Lecture 11 SLIDES

# Still 31 ratio for each character f1 generation

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Unformatted text preview: ion, four phenotype combinations result in ratio of 9:3:3:1 Note! still 3:1 ratio for each character! F1 Generation Hypothesis of dependent assortment YyRr Hypothesis of independent assortment Sperm Sperm 1⁄ 2 1⁄ 4 “Each pair of alleles segregates independently of other pairs of alleles during gamete formation” YR 1⁄ 4 Yr 1⁄ 4 yR 1⁄ 4 yr YR 1⁄ 2 yr 1⁄ 4 Eggs YR YYRR Yr YYRr YyRR YyRr F2 Generation (predicted offspring) Eggs 1⁄ 2 YR YYRR YyRr 1⁄ 4 YYrr YYrr YyRr Yyrr 1⁄ 2 yr YyRr 3⁄ 1⁄ yyrr 1⁄ 4 yR YyRR yr YyRr 9⁄ 16 3⁄ 16 YyRr yyRR yyRr 4 4 1⁄ 4 Phenotypic ratio 3:1 Yyrr 3⁄ 16 yyRr 1⁄ yyrr 16 e.g. in previous example, factors for pea color traits segregated independently of factors for pea shape traits. Phenotypic ratio 9:3:3:1 315 108 101 32 Phenotypic ratio approximately 9:3:3:1 This sounds familiar from what you already learned about meiosis! Independent Assortment contributes to genetic variation Homologous pairs of chromosomes orient randomly during meiosis I. Each chromosome pair sorts independently of other homologous pairs. Possibility 1 Possibility 2 Mendel’s Laws Reflect Rules of Probability for specific combination of outcomes ! The Multiplication Rule The probability that independent events will occur together is Two probable arrangements of chromosomes at metaphase I product of their individual probabilities. P2 Ex: Probability offspring will be P1 p2 ? 1/ 2 1⁄ 2 P1 1⁄ 2 Sperm 1⁄ 2 p1 r R R R 1⁄ 4 R R r 1⁄ 4 Eggs r (P1) X 1/2 (p2 ) = 1/4 P1 p2 1⁄ 2 r R 1⁄ 4 r r 1⁄ 4 p2 Combination 1 Combination 2 Combination 3 Combination 4 Figure 13.10 Figure 14.9 25% probability of heterozygous outcome Mendel’s Laws Reflect Rules of Probability for exclusive outcomes ! Solving Complex Genetics with the Rules of Probability The Rule of Addition The probability that exclusive events occur is the sum of individual probabilities. 1⁄ 2 P1 R Sperm 1⁄ 2 p1 r Review in Textbook and in Sections Example: Q: PpYy x Ppyy P2 Ex: Probability offspring will be heterozygous ? 1/ 4 1⁄ 2 R R 1⁄ 4 R R r 1⁄ 4 Eggs r 1⁄ 2 Probability of recessive phenotypes for both characters? Probabilty of pp 1/2 x 1/2 = 1/4 Probability of yy 1/2 x 2/2 = 2/4 Probability of ppyy 1/4 x 2/4 = 2/16 (P1 p2 ) + 1/4 (p1 P2 ) = 2/4 P p r R 1⁄ 4 r r 1⁄ 4 A: p2 50% probability of heterozygous outcome Figure 14.9 Exceptions to Mendelian Inheritance • Linkage Exceptions to Mendelian Inheritance P Generation YYRR yyrr Gametes YR ! yr • Linkage • Incomplete Dominance • Codominance • Multiple A...
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