2 Gregor Mendel Austrian Monk 1800’s Observed some traits disappeared in one generation, only to reappear in the nextHypothesis: Some traits are stronger than others.Experimental DesignNeeded 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 timeDoes not require much spacePeas undergo self-fertilizationPollinate themselvesCould also pry open petal to make specific crosses between plants.Many visible traits:Flower colorSeed colorSeed shapePlant height
5 Mendel’s Experimental Design Parental GenerationFor each trait studied he wanted a true breeding plant to begin the experiment.True BreedingGeneration after generation breeds true to a single visible trait.ExperimentCross 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 generationRecorded The Numbers & Traits Of All GenerationsF1—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 ExperimentIn this cross Mendel used Parental Tall (TT) and short (tt)All the F1 generation plants were TallThese F1 plants self fertilized to produce the F2 plantsThe 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 colorGenotype – the sum of all alleles in an individualPhenotype – the physical representation of the genotype (what the individual looks like)
8 TerminologyHomozygous – the two alleles for a given trait are the same in an individualIn our flowers PP homozygous purple or pp homozygous for whiteHeterozygous – the two alleles for a given train are different in an individualIn our flowers Pp heterozygous for flower color purple
9 TerminologyDominant – the allele expressed as the phenotype in a heterozygote individualDenoted by using the upper case of the letter for the traitThe Pp plant will be PurpleRecessive – the allele not expressed in a heterozygoteDenoted by using the lower case of the letter for the traitThe lower case p denotes the recessive white allele
10 Terminology Examples Pea plants have either white or purple flowers P – purple p – whitePurple is dominant to whiteA purple plant can be homozygousPP for flower colorOR heterozygousPp for flower color
11 Terminology ExamplesA white flower plant must be homozygous (pp) for flower colorThe 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 gametesEach gamete will have a single copy of each gene
13 Determining Possible Gametes Start with the genotype of each parentParent 1 PP; Parent 2 ppEach of parent 1’s gametes will contain 1 PEach of parent 2’s gametes will contain 1 pParent 1 GametesParent 2 GametesPPpp
14 Determining Possible Gametes The parental generation cross between these two plants PP x pp will result in all progeny being PpThey will be phenotypically purpleThey are all heterozygous for flower colorPp
15 Determining Possible Gametes These first generation (F1) plants will then self fertilize to create the second (F2) generationStart with the genotype of each parentParent 1 Pp; Parent 2 PpHalf of parent 1’s gametes will contain P the other half will contain pHalf of parent 2’s gametes will contain P the other half will contain pPpPp
16 Punnett Squares Graphical representation of each parents gametes Allows for prediction of possible progeny for a given set of parentsDetermine the gametes of the parentsArrange the gametes on a grid to predict the genotypes and phenotypes of the next generation
17 Punnett SquaresArrange the gametes of one parent across the top of the gridPp
18 Punnett SquaresArrange the gametes of the other parent down the side of the gridPpPp
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 boxesPpPPppPp
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 boxesPpPPPPpPpppp
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 PpPpPPPPpPpppp
22 Punnett Squares P p P P P P p P p p p p Genotypically Phenotypically 25% Homozygous Purple50% Heterozygous Purple25% Homozygous WhitePhenotypically75% Purple25% WhiteHomozygousHeterozygousPPPPpPurplePurpleHeterozygousHomozygousPppppWhitePurple
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 plantIf it is homozygouspurpleIf it is heterozygouspurple
26 Two Factor CrossesA 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 squareKeep 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 shortA plant can be heterozygous for either or both of these traits (RrTt, RRTt, RrTT,)round tall for all of them
28 Two Factor CrossesWe’ll cross two double heterozygous plants (RrTt x RrTt)Each gamete will have one allele of each traitThe gametes produced by each of these individuals are as followsRTRtrTrt
29 Two Factor Crosses R t r T Arrange the gametes of one parent across the top of the gridArrange the gametes of the other parent down the side of the gridRrTt
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 boxesRrtTRtRrtTRTRrtTTRrtTrtrRrtT
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 boxesRrtTRrTtRrRrtTRrTtRrRrTtTtTtRrtTRrtTRrtT
32 Two Factor CrossesDetermine all of the possible phenotypes and genotypes for the progeny
33 Two Factor Crosses Phenotypes Genotypes Round and Tall Round and Short Wrinkled and TallWrinkled and ShortGenotypesRound / TallRRTtRrTtRRTTRound / ShortRRttRrttWrinkled / TallrrTTrrTtWrinkled / Shortrrtt
34 Standard DominanceIn the heterozygote, the dominant allele is expressed as the phenotypePurple flower color is dominant to whiteA Pp plant will be purple
35 Incomplete DominanceThe heterozygote individual has a phenotype in between the two phenotypesFlower color of rosesR redr whiteRR = redRr = pinkRr = white
36 Incomplete Dominance RR x rr F1 all Rr = pink Cross the F1s Rr x Rr F2 25% Red50% Pink25% White
37 CoDominance Neither allele is dominant Heterozygote expresses both alleles equallyIn flowers a red and white flowerABO blood groups
38 CoDominance There are 3 alleles for blood type IA – AIB – Bi – OEach person has 2 allelesThe combination of the 2 alleles determines the individual’s blood type
39 CoDominance Phenotype A has 2 possible genotypes IA IAIA iPhenotype B has 2 possible genotypesIB IBIB iPhenotype O has only 1 possible genotypeii
40 CoDominanceIf mom is IA IA and dad is IB IB all of their children will be type IA IBIf mom is IA i and dad is IB i what are the possible blood types of their children?
