LECTURE 2MENDELIAN GENETICS IIDIHYBRID CROSS, INDEPENDENT ASSORTMENT, FORKED-LINE METHOD, PEDIGREESAUGUST 22, 2014Reading: Chapter 3.3, 3.4, 3.6, 3.7. Chapter 1.1, 1.3.Homework: Chapter 3 Q 7, 8, 9, 12, 13, 16, 26, 28, 30. This homework is to be turned in to the teaching assistants during discussion the week of Aug 25. A job nicely done and delivered on time will receive 15 points.A. Follow-up from Monohybrid cross Five hypotheses underlying Mendelian genetics:A trait is determined by a geneThe gene exists in multiple discrete variants (alleles)One individual has two copies of the gene, two alleles (diploid genotype)The two alleles together determine the trait (phenotype)Each parent randomly passes one allele to progeny via the (haploid) gametePunnett squareTestcross: Testing predictions made by Mendel's hypothesis.Cross a plant of unknown genotype (RR or Rr) to a homozygous recessive (rr) 'tester'.i) to confirm that F1s are heterozygousii) confirm that the 3:1 phenotype ratio in the F2 has an underlying 1:2:1 genotype ratio.Five steps to solving a Transmission Genetics problem. determine:Phenotype of parentsGenotype of parentsGenotypes of gametesGenotypes of progenyPhenotype of progenyB. Scientific Method Observation –> Hypothesis –> Prediction -> Experimental test -> New observationi) Confirm hypothesis ORii) Does not confirm (refutes) hypothesis -> Infer new hypothesisQ: Describe how Mendel followed the scientific method.Model: Usually a set of hypotheses (often drawn as a figure)Law/Theory: A hypothesis/model that has been validated over and over again; often using many different approaches; broadly accepted as true by most that use logical reasoning.C. Dihybrid cross – Mendel’s 2nd law Aa; Bb x Aa; Bb1. Mendel’s 2nd law states that alleles from multiple genes reassort independently.-> 'Occam's razor' = The logical concept to always test the simplest possible hypothesis first. 2. 1:1:1:1 segregation of allele combinations among F1 gametes.3. 9:3:3:1 phenotypic reassortment in the F2. 4. The F2 has 2 allele combinations that are parental; and 2 others that are recombinant.-> We now know that there is an important exception to this law – linkage.Q: Given YY; rr x yy; RR. Our hypothesis was that the genes for seed color (Y/y) and seed shape (R/r) will segregate independently. This led us to predict phenotypic ratios of 9:3:3:1. The data were consistent with this hypothesis. Now, alternatively, we could have started with a different hypothesis: The R allele might be linked to the y allele and r might be linked to Y. What would be the predicted phenotypic ratios in the F2 in this case? Reminder: Start with RR yy x rr YY to derive the F1.Q: Are you keeping straight what is meant by the term gene and what is meant by the term allele? Strictly speaking, a gene is responsible for a trait (e.g. size or eye color; grammatically a noun) and the alleles of thegene determine the specific characteristic in which the trait expresses itself (e.g. small, blue; i.e. often an adjective). However, in reality you will often hear the word 'gene' used instead of 'allele'.D. Using the forked line method or branch diagram to predict phenotypic/genotypic ratios.Application of the ‘product rule’; saves time compared to the Punnett square. Example: When a dihybrid F1 (Yy; Rr) is selfed to derive an F2, the fraction of progeny with the Y and R phenotypes will be 3/4 * 3/4 = 9/16.Alert: This method only works when the two genes in question (Y and R) segregate and reassort independently. The forked line method shines when three or more genes are involved. E. Trihybrid crosses and the 2n rule The 2 n rule: If the number of hybrid genes is given by n, then there are two to-the-power-of n possible gametes.Given n=3How many gametes? 8How many boxes in the Punnett square (F1->F2)? 64How many phenotype entries in the forked line diagram? 8How many genotype entries in the forked line diagram? 27F. Human pedigreesRules of thumb for determining patterns of inheritance:1. Check various hypotheses in the order of likelihood: recessive? -> dominant? -> sex-linkage -> no genetic basis? -> cytoplasmic inheritance? -> other?2. Recessive traitphenotype ‘skips’ generations (as you go back in time, from bottom to top)if two affected individuals have children, 100% are affected. if two carriers have children, ~1/4 of children are affected. * assume that ‘carriers’ are rare in the general population.3. Dominant traitphenotype does not skip generations (going back in time)children of 'affected x healthy' are not necessarily affected (p = 50%)4. Sex (X)-linked recessive traits (more later)mostly observed in malesmostly inherited via heterozygous females5. Assign likely genotypes (AA, Aa, aa, A-, a-)6. Check your hypothesis against ALL individuals of the pedigree.7. Check the pedigree bottom-up and top-down. G. Probabilities in genetic counseling. “Will my child be affected?” (more later)Example: rare recessive trait: 1. What is the probability for the mother to be a carrier?2. What is the probability of the father being a carrier?3. What is the probability of the future child being affected if both parents were carriers? The combined probability will be the product of the three