Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1 1 Unit 3 Genetics

2 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 3 Gregor Mendel Click on the image for the animation

4 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 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 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 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 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 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 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 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 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 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 PPpp Parent 1 Gametes Parent 2 Gametes

14 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 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 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 17 Punnett Squares Arrange the gametes of one parent across the top of the grid

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

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

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

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

22 22 Punnett Squares Genotypically  25% Homozygous Purple  50% Heterozygous Purple  25% Homozygous White Phenotypically  75% Purple  25% White Homozygous Purple Heterozygous Purple Homozygous White

23 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 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 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 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 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 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 29 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 30 Two Factor Crosses Fill in the boxes with the gametes from the top down through the boxes

31 31 Two Factor Crosses Fill in the boxes with the gametes from the side across through the boxes

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

33 33 Two Factor Crosses 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

34 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 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 36 Incomplete Dominance RR x rr F1 all Rr = pink Cross the F1s Rr x Rr F2  25% Red  50% Pink  25% White

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

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

39 39 CoDominance Phenotype A has 2 possible genotypes  I A I A  I A i Phenotype B has 2 possible genotypes  I B I B  I B i Phenotype O has only 1 possible genotype  ii

40 40 CoDominance If mom is I A I A and dad is I B I B all of their children will be type I A I B If mom is I A i and dad is I B i what are the possible blood types of their children?

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

42 42 CoDominance Fill in the boxes as before

43 43 CoDominance Genotypes  AB  Bi  Ai  ii Phenotypes  AB  B  A  O

44 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 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 46 Epistasis Two genes are required to produce a particular phenotype We will study coat colors of Labrador retrievers

47 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 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 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 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_

51 51 Examples The black puppies have at least one B and one E The yellow puppies must be homozygous recessive for deposition (ee) but you don’t know what their pigment alleles are

52 52 Examples Parents: B _ E _ x bbE _ Puppies: B _E _ and _ _ ee  Remember the question is what are the genotypes of the parents …

53 53 Examples Parents: B _ E _ x bbE _ Puppies: B _E _ and _ _ ee To have any yellow puppies, each parent must have a recessive (e) allele to give To not have any brown puppies, all puppies must be receiving a dominant B

54 54 Examples Parents: B B E e x bbEe Puppies: B _E _ and _ _ ee  The red indicates the alleles you figured out!  The black dog parent must be homozygous because there were no brown puppies When we set up the punnett square we see we’re right …

55 55 Examples Parents: B B E e x bbEe  Gametes: BE, Be and bE, be

56 56 Examples Parents: B B E e x bbEe  Gametes: BE, Be and bE, be Black Yellow

57 57 Examples What are the genotypes of the Brown and Yellow parents who only produce Black and Yellow puppies? First determine what you know about the genetics of the parent dogs and their puppies

58 58 Examples You know the yellow lab parent must be homozygous recessive for deposition (ee) 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 (bbE_)

59 59 Examples The black puppies have at least one B and one E The yellow puppies must be homozygous recessive for deposition (ee) but you don’t know what their pigment alleles are

60 60 Examples Parents: _ _ ee x bbE _ Puppies: B _E _ and _ _ ee  Remember the question is what are the genotypes of the parents …

61 61 Examples Parents: _ _ ee x bbE _ Puppies: B _E _ and _ _ ee To have any yellow puppies, each parent must have a recessive (e) allele to give To not have any brown puppies, all puppies must be receiving a dominant B

62 62 Examples Parents: B B ee x bbEe Puppies: B _E _ and _ _ ee  The red indicates the alleles you figured out! When we set up the punnett square we see we’re right …

63 63 Examples Parents: B B e e x bbEe  Gametes: BE, Be and bE, be

64 64 Examples Parents: B B e e x bbEe  Gametes: BE, Be and bE, be Black Yellow Black

65 65 Environmental Factors Influence Gene Expression  Gene products are proteins.  Proteins are affected by temp, PH, salt concentration, etc.  Some gene products appear differently in different environments.

66 66 Environmental Gene Interaction Artic Fox Protein production (Pigment) temperature regulated. Warm  Produce pigment  Brown. Cold  No pigment  White. Summer Coat Winter Coat

67 67 Environmental Gene Interaction Siamese Cat Cooler areas produce pigment. Ears/nose & extremities. If cat gets obese the layer of fat will insulate the skin from the body heat, entire cat will now be pigmented.

68 68 Genetic Issues Recessive does not denote good … Dominant does not denote good  Perfect Vision  Recessive  Huntington’s disease  Dominant Meiosis can make errors in gamete formation

69 69 Meiotic Error Non-Disjunction The failures of chromosomes to separate during Anaphase I or II. Results in one gamete with too many chromosomes & one gamete with too few.

70 70 Non-Disjunction Examples Down’s Syndrome: Trisomy 21 Individual has three copies of chromosome 21. More common in older mothers. Males produce sperm throughout their physically mature lifetime Females are born with all their eggs. Eggs are stalled at the end of meiosis I. Upon physical maturity, 1-2 eggs per month complete meiosis, two are ovulated  Older mothers….older eggs. Any damage accumulates over the life-time and can result in non disjunction in meiosis 2. Nondisjunction In Sex Chromosomes  Can occur in both male and females.  XXX Sterile female  XXY Sterile male  XYY Fertile male  X  Sterile female  Y  Non viable

71 71 Dominant Diseases Huntington’s disease is a dominant progressive neurological disorder resulting in death How many alleles are required for Huntington’s disease? How many alleles are required for a recessive disorder?

