BIOL226 Topic 1: Fundamental Principles of Inheritance PDF

BIOL226 2025 topic 1 The Fundamental Principles of Inheritance  PERSPECTIVE...

BIOL226 2025 topic 1 The Fundamental Principles of Inheritance  PERSPECTIVE https://doi.org/10.1038/s41588-022-01109-9 How did Mendel arrive at his discoveries? Peter J. van Dijk! !1 ✉, Adrienne P. Jessop2 and T. H. Noel Ellis! !3 There are few historical records concerning Gregor Johann Mendel and his work, so theories abound concerning his motivation. These theories range from Fisher’s view that Mendel was testing a fully formed previous theory of inheritance to Olby’s view “Uniquely among nineteenth-century scientists, Mendel con- that Mendel was not interested in inheritance at all, whereas textbooks often state his motivation was to understand inheri- tance. In this Perspective, we review current ideas about how Mendel arrived at his discoveries and then discuss an alternative ducted a coordinated set of quantitative experiments and scenario based on recently discovered historical sources that support the suggestion that Mendel’s fundamental research on the inheritance of traits emerged from an applied plant breeding program. Mendel recognized the importance of the new cell concluded that inheritance was non-blending. Even after its theory; understanding of the formation of reproductive cells and the process of fertilization explained his segregation ratios. This interest was probably encouraged by his friendship with Johann Nave, whose untimely death preceded Mendel’s first 1865 publication, his work was not understood and remained lecture by a few months. This year is the 200th anniversary of Mendel’s birth, presenting a timely opportunity to revisit the events in his life that led him to undertake his seminal research. We review existing ideas on how Mendel made his discoveries, before presenting more recent evidence. neglected for 35 years. “ T he Augustinian friar Gregor (Johann) Mendel (1822–1884) is Perspective gives a new picture of how Mendel arrived at his discov- the founder of the science of genetics. His crossbreeding experi- eries. We review existing ideas on how Mendel made his discoveries ments with peas, reported in two lectures in the spring of 1865 before presenting the recent evidence. However, we begin with what and published in 1866, are so instructive that they are still used to Mendel himself writes about it in the 1866 paper. introduce genetics. Textbooks simply state that Mendel conducted his Text pea crosses to study the rules of inheritance. However, this obscures how little we really know about Mendel. The impetus to study the What does the 1866 paper reveal about Mendel’s aims and working method? rules of inheritance was less evident in Mendel’s time than it seems Mendel’s introduction is brief. It starts by mentioning the recur- today. Uniquely among nineteenth-century scientists, Mendel con- rence of the same hybrid form when ornamental plants are crossed, ducted a coordinated set of quantitative experiments and concluded that is, that the F1 from two inbreds is uniform. Mendel gives this that inheritance was nonblending. Even after its publication, his work observation as his impetus and explains that he aimed to ‘follow up was not understood and remained neglected for 35 years. the development of the hybrids in their progeny’, an expression that The publications by Hugo De Vries, Carl Correns and Erich von recalls Franz Unger, his botany professor in Vienna, who saw the Tschermak in 1900 mark the beginning of the broad appreciation of parental plant and its offspring as a unity connected by descent6 Mendel’s work. Mendel had been dead for 16 years, and his notes no (Unger, 1852). According to Mendel, so far no study had classified longer existed. What remained was the article Experiments on Plant all offspring of hybrids and established their numerical relation- Hybrids1 (Versuche über Pflanzen-Hybriden, hereafter the 1866 ships. Thus, Mendel set out to study the composition of the F2 and paper), which was published at the end of 1866 in the Proceedings of possibly later generations. the Natural Science Society (NSS) of Brünn, the capital of Moravia Then follows a whole section on the selection of the experimen- (now Brno, Czech Republic; Supplementary Note). Searches for tal plants, where Mendel states: “it cannot be immaterial which other documentation shortly after the rediscovery yielded only a plant species were chosen as support for the experiments and in few letters that Mendel had written after 1866 to Carl Nägeli (1817– which way these experiments were carried out … On account of 1891), professor of botany in Munich, together with fragments their particular flower structure, particular attention was paid to of Nägeli’s responses published by Correns in 1905 (ref. 2). Until the Leguminosae right from the start. Experiments which were per- recently, no historical documents were known concerning his pea formed on several members of this family led to the result that the experiments from the ten years when these were conducted. genus Pisum sufficiently meets the posited requirements”1,7.  we are now able to clone and map 4/7 genes that Mendel has mapped. dominant Color of seed (I) green One of Mendel's seven pea traits "To the difference in the color of the seed albumen (endosperm). (i) The albumen of the ripe seeds is either pale yellow, bright yellow and orange colored, or it possesses a more or less intense green tint. This difference of color is easily seen in the seeds as their recessive coats are transparent." yellow J. Mendel - Experiments in Plant Hybridization, 1865 (I) staygreen: enzyme that is important to break down chlorophyll. degredation of these pigments break this down. enzymes recycle. yellow seeds cannot breakdown chlorophyll 1865 - 2007 142 years later... staygreen gene sgr WT sgr (I/i or I/I) mutated sgr (i/i) Retention of ‘greenness’ in leaves Normal protein functions in the catabolism of chlorophyll in senescence photosynthetic tissue in plants, results in yellow tissue. Mutations in sgr impair chlorophyll degradation, delay/prevent nutrient re-absorption in senescing