Complementation_Problems_KEY - Biology 202 Complementation...

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Biology 202 Complementation Problems 1/28/11 This problem set is optional – it is available as a study aid for Midterm 1. It is not to be handed in. From Griffiths et al.: Chapter 6, problems 2, 19a, 22, 25 2) Assuming homozygosity for the normal gene, the mating is A/A . b/b x a/a . B/B. The children would be normal, A/a . B/b. 19a) The mutations are in two different genes as the heterokaryon is prototrophic (the two mutations complemented each other). 22) The suggestion from the data is that the two albino lines had mutations in two different genes. When the extracts from the two lines were placed in the same test tube, they were capable of producing color because the gene product of one line was capable of compensating for the absence of a gene product from the second line. a) Grind a sample of each specimen separately (negative control) to ensure that the grinding process does not activate or release an enzyme from some compartment that allows the color change. Another control is to cross the two pure-breeding lines. The cross would be A/A ; b/b x a/a ; B/B. The progeny will be A/a ; B/b, and all should be reddish purple. b) The most likely explanation is that the red pigment is produced by the action of at least
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Complementation_Problems_KEY - Biology 202 Complementation...

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