Ppt genetics problems


Published on

Published in: Education, Lifestyle, Technology
  • Be the first to comment

  • Be the first to like this

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Ppt genetics problems

  2. 2. Question #1• How many different kinds of gametes could the following individuals produce? 1. aaBb 2. CCDdee 3. AABbCcDD 4. MmNnOoPpQq 5. UUVVWWXXYYZz
  3. 3. Question #1• Remember the formula 2n• Where n = # of heterozygous 1. aaBb = 2 2. CCDdee = 2 3. AABbCcDD = 4 4. MmNnOoPpQq = 32 5. UUVVWWXXYYZz = 2
  4. 4. Question #2• In dogs, wire-haired is due to a dominant gene (W), smooth-haired is due to its recessive allele (w).• WW, Ww = wire haired• ww = smooth haired
  5. 5. Question #2A• If a homozygous wire-haired dog is mated with a smooth-haired dog, what type of offspring could be produced. W W w w
  6. 6. Question #2A W W w Ww Wwfg F1 generation w Ww Ww all heterozygous
  7. 7. Question #2B• What type(s) of offspring could be produced in the F2 generation?• Must breed the F1 generation to get the F2.• Results of F1 Cross: Ww x Ww
  8. 8. Question #2B W wW WW Ww F2 generationw Ww ww genotype: 1:2:1 ratio phenotype: 3:1 ratio
  9. 9. Question #2C• Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth- haired pup.• If these, two wire-haired dogs mate again, what are the chances that they will produce another smooth-haired pup?• What are the chances that the pup will wire- haired pup?
  10. 10. Question #2C W wW WW Ww F2 generationw Ww ww- 1/4 or 25% chance for smooth-haired- 3/4 or 75% chance for wire-haired
  11. 11. Question #2D• A wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smooth-haired.• What are the phenotypes and genotypes of the pups they could produce?• Show the results of crossing: Ww x ww
  12. 12. Question #2D W ww Ww www Ww ww phenotypes: 1:1 ratio genotypes: 1:1 ratio
  13. 13. Question #3• In snapdragons, red flower (R) color is incompletely dominant over white flower (r) color.• The heterozygous (Rr) plants have pink flowers. RR - red flowers Rr - pink flowers rr - white flowers
  14. 14. Question #3A• If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the plants F1 generation?• RR x rr
  15. 15. Question #3A R Rr Rr Rr F1 generationr Rr Rr phenotypes: 100% pink genotypes: 100% heterozygous
  16. 16. Question #3B• What genotypes and phenotypes will be produced in the F2 generation?• Rr x Rr
  17. 17. Question #3B R rR RR Rr F2 generationr Rr rr phenotypes: 1:2:1 ratio genotypes: 1:2:1 ratio
  18. 18. Question #3C• What kinds of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant?• RR x Rr
  19. 19. Question #3C R RR RR RRr Rr Rr 50%: red flowered 50%: pink flowered
  20. 20. Question #3D• What kind of offspring is/are produced if a pink-flowered plant is crossed with a white-flowered plant?• Rr x rr
  21. 21. Question #3D R rr Rr rrr Rr rr 50%: white flowered 50%: pink flowered
  22. 22. Question #4• In humans, colorblindness (cc) is a recessive sex-linked trait.• Remember: XX - female XY - male
  23. 23. Question #4A• Two normal people have a colorblind son.• What are the genotypes of the parents?• XCX_? x XCY• What are the genotypes and phenotypes possible among their other children?
  24. 24. Question #4A XC Y ← parentsXC XCXC XCYXc XCXc XcY 50%: female (one normal, one a carrier) 50%: male (one normal, one colorblind)
  25. 25. Question #4B• A couple has a colorblind daughter.• What are the possible genotypes and phenotypes of the parents and the daughter?
  26. 26. Question #4B Xc YXC XCXc XCYXc XcXc XcY parents: XcY and XCXc or XcXc father colorblind mother carrier or colorblind daughter: XcXc - colorblind
