1. Mendelian genetics It’s all about jargon, ratios, and nomenclature

3. fig. 2-3 advantages of peas: self pollination and cross-pollination - chose characters with only two states, or phenotypes - created pure-bred lines Mendel’s peas

4. figs. 2-4 and 2-5 -cross-pollinated parental generation (P) to create first filial (F1) generation -also did reciprocal crosses first filial (F1) generation were all the same

5. self-crossed F1 to obtain second filial (F2) generation – ‘missing’ type reappeared – all progeny in 3:1 ratio deduced presence of dominant and recessive traits

6. Mendel’s postulates 1. there are hereditary ‘particles’ that determine traits (genes) 2. genes are in pairs (alleles) 3. members of a gene pair segregate equally into the gametes 4. each gamete carries only one allele 5. gametes combine at random with respect to allele type heterozygotes have different alleles homozygotes have same alleles

7. a bit of notation… dominant trait takes capital letter; recessive is lower case A/a letter is determined by phenotype of the dominant trait yellow peas are dominant, therefore Y/y

8. a bit of notation… dominant trait takes capital letter; recessive is lower case A/a letter is determined by phenotype of the dominant trait yellow peas are dominant, therefore Y/y Alternative notation wild type trait is +; mutant is -a + /a - ora’/a dominant genotype (when we don’t know what the second allele is) can be abbreviated a + /– this includes a + /a + and a + /a - genotypes

9. What causes dominance? Genes code for proteins Proteins are either structural or functional (enzymes) An ‘error’ in the genetic code may yield a non-functional enzyme, or no protein, or a less efficient enzyme

10. Figure 2-6 Punnett square is used to work out progeny of a cross P Gametes F 1 Overall F 2 ratio

11. phenotypic ratio yellow green Y y yellow Y/Y Y/y Y green Y/y y/y y ratio = 3 yellow : 1 green genotypic ratio yellow green Y y yellow Y/Y Y/y Y green Y/y y/y y ratio = 1 YY: 2 Y/y : 1 y/y note: Y/y = y/Y

13. Probability rules: Multiplication rule: the probability of two independent outcomes occurring simultaneously is equal to the product of each of the two outcomes taken separately. independent probabilities can be multiplied what is the probability of a litter of three being all male?

14. Probability rules: Multiplication rule: the probability of two independent outcomes occurring simultaneously is equal to the product of each of the two outcomes taken separately. independent probabilities can be multiplied what is the probability of a litter of three being all male? probability than any one offspring will be male = ½ probability that all three will be male – ½ x ½ x ½ = 1/8

15. Probability rules: Addition rule: the overall probability of any combination of mutually exclusive outcomes is equal to the sum of the probabilities of the outcomes taken separately. mutually exclusive probabilities can be added What is the probability that a litter of three will have at least one female? 0 females = ½ x ½ x ½ = 1/8 1 female= ½ x ½ x ½ = 1/8 x 3 = 3/8 2 females= ½ x ½ x ½ = 1/8 x 3 = 3/8 3 females = ½ x ½ x ½ = 1/8 sum of probabilities containing a female = 7/8

16. dihybrid cross (more ratios) F1 – round, yellow X wrinkled, green = R/R, Y/Y X r/r, y/y round, yellow R/r, Y/y

17. dihybrid cross (more ratios) R/YR/yr/Yr/y R/Y R/y r/Y r/y R/r, Y/y R/r, Y/y

18. dihybrid cross (more ratios) R/YR/yr/Yr/y R/Y R/y r/Y r/y R/r, Y/y R/r, Y/y R/R:Y/Y R/R:Y/y R/r:Y/Y R/r:Y/y R/R:Y/y R/R,y/y R/r:Y/y R/r:y/y R/r:Y/Y R/r:Y/y r/r:Y/Y r/r:Y/y R/r: Y/y R/r,y/y r/r:Y/y r/r:y/y round, yellow round, green wrinkled, yellow wrinkled, green

