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| Inquiry Questions FIGURE 19.22 Gompertz curves. While human populations may differ 25-fold in their mortality rates before puberty, the slopes of their Gompertz curves are about the same in later years. How do you explain this difference? Answer: The differences in mortality rates between the different populations are due primarily to the effects of improved nutrition, living conditions, and advances in health care on mortality before puberty. Once a person, any person, passes puberty, his or her likelihood of death is not so markedly influenced by these improvements. FIGURE 19.23 Hayflick's experiment. Fibroblast cells stop growing after about 50 doublings. Growth is rapid in phases I and II, but slows in phase III, as the culture become senescent, until the final doubling. Cancer cells, by contrast, do not "age." How do cancer cells overcome this 50-division "Hayflick limit"? Answer: Cancer cells are able to overcome the Hayflick limit because of mutations disabling the genes that normally act to repress the production of telomerase, an enzyme that adds to telomeres so that their length is not reduced as the cell line proliferates. If their telomeres do not shorten, the cells will keep dividing long after 50 divisions. Self Test 1). Which of the following series of events represents the path of vertebrate development? a). formation of blastula, cleavage, neurulation, cell migration, gastrulation, organogenesis, growth b). formation of blastula, cleavage, gastrulation, neurulation, cell migration, organogenesis, growth c). cleavage, formation of blastula, gastrulation, neurulation, cell migration, organogenesis, growth d). cleavage, gastrulation, formation of blastula, neurulation, cell migration, organogenesis, growth Answer: c 2). Which of the following statements about Drosophila development is FALSE? a). Drosophila go through four larval instar stages before undergoing metamorphosis. b). During the syncytial blastoderm stage, nuclei line up along the surface of the egg. c). Imaginal discs are groups of cells set aside that will give rise to key parts of the adult fly. d). Maternal, rather than zygotic, genes govern early Drosophila development. Answer: a 3). If a plant embryo failed to form enough ground tissue, what function(s) would likely be directly affected in the corresponding mature plant? a). seed formation b). meristem development c). cotyledon formation d). food and water storage Answer: d 4). C. elegans is a powerful developmental model because a). these nematodes are very small, so it is easy to maintain a large population in a laboratory. b). the fate of every cell has been mapped. c). the fate of cells that will become eggs and sperm are predetermined. d). these nematodes have the same amount of DNA as Drosophila. Answer: b 5). Which of the following best describes a morphogen? a). a cell that secretes diffusible signaling molecules that play a role in specifying cell fate b). a diffusible signaling molecule that plays a role in specifying cell fate c). a protein that helps mediate direct cell–cell interaction d). a protein that enables cells to become totipotent Answer: b 6). What would happen as a result of a transplantation experiment in a chick embryo in which cells determined to become a forelimb were replaced by some cells determined to become a hindlimb? a). A hindlimb would form in the region where the forelimb should be. b). A forelimb would form in the region where the hindlimb should be. c). Nothing; the forelimb would form normally. d). Neither a forelimb nor a hindlimb would form because the cells were already determined. Answer: a 7). Which group of genes, identified by Nusslein-Volhard and Caroll, is responsible for the final stages of segmentation in Drosophila embryos? a). morphogen gradient genes b). gap genes c). segment-polarity genes d). pair-rule genes Answer: c 8). Suppose that during a mutagenesis screen to isolate mutations in Drosophila, you came across a fly with legs growing out of its head. What gene cluster is likely affected? a). Bicoid b). Hunchback c). Bithorax d). Antennapedia Answer: d 9). What would be the likely result of a mutation of the bcl-2 gene on the level of apoptosis? a). no change b). a decrease in apoptosis c). an increase in apoptosis d). First it would increase, but later it would decrease. Answer: c 10). The gene clock hypothesis is best described by which of the following explanations? a). Mutations accumulate partially through the addition of an –OH group to the base guanine. b). Specific genes exist to promote longevity. c). Free radicals can cause genetic mutations, particularly when we are sleeping. d). Calorie restriction leads to an increased life span. Answer: b Test Your Visual Understanding 1). Hayflick's experiment revealed that noncancerous cells have a definitive life span, whereas cancer cells do not have the same restrictions. Draw in curves that represent the growth patterns of each cell type. Answer: The curves should resemble those depicted in figure 19.23. Apply Your Knowledge 1). You have generated a cell line that expresses an altered form of cadherin. This mutant cadherin has the 110-amino-acid extracellular domain that is required for interaction with other cadherins, but lacks a transmembrane domain. If you were to mix this cell population with other cells expressing a wild-type, or normal, form of cadherin, would you expect these two cell populations aggregate with each other? Why or why not? Would the mutant cells be able to aggregate with other mutant cells? Answer: Although the mutant cells express the cadherin interaction domain, this altered form of the protein would not be present on the cell surface because it lacks a transmembrane domain. Therefore, the mutant cells would be unable to interact and aggregate with each other or with wild-type cells. |