Presentation on theme: "Genotypes & Phenotypes"— Presentation transcript:
1 Genotypes & Phenotypes Capital letters used for dominant allelelowercase of the same letter used for recessiveex. round seeds are dominant over wrinkled.R = round, r = wrinkledGenotypethe alleles for an individual, ex. for seeds:- RR (homozygous round)- rr (homozygous wrinkled)- Rr (heterozygous round)Phenotypethe expressed trait for an individual (what is seen) ex. for seeds:- round- wrinkledHeterozygous genotypes (Rr) always have the phenotype of the dominant allele.
2 Plant Height Experiments F1 genotype:- 4 Tt (heterozygous)F1 phenotype:- 4 tallF2 genotypes:- 1 TT (homozygous dominant)- 2 Tt (heterozygous)- 1 tt (homozygous recessive)- ratio 1:2:1F2 phenotypes:- 3 tall- 1 short- ratio 3:1ttP GenerationF1 GenerationF2 GenerationTrue-breeding parentsTall = dominant = T Short = recessive = tTTHybrid offspring All heterozygousall TtMonohybrid cross from F1 generationTtTallall TallshortGenotypes 1 TT : 2 Tt : 1 ttPhenotypes 3 Tall : 1 short
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
5 Practice Problem #1A 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?Giveninflated pod (I) - dominantconstricted pod (i) - recessiveII or Ii = Inflated podsii = constricted podsheterozygous inflated x constricted = Ii x iiRequiredPunnet squaregametes from the crossexpected proportions of F1 plants with genotypes II, Ii, and iiexpected proportions of phenotypes (inflated pods and constricted)
6 Solution The Ii parent produces gametes I and i. Paraphrase1/2 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. Paraphrase½ the offspring would be heterozygous (Aa) and ½ would be aa. Therefore, ½ of the offspring would be brown, and ½ would be albino3. SolutionThe 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. SolutionThe 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.DDDdDdTtTTTttt
8 Test Crosses how genotype can be determined from phenotype breed an individual with the dominant phenotype with a homozygous recessiveIf offspring are all dominant phenotype, the parent was homozygous dominantIf offspring are half dominant phenotype, and half recessive phenotype, the parent was heterozygous dominant.
9 Dihybrid Crossesa cross between two individuals that differ in two traitsMendel used to determine whether or not the inheritance of one characteristic influenced the inheritance of anotherex: 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 generationex. round yellow seeds x wrinkled greencrossed two dihybrid F1 to get F2 (dihybrid cross)Results:all F1 were round yellowF2 had phenotypic ratio of 9:3:3:1
11 Flower location: A = Axial a = terminal Dihybrid CrossesState allele symbolsFollow steps belowStep1. 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.ExampleFlower location: A = Axial a = terminalHeight:T = Tall t = short
12 Practice Problem #1In 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?GivenLong 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 BbLlRequiredDetermine the gametes using FOIL.make a Punnett squarefind 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 otherlaws apply to all living things, not just plantsmany, but not all, traits follow Mendel’s laws