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UT Knoxville BIOL 240 - 4.williams_lecture.dihybridcross

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PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 54Slide 55Slide 56Slide 57Slide 581Biology 240 - General GeneticsSpring 2014Dr. Joe WilliamsContact: 974-6202, [email protected], January 20thUnless otherwise indicated, photo credits are from:A.Klug WS, Cummings MR, Spencer CA, Palladino MA, Killian D. 2014. Concepts of Genetics. Boston: Pearson.B.Griffiths, S. Wessler, R. Lewontin, and S. Carroll. 2008. Introduction to Genetic Analysis, Eighth Edition. W. H. Freeman & Company.2Class businessPlease read ALL OF CHAPTER 3 and Chapter 4.1-4.9 (THIS IS DIFFERENT FROM SYLLABUS)First quiz- Thursday at end of class (10 minutes)Quiz will cover chapters 1-3 and all lectures including today’s but NOT Thursday’sobservedphenotypes1:2:1 3:1inferred“genotypes”41. There must be physical particles of inheritance, or genes. 2. Genes must be in pairs (alleles) – one from each parent. One allele can determine the phenotype (dominant vs. recessive alleles3. Each gamete must possess only a single allele.4. Equal segregation – the two alleles in each individual must be passed to gametes in equal proportions.5. Random fertilization –gametes carrying one allele must have the same chance of fertilization as gametes carrying the other allele.Five things that must be true based on Mendel’s observations(each is a testable prediction)Textbook1235phenotypegenotype34Particles in pairsOne to gameteand so to zygoteEqual proportionsEqual chances offertilization6Probability rules:Flip a coin. If the coin is “fair”, equal probability of heads or tails“If P {heads} = P {tails},” then P {heads} = ½Roll the dice. If die is fair, then 1, 2, 3, 4, 5, or 6 are equi-probable“If P {1} = P {2} = … P{6},” thenP {roll a 4}= 1/67Probability rules:“If P {heads} = P {tails}, then P {heads} = ½”“If P {1, 2, 3, 4, 5, or 6} is equi-probable, then P {roll a 4}= 1/6”If probability of an A or an a allele passing to gamete is equal, thenP {sperm with A allele} = ½What is the probability of A allele of sperm and A allele of egg meeting?P {sperm with A allele} = ½P {egg with A allele} = ½8Probability rules:Product rule:If two probabilities are independent, then the probability of them bothoccurring is the product of the two probabilities.What is the probability of A allele of sperm and A allele of egg meeting?P {sperm with A allele} = ½ (no matter what happens with eggs)P {egg with A allele} = ½ (no matter what happens with sperm)So P {zygote = A/A} = ½ x ½ = 1/49Product rule3410If the model is true, then use it to make predictions.Test the predictions and see if they hold.Concept 4: Equal segregation. Predicts that each allele has the same probability of passing to the next generation.Parent is a known heterozygote Point 4If each gamete gets 1 alleleand ½ of gametes carry each allele and fertilization is randomthen only ¼ of offspring willnot inherit a dominant alleleand only ¼ of offspring willinherit both dominant alleles11But how were we sure that the F1 was really a heterozygote?Y/y12XPure parentsXThis is called a testcross:It tests: Parent = heterozygoteIt tests: Equal segregation13XPure parentsX5852Results14Concept # 4: equal segregation =Mendel’s first law: The two members of a gene pair segregate from each other into the gametes;so half the gametes carry one member of the pair and the other half carries the other member.15Concept # 4 (and 5), equal segregation =Mendel’s first law: The two members of a gene pair segregate from each other into the gametes;so half the gametes carry one member of the pair and the other half carries the other member.To see that equal segregation has occurredyou have to also have random fertilization (#5)(but we now know that you usually do)16XPF1F2We now can move from phenotypes tothe underlying genotypesXY/Y y/y Y/yY/Y Y/y Y/y y/yPhenotypesGenotypes17XPF1F2F2 3:1 ratio of yellow to green Phenotypic ratioF2 1:2:1 ratio of Y/Y to Y/y to y/y Genotypic ratioTerminology reviewY/Y y/y Dominant or recessive homozygotesY/y Heterozygote, or F1 hybridY/Y Y/y Y/y y/yPhenotypeGenotypeMonohybrid cross = a single trait is being followed18Single-gene inheritanceIn diploid monohybrid crosses:3:1 phenotypic ratio if dominance1:2:1 genotypic ratioIn haploid inheritance:all progeny have one copy and it is expressedalso true when gene is hemizygous = there is only one copy present even though diploideg. X-linked genesNEXT TOPIC: Multiple-gene inheritance and Mendel’sfinding of independent assortment of genes20Dihybrid crossAre these characters independentof each other?All four combinations were seen:Round + YellowRound + GreenWrinkled + YellowWrinkled + GreenFour phenotypes now21Dihybrid crossMendel’s model predicted that each character “allele” would pass to a gamete with a probability of ½ and the test cross verified this.Do alleles of different genes also get transmitted with equal probability?22Dihybrid crossAlternative hypotheses: 1) Both “genes” are transmitted by a single particle – not independent2) Each “gene” is transmitted by separate particles – independentIf genes are independent, then all combinations should be seen Four combinations possible: probability of passing to gamete:Round + Yellow ?Round + Green ?Wrinkled + Yellow ?Wrinkled + Green ?23TerminologyCapital = dominant; lower case = recessiveGene/allele abbreviations are italicized as wellRound is dominant to wrinkled (3:1 ratio), so R and rYellow is dominant to green (3:1 ratio), so Y and y24Double heterozygote genotypeDoubly dominant phenotype25Why 9:3:3:1 ratio?Dihybrid cross26Why 9:3:3:1 ratio?Dihybrid crossfirst make all possible gamete types: Y R (gene 1 has Y, gene 2 has R)Y ry Ry rThere are only four ways to combine these allelesin a gamete27“Punnett square”Use to calculateprobabilities and visualizeall possible classes(genotypic and phenotypic)If genes are independent, then their alleles will be transmitted to gametesat random2812:4=3:1Equal segregation holdsfor round:


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