Genetic Terminology Gene Segment of DNA that codes

Genetic Terminology § Gene - Segment of DNA that codes formation of a protein § Locus – Position of gene on a chromosome § Trait - any characteristic that can be passed from parent to offspring § Heredity - passing of traits from parent to offspring § Genetics - study of heredity 1

Alleles • A gene that controls one function can exist in different forms. These different forms are called alleles. • Each different allele is identified by its specific phenotypic action. • Alleles are commonly represented by letters of the alphabet. o Eg. The gene LDLR controls blood cholesterol levels. Located on chromosome 19, it has two allelic forms: B = abnormally high cholesterol levels b = normal range

Autosomal Genotype • Remembering that nonsex chromosomes occur in homologous pairs in the diploid cell – there are two copies of each gene. • The double set of genetic instructions present makes up the genotype. • The number of possible genotypes depends on the number of allelic forms of the gene.

Genotype terminology Homozygous (pure) genotype - gene combination involving 2 dominant or 2 recessive genes (e. g. RR or rr) Heterozygous (hybrid) genotype - gene combination of one dominant & one recessive allele (e. g. Rr)

Phenotype • The visible expression of the genotype is called the phenotype. The expression may be a physical, biochemical or physiological trait. – Dominant trait: require only a single copy of the responsible allele for its phenotypic expression – Recessive trait: refers to a trait that is not expressed in a heterozygote – Co-dominant trait: both alleles in the heterozygote are expressed in the phenotype

e. g. 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 6

Genes and Environment Determine Characteristics 7

The relationship between genotype and phenotype is rarely simple! Phenotype = Genotype + Environmental Factors Hydrangeas: pink or blue? Both plants have the pigment for colour called anthocyanin. In acidic soils (low p. H) the flowers are blue. In alkaline soils (high p. H) the flowers are pink). Same genotype – different phenotype

Allele Combination AA BB AB or or Phenotype Ai Bi A B AB Multiple Alleles • For some genes, three or more alleles can be present in the population. • You will only inherit two alleles (one on each chromosome) • The combination of any two alleles determines the final phenotype. • The ABO blood groupings in humans is an example. ii O • Three alleles are involved in controlling blood group IA, IB and i.

Monogenic Traits • Monogenic traits are due to the action of a single gene with two or more allelic forms. • These traits show discontinuous variation the members of the population can be grouped into a few discrete and non-overlapping classes. • E. g. blood types

Polygenic Traits • Polygenic traits are due to the actions of many genes (and their allelic forms). These traits show continuous variation (e. g. height).

Human Sex Chromosomes • Traits (genes) located on the sex chromosomes – XX genotype for females – XY genotype for males • The X chromosome may carry up to 1500 genes. Genes located on the X chromosome are said to be X-linked. • Females have two alleles of a particular gene whereas males have only one (hemizygous genotype). This accounts for why many X-linked diseases show up more frequently in males than in females. • The Y chromosome has less than 300 genes. Genes located on the Y chromosome are said to be Y-linked. Males are also hemizygous for Y-linked genes.

X Inactivation in Female Mammals • Females who have two X chromosomes but one is subject to inactivation. • 75% of alleles on one X chromosome are switched off in early embryonic development. • 15% remain activated with another 10% altering their activation state in different females and in different cells within the same female.

Mendel’s Pea Plant Experiments 14

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 • 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" 15

Site of Gregor Mendel’s experimental garden in the Czech Republic 16

Particulate Inheritance • Mendel stated that physical traits are inherited as “particles” • Mendel did not know that the “particles” were actually chromosomes & DNA 17

Reproduction in Flowering Plants • Pollen contains sperm –Produced by the stamen • Ovary 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 18

How Mendel Began Mendel produced pure strains by allowing the plants to selfpollinate for several generations 19

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Mendel’s Experimental Results 22

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 the results approximate to theoretical ratio 23

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 24

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 25

Monohybrid Cross • A trait determined by one gene with two or more allelic forms.

