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Set up Cornell Notes on pg. 87 Topic: 7.4 Human Genetics and Pedigrees Essential Questions :

7.4 Human Genetics and Pedigrees. 2.1 Atoms, Ions, and Molecules. Set up Cornell Notes on pg. 87 Topic: 7.4 Human Genetics and Pedigrees Essential Questions : 1. NO EQ. KEY CONCEPT A combination of methods is used to study human genetics. Autosomal Dominant Pedigree. 86.

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Set up Cornell Notes on pg. 87 Topic: 7.4 Human Genetics and Pedigrees Essential Questions :

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  1. 7.4 Human Genetics and Pedigrees 2.1 Atoms, Ions, and Molecules Set up Cornell Notes on pg. 87 • Topic: 7.4 Human Genetics and Pedigrees • Essential Questions: 1. NO EQ KEY CONCEPT A combination of methods is used to study human genetics.

  2. Autosomal Dominant Pedigree 86 Autosomal Recessive Pedigree Autosomal Recessive Pedigree #2

  3. KEY CONCEPT A combination of methods is used to study human genetics.

  4. The basic principles of genetics are the same in all sexually reproducing organisms. • Inheritance of many humantraits is complex • Single-gene traits areimportant in understandinghuman genetics. • Ex: widow’s peak Widow’s peak

  5. A pedigree is a chart for tracing genes in a family. • Phenotypes are used to infer genotypes Please copy on Pg. 87 Pedigree Key: Boxes = males Circles = females Shaded = they show the trait White = does not show trait Half shaded = carrier (Carrier= Only for recessive disorders) Offspring is dead=

  6. How many females are in this family? • How many carriers? • How many children were in generation two? • How many offspring in generation three are affected by the trait? Pg. 86 top margin Generation 1 Generation 2 Generation 3 Generation 4

  7. How many females are in this family? 11 • How many carriers? 7 • How many children were in generation two? 5 • How many offspring in generation three are affected by the trait? 2 Generation 1 Generation 2 Generation 3 Generation 4

  8. In pedigrees that show Autosomal Dominant Traits/Disorders- Many family members will be shaded Widow’s peak No Widow’s peak

  9. On the top of pg. 86, please create this pedigree and answer the following questions. Always include GENOTYPE when possible:  Jamie and Joe married in 1912. Joe was homozygous dominant for a Widow’s Peak, while Jamie did not have a Widow’s Peak. They had two children: A son named Kyle and a daughter named Marie. • Kyle married a woman who had no Widow’s Peak. • Marie never married. • Kyle and his wife had three children: Two boys and a girl. • Do Kyle and Marie have Widow’s Peaks? • What percent of Kyle’s children can we expect to have Widow’s Peaks? • What percent of Kyle’s children can we expect to have NO Widow’s Peaks?

  10. Joe Jamie WW ww Marie Kyle

  11. Joe Jamie WW ww • WW x ww • 100% Ww Marie Kyle Ww Ww • Yes, Kyle and Marie have Widow’s Peaks

  12. Joe Jamie WW ww • WW x ww • 100% Ww Marie Kyle Ww ww Ww • Wwx ww • 50% Wwand 50% ww ? ? ? • Yes, Kyle and Marie have Widow’s Peaks • 50% should have Widow’s Peaks • 50% should have NO Widow’s Peaks

  13. In pedigrees that show Autosomal Recessive Traits/Disorders only a few organisms, will be shaded in. • Only will show up if heterozygotes mate, may skip a few generations before reappearing. • REMEMBER: if it is an autosomal recessive disorders CARRIERS will be present and must be half shaded in.

  14. On the middle of pg. 86, Fill in the genotypes of this incomplete pedigree. Shade if necessary. Albinism is an autosomal RECESSIVE disorder aa

  15. You need to complete two Punnet sqs. to find out the answer Aa or AA? aa aa aa aa

  16. AA x aa • 100% Aa? • Aa x aa • 50% Aa--50% aa Aa or AA? aa aa aa aa

  17. AA x aa • 100% Aa? • Aa x aa • 50% Aa--50% aa Aa or AA? aa aa aa aa

  18. Aa aa Aa Aa aa Aa aa aa

  19. Can use previous Punnett Sq to answer this Aa aa Aa or AA? Aa Aa aa Aa aa aa

  20. Aa aa Aa Aa Aa aa Aa Aa aa aa

  21. AA x aa • 100% Aa? • Aa x aa • 50% Aa--50% aa Aa aa Aa or AA? Aa Aa Aa aa Aa Aa aa aa Anyway to know for sure?

  22. Aa aa Aa or AA ? Aa Aa Aa aa Aa ? ? Aa aa aa

  23. SHADE AS NECESSARY Aa aa Aa or AA ? Aa Aa Aa aa Aa ? ? Aa or AA Aa aa aa

  24. SHADE AS NECESSARY Aa aa Aa or AA ? Aa Aa Aa aa Aa ? ? Aa or AA Aa aa aa

  25. Bottom of Pg. 86 Mary and Joe were married in 1950. Both Mary and Joe were carriers for a fatalrecessive disorder called Cystic Fibrosis. They had three children: A son named Pete who was a carrier, a son named Charles who was not a carrier and did not have the disease, and a girl named Isabel who died from Cystic Fibrosis. Pete married a woman who was homozygous dominant. Pete and his wife are worried about having a child with Cystic Fibrosis. Should they worry? What are the chances of any of their children having Cystic Fibrosis? Carriers?

