2005MT2_Key_Pts_v2

2005MT2_Key_Pts_v2 - BICD100 Genetics Fall 2005 Midterm 2...

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BICD100 Genetics Fall 2005 Midterm 2 Updated key with partial credit explained REGRADE REQUESTS must be turned in by 3:30PM on Thursday December 1 st , 2005. Grading is not curved in this class, but to give you an idea how you fared relative to your peers, here is the grade distribution. It is not meaningful to report a mean and std because this distribution is not at all normal (Gaussian). 6 8 10 12 14 16 18 20 22 24 26 0 5 10 15 20 25 30 35 40 45 50 F D C- C C+ B- B- B B B+ B+ A- A A+ Statistics for Midterm 2 A+,A,A- 109 36% B+,B,B- 126 41% C+,C,C - 39 13% D 7 2% F 26 8% other 16 5% total 323 Perfect (A+) 44 14% There are still 22 students (7% of the class) going into the final exam with a total score of 50 or higher on the two midterms. Congratulations!
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MIDTERM 2 VERSION A KEY 1. You are studying three autosomal recessive mutations in Drosophila that are linked: rosy, hairy, and wobbly. You cross a fly from a pure-breeding wobbly strain to a fly from a pure-breeding hairy, rosy strain to obtain an F1. You then cross the F1 flies to flies from a pure-breeding rosy, hairy, wobbly strain to obtain the following F2: 12 rosy wobbly 84 hairy rosy wobbly 407 hairy wobbly 1941 wobbly 2031 hairy rosy 12 hairy 408 rosy 105 wild type TOTAL 5000 Derive the most accurate map you can of the order of these genes and the distances between them. State the amount of interference, if any. SHOW ALL YOUR WORK. Parents: wild type x rosy hairy wobbly wobblyrosy hairy 4.3 16.8 Interference = 0.33 (expected 35.7 double xo, observed 24) 1 point for calculation of distance wobbly-rosy = 4.3 cM 1 point for calculation of distance rosy-hairy = 16.8 cM 1 point for calculation of distance wobbly-hairy = 20.1 cM OR for saying rosy is in the middle based on identifying the rare DCO class In which case you are not required to also compute the outer distance 1.5 point for drawing correct gene order and all distances on map 1.5 for interference (+0.5, if correct formula was used with wrong number, -0.5 if the interference was multiplied by 100%)
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2. You cross an arg3 haploid yeast strain to an ind2 haploid yeast strain to obtain a diploid that is heterozygous for both genes: arg3 + x + ind2 arg3 + diploid + ind2 The yeast strain produces unordered tetrads. For each of the following situations, diagram the 4-chromatid stage of a meiosis producing a Tetratype tetrad. The diagram should show the location(s) of crossover(s) if any, and clearly indicate which chromatids segregate together in order to explain the genotypes of the 4 spores. For full credit you must show the minimum number of crossovers that could explain the tetrad type. If there is more than one way to obtain the specified tetrad type using the same number of crossover events, show only one example. a) if arg3 and ind2 are linked:
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This note was uploaded on 08/30/2008 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.

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2005MT2_Key_Pts_v2 - BICD100 Genetics Fall 2005 Midterm 2...

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