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Unit 3

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  1. Unit 3 Genetics

  2. Gregor Mendel • Austrian Monk 1800’s • Observed some traits disappeared in one generation, only to reappear in the next • Hypothesis: Some traits are stronger than others. • Experimental Design • Needed something which could be easily manipulated. • Something with a variety of visible characteristics.

  3. Gregor Mendel Click on the image for the animation http://www.youtube.com/watch?v=MWkFxWXHTnw&playnext_from=TL&videos=c5-UkltqOnI

  4. Mendel Chose the Garden Pea • Quick generation time • Does not require much space • Peas undergo self-fertilization • Pollinate themselves • Could also pry open petal to make specific crosses between plants. • Many visible traits: • Flower color • Seed color • Seed shape • Plant height

  5. Mendel’s Experimental Design • Parental Generation • For each trait studied he wanted a true breeding plant to begin the experiment. • True BreedingGeneration after generation breeds true to a single visible trait. • Experiment • Cross two different parental generation plants (each One displays One version of the trait) • Parental white flower x purple flower cross them and plant the seeds to get the first generation • Recorded The Numbers & Traits Of All Generations • F1—First Filial (Latin for son/daughter) all purple 100%. • Allowed F1 to self-fertilize. • Counted the F2 for number of each trait. • 75% purple, 25% white.

  6. Mendel’s Experiment • In this cross Mendel used Parental Tall (TT) and short (tt) • All the F1 generation plants were Tall • These F1 plants self fertilized to produce the F2 plants • The F2 generation had 75% Tall and 25% short

  7. Terminology • Allele – different versions of a trait • Example: Pea plants have white or purple flowers. • White and Purple are different alleles for the trait of flower color • Genotype – the sum of all alleles in an individual • Phenotype – the physical representation of the genotype (what the individual looks like)

  8. Terminology • Homozygous – the two alleles for a given trait are the same in an individual • In our flowers PP homozygous purple or pp homozygous for white • Heterozygous – the two alleles for a given train are different in an individual • In our flowers Pp heterozygous for flower color purple

  9. Terminology • Dominant – the allele expressed as the phenotype in a heterozygote individual • Denoted by using the upper case of the letter for the trait • The Pp plant will be Purple • Recessive – the allele not expressed in a heterozygote • Denoted by using the lower case of the letter for the trait • The lower case p denotes the recessive white allele

  10. Terminology Examples • Pea plants have either white or purple flowers • P – purple p – white • Purple is dominant to white • A purple plant can be homozygous • PP for flower color • OR heterozygous • Pp for flower color

  11. Terminology Examples • A white flower plant must be homozygous (pp) for flower color • The recessive trait will only be seen as the phenotype when the individual has two copies of that recessive allele

  12. Meiosis • Specialized cell division for sexual reproduction • Results in the production of haploid gametes • Each gamete will have a single copy of each gene

  13. Determining Possible Gametes • Start with the genotype of each parent • Parent 1 PP; Parent 2 pp • Each of parent 1’s gametes will contain 1 P • Each of parent 2’s gametes will contain 1 p Parent 1 Gametes Parent 2 Gametes P P p p

  14. Determining Possible Gametes • The parental generation cross between these two plants PP x pp will result in all progeny being Pp • They will be phenotypically purple • They are all heterozygous for flower color Pp

  15. Determining Possible Gametes • These first generation (F1) plants will then self fertilize to create the second (F2) generation • Start with the genotype of each parent • Parent 1 Pp; Parent 2 Pp • Half of parent 1’s gametes will contain P the other half will contain p • Half of parent 2’s gametes will contain P the other half will contain p P p P p

  16. Punnett Squares • Graphical representation of each parents gametes • Allows for prediction of possible progeny for a given set of parents • Determine the gametes of the parents • Arrange the gametes on a grid to predict the genotypes and phenotypes of the next generation

  17. Arrange the gametes of one parent across the top of the grid Punnett Squares P p

  18. Arrange the gametes of the other parent down the side of the grid Punnett Squares P p P p

  19. Fill in the boxes of the parental gametes from the top down through the boxes Punnett Squares P p P P p P p p

  20. Fill in the boxes of the parental gametes from the side across through the boxes Punnett Squares P p P P P P p P p p p p

  21. The interior of the boxes now has the possible genotypes of the progeny from this parental cross of Pp x Pp Punnett Squares P p P P P P p P p p p p

  22. Genotypically 25% Homozygous Purple 50% Heterozygous Purple 25% Homozygous White Phenotypically 75% Purple 25% White Punnett Squares P p Homozygous Heterozygous P P P P p Purple Purple Heterozygous Homozygous P p p p p White Purple

  23. Test Cross • Used to determine an unknown individual’s genotype. • Cross unknown with a known homozygous. • Which homozygous? A dominant or recessive? … see next slide for answer

  24. Test Cross • *Homozygous Recessive! • Allows you to deduce the genotype by offspring produced. • If any offspring are recessive then the unknown must have been Heterozygous. • If all offspring are dominant the unknown purple plant is homozygous dominant

  25. Test Cross • You found a purple pea plant … • Is it homozygous or heterozygous for flower color? • Test cross is with a white pea plant If it is homozygous purple If it is heterozygous purple

  26. Two Factor Crosses • A two factor cross looks at two different traits at the same time for example seed shape (round or wrinkled) AND plant height (tall or short) • Set up is the same for the punnett square • Keep the alleles for each trait together in the boxes!!

