gene key 5

gene key 5 - GENE 3200 Bedell - MWF Final Exam Dec 10, 2009...

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Unformatted text preview: GENE 3200 Bedell - MWF Final Exam Dec 10, 2009 ID no. 810____________ KEY ________________ 1 Write your ID number (810 number, NOT SSN) on every page. Write your initials on the first page only. All of the multiple choice questions have only one correct answer. The exam should be completed in ink. Regrades will not be given if you write in pencil. All requests to re-grade an exam must be made in writing to Dr. Bedell within 24 hours of the grades being posted on WebCT. 1. (2 pts) Calculate the probability of the production of a homozygous recessive genotype for the following cross: AaBbccddEe AaBbCcddee 1/4 1/4 1/2 1 1/2 = 1/64 or 0.0156 2. You have a pure-breeding strain of mice with short tails and you cross that with another pure- breeding strain that has long tails. All of the F1 progeny of the cross between the two parental strains had short tails. In the F2 progeny 91 mice had short tails and 29 had long tails. 2 (cont) (2 pts) Write the phenotypes and genotypes of the two parental strains. Use T for the dominant allele and t for the recessive allele. TT have short tails tt have long tails 3. (2 pts, 0.5 pt for each value). In tomatoes, cut leaf and potato leaf are alternative characters, with cut (C) dominant to potato (c). Purple stem and green stem are another pair of alternative characters, with purple (P) dominant to green (p). A true-breeding cut, green tomato plant was crossed with a true-breeding potato, purple plant, and the F1 plants were intercrossed. Shown below are the phenotypes and observed numbers of F2 progeny. 3. (cont) With these observed F2 progeny and phenotype ratios, make a hypothesis about the inheritance pattern and then fill in the table below with the expected numbers of F2 offspring of each phenotypic class. Show all equations needed to work this problem and give a brief justification for each answer. Phenotype Number of F2 progeny observed Number of F2 progeny expected cut, purple 189 180 cut, green 67 60 potato, purple 50 60 potato, green 14 20 Total number of F2 progeny is 320, hypothesis is that the two genes are segregating independently so expect 9:3:3:1 segregation. 320/16 = 20 (for potato, green = homozygous recessive for both) 320/16 x 9 = 180 (for cut, purple = C-; P- 320/16 x 3= 60 (for cut, green = C-; pp and potato, purple = cc; P-) GENE 3200 Bedell - MWF Final Exam Dec 10, 2009 ID no. 810____________ KEY ________________ 2 4. Sally (IV-1) and Jim (IV-2) both had relatives with hypercholesterolemia, a rare autosomal recessive disorder. To the right is a pedigree with all of the relatives of Sally and Jim (assume complete penetrance). They want to know the chances that a child of theirs would have this disorder....
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gene key 5 - GENE 3200 Bedell - MWF Final Exam Dec 10, 2009...

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