Genetic Terminology Gene - Segment of DNA that codes for 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

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. 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 non-sex 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 RED YELLOW

Genes and Environment Determine Characteristics

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 pH) the flowers are blue. In alkaline soils (high pH) the flowers are pink). Same genotype – different phenotype

Multiple Alleles or B AB O Allele Combination Phenotype or A 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. Three alleles are involved in controlling blood group IA, IB and i. AA Ai BB Bi AB ii

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. See page 315 for case study

Mendel’s Pea Plant Experiments

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 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"

Site of Gregor Mendel’s experimental garden in the Czech Republic Mendelian Genetics 9/21/2018 Site of Gregor Mendel’s experimental garden in the Czech Republic Fig. 5.co

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

Reproduction in Flowering Plants Mendelian Genetics 9/21/2018 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

How Mendel Began Mendel produced pure strains by allowing the plants to self-pollinate for several generations

Mendel’s Experimental Results

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 the results approximate to the theoretical ratio

Generation “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 generation F2 generation = the second-generation offspring in a breeding experiment. (2nd filial generation) From breeding individuals from the F1 generation

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

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

Punnett Square Used to help solve genetic problems

P1 Monohybrid Cross r r Rr Rr R R Rr Rr Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Round seeds x Wrinkled seeds RR x rr Genotype: Rr Phenotype: Round Genotypic Ratio: All alike Phenotypic Ratio: All alike r r Rr Rr R R Rr Rr

P1 Monohybrid Cross Review Homozygous dominant x Homozygous recessive Offspring all Heterozygous (hybrids) Offspring called F1 generation Genotypic & Phenotypic ratio is ALL ALIKE

F1 Monohybrid Cross R r RR Rr R r Rr rr Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Round seeds x Round seeds Rr x Rr R r Genotype: RR, Rr, rr G.Ratio: 1:2:1 Phenotype: 3 Round & 1 wrinkled P.Ratio: 3:1 RR Rr R r Rr rr

F1 Monohybrid Cross Review Heterozygous x heterozygous Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr Offspring called F2 generation Genotypic ratio is 1:2:1 Phenotypic Ratio is 3:1

…And Now the Test Cross Mendel then crossed a pure & a hybrid from his F2 generation This is known as an F2 or test cross

F2 Monohybrid Cross (1st) Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Round seeds x Round seeds RR x Rr Genotype: RR, Rr Phenotype: Round Genotypic Ratio: 1:1 Phenotypic Ratio: All alike R r RR Rr R R RR Rr

F2 Monohybrid Cross (2nd) Trait: Seed Shape Alleles: R – Round r – Wrinkled Cross: Wrinkled seeds x Round seeds rr x Rr R r Genotype: Rr, rr Phenotype: Round & Wrinkled G. Ratio: 1:1 P.Ratio: 1:1 Rr rr r r Rr rr

F2 Monohybrid Cross Review Homozygous recessive x heterozygous(hybrid) Offspring: 50% Homozygous rr 50% Heterozygous Rr Phenotypic Ratio is 1:1

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

1. Breed the P1 generation tall (TT) x dwarf (tt) pea plants t T

2. Breed the F1 generation tall (Tt) vs. tall (Tt) pea plants T t T t

1. Solution: tall (TT) vs. dwarf (tt) pea plants t produces the Tt T All Tt = tall (heterozygous tall) produces the F1 generation Tt

2. Solution: tall (Tt) x tall (Tt) pea plants T t TT Tt tt produces the F2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype TT Tt tt

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

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)

Law of Dominance

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.

Applying the Law of Segregation

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.

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: 2n (n = # of heterozygotes)

Remember: 2n (n = # of heterozygotes) Question: How many gametes will be produced for the following allele arrangements? Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq

Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes

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

Dihybrid Cross Try filling in the punnet square and work out ratios RY

Dihybrid Cross RY Ry rY ry Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 phenotypic ratio RRYY RRYy RrYY RrYy RRyy Rryy rrYY rrYy rryy

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

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

Test Cross Possible results: bC b___ bc bbCc C bC b___ bc bbCc bbcc or

Summary of Mendel’s laws 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

Incomplete Dominance F1 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

Incomplete Dominance r produces the R Rr F1 generation All Rr = pink (heterozygous pink) produces the F1 generation Rr

Incomplete Dominance

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

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

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

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)

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