1. Genetics *transmission of traits – heredity *variation *genetics

2. Two main hypotheses on how traits were transmitted: *blending inheritance *particulate inheritance Gregor Johann Mendel 1822-1884 The father of transmission genetics:

3. Mendel tracked heritable characters for three generations *P - parental generation *F 1 – first filial generation *F 2 – second filial generation -Example: X TallDwarf P F 1 – all Tall Tall F2F2

4. Mendel’s hypotheses (to explain his results) *genes and alleles 1. Alternative versions of genes (alleles) account for variation in inherited characters 2. For each character, an organism inherits two alleles, one from each parent

5. 3. If two alleles differ, one is dominant, the other recessive X TallDwarf P: DD dd F 1 – all Tall Tall Dd 4. The two alleles for each character segregate (separate) during gamete production. Mendel’s Law of Segregation

6. Punnett Square predicts the results of a genetic cross between individuals of known genotype Ddd TallDwarf P: DD dd X Gamete formation: D *genotype *phenotype *Homozygous *Heterozygous

7. Mendel’s Law of Independent Assortment *What happened when he looked at two characters? If they segregate together:If they segregate independently:

9. Dihybrid cross- A genetic cross between two individuals involving two characters F1F1 All yellow, round GGWWggww Example: P1P1 yellow, roundgreen, wrinkled X Punnett square and the law of independent assortment: GW GW gw GgWw

10. F1F1 All yellow, round GgWw F1F1 All yellow, round GgWw X F2F2 9/16 yellow, round 3/16 yellow, wrinkled 3/16 green, round 1/16 green, wrinkled Punnett square and the law of independent assortment: GW Gw gW gw gW GW gw Gw GGWWGGWwGgWWGgWw GGww GGWw GgWwGgww GgWWGgWw ggWWggWw GgWw Ggww ggWw ggww 9:3:3:1 Phenotypic ratio; Genotypic ratio as follows: 1/16 GGWW, 2/16 GGWw, 2/16 GgWW, 4/16 GgWw 1/16 GGww, 2/16 Ggww 1/16 ggWw, 2/16 ggWw 1/16 ggww

11. Mendelian inheritance is based on probability Example- coin toss *1/2 chance landing heads *Each toss is an independent event *Coin toss, just like the distribution of alleles into gametes *The rule of multiplication – determines the chance that two or more independent events will occur together ½ x ½ = ¼

12. Sample problem Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and offspring for the following families. When two alternative genotypes are possible, list both. (A) Two non albino (normal) parents have five children, four normal and one albino. (B) A normal male and an albino female have six children, all normal. A=normal (not albino) a=albino 1) establish gene symbols:2) Establish: genotypephenotype AAnormal Aanormal aaalbino Move on to part (A): Parents are both phenotypically normal, genotypes could be EITHER AA or Aa, an albino phenotype could only result from an aa genotype. *One a had to come from the mother and one a had to come from the father, so, the parents must be genotypically Aa. A a Aa AA Aa aa Answer to (A): Genotype of the parents = Aa Genotype of normal children = AA or Aa Genotype of albino child = aa

13. 2. (B) A normal male and an albino female have six children, all normal. 1)The female is phenotypically albino; genotype can only be aa 2)The male is phenotypically normal; genotype can be AA or Aa 3) Since all children are normal one might assume the male to be AA AA a a Aa F1: Genotype: all Aa Phenotype: all normal 4) BUT male COULD also be Aa ! a a Aa Aa aa F1: Genotype: ½ Aa, ½ aa or a 1:1 ratio Phenotype: ½ normal, ½ albino or a 1:1 ratio *IF the father was genotypically Aa, then what is the likelihood (chance or probability) of this couple having 6 normal children? Recall the product law! ½ x ½ x ½ x ½ x ½ x ½ = 1/64