41 CoDominanceSet up a punnett square as normal with mom’s alleles across the top and dad’s alleles down the sideAIiBIi
42 CoDominance I i I I I I i I i i i i Fill in the boxes as before A A B
44 CoDominance Questions to ponder Female blood type A, Male blood type ABWhat 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 ABWhat blood type(s) is/are not possible for their children? … type O is not possibleBecause Dad is AB he does not have a copy of the recessive i to pass down to any of his childrenMale blood type OWhat blood type(s) is/are not possible for their children? … type AB is not possibleTo 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 = BrownDepositionE = yes, e = nowhether 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 EBBEE, BBEe, BbEE, BbEeBrown or chocolate LabsMust have 2 bb and at least one EbbEE, bbEeYellow LabsAre homozygous recessive for deposition eeThey cannot put the pigment into their furBBee, Bbee, bbee
49 ExamplesWhat 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 ExamplesYou 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 giveTo 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 puppiesWhen 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 bbEeGametes: BE, Be and bE, beBEBebEBbEEBbEebBbEeBbeee
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 bbEeGametes: BE, Be and bE, beBEBebEBbEEBbEeBlackBlackbBbEeBbeeeBlackYellow
57 ExamplesWhat 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 ExamplesYou 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 giveTo 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 bbEeGametes: BE, Be and bE, beBBeebEBbEeBbEebBbeeBbeee
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 bbEeGametes: BE, Be and bE, beBBeebEBbEeBbEeBlackBlackbBbeeBbeeeYellowYellow
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 WinterCoatArtic FoxProtein production (Pigment) temperature regulated.WarmProduce pigmentBrown.ColdNo pigmentWhite.Summer Coat
67 Environmental Gene Interaction Siamese CatCooler 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 IssuesRecessive does not denote good … Dominant does not denote goodPerfect VisionRecessiveHuntington’s diseaseDominantMeiosis 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 21Individual has three copies of chromosome 21.More common in older mothers.Males produce sperm throughout their physically mature lifetimeFemales 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 ovulatedOlder mothers….older eggs. Any damage accumulates over the life-time and can result in non disjunction in meiosis 2.Nondisjunction In Sex ChromosomesCan occur in both male and females.XXX Sterile femaleXXY Sterile maleXYY Fertile maleX Sterile femaleY Non viable
71 Dominant DiseasesHuntington’s disease is a dominant progressive neurological disorder resulting in deathHow many alleles are required for Huntington’s disease?How many alleles are required for a recessive disorder?
72 Dominant DiseasesHuntington’s disease is a dominant progressive neurological disorder resulting in deathHow many alleles are required for Huntington’s disease?It’s dominant so you only need one allele to have the disorderHow 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 chromosomesThese traits are inherited equally among males and femalesSex LinkedTrait is on either the X or Y chromosomeThese traits are inherited differently based on the person’s genderCarrierindividual has one copy of the recessive allele but is not afflicted by the diseaseThere 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 traitsSex Chromosomes X, YXX=FemaleXY=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 chromosomeMales 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
77 Pedigrees Horizontal Line Represents Marriage Carrier Dad Carrier Mom Vertical Line Downto their childrenAfflictedDaughterNormalSonCarrierson
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 twoCarriers … from this information dad is most probablynormal
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 disorderXcYXcXXcXcXcYXYXcXWhat is the father’s most probable genotype?A daughter is afflicted. Since this is sex linked and recessiveWe know that dad must be afflicted because the daughter receiveda recessive X from each of her parents.
82 How do our genes make us who we are? Genes are the construction plans for proteinsDNA transcribed into RNA (Single stranded)Called mRNA—Messenger RNARibosomes 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 DNABinds to promoter at the beginning of each geneRead DNA one base at a timeTranscriptionRNA Polymerase binds to promoterReads one base at a time synthesizing single stranded mRNA from the DNA templatemRNA 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, CytosineRNA 4 basesAdenine, Uracil, Guanine, Cytosine20 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: AUGStop: UAA, AAG, UGA
88 Find AUGThe first letter of the codon is A … So you know you start with A as theFirst Position
89 Find AUGThe second letter of the codon is U … So you know you’ve narrowed it downTo these 4 with the Second Position
90 Find AUGThe third letter of the codon is G … So you now you know it isMet (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 RNAMolecules 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 timeadding 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 chainContinues 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 apartCompleted 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.ActivatorsIncrease 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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