2Responsible for the Laws governing Inheritance of Traits Mendelian Genetics4/15/2017Gregor Mendel ( )Responsible for the Laws governing Inheritance of Traits
3Gregor Johann Mendel Austrian monk Mendelian Genetics4/15/2017Gregor Johann MendelAustrian monkStudied the inheritance of traits in pea plantsDeveloped the laws of inheritanceMendel's work was not recognized until the turn of the 20th century
4Mendelian Genetics4/15/2017Gregor Johann MendelBetween 1856 and 1863, Mendel cultivated and tested some 28,000 pea plantsHe found that the plants' offspring retained traits of the parentsCalled the “Father of Genetics"
61. Genes In Pairs: Genetic characters are controlled by genes that exist in pairs of alleles in individual organisms and are passed from parents to their offspring. When two organisms produce offspring, each parent gives the offspring one of the alleles from each pair.
72. Dominance and Recessiveness: When two unlike alleles responsible for a single character are present in a single individual, one allele can mask the expression of another allele. That is, one allele is dominant to the other. The latter is said to be recessive.
83. The Law of Segregation: During the formation of gametes, the paired alleles separate (segregate) randomly so that each gamete receives one allele or the other.
94. The Law of Independent Assortment: During gamete formation, segregating pairs of alleles assort independently of each other. Example: genes on different chromosomes will segregate independently. Linked genes (close together on one chromosome) do not follow this law.
10Rule of DominanceThe trait that is observed in the offspring is the dominant trait (uppercase)The trait that disappears in the offspring is the recessive trait (lowercase)
11Law of SegregationThe two alleles for a trait must separate when gametes are formedA parent randomly passes only one allele for each trait to each offspring
12Law of Independent Assortment The genes for different traits are inherited independently of each other.
13Site of Gregor Mendel’s experimental garden in the Czech Republic Mendelian Genetics4/15/2017Site of Gregor Mendel’s experimental garden in the Czech RepublicFig. 5.co
14Particulate Inheritance Mendelian Genetics4/15/2017Particulate InheritanceMendel stated that physical traits are inherited as “particles”Mendel did not know that the “particles” were actually Chromosomes & DNA
15Mendelian Genetics4/15/2017Genetic TerminologyTrait - any characteristic that can be passed from parent to offspringHeredity - passing of traits from parent to offspringGenetics - study of heredity
16Types of Genetic Crosses Mendelian Genetics4/15/2017Types of Genetic CrossesMonohybrid cross - cross involving a single trait e.g. flower colorDihybrid cross - cross involving two traits e.g. flower color & plant height
17Designer “Genes” Alleles - two forms of a gene (dominant & recessive) Mendelian Genetics4/15/2017Designer “Genes”Alleles - two forms of a gene (dominant & recessive)Dominant - stronger of two genes expressed in the hybrid; represented by a capital letter (R)Recessive - gene that shows up less often in a cross; represented by a lowercase letter (r)
18Mendelian Genetics4/15/2017More TerminologyGenotype - gene combination for a trait (e.g. RR, Rr, rr)Phenotype - the physical feature resulting from a genotype (e.g. red, white)
19Here are the steps used to solve a Monohybrid Mendelian Genetic Problem: Figure out the genotypes of the parents (Pure Dom, Hybrid, Pure Rec) Use the Capital letter of the DOMINANT trait and then its lower case form to represent the recessive trait.Step 2:Figure out what kinds of gametes the parents can produce.Step 3:Set up a Punnett Square for your mating.Step 4:Fill in the babies inside the table by filling each square by going Down and to the Left.Step 5:Figure out the genotypic ratio for your predicted babies.How Many of the offspring are--Pure Dom: Hybrid: Pure RecStep 6:Figure out the phenotypic ratio for your predicted babies.How Many of the offspring are-- Dom : RecStep 7:Answer the question you've been asked.
20Solving Genetics Problems I: Monohybrid Crosses Classical genetics is a science of logic and statistics. While many find the latter intimidating, the mathematical side of most classical genetics puzzles is relatively simple — and there are actually ways to get around most of the math. The logic part is inescapable. All genetics problems are solved using the same basic logic structure. If you learn the sense of the approach, you can solve virtually any genetics problem, provided you are given enough basic information.
