Lecture 5,6_Genetic Mapping

Lecture 5,6_Genetic Mapping - Ch. 5. Gene Linkage and...

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Ch. 5. Gene Linkage and Genetic Mapping Independent assortment of 2 genes occurs because they are on different chromosomes. Since organisms have thousands of genes and only a few chromosomes, each chromosome has many genes. For example, Drosophila has 4 chromosomes and approximately 13,000 genes. Therefore, an average chromosome has more than 3,000 genes. In humans, there are 23 chromosomes and ~25,000 genes. Each chromosome, on the average, would have ~1,100 genes. Genes on the same chromosome are inherited together. They are said to be " Linked " or to show " Genetic linkage " or they are said to belong to the same Linkage Group . In a cross between AB x ab, 4 kinds of progeny are possible- AB and ab are the parental combinations while Ab and aB are recombinants. Genes on different chromosomes assort independently because there are two alternative ways in which the homologous chromosomes of two pairs can line up during the metaphase in meiosis I ( Fig. 5.1 ). One orientation gives the gametes AB and ab, while the second orientation gives the aB and Ab gametes. The combined outcome is equal amounts of parental (AB, ab) and recombinant (aB and Ab) gametes. In contrast, linked genes produce less than 50% recombinant gametes [(AB + ab) > (Ab + aB)]. Recombinant gametes are formed due to breakage and reunion of chromatids ( crossing-over ) between 2 genes on the same chromosome ( Fig. 5.3 ). Discovery of Genetic Linkage. It was discovered in 1905 in sweet pea. A cross between a Purple flower, long Pollen plant and a Red Flowers, round Pollen plant produced a F 1 that was all Purple, Long. The F 2 data did not fit the expected 9 : 3 : 3 : 1 ratio. F2 Phenotype Expected Observed Purple Long (P_L_) 56.2% 69.5% Purple Round (P_l l) 18.7% 5.6% Red Long (ppL_) 18.7% 5.6% Red round (pp l l) 6.2% 19.3% Both parental classes were more frequent than expected and the 2 recombinant classes were less frequent than expected. The data fail the X 2 test for a fit to a 9 : 3 : 3 : 1 ratio. The observed results can be explained if it is assumed that the parental gametes (PL and pl) are produced 44% each and the recombinant gametes (pL and Pl) are produced 6% each (instead of 25% each, as in the case of independent assortment. The linkage in this case is not complete and one sees 12% recombinant gametes. When linkage is complete, no recombinant progeny are produced. The F 2 progeny consist of 3/4 double-dominant phenotype and 1/4 double- recessive phenotype (i.e., only the 2 parental phenotypes are seen). Calculating linkage distances from F 2 data is somewhat complicated. As a result , all genetic mapping is done by crossing a heterozygote to a double recessive organism (a "testcross"). Linkage Studies in Drosophila
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Lecture 5,6_Genetic Mapping - Ch. 5. Gene Linkage and...

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