7.06_2004_PS7key - 7.06 Spring 2004 PS 7 Key 1 of 9 Problem...

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7.06 Spring 2004 PS 7 Key 1 of 9 Problem set 7 KEY You are a brilliant yeast biologist that has developed a method with which to follow the segregation of chromosomes during meiosis. In your system, chromosomes can be visualized by “marking” them with a GFP fluorescenct signal (a Tet-operator DNA sequence has been integrated into a single chromosome and a Tet-Repressor (Tet- operator binding protein) fused to GFP binds to this DNA element, thereby producing a single fluorescent dot on the Tet-operator containing chromosome). In your diploid yeast strain one homologue of chromosome 5 has the Tet-operator while the other homologue does not. This will allow you to follow the fate of a single homologue during meiosis. During pre-meiotic S phase each homologue is replicated. However, because each pair of two sisters is tightly associated you cannot resolve the two Tet-Repressor-GFP dots on each sister of the “marked” chromsome 5(remember the other homologue (two sisters) are not marked). Therefore, in metaphase 1, you can see only a single GFP dot in the nucleus, corresponding to the paired sisters of one homologue of chromosome 5. After the reductional division of anaphase I, two nuclei are produced, only one of which has the GFP dot. However, after anaphase II, the two sisters are separated and now you can resolve two dots. Therefore, after anaphase II, four spores are produced, two of which have GFP dots because they received a chromatid corresponding to the marked chromosome 5 while the other two spores received a chromatid from the unmarked chromosome 5. S phase Ana I Ana II
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7.06 Spring 2004 PS 7 Key 2 of 9 a) In a spo11 null mutant, DSB are never made and so the homologues never pair. Co-orientation of kinetochores is unaffected, however. Draw what you predict the Anaphase I and the Anaphase II two cells would look like in this mutant. In spo11 mutants the homologues will segregate randomly during anaphase I, but sister chromatids will segregate together because co-orientation is maintained. In anaphase II, sisters will segregate to opposite poles, as usual. However, the resulting nuclei are likely aneuploid. b) Would a spo11 mutant have increased rates of chromosome mis-segregation? Why or why not? A spo11 mutant would have high rates of chromosome mis-segregation. At metaphase I, homologues will ot be joined by chiasmata. As a result they will line up independently on the metaphase plate. During anaphase I, homologues will then segregate independently of each other – that is, in some cases one nucleus will get both copies of a homologue pair while the other nucleus will get neither and sometimes they will randomly go to different nuclei. During Anaphase II, the sisters will segregate correctly. Therefore, spo11 mutants will produce disomic spores and spores missing a chromosome. c)
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This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore.

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7.06_2004_PS7key - 7.06 Spring 2004 PS 7 Key 1 of 9 Problem...

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