Presentation on theme: "1 Mendelelian Genetics 2 Gregor Mendel (1822-1884) Responsible for the Laws governing Inheritance of Traits."— Presentation transcript:
1 Mendelelian Genetics
2 Gregor Mendel ( ) Responsible for the Laws governing Inheritance of Traits
3 Gregor Johann Mendel Austrian monk Studied the inheritance of traits in pea plants Developed the laws of inheritance Mendel's work was not recognized until the turn of the 20th century
4 Gregor Johann Mendel Between 1856 and 1863, Mendel cultivated and tested some 28,000 pea plants He found that the plants' offspring retained traits of the parents Called the “Father of Genetics"
Mendel’s Laws of Inheritance
1. 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.
2. 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.
3. 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.
4. 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.
Rule of Dominance The trait that is observed in the offspring is the dominant trait (uppercase) The trait that disappears in the offspring is the recessive trait (lowercase)
Law of Segregation The two alleles for a trait must separate when gametes are formed A parent randomly passes only one allele for each trait to each offspring
Law of Independent Assortment The genes for different traits are inherited independently of each other.
13 Site of Gregor Mendel’s experime ntal garden in the Czech Republic
14 Mendel stated that physical traits are inherited as “particles” Mendel did not know that the “particles” were actually Chromosomes & DNA Particulate Inheritance
15 Genetic Terminology Trait - any characteristic that can be passed from parent to offspring Heredity - passing of traits from parent to offspring Genetics - study of heredity
16 Types of Genetic Crosses Monohybrid cross - cross involving a single trait e.g. flower color Dihybrid cross - cross involving two traits e.g. flower color & plant height
17 Designer “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)
18 More Terminology Genotype - gene combination for a trait (e.g. RR, Rr, rr) Phenotype - the physical feature resulting from a genotype (e.g. red, white)
Here are the steps used to solve a Monohybrid Mendelian Genetic Problem: Step 1: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. Pure Dom: Hybrid: Pure Rec How Many of the offspring are--Pure Dom: Hybrid: Pure Rec Step 6:Figure out the phenotypic ratio for your predicted babies. Dom : Rec How Many of the offspring are-- Dom : Rec Step 7:Answer the question you've been asked. 19
Solving 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. 20
Sample 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: 21
Each of our parent gerbils is heterozygous for this gene. So here is our mating: 23 Step One: Figure out the genotypes of the parents.
Step 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. 24
Ritz is also a heterozygote, so he can produce two kinds of sperm: B sperm and b sperm. Something like this: 25
Step 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. 26
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: 27 Step 4: fill in the Punnett Square, down and to the left BB Bb bb Bb
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. 28 Step Five: Figure out the genotypic ratio for your predicted babies.
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. 29
Step 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. 30
So the answer to our question is, 3/4 of the babies should be brown. 31
Step 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). 32
33 Punnett Square Used to help solve genetics problems
35 Genotype & Phenotype in Flowers Genotype of alleles: R = red flower r = yellow flower All genes occur in pairs, so 2 alleles affect a characteristic Possible combinations are: GenotypesRR Rrrr PhenotypesRED RED YELLOW
36 Genotypes Homozygous genotype - gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure Homozygous 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
