11 EXPERIMENT P Generation (true-breeding parents) Purple flowers White FigEXPERIMENTP Generation(true-breedingparents)PurpleflowersWhiteflowersFigure 14.3 When F1 hybrid pea plants are allowed to self-pollinate, which traits appear in the F2 generation?
12 F1 Generation (hybrids) FigEXPERIMENTP Generation(true-breedingparents)PurpleflowersWhiteflowersF1 Generation(hybrids)Figure 14.3 When F1 hybrid pea plants are allowed to self-pollinate, which traits appear in the F2 generation?All plants hadpurple flowers
13 EXPERIMENT P Generation (true-breeding parents) Purple flowers White FigEXPERIMENTP Generation(true-breedingparents)PurpleflowersWhiteflowersF1 Generation(hybrids)Figure 14.3 When F1 hybrid pea plants are allowed to self-pollinate, which traits appear in the F2 generation?All plants hadpurple flowersF2 Generation705 purple-floweredplants224 white-floweredplants
30 What does that mean in terms of meiosis? Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes.So a AaBb parent can produce gametes:AB, aB, Ab, ab
31 Predicted outcomes of heterozygous dihybrid cross: AaBb x AaBb 9 both dominant traits 3 one dominant, one recessive trait 3 other dominant, other recessive trait 1 both recessive traits
32 EXPERIMENT RESULTS Fig. 14-8 P Generation F1 Generation Hypothesis of YYRRyyrrGametesYRyrF1 GenerationYyRrHypothesis ofdependentassortmentHypothesis ofindependentassortmentPredictionsSpermorPredictedoffspring ofF2 generation1/4YR1/4Yr1/4yR1/4yrSpermFigure 14.8 Do the alleles for one character assort into gametes dependently or independently of the alleles for a different character?1/2YR1/2yr1/4YRYYRRYYRrYyRRYyRr1/2YRYYRRYyRr1/4YrEggsYYRrYYrrYyRrYyrrEggs1/2yrYyRryyrr1/4yRYyRRYyRryyRRyyRr3/41/41/4yrPhenotypic ratio 3:1YyRrYyrryyRryyrr9/163/163/161/16Phenotypic ratio 9:3:3:1RESULTS31510810132Phenotypic ratio approximately 9:3:3:1
33 Fig. 14-UN11REMEMBER!Each gamete has tohave one of each allele!
36 The laws of probability govern Mendelian inheritance Probability of an event occurring = expected over possibilitiesEX: Probability of throwing a tail with a coin is?1/2Segregation in a heterozygous plant is like flipping a coin: Each gamete has a 1/2 chance of carrying the dominant allele and a 1/2 chance of carrying the recessive allele
37 1/2 1/2 1/2 1/4 1/4 1/2 1/4 1/4 Rr Rr Segregation of Segregation of Fig. 14-9RrRrSegregation ofalleles into eggsSegregation ofalleles into spermSperm1/2R1/2rFigure 14.9 Segregation of alleles and fertilization as chance eventsRR1/2RRr1/41/4Eggsrr1/2Rrr1/41/4
38 Mendel’s laws of segregation and independent assortment reflect the rules of probability The multiplication rule states that the probability that two or more independent events will occur together is the product of their individual probabilities
39 What is the probability of getting an F2 (Aa x Aa) offspring that is homozygous (aa) for both traits?½ x ½ = 1/4
41 Using the addition rule: In the cross of Rr x Rr, what is theprobability of Rr in offspring?If sperm is R and egg is r = ½ x ½ = ¼If sperm is r and egg is R = ½ x ½ = ¼So adding those together is ¼ + ¼ = 1/2
42 Solving Complex Genetics Problems with the Rules of Probability We can apply the multiplication and addition rules to predict the outcome of crosses involving multiple charactersA dihybrid or other multicharacter cross is equivalent to two or more independent monohybrid crosses occurring simultaneouslyIn calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied together
43 ExamplesAaBb X AabbWhat is probability of obtaining a Aabb offspring?Aa = ½bb = ½Answer = 1/4
44 Probability of obtaining AabbCc? AaBbCc x aabbCcProbability of obtaining AabbCc?For Aa from Aa x aa = ½For bb from Bb x bb = ½For Cc from Cc x Cc = ½The answer is ½ x ½ x ½ = 1/8
45 AaBbcc x AabbCC a) Probability of offspring that are A_B_C_? A_ = ¾, B = ½, C = 1/1 answer 3/8b) A_bbC_aaB_C_A_B_C_c) aabbC_
56 Multiple AllelesMost genes exist in populations in more than two allelic formsFor 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: A, B, AB, O. O is recessive.The enzyme encoded by the A allele adds the A carbohydrate, whereas the enzyme encoded by the B allele adds the B carbohydrate; the enzyme encoded by the O allele adds neither
57 You may see IA , IB or i for the alleles. FigAlleleCarbohydrateYou may see IA , IB or i for the alleles.AABBOnone(a) The three alleles for the ABO blood groupsand their associated carbohydratesRed blood cellappearancePhenotype(blood group)GenotypeFigure Multiple alleles for the ABO blood groupsAA or AOABB or BOBABABOOO(b) Blood group genotypes and phenotypes
58 If Mary has A blood and Ben has B blood, can they have a child with O blood?
64 FigAaBbCcAaBbCcSperm1/81/81/81/81/81/81/81/81/81/81/8Figure A simplified model for polygenic inheritance of skin color1/8Eggs1/81/81/81/8Phenotypes:1/646/6415/6420/6415/646/641/64Number ofdark-skin alleles:123456
70 What is the dominant trait? ____________________________ EXPERIMENTAL DATAIn a cross of true-breeding green seed coat peas with true-breeding yellow seed coat peas, all of the F1 peas had green-seed coats. In the F2, you counted 122 green ones and 33 yellow seeded peas. The total number of individuals you counted, N, is 155.What is the dominant trait? ____________________________What is your hypothesis for the F2 population? Show a Punnett square to show the proposed results. Fill in the tables below with your observed data, calculate the expected result according to your hypothesis. Determine the Χ2 value for each experiment, and use the table of probabilities to accept or reject the null hypotheses.Monohybrid crossPhenotypes Observed Expected d = o –e d d2/e Total (X2 )_____ X2 = _____ degrees of freedom: _____Range of probability: ______________Accept or reject null hypothesis? __________
71 Experimental Data for Chi Square Determination In a cross of true-breeding rough seed coat peas with true-breeding smooth seed coat peas, all of the F1 peas had rough-seed coats. In the F2, you counted 79 rough coats and 33 smooth coats. The total number of individuals you counted is 112.What is the dominant trait? ____________________________What is your hypothesis for the F2 population? Show a Punnett square to show the proposed results.
72 Fill in the tables below with your observed data, calculate the expected result according to your hypothesis. Determine the Χ2 value for each experiment, and use the table of probabilities to accept or reject the null hypotheses.Monohybrid crossPhenotypes Observed (o) Expected (e) d = o – e d d2/eRoughSmoothTotal _____Χ2 = _____ degrees of freedom: _____Range of probability: ______________Accept or reject null hypothesis? __________