BIO 305 1st Edition Lecture 3 Outline of Last Lecture I. Brief History of Mendel and DarwinII. Mendel’s ExperimentIII. Mathematical RulesIV. StatisticsOutline of Current Lecture I. PedigreesII. Extension of Mendelian GeneticsIII. Different Types of DominanceIV. BloodtypesV. Multiple Gene InteractionVI. Vocabulary and Sample QuestionsCurrent LectureI. PedigreesPedigrees analysis – scrutinizing medical records of matings throughout generations, used to understand single gene inheritanceBe able to interpret pedigree symbols using the key in the textbook:An autosomal recessive pedigree is characterized by a rarity of affected individuals, while in an autosomal dominant pedigree; there is a higher prevalence of affected individuals in more generationsII. Extension of Mendelian GeneticsWhat is the genetic basis for the contrasting phenotypes: round vs. wrinkled peas?SBE1 (starch branching enzyme 1) – the gene that encodes for a functional protein that convertsamylose (unbranched starch) into amylopectin (branched starch) The recessive mutation of this gene results in the deactivation of the enzyme, which results in an accumulation of sucrose that, by osmosis, causes water to exit the pea. Thus the pea appearswrinkledNull allele – a special case of a “loss-of-function mutation”, in which there is a loss (but not complete loss!) of function Neutral mutation – a mutation that doesn’t affect the fitness of the organismIII. DominanceMust be able to specify the type of dominance, which depends on the context of the phenotypeFull dominance – allele that is expressed when only one copy is present, as in a heterozygote, whereas the alternative allele will be fully recessivePartial (incomplete) dominance – the phenotype of a heterozygote is measurable as an intermediate between those of the two homozygotes (Ex: AA = red, aa = white, Aa = pink)A 1:2:1 ratio is characteristic of a partial dominance (instead of a 3:1 ratio)Co-dominance – there is expression of both alleles of a heterozygote (ex: blood-type)Ex: In a dihybrid cross of a blood-type and pigmentation (AaIAIB x AaIAIB for example), you would see a 6:3:3:2:1:1 ratio instead of 9:3:3:1 (but shows 3:1 and 1:2:1 when these phenotypes are considered individuallyImportant Concept: Type of dominance depends on what phenotype is being described.Ex: Tay-SachsAt the individual level, there is full dominance in that TT and Tt individuals are normal. At the mRNA level, it is co-dominant because a Tt genotype expresses both levels here. But in terms of enzyme activity, partial-dominance is seen in that Tt results in 50% of the enzyme activity as TT individuals do.IV. Blood-typesThe Bambay Phenotype – caused by the absence of a gene (FUT1) that encodes for an intermediate in the biochemical pathway converting a precursor (called H-substance) into A or BantigensThe secretor locus – people with the se/se mutation (about 20% of humans) have A and B antigens that are only present in the red-blood cell (as oppose to also being in body fluids)Ex: Semen with A/B antigens cannot be used in a case against an individual with se/seV. Multiple Gene InteractionIf one gene (A) is needed to express another gene (B), these are said to be complementary to each other. Also, A is epistatic to B and B is hypostatic to A.Example: Without H-substance, blood cannot form A or B antigens and phenotype is O. This means H gene is epistatic to ABO genes. Thus, for a cross of (IAIBHh x IAIBHh): Other Examples of Modified Dihybrid Ratio:Case 1) 9 : 3 : 4Agouti: black is a 9:3 = 3:1 ratio(Agouti + black) : albino = 12:4 = 3:1Thus, we decide which gene is epistatic to the other by devising this path:Precursor molecule (colorless)  (B/-)  Black Pigment  (A/-)  Agouti PatternCase 2) 12 : 3 : 1Yelow : Green is the 3:1 ratio among the phenotypesWhite: (yellow + green) is then 12:4, thus also a 3:1 ratioNote: White phenotype is the dominant allele, masking the expression of the othersCase 3) 9 : 7Notice that purple is 9/16, thus it’s genotype must be A_B_Precursor molecule (colorless)  (A/-)  Intermediate  (B/-)  PurpleRatio shows that two gene products are necessary to be purple: complementary to each otherCase 4) 9 : 6 : 1Similar to Case 5, as two gene products are needed to produce disc phenotypeHowever, only one gene product results in the sphere phenotype, thus:Disc = A_B_ ; Sphere = A_bb, aaB ; Long: aabbCase 5) 13 : 3Notice that colored occurs with a genotype: A_bb, thus:A dominant allele at Gene B (for white phenotype) masks expression of alleles of Gene BCase 6) 10: 3 : 3For expression of white spotted to be 10: A_B_ and aabbThus, white is A_bb and colored is aaB_Case 7) 15 : 1The dominant phenotype’s condition is that it has at least one A or one B alleleCase 8) 6 : 3 : 3 : 4Recall that incomplete dominance has a ratio of 1:2:1 instead of 3:1, thus:sooty : red : jet = 3: 6 : 3, (sooty + red + jet) : black = 12 : 4 = 3:1In all cases, Mendel’s two laws have been followed. When the F2 phenotype ratio is expressed in sixteenths, you will know that it is the action of two gene pairs.Vocabulary:Wild-type (+) – allele that appears the most often in a population, usually dominantMutant allele – has modified genetic information, usually results in an altered productFitness – survivability; the ability to adapt to one’s environmentLethal alleles – alleles that result in the death of the organismEpistasis – the effect of one gene masks or modifies the effect of anotherComplementation Analysis - used to determine if two mutations occur on the same geneComplementation – mutations are in two different genes if parents both affected but no children are affectedNon-complementation – mutations in the same gene if both parents and children are all affectedPleiotropy – the expression of a single gene with multiple phenotypic effectsMarfan syndrome – an example of pleiotropy where there is lens dislocation, aortic aneurysm, and lengthened bones in limbs caused by a mutation in a gene encoding for connective tissue: fibrillinSample Questions:1) The genotype of II-1 must be: a) AA or Aa b) AA or aa c) Aa or aa d) AA, Aa, or aa e) None of the above2) What is the probability that if these parents had a fourth child, the child is healthy?3) What is the probability that II-3 is heterozygous for the dominant and recessive alleles?ANS: A, 3/8,