protein) = no protein Blood Type Possible Genotypes Protein Antigen A glycoprotein B glycoprotein AB and glycoprotein O glycoprotein 18

19 Blood Testing: Blood Transfusions: Universal Donor: Universal Recipient: Let s Practice: 1) Determine all the possible genotypes of the children of a mother with type O blood and a father with type AB blood. 19

20 2) Show how two parents who do not have blood type O can produce a child with type O blood. 3) Mrs. Frank and Mrs. Jones both had babies the same day in the same hospital. Mrs. Frank took home her son Bobby and Mrs. Jones took home her son Gene. Mrs. Jones began to suspect that her son had been switched. Blood tests were made: Mr. Frank was type A, Mrs. Frank was type B, Mr. Jones was type A, and Mrs. Jones was type A. Bobby had type O blood and Gene has type B blood. Has a mix-up occurred? Explain. 20

21 D. Pleiotropy One gene influences Example: Sickle cell anemia N=normal red blood cell n=sickle red blood cell E. Gene Interactions: More than is determining a trait Epistasis the phenotypic expression of another GENE is the stronger gene Example Coat Color in Labrador Retrieves Epistatic gene determines the production of pigment Second gene determines the color of the pigment Key E=pigment deposited in hair e=no pigment in hair B= black pigment b = brown pigment A 9:3:3:1 ration was NOT obtained 21

22 Collaboration Two genes interact to form a phenotype that neither gene can express alone Example: Comb Type in chickens Key R= rose r=single P=pea p=single = walnut Single Pea Rose Walnut Polygenic Inheritance (Quantitative Characters) effect of two or more genes on a single phenotype character Example: Three genes determining pigment production Each dominant allele gives one unit of skin pigmentation Note the bell shaped curve 22

23 Gene Gene which can alter the expression of another gene Example: Human eye color Gene -Pigment Production B=brown b=blue Modifier Gene- Pigment distribution Modifier Gene-Pigment Tone Beagle Spots F. Environmental Influence Degree to which the can influence the expression of a gene Examples: Color in a Himalayan rabbit Experiment Expression of the black pigmentation was influenced by Color of hydrangea flowers Humans: Nature verses Nurture Sun tanning & skin color Nutrition & height Note: The expressions of many traits are - involving genetic and environmental factors 23

24 III. Pedigree Charts Pedigree =family tree that diagrams the relationships among parents and children across generations and that shows the inheritance pattern of a particular phenotypic characteristic. Symbols Square= Circle= Horizontal line connecting male and female= has occurred; are listed below in birth order symbols indicate individuals showing the trait being traced Tracing Pedigrees 24

25 Example-Determine all possible phenotypes Key: E=Attached ears e=unattached ears SUMMARY: Parent 1= 2= F1 1= 2= 3= 4= F2 1= 2= 3= 4= 5= 6= 7= 8= 9= 10= 11= 12= F3 25

26 IV. Chromosomal Basis of Inheritance A. Background Gene map of Drosophila melanogaster (alias-the Fruit Fly) Genes can be same chromosome Linked genes will during Independent Assortment Linked genes can only be separated by on the 26

27 B. Sex Linked Inheritance 1. Morgan studies: a. Why did Morgan choose the Fruit Fly? They are easy to culture in the lab They are prolific breeders They have a short generation time They have only four pairs of chromosomes which are easily seen under the microscope. b. A Special Cross 2. Sex Chromosomes All F 1 offspring have eyes F 2 offspring have a 3:1 ratio BUT ALL white eyed flies are Is eye color linked to the sex chromosome? 27