Punnett Square Used to help solve genetic problems 27

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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: 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 G. Ratio: 1: 2: 1 Phenotype: 3 Round & 1 wrinkled 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

…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 33

F 2 Monohybrid Cross (1 st) • Trait: Seed Shape • Alleles: R – Round r – Wrinkled • Cross: Round seeds x Round seeds • RR x Rr R RR Rr Genotype: RR, Rr Phenotype: Round Genotypic Ratio: 1: 1 Phenotypic Ratio: All alike 34

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: Round & Wrinkled G. Ratio: 1: 1 P. Ratio: 1: 1 35

F 2 Monohybrid Cross Review § Homozygous recessive x heterozygous(hybrid) § Offspring: 50% Homozygous rr 50% Heterozygous Rr § Phenotypic Ratio is 1: 1 36

Test crosses • Can be used to determine if an individual of dominant phenotype is homozygous or heterozygous • There are two possible testcrosses: Homozygous dominant x Homozygous recessive = All heterozygous dominant Hybrid x Homozygous recessive = Mix of dominant and recessive phenotypes

Monohybrid cross Practice Problems 38

1. Breed the P 1 generation • tall (TT) x dwarf (tt) pea plants t t T T 39

2. Breed the F 1 generation • tall (Tt) vs. tall (Tt) pea plants T t 40

1. Solution: tall (TT) vs. dwarf (tt) pea plants t t T Tt Tt produces the F 1 generation T Tt Tt All Tt = tall (heterozygous tall) 41

2. Solution: tall (Tt) x tall (Tt) pea plants T t T TT Tt 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 42

Results of Monohybrid Crosses • Inheritable factors or genes are responsible for all heritable characteristics • Phenotype is based on genotype • Each 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 same 43

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) 44

Law of Dominance 45

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. 46

Applying the Law of Segregation 47

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. 48

Dihybrid Cross • A breeding experiment that tracks the inheritance of two traits. • Mendel’s “Law of Independent Assortment” – a. Each pair of alleles segregates independently during gamete formation – b. Formula: 2 n (n = # of heterozygotes) 49

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 50

Answer: 1. Rr. Yy: 2 n = 22 = 4 gametes RY Ry r. Y ry 2. Aa. Bb. CCDd: 2 n = 23 = 8 gametes ABCD ABCd Ab. CD Ab. Cd a. BCD a. BCd ab. CD 3. Mm. Nn. Oo. PPQQRrss. Tt. Qq: 2 n = 26 = 64 gametes 51

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 52

Dihybrid Cross RY Ry r. Y ry Try filling in the punnet square and work out ratios 53

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 54

Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9: 3: 3: 1 55

Test Cross • A mating between an individual of unknown genotype and a homozygous recessive individual. • Example: bb. C__ x bbcc • • • BB = brown eyes Bb = brown eyes bb = blue eyes • • • CC = curly hair Cc = curly hair cc = straight hair b. C b___ bc 56

Test Cross • Possible results: bc b. C b___ C bb. Cc or bc b. C b___ c bb. Cc bbcc 57

Summary of Mendel’s laws LAW DOMINANCE SEGREGATION INDEPENDENT ASSORTMENT PARENT CROSS OFFSPRING TT x tt tall x short 100% Tt tall Tt x Tt tall x tall 75% 25% Rr. Gg x Rr. Gg round & green x round & green 9/16 3/16 1/16 tall short round seeds & green pods round seeds & yellow pods wrinkled seeds & green pods wrinkled seeds & yellow pods 58

Incomplete Dominance • F 1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. • Example: snapdragons (flower) red (RR) x white (rr) • RR = red flower • rr = white flower r r R R 59

Incomplete Dominance r r R Rr Rr produces the F 1 generation All Rr = pink (heterozygous pink) 60

Incomplete Dominance 61

Codominance • Two alleles are expressed (multiple alleles) in heterozygous individuals. • Example: blood type • • 1. 2. 3. 4. type A type B type AB type O = = IAIA or IAi IBIB or IBi I AIB ii 62

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 63

Another Codominance Problem • Example: male Type O (ii) x AB (IAIB) female type IA IB i I Ai I Bi 1/2 = IAi 1/2 = IBi 64

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) X girl - type AB (IAIB) 65

Codominance Answer: IA IB i i I AI B ii Parents: genotypes = IAi and IBi phenotypes = A and B 66