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Dihybrid Crosses Incomplete Dominance Codominance Multiple Alleles

Dihybrid Crosses Incomplete Dominance Codominance Multiple Alleles. In a nutshell!. PP = purple. Pp = purple. pp = white. Review: Dominant/Recessive. One allele is dominant over the other (capable of masking the recessive allele) Example: Purple flowers are dominant over white. PP. Pp.

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Dihybrid Crosses Incomplete Dominance Codominance Multiple Alleles

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  1. Dihybrid CrossesIncomplete DominanceCodominanceMultiple Alleles In a nutshell!

  2. PP = purple Pp = purple pp = white Review: Dominant/Recessive • One allele is dominant over the other (capable of masking the recessive allele) • Example: Purple flowers are dominant over white

  3. PP Pp Pp pp Review Problem: Dominant/Recessive • In pea plants, purple flowers (P) are dominant over white flowers (p) show the cross between two heterozygous plants. P p GENOTYPES: - PP (1); Pp (2); pp (1) - ratio 1:2:1 P p PHENOTYPES: - purple (3); white (1) - ratio 3:1

  4. Dihybrid Cross:a cross that shows the possible offspring for two traits Coat Texture: R: Rough r: Smooth Fur Color: B: Black b: White In this example, we will cross a heterozygous individual with another heterozygous individual.

  5. Step 1: Determine the parents’ genotypes Step 2: You must find ALL possible gametes that can be made from each parent. Remember, each gamete must have one B and one R. Parent 1: BbRr Parent 2: BbRr

  6. Parent 1 Possible gametes: BR Br bR br Parent 2 Possible gametes: BR Br bR br Parent 1: BbRr Parent 2: BbRr F O I L

  7. Step 3: arrange all possible gametes for one parent along the top of your Punnett Square, and all possible gametes for the other parent down the side of your Punnett Square…

  8. BR Br bR br BR Br bR br BbRr x BbRr Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth Step 4: Find the possible genotypes of the offspring

  9. BR Br bR br BR BBRR BBRr BbRR BbRr Br BBRr BBrr BbRr Bbrr bR BbRR BbRr bbRR bbRr br BbRr Bbrr bbRr bbrr BbRr x BbRr Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth

  10. BR Br bR br BR BBRR BBRr BbRR BbRr Br BBRr BBrr BbRr Bbrr bR BbRR BbRr bbRR bbRr br BbRr Bbrr bbRr bbrr 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? Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth

  11. BR Br bR br BR BBRR BBRr BbRR BbRr Br BBRr BBrr BbRr Bbrr bR BbRR BbRr bbRR bbRr br BbRr Bbrr bbRr bbrr How many of the offspring would have 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 9:3:3:1 Fur Color: B: Black b: White Coat Texture: R: Rough r: Smooth

  12. On a blank sheet of paper… In pea plants, yellow seeds (Y) are dominant over green seeds (y), and rounded peas (R) are dominant over wrinkled peas (r). Cross a plant that is heterozygous for both traits with a plant that is homozygous recessive for both traits. Draw a Punnett square to show all possible offspring, and determine the phenotypic ratio.

  13. YR Yr yR yr yr YyRr Yyrr yyRr yyrr yr YyRr Yyrr yyRr yyrr yr YyRr Yyrr yyRr yyrr yr YyRr Yyrr yyRr yyrr YyRr X yyrr

  14. RR = red rr = white Rr = pink Incomplete Dominance • A third (new) phenotype appears in the heterozygous condition. • Flower Color in 4 O’clocks

  15. R r r r Rr Rr rr rr Problem: Incomplete Dominance • Show the cross between a pink and a white flower. GENOTYPES: - Rr (2); rr (2); RR (0) - ratio 2:2:0 PHENOTYPES: - pink (2); white (2) Red (0) - ratio 2:2:0

  16. Humans have 23 pairs of chromosomes. 22 of the pairs are called autosomes. One pair of chromosomes is related to the sex of an individual , these chromosomes are called sex chromosomes XX = Female XY = Male Sex Linked Traits

  17. Genes that are located on the X chromosome are called sex-linked genes. Traits determined by sex-linked genes are called sex-linked traits  Ex: Color Blindness (c = colorblind, C = normal) Sex Linked Traits

  18. Color Blindness is recessive. FEMALES: both X chromosomes must have the gene in order for the trait to be expressed. MALES: only have one X chromosome. If that X chromosome has the colorblind allele, then the trait will be expressed. Colorblindness affects 1 in 100 females; 1 in 10 males Sex Linked Traits

  19. X-linked Recessive Inheritance One copy of an altered gene on the X chromosome causes the disease in a male. = Hemizygote Male Female X Y X X An altered copy on one of the X chromosome pair causes carrier status in a female. = Heterozygote

  20. XCXC – Normal Female XCXc– Carrier Female XcXc– Color Blind Female XCY– Normal Male XcY– Colorblind Male Sex Linked Traits

  21. 1. A colorblind male marries a normal female. What are the offspring genotypes and phenotypes? (C = normal, c = colorblind) Xc Y XC XC Sex Linked Punnett Squares

  22. 1. A colorblind male marries a normal female. What are the offspring genotypes and phenotypes? (C = normal, c = colorblind) Xc Y XC XC Xc XC Y XC XC XcXC Y Sex Linked Punnett Squares Genotypic Ratio: __________ Phenotypic Ratio: __________ What percentage of sons with colorblindness? _______ daughters? ______

  23. Multiple Alleles • There are more than two alleles for a trait • Blood type in humans • Blood Types? • Type A, Type B, Type AB, Type O • Blood Alleles? • A, B, O

  24. Rules for Blood Type • A and B are codominant • A and B are dominant over O

  25. AO AO BO BO Problem: Multiple Alleles • Show the cross between a mother who has type O blood and a father who has type AB blood. O O GENOTYPES: • AO (2) BO (2) • ratio 2:2:0:0:0:0 A B PHENOTYPES: - type A (2); type B (2) - ratio 2:2:0:0

  26. BO AB AO OO Problem: Multiple Alleles • Show the cross between a mother who is heterozygous for type B blood and a father who is heterozygous for type A blood. GENOTYPES: A O • AB (1); BO (1); • AO (1); OO (1) • - ratio 1:1:1:1:0:0 B O PHENOTYPES: • type AB (1); type B (1) • type A (1); type O (1) • - ratio 1:1:1:1

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