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Unformatted text preview: 252 Chapter 14 classes allow you to calculate the map distance between the st and on genes in the translocation. The rf = 10 mu. N2 an 671+ T1 N2 071 671+ T] S 1; st L L st+ st Lil T2 N1 T2 N1 Lily i AND AND ‘ ‘31; 3L: l 1L1 iii} N2 C” ”1+ T1 N2 0" 01* T1 iii stT :1“ St l ‘ stJr st l T2 N1 T2 N1 L 14-19. The semisterile F1 is a translocation heterozygote and will produce 1/2 fertile : 1/2 semisterile ~ ‘2 progeny from alternate segregation. Products of adj acent—l or adj acent—2 segregation are imbalanced ‘1 1‘ and therefore inviable—this is the basis of the semisterility. Because the only viable gametes are the EL result of alternate segregation, genes that are on the chromosomes involved in the translocation will ill; not show independent assortment. Instead, if the genes are located very close to the translocation breakpoints, they will show only the parental classes; that is, the genes will display pseudolinkage. Genes that are on any other chromosome will assort independently of the translocation (i.e., such genes will assort independently from fertility/semisterility). Ml a. If the yg gene is on a different chromosome than those involved in the translocation, the traits 91‘ will assort independently. The product rule says you can cross multiply the two monohybrid 1" ratios: 1/2 ngF (normal leaf color); 1/2 yg (yellow green) and 1/2 fertile : 1/2 semisterile to give: l LrL 1/4 fertile yg+ : 1/4 fertile yg : 1/4 semisterileyg+ : 1/4 semisterile yg. l b. If the translocation involved chromosome 9, the fates of the fertility and leaf color phenotypes are connected—these genes will show pseudolinkage. The original cross was: semisterile yg+ x fertile yg —> F1 semisterile x fertile yg. This means the normal, nontranslocated chromosome 9 has the yg allele, while the translocated chromosome 9 has the yg+ allele. Thus, the chromosomes of the heterozygous F 1 at meiosis I would look like: ...
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