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Capítulo 14
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1. Both major types of reproduction-asexual and sexual-require cell division as an obligatory step. Every DNA strand of an offspring organism can be traced through either mitosis or meiosis to the DNA strands of its parents.

2. Examples would include the synthesis of enzymes involved in production of deoxyribonucleoside triphosphates and the synthesis of cyclins that activate Cdks required for the G 1 -S transition.

3. One-third of the cells, because one-third of the cells were in S phase during the 15-minute pulse. None of the mitotic cells would be labeled. Nearly 6 hours for the cells that were at the end of S to pass through G 2 . None of the cells, because 18 hours is one generation time; the cells in mitosis after an 18-hour chase would not have been in S phase during the 15-minute pulse.

4. The mininum amount of time of labeling would be 12 hours. This time would be needed for the cells that had just entered G 2 at the start of the labeling period to reenter S phase. You could determine the time it took for the number of cells in the population to double..

5. A G 2 -phase cell is not activated to initiate DNA synthesis because it has already duplicated its DNA and will not do so again. Each replicon can only undergo one round of replication during a single cell cycle.

6. In G1, the cells would not divide because the Cdk was never phosphorylated on Thr 161, which is required for activity. In G2, the cells would behave normally that particular cell cycle because the Cdk was phosphorylated prior to temperature elevation and there is no phosphatase that removes this phosphate late in the cell cycle.

7. Removal of activating phosphate, addition of inhibitory phosphates, degradation of cyclin, and binding by CKI.

8. By cell fusion and by mitosis without cytokinesis. That mitosis and cytokinesis are independent processes that do not have to occur together.

9. Inject fluorescently labeled tublin into the cell. Because the polar microtubules are presumed to be responsible for spindle elongation during anaphase B, you would expect to see tubulin subunits being added at their plus ends, which would be located away from the poles, such as at the cell equator.

10. 100 percent; 25 percent.

11. 46; 46; 23; 23; 23 number of chromosomes. 92; 92; 46; 46; 0 number of chromatids. 4C; 4C; 2C; 2C; 1C amount of DNA.

13. Four.

14. The fetuses should show evidence of secondary nondisjunction rather than primary nondisjunction, i.e., evidence that the cells contain sister chromatids rather than both homologues from the mother. Evidence contradicts this conclusion since most mistakes occur during meiosis I.

15. That it required the synthesis of DNA, presumably associated with DNA repair mechanisms.

16. That the laser destroyed the MAD2 protein molecules present at the kinetochore of the monoattached chromosome, thereby eliminating the wait signal these proteins normally provide

17. You did receive half of your chromosomes from each parent at fertilization but because of independent assortment, you could theoretically have received anywhere from zero to 23 chromosomes from a particular grandparent. Crossing over doesn't really change the answer, but means that the chromosomes from your grandparents are no longer strictly maternal or paternal, but have portions from each grandparent.

18. Once an origin is fired, it cannot fire again until it has been converted into a prereplication complex, which can only occur during late M-G 1 . The S phase cell does not contain the factors necessary for pre-replication complex formation.

19. The Cdk would be 1) constitutively active, 2) an inactive mutant,.

20. Before DNA replication can begin, a number of events must take place that are not dependent upon cyclin E-Cdk2, so they cannot be triggered by activation of the kinase. Centrosome duplication, in contrast, is more directly activated by this kinase.

21. 1) would fail to arrest progress of mitosis in the presence of an unattached chromosome, 2) would make the APC unable to ubiquinate the anaphase inhibitor so the cell would not be able to exit metaphase, 3) the alternate substrate-targeting subunit (Cdh1) would not be able to bind to the APC so the cell would not be able to exit mitosis and enter G 1 .







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