37 Genes and Environment Determine Characteristics
38 Mendel’s Pea Plant Experiments
39 Why peas, Pisum sativum? Can be grown in a small area Produce lots of offspring Produce pure plants when allowed to self- pollinate several generations Can be artificially cross-pollinated
40 Reproduction in Flowering Plants Pollen contains spermPollen contains sperm –Produced by the stamen Ovary contains eggsOvary contains eggs –Found inside the flower Pollen carries sperm to the eggs for fertilization Self-fertilization can occur in the same flower Cross-fertilization can occur between flowers
41 Mendel’s Experimental Methods Mendel hand-pollinated flowers using a paintbrushMendel hand-pollinated flowers using a paintbrush –He could snip the stamens to prevent self- pollination –Covered each flower with a cloth bag He traced traits through the several generationsHe traced traits through the several generations
42 How Mendel Began Mendel produced pure strains by allowing the plants to self- pollinate for several generations
43 Eight Pea Plant Traits Seed shape --- Round (R) or Wrinkled (r)Seed shape --- Round (R) or Wrinkled (r) Seed Color ---- Yellow (Y) or Green ( y )Seed Color ---- Yellow (Y) or Green ( y ) Pod Shape --- Smooth (S) or wrinkled ( s )Pod Shape --- Smooth (S) or wrinkled ( s ) Pod Color --- Green (G) or Yellow (g)Pod Color --- Green (G) or Yellow (g) Seed Coat Color ---Gray (G) or White (g)Seed Coat Color ---Gray (G) or White (g) Flower position---Axial (A) or Terminal (a)Flower position---Axial (A) or Terminal (a) Plant Height --- Tall (T) or Short (t)Plant Height --- Tall (T) or Short (t) Flower color --- Purple (P) or white ( p )Flower color --- Purple (P) or white ( p )
46 Mendel’s Experimental Results
47 Did the observed ratio match the theoretical ratio?Did the observed ratio match the theoretical ratio? The theoretical or expected ratio of plants producing round or wrinkled seeds is 3 round :1 wrinkled Mendel’s observed ratio was 2.96:1 The discrepancy is due to statistical error The larger the sample the more nearly the results approximate to the theoretical ratio
48 Generation “Gap” Parental P 1 Generation = the parental generation in a breeding experiment.Parental P 1 Generation = the parental generation in a breeding experiment. F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation)F 1 generation = the first-generation offspring in a breeding experiment. (1st filial generation) –From breeding individuals from the P 1 generation F 2 generation = the second- generation offspring in a breeding experiment. (2nd filial generation)F 2 generation = the second- generation offspring in a breeding experiment. (2nd filial generation) – From breeding individuals from the F 1 generation
49 Following the Generations Cross 2 Pure Plants TT x tt Results in all Hybrids Tt Cross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt
50 Monohybrid Crosses
51 Trait: Seed ShapeTrait: Seed Shape Alleles: R – Roundr – WrinkledAlleles: R – Roundr – Wrinkled Cross: Round seeds x Wrinkled seeds: P 1 = RR x rrCross: Round seeds x Wrinkled seeds: P 1 = RR x rr P 1 Monohybrid Cross R R rr Rr Genotype:Rr Genotype: Rr PhenotypeRound Phenotype: Round Genotypic Ratio:All alike Genotypic Ratio: All alike Phenotypic Ratio: All alike
52 P 1 Monohybrid Cross Review Homozygous dominant x Homozygous recessive Offspring all Heterozygous (hybrids) Offspring called F 1 generation Genotypic & Phenotypic ratio is ALL ALIKE
53 Trait: Seed ShapeTrait: Seed Shape Alleles: R – Roundr – WrinkledAlleles: R – Roundr – Wrinkled Cross: Round seeds x Round seeds: P 1 = Rr x RrCross: Round seeds x Round seeds: P 1 = Rr x Rr F 1 Monohybrid Cross R r rR RR rrRr Genotype:RR, Rr, rr Genotype: RR, Rr, rr PhenotypeRound & wrinkled Phenotype: Round & wrinkled G.Ratio:1:2:1 G.Ratio: 1:2:1 P.Ratio: 3:1
54 F 1 Monohybrid Cross Review Heterozygous x heterozygous Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr Offspring called F 2 generation Genotypic ratio is 1:2:1 Phenotypic Ratio is 3:1
