Discussion 3 key Summer 2009

Discussion 3 key Summer 2009 - The number of recombinants...

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GENETICS (BIO F325) Summer 2009 Discussion 3 key _____________________________________________________________________________________ 1. (i) (d) (ii) (c) 2. 2 23 different gametes. 3. (a) chromatin (b) homologous (c) scaffold (d) mammals 4. χ 2 = 16.10; the hypothesis is rejected. 5. (a) Cis conformation (b) Trans conformation 6. (a) 25% (b) 50% (c) 45% (d) 38% 7. (c) 8. (a) The proportions of gametes produced by parent 1 were: 38% a + ; b + 37% a;b 12% a + ; b 13% a;b + (b) No, because we expect 25% of each under independent assortment. (c) The two gene pairs must be linked; the %R (RF) = 12 + 13 = 25%, and the two loci are 25 map units apart on the same chromosome. (d) a + b + /a b and a b/a b 9. This is a testcross, so the ratio of meiotic products will be revealed in the phenotypes of the progeny. The progeny will be: 43% wild type 43% scarlet, spineless 7% scarlet 7% spineless 10. 32% 11. The class asked for is one-half of one of the single-crossover classes.
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Unformatted text preview: The number of recombinants in the a – b region, including doubles, will be 10%, or 100 progeny. Of these, 20 will be double crossovers, leaving 80 in the single crossover category a + b + d + a b d + . Of these 80, 40 will be a + b + d . 12. First, if there is no interference, the number of double-crossover progeny can be predicted simply from the map distances given (10% and 20%). When converted to frequencies and multiplied, we have the expected number of double crossovers, 20 of 1000 progeny. Of these 20 double crossovers, 10 are RdY . If the coefficient of coincidence were 0.3, the total number of double crossovers will be 20 x 0.3 = 6 out of 1000, and of these 3 will have the RdY genotype. 2...
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This note was uploaded on 06/15/2009 for the course BIO 89165 taught by Professor Saxena during the Summer '09 term at University of Texas.

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Discussion 3 key Summer 2009 - The number of recombinants...

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