21Sample Problem using Steps: This problem involves two gerbils named Honey and Ritz. The gene in question is a fur color gene which has two alleles — dominant brown (B) and recessive black (b). It's a very good idea to write down the information you are given in a problem so that it will be easy for you to refer to it when necessary. So begin by writing something like this at the top of your work page:
23Step One: Figure out the genotypes of the parents. Each of our parent gerbils is heterozygous for this gene. So here is our mating:Step One: Figure out the genotypes of the parents.
24Step One: Figure out what kinds of gametes the parents can produce.. Once you've got that settled, you need to address the question of all of the possible kinds of babies they could produce. Before any parent makes babies, of course, that parent makes gametes. So in order to find what kinds of babies they can have, you must first determine what kinds of gametes they can produce. Since Honey is a heterozygote (and paying attention to Rule #1), she can produce two kinds of eggs: B eggs and b eggs.
25Ritz is also a heterozygote, so he can produce two kinds of sperm: B sperm and b sperm. Something like this:
26Step Three: Set up a Punnett Square for your mating Now you need to determine all the possible ways that his sperm can combine with her eggs. There are several different techniques used for this operation. The most popular among students is the Punnett Square. Punnett Squares are probability tables — a way to do statistics while avoiding as much math as possible.
27Step 4: fill in the Punnett Square, down and to the left Setting up a Punnett Square is easy. You need to create a chart with one column for each of the female's egg types, and one row for each of the male's sperm types. For Honey and Ritz, your table would look like this, then fill in the babies genotype by going down and to the left:BbBBBbbb
28Step Five: Figure out the genotypic ratio for your predicted babies. So we have now figured out that, if Honey and Ritz have a lot of babies, we can predict that 1/4 of them should be BB, 1/2 of them (2/4) should be Bb, and 1/4 should be bb.
29This conclusion is often expressed as a genotypic ratio:1BB:2Bb:1bb This conclusion is often expressed as a genotypic ratio:1BB:2Bb:1bb. This means that we are predicting that, for every BB baby, they should have 2 Bb babies (twice as many), and one bb baby.
30Step Six: Figure out the phenotypic ratio for your predicted babies. To do this, you need to ask yourself one question: do any of these different genotypes produce the same phenotype? In other words, do any of these babies look alike? This is where dominance enters the picture. If B is completely dominant to b, all gerbils with at least one B will look pretty much alike, no matter whether their second allele is B or b. So BB and Bb have the same phenotype, and we can add them together. Thus, our phenotypic ratio is 3 Brown:1 Black. Or, there should be three times as many brown babies as black babies.
31So the answer to our question is, 3/4 of the babies should be brown.
32Step Seven: Answer the question you've been asked. The mating scheme we've just worked through is called a monohybrid cross. This means that we were paying attention to only one gene (mono=1), and both of our parents were heterozygous for that gene (hybrid=heterozygous).
