Introduction to Genetic Analysis 126

Introduction to Genetic Analysis 126 - 44200_04_p115-150...

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is approximately 11 map units. So, in the progeny from a testcross of a female dihybrid in cis conformation ( pr vg / pr ± vg ± ) heterozygote, we know that there will be 11 percent recombinants. These recombinants will con- sist of two reciprocal recombinants of equal frequency: thus, percent will be pr vg ± / pr vg and percent will be pr ± vg / pr vg . Of the progeny from a testcross of a di- hybrid in trans conformation (female pr vg ± / pr ± vg ), heterozygote, percent will be pr vg / pr vg and per- cent will be pr ± vg ± / pr vg . There is a strong implication that the “distance” on a linkage map is a physical distance along a chromosome, and Morgan and Sturtevant certainly intended to imply just that. But we should realize that the linkage map is an entity constructed from a purely genetic analysis. The linkage map could have been derived without even knowing that chromosomes existed. Furthermore, at this point in our discussion, we cannot say whether the “genetic distances” calculated by means of recombinant
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This note was uploaded on 01/10/2011 for the course BIOL BIOL taught by Professor Johnson during the Spring '08 term at Aberystwyth University.

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