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Berkeley MCELLBI 140 - Lecture Notes

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LECTURE 2: MENDEL’S LAWS and EXTENSIONS TO MENDELReading: Ch. 2, p. 13-28; Ch. 3, p. 43-53Supplemental Reading: Mendel’s 1865 paper (use the ESP link off the 140 website)Problems: Ch. 2, solved problems I, II; 2-4, 2-5, 2-7, 2-8, 2-13, 2-14, 2-16 – 2-20; Ch. 3, 3-2 –3-5, 3-7 – 3-9, 3-11, 3-13, 3-14Website: http://mcb.berkeley.edu/courses/mcb140/In the late 1850’s when Mendel was making his discoveries, no one had yet describedmeiosis or genes or even chromosomes. Two theories of inheritance were prevalent at the time;(1) blended inheritance, in which traits from the parents become mixed and forever changed and(2) “uni-parental” inheritance, in which one parent contributes most to the inherited features ofprogeny. Mendel performed a very careful, controlled, and quantitative analysis to examine therules of inheritance. In his own words, he sought “to deduce the law according to which [traits]appear in successive generations.Why did Mendel succeed?- He used an experimentally tractable system, Pisum sativum (garden peas). Peas have shortgeneration time and can be self- or cross-fertilized.- He studied easily scorable (unambiguous) traits (discrete vs continuous traits).- He used pure-breeding lines.- He carefully controlled his matings, so that he could be sure that the progeny observed werereally a result of the cross he performed. He realized that “accidental” pollinations wouldcompletely confuse his analyses.- Quantitative analysis! Mendel worked with large numbers of plants and used statistics toanalyze the data.The monohybrid crossA cross between plants that exhibit differences in one distinct trait (seed color, seed shape,flower color, etc), i.e. plants that produce yellow peas x plants that produce green peas.GenerationParental (P) yellow x green (cross-fertilize)First filial (F1) all yellow (self-fertilize)Second filial (F2) 3 yellow: 1 greenWhat did Mendel deduce from these crosses?- “Blending” did not occur. Green peas were recovered in the F2 generation.- Yellow was dominant to green.- There must be two types of yellow peas, those that breed true (like the yellow P generationplants) and those that carry latent information for green peas (like the F1 plants). Heconfirmed this by selfing some of the F2 plants that gave yellow peas: 1/3 of these breed true(all yellow) and 2/3 gave a 3:1 ratio of yellow to green.- To account for the 3:1 ratio, Mendel deduced that each plant carries two “differentiatingcharacters” (genes), one inherited from the maternal parent and one from the paternal parent.Each gene can have alternative forms (alleles); if the two alleles are the same, the parentaltype results and if the two alleles are different, a hybrid type results.LAW OF SEGREGATION: “The two alleles for each trait segregate during gamete formation,then unite at random, one from each parent, at fertilization”.To represent the monohybrid cross a different way:Genotype PhenotypeP YY x yy yellow x greenF1 Yy yellowF2 1 YY: 2 Yy: 1 yy 3 yellow: 1 greenThe Punnett Square allows us to visualize a cross by examining the possible combinations ofgametes from the parents. A cross between two Yy heterozygotes:YyYYY(yellow)yY(yellow)yYy(yellow)yy(green)The Law of the Product states “The probability of two or more independent events occurringtogether is the product of the probabilities that each event will occur by itself.”-- e.g, probability of yy progeny from a cross of two Yy heterozygotes = 1/2 x 1/2 = 1/4The Law of the Sum states “The probability of either of two such mutually exclusive eventsoccurring is the sum of their individual probabilities”.-- What’s the probability of yellow progeny from a cross of two Yy heterozygotes?Probability of YY + Yy + yY = 1/4 + 1/4 + 1/4 = 3/4THE LAW OF INDEPENDENT ASSORTMENT: “During gamete formation, different pairsof alleles segregate independently of one another.”The dihybrid cross:Generation Phenotype GenotypeP yellow; round x green; wrinkled YYRR x yyrrF1 all yellow; round YyRrF2 yellow; round (9 parental type) Y-R-yellow; wrinkled (3 recombinant type) Y-rrgreen; round (3 recombinant type) yyR-green; wrinkled (1 parental type) yyrrThe 9:3:3:1 ratio observed derives from two separate 3:1 phenotypic ratios: the ratio of yellow togreen is 12:4 (or 3:1) and of round to wrinkled is 12:4 (or 3:1).How could you determine the genotype of the yellow; round plants? You could perform a TestCross by crossing an individual showing a dominant phenotype (Y-R-) to an individual with therecessive phenotype (yyrr) to reveal the genotype behind the dominant phenotype. The test crossis a very important tool for the geneticist; this is especially true when analyzing inheritance inorganisms that cannot self-fertilize!If Y-R- is YYRR, then all test cross progeny will be yellow; round.If Y-R- is YYRr, then half will be yellow; round and half will be yellow; wrinkled.If Y-R- is YyRR, then half will be yellow; round and half will be green; round.If Y-R- is YyRr, then all four combinations occur with equal probability (1:1:1:1).Bottom line of Independent Assortment:During Meiosis I, different alleles of two genes on different chromosomes will move to oppositepoles independently of one another.- Only true for genes that lie on different chromosomes or for genes that lie very far apart onthe same chromosome.- Genes on the same chromosome exhibit “linkage”, that is they tend to assort together. We’lldiscuss linkage in more detail next week.EXTENSIONS TO MENDELIn determining the laws of inheritance, Mendel used a set of guidelines: (1) one of the two allelesof a given gene showed complete dominance over the other, (2) there are only two alleles of anygiven gene, (3) genes determine one specific trait, and (4) all genotypes are equally viable. Let’slook at some exceptions to these general “rules”.What if the alleles show incomplete dominance? If this is the case, a novel phenotype, unlikethat of either parent shows up in the F1. Color in snapdragons is an excellent example.P: pure-breeding red flowers x pure-breeding white flowersF1: all pink -- NEW PHENOTYPE (NOT LIKE EITHER PARENT!)F2: red:pink:white (1:2:1). This ratio is a good indication of the lack of complete dominance.The phenotypic ratios are an exact reflection of the genotypic ratios. Molecularly, one can thinkof the combinations as follows: CR CR gives 2 doses of gene expression and CW CW gives nodose. CR CW has one dose of gene


Berkeley MCELLBI 140 - Lecture Notes

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