leaves normal delayed clorophyll catabolism clorophyll catabolism  The fate of Mendel's genes: molecular cloning of the mutations for 4/7 traits so far Text R (starch branching enzyme 1) I (stay green enzyme) A (bHLH transcription factor) FA (?) LE (GA 3-oxidase) V (?) GP (?) (mapped in a chromosomal region in 2021) GP gene mapping update Single Gene Inheritance Chapter 3 Text  Topic Outline o Mendel and the Patterns of Inheritance o The First and Second Mendelian Laws o Mendelian Inheritance in Humans understanding in context of human What are you expected to be able to do by the end if this topic? Understand conceptually the rationale, conclusions and practical consequences the two Mendelian Laws Be able identify in a cross scheme the signature of Mendelian laws (segregation, dominance) Be able to derive progeny frequencies predicted by these laws Predict simple probabilities of inheritance in progenies and pedigrees Be able to statistically test against a Mendelian ratio Be able to work out inheritance patterns in human pedigrees Study topics  MUTANT IDENTIFICATION IS THE CRUX OF GENETIC ANALYSIS goal of geneticists is to find a difference in the phenotype. In whole organisms (growth as a phenotype) function of genes, mutations, variations, differences, difference of fungi in. a petri dish. why things mutate, Neurospora crassa mutants that affect growth patterns  Or in the development of complex organs in higher eukaryotes, for example why do these mutants change, why do outcomes change Mutants in floral development in Arabidopsis thaliana  Mendel’s Central Question: can you predict a cross. What happens when two different plant varieties are crossed to form a hybrid?  The mighty pea plant “It cannot be immaterial which plant species were chosen as support for the experiments and in which way these experiments were carried out... On account of their particular flower structure, particular attention was paid to the Leguminosae right from the start. Experiments which were performed on several members of this family led to the result that the genus Pisum sufficiently meets the posited requirements” G. Mendel his choice of model helped him suceed. easy to access models are easier to work withPisum sativum he could cross it the way he wanted. and there was many different types of peas, and he could start deciding how he wanted to cross them. he selected traits that allowed him that were segregating in the parent. Mendel’s Approach why mendel chose the pea plant. allows for cross fertilization, the variables were key as to not be able to mess with the crosses, if he didnt it couldve messed with offsrping and he wouldnt have been able to predict Chose the garden pea – Easy to grow, short life cycle, and mating (pollination) could be controlled – Different varieties were readily available – Mendel chose varieties that were true breeding. they were uniformly homozygous. Mendel carefully chose which characters to study. – There were two easily distinguishable states (e.g., tall or short).  the crosses that mendel had used, we wont talk about all of these but lots The 7 Characters Mendel Studied  Cross-polinization and selfing are two types of crosses can control who crosses with who. there is cross progeny and self progeny,  he realized it only makes sense to be able to think of the unknown. if he was able to predict. made plants homozygous ( several generations) so that they only contained one gene that he was Breeding a line true looking for. he segregated genes so that he could only have one gene (pure plant) Before he began his experiments, Mendel crossed plants that generated wrinkled seeds between themselves 𝑠𝑒𝑙𝑓𝑖𝑛𝑔 𝑠𝑚𝑜𝑜𝑡ℎ 𝑙𝑖𝑛𝑒 𝑠𝑒𝑙𝑓𝑖𝑛𝑔 𝑟𝑜𝑢𝑔ℎ 𝑙𝑖𝑛𝑒 multiple times until the only possible seed that was produced in the next generation was wrinkled. He did the same in parallel with plants that produced smooth seeds 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 1 and for all other traits in his different crosses. This generated what geneticists call ‘true breeding’ lines, 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 2 or, in modern terms, homozygous for the respective loci. Text Mendel’s rationale was that it was critical to know with 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 3 certainly the result of selfing (crosses within lines) before he addressed how these traits would segregate when 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 4 crossing plants that vary in each trait with each other. 𝑃0 𝑐𝑟𝑜𝑠𝑠 𝑥 𝑡𝑟𝑢𝑒 𝑏𝑟𝑒𝑒𝑑𝑖𝑛𝑔 This control step explains Mendel’s success in arriving at the reproducible distributions of phenotypes in F2 progenies, the foundation of the inheritance laws.  Mendel’s Approach Designed crosses carefully and kept detailed records of each (reproducible). Counted the number of offspring that had the traits he was following (math background). Text Kept track of generations (parents vs. offspring) and followed inheritance over several generations. Asked very specific questions and made testable predictions from results. he was careful and made sure that he could collect the progeny that he was crossing for. he further down to other crosses that he collected generations for control crosses:  the ability to start with two breeding generations where you know the genotype of this. the offspring will contain genes from their parents. then you can predict the genotype of schematics of a Mendelian (controlled) cross offspring since they will be heterozygous hybrids. Mendel introduced the concept of controlled crosses Controlled crosses start from the vantage point of knowing the genotype P0 of parentals (true breeding or P0 generation). A scientific and reproducible way to track down the segregation and inheritance of traits down generations. Controlled crosses in Mendelian genetics stipulates generations that F1 inherit alleles in a predicted manner and have a predictable disposition of genotypic and phenotypic frequencies for alleles with clear recessive/dominant relationships: F1 (familial 1, the first hybrid generation ) F2 (familial 2, the second hybrid or re-segregation generation ) As we will see, Mendelian Laws allow the prediction