1 Mendelelian Genetics 2 Gregor Mendel 1822 1884

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Mendelelian Genetics 2

Gregor Mendel (1822 -1884) 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 20 th century 4

§ § Inheritance Mendel stated that physical traits are inherited as “particles” Mendel did not know that the “particles” were actually Chromosomes & DNA 6

§ § § 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 7

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 8

Punnett Square Used to help solve genetics problems 9

§ § § 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) 11

§ § 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) 12

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: Genotypes RR Rr rr Phenotypes RED YELLOW 13

§ Genotypes 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 § 14

Eight Pea Plant Traits Seed 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) 21

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Mendel’s Experimental Results 24

Did the observed ratio match 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 theoretical ratio 25

Generation “Gap” Parental P 1 Generation = the parental generation in a breeding experiment. F 1 generation = the first-generation offspring in a breeding experiment. (1 st filial generation) From breeding individuals from the P 1 generation F 2 generation = the second-generation offspring in a breeding experiment. (2 nd filial generation) From breeding individuals from the F 1 generation 26

Following the Generations Cross 2 Results in Cross 2 Hybrids Pure all get Plants Hybrids 3 Tall & 1 Short TT x tt Tt TT, Tt, tt 27

Monohybrid Crosses 28

P 1 Monohybrid Cross Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Round seeds x Wrinkled seeds RR x rr r r R Rr Rr Genotype: Rr Phenotype: Phenotype Round Genotypic Ratio: All alike Phenotypic Ratio: All alike 29

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 30

F 1 Monohybrid Cross Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Round seeds x Round seeds Rr x Rr R RR Rr rr Genotype: RR, Rr, rr Phenotype: Phenotype Round & wrinkled G. Ratio: 1: 2: 1 P. Ratio: 3: 1 31

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 32

What Do the Peas Look Like? 33

…And Now the Test Cross Mendel then crossed a pure & a hybrid from his F 2 generation This is known as an F 2 or test cross There are two possible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid 34

F 2 Monohybrid Cross st (1 ) Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Round seeds x Round seeds RR x Rr R RR Rr Genotype: RR, Rr Phenotype: Phenotype Round Genotypic Ratio: 1: 1 Phenotypic Ratio: All alike 35

F 2 Monohybrid Cross (2 nd) Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Wrinkled seeds x Round seeds rr x Rr R r r Rr Rr r rr rr Genotype: Rr, rr Phenotype: Phenotype Round & Wrinkled G. Ratio: 1: 1 P. Ratio: 1: 1 36

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 37

Practice Your Crosses Work the P 1, F 1, and both F 2 Crosses for each of the other Seven Pea Plant Traits 38

Mendel’s Laws 39

Law of Segregation 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 offspring. 43

Applying the Law of Segregation 44

Law of Independent Assortment Alleles for different traits are distributed to sex cells (& offspring) independently of one another. This law can be illustrated using dihybrid crosses. 45

Dihybrid Cross A breeding experiment that tracks the inheritance of two traits. Mendel’s “Law of Independent Assortment” a. Each pair of independently b. Formula: 2 n alleles segregates during gamete formation (n = # of heterozygotes) 46

Question: How many gametes will be produced for the following allele arrangements? Remember: 2 n (n = # of heterozygotes) 1. Rr. Yy 2. Aa. Bb. CCDd 3. Mm. Nn. Oo. PPQQRrss. Tt. Qq 47

Answer: 1. Rr. Yy: 2 n = 22 = 4 gametes RY Ry r. Y ry 2. Aa. Bb. CCDd: 2 n ABCD ABCd a. BCD a. BCd = 23 = Ab. CD ab. CD 8 gametes Ab. Cd ab. CD 3. Mm. Nn. Oo. PPQQRrss. Tt. Qq: 2 n = 26 = 64 gametes 48

In guinea pigs, the allele for black fur (B) is dominant. The allele for brown fur (b) is recessive. Two guinea pigs were crossed as shown in the Punnett square below. Numbers 1, 2, 3, and 4 represent the types of offspring produced from the cross. Describe the phenotype of The parents. 49

Info Mock HAS test March 16 th Next Wednesday Test Friday on Genetics/Punnett Squares 50

Dihybrid Cross Traits: Seed shape & Seed color Alleles: R round r wrinkled Y yellow y green Rr. Yy RY Ry r. Y ry x Rr. Yy RY Ry r. Y ry All possible gamete combinations 51

Dihybrid Cross RY Ry r. Y ry 52

Dihybrid Cross RY RY RRYY Ry RRYy r. Y Rr. YY ry Rr. Yy Ry r. Y ry RRYy Rr. YY Rr. Yy RRyy Rr. Yy Rryy Rr. Yy rr. YY rr. Yy Rryy rr. Yy rryy Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9: 3: 3: 1 phenotypic ratio 53

Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9: 3: 3: 1 54

Incomplete Dominance and Codominance 55

Codominance Two alleles are expressed (multiple alleles) in heterozygous individuals. Example: blood type 1. 2. 3. 4. type A B AB O = = IAIA or IAi IBIB or IBi I AI B ii 56

Codominance Problem Example: homozygous male Type B (IBIB) x heterozygous female Type A (IAi) IA i IB I AI B I Bi 1/2 = IAIB 1/2 = IBi 57

Another Codominance Problem • Example: male Type O (ii) x female type AB (IAIB) IA IB i I Ai I Bi 1/2 = IAi 1/2 = IBi 58

Antigen-is a foreign molecule that, when introduced into the body, triggers the production of an antibody by the immune system 59

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? boy - type O (ii) AB (IAIB) X girl - type 60

Codominance Answer: IA IB i i I AI B ii Parents: genotypes = IAi and IBi phenotypes = A and B 61

Sex-linked Traits (genes) located on the sex chromosomes Sex chromosomes are X and Y XX genotype for females XY genotype for males Many sex-linked traits carried on X chromosome 62

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Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes fruit fly eye color XX chromosome - female Xy chromosome - male 64

Sex-linked Trait Problem Example: Eye color in fruit flies (red-eyed male) x (white-eyed female) X RY x X r Remember: the Y chromosome in males does not carry traits. Xr Xr RR = red eyed Rr = red eyed R X rr = white eyed XY = male Y XX = female 65

Sex-linked Trait Solution: Xr XR XR Xr Y Xr XR Xr Xr Y 50% red eyed female 50% white eyed male 66

Female Carriers 67

In humans, the allele for having feet with normal arches is dominant (A). The allele for flat feet is recessive (a). The pedigree below shows the occurrence of normal arches and flat feet in four generations of a family. In the pedigree, individuals are identified by the generation and individual numbers. For example, Individual 2 in Generation I is identified as I-2. 68

normal arches is dominant (A). The allele for flat feet is recessive (a). 69

Codominance The lubber grasshopper is a very large grasshopper, and is black with red and yellow stripes. Assume that red stripes are expressed from homozygous RR genotype, yellow stripes from YY genotype and both from Heterozygous genotype. Cross two grasshoppers, both with Red and yellow stripes. What are the phenotypic and genotypic ratios? 70

Blood typing What are the possible genotypes of the offspring the resulted from a type B parent crossed with a type O parent. 71