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Discussion 4 key Fall 2009

Discussion 4 key Fall 2009 - a b d e g h k l all affect one...

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GENETICS (BIO 325) Fall 2009 Discussion sheet 4 Key _____________________________________________________________________________________ 1. (a) Two (b) Group 1 – mutants 1,2,5 Group 2 – mutants 3,4 2. (a) Yes (b) Yes (c) No (d) No (e) Yes 3. In incomplete dominance, the phenotype of the heterozygote lies in the range between the phenotypes of the homozygous genotypes. For example, a cross of homozygous plants with red and white flower color results in plants with pink flowers. In codominance, the phenotype of the heterozygote is a mixture of the phenotypes of both of the corresponding homozygous genotypes. The heterozygous phenotype is not an intermediate of the homozygous genotypes. For example, blood groups in humans. 4. (a) 12:3:1 (b) 9:3:4 (c) 15:1 (d) 9:7 (e) 13:3 5. The pathway is Y to X to Z, with d blocking the synthesis of Y, and c the synthesis of X from Y. 6. (a) Five;
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Unformatted text preview: a, b, d, e, g, h, k, l all affect one gene. Mutations c, f, i, j affect four distinct genes, all different from the first. (b) 1 → 3 → 2 → 4 → uridine The order of compounds is determined from accumulation phenotypes, not growth phenotypes. (c) In the above sequence, the gene represented by mutation a blocks before compound 1, f between 1 and 3, j between 3 and 2, c between 2 and 4, and i between 4 and uridine. 7. Growth on minimal medium and A B C D E –––––––––––––––––––––– Mutant 1 + + 2 + + 3 + 4 5 + + 8. (a) and (c) (b) Mutant strain At least some accumulation of: 1 D, E, (A, B, C, threonine, if threonine cannot by utilized) 2 A,B,D,E (threonine, if threonine cannot be utilized) 1 3 D,E,F,G 4 no accumulation (provided methionine is utilized) 9. 2...
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