of the distribution of phenotypes for (n) number of generations after F2. F2 Mendel’s laws apply to any genetic cross/pedigrees, but these generational signatures are only readily identifiable in controlled crosses. Geneticists today still use the structure of controlled crosses to make sense of how genes are being transmitted, to design crosses, predict frequencies and engineer new genotypes. Good choice or organism, lucky choice of traits Text Traits controlled by single genes High number of progenies Carefully development of a crossing system (‘controlled crosses’) Use of true breeding parentals (homozygotes)  Text MENDEL'S EXPERIMENT #1 EVIDENCE FOR A DUAL NATURE OF GENETIC INFORMATION He then took pollen from pea plants with wrinkled seeds and put it on the stigma of flowers from plants with smooth seeds. he is looking at the texture of skin, x LOOK AT SEEDS OF FILIAL GENERATION 1, OR F1  monohybrid experiments: crossing single trait organisms dihybrid: Types of Mendelian Crosses: tri: tetra Monohybrid Crosses Monohybrid Crosses – crosses of two varieties of true-breeding plants that differed in only one character Examples: Seed shape or seed color. – Plants with smooth (round) seed crossed with plants with wrinkled seeds. – Plants with yellow seeds crossed with plants with green seeds. In every case, one trait was dominant and one trait was recessive.  MENDEL'S EXPERIMENT #1 EVIDENCE FOR A DUAL NATURE OF GENETIC INFORMATION Text ALL F1 SEEDS WERE SMOOTH Mendel called "smooth", a dominant phenotype  MENDEL'S EXPERIMENT #1 EVIDENCE FOR A DUAL NATURE OF GENETIC INFORMATION In the F1 progeny: ALL F1 SEEDS WERE SMOOTH His first insight: Did the information for wrinkled disappeared in the F1 plant? for the frist time he realized that even if looks identical. they may not be identical. aka the genotype appears and realizes there is a recessive trait in this genotype. information absence must come from the alleles in which one expresses and the other one does not  MENDEL'S EXPERIMENT #1 EVIDENCE FOR A DUAL NATURE OF GENETIC INFORMATION Mendel then crossed F1 smooth seed plants. Plants that in that respect look exactly like the true breeding smooth P0 progenitor. dominant phenotype will show if it is present. x LOOK AT SEEDS OF FILIAL GENERATION 2, OR F2  MENDEL'S EXPERIMENT #1 In the F2 progeny: 1850 5474 The pea seeds were not all smooth: From 253 plants allowed to self-cross 5474 seeds were smooth and 1850 were wrinkled information wasnt lost. it was hidden due to the genotype. a ratio of 2.96 to 1 (roughly 3 : 1) the appearance of the phenotype is not random. reproduces this 6 times. appears to be a predictive ratio. The wrinkled seed trait was not present in the F1 seeds BUT THE INFORMATION FOR WRINKLED MUST HAVE BEEN, since the trait reappeared in F2 generation. There is therefore a difference between the expression of a trait (phenotype) in peas and the presence of the information for that trait (genotype). Mendel obtained similar results for all 7 characters that he studied Text very close to 3:1 ratio  Two different ways in which information can be combined: One phenotype, one possible genotype (homozygous recessive) yy (mutant) One phenotype, two possible genotypes (homozygous dominant, heterozygous) YY Yy (WT) first set is that there is different phenotypes predict genotypes which are different. green has only one genotype.  Mendel devised a simple genetic cross to tell apart homozygous x heterozygous displaying the dominant phenotype. He named it test cross. We use this strategy to this day. Who am I? TEST CROSS (known genotype) (unknown genotype) true-breeding recessive F1 hybrid (Yy) or (tester ) true-breeding dominant (YY) In a test cross, the F1 hybrid (or the organism with yy ? the dominant phenotype and unknown genotype) is crossed with a recessive individual, called the tester. x A tester is ALWAYS homozygous recessive to the information being considered. Notice that by virtue of being recessive, the trait displayed by the test cross progeny will be entirely determined by the information inherited from the plant with unknown genotype (the contribution from the tester will never dominate) TEST CROSS PROGENY segregation means that the test cross progeny reflects the information. that you get a recessive and a dominant allele. segeragates dominant will show dominant allele. segeragates recessiuve it will show the recessive allele  When the tested plant is homozygous dominant for the trait: Who am I? TEST CROSS cross with dominant and recessive yy menas that you will get a 100% YY dominant phenotype. x tester gametes y tester cross progeny Y Yy F1 gametes 100% TEST CROSS PROGENY 22 0 Only Yy, 100% dominant yellow  When the tested plant is a hybrid (heterozygous for the trait): Who am I? TEST CROSS yy Yy x tester gametes y tester cross progeny Y Yy F1 gametes 50% y yy 50% TEST CROSS PROGENY 11 11 Yy and YY (50% : 50%)  If the information for a trait can be silenced (hidden), genetic blending cannot apply. "blending" smooth seeded plants (F1) should never produce wrinkled seeds Mendel suggested instead a particulate theory of genetics: Genetic information comes as two separate and independent particles where: o One particle could be dominant over another. o The combination of these particles change Text in F1 and F2 plants (the particles must be able to be shuffled and segregate in different combinations in the gametes). o This segregation events and the combination of information in gametes must be random. o Randomness alone generates a distribution of dominant and recessive information in the gametes that explains the phenotypic ratios found in F2 progenies. if hidden in one genotype it can be unhidden in another  A FEW DEFINITIONS BEFORE WE GO AHEAD Mendel called the segregating agents behind his pea traits ‘ elemente ‘ (elements). He never used the term ‘gene’ or ‘locus’ or ‘allele’. The common classical genetics jargon had not yet been coined. "ADDRESS" LOCUS the physical region in chromosomes where genes are. "PARTICLE" GENE unit of hereditary information. VERSION OF THE PARTICLE THAT ALLELE a version of a gene that occupies a locus. SEGREGATES TWO IDENTICAL HOMOZYGOUS when both alleles of a gene (locus) are the same in a 2n organism. PARTICLES TOGETHER TWO DIFFERENT HETEROZYGOUS when two alleles of a gene (locus) in a 2norganism differ. PARTICLES TOGETHER  MENDEL'S HYPOTHESIS IN MODERN TERMS 1. Alternative versions of genes (with different alleles) account for variation in inherited characteristics 2. Diploid organisms inherit two alleles, one from each parent 3. If two alleles are different, one may be dominant (different alleles in a pool of alleles for a given loci present in a population have hierarchical dominant vs recessive relationships) 4. Each haploid gamete carries only one allele of a given trait because they segregate from one another during meiosis. In the zygote, the ploidy is restored and the dominant allele determines the phenotype.  