Genotypes & Phenotypes

1. Genotypes & Phenotypes • Capital letters used for dominant allele • lowercase of the same letter used for recessive • ex. round seeds are dominant over wrinkled. • R = round, r = wrinkled Genotype • the alleles for an individual, ex. for seeds: - RR (homozygous round) - rr (homozygous wrinkled) - Rr (heterozygous round) Phenotype • the expressed trait for an individual (what is seen) ex. for seeds: - round - wrinkled • Heterozygous genotypes (Rr) always have the phenotype of the dominant allele.

2. P Generation Tall short True-breeding parents Tall = dominant = TShort = recessive = t TT tt F1 Generation Hybrid offspringAll heterozygous all Tall all Tt F2 Generation Genotypes1 TT : 2 Tt : 1 tt Phenotypes3 Tall :1 short Monohybrid crossfrom F1 generation short Tall Tall Tall tt TT Tt Tt Plant Height Experiments • F1 genotype: - 4 Tt (heterozygous) • F1 phenotype: - 4 tall • F2 genotypes: - 1 TT (homozygous dominant) - 2 Tt (heterozygous) - 1 tt (homozygous recessive) - ratio 1:2:1 • F2 phenotypes: - 3 tall - 1 short - ratio 3:1

3. Probability & Punnet Squares Probability – the likelihood of an outcome Punnett square - used to determine the probabilities of allele combinations when the genotypes of the parents are known

4. Making Punnet Squares

5. Practice Problem #1 • Ageneticist crosses two pea plants. One of the plants is heterozygous for the dominant inflated pea pod trait, and the other plant has constricted pea pods. What would be the expected genotypic and phenotypic proportions of the offspring? Given • inflated pod (I) - dominant • constricted pod (i) - recessive • II or Ii = Inflated pods • ii = constricted pods • heterozygous inflated x constricted = Ii x ii Required • Punnet square • gametes from the cross • expected proportions of F1 plants with genotypes II, Ii, and ii • expected proportions of phenotypes (inflated pods and constricted)

6. Solution • The Ii parent produces gametes I and i. • The ii parent produces gametes i and i. Paraphrase 1/2 of the offspring would be heterozygous (Ii) and 1/2 would be ii. Therefore, 1/2 of the offspring would have inflated pods, and 1/2 would have constricted pods.

7. Solutions #2- 4 2. Solution • The Aa parent produces gametes A and a. • The aa parent produces gametes a and a. Paraphrase • ½ the offspring would be heterozygous (Aa) and ½ would be aa. Therefore, ½ of the offspring would be brown, and ½ would be albino 3. Solution • The DD parent produces gametes D and D. • The Dd parent produces gametes D and d. Paraphrase • ½ of the offspring would be heterozygous (Aa) and ½ would be homozygous (AA). Therefore, all of the offspring would have dimples. 4. Solution • The Tt parents produce gametes T and t. Paraphrase • ½ of the offspring would be heterozygous (Tt), ¼ would be homozygous dominant (TT) and ¼ would be homozygous recessive (tt). Therefore, ¾ of the offspring would taste PTC, and ¼ would not.

8. Test Crosses • how genotype can be determined from phenotype • breed an individual with the dominant phenotype with a homozygous recessive • If offspring are all dominant phenotype, the parent was homozygous dominant • If offspring are half dominant phenotype, and half recessive phenotype, the parent was heterozygous dominant.

9. Dihybrid Crosses • a cross between two individuals that differ in two traits • Mendel used to determine whether or not the inheritance of one characteristic influenced the inheritance of another • ex: Does pea shape influence the inheritance of pea colour? Method: • choose plants homozygous for 2 traits (colour & shape) • crossbred 2 different ones to get dihybrid F1 generation • ex. round yellow seeds x wrinkled green • crossed two dihybrid F1 to get F2 (dihybrid cross) Results: • all F1 were round yellow • F2 had phenotypic ratio of 9:3:3:1

10. F1 Generation F2 Generation P Generation

11. Dihybrid Crosses State allele symbols Follow steps below

12. Practice Problem #1 • In mice, the normal long-tail phenotype is dominant to the short-tail trait, and black coat colour is dominant to brown coat colour. If two long-tailed black mice, heterozygous for both traits, are mated, what proportion of their offspring will be brown with short tails? Given • Long tail (L) is dominant. Short tail (l) is recessive. • LL or Ll - long tails, ll - short tails. • Black coat colour (B) is dominant. Brown coat colour (b) is recessive. • BB or Bb - black coat colour, bb - brown coat colour. • The cross is BbLl X BbLl Required • Determine the gametes using FOIL. • make a Punnett square • find the proportion of offspring that will be brown with short tails

13. Solution • The BbLl parents produce gametes BL, Bl, bL, bl. Paraphrase Mice with brown coats and short tails have the genotype bbll. 1/16 of the offspring will be brown with short tails.

14. Law of Independent Assortment • Mendel found that this phenotypic ratio (9:3:3:1) was always found in the offspring of such dihybrid crosses. • law states that the inheritance of alleles for one trait does not affect the inheritance of alleles for another trait. • Different pairs of alleles are passed to the offspring independently of each other • laws apply to all living things, not just plants • many, but not all, traits follow Mendel’s laws