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Unformatted text preview: Map distance tells you the actual % or recombinant gametes. So, if A-B is 10 m.u. apart, then 10% progeny of test cross are recombinants. That means that 90% will have parental genotype. There are 2 parental genotypes: AB and ab, each of those will occur 45% of the time (90%/2). Thus, occurrence of aabb will be 45% in this case. You can work out 24 m.u. in the same way. b) However, if genes are completely linked, then only parental genotypes are possible (AB or ab gametes) and test cross will result in 50% of each AaBb and aabb progeny....
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This note was uploaded on 01/29/2012 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas at Austin.
- Spring '08