Mendel’s Explanation of Monohybrid Cross Results When an individual (a homozygote) has two identical factors, the trait breeds true, but when the two factors are different (in a heterozygote), the phenotype will be the phenotype of the dominant factor. Because the non-dominant (recessive) traits re-appears in F2, the two alleles of the same gene in F1 (and consequently in any other generation) must segregate from each other during germ cell (gamete) formation = Mendel’s First Law or Law of Segregation At fertilization, gametes fuse randomly. The 3:1 ratios represents the mathematical chance of dominant and recessive alleles meeting in a genotype.  THE FIRST MENDELIAN LAW Law of Segregation: Alleles of a gene separate independently from each other during transmission from parent to offspring (through meiosis). The dominant phenotype appears at 100% in the F1 (hybrid genotype). The phenotypic frequencies in F2 conforms to : 3 : 1 dominant recessive (this is the mathematical prediction built in the First Law) homozygous still segregates, but you dont see it because the information is the same. which menas that the evidence of segreation where do you see it? you only see it in progeny from hybrids  Mendel’s Principles: derivation of the first law The Principle of Dominance: In a heterozygote, one allele may conceal the presence of another. The Principle of Segregation: In a heterozygote, two different alleles segregate from each other during the formation of gametes. Writing Mendel's first law in cross format Two Dominant Two Recessive Factors Factors Green and yellow information will be equally distributed in the gametes Only plants with two recessive factors , y/y, will be green  shows you the information inside for phenotype and genotype. you have Two Frequencies in F2 gametes from Punnett Square (Reginald Punnet) YY Yy yy 25% 50% 25% 75% 25%  YY Yy yy Genotypic frequency 25% 50% 25% (inferred by Mendel) (1) 1:2:1 (1) (2) Genotypic frequency = Phenotypic frequency 3:1 Phenotypic frequency 75% 25% (found by Mendel) (3) (1) The practical use of Mendel’s laws Text In an F2 progeny with 120 peas segregating yellow and green seed colour, 60 should homozygous e 60 heterozygous for seed colour information. heterozygous YY Yy yy homozygous 50% homozygous 50%  heText Mendel’s Results & Rules of Probability How can we explore the predictive power of Mende’s first law? If alleles segregate at random during gamete formation, and fertilization is also random, offspring ratios should follow the rules of Probability: – Probability of occurrence of two independent events = probability of 1st event times the probability of 2nd event. – Probability of a A/A genotype (in offspring) = probability a male gamete carrying allele A x probability of a female gamete with the same allele.  Application of Probability Rules Suppose an Yy F1 plant (with yellow seeds) was allowed to self-fertilize: Probability of YY offspring = probability of Y from ♀ X probability of Y from ♂ ! ! !. = or 25% " " # heterozygote, Probability of yy offspring = Probability of Yy offspring =  Application of Probability Rules Suppose an Yy F1 plant (with yellow seeds) was allowed to self-fertilize: Probability of YY offspring = probability of Y from ♀ X probability of Y from ♂ ! ! !. = or 25% " " # Probability of yy offspring = ! ! !. = or 25% Text " " # Probability of Yy offspring = ??  Application of Probability Rules Suppose an Yy F1 plant (with yellow seeds) was allowed to self-fertilize: Probability of YY offspring = probability of Y from ♀ X probability of Y from ♂ ! ! !. = or 25% " " # Probability of yy offspring = ! ! !. = or 25% " " # Probability of Yy offspring = ! ! ! 2.. = or 50% " " " how you do the ratio: segregating will be these frequencies  explained by the first law of segregation YY Yy yy Probability of yy genotype = Product of the probabilities of gametes carrying the y allele.  Predictions of Mendel’s 1st Law 1. Reciprocal crosses should yield identical results. o Randomness in allele segregation is independent on the source of the allele (male or female germline): – Pollen from plant with dominant trait x Ova from plant with recessive à 3:1 ratio in F2 – Pollen from plant with recessive trait X Ova from plant with dominant trait à still 3:1 ratio in F2  Predictions of Mendel’s 1st Law selfing gave segregation made into heterozygote 2. Of the plants with yellow seeds in the F2, 2/3 should be Yy and 1/3 should be YY. F2: 1 YY 2 Yy 1 yy yellow green How to test? Mendel allowed individual F2 yellow plants to self-pollinate and kept track of the offspring of each F2 plant separately. Result: 166 F2 plants produced only yellow F3 seeds (YY homozygotes) 353 F2 plants produced both yellow and green F3 seeds (Yy heterozygotes) ~1/3 YY homozygotes: ~2/3 Yy heterozygotes  Predictions of Mendel’s 1st Law 3. Gametes produced by a heterozygote should come at a ratio of 1:1. Test Cross: cross of plant with unknown genotype to the homozygous recessive parent (tester). If unknown is YY, we predict: YY x yy à all Yy (yellow) offspring If unknown is an F1 plant (Yy), which should produce Y and y gametes with equal frequency: Yy x yy à ½ Yy (yellow), ½ yy (green)  Test cross is used in Test Cross genetics to follow test cross (TC) segregation in the hybrid meiosis tester Contributes only with 1 type of gamete (always the recessive type). Phenotypes in TC progeny reflect information and frequency of the F1 gamete F2 progeny TC progeny F2 dominant and recessives. frequency takes over recessive so you dont see it, recessive is hidden in dominant. but test cross, one of the individuals only provides recessive traits so no segregation occurs. test cross eliminates recessive traits. test cross 1:1 is mendelian. 