Transcript Chapter 2
Chapter 2 classical Mendelian genetics © 2006 Jones & Bartlett Publishers Gregor Mendel monk gardener careful observer experimenter http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://biology.clc.uc.edu/Fankhauser/Travel/Berlin/for_web/Mendel_in_Brno.html Mendel at that time (1850’s-1860’s) it was thought that the traits passed on by the parents were blended in the offspring pink red + white = ? purple + white = purple or white Mendel parents contributed to offspring factors remained unchanged factors set out to trace their movements looked at phenotypeappearance looked at ratio’s Mendel used peas access to different varieties usually self-pollinated could manipulate pollinization female male Mendel started with true-breeding plants parent 1 round x parent 1 offspring round round wrinkled x wrinkled wrinkled Mendel started with true-breeding plants seed shape round vs. wrinkled seed color yellow vs. green Mendel seed shape round vs. wrinkled Ppollen Pflower F1 round x wrinkled round wrinkled x round round Fig. 2.2. Reciprocal crosses of true-breeding pea plants © 2006 Jones & Bartlett Publishers Mendel seed shape round vs. wrinkled Ppollen Pflower F1 round x wrinkled round wrinkled x round round dominant vs. recessive hybrids recessive dominant Fig. 2.3. Traits studied in peas by Mendel © 2006 Jones & Bartlett Publishers F1 cross round hybrid X ? round hybrid Fig. 2.7. © 2006 Jones & Bartlett Publishers F1 cross round hybrid X round hybrid each hybrid has two parents therefore each hybrid has two “factors” F1 cross round hybrid X round hybrid To solve: define terms (use consistently) determine parent genotype determine gamete genotype Punnett square To solve: R define terms = round capital letter=dominant ?r = wrinkled lower case letter-recessive R and r are alleles: alternative forms of a gene To solve: determine parent genotypes original cross: round vs. wrinkled (pure-breeding) parents: RR (homozygous) rr To solve: determine gamete genotypes original cross: round vs. wrinkled (pure-breeding) parents: RR rr gametes: R or R r or r Punnett square To solve: R R r Rr Rr r Rr Rr all F1 would be Rr (heterozygous) back to the F1 cross round hybrid X round hybrid parents: Rr Rr gametes: R or r R or r #1 Punnett square R r R RR Rr r Rr rr What would the ratio of round to wrinkled be? Table 2.1. Results of Mendel’s monohybrid experiments © 2006 Jones & Bartlett Publishers Mendel’s first law… …The Law of Segregation The two “factors” in the adult separate from each other during the production of the gametes. #2 homologous pairs of chromosomes separate during meiosis I. A problem: Suppose you have a plant with purple flowers, but unknown ancestry. What is its’ genotype? Is it homozygous dominant or heterozygous? PP Pp 2 P’s or not….. A solution: cross it with homozygous recessive P? x pp do it! A solution: if PP x pp all offspring would be Pp (heterozygous) (purple flowers) if Pp x pp some offspring would be Pp (purple) some offspring would be pp (white) The solution: #3 Geneticists call this a “test cross” (pg. 44) pause for ? Mendel also looked at two traits at once round yellow vs wrinkled vs green RR, rr YY, yy pure-breeding round and yellow pure-breeding green and wrinkled pure-breeding round and yellow pure-breeding green and wrinkled parental genotype RRYY rryy gametes genotype RY ? ry ? ry RY RrYy all heterozygous; phenotype =? all F1 were round, yellow hybrids RrYy (dihybrids) Mendel then did a dihybrid cross (F1 cross) parental genotype RrYy RrYy gametes genotype ? ? split class R Y R y r y r Y metaphase I cell RY ry Ry rY RY RRYY RrYy Ry RRyy RrYy ry rryy rY RrYy rrYY RrYy 3 round, yellow 1 wrinkled, green 1 round, green 2 round, yellow 1 wrinkled, yellow Mendel’s results: (dihybrid cross) round, yellow 315 9 round, green 108 3 wrinkled, yellow 101 3 wrinkled, green 32 1 556 #4 Fig. 2.11. Independent segregation of the Ww and Gg allele pairs © 2006 Jones & Bartlett Publishers Fig. 2.12. Dihybrid cross Mendel’s second law… …The Law of Independent Assortment (in modern language) #5 How a pair of homologous chromosomes align at Metaphase I is independent of how all the other pairs of chromosomes align at metaphase I RY Ry R Y R y r y r Y rY ry metaphase I cell 2.4 probability a fraction (ratio; like 1/4) between 0 will never happen and 1 will always happen yellow yellow Yy x Yy Y y Y YY Yy y Yy yy yellow yellow Yy x Yy What is the probability of getting plants with green seed ? Y y Y YY Yy y Yy yy 2001 green 6022 yellow 2001 (6022 + 2001) = 1 4.01 yellow yellow Yy x Yy Y y Y YY Yy y Yy yy What is the probability of getting plants with yellow peas? YY Yy 1/4 2/4 prob [YY] [YY] ++ prob prob [Yy] [Yy] prob [YY or Yy] = prob = 1/4 + 2/4 = 3/4 #6 R r R RR Rr r Rr rr RrYy x RrYy Y y Y YY Yy y Yy yy What is the probability of getting plants with round, yellow peas? YY or Yy and RR or Rr 3/4 3/4 prob [round and yellow] = prob [round] x prob [yellow] = 3/4 x 3/4 = 9/16 #7 ? 