ZOL 141 1st Edition Lecture 10Outline of Last Lecture I. Gregor Johann MendelA. Study of single traitsB. Experimental setupC. ConclusionsD. Mendel’s first lawOutline of Current Lecture I. Experimental SetupII. Experiments and ResultsIII. Branch Diagrama. Probability RulesIV. ProbabilityCurrent LectureTransmission of Genes from Generation to Generation: Part TwoQuestion addressed: does the inheritance of one characteristic affect the inheritance of a different characteristic?Experimental Setup- From previous crosses with peas:o Smooth is dominant to wrinkledo Yellow is dominant to green- Symbolso Seed shape: S=smooth, s=wrinkledo Seed color: Y=yellow, y=greenExperiments and Results- Smooth yellow x wrinkled greeno all smooth yellow- Dihybrid crossed smooth yellow x smooth yellowo 9/16 smooth yellow, 3/16 wrinkled yellow, 3/16 smooth green. 1/16 wrinkled greeno 9:3:3:1…both parental phenotypes appeared + two new phenotypes o traits are not linked together- conclusiono alleles of one gene pair assort into gametes independently of alleles belonging to other gene pairso principle of independent assortment Mendel’s second law- Mendel looked at all traits individually: seed shape, seed colorDetermining types of gametes- Pea plant has genotype SsYy. What types of gametes can this plant produce?o Each gamete has to receive one copy of each geneo Equal chance that either allele is included in the gameteo Number of gametes is 2^n, n=the number of heterozygous traitsBranch Diagram- Use probability to consider the inheritance of multiple traits simultaneously by combining the individual probabilities of each trait- Probabilityo Rule 1: Product rule-the probability of independent events occurring together is the product of the probabilities of the individual events Two dice: what is probability of rolling a four and a five? 1/6 x 1/6, so probability of rolling a four with the first AND a five with the second: 1/36o Rule 2: Sum rule- the probability of either of two mutually exclusice events occurring is the sum of their individual probabilities One die: what is probability of rolling a four OR a five 1/6+1/6= 2/6 or 1/3- steps for branch diagram--genotypeo 1. Visualize the dihybrid cross as two monohybrid crosseso 2. List probability of genotype for one traito 3. List probability of each genotype for the second trait next to each of the genotypes for the first traito 4. Follow each branch and multiply probabilities (product rule)- Steps for branch diagram-phenotypeo 1. Visualize dihybrid cross as two monohybrido 2. listprob of each phenotype for one traito 3. listprob of each phenotype for the second trait next to each of the phenotypes for the first traito 4. follow each branch and multiply probabilities Probability- four traits- what is prob of having genotype AaBBccDd?o i.e. Aa and BB and cc and Ddo Aa=1/2 and BB=1/4 and cc=1/4 and Dd=1/2o So ½ x ¼ x 1/4 x