3:1 is F2 cross  In testcrosses, genotypic and phenotypic frequencies are identical tester gametes Genotypic Phenotypic frequency frequency Yy hybrid gametes 50% 50% yy 50% 50% The Frequency of the hybrid recessive allele = Frequency of the recessive Phenotype The Frequency of the hybrid dominant allele = Frequency of the dominant Phenotype Finding evidence of Mendel's conclusions in genetic crosses  Which generation directly shows evidence for dominance? 𝑝𝑎𝑟𝑒𝑛𝑡𝑎𝑙 𝑃0 𝑥 𝑃0 𝑓𝑖𝑙𝑖𝑎𝑙 1 F1 is generally the one that shows dominance 𝐹1 𝑥 𝐹1 if dominance is present 𝑓𝑖𝑙𝑖𝑎𝑙 2 𝐹2  The F1 (hybrid) generation directly suggests dominance 𝑝𝑎𝑟𝑒𝑛𝑡𝑎𝑙 𝑃0 𝑥 𝑃0 𝑓𝑖𝑙𝑖𝑎𝑙 1 𝐹1 𝑥 𝐹1 𝑓𝑖𝑙𝑖𝑎𝑙 2 𝐹2 In a Mendelian cross, the F1 generation will consist exclusively of heterozygotes displaying the dominant phenotype for the allele combination in the locus in being examined.  Which generation directly provides evidence for the independent segregation of alleles? 𝑝𝑎𝑟𝑒𝑛𝑡𝑎𝑙 𝑃0 𝑥 𝑃0 𝑓𝑖𝑙𝑖𝑎𝑙 1 𝐹1 𝑥 𝐹1 𝑓𝑖𝑙𝑖𝑎𝑙 2 𝐹2  hidden information reappears bc of segregation of the F1 gamets The F2 generation directly suggests free segregation of alleles 𝑝𝑎𝑟𝑒𝑛𝑡𝑎𝑙 𝐴/𝐴 𝑥 𝑎/𝑎 𝑓𝑖𝑙𝑖𝑎𝑙 1 𝐴/𝑎 𝑥 𝐴/𝑎 𝑓𝑖𝑙𝑖𝑎𝑙 2 𝐴/𝐴 𝐴/𝑎 𝑎/𝑎 In a Mendelian cross, the F2 generation shows the outcome of segregation of alleles in the F1 gametes  𝐴/𝐴 𝑥 𝑎/𝑎 𝑔𝑎𝑚𝑒𝑡𝑒𝑠 𝐴 𝑎 𝑔𝑎𝑚𝑒𝑡𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑃0 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑠 𝑡ℎ𝑒 ℎ𝑦𝑏𝑟𝑖𝑑 𝑔𝑒𝑛𝑜𝑡𝑦𝑝𝑖𝑐 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑖𝑛 𝐹1 (100%) 100% 100% 𝐴/𝑎 𝑥 𝐴/𝑎 𝑔𝑎𝑚𝑒𝑡𝑒𝑠 𝐴 𝑎 𝐴 𝑎 𝑔𝑎𝑚𝑒𝑡𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝐹1 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑠 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓𝑔𝑒𝑛𝑜𝑡𝑦𝑝𝑒𝑠 𝑖𝑛 𝐹2 50% 50% 𝐴/𝐴 𝐴/𝑎 𝑎/𝑎 25% 50% 25% The random distribution of F1 alleles in its gametes will generate F2 phenotypic proportions of 75% dominants to 25% recessives, according to the First Law. Applying the First Law  Two basic probability rules: landscape of possible events (1) What is the probability that two independent events, A and B, will occur together? (2) What is the probability that at least one of the two events, A or B, will occur at all?  The Multiplicative Rule If the events A and B are independent, the probability that they will occur together, P(A and B), is their product: P 𝐴&𝐵 = P𝐴.P𝐵 Example: probability of drawing the ace of hearts in a card deck, P(A & H) 4 1 P(A) = 52 P(H) = 4 (52 total number of cards) P (4 types of card in a card deck) # ! # P(A). P(H) = = = ~2% @" # "AB  What is the probability of two independent events will occur at the same time? ex. In any given population, half of the individuals are men and 1/10 of them belong to blood group B. What is the probability of choosing an individual at random who is both man and belongs to blood group B? 1 event one chances of being (sex ratio in human pop) 2 1 event two chances of being blood type B 10 1 1 1 P( ). P(typeB) =. = = (5%) 2 10 20 Probability of two independent events occurring together is the PRODUCT of the separate two probabilities  The Additive Rule If the events A and B are independent, the probability that only one of them occurs, denoted P(A or B), is: P 𝐴 + P 𝐵 - [P 𝐴. P 𝐵 ] when a or b happens independtly of eachother, the proobability that one of them occurs Example: Probability of drawing an ace OR a heart, P(A or H) 4 1 P(A) = 52 P(H) = 4 ! $ ! $ $% P(A or H) = "# + ! -[ "#. ! ]= "# = ~30%  The Additive Rule If the two events do not overlap in the sample space, they are said to be mutually exclusive. In this case: P 𝐴. P 𝐵 = 0 and P 𝐴 𝑜𝑟 𝐵 = P 𝐴 + P 𝐵 Example: probability of drawing an ace or a king, P(A or K) 4 4 P(A) = 52 P(K) = 52 ! ! & P(A or K) = + = ~15% "# "# "#  What is the probability of either one or another of two mutually exclusive events happening? ex. What is the probability that a child will inherit the two alleles (not necessarily the same allele) of a given locus from the two grandfathers (F) or from the two grandmothers (M) and not one allele from a grandpa and one from a grandma (non-overlapping probabilities)? grandfather event 1 1 2 2 ? event one 1 1 1 P 𝐹.P 𝐹 = 2. 2 = = (25%) 4 (multiplicative rule)  What is the probability of either one or another of two mutually exclusive events happening? grandmother event 1 1 2 2 ? event two 1 1 1 P 𝐹.P 𝐹 = 2. 2 = = (25%) 4 (multiplicative rule)  grandfather event + grandmother event ! ! ! P(F+M) = + = = (50%) # # " Probability of one or another event occurring is the SUM of the separate probabilities Independent Assortment of Genes Chapter 3  Mendel rationalized the independence of alleles from different traits (the basis for the Second Law): segregation of multiple traits at the same time—dihybrid crosses etc. “If then I extend this combination of simple series to any number of differences between the two parental plants, I have indeed entered the rational domain. This seems permissible, however, because I have proved by previous experiments that the development of any two differentiating characteristics proceeds independently of any other differences”. G. Mendel, 1866 How the distribution of phenotypes of two traits segregate relative to each other? PURPLE FLOWER, white flower, TALL dwarf Does the Law apply to alleles of different loci? MENDEL CHOSE TO ANALYSE SEED COLOUR (YELLOW OR GREEN) AND SEED SHAPE (ROUND OR WRINKLED) SHAPE trait 1 trait 2 COLOUR Types of Mendelian Crosses: Dihybrid Crosses Monohybrid Crosses – crosses of two varieties of true-breeding plants that differed in two characters. In every character, one trait is dominant and one trait was recessive. P Round, yellow x wrinkled, green (RR YY) (rr yy) Possible Outcomes: With respect to transmission: Hypothesis 1a: Factors for seed shape and seed color are transmitted