55 What Do the Peas Look Like?
56 …And Now the Test Cross Mendel then crossed a pure & a hybrid from his F 2 generationMendel then crossed a pure & a hybrid from his F 2 generation This is known as an F 2 or test crossThis is known as an F 2 or test cross There are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x HybridThere are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid
57 Trait: Seed ShapeTrait: Seed Shape Alleles: R – Roundr – WrinkledAlleles: R – Roundr – Wrinkled Cross: Pure Round seeds x Hybrid Round seedsCross: Pure Round seeds x Hybrid Round seeds P =RR x RrP =RR x Rr F 2 Monohybrid Cross (1 st ) R R rR RR RrRR Rr Genotype:RR, Rr Genotype: RR, Rr PhenotypeRound Phenotype: Round Genotypic Ratio:1:1 Genotypic Ratio: 1:1 Phenotypic Ratio: All alike
58 Trait: Seed ShapeTrait: Seed Shape Alleles: R – Roundr – WrinkledAlleles: R – Roundr – Wrinkled Cross: Wrinkled seeds x Hybrid Round seeds = rr x RrCross: Wrinkled seeds x Hybrid Round seeds = rr x Rr F 2 Monohybrid Cross (2nd) r r rR Rr rrRr rr Genotype:Rr, rr Genotype: Rr, rr PhenotypeRound & Wrinkled Phenotype: Round & Wrinkled G. Ratio:1:1 G. Ratio: 1:1 P.Ratio: 1:1
59 F 2 Monohybrid Cross Review Homozygous x heterozygous(hybrid) Offspring: 50% Homozygous RR or rr 50% Heterozygous Rr Phenotypic Ratio is 1:1 Called Test Cross because the offspring have SAME genotype as parents
60 Mendel’s Laws
61 Results of Monohybrid Crosses Inheritable factors or genes are responsible for all heritable characteristicsInheritable factors or genes are responsible for all heritable characteristics Phenotype is based on GenotypePhenotype is based on Genotype Each trait is based on two genes, one from the mother and the other from the fatherEach trait is based on two genes, one from the mother and the other from the father True-breeding individuals are homozygous ( both alleles) are the sameTrue-breeding individuals are homozygous ( both alleles) are the same
62 Law of Dominance In 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)
63 Law of Dominance
64 Law of Segregation During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other.During 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 offspringAlleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring.
65 Applying the Law of Segregation
66 Law of Independent Assortment Alleles for different traits are distributed to sex cells (& offspring) independently of one another.Alleles for different traits are distributed to sex cells (& offspring) independently of one another. This law can be illustrated using dihybrid crosses.This law can be illustrated using dihybrid crosses.
67 Sex-linked Traits Traits (genes) located on the sex chromosomesTraits (genes) located on the sex chromosomes Sex chromosomes are X and YSex chromosomes are X and Y XX genotype for femalesXX genotype for females XY genotype for malesXY genotype for males Many sex-linked traits carried on X chromosomeMany sex-linked traits carried on X chromosome
68 Sex-linked Traits Sex Chromosomes XX chromosome - femaleXy chromosome - male fruit fly eye color Example: Eye color in fruit flies
69 Sex-linked Trait Problem Example: Eye color in fruit flies (red-eyed male) x (white-eyed female) X R Y x X r X r Remember: the Y chromosome in males does not carry traits. RR = red eyed Rr = red eyed rr = white eyed XY = male XX = female XRXR XrXr XrXr Y
70 Sex-linked Trait Solution: X R X r X r Y X R X r X r Y 50% red eyed female 50% white eyed male XRXR XrXr XrXr Y
71 Female Carriers
72 Incomplete Dominance and Codominance
73 Incomplete Dominance F1 hybrids in betweenphenotypesF1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example:snapdragons (flower)Example: snapdragons (flower) red (RR) x white (WW) RR = red flowerRR = red flower WW = white flower R R WW
74 Incomplete Dominance RWRWRWRW R RW All RW = pink (heterozygous pink) produces the F 1 generation W
75 Incomplete 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 P 1, the Punnett Square, and give the genotypic and phenotypic ratios for this cross.
