Genetics

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

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

3. *P - parental generation *F1 – first filial generation *F2 – second filial generation Mendel tracked heritable characters for three generations -Example: F2 P X Tall Dwarf F1 – all Tall Tall

4. *genes and alleles Mendel’s hypotheses (to explain his results) 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 4. The two alleles for each character segregate (separate) during gamete production. P: X Tall Dwarf DD dd F1 – all Tall Tall Mendel’s Law of Segregation Dd

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

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

9. Example: P1 X GgWw GgWw gw GgWw GgWw yellow, round green, wrinkled GgWw GgWw GgWw GgWw gw GGWW ggww GgWw GgWw GgWw GgWw gw F1 GgWw GgWw GgWw GgWw gw All yellow, round Dihybrid cross-A genetic cross between two individuals involving two characters Punnett square and the law of independent assortment: GW GW GW GW GgWw

10. F1 F1 X GGWW GGWw GgWW GgWw GW All yellow, round All yellow, round Gw GGWw GGww GgWw Ggww GgWw GgWw ggWW ggWw gW GgWW GgWw gw 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 F2 9/16 yellow, round 1/16 GGww, 2/16 Ggww 3/16 yellow, wrinkled 1/16 ggWw, 2/16 ggWw 3/16 green, round 1/16 ggww 1/16 green, wrinkled Punnett square and the law of independent assortment: GW Gw gW gw

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. 1) establish gene symbols: A=normal (not albino) a=albino A a A AA Aa Aa aa a 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. 2) Establish: genotypephenotype AA normal Aa normal aa albino Move on to part (A): Parents are both phenotypically normal, genotypes could be EITHERAA 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. Answer to (A): Genotype of the parents = Aa Genotype of normal children = AA or Aa Genotype of albino child = aa

13. A A a A a a Aa Aa Aa aa Aa Aa aa Aa a a 2. (B) A normal male and an albino female have six children, all normal. • The female is phenotypically albino; genotype can only be aa • 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 4) BUT male COULD also be Aa ! *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 F1: Genotype: ½ Aa, ½ aa or a 1:1 ratio Phenotype: ½ normal, ½ albino or a 1:1 ratio F1: Genotype: all Aa Phenotype: all normal