Name. More Complicated Inheritance Patterns. INCOMPLETE DOMINANCE: Remember, here the different alleles cause a blended new phenotype.

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1 Name More Complicated Inheritance Patterns INCOMPETE DOMINANCE: emember, here the different alleles cause a blended new phenotype. 1. In four o'clock flowers, red () is incompletely dominant over white (). Cross two pink () parents. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities indicated. Genotypes of parents: x Genotypic ratio: 1 : 2 : 1 ed 1/4 hite 1/4 Phenotypic ratio: 1 red: 2 pink: 1 white Pink 1/2 2. Now cross a red flower with a white flower. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities. Genotypes of parents: x Genotypic ratio: 100% ed 0 hite 0 Phenotypic ratio: 100% pink Pink 100% 3. hat cross would result in a phenotypic ratio of 1 red: 2 pink: 1 white? Genotypes of parents x 4. hat cross would result in a phenotypic ratio of 1 red: 1 pink? Genotypes of parents x

2 5. In squash plants, round squash () are incompletely dominant over long squash (). Cross 2 oval () parents. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities indicated. Genotypes of parents: x Genotypic ratio: 1 : 2 : 1 ound 1/4 ong 1/4 Phenotypic ratio: 1 round: 2 oval: 1 long Oval 1/2 6. Now cross a round squash with an oval squash. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities. Genotypes of parents: x Genotypic ratio: 1 : 1 ound 1/2 ong 0 Phenotypic ratio: 1 round: 1 oval Oval 1/2 7. hat cross would result in offspring that are all oval? Genotypes of parents x 8. hat cross would result in a phenotypic ratio of 1 oval: 1 long? Genotypes of parents x

3 CODOMINANCE: emember, here two alleles can both be dominant and express or show themselves equally. 9. In chickens, black feathers () are codominant with white feathers (). The heterozygous genotype () results in erminette, or speckled black AND white feathers. Cross two erminette chickens. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities indicated. Genotypes of parents: x Genotypic ratio: 1 : 2 : 1 lack 1/4 hite 1/4 Phenotypic ratio: 1 black: 2 erminette: 1 white Erminette 1/2 10. Now cross an erminette chicken with a white chicken. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities indicated. Genotypes of parents: x Genotypic ratio: 1 : 1 lack 0 hite 1/2 Phenotypic ratio: 1 erminette: 1 white Erminette 1/2 11. hat cross would result in a phenotypic ratio of 1 black: 1 erminette? Genotypes of parents x 12. hat cross would result offspring that are all erminette? Genotypes of parents x

4 13. In shorthorn cattle red coat color () is codominant with white coat color (). The heterozygous genotype () results in a roan, or red AND white coat color. Cross a red bull (male) with a white cow (female). rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities. Genotypes of parents: x Genotypic ratio: 100% ed 0 hite 0 Phenotypic ratio: 100% roan oan 100% 14. Now cross a roan bull with a red cow. rite the genotypes of the parents, show a Punnett square, complete the genotypic and phenotypic ratios of the offspring, and determine the probabilities. Genotypes of parents: x Genotypic ratio: 1 : 1 ed 1/2 hite 0 Phenotypic ratio: 1 red: 1 roan oan 1/2 15. hat cross would result in a phenotypic ratio of 1 roan: 1 white? Genotypes of parents x 16. hat cross would result in a phenotypic ratio of 1 red: 2 roan: 1 white? Genotypes of parents x

5 MUTIPE AEES: Many traits involve more than just two alleles. For example, in humans, O blood types are determined three alleles; A,, and O. (Furthermore, A and are codominant over the recessive allele O.) 17. lood types display codominance in the form of multiple alleles. Complete the following crosses. Show a Punnett square and then write the genotypic and phenotypic ratios of the offspring for each. (a) Type A (homozygous) x Type (homozygous) (b) Type A (homozygous) x Type (heterozygous) (c) Type A (heterozygous) x Type (homozygous) a. I A I A X I I b. I A I A X I i c. I A i X I I Genotype atio: Genotype atio: Genotype atio: 100% I A I 1 I A I : 1 I A i 1 I A I : 1 I i Phenotype atio: Phenotype atio: Phenotype atio: 100% 1 : 1 A 1 : Complete the following crosses. Show a Punnett square and then write the genotypic and phenotypic ratios of the offspring for each. (a) Type x Type (b) Type x Type O (c) Type A (heterozygous) x Type (heterozygous) a. I A I X I A I b. I A I X ii c. I A i X I i AA Genotype atio: Genotype atio: Genotype atio: 1 I A I A : 2 I A I : 1 I I 1 I A i: 1 I i 1 I A I : 1 I A i: 1 I i: 1 ii Phenotype atio: Phenotype atio: Phenotype atio: 1 A : 2 : 1 1 A : 1 1 : 1 A : 1 : 1 O ii 19. Could a parent with type blood and a parent with type O blood have a child with type O blood? Explain why or why not. Draw a Punnett square to support your answer. NO offspring will all be A or I A I X ii

6 POYGENIC: This is a situation where two or more gene pairs, probably on different chromosomes, work together to create many variations for a specific trait. Most human traits are thought to be polygenic. In humans, at least 8 different genes determine eye color, and is thus considered polygenic. For simplicity, consider the influence of 3 genes; A,, and C. The greater the number of dominant alleles, the darker the eye color. 20. Cross a male with the genotype AaCc with a female with the genotype Abcc ist the genotypes of the gametes produced : The father: C, c, ac, ac The mother: c, Abc Create a Punnett square. Then list the genotypes of each child in order from lightest to darkest eye coloration based on the number of dominant alleles. ightest Aabcc (2 dominant alleles) C c ac ac c Abc Aacc (3 dominant alleles) ACc 5 AbCc 4 AabCc (3 dominant alleles) Acc 4 Abcc 3 Abcc (3 dominant alleles) AaCc (4 dominant alleles) AaCc 4 AabCc 3 Acc (4 dominant alleles) Aacc Aabcc AbCc (4 dominant alleles) 3 2 Darkest ACc (5 dominant alleles)

7 In parakeets, two pairs of alleles (Yellow and lue) work together to determine a bird s color. Parakeets that are green in color are determined the genotype Y. Parakeets that are blue in color are determined the genotype yy. Parakeets that are yellow in color are determined b. Parakeets that are white in color only exist as the genotype by. ist the possible genotypes for each of the colors given. Green lue: Yellow: hite: 1. YY 2. y 3. Yy 4. Y 1. yy 2. y 1. by 2. by 1. by Cross the following parakeets. rite the genotypes of the parents, complete a Punnett square, and write the fraction of offspring that could have each color heterozygous green parakeets y X y 9/16 green 3/16 blue 3/16 yellow 1/16 white Y y Y y YY Yy Y y Yy yy yyy y Y y by by yyb y by by 22. Heterozygous yellow x homozygous yellow by X by 0 green 0 blue 100% yellow 0 white by by

8 23. Homozygous blue x heterozygous yellow yy X by 1/2 green 1/2 blue 0 yellow 0 white y y y 24. Heterozygous blue x white y X by 0 green 1/2 blue 0 yellow 1/2 white y y by 25. Homozygous yellow x heterozygous blue by X y 1/2 green 0 blue 1/2 yellow 0 white y y by