  26. Joe Mary Ff Ff Isabel Charles Pete Ff FF FF ff ?

  27. Joe Mary Ff Ff Isabel Charles Pete Ff FF FF ff • FF x Ff • O%- No chance of any of their children having the disorder • 50% will be carriers They DO NOT need to be worried about having a child with CF

  28. 7.4 Human Genetics and Pedigrees: Sex-linked traits 2.1 Atoms, Ions, and Molecules Set up Cornell Notes on pg. 89 • Topic: 7.4 Human Genetics and Pedigrees • Essential Questions: 1. NO EQ

  29. Y X Females can carry sex-linked genetic disorders. • Males (XY) express all of their sex linked genes. • Expression of the disorder depends on which parent carries the allele and the sex of the child.

  30. In pedigrees showing Sex-linked traits: More males will be shaded in because they do not have another X to mask the disorder Ex: Color blindness males females

  31. XMXM= Normal XMXm= carrier XmXm= CB XMY= Normal XmY= CB X-linked Color Blindness- Recessive XmY XMXm XMXm XMXm XMY XMXm XmY On pg. 88, Fill in the genotypes of this incomplete pedigree. Shade if necessary. ? XmY XMXM or XMXm XmY

  32. X Y • A karyotype is a picture of all chromosomes in a cell.

  33. Karyotypes can show changes in chromosomes. • deletion of part of a chromosome or loss of a chromosome • large changes in chromosomes • extra chromosomes or duplication of part of a chromosome

  34. In down syndrome a person has an extra copy of chromosome 21. • In Klinefelter’s syndrome a male has an extra X (XXY).

  35. Pedigree Practice Worksheet

  36. How many males? • How many males have hemophilia? • How many females? • How many females have hemophilia? 5 4 6 8 7 9 10 11 12 14 13 15 16

  37. How many males? 8 • How many males have hemophilia? 3 • How many females? 8 • How many females have hemophilia? 2 5 4 6 8 7 9 10 11 12 14 13 15 16

  38. How many marriages are there? 5 4 6 8 7 9 10 11 12 14 13 15 16

  39. How many marriages are there? 3 5 4 6 8 7 9 10 11 12 14 13 15 16

  40. How many children did the first couple have? • How many children did the third couple have? • How many generations are there? • How many members in the 4th generation? 5 4 6 8 7 9 10 11 12 14 13 15 16

  41. How many children did the first couple have? 2 • How many children did the third couple have? 7 • How many generations are there? 4 • How many members in the 4th generation? 7 5 4 6 8 7 9 10 11 12 14 13 15 16

  42. Now the harder part… Determine the GENOTYPES for as many of the family members as possible. You can find all genotypes except for one. 5 4 6 8 7 9 10 11 12 14 13 15 16

  43. Fill in the OBVIOUS genotypes first (the affected individuals) 5 4 6 8 7 9 10 11 12 14 13 15 16

  44. Fill in the OBVIOUS genotypes first • the affected individuals • The non-affected males XnXn 5 4 6 8 7 9 XnY 10 11 12 14 13 15 16 XnXn XnY XnY

  45. Then start at the top, and using Punnett Sqs. Determine the possible genotypes of the female offspring in each generation. You may need to complete more than one!!!!! XnXn 5 4 XNY 6 8 7 9 XNY XnY 10 11 12 14 13 15 16 XnXn XnY XNY XNY XnY

  46. / XnXn XNY XnXn XNY • 100% XNXn 5 4 XNXn XNY XNXn 6 8 7 9 XNY XnY 10 11 12 14 13 15 16 XnXn XnY XNY XNY XnY

  47. / XNXn XNY XnXn XNY • 50% • 50% XNXn XNXN 5 4 XNXn XNY XNXn 6 8 7 9 XNY XnY 10 11 12 14 13 15 16 XnXn XnY XNY XNY XnY

  48. XnXn XNY 5 4 XNXn XNY XNXn XNXn or XNXN 6 8 7 9 ? XNY XnY 10 11 12 14 13 15 16 XnXn XnY XNY XNY XnY

  49. XnXn XNY 5 4 XNXn XNY XNXn You need to complete 2 test crosses to figure out what mom is. 6 8 7 9 ? XNXn or XNY XnY XNXN 10 11 12 14 13 15 16 XnXn XnY XNY XNY XnY

  50. / XnY XNXn XnXn XNY and / 5 XnY XNXN 4 XNXn XNY XNXn 6 8 7 9 ? XNXn or XNY XnY XNXN 10 11 12 14 13 15 16 XnXn XnY XNY XNY XnY

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