1 / 97

Unit 3

Unit 3. Genetics. 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.

Mercy
Download Presentation

Unit 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 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_

More Related