Bios 208 1st Edition Lecture 31Outline of Last Lecture I. Gregor MendelII. A Genetic CrossIII. Genotype versus phenotypeIV. Useful Genetic VocabularyOutline of Current Lecture I. Test CrossII. Mendel’s 1st Law of segregationIII. A Dihybrid CrossIV. Law of Independent Assortment: Mendel’s 2nd LawV. Extending Mendelian Genetics for a Single GeneVI. Multiple AllelesVII. EpistasisCurrent LectureI. Test CrossA. Dominant phenotype with unknown genotype; Could be PP (homo. dom.) or Pp (hetero.)B. Cross unknown dominant to pp (homo. rec.)a) if unknown is PP, offspring will be 100% dominant phenotypeb) if unknown is Pp, offspring will be 50% Pp (Dom. Pheno.) and 50% pp (rec. pheno.); 1:1 ratioII. Mendel’s 1st Law of segregationA. Parents F1 F2B. AA x aa > Aa x Aa > ¼ AA , ½ Aa , ¼ aaC. Genotypes 1:2:1D. Phenotypes 3:1E. ¾ A- (Dom.) ¼ aa (rec.)F. During the formation of reproductive cells (gametes), pairs of hereditary factors (genes) for a specific trait separate so that offspring receive one factor from each parent.G. The trait which disappears in the F1 phenotype (white flowers) is not lost; it is masked, and reappears in the F2.H. The laws of probability govern Mendelian inheritanceThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.I. Mendel’s laws of segregation and independent assortment reflect the rules of probabilityJ. When tossing a coin, the outcome of one toss has no impact on the outcome of the nexttossK. In the same way, the alleles of one gene segregate into gametes independently of another gene’s allelesIII. A Dihybrid CrossA. A dihybrid cross (e.g., seed shape and seed color).B. What is the fate of traits that began together in the parents?C. Here, P1 is yellow/round, YYRR; P2 is green/wrinkled, yyrr:D. Remain together? 2) Assort independently?a) YYRR x yyrr 100% YyRrb) P1 P2 F1E. YyRr x YyRr produce F2a) F1 F1a) 9/16 Yellow, Round YYRR Parentalb) 3/16 Yellow, wrinkled YY rr non parentalc) 3/16 green, Round yy RR non parentald) 1/16 green, wrinkled yy rr Parentale) phenotype genotypeF. Mendel identified his second law of inheritance by following two characters at the same timeG. Crossing two true-breeding parents differing in two characters produces dihybrids in the F1 generation, heterozygous for both charactersH. A dihybrid cross, a cross between F1 dihybrids, can determine whether two characters are transmitted to offspring as a package or independentlyIV. Law of Independent Assortment: Mendel’s 2nd LawA. Genes that influence different traits are inherited independently.B. Recombinants (non-parental combinations of traits) appear in the F2 generationC. Inheritance patterns are often more complex than predicted by simple Mendelian geneticsD. The relationship between genotype and phenotype is rarely as simple as in the pea plantcharacters Mendel studiedE. Many heritable characters are not determined by only one gene with two allelesF. However, the basic principles of segregation and independent assortment apply even to more complex patterns of inheritanceV. Extending Mendelian Genetics for a Single GeneA. Inheritance of characters by a single gene may deviate from simple Mendelian patterns in the following situations:a) When alleles are not completely dominant or recessiveb) When a gene has more than two allelesc) When a gene produces multiple phenotypesB. Beyond Mendel:a) incomplete dominanceb) epistasisc) multiple alleles (ABO blood groups)d) co-dominance (ABO blood groups)e) pleiotropyf) polygenic traitsg) environmental influencesVI. Multiple AllelesA. Most genes exist in populations in more than two allelic formsB. For example, the four phenotypes of the ABO blood group in humans are determined by three alleles for the enzyme (I) that attaches A or B carbohydrates to red blood cells: IA, IB, and i.C. The enzyme encoded by the IA allele adds the A carbohydrate, whereas the enzyme encoded by the IB allele adds the B carbohydrate; the enzyme encoded by the i allele adds neitherVII. EpistasisA. In epistasis, a gene at one locus alters the phenotypic expression of a gene at a second locusB. For example, in Labrador retrievers and many other mammals, coat color depends on two genesC. One gene determines the pigment color (with alleles B for black and b for brown)D. The other gene (with alleles C for color and c for no color) determines whether the pigment will be deposited in the hairE. Epistasis: A single trait is influenced by the interaction of 2 or more genes; Coat color in animal fur; Golden retriever dogs.F. BbCc x BbCc >a) 9 black 9/16 B- C-b) 3 chocolate 3/16 bb C-c) 4 yellow 3/16 B- ccd) 1/16 bb ccG. What’s going on?a) Need C- to get any colorb) Also need B- (i.e., B-C-) to get blackc) B and C encode enzymes that are part of a pigment synthesis