Name: ___________________________ Lab room, day, time: __________________TA name: _______________Part 1: Sordaria ResultsReview the lab manual and lecture.1. Your goal is to determine the distance of the Tan locus from its centromere.a. (1 point) Define a map unit (also referred to as a centimorgan). This should be a complete sentence and not the abbreviation given in lecture (ifyou pull a definition from a source, you should cite it).Something along the lines of: A map unit is the distance between a gene and its centromere (or another gene) which results in one percent crossover rates during meiosis.b. (1 point) Your section data has a total number of scored asci showing an MI segregation pattern and the total number of scored asci showing an MIIpattern. Write your own formula for calculating map units using that data. Remember that Sordaria has a particular quirk in this regard due to the spores being contained in asci.[Percent crossover rates is equal to m.u., so they just need to have a formula to calculate the percent of crossovers in their samples. The quirk is that since a recombinant (MII) ascus shows both a set of recombinant and non-recombinant spores they should take half of the % of MII asci).]Examples:Tan m.u. = ½ (# M2)/(# M1 + # M2)Tan m.u. = ((#M2 asci)/(total asci))/2They should have more than just “percentage of M2.”c. (1 point) Calculate the distance in map units of the Tan locus from its centromere.Answers will depend on each section’s data but should be the same for all students within a given section.Patterns of inheritance:This semester you are working to figure out the pattern of inheritance of 2 different lines of flies. This task and its associated statistical tests are much easier if you can recognize or reason through the patterns you are seeing. Last week you were asked to consider the hybrid outcomes of the F1 generation. This week you will consider the patterns observed in the F2 resulting from an F1 sibling cross. For the following questions you will get the “safest” results by doing the punnet square rather than just guessing. When in doubt, write it out.For all of the following questions assume the patterns seen are the results of crossing heterozygous siblings in the F1 and that the P generation was crossed just like you did in the lab (the P generation males showed the mutation and females were wild type).2. The simplest patterns are those dealing with single, dimorphic, and autosomal genes/traits (the specific trait doesn’t matter, these are generalized patterns).a. (0.5 points) What are the phenotype proportions (you can work with % or ratios) of individuals that are wild type vs. mutant for a recessive mutation? Make sure you provide labels and not just values.3 wild-type : 1 mutant (75% wild type, 25% mutant).b. (0.5 points) What are the phenotype proportions if the mutation is dominant?3 mutant: 1 wild typec. (1 point) What if it was a recessive mutation on the X chromosome rather than being autosomal? Work this one out by assuming the P generation was between affected males (which are technically hemizygous) and homozygous wild-type females.For “all” flies it would be – 2 female wild type : 1 male mutant : 1 male wild typeThey can also split it into females and malesFemales: 1 wild type : 0 mutant (i.e. 100% of females are wild type)Males: 1 wild type : 1 mutantd. (1 point) How do the patterns between a, b, and c compare and contrast? What features stand out most clearly to differentiate the three patterns?They are all clearly different. In each case there is a mix of both mutant and wild type phenotypes (no pattern has 100% of just one unless you consider the females in1c separately). For recessive mutations, wild type is most common, for dominant, themutation is most common, and finally for X-linked recessive the stand out feature is that no females are affected and only (half of) the males show the phenotype.3. Things get more complex when looking at a cross of individuals heterozygous for two traits. This one can be helpful to consider actual phenotypes rather than just wild type vs. mutant. Refer to the gene/mutation chart on p. 113 of the lab manualfor these questions.a. (1 point) What would be the F2 phenotypic proportions for a dihybrid (e.g.heterozygous for both traits) cross involving the apterous and ebony mutations?Dihybrids are “easy” in that they follow a 9:3:3:1 ratio pattern if they are unlinked.9 wild type: 3 wingless : 3 black-bodied :1 wingless and black-bodied.b. (1 point) What are the F2 proportions for the same type of cross but with curled and ebony?Dihybrids with 1 dominant are still 9:3:3:1, but the categories are distributed differently. It’s safest to do a punnet square here.9 curly wings : 3 curly wings and black-bodied : 3 wild type : 1 black-bodied.c. (1 point) What if the genes were linked due to being on the same chromosome, as in the case of apterous and pink? Calculate assuming that crossing over does not occur (100% linked).Things get complicated here because the traits sort as a unit. There are only 4 possible gametes that can be produced (2 for each parent) so the punnet square will only be 2x2. Without crossing over there will never be “non-parental” phenotypes.+ + ap p+ + +/+ +/+ +/ap +/pap p ap/+ p/+ ap/ap p/p3 wild type : 1 dark-eyed and black-bodiedd. (1 point) How do the patterns between parts a, b, and c compare and contrast? What features stand out most clearly to differentiate the three patterns?There is less comparison between these three inheritance types beyond the fact that they show the presence of both wild type and mutant traits. 2a and 2b have the same9:3:3:1 ratios. 2c is very different than 2a or 2b because it only has a 3:1.In terms of stand-out features, 2a and 2b probably most obviously differ by what is in the most common (9) category (is it wild type or a dominant mutation?). 2c standsout because it has 2 traits, but neither of the recessive mutations appear singly; if it shows one recessive phenotype it ends up showing the other. [The “parental” phenotypes (P generation) were wild type and wingless with pink bodies. That ends up being the only phenotypes seen in the F2 as well, so the “non-parental” phenotypes are not
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