together (associated somehow). Hypothesis 1b: Factors for seed shape and seed color are transmitted independently of each other, as alleles of the same trait. Possible Outcomes With respect to expression: Hypothesis 2a: Different characteristics of seeds influence each other or interact. Hypothesis 2b: Different characteristics of seeds are expressed independently of each other. Two monohybrid crosses hypothesis b was correct CROSS 1 CROSS 2 “shape” locus “colour” locus R/R r/r y/y Y/Y P0 x x R/r Y/y F1 R r Y y R Y R/R R/r Y/Y Y/y F2 r y R/r r/r Y/y y/y 3 : 1 3 : 1  YELLOW SMOOTH DOMINANT OVER GREEN DOMINANT OVER WRINKLED F1 F1 Y/y R/r Y - YELLOW R - ROUND y - green r - wrinkled  “shape” locus “colour” locus R/R r/r Y/Y y/y P0 x x R/r Y/y F1 R r Y y R Y R/R R/r Y/Y Y/y F2 r y R/r r/r Y/y y/y 3 : 1 3 : 1 EXPECTED RATIOS ACCORDING TO THE FIRST LAW Combining monohybrid analyses MENDEL'S EXPERIMENT #2 COLOUR SHAPE x CROSSING TRUE-BREEDING PLANTS FOR BOTH LOCI TWO TRAITS INVOLVED, TWO DOMINANT AND TWO RECESSIVE PHENOTYPES POSSIBLE  x What is the F1 phenotype (s) ? ? ? ? ?  x F1 Y/y; R/r YELLOW, ROUND PEAS (double dominant)  DIHYBRID CROSSES seed colour and seed shape P0 true-breeding x y/y; r/r Y/Y; R/R locus 1 locus 2 locus 1 locus 2 F1 dihybrid Y/y; R/r 100%  Y/y; R/r Conclusion 1: Dominance relationships are not altered by the presence of a second pair of factors. P0 is always the true breed  Y/Y; R/R y/y; r/r DIHYBRID CROSSES P0 seed colour and seed shape x As predicted by the first Law, alleles of the F1 hybrid will Y/y; R/r randomly segregate to meet in F1 the gametes. In a dihybrid cross, 2 genes, 4 alleles are in play. GAMETES Random segregation will lead to Y,R Y,r y,R y,r 4 possible gamete combinations for the two genes, as indicated Y,R Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R in the x and y axis of the Punnet F2 Y,r square on the left. GAMETES Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r Note that the F2 progeny will also be different from a y,R monohybrid F2 generation. Y/y; R/R Y/y; r/R y/y; R/R y/y; r/R y,r Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r  Y/Y; R/R y/y; r/r Two segregating genes create new P0 x phenotypic combinations A signature of a dihybrid F2 is the appearance Y/y; R/r of two novel phenotypes, absent from the P0 F1 and F1 generation. To identify that, first note the two parental phenotypes, those that the P0 individual GAMETES Y,R Y,r y,R y,r display in the cross. Y,R Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R F2 Y,r PARENTAL PHENOTYPES GAMETES Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r y,R Y/y; R/R Y/y; r/R y/y; R/R y/y; r/R YELLOW; GREEN; y,r ROUND WRINKLED Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r  Y/Y; R/R y/y; r/r Segregation of alleles of two genes in the P0 x F1 creates recombination in the F2 generation Y/y; R/r F1 The random ‘assortment’ of dominant and recessive allele in the F1 gametes will produce two GAMETES F2 phenotypes that were unseen in the P0 and F1. Y,R Y,r y,R y,r These are referred as recombinant phenotypes. Y,R Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R F2 Y,r GAMETES Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r y,R Y/y; R/R Y/y; r/R y/y; R/R y/y; r/R GREEN; YELLOW; ROUND WRINKLED y,r Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r RECOMBINANT PHENOTYPES MENDEL'S F2 PHENOTYPE DATA Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R Similar to his monohybrid experiments, Mendel noticed that the proportions of the 4 (2 parental Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r and two recombinant) F2 phenotypic classes appear in reoccurring frequencies. Y/y; R/R This becomes the basis for the mathematical Y/y; r/R y/y; R/R y/y; r/R prediction of the Second Mendelian Law. Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r 315 108 101 32 (x/32) + + + = 556 F2 SEEDS 9 : 3 : 3 : 1 Phenotypic proportions in F2 of a dihybrid cross law of independent assortment: mechanism for frequencies for parental and recombinant is random  THE SECOND MENDELIAN LAW Law of Independent Assortment: Alleles of two (or more) genes (loci) segregate independently during transmission from parent to offspring. The two dominant phenotypes appear at 100% in the F1 (hybrid genotype). In the F2, 4 phenotypic classes are present: o two parental phenotypes present in the P0 generation o two new phenotypes (recombinants) absent from P0 or F1. The expected F2 frequency for phenotypes generated by alleles of two loci is: 9 : 3 : 3 : 1 double dominant; recessive; double dominant recessive dominant recessive (DD) (DR) (RD) (RR) (this is the mathematical prediction built in the Second Law) Conclusion 2: The segregation of one pair of factors is independent of the segregation of another pair. Mathematical prediction of the second law: Dominant and recessive phenotypes for both genes will recombine such that 9/16 will be double dominants, 1/16 double recessive and 6/16 dominant for a gene and recessive for the other, respectively.  Dissecting the 9:3:3:1 ratio Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r Y/y; R/R Y/y; r/R y/y; R/R y/y; r/R Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r 315 108 101 32 + + + = 556 F2 SEEDS YELLOW : 315 + 101 = 416 GREEN : 108 + 32 = 140  YELLOW 416 + + + = 556 F2 SEEDS 140 GREEN 3 : 1 YELLOW GREEN PREDICTED BY THE FIRST MENDELIAN LAW FOR THE COLOUR LOCUS  Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r Y/y; R/R Y/y; r/R y/y; R/R y/y; r/R Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r 315 108 101 32 + + + = 556 F2 SEEDS ROUND WRINKLED 315 + 108 = 423 101 + 32 = 133  ROUND 423 + + + = 556 F2 SEEDS 133 WRINKLED 3 : 1 ROUND WRINKLED PREDICTED BY THE FIRST MENDELIAN LAW FOR THE SHAPE LOCUS  The Second Mendelian Law (SML) is the mathematical expression of the First Law applied to alleles of two or more loci (genes). SHAPE LOCUS 3 : 1 9 : 3 : 3 : 1 COLOUR LOCUS 3 : 1 SML applies therefore not only to dihybrids, but for segregation of alleles of (n) genes (trihybrid, tetrahybrid, pentahybrid…etc) for which a derivation of the predicted phenotypic ratios can be derived, as long as alelles segregate freely. Predictions based on Rules of Probability (for Phenotypes) For seed shape: expected F2 frequencies are ¾ round (D), ¼ wrinkled (R) For seed color: expected F2 frequencies are ¾ yellow (D), ¼ green (R) Therefore, if inheritance is independent, the expected frequencies of the 4 F2 phenotypes are their products: genotypic frequencies ! ! 