2.6 dominance ? RR same phenotype ? Rr R gene codes for starch branching enzyme 1 (SBEI) dominance is not necessarily all or none Fig. 2.20. Three attributes of phenotype affected by Mendel's alleles W and w, which determine round versus wrinkled seeds. © 2006 Jones & Bartlett Publishers RR and Rr same phenotype in peas, but really different at the molecular level Are there some cases where homozygous dominant (RR) looks different than heterozygous (Rr) ? Fig. 2.21. Incomplete dominance in snapdragons. © 2006 Jones & Bartlett Publishers Incomplete dominance When the heterozygote is intermediate between the homozygous phenotypes Seen with traits that are quantitative (can be measured on a continuous scale) as opposed to a discrete trait (appears to be all or none) Blood typing (ABO) Codominance both traits are expressed Human blood types: A B AB O Blood typing (ABO) three alleles (multiple alleles) IA IB i make “A” carbohydrate make “B” carbohydrate make neither carbohydrate Fig. 2.22. The ABO antigens on the surface of human red blood cells are carbohydrates. Blood typing (ABO) Human blood types: phenotype: A B AB O genotype: IAIA, IAi IBIB, IBi codominant IAIB ii Blood typing (ABO) Our immune system makes proteins called antibodies to attack foreign molecules called antigens. Blood typing (ABO) For someone with type A blood (someone with the IA allele): A carbohydrate “self” B carbohydrate “non-self” or foreign antigen Blood typing (ABO) For someone with type B blood (someone with the IB allele): A carbohydrate “non-self” or foreign antigen B carbohydrate “self” Blood typing (ABO) For someone with type O blood (someone with ii alleles): A carbohydrate “non-self” or foreign antigen B carbohydrate “non-self” or foreign antigen Blood typing (ABO) For someone with type AB blood (someone with IA and IB alleles): A carbohydrate “self” B carbohydrate “self” Blood typing (ABO) What kind of antibodies would they make? Type A B AB O - B antibodies - A antibodies - neither A nor B antibodies - A and B antibodies Table 2.3. Genetic control of the ABO blood groups AB universal recipient O universal donor © 2006 Jones & Bartlett Publishers Fig. 2.23. Antibody against typeA antigen will agglutinate red blood cells carrying the type-A antigen. © 2006 Jones & Bartlett Publishers Other reasons why Mendel rules aren’t always observed incomplete dominance multiple alleles polygenic traits ? variable expressivity same mutation-different results penetrance (phenotype = expected) complete 100% lung incomplete < 100% cancer example 2.7 Epistasis non-allelic genes interacting to affect the same trait e.g., When the expected 9:3:3:1 ratio of a dihybrid cross is altered Merriam Webster: suppression of the effect of a gene by a nonallelic gene 2.7 Epistasis Given: Two different genes C and P for flower color C c purple flowers white flowers P p purple flowers white flowers plants with C- and P- genotypes have purple flowers (wt) plants with cc or pp have white ? flowers 2.7 Epistasis aside: How could that happen? (hypothetically) E1 A E2 B mutation #1 E3 C E4 D mutation #3 purple pigment 2.7 Epistasis CC pp phenotype gamete genotype F1 genotype F1 phenotype x cc PP white flowers Cp cP CcPp all purple F1 cross what gametes ? Fig. 2.24. A cross showing epistasis in the determination of flower color in peas. © 2006 Jones & Bartlett Publishers CP Cp cP cp CP CCPP CCPp CcPP CcPp Cp CCPc CCpp CcPp Ccpp cP CcPP CcPp ccPP ccPp cp CcPp Ccpp ccpp ccPp epistasis When the expected 9:3:3:1 ratio of a dihybrid cross is altered 9:7 ratio purple to white Fig. 2.25. Modified F2 dihybrid ratios. © 2006 Jones & Bartlett Publishers 2.8 Complementation mutations c and p show complementation CCpp white x CcPp purple ccPP white Fig. 2.26. Complementation reveals whether two recessive mutations are alleles of different genes. © 2006 Jones & Bartlett Publishers 2.8 Complementation means that mutations affect different genes Lack of complementation means that mutations affect the same gene Fig. 2.27. Results of complementation tests among six mutant strains of peas. © 2006 Jones & Bartlett Publishers Fig. 2.28. A method for interpreting results of complementation tests.