33Punnett Square Used to help solve genetics problems Mendelian Genetics 4/15/2017Punnett SquareUsed to help solve genetics problems
35Genotype & Phenotype in Flowers Mendelian Genetics4/15/2017Genotype & Phenotype in FlowersGenotype of alleles: R = red flower r = yellow flowerAll genes occur in pairs, so 2 alleles affect a characteristicPossible combinations are:Genotypes RR Rr rrPhenotypes RED RED YELLOW
36Mendelian Genetics4/15/2017GenotypesHomozygous genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure Heterozygous genotype - gene combination of one dominant & one recessive allele (e.g. Rr); also called hybrid
37Genes and Environment Determine Characteristics Mendelian Genetics4/15/2017Genes and Environment Determine Characteristics
38Mendel’s Pea Plant Experiments Mendelian Genetics4/15/2017Mendel’s Pea Plant Experiments
39Why peas, Pisum sativum? Can be grown in a small area Mendelian Genetics4/15/2017Why peas, Pisum sativum?Can be grown in a small areaProduce lots of offspringProduce pure plants when allowed to self-pollinate several generationsCan be artificially cross-pollinated
40Reproduction in Flowering Plants Mendelian Genetics4/15/2017Reproduction in Flowering PlantsPollen contains spermProduced by the stamenOvary contains eggsFound inside the flowerPollen carries sperm to the eggs for fertilizationSelf-fertilization can occur in the same flowerCross-fertilization can occur between flowers
41Mendel’s Experimental Methods Mendelian Genetics4/15/2017Mendel’s Experimental MethodsMendel hand-pollinated flowers using a paintbrushHe could snip the stamens to prevent self-pollinationCovered each flower with a cloth bagHe traced traits through the several generations
42Mendelian Genetics4/15/2017How Mendel BeganMendel produced pure strains by allowing the plants to self-pollinate for several generations
43Eight Pea Plant Traits Seed shape --- Round (R) or Wrinkled (r) Mendelian Genetics4/15/2017Eight Pea Plant TraitsSeed shape --- Round (R) or Wrinkled (r)Seed Color ---- Yellow (Y) or Green (y)Pod Shape --- Smooth (S) or wrinkled (s)Pod Color --- Green (G) or Yellow (g)Seed Coat Color ---Gray (G) or White (g)Flower position---Axial (A) or Terminal (a)Plant Height --- Tall (T) or Short (t)Flower color --- Purple (P) or white (p)
47Did the observed ratio match the theoretical ratio? Mendelian Genetics4/15/2017Did the observed ratio match the theoretical ratio?The theoretical or expected ratio of plants producing round or wrinkled seeds is 3 round :1 wrinkledMendel’s observed ratio was 2.96:1The discrepancy is due to statistical errorThe larger the sample the more nearly the results approximate to the theoretical ratio
48Mendelian Genetics4/15/2017Generation “Gap”Parental P1 Generation = the parental generation in a breeding experiment.F1 generation = the first-generation offspring in a breeding experiment. (1st filial generation)From breeding individuals from the P1 generationF2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation)From breeding individuals from the F1 generation
49Following the Generations Mendelian Genetics4/15/2017Following the GenerationsCross 2 Pure Plants TT x ttResults in all Hybrids TtCross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt
51P1 Monohybrid Cross r r Rr Rr R R Rr Rr Trait: Seed Shape Mendelian Genetics4/15/2017P1 Monohybrid CrossTrait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Wrinkled seeds: P1 = RR x rrGenotype: RrPhenotype: RoundGenotypic Ratio: All alikePhenotypic Ratio: All alikerrRrRrRRRrRr
52P1 Monohybrid Cross Review Mendelian Genetics4/15/2017P1 Monohybrid Cross ReviewHomozygous dominant x Homozygous recessiveOffspring all Heterozygous (hybrids)Offspring called F1 generationGenotypic & Phenotypic ratio is ALL ALIKE
53F1 Monohybrid Cross R r RR Rr R r Rr rr Trait: Seed Shape Mendelian Genetics4/15/2017F1 Monohybrid CrossTrait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Round seeds: P1 = Rr x RrRrGenotype: RR, Rr, rrPhenotype: Round & wrinkledG.Ratio: 1:2:1P.Ratio: 3:1RRRrRrRrrr
54F1 Monohybrid Cross Review Mendelian Genetics4/15/2017F1 Monohybrid Cross ReviewHeterozygous x heterozygousOffspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rrOffspring called F2 generationGenotypic ratio is 1:2:1Phenotypic Ratio is 3:1
55What Do the Peas Look Like? Mendelian Genetics4/15/2017What Do the Peas Look Like?
