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12 GENETIC DATA ANALYSIS2.1 Strategies for learning geneticsWe will begin this lecture by discussing some strategies for learning genetics. Genetics isdifferent from most other biology courses you have taken in that memorization is not veryimportant. You are expected to learn vocabulary and some examples of genetic disorders,formulae, etc. But learning and applying concepts is much more important. In particular, youneed to be able to think about and analyze genetic data. Almost all the homework and examquestions in this part of the course will require you to examine and analyze data, and to makesome conclusion based upon your analysis. In this way, the work you do in this course is muchmore similar to work done by real geneticists. This is one reason why students who have always done well in biology sometimes do poorly ingenetics. Learning how to approach a set of genetic data in a logical and consistent way takespatience and a good deal of practice. I will introduce you to ways to think about problems during lectures. It is therefore veryimportant that you attend lecture. I will use several techniques of “active learning”, becausethese techniques have been proven to be more effective in increasing student learning than thetraditional lecture-only format. I think these techniques also make class time more interestingfor students. So, please do the following:1) Come to class!2) Bring a calculator and paper with you.3) Read assigned reading and glance over the lecture notes before coming to class.4) During class, listen and think about the material that is being presented.5) Try to answer questions that are posed.6) Ask questions about anything that is not clear!7) The very best way to learn something is to teach it to someone else. Get together withclassmates, trade off trying to explain a concept or the solution to a particular problem. And please don’t do the following:1) Don’t spend class time furiously taking notes. All lecture notes and slides will beprovided to you on the web page. 2.2 Genetic Data Analysis: Useful Rules of ProbabilitySum rule: The combined probability of two events that are mutually exclusive is the sum of the individualprobabilities.Genetic example of the sum rule:Parental genotypes (monohybrid cross): GG x gg ‘x’ indicates a genetic crossF1 genotype: Gg F2 (produced by mating F1 individuals): 1/4 GG: 1/2 Gg: 1/4 gg (1:2:1 genotypic ratio)2Q: What is the probability that an F2 offspring of a monohybrid cross has thedominant phenotype (is either GG OR Gg)?A: P[dominant phenotype] = P[GG] + P[Gg] = ____ + ____ = ____(Note: the P[x] stands for “Probability of x”)Product rule: The probability of both of two independent events is the product of the individualprobabilities.Genetic example using product rule and sum rule:Parental genotypes (dihybrid cross): GGww x ggWWF1: GgWwF2: 9/16 G–W– : 3/16 G–ww : 3/16 ggW– : 1/16 ggww(Note: the – indicates either the dominant or recessive allele G– indicates an individual with thedominant phenotype: either GG or Gg)Q: What is the probability that an F2 offspring of the following dihybrid crosswill have at least one dominant allele for each trait (i.e. that it will be G–W–)? Solution: First note that segregation of the G locus is independent ofsegregation at the W locus (Mendel’s law of independent assortment) So thatP[G–W–] = P[G–] * P[W–]As above, GG and Gg are mutually exclusive (an individual can be GG or Gg, butnot both). WW and Ww are also mutually exclusive.P[G-] = P[Gg] or P[GG] = P[Gg] + P[GG]P[Gg] = 1/2 P[GG] = 1/4So: P[G-] = 1/2 + 1/4 = 3/4P[W-]= P[Ww] + P[WW] = 1/2 + 1/4 =3/4P[G-W-] = P[G-] * P[W-] = 3/4 * 3/4Answer: iii. Complex genetic problem--6 independent genes: Parental genotypes: AA bb CC DD ee ff x aa BB cc dd EE FFF1: Aa Bb Cc Dd Ee Ff x Aa Bb Cc Dd Ee FfQ: What proportion of F2 progeny will be AA bb Cc DD ee Ff ?3A:iv. Genetic example using conditional probabilityQ: In the F2 progeny from monohybrid cross, what is the proportion ofheterozygotes among dominant progeny? A: P[heterozygous] / P[dominant] = P [Gg] / P[GG or Gg] =1/2 / [1/4 + 1/2] =1/2 / 3/4 = 2/32.3 Genetic Data Analysis: Goodness of Fit What if the F2 generation of a monohybrid cross has a phenotypic ratio close to 3:1, but not exactly 3:1?Q: How do we tell if deviations from expected proportions in a geneticexperiment are due to chance or due to the fact that our genetic hypothesis iswrong?A: Repeat an experiment many times. E.g, the experiment is: toss a fair coin100 times and count how many heads are throw. Repeat this experiment 100times. Q: What is the most likely outcome of a single experiment?A: Q: Will all 100 experiments have the same outcome if the coin is fair?A: Coin Toss Experiment01234567891 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97Number of HeadsNumber of Times/1004Getting 65 H and 35 T is an unlikely result, but not impossible. How unlikely is it? The probability isactually 0.0017. In other words, this result is expected to occur 17 times out of every thousand times theexperiment is repeated (or 1.7 times out of every on hundred trials).In same way, consider a cross of Aa to an individual of unknown genotype that yields 100 progeny. Wehypothesize that the unknown individual has genotype aa.Q: If the progeny contain 65 of the dominant phenotype, and 35 of therecessive phenotype, can we reject our hypothesis that the unknown parentwas aa and that the cross was therefore Aa x aa? Another way to phrase thequestion is, “how likely are we to get a result this different from the expectedresult if our hypothesis is true?”First, what is the expected result? What is the observed result? How could wemeasure the difference between the expected and observed results?A: A test commonly used to measure the goodness of fit is the CHI-SQUARE TEST (pronouncedwith a hard ‘C’ and to rhyme with ‘pie’).The observed number (O) in each category is compared


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UIUC IB 201 - GENETIC DATA ANALYSIS

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