  27. 27. Question #5• In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f).• Dimpled cheeks (D) are dominant to non-dimpled cheeks (d).
  28. 28. Question #5A• Two persons with freckles and dimpled cheeks have two children: one has freckles but no dimples and one has dimples but no freckles.• What are the genotypes of the parents? Parents: F__D__ x F__D__ Children: F__dd x ffD__
  29. 29. Question #5B• What are the possible phenotypes and genotypes of the children that they could produce?• Cross: FfDd x FfDd• This is a dihybrid cross
  30. 30. Question #5B• Possible gametes for both: FD Fd fD fd FD Fd fD fdFD FFDD FFDd FfDD FfDdFd FFDd FFdd FfDd FfddfD FfDD FfDd ffDD ffDdfd FfDd Ffdd ffDd ffdd
  31. 31. Question #5BPhenotype : Freckles/Dimples: 9 Freckles/no dimples: 3 no freckles/Dimples: 3 no freckles/no dimples: 1 Phenotypic ratio will always been 9:3:3:1 for all F1 dihybrid crosses.
  32. 32. Question #5BGenotypic ratio: FFDD - 1 FFDd - 2 FFdd - 1 FfDD - 2 FfDd - 4 Ffdd - 2 ffDD - 1 ffDd - 2 ffdd - 1
  33. 33. Question #5C• What are the chances that they would have a child whom lacks both freckles and dimples?• This child will have a genotype of ffdd• Answer: 1/16
  34. 34. Question #5D• A person with freckles and dimples whose mother lacked both freckles and dimples marries a person with freckles but not dimples whose father did not have freckles or dimples.• Cross: FfDd x Ffdd• Possible gametes: FD Fd fD fd x Fd fd
  35. 35. Question #5D• What are the chances that they would have a child whom lacks both freckles and dimples? FD Fd fD fdFd FFDd FFdd FfDd Ffddfd FfDd Ffdd ffDd ffddAnswer: 1/8
  36. 36. Question #6• Sixteen percent of the human population is known to be able to wiggle their ears.• This trait is determined to be a recessive gene.• These is a population genetics question.• Use the following equation: 1 = p2 + 2pq + q2
  37. 37. Question #6A• What of the population is homozygous dominant for this trait?• q2 = 16% or .16: √ q2 = √ .16 q = .4• then use : 1 = p + q 1 = p + .4 1- .4 = p p = .6• Now use p2 for answer: .62 = .36 or 36%
  38. 38. Question #6B• What of the population is heterozygous for this trait?• We know that q = .4 and p = .6• Now use 2pq for answer: 2(.6)(.4) = .48 or 48%
  39. 39. Question #7• In dogs, the inheritance of hair color involves a gene B for black hair and gene b for brown hair b.• A dominant C is also involved. It must be present for the color to be synthesized.• If this gene is not present, a blond condition results.BB, Bb - black hair CC, Cc - colorbb - brown hair cc - blond
  40. 40. Question #7A• A brown haired male, whose father was a blond, is mated with a black haired female, whose mother was brown haired and her father was blond. Male: bbCc (gametes: bC bc) Female: BbCc (gametes: BC Bc bC bc)• What is the expected ratios of their offspring?
  41. 41. Question #7A BC Bc bC bcbC BbCC BbCc bbCC bbCcbc BbCc Bbcc bbCc bbccOffspring ratios: Black: 3/8 Brown: 3/8 Blond: 2/8 or 1/4
  42. 42. Question #8• Henry Anonymous, a film star, was involved in a paternity case. The woman bringing suit had two children, on whose blood type was A and the other whose blood type was B.• Her blood type was O, the same as Henry’s!• The judge in the case awarded damages to the woman damages to the woman, saying that Henry had to be the father of at least one of the children.
  43. 43. Question #8A• Obviously, the judge should be sentenced to Biology. For Henry to have been the father of both children, his blood type would have had to be what? IA IB ← Answer i IAi IBi i IAi IBi