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.

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 tt P Generation F1 Generation F2 Generation True-breeding parents Tall = dominant = T Short = recessive = t TT Hybrid offspring All heterozygous all Tt Monohybrid cross from F1 generation Tt Tall all Tall short Genotypes 1 TT : 2 Tt : 1 tt Phenotypes 3 Tall : 1 short

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

Making Punnet Squares

Practice Problem #1 A geneticist 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)

Solution 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.

Solutions #2- 4 2. Solution 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. 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. 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. D DD d Dd T t TT Tt tt

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.

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

F2 Generation F1 Generation P Generation

Flower location: A = Axial a = terminal Dihybrid Crosses State allele symbols Follow steps below Step 1. Draw a square with a 4 by 4 grid. 2. Consider all possible gametes produced by the first parent. (Use FOIL) Write the alleles for these gametes across the top of the square. 3. Consider all possible gametes produced by the second parent. Write the alleles for these gametes down the side of the square. 4. Complete the square by writing all possible allele combinations from the cross. 5. Determine the genotypic and phenotypic proportions of the offspring. Example Flower location: A = Axial a = terminal Height: T = Tall t = short

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

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.

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