19. dihybrid cross (more ratios) R/YR/yr/Yr/y R/Y R/y r/Y r/y R/r, Y/y R/r, Y/y R/R:Y/Y R/R:Y/y R/r:Y/Y R/r:Y/y R/R:Y/y R/R,y/y R/r:Y/y R/r:y/y R/r:Y/Y R/r:Y/y r/r:Y/Y r/r:Y/y R/r: Y/y R/r,y/y r/r:Y/y r/r:y/y round, yellow round, green wrinkled, yellow wrinkled, green phenotype ratio: 9 round, yellow 3 round, green 3 wrinkled, yellow 1 wrinkled, green

20. Fig. 2-10

21. Gene interactions one gene may affect several phenotypes (pleiotropy) - PKU (phenylketonuria): single gene for enzyme phenylalaninephenylketonuria hydroxylase) that converts phenylalanine to tyrosine;phenylalaninetyrosine - loss of function results in mental retardation, lower pigmentation, additional traits several genes may produce the same phenotype - importance of duplicate genes

22. incomplete dominance – heterozygote has intermediate phenotype RR = red flower Rr = pink flower rr = white flower usually found when phenotype is a continuous trait (quantitative) e.g., weight, height, fecundity, amount of enzyme produced alternative is discrete traits e.g., round vs. wrinkled, yellow vs. green peas Gene interactions

23. co-dominance – intermediate phenotype is formed when two dominant alleles are present in heterozygote - characterized by having three phenotypes genotypeblood type A/A and A/i A produces A antigen B/B and B/i Bproduces B antigen i/i Oproduce neither antigen A/B ABproduce both antigens Gene interactions

24. Note that terms are somewhat arbitrary: depend on level of analysis RR = red flower Rr = pink flower rr = white flower Incomplete dominance if R allele produces pigment, r does not Co-dominance if R produces red pigment, r produces white pigment A better example is roan cattle: RR – red hair rr – white hair Rr – red AND white hairs Gene interactions

25. epistasis – two or more genes interact to form a phenotype genotype flower color 9 W/-, M/- blue 9 3 W/-, m/m magenta 3 3 w/w, M/- white 1 w/w, m/m white 4 Fig. 4-13 Gene interactions

26. sickle-cell anemia: - single gene mutation (SNP) -affects hemoglobin configuration, which distorts RBCs -malaria parasite cannot digest altered hemoglobin genotypephenotype A/Ano anemia; normal blood cells A/Sno anemia; sickle only with low O 2 S/Ssevere or fatal anemia with sickle cells sickle cell anemia malaria Gene interactions

27. sickle-cell anemia: - single gene mutation (SNP) -affects hemoglobin configuration, which distorts RBCs -malaria parasite cannot digest altered hemoglobin genotypephenotype A/Ano anemia; normal blood cells A/Sno anemia; sickle only with low O 2 S/Ssevere or fatal anemia with sickle cells dominant:reccessive for expression of anemia incomplete dominance for cell shape codominance for production of hemoglobin

28. ratios so far: 1:3 (phenotypic result of a mono-hybrid cross in dominance : recessive traits) 1:2:1 (genotypic result of a mono-hybrid cross) and phenotypic result for incomplete dominant traits 9:3:3:1 (dihybrid cross in dominance : recessive traits) 1:2 if lethal allele is present A/A ¼ A/a 2/4 a/a ¼ - but they are all dead 15:1 (dihybrid cross with duplicate genes)

29. penetrance = whether genotype is expressed in phenotype due to modifiers, epistatic genes, suppression expressivity = degree to which genotype is expressed in the phenotype due to other alleles, or environment Fig. 4-23

30. was Mendel honest? all dominant: recessive traits only two alleles at each gene no gene interactions no sex-linked traits all independent traits (no linkage; pea has 7 chromosomes)

31. How do we evaluate all this???? (how close must the data be to the ratios?) Chi square = Χ 2 = (observed – expected) 2 expected  Monohybrid cross with incomplete dominance: dev. from exp. (obs-exp) 2 phenotype observedexpected (obs-exp) exp red 19 25-6 1.44 pink 57 50 70.98 white 24 25-10.04 total100 1002.46 = Χ 2 degrees of freedom = 2 (= N – 1)

32. p = probability of obtaining the statistic by random chance

33. Χ 2 = 2.46