56Mendelian Genetics4/15/2017…And Now the Test CrossMendel then crossed a pure & a hybrid from his F2 generationThis is known as an F2 or test crossThere are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid
57F2 Monohybrid Cross (1st) Mendelian Genetics4/15/2017F2 Monohybrid Cross (1st)Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Pure Round seeds x Hybrid Round seedsP =RR x RrGenotype: RR, RrPhenotype: RoundGenotypic Ratio: 1:1Phenotypic Ratio: All alikeRrRRRrRRRRRr
58F2 Monohybrid Cross (2nd) Mendelian Genetics4/15/2017F2 Monohybrid Cross (2nd)Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Wrinkled seeds x Hybrid Round seeds = rr x RrRrGenotype: Rr, rrPhenotype: Round & WrinkledG. Ratio: 1:1P.Ratio: 1:1RrrrrrRrrr
59F2 Monohybrid Cross Review Mendelian Genetics4/15/2017F2 Monohybrid Cross ReviewHomozygous x heterozygous(hybrid)Offspring: 50% Homozygous RR or rr 50% Heterozygous RrPhenotypic Ratio is 1:1Called Test Cross because the offspring have SAME genotype as parents
61Results of Monohybrid Crosses Mendelian Genetics4/15/2017Results of Monohybrid CrossesInheritable factors or genes are responsible for all heritable characteristicsPhenotype is based on GenotypeEach trait is based on two genes, one from the mother and the other from the fatherTrue-breeding individuals are homozygous ( both alleles) are the same
62Mendelian Genetics4/15/2017Law of DominanceIn a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation.All the offspring will be heterozygous and express only the dominant trait.RR x rr yields all Rr (round seeds)
64Mendelian Genetics4/15/2017Law of SegregationDuring the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other.Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring.
65Applying the Law of Segregation Mendelian Genetics4/15/2017Applying the Law of Segregation
66Law of Independent Assortment Mendelian Genetics4/15/2017Law of Independent AssortmentAlleles for different traits are distributed to sex cells (& offspring) independently of one another.This law can be illustrated using dihybrid crosses.
67Sex-linked Traits Traits (genes) located on the sex chromosomes Mendelian Genetics4/15/2017Sex-linked TraitsTraits (genes) located on the sex chromosomesSex chromosomes are X and YXX genotype for femalesXY genotype for malesMany sex-linked traits carried on X chromosome
68Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes Mendelian Genetics4/15/2017Sex-linked TraitsExample: Eye color in fruit fliesSex ChromosomesXX chromosome - femaleXy chromosome - malefruit flyeye color
69Sex-linked Trait Problem Mendelian GeneticsSex-linked Trait Problem4/15/2017Example: Eye color in fruit flies(red-eyed male) x (white-eyed female) XRY x XrXrRemember: the Y chromosome in males does not carry traits.RR = red eyedRr = red eyedrr = white eyedXY = maleXX = femaleXRXrY
70Sex-linked Trait Solution: Mendelian Genetics4/15/2017Sex-linked Trait Solution:XRXrY50% red eyed female50% white eyed maleXR XrXr Y
72Incomplete Dominance and Codominance Mendelian Genetics4/15/2017Incomplete Dominance and Codominance
73Mendelian Genetics4/15/2017Incomplete DominanceF1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties.Example: snapdragons (flower)red (RR) x white (WW)RR = red flowerWW = white flowerWR
74Incomplete Dominance W R W All RW = pink (heterozygous pink) Mendelian Genetics4/15/2017Incomplete DominanceWRWAll RW = pink(heterozygous pink)produces theF1 generationRW
75Incomplete Dominance Problem: In cattle when a red bull(RR) is mated with white(WW) cow the offspring are roan(RW) a blending of red and white. Mate a red bull with a roan cow. Show the P1, the Punnett Square, and give the genotypic and phenotypic ratios for this cross.
