NAME____________________________________ BIOLOGY 52 - - CELL AND DEVELOPMENTAL BIOLOGY - - FALL 1999 Fourth Examination - - December 1999 Answer each question, noting carefully the instructions for each. Repeat- Please read the instructions for each question before answering!!! Be specific in each answer, and print name on top of each page! (I mean it!) --------------------------------------------------------------------------------------------------------------------1. The following 6 true-or-false questions are worth 10 points in total. Mark each as T or F and for any that is/are false, indicate briefly why in the space below. a. T F Isolated animal pole cells make endoderm. b. T F When animal pole cells are cultured with vegetal pole cells, the animal pole cells make c. T F Isolated vegetal pole cells make ectoderm. b. T F When vegetal pole cells are cultured with animal pole cells, the vegetal pole cells make e. T F Blood borne hormones regulate insect metamorphosis. f. T F The receptors for steroid hormones are transmembrane receptor tyrosine kinases. ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ 2. (circle all correct answers- 2 pts) Which genes contain binding sites for myoD in their promotor? A. Skin keratin. D. A bone specific-ECM component. B. A brain-specific tubulin. E. Fruit fly Engrailed. C. A skin-specific hemidesmosome protein. F. Muscle-specific actin. (2 points) Which of the three germ layers should express myoD? ________________________ 3. (4 points) We discussed several mechanisms by which cells remember the decisions they make. What mechanism is affected by treating cells with azacytidine? _________________________ When cultured fibroblasts were grown for several cell generations in azacytidine, they did something remarkable. What? _____________________________________________________________ ______________________________________________________________________________ 4. The following four true-or-false questions are worth 5 points in total. Mark each as T or F and for any that is/are false, indicate briefly why in the space below. a. T F Ecdysone is a steroid hormone. b. T F Fruit fly development begins with many nuclei in one cell. c. T F In embryos mutant for Notch, all ventral ectodermal cells become neural cells. d. T F Engrailed encodes a receptor tyrosine kinase that is the receptor for wingless. 1 NAME____________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ 5. (2 points- Put letter corresponding to all correct answers in the blanks) A particular equatorial cell in a frog embryo undergoes the following steps in its development. A B C Animal pole ---> Ventral/intermediate mesoderm -----> ventral mesoderm ---> muscle cell At which step(s) is the cell making determination decisions ____________________ At which step(s) is the cell differentiating ____________________ 6. (8 points) Use the choices below to fill in each blank: a. b. c. d. Secreted or membrane-bound cell-cell signal. Transmembrane receptor. G protein. Transcription factor. Hunchback protein acts as a _____________ Lin-12 protein acts as a ___________________ Engrailed protein acts as a _____________ Notch protein acts as a __________________ Wingless protein acts as a _____________ Steel protein acts as a _________________ ras protein acts as a __________________ MyoD protein acts as a ___________________ 7. (4 points) (Choose the most correct of the following answers for each question) a. b. c. d. e. f. g. h. i. The second thoracic segment is transformed into a third thoracic segment. All of the structures in the animal develop into head structures. The animal lacks a normal head. The third thoracic segment is transformed into the second thoracic segment. All segments in the embryo develop as second thoracic segments. The animal lacks a normal tail. The flies go to Duke, because they cannot tell their heads from their tails. All segments in the embryo develop as third thoracic segments. The embryo is normal What happens to embryonic development: If an embryo contains no Bicoid protein?: _____________ If all nuclei in an embryo contain high levels of Bicoid protein? _________________ If an embryo completely lacks Ultrabithorax protein? ______________________ If an embryo expresses normal levels of Ultrabithorax protein in T3 and A1? _________________ 8. (2 points) Assume that Hunchback, one of the gap genes, is normally expressed in the segments that make up the head. (Circle one) Is it likely to be turned ON or OFF by high levels of bicoid? 2 NAME____________________________________ (Circle one) In a Hunchback homozygous mutant embryo, a. b. c. c. d. e. f. The second abdominal segment is transformed into the first abdominal segment. Development proceeds normally. All segments are transformed into head segments. All segments are transformed into second abdominal segments. The second through seventh abdominal segments are missing. All segments are transformed into eighth abdominal segments. The head segments are missing. 