28 3. Sex Linked Inheritance Traits that are linked (coded for) on the. Human sex linked traits Sex Determination Male Female Human Birds XX XY Bees XO XX 4. Sex Linked Traits Shape-the X chromosome is than the Y chromosome Males have two bottoms but only one top Sex linked trait are those that have their on the top of the X chromosomes. Males will have only allele for the trait.(hemizygous) Males need only recessive allele at that locus to express the trait. Hybrid female is a. She will not express the trait but can pass it on to her male offspring 5. Human Barr Body Females have doses of alleles on the X chromosome is achieved by inactivating one of the X chromosomes The inactive X is and is attached to the nuclear membrane Determine the number of Barr bodies in the following: Genotype Sex Number of Barr Bodies XX XY XO XXX XXY 28

29 Barr Body Inactivation of the X chromosome is random and occurs independently in embryonic cells giving rise to a (having clusters of cells derived from the active X inherited from the father and the active X inherited from the mother. Note: chromosome traits are associated with the Y 29

30 6. Sex Linked Inheritance Key: X N = normal allele X n =recessive allele Y Example: X N X N =Normal Female X N X n =carrier female X N Y=normal male X n Y=male expressing the recessive allele Determine the phenotypes of a cross between a normal male and female carrier for color blindness KEY: X N =normal female X n =colorblind Y= male X N Y Normal Male X Carrier Female X N Y X X N X n X N X N X N X N Y X n X N X n X n Y 25% normal female 25% carrier female 25% normal male 25% colorblind male Let s Practice: 1. Determine all the possible genotypes of the children of a mother who is a carrier for hemophilia and a normal father. 30

31 2. Determine the percentage of colorblind females which could result from the cross between a colorblind male and a carrier C. Linked Gene Inheritance 1. History Punnett and Baleson Key: P=purple p=white Cross: Parent Generation PPLL F 1 Generation 100% hybrid purple long L=long L=short Homozygous purple long X Homozygous white short ppll PpLl (purple long) PpLl X PpLl F 2 Generation PL Pl pl pl Results: Expected (9:3:3:1) PL A 9:3:3:1 ratio was obtained Pl PPLL PPLl PPLl PPll PpLL PpLl Ppll Ppll pl PpLL PpLl ppll ppll pl PpLl Ppll ppll ppll 31

32 To try to explain the results they performed a test cross Hybrid purple long X Homozygous white short Results: Expected (4:4:4:4) Actual: 7:1:1:7 The majority looked Purple long and white short pl pl pl PL PpLp PpLp PpLp Pl Ppll Ppll Ppll pl ppll ppll ppll pl ppll ppll ppll pl PpLp Ppll ppll ppll Explanation: The gene for color and the gene for length are on the same chromosome. They will follow the Law of Independent Assortment. They can only be separated by The recombinant allele combinations do not occur as frequently D. Recombination Frequency The greater the distance between linked genes, the chance (percentage) that crossover will occur. Percent Recombination= the One percent crossover or will equal one 32

33 The genes are apart E. Chromosome Mapping The percent recombination can lead to on a chromosome One map unit= 1% recombination frequency Procedure Establish the relative distances between those genes farthest apart or with the highest recombination frequency Determine the recombination frequency between the third gene and the first. Consider all possibilities Determine the recombination frequency between the third gene and the second. Eliminate the incorrect sequence Example: Loci Recombination Frequency Map Units b v 17.0% 17 c b 9.0% 9 c v 9.5% 9.5 (1) b v Possibilities: c b v or b c v 33

34 (2) c b Correct Sequence: b c v (3) c v Let s Practice 1. A tomato breeder wants to determine if the genes for plant height and plant color are on the same chromosome. The allele for tall (T) is dominant over dwarf (t), and the allele for green (G) is dominant over light green(g). A homozygous tall green plant was crossed with a homozygous recessive plant and the F 1 progeny were then test crossed. The test cross results are summarized in the table below. Determine if the genes are linked, and if so, how many map units apart are they? Phenotype Number of test crossed progeny Tall-green 434 Dwarf-light 466 green Tall-light 51 green Dwarfgreen Three genes have the following percent recombination. Determine the order of the genes. A-B=20% B-C=10% A-C=30 % 34

35 NOTES: 35

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