Exam 2 - GENE 3200 Fall 2010 Bedell Exam 2 Oct 7 ID no...

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GENE 3200 – Fall 2010 – Bedell Exam 2 – Oct 7 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 48 hours of the grades being posted on WebCT. 1. The best football players have dominant alleles for three genes: big (B), smart (D), and fast (F). Recessive alleles for each of these genes result in being small (b), dumb (d), and slow (f). Data from a (hypothetical) three-pt testcross are shown below. Phenotype Genotype Number Big, smart, fast B, D, F 324 Nonrecombinant Small, smart, fast b, D, F 49 SCO Big, dumb, fast B, d, F 3 DCO Big, smart, slow B, D, f 42 SCO Small, dumb, fast b, d, F 32 SCO Small, smart, slow b, D, f 5 DCO Big, dumb , slow B, d, f 51 SCO Small, dumb, slow b, d, f 309 Nonrecombinant Total: 815 1A. (2 pts) What is the order of the three genes? Give a brief explanation for your answer. To deduce which gene is in the middle it is necessary to find out the double recombinant classes and compare them to the parental types. Double recombinants are the least common, in this case b D f and B d F. Only the alleles of D have changed their position with respect to the parental combination, therefore the order of the genes is B D F (or F D B). 1B. (6 pts) What are the distances in map units between the three genes? Show your calculations. B – D distance = (51 + 49 + 5 + 3)/815 = 108/815 = 0.133, or 13.3% D – F distance = (42 + 32 + 5 + 3)/815 =82/815 = 0.101, or 10.1% B – F distance = 13.3 + 10.1 = 23.3 Or B- F distance = ((51 + 49 + 42 + 32 + 5 + 5 + 3 + 3) = 190/815 = 23.3 1C. (2 pts) How much crossover interference (if any) occurred? Show your calculations. Expected double crossovers = 0.133 x 0.101 x 815 = 11 Interference = I = 1 – observed DCO = 1 - 8/11= 1 – 0.73 = 0.27 expected DCO
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GENE 3200 – Fall 2010 – Bedell Exam 2 – Oct 7 ID no. 810__________ KEY ________________ 2 2. (2 pts) In Griffith's transformation experiments the mice die when they were injected with which of the following kinds of bacteria? (a) heat-killed, S strain (b) live, R strain (c) live, R strain + heat-killed, S strain (d) heat-killed, R strain (e) live, R strain plus heat-killed, R- strain 3. (2 pts) What is the name of the structure to the right? (a) Alanine (b) Uracil (c) Adenine (d) Guanine (e) Methionine ANSWER 3B. (2 pts) Number the positions of the carbon and nitrogen atoms in the structure on the right. 3C.
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Exam 2 - GENE 3200 Fall 2010 Bedell Exam 2 Oct 7 ID no...

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