78 Codominance Two alleles are expressed (multiple alleles) in heterozygous individuals.Two alleles are expressed (multiple alleles) in heterozygous individuals. Example: blood typeExample: blood type 1.type A= I A I A or I A i1.type A= I A I A or I A i 2.type B= I B I B or I B i2.type B= I B I B or I B i 3.type AB= I A I B3.type AB= I A I B 4.type O= ii4.type O= ii
79 Codominance Problem Example:homozygous male Type B (I B I B ) x heterozygous female Type A (I A i) IAIBIAIB IBiIBi IAIBIAIB IBiIBi 1/2 = I A I B 1/2 = I B i IBIB IAIA i IBIB
80 Another Codominance Problem Example:Example: male Type O (ii) x female type AB (I A I B ) IAiIAiIBiIBi IAiIAiIBiIBi 1/2 = I A i 1/2 = I B i i IAIA IBIB i
81 Codominance Question: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents?Question: 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 (I A I B )boy - type O (ii) X girl - type AB (I A I B )
82 Codominance Answer:Answer: IAIBIAIB ii Parents: genotypes genotypes = I A i and I B i phenotypes phenotypes = A and B IBIB IAIA i i
83 Dihybrid Cross A breeding experiment that tracks the inheritance of two traits.A breeding experiment that tracks the inheritance of two traits. Mendel’s “Law of Independent Assortment”Mendel’s “Law of Independent Assortment” a. Each pair of alleles segregates independently during gamete formationa. Each pair of alleles segregates independently during gamete formation b. Formula: 2 n (n = # of heterozygotes)b. Formula: 2 n (n = # of heterozygotes)
84 Question: How many gametes will be produced for the following allele arrangements? Remember: 2 n (n = # of heterozygotes)Remember: 2 n (n = # of heterozygotes) 1.RrYy1.RrYy 2.AaBbCCDd2.AaBbCCDd 3.MmNnOoPPQQRrssTtQq3.MmNnOoPPQQRrssTtQq
85 Answer: 1. RrYy: 2 n = 2 2 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2 n = 2 3 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2 n = 2 6 = 64 gametes
86 Dihybrid Cross Traits: Seed shape & Seed colorTraits: Seed shape & Seed color Alleles:Alleles: R round r wrinkled Y yellow y green RrYy x RrYy RY Ry rY ry All possible gamete combinations
87 Dihybrid Cross RYRyrYry RYRy rY ry
88 Dihybrid Cross RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 phenotypic ratio RYRyrYryRY Ry rY ry
90 Test Cross A mating between an individual of unknown genotype and a homozygous recessive individual.A mating between an individual of unknown genotype and a homozygous recessive individual. Example: bbC__ x bbccExample: bbC__ x bbcc BB = brown eyesBB = brown eyes Bb = brown eyesBb = brown eyes bb = blue eyesbb = blue eyes CC = curly hairCC = curly hair Cc = curly hairCc = curly hair cc = straight haircc = straight hairbCb___bc
91 Test Cross Possible results:Possible results:bCb___bcbbCc CbCb___bc bbccor c
92 Summary of Mendel’s laws LAW PARENT CROSS OFFSPRING DOMINANCE TT x tt tall x short 100% Tt tall SEGREGATION Tt x Tt tall x tall 75% tall 25% short INDEPENDENT ASSORTMENT RrGg x RrGg round & green x round & green 9/16 round seeds & green pods 3/16 round seeds & yellow pods 3/16 wrinkled seeds & green pods 1/16 wrinkled seeds & yellow pods
93 Genetic Practice Problems
94 Breed the P 1 generation tall (TT) x dwarf (tt) pea plantstall (TT) x dwarf (tt) pea plants T T tt
95 Solution: T T tt Tt All Tt = tall (heterozygous tall) produces the F 1 generation tall (TT) vs. dwarf (tt) pea plants
96 Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plantstall (Tt) vs. tall (Tt) pea plants T t Tt
97 Solution: TT Tt tt T t Tt produces the F 2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype 3:1 phenotype tall (Tt) x tall (Tt) pea plants