9. (2 points) C. elegans lin-12 encodes a protein related to which fly protein? _______________ 10. The following four true-or-false questions are worth 5 points in total. Mark each as T or F and for any that is/are false, indicate briefly why in the space below. a. T F Levels of fruit fly Bicoid are highest at the tail end of the embryo. b. T F All neural cells arise from a single lineage in the nematode. c. T F All germ cells arise from a single lineage in the nematode. d. T F Mutations in ced-3 and ced-4 eliminate all programmed cell death. ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ 11. (1 point; pick the best choice) By the current model, which of the following cells must express on their cell surface the receptor for the anchor cell signal? A. The anchor cell. D. The gonadal cells. B. The centralmost vulval precursor cell. E. The central three vulval precursor cells. C. The outside three vulval precursor cells. F. All six vulval precursor cells. 12. (2 points) White-spotting mutations affect the blood system, the germ cells, and melanocytes. What do all of these tissues have in common that makes them targets of this mutation? ____________________________________________________________________________ (Circle one) Which cell type expresses the product of the Steel gene? Blood precursor cells // bone marrow stromal cells (Circle one) Which cell type expresses the product of the White-spotting gene? Blood precursor cells // bone marrow stromal cells 13. (2 points) When doing the "knockout" of Wnt1, among the progeny of the original mosaic founder were mice heterozygous for the mutant gene (i.e., with one mutant copy of Wnt1 and one wild-type copy). If a heterozygous male was crossed to a heterozygous female sister, what fraction of their progeny will show the mutant phenotype (assume Wnt1 is recessive)? 3 NAME____________________________________ ________________________________________________________________________ 14. The following six true-or-false questions are worth 8 points in total. Mark each as T or F and for any that is/are false, indicate briefly why in the space below. a. T F Human eggs implant in the uterus just prior to fertilization. b. T F The mouse is scientists favorite model for studying mammalian development. c. T F Steel mutant cells can be rescued by wild-type neighbors. d. T F White spotting mutant cells can be rescued by wild-type neighbors. e. T F The mouse wingless homolog wnt-1 is required for continued expression of the mouse f. T F Fruit fly eyeless is related to the mouse Pax gene small eye. ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ 15. (10 points) The diagram shows the cell fates chosen by different cells of the vulval equivalence group in a wild-type worm. How many 1° cells, now many 2° cells, and how many 3° cells will be made in the cases listed below? Hint: in wild-type (total =six) the answer is 1° _____1_____ 2° _____2_______ 3° ______3______ anchor cell gonad 3° cell 3° cell 2° cell 1° cell 2° cell Normal hypoderm = skin 3° cell Normal hypoderm = skin Make vulva In a mutant in which the gonad is ablated but the anchor cell remains (total =six) 1° ____________ 2° _____________ 3° _____________ In a worm homozygous mutant for the ras G-protein which is downstream of the receptor for the anchor cell signal (total=six) 1° ____________ 2° _____________ 3° _____________ In a mutant homozygous mutant for the worm EGF receptor, which is the receptor for the anchor cell signal (total=six) 1° ____________ 2° _____________ 3° _____________ In an animal in which both the anchor cell and the outside three vulval precursor cells were killed with a laser beam (total =3) 1° ____________ 2° _____________ 4 3° _____________ en NAME____________________________________ In an animal which is mutant for lin-12 and in which all vulval precursor cells are genetically engineered to make high levels of the anchor cell signal (total =6) 1° ____________ 2° _____________ 3° _____________ 16. (8 points) You are "knocking-out" wnt1. The ES cell line which is crucial for this procedure was derived from what source? _______________ These cells differ from those of most other cultured cells in a way which is critical for this procedure; what can they do that most cultured cells cannot? ____________________________________________________________________________ ____________________________________________________________________________ You add to these cells cloned DNA carrying the wnt1 gene. You modified the wnt1 gene by adding a second gene in the middle of it. From what group of organisms was the second gene originally derived? ___________________________________________ Most cells do not incorporate the second gene into their own DNA-- only a very small percentage do so. How do you select these cells? ___________________________________________________ _______________________________________________________________________________ Having a pure population of cells that are heterozygous mutant, what is your next step (the very next step, not the whole process!)