𝟗 Round, yellow = DD " " 𝟏𝟔 ! & 𝟑 DR Round, green: " " = 𝟏𝟔 ! & 𝟑 RD Wrinkled, yellow: " " = 𝟏𝟔 RR Wrinkled, green: & " & " = 𝟏 𝟏𝟔 Hence, 9:3:3:1  What if I am following the segregation of more than 2 traits? The Forked-Line Method for a F2 progeny of a trihybrid cross Locus 1 Locus 2 Locus 3 Phenotype 3dominant … Phenotype 2 dominant Phenotype 3recessive … Phenotype 1 class Phenotype 2 recessive Phenotype 3dominant … Phenotype 3 recessive … trihybrid ratios  Derivations of the Second Law For alleles of the two gene pairs to assort independently into gametes, they have to be on different pairs of chromosomes. 2nd Law applies to trihybrid crosses (& beyond) as long as they are in different chromosomes. – 3:1 segregation with respect to each trait independently of segregation with respect to the other traits. – Polynomial equation: 𝑛 3 1 𝐷+ 𝑅 = 4 4 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑒𝑛𝑒𝑠 𝑖𝑛𝑣𝑜𝑙𝑣𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 𝐷 = 𝑑𝑜𝑚𝑖𝑛𝑎𝑛𝑡 𝑝ℎ𝑒𝑛𝑜𝑡𝑦𝑝𝑖𝑐 𝑐𝑙𝑎𝑠𝑠𝑒𝑠 𝑅 = 𝑟𝑒𝑐𝑒𝑠𝑠𝑖𝑣𝑒 𝑝ℎ𝑒𝑛𝑜𝑡𝑦𝑝𝑖𝑐 𝑐𝑙𝑎𝑠𝑠𝑒𝑠  𝑛 Expected Phenotypic Ratios in Crosses 3 1 𝐷+ 𝑅 = 4 4 𝑛=1 1 3 1 Monohybrid Cross (1 pair of traits): 4 𝐷+ 𝑅 4 3/4 D + 1/4 R 2 phenotypes in F2 offspring (21) 𝑛=2 3 1 2 Dihybrid Cross (2 pairs of traits): 4 𝐷+ 𝑅 4 9/16 D1D2 + 3/16 D1R2 + 3/16 R1D2 + 1/16 R1R2 4 phenotypes in F2 offspring (22) 𝑛=3 3 3 1 𝐷+ 𝑅 Trihybrid Cross (3 pairs of traits): 4 4 27/64 D1D2D3 + 9/64 D1D2R3 + 9/64 D1R2D3 + 9/64 R1D2D3 + 3/64 D1R2R3 + 3/64 R1D2R3 + 3/64 R1R2D3 + 1/64 R1R2R3 8 phenotypes in F2 offspring (23) …  Calculating the frequency of genotypes in crosses Mendelian Laws reflect the random distribution of alleles segregated into gametes of a diploid cell and therefore probability laws apply. We can use probability to determine the expected frequencies of genotypes in a progeny based on this assumption (of equal segregation of alleles and independent assortment of alleles) First Mendelian Law Second Mendelian Law  second law bhevaves independently. with frequencies Applying the Second Law to predict frequencies In the following tetrahybrid cross: Aa Bb Cc Dd x Aa Bb Cc Dd Locus A Locus B Locus C Locus D Locus A Locus B Locus C Locus D What is the proportion of the progeny with the genotype AA bb Cc Dd ? Consider every locus separately (as their alleles will segregate independently from the other loci). For each locus, there are always two alleles: either they are different, or the same allele twice. The First Law states that alleles of a single locus will equally segregate to the gametes, so each has a $ 50% change # of being represent in the gametes. If these alleles are the same (as in the homozygote), then the only allele type will account for 100% of the information for that locus in the gametes. Gametic frequencies 0.5 0.5 If you find it necessary, use a Punnet square to visualize this. 𝐴 𝑎 𝐴/𝐴 𝑎/𝑎 Gametic frequencies 𝐴/𝑎 1 0.5 𝐴 𝐴/𝐴 4 𝐴/𝑎 homozygous heterozygous 1 Genotypic (0.5 x 0.5) 2 (0.5 x 0.5) 2 Frequencies 0.5 𝑎 𝐴/𝑎 𝑎/𝑎 1 =0.25 (1/4) =0.5 (1/2) 4  F1 F1 Aa Bb Cc Dd x Aa Bb Cc Dd What is the proportion of the progeny with the genotype 0.5 0.5 AA bb Cc Dd? 𝐴 𝑎 1 0.5 𝐴 𝐴/𝐴 4 𝐴/𝑎 AA bb Cc Dd 1 1 1 1 1 1 2... = = (1.5%) 0.5 𝑎 𝐴/𝑎 𝑎/𝑎 1 4 4 2 2 64 4 Homozygous loci Heterozygous loci 1.5% of the F2 progeny will have the genotype AA bb Cc Dd A pea plant is of the genotype AaBb, where A and B are dominant alleles. What are the probabilities that: i) a polen from this plant carries the (a) allele? ii) when selfed, a (aaBb) plant is produced? Aa x Aa Aa Bb AB Ab aB ab 50% A a 50% A 50% a A a 50% 50% i.) 50%, first law. ii.) 12.5% A pea plant is of the genotype AaBb, where A and B are dominant alleles. What are the probabilities that: i) a polen from this plant carries the (a) allele? ii) when selfed, a (aaBb) plant is produced? Aa Bb x Aa Bb 0.5 0.5 𝐴 𝑎 aa Bb 1 0.5 𝐴 𝐴/𝐴 4 𝐴/𝑎 1 1 1 1 2. = (12.5%) 0.5 𝑎 𝐴/𝑎 𝑎/𝑎 1 4 2 8 4  A pea plant is of the genotype AaBb, where A and B are dominant alleles. What is the probability that a (aabb) genotype will be produced in the progeny? Aa Bb x Aa Bb aa bb 1 1 1. = (6.2%) 4 4 16 Genotypic frequency Phenotypic 9 : 3 : 3 : 1 frequency 16 16 16 16 The double recessive is the only F2 class where phenotypic frequency equals genotypic frequency aa bb IS PARENTAL OR RECOMBINANT PHENOTYPES ALWAYS THE SAME ?  Y/Y; R/R y/y; r/r P0 x recombinants appear with shuffling parental genotypes Y/y; R/r F1 Parental and Recombinant phenotypes will vary, GAMETES depending on the cross Y,R Y/Y; R/R Y/Y; r/R y/Y; R/R y/Y; r/R F2 Y,r GAMETES Y/Y; R/r Y/Y; r/r y/Y; R/r y/Y; r/r y,R Y/y; R/R Y/y; r/R y/y; R/R y/y; r/R y,r Y/y; R/r Y/y; r/r y/y; R/r y/y; r/r Y/Y; R/R y/y; r/r x IN THE CROSS WE HAVE BEEN ANALYZING... Y/y; R/r parentals recombinants Y/Y; r/r y/y; R/R x Y/y; R/r F1 ? In the cross on the left, the two recessive phenotypes and the two dominant phenotypes are present in single P0s. Would the predicted F1 phenotype change? Y/Y; r/r y/y; R/R x Y/y; R/r No, dominance does not vary with the origin of the gamete.  CROSS 2 CROSS 1 dominant; recessive recessive; dominant double dominant double recessive Y/Y; r/r y/y; R/R Y/Y; R/R y/y; r/r x x Y/y; R/r F1 Y/y; R/r = Y/y; R/r =  CROSS 2 CROSS 1 dominant; recessive recessive; dominant double dominant double recessive Y/Y; r/r y/y; R/R Y/Y; R/R y/y; r/r x x Y/y; R/r F1 Y/y; R/r = Y/y; R/r = Will F2 phenotypic ratios changed?  No! As long as alleles assort independently, the F2 phenotypes will be determine by the random combination of dominant and recessive alleles resulting in the predicted Mendelian ratio, independently of the parent from which each allele originally came from. P0 x x F1 315 108 101 32 F2 + + + = 556 F2 SEEDS 9:3:3:1  CROSS 2 CROSS 1 dominant; recessive recessive; dominant double dominant double recessive Y/Y; r/r y/y; R/R Y/Y; R/R y/y; r/r x x Y/y; R/r F1 Y/y; R/r = Y/y; R/r = 9 : 3 : 3: 1  Does anything change at all?  