72 72 Dominant Diseases Huntington’s disease is a dominant progressive neurological disorder resulting in death How many alleles are required for Huntington’s disease?  It’s dominant so you only need one allele to have the disorder How many alleles are required for a recessive disorder?  Recessive traits are only expressed if the individual is homozygous so you need 2 alleles, one from each parent

73 73 Disease Terminology Autosomal  Trait is on one of the 22 pairs of non-sex chromosomes  These traits are inherited equally among males and females Sex Linked  Trait is on either the X or Y chromosome  These traits are inherited differently based on the person’s gender Carrier  individual has one copy of the recessive allele but is not afflicted by the disease  There are no carriers for dominant traits

74 74 Sex Linked Traits These traits are encoded on the X or Y chromosomes The gender of the individual is linked to the expression of these traits Sex Chromosomes X, Y  XX=Female  XY=Male. Male sperm carry either X or Y  determines gender of offspring. Female eggs only carry an X for sex chromosome. Since female have two X chromosomes, they follow standard dominance patterns for genes carried on X chromosome.  Females cannot inherit genes on the Y chromosome because they don’t have a Y chromosome Males only have one X chromosome. Any genes on their one chromosome are automatically expressed.

75 75 Sex Linked Traits Click on the image for the animation

76 76 Pedigrees Males Females NormalCarrierAfflicted

77 77 Pedigrees Carrier Dad Carrier Mom Afflicted Daughter Normal Son Carrier son Horizontal Line Represents Marriage Vertical Line Down to their children

78 78 Pedigree Example This is an autosomal recessive disorder ? What is the father’s most probable genotype?

79 79 Pedigree Example This is an autosomal recessive disorder What is the father’s most probable genotype? Mom is a carrier, they have 2 normal kids and two Carriers … from this information dad is most probably normal DD Dd DD

80 80 Pedigree Example This is a sex linked recessive disorder What is the father’s most probable genotype? ?

81 81 Pedigree Example This is a sex linked recessive disorder What is the father’s most probable genotype? A daughter is afflicted. Since this is sex linked and recessive We know that dad must be afflicted because the daughter received a recessive X from each of her parents. XcYXcY XcXXcX XcXXcX XcXcXcXc XY XcYXcY

82 82 How do our genes make us who we are? Genes are the construction plans for proteins DNA transcribed into RNA (Single stranded)  Called mRNA—Messenger RNA Ribosomes can only read mRNA

83 83 Transcription Click on the image for the animation

84 84 Transcription RNA Polymerase  Enzyme which reads DNA & makes the mRNA copy.  mRNA copy made by complimentary base pairing to the DNA  Binds to promoter at the beginning of each gene  Read DNA one base at a time Transcription  RNA Polymerase binds to promoter  Reads one base at a time synthesizing single stranded mRNA from the DNA template  mRNA transported to the cytoplasm through the nuclear pores

85 85 Translation Click on the image for the animation

86 86 The Genetic Code DNA 4 bases  Adenine, Thymine, Guanine, Cytosine RNA 4 bases  Adenine, Uracil, Guanine, Cytosine 20 Amino Acids (A.A,).  How to specify each individual A.A.  Use a distinct, unique set of 3 bases.  Called a codon.  Each A.A. is coded for by at least one codon.  Special codons.  Start/F metheonine: AUG  Stop: UAA, AAG, UGA

87 87 The Codon Table

88 88 Find AUG The first letter of the codon is A … So you know you start with A as the First Position

89 89 Find AUG The second letter of the codon is U … So you know you’ve narrowed it down To these 4 with the Second Position

90 90 Find AUG The third letter of the codon is G … So you now you know it is Met (methionine) or start if it’s the first codon of the sequence

91 91 Translation Protein synthesis  Ribosomes read the mRNA.  Assemble A.A. in order by reading one codon at a time. How do the A.A. get to the ribosomes?  tRNA—Transfer RNA  Molecules which bring A.A. to ribosomes.  Has Anti codon at one end to temporarily base pair to mRNA.  Has corresponding A.A. at other end

92 92 Translation Steps Initiation  Small ribosome subunit binds to mRNA at start codon AUG.  Large subunit then binds to complex.

93 93 Translation Steps Elongation  Ribosome moves down mRNA one codon at a time  adding one amino acid at a time.  tRNA comes in and binds by complimentary base pairing.  Peptide bond is formed between the new amino acid and the peptide chain  Continues down mRNA until stop codon reached.  Transfer RNA only carries specific A.A.

94 94 Translation Steps Termination  Stop codon; signal the two ribosome subunits to break apart  Completed protein released

95 95 Genetic Control Regulation Of Gene Expression Differentiation requires that some of our genes be turned on or off in specific cells. Remember, each and every cell has Full set of DNA. Cells express different genes according to their functions.

96 96 Genetic Control Repressors  Prevents transcription.  Bind over Promoter.  Blocks RNA ploymerase from binding. Activators  Increase rate of transcription.  Bind upstream of Promoter.  Assist RNA polymerase in binding to DNA.

97 97 DNA / RNA and Genetic Control Click on the image for the animation


Download ppt "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."

Similar presentations


Ads by Google