78Mendelian Genetics4/15/2017CodominanceTwo alleles are expressed (multiple alleles) in heterozygous individuals.Example: blood type1. type A = IAIA or IAi2. type B = IBIB or IBi3. type AB = IAIB4. type O = ii
79Mendelian Genetics4/15/2017Codominance ProblemExample: homozygous male Type B (IBIB) x heterozygous female Type A (IAi)IBIAiIAIBIBi1/2 = IAIB1/2 = IBi
80Another Codominance Problem Mendelian Genetics4/15/2017Another Codominance ProblemExample: male Type O (ii) x female type AB (IAIB)iIAIBIAiIBi1/2 = IAi1/2 = IBi
81Mendelian Genetics4/15/2017CodominanceQuestion: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents?boy - type O (ii) X girl - type AB (IAIB)
82Codominance Answer: IB IA i IAIB ii Parents: genotypes = IAi and IBi Mendelian Genetics4/15/2017CodominanceAnswer:IBIAiIAIBiiParents:genotypes = IAi and IBiphenotypes = A and B
83Mendelian Genetics4/15/2017Dihybrid CrossA breeding experiment that tracks the inheritance of two traits.Mendel’s “Law of Independent Assortment”a. Each pair of alleles segregates independently during gamete formationb. Formula: 2n (n = # of heterozygotes)
84Mendelian Genetics4/15/2017Question: How many gametes will be produced for the following allele arrangements?Remember: 2n (n = # of heterozygotes)1. RrYy2. AaBbCCDd3. MmNnOoPPQQRrssTtQq
85Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry Mendelian Genetics4/15/2017Answer:1. RrYy: 2n = 22 = 4 gametesRY Ry rY ry2. AaBbCCDd: 2n = 23 = 8 gametesABCD ABCd AbCD AbCdaBCD aBCd abCD abCD3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
86All possible gamete combinations Mendelian Genetics4/15/2017Dihybrid CrossTraits: Seed shape & Seed colorAlleles: R round r wrinkled Y yellow y greenRrYy x RrYyRY Ry rY ryRY Ry rY ryAll possible gamete combinations
88Dihybrid Cross RY Ry rY ry Round/Yellow: 9 Round/green: 3 Mendelian Genetics4/15/2017Dihybrid CrossRYRyrYryRound/Yellow: 9Round/green: 3wrinkled/Yellow: 3wrinkled/green: 19:3:3:1 phenotypic ratioRRYYRRYyRrYYRrYyRRyyRryyrrYYrrYyrryy
90Mendelian Genetics4/15/2017Test CrossA mating between an individual of unknown genotype and a homozygous recessive individual.Example: bbC__ x bbccBB = brown eyesBb = brown eyesbb = blue eyesCC = curly hairCc = curly haircc = straight hairbCb___bc
91Test Cross Possible results: bC b___ bc bbCc C bC b___ bc bbCc bbcc or Mendelian Genetics4/15/2017Test CrossPossible results:bCb___bcbbCcCbCb___bcbbCcbbccorc
92Summary of Mendel’s laws Mendelian Genetics4/15/2017Summary of Mendel’s lawsLAWPARENT CROSSOFFSPRINGDOMINANCETT x tt tall x short100% Tt tallSEGREGATIONTt x Tt tall x tall75% tall 25% shortINDEPENDENT ASSORTMENTRrGg x RrGg round & green x round & green9/16 round seeds & green pods 3/16 round seeds & yellow pods 3/16 wrinkled seeds & green pods 1/16 wrinkled seeds & yellow pods
93Genetic Practice Problems Mendelian Genetics4/15/2017Genetic Practice Problems
94Breed the P1 generation tall (TT) x dwarf (tt) pea plants t T Mendelian Genetics4/15/2017Breed the P1 generationtall (TT) x dwarf (tt) pea plantstT
95tall (TT) vs. dwarf (tt) pea plants Mendelian Genetics4/15/2017Solution:tall (TT) vs. dwarf (tt) pea plantsTtAll Tt = tall(heterozygous tall)produces theF1 generationTt
96Breed the F1 generation tall (Tt) vs. tall (Tt) pea plants T t T t Mendelian Genetics4/15/2017Breed the F1 generationtall (Tt) vs. tall (Tt) pea plantsTtTt
97tall (Tt) x tall (Tt) pea plants Mendelian Genetics4/15/2017Solution:tall (Tt) x tall (Tt) pea plantsTtproduces theF2 generation1/4 (25%) = TT1/2 (50%) = Tt1/4 (25%) = tt1:2:1 genotype3:1 phenotypeTTTttt