Genetics assign2-draft
- 1. Question1a. TrihybridCross – Linkage ina Three-PointTestcross Parental gene sequence: ACb = Acb and aCB acB F1 - 43% AaCcbb – 86% Parental (ACb/acB) - 43% aaccBb - 4% aaccbb – 8% Single crossover - 4% AaCcBb - 3% AaccBb – 6% Single crossover - 3% aaCcbb - 0.5% aaCcBb – 1% double crossover - 0.5% Aaccbb Parental gametes=ACb (2480) andacB (2429) Double crossovers= Acb (13) and aCB (19) Equationusedtowork outthe distance betweensingle crossovers: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴 − 𝐶 = 𝑁𝑜. 𝑜𝑓 𝑟𝑒𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑛𝑡 𝑇𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑝𝑟𝑜𝑔𝑎𝑛𝑦 × 100 Distance betweenA –C is 8.44cM 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴 − 𝐶 = 453 + 32 5742 × 100 = 8.44 Distance betweenC –B is6.615cM 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐶 − 𝐵 = 348 + 132 5742 × 100 = 6.615 Distance betweenA –B is 15.06cM 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴 − 𝐵 = 453 + 348 + +2(32) 5742 × 100 = 15.06 8.44cM + 6.615cM = 15.06cM ∴ It can be stated that the calculations are correct. A 8.44cM C 6.62cM B 15.06cM
- 2. Question1b. To calculate coincidence andinterference,the follow equationwasused: 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐶𝑜𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 = % 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟𝑠 % 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟𝑠 A – C = 8.44cM (or 0.0844%), C – B = 6.62cM (or 0.0662%) 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 % = (0.0844 × 0.0662) × 100 = 0.5587% Double cross= 0.5587% 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 % = 19 + 13 5742 × 100 = 0.5572 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐶𝑜𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 = 0.5572 0.5587 = 0.9973 Coincidence =99.73% Interference1 − 0.9973 = 0.0027 0.27% of double crossoversdidnotformdue to interference. Question2 Answer:a) The pedigree isX-linkedrecessive,thisisdue tomode of inheritance inwhichamutation (rare disease) isinagene on the X chromosome whichcausesthe phenotype toonlybe expressedin maleswhoare hemizygousforthisgene mutation.ItisnotX-linkeddominantasX-linkeddominance requiresonlyone copyof the allele tocause a disorderwheninheritedfromaparentwhohas the disorder,howeverinthispedigree,thisisnotthe case. Answer:b) A = normal,a = affected XA XA = normal female,XA Xa = female carrier,Xa Xa = affectedfemale XA Y = normal male and Xa Y = affectedmale Animal 1 possible genotypes:XA XA or XA Xa Animal 3 possible genotypes:XA Xa Answer:c) Possible parentgenotype,male =Xa Y and female =XA XA or XA Xa Firstcross: Xa Y x XA XA
- 3. XA XA Xa XA Xa XA Xa Y XA Y XA Y None of the childrenare effected,4/4are normal.100% chance of femalestobe carriersand 100% chance of malesto be normal. Secondcross: Xa Y x XA Xa XA Xa Xa XA Xa Xa Xa Y XA Y Xa Y The probabilityof the childrenbeingaffectedis50% (2/4). The probabilityof femalesbeingaffected is50%, and the probabilityof malesbeingaffectedis50%.