THE "NEW" RECOMBINANT INFORMATION IS DIFFERENT BETWEEN THE TWO CROSSES WITH DIFFERENT PARENTAL GENOTYPES P0 x x F1 F2 9:3:3:1 9:3:3:1 While this will not impact frequencies in a case of Mendelian segregation, determining Parental and Recombinant phenotypes is critical to assess frequencies of recombination in cases of linkage.  Does it matter what is P and what is R phenotypes ? Frequency of Gametes Parental = Recombinant IF follows Mendel’s Laws Meaning the predicted frequency of recombination in the cross (assortment) does not need to be calculated, the progeny shows a Mendelian ratio in accordance to the Second Law (f=50%), identifying recombinants is not necessary. Parental ≠ Recombinant IF violates Mendel’s Laws Meaning the frequency of recombinant gametes depends on (linkage) factors other than assortment (crossover) which does not comply with the Second Law predictions and need to be calculated for every cross separately. For that, identifying recombinant and parental phenotypes in the progeny is key.  How do I statistically prove (test) whether progeny numbers in a generation conform with predictions in Mendelian Laws?  Formulating and Testing Genetic Hypotheses Hypothesis: a well-formulated scientific idea Data collected from observations or from experimentation enable scientists to test hypotheses. Genetics: Are the results of a cross consistent with a hypothesis?  Using the F2 progenies in a Mendelian Dihybrid Cross =556 Real proportions 9.8 3.4 3.1 1 How significant are these differences? Sampling error? Mendel’s proportions 9 3 3 1  What are the chances that, if I repeat Mendel's cross for the color/skin shape in peas, I would get F2 phenotypic proportions consistent with a 9:3:3:1 ?  The Chi-Square (c2) Test Step1. Predict Expected numbers based on hypothesis. Step2. Calculate how well our data fit our hypothesis (the c2 statistic). Step3. Determine the degrees of freedom. Step4. Accept or reject our hypothesis. (Compare the c2 statistic to the critical value)  The Chi-Square (c2) Test prediction of the second law Step1. Predict Expected numbers based on hypothesis. 9:3:3:1 Step 2. Calculate how well our data fit our hypothesis (the c2 statistics). O E 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 − 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑥2 = 3 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 E Where “O” and “E” are the total number of progenies in each phenotypic class  The Chi-Square (c2) Test c2 tests whether the sample collected (observed dataset) can be used to support a hypothesis for which a prediction can be made (expressed in the expected values). Often, this will test against the hypothesis (null hypothesis) when it is suspected that the sample data significantly differs from the testable expectation. In Classical Genetics, this prediction is often a Mendelian ratio. The test does not prove the hypothesis per se, it simply indicates whether the hypothesis can be explained by the data set (how well your results fits the hypothesis). (!"#$%&$' " E)*$+,$') ! ! =# ! E)*$+,$' Calculating expected frequencies and the Chi-square value Mendel’s predicted numbers distribution (9) (3) (3) (1) Total # offspring x expected frequency of the phenotype 556 x 9/16= 312.75  The Chi-Square (c2) Test accept critical value reject Step3. Determine the degrees of freedom. Number of phenotypic classes -1, e.g. 4-1=3 The critical value (cv) defines the confidence interval (95%). By convention, the cv of 5% represents the maximum accepted chances of second experiment giving out the the same deviation to the prediction as in the observed data (explained solely by sampling error). Step4. Accept (‘fail to reject’) or reject our hypothesis. (Compare the c2 statistic to the critical value) A c2 value greater than critical value (associated to P=0.05) rejects the hypothesis A c2 value smaller than critical value (associated to P=0.05) fails to reject the hypothesis (the observed data cannot be used to validate the hypothesis) c2 = 0.51 0.05 Difference between observed and predicted values is significant and likely not due to sampling error (>5% probability that error explains the data). sample cannot support hypothesis  Comparison to the Critical Value DeVries’ Dihybrid Cross: § c2 = 22.91 § Degrees of Freedom = 4 - 1 = 3 § Critical Value = 7.815 § 22.91>>7.815 (higher probability associated) § Reject the null hypothesis  When do I use this? A typical use of chi-square is testing against a Mendelian ratio : Since the prediction for phenotype distributions in a non-Mendelian case depends on previously knowing the frequencies of parentals and recombinants (derived, for example, from the genetic map of the genes involved in case of linkage), statistically testing for these exceptions is often not possible. Instead, using a non-variant Mendelian ratio as a null hypothesis offers a more practical way to determine the inheritance pattern. More often, the chi-test is used to rule out Mendelian inheritance for the genes in question, such that calculating their recombination frequency can be logically used to understand their disposition on the same chromosome. 3:1 (monohybrid) Examples of Mendelian ratios used to test against null hypotheses: F2 9:3:3:1 (dihybrid) 27:9:9:9:3:3:3:1 (trihybrid) used to test against something 1:1:1:1: (test cross)  Testing genetic hypotheses The garden flower Salpiglossis sinuata comes in two colors. A cross is made between true-breeding parental plants to produce F1 plants, which are in turn self-fertilized to produce F2 progeny. Red x blue à F1 All redà F2 72 red and 29 blue. State and test the most plausible hypothesis explaining the inheritance of this trait. P0 P0 F1 x = Red dominant F2 x = F1 F1 21 72 2, not 4 phenotypes in F2 Flower color phenotype determined by alleles of the same locus rr RR Rr x = F2 x = Rr Rr rr Rr + RR 29 72 Mendelian ratio 25% (1) 75% (3) Hypothesis: R and r are alleles of a single gene Prediction: F2 ratios should be 3 red (dom) : 1 purple (rec) O E (!"#$%&$' " E)*$+,$') ! RR !! = # E)*$+,$' Rr 72 76 29 25 df= 2-1 =1 rr total 101 101 c2 = 0.02