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Unit 3 Genetics.

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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

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

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

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

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

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

22 Punnett Squares P p P P P P p P p p p p Genotypically Phenotypically
25% Homozygous Purple 50% Heterozygous Purple 25% Homozygous White Phenotypically 75% Purple 25% White 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 Two Factor Crosses R t r T
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 R r T t

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

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

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

33 Two Factor Crosses Phenotypes Genotypes 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 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 IB i 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 CoDominance I i I I I I i I i i i i Fill in the boxes as before A A B

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

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

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 Brown or chocolate Labs Yellow 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_

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

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

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

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

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

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

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

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 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 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 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 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 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 Disease Terminology Autosomal Sex Linked Carrier
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 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 Sex Linked Traits Click on the image for the animation

76 Pedigrees Males Normal Carrier Afflicted Females

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

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

79 Pedigree Example This is an autosomal recessive disorder DD Dd DD Dd
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

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

81 Pedigree Example XcY XcXc XcY XY XcX
This is a sex linked recessive disorder XcY XcX XcXc XcY XY XcX 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.

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 Transcription Click on the image for the animation

84 Transcription RNA Polymerase Transcription
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 Translation Click on the image for the animation

86 The Genetic Code DNA 4 bases RNA 4 bases 20 Amino Acids (A.A,).
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 The Codon Table

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

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 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 Translation Protein synthesis How do the A.A. get to the ribosomes?
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 Translation Steps Initiation
Small ribosome subunit binds to mRNA at start codon AUG. Large subunit then binds to complex.

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

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


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