test2 - GENE 3200 – Summer 2010 Exam 2 – Bedell –...

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Unformatted text preview: GENE 3200 – Summer 2010 Exam 2 – Bedell – July 1 ID no. 810_______ KEY-Revised 7/7/10 ________ 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 48 hours of the grades being posted on WebCT. Answers to numbers 25 and 26 were not correct on original key. 1. Construct a genetic map based on the progeny genotypes from the three-point cross below in which an F 1 Aa Bb Dd was crossed to aa bb dd. Show all equations needed to work this problem and give a brief justification for each answer. A B D 1242 A B d 7 A b D 128 a B D 206 A b d 218 a B d 114 a b D 15 a b d 1216 Total: 3146 1A. (2 pts) What is the order of the three genes? Give a brief explanation for your answer. a d b or b d a (or ADB or BDA) First have to figure out the configuration of F1 alleles. The most common classes of offspring are nonrecombinants, in this case A B D and aa bb dd, so the configuration of the F1 is the genotype in question is ABD/abd. Then, use least frequent class of progeny to deduce middle gene because they have to result from double crossovers. ABd and abD result from double crossovers and only d alleles have changed with respect to parental configuration. So d must be in the middle. 1B. (5 pts) What are the distances (in m.u.) between the three genes? Show your calculations . a and d = (206 + 218 + 15 + 7)/3146 = 0.142, or 14.2%. d and b = (128 + 114 + 15 + 7) = 0.084, or 8.4%. a – b = 14.2 + 8.4 = 22.6 Sum the number of recombinants in each class of progeny, i.e. recombinants between a and d are aBD, Abd, ABd, and abD; recombinants between d and b are AbD, aBd, ABd, and abD. a - b distance is sum of the a - d and d – b distances. 1C. (3 pts) How much crossover interference (if any) occurred? Show your calculations. Expected double crossovers = 0.142 x 0.084 x 3146 = 37 Interference = I = 1 – observed DCO = 1 - 22/37 = 1 – 0.59 = 0.41 expected DCO 2. (3 pts) Which of the following statements are true of double-stranded DNA? Circle all correct answers. A. A/G = T/C B. A + T = G + C C. A/G = C/T D. (C+A) / (G + T) = 1 E. A + G = C + T GENE 3200 – Summer 2010 Exam 2 – Bedell – July 1 ID no. 810 _______KEY-Revised 7/7/10 ________ 2 Questions 3 - 7 refer to the structure at the right. 3. (2 pt). What is the name of the nitrogenous base shown in the figure? A. Guanine D. Thymine B. Adenine E. Uracil C. Cytosine F. None of the above 4. (2 pts). Put the correct number on each carbon and nitrogen atom in the central ring structure of the nitrogenous base....
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This note was uploaded on 09/01/2010 for the course MIBO MIBO 3500 taught by Professor Dustman during the Summer '09 term at UGA.

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test2 - GENE 3200 – Summer 2010 Exam 2 – Bedell –...

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