  27. Two Factor Crosses • Peas can be either round (R) or wrinkled (r) • Plants can be tall (T) or short (t) • A plant can be homozygous for either or both of these traits (RRTT, RRtt, rrTT,rrtt) • round tall, round short, wrinkled tall, wrinkled short • A plant can be heterozygous for either or both of these traits (RrTt, RRTt, RrTT,) • round tall for all of them

  28. Two Factor Crosses • We’ll cross two double heterozygous plants (RrTt x RrTt) • Each gamete will have one allele of each trait • The gametes produced by each of these individuals are as follows • RT • Rt • rT • rt

  29. R R T T t t r r R t R T T r t r Two Factor Crosses • Arrange the gametes of one parent across the top of the grid • Arrange the gametes of the other parent down the side of the grid

  30. R R R R R R R R R R T T T T T T T T T T t t t t t t t t t t r r r r r r r r r r Two Factor Crosses • Fill in the boxes with the gametes from the top down through the boxes R t R T T r t r

  31. R R R R R R R R R R T T T T T T T T T T t t t t t t t t t t r r r r r r r r r r R R R R R t t t t t R R R R R T T T T T T T T T T r r r r r t t t t t r r r r r Two Factor Crosses • Fill in the boxes with the gametes from the side across through the boxes

  32. Two Factor Crosses • Determine all of the possible phenotypes and genotypes for the progeny

  33. Phenotypes Round and Tall Round and Short Wrinkled and Tall Wrinkled and Short Genotypes Round / Tall RRTt RrTt RRTT Round / Short RRtt Rrtt Wrinkled / Tall rrTT rrTt Wrinkled / Short rrtt Two Factor Crosses

  34. Standard Dominance • In the heterozygote, the dominant allele is expressed as the phenotype • Purple flower color is dominant to white • A Pp plant will be purple

  35. Incomplete Dominance • The heterozygote individual has a phenotype in between the two phenotypes • Flower color of roses • R red • r white • RR = red • Rr = pink • Rr = white

  36. Incomplete Dominance • RR x rr • F1 all Rr = pink • Cross the F1s Rr x Rr • F2 • 25% Red • 50% Pink • 25% White

  37. CoDominance • Neither allele is dominant • Heterozygote expresses both alleles equally • In flowers a red and white flower • ABO blood groups

  38. CoDominance • There are 3 alleles for blood type • IA – A • IB – B • i – O • Each person has 2 alleles • The combination of the 2 alleles determines the individual’s blood type

  39. CoDominance • Phenotype A has 2 possible genotypes • IA IA • IA i • Phenotype B has 2 possible genotypes • IB IB • IB i • Phenotype O has only 1 possible genotype • ii

  40. CoDominance • If mom is IA IA and dad is IB IB all of their children will be type IA IB • If mom is IA i and dad is IBi what are the possible blood types of their children?

  41. CoDominance Set up a punnett square as normal with mom’s alleles across the top and dad’s alleles down the side A I i B I i

  42. A A A I I I B B B I I I CoDominance Fill in the boxes as before i i i i i i

  43. Genotypes AB Bi Ai ii Phenotypes AB B A O i A A A I I I i B B B I I I i i i i CoDominance

  44. CoDominance • Questions to ponder • Female blood type A, Male blood type AB • What blood type(s) is/are not possible for their children? • Male blood type O • What blood type(s) is/are not possible for their children?

  45. CoDominance • Questions to ponder • Female blood type A, Male blood type AB • What blood type(s) is/are not possible for their children? … type O is not possible • Because Dad is AB he does not have a copy of the recessive i to pass down to any of his children • Male blood type O • What blood type(s) is/are not possible for their children? … type AB is not possible • To have blood type O Dad must be homozygous ii so all of his children will get one copy of i from him thus AB blood type is not possible

  46. Epistasis • Two genes are required to produce a particular phenotype • We will study coat colors of Labrador retrievers

  47. Epistasis • The interaction of 2 genes determine coat color • Pigment • B = Black, b = Brown • Deposition • E = yes, e = no • whether the pigment can be integrated into the fur

  48. Epistasis • Black Labs • Must have at least one B and one E • BBEE, BBEe, BbEE, BbEe • Brown or chocolate Labs • Must have 2 bb and at least one E • bbEE, bbEe • Yellow Labs • Are homozygous recessive for deposition ee • They cannot put the pigment into their fur • BBee, Bbee, bbee

  49. Examples • What are the genotypes of the Brown and Black parents who only produce Black and Yellow puppies? • First determine what you know about the genetics of the parent dogs and their puppies

  50. Examples • You know the black lab parent has at least one B allele and one E allele B_E_ • You know the brown parent must be homozygous for brown pigment because it is recessive bb and that the pigment is deposited so at least one E so it is bbE_