Gregor Mendel

*became an ordained priest in 1847, and * Born in Heinzendorf, Austria on July 22, 1822. * He died in Brno, Austria January 6, 1884. * Austrian monk who found actual proof of the existence of genes...the father of genetics, *work was relatively unappreciated until the early 1900's. *became an ordained priest in 1847, and in 1851entered into the University of Vienna to train to be a teacher of Mathematics and Biology. Failed the elementary teachers exam and returned to Brno in 1845 ,became a teacher at a technical high school, where he taughtmathematics, natural science and general science.

* 1868, promoted in the monastery to Abbot. * During the middle of his life, did work into the theories of heredity. Using pea pod plants, studied seven basic characteristics, traced these and discovered three basic laws which governed the passage of a trait from one member of a species to another member of the same species. First presentation was on his eight years of experimentation with artificial plant hybridization. *first two communications were published in 1853 to 1854. Contained information about damage to plants by insects. * 1856 to 1863 cultivated, tested almost 28,000 plants

*Traits united through fertilization *Rediscovery of his works brought us close to an era of speculation on heredity. *Opened a new pathway of study on heredity to reveal a new mechanism operating in the sense of evolution. work and theories, later became the basis for the study of modern genetics and are still recognized and used today. * His work led to the discovery of particulate inheritance, dominant and recessive traits, genotype and phenotype, and the concept of heterozygousity and homozygousity.

Mendel’s Principles ( see other sheets) LAW of SEGREGATION LAW of INDEPENDENT ASSORTMENT

Types of Crosses 1. Monohybrid Cross (A cross of “one” trait) A trait is controlled by two alleles, one from each parent. Each can be: Dominant: TT (homozygous) Tt (Heterozygous) Recessive: tt

t P1= TT x tt T T Tt Tt Tt Tt Test Cross using a Punnett square: Homozygous parents: TT for Tall plants x tt (short) P1= TT x tt F1 cross (first filial,first generation ) T T t Tt Tt Tt Tt

Genotypic ratio( alleles): 4:0 All Heterozygous dominant All Tt Phenotypic ration( expression): 4:0 All Tall

P2 Tt x Tt (previous F1) T t T t F2 cross TT Tt Tt tt

Genotypic ratio: 1:2:1 1 TT: 2Tt: 1tt Phenotypic ration: 3:1 3 Tall : 1 short

2. Dihybrid Cross A cross of two traits (2 sets of alleles) P1 TTYY Tall plants, yellow flowers ttyy Short plants, white flowers All Heterozygous Punnett square of ?

TY TY TY TY TtYy TtYy TtYy TtYy ty F1

Genotypic ratio: 16:0 All TtYy Heterozygous dominant Phenotypic ration:16:0 All tall, yellow

F2 Cross : P2 TtYy x TtYy TtYy TtYy TY,Ty tY, ty

TY Ty tY ty TY Ty tY ty TTYY TTYy TtYY TtYy TTYy TTyy TtYy Ttyy TtYY F2

Genotypic ratio: Many different Phenotypic ration: 9:3:3:1 9 Tall, Yellow 3 Tall, white 3 Yellow, short 1 short, white

3. Incomplete Dominance A heterozygous genotype is a blending of the dominant and recessive allele (new phenotype) The dominant allele isn’t expressed over the recessive.

r Genotype: All Rr Phenotype: All PINK F1 cross P1 RR red flowers, rr white flowers R R r Rr Rr Rr Rr Genotype: All Rr Phenotype: All PINK

R r F2 Cross: P2 Rr (pink) x Rr(pink) Rr rr Rr Genotype: 1:2:1 1 RR: 2 Rr: 1rr Phenotype: 1:2:1 1 Red: 2 Pink: 1white

4. Codominance: There is more than one dominate allele (must be at least three alleles for a trait) If together, they are both expressed. Example: Inheritance of Blood Types

Since Blood type is codominant, we use a common letter for all. Use I for dominant A and B, use i for recessive o. Phenotypes Genotypes Type A IAIA, IA io Type B IBIB, IBio Type O io io Type AB IA IB

If a woman is heterozygous for type A Sample problem: If a woman is heterozygous for type A blood and her husband is homozygous for type B, what blood types could their children be? Woman: IA io Man: IBIB IA io AB Bo IB AB Bo

Possible Phenotypes: ½ type AB, ½ type B Possible Genotypes: ½ IAIB, ½ IBio

5. Sex Linked: *The trait is carried on the sex chromosome…X- linked *Must show XX for female XY for male *Only “X” carries the gene

Female, homozygous dominant XAXA Female, recessive XaXa Female, heterozygous, dominant XAXa “Carrier” Male dominant XAY Male, recessive XaY

If a woman is heterozygous for a Sample problem: If a woman is heterozygous for a Reccessive trait, Red-green colorblindness, and her husband is not color blind, What portion of their children could be Color-blind? What portion of their sons? Daughters? Woman: XAXa Man: XAY XAXA XAXa XA XAY XaY Y XA Xa

Possible Phenotypes: ½ normal female Possible Genotypes: ¼ normal males ¼ colorblind males

red flowers is crossed with a plant recessive for white flowers. Practice Problem #1 A plant heterozygous dominant for red flowers is crossed with a plant recessive for white flowers. Give the expected phenotype and genotype for the F1generation. Show the cross.

Practice #2 A strawberry plant that is heterozygous for red, seeded fruits is crossed with a strawberry plant that produces pink, seedless fruits. What is the expected genotype and phenotype of the F1 generation? Show the cross. What proportion of the strawberry plants will produce red, seedless fruits? (Red is dominant Seeded is dominant)

Practice #3 Coat color in Labs is inherited by incomplete Dominance. Black is dominant, yellow is Recessive and Chocolate is the heterozygous Phenotype. If a yellow lab is crossed with a chocolate lab, what proportion of the offspring will be chocolate? Show your cross.

Homework *Read chapter 13 *Rest of problems on sheet and textbook assigned Pg. 277-278 # 3, 7, 8, 9, 13,14,17-21