? 5 NAME____________________________________ 17. (4 points; choose the best answer-- use each answer only once, and pick a property of that cell, not its descendents) We discussed how virtually all of the properties of cancer cells are also possessed by normal cells. Choose from the list ONE normal cell type that has the listed cell property: Divides indefinitely without differentiating: ___________________ Migrates to a new cellular location: ____________________ Divides rapidly even though in contact with neighbors: ___________________ Can penetrate the walls of blood vessels and thus move to distant locations: _________________ A. B. C. D. E. Skin cells Bone marrow stem cells. Kidney cells. Neurons of the adult brain. T lymphocytes of the immune system. F. G. H. I. Neural crest cells. Preblastoderm embryonic cells. Ectoderm. Liver cells. 18. The following four true-or-false questions are worth 5 points in total. Mark each as T or F and for any that is/are false, indicate briefly why in the space below. a. T F Metastatic tumors generally gain the ability to digest and thus penetrate the basal lamina. b. T F Activation of the Src kinase stimulates skin cells to migrate. c. T F Metastatic tumors generally lose expression or function of integrins. d. T F Retroviral activation of the mouse wingless homolog Wnt-1 can cause breast cancer in mice. ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ ____________________________________________________________________________ 19. (2 points- fill in the blanks) choose the correct genotype from among those listed: A. + / + B. Rbmutant / + C. Rbmutant / Rbmutant A kidney cell from a normal individual? ___________ A cell in a sporadic retinoblastoma tumor that arose in a normal individual? ___________ A kidney cell from an individual with a familial predisposition to retinoblastoma? _________ A retinoblastoma tumor cell from an individual with a familial predisposition to retinoblastoma? _____ 6 NAME____________________________________ (One point: Circle one) The cell biological function of the normal Retinoblastoma protein is to: A. Repair damaged DNA. C. Phosphorylate integrins, promoting cell motility. B. Turn on genes required for cell proliferation. D. Turn off genes required for cell proliferation. 7 NAME____________________________________ 20. (7 points) Skin cells normally proliferate at a slow rate in the body, replacing those lost due to normal attrition. When the skin is wounded, what happens to the cell proliferation rate of skin cells? (Choose ONE) It increases / It remains the same / It decreases. When wounded, skin cells are exposed to PDGF. What cell type makes PDGF? ____________ What cell type has cell surface receptors for PDGF? __________________ Imagine two different types of mutation in the PDGF receptor tyrosine kinase. Mutation 1 (PDGFRACTIVATED) renders it constitutively active. Mutation 2 (PDGFR-INACTIVE) kills the normal function. Normal PDGF receptor is designated +. What would be the rate of cell proliferation in skin cells of the following genotypes, in the absence and presence of PDGF? In each blank, fill in the rate of proliferation, either SLOW or FAST : Absence of PDGF Presence of PDGF + / + _____________ _____________ PDGFR-ACTIVATED / + _____________ _____________ PDGFR-ACTIVATED / PDGFR-ACTIVATED _____________ _____________ PDGFR-INACTIVE / PDGFR-INACTIVE _____________ _____________ 21. (3 points) If a person suffers from Xeroderma pigmentosum, they are excessively sensitive to a particular mutagen. Which one? _______________ What biochemical/cell biological process is impaired in these individuals? _______________________________ The incidence of which disease is thus increased in these individuals? ________________________ 22. (2 points) The p53 protein is often called the "guardian of the genome". It carries out this role by acting as part of a checkpoint in the cell cycle. Choose from the list the most likely immediate response of the following cells if DNA is damaged: Normal (i.e., wild-type) cell ________ Cell which lacks p53 protein A. Metastasize to a distant place in the body. B. Halt the cell cycle until the DNA is repaired. ________ C. Lose contact inhibition of growth. D. Continue the cell cycle without halt. I certify that I have performed my work on this examination in full conformity with the provisions of the Honor Code. Signature___________________________________ 8