cell bio final test bank questions chp 19-21

In the following simplified diagram of cell divisions in a multicellular species, the germ-cell and somatic-cell lineages are depicted. Which of the following cells (1 or 2) represents the germ line? Write down 1 or 2 as your answer.

1 Only the germ line contributes to the organism in the next generation; however, by sacrificing their survival to support the germ cells, the somatic cells help to propagate copies of their own genome.

42. CFSE (carboxyfluorescein succinimidyl ester) is a cell-tracing molecule that, once inside a cell, is modified to yield a highly fluorescent molecule that is retained within the cell. Due to its stability, it can be used to trace cells and follow their divisions: each division dilutes the fluorescent dye twofold. You label T cells from a healthy mouse with CFSE and inject them into either a wild-type strain of mouse or a strain with T cell deficiency, both of which are immunologically compatible with the donor mouse. You later collect and analyze peripheral T cells from the recipient mice and draw the following histograms according to the fluorescent intensity per cell. Based on these results, do you think the total number of T cells in the mouse body is (1) controlled by intracellular programs (as in the thymus, for example) or is (2) regulated as a whole (as in the spleen)? Write down 1 or 2 as your answer.

2 The injected T cells have divided (the CFSE dye is diluted) in the T-cell-deficient mouse but not in the wild-type mouse, supporting the idea that the total population of these cells is regulated as a whole.

13. In a developing Drosophila melanogaster embryo, a hierarchy of gene regulatory interactions subdivides the embryo to regulate progressively finer details of patterning. For each of the following proteins, indicate which expression pattern (1 to 4) in the schematic drawing below is more appropriate. Your answer would be a four-digit number composed of digits 1 to 4 only, with each digit used once, e.g. 3412. ( ) Hunchback (product of a gap gene) ( ) Engrailed (product of a segment-polarity gene) ( ) Bicoid (product of an egg-polarity gene) ( ) Even-skipped (product of a pair-rule gene)

2413 In the regulatory hierarchy of egg-polarity and segmentation genes, the gapgene products provide a further tier of positional signals by controlling the expression of other genes, including the pair-rule genes. The pair-rule genes, in turn, collaborate with one another and with the gap genes to set up a regular, periodic pattern of expression of the segment-polarity genes, which collaborate with one another to define the internal pattern of each individual segment.

Myelomas are cancers of blood plasma cells—white blood cells that are normally responsible for producing large quantities of antibodies. In the following "Circos plot" for myelomas in a hypothetical mammalian genome, the interchromosomal rearrangements are indicated by red lines and variations in copy numbers are indicated in blue. The positions of Mutation groups P erce nta g e of total g e n es stu die d a b 0 100 1 2 1 2 named genes are indicated with arrows. On which chromosome do you expect to find the antibody genes? Write down the chromosome number (1 to 5) as your answer.

5 Translocation between immunoglobulin loci and proto-oncogene loci activates the proto-oncogenes and can transform the cell. These translocations are common in myelomas, where each one takes advantage of the powerful transcription potential of the immunoglobulin genes.

15. Which of the following schematic drawings better depicts the regulatory network that maintains Engrailed (en), Hedgehog (hh), and Wingless (wg) expression following cellularization in a developing Drosophila embryo?

A (most anterior); Wg ----> En --> Hh ----> In each segment, cells that express the signal protein Wingless induce their neighboring cells to express the transcription regulator Engrailed, which activates the expression of the signal protein Hedgehog in these cells. This in turn activates the expression of Wingless in the first cell, maintaining the fine patterning within the segment.

Compared to cells of a normal tissue, which of the following occurs less frequently in cells within a solid tumor?

A. Apoptosis Although cancer cells tend to avoid apoptosis, they still die on a massive scale by necrosis in solid tumors.

You have karyotyped cells from two colorectal tumor samples, one from a hereditary nonpolyposis colorectal cancer (HNPCC) patient, and the other from a familial adenomatous polyposis coli (FAP) patient. One group of the karyotypes shows gross chromosomal abnormalities with extra or deleted chromosomes and several translocations and deletions. The other group, however, is almost normal, and comparable to noncancerous samples. Which group would you expect to have loss-of-function mutations in the DNA mismatch repair system genes MSH2 and MLH1 as their primary driver mutations?

A. HNPCC, which has an almost normal karyotype HNPCC tumor cells show an almost normal karyotype, in sharp contrast to FAP tumor cells.

49. Consider the embryonic cerebral cortex of a mammal. Near which side of the cortex are the cell bodies of radial glial cells located? To which side of the cortex do the first-born neurons eventually migrate?

A. Inner surface; inner surface The radial glial cells form a scaffold in the developing cortex to guide the migrating neurons to their destined layer. The first-born neurons migrate and settle closest to the lumenal (inner) surface of the cortex, whereas last-born neurons crawl past them and settle in outer layers. The radial glial cells divide near the lumenal surface (where their cell bodies are located) to give rise to neurons and glia.

Mutations in two important cancer-critical genes, encoding p53 and Rb, respectively, are commonly found in cancers. What type of mutations are these expected to be?

A. Loss-of-function mutations in both genes Both p53 and Rb are coded by tumor suppressor genes, and their inactivation promotes cancer.

Among the following cancers, one is currently leading to the most number of deaths in the United States and in the rest of the world. In the United States, it contributes to cancer mortality more than the next three killing cancers combined. Worldwide, it claims more than 1.5 million lives every year. Which cancer is this?

A. Lung cancer

You have analyzed a large set of human cancer-critical genes for a selected group of carcinomas, classifying each of the genes based on whether they are known to undergo somatic or germ-line mutations, as well as based on whether they are dominant or recessive. You then group them and plot the statistics in the following histograms. Which group (a or b) do you think represents somatic, as compared to germ-line, mutations? Which group (1 or 2) do you think represents dominant, as compared to recessive, mutations? (Note that the sum of percentages of somatic and germ-line mutations is more than 100%, since some genes are mutated in both somatic and germ cells).

A. a; 1 Most cancers arise from dominant somatic mutations in cancer genes; however, most germ-line mutations are recessive, which enables them to be carried through generations.

Carcinoma of the uterine cervix in humans ...

A. can be largely prevented by vaccination. Cervical cancer is mostly a result of papillomavirus infection transmitted sexually. Accidentally, genes in this DNA virus that interfere with p53 function are expressed in basal epithelial cells of the uterine cervix, leading to lesions that can very slowly develop into cancer.

For each of the following genes involved in regulation of cell growth through the mTOR pathway, indicate whether the gene is activated (A) or inactivated (I) in cancer cells compared to normal healthy cells. Your answer would be a four-letter string composed of letters A and I only, e.g. AAAA. ( ) mTOR ( ) Akt ( ) PTEN ( ) PI3K

AAIA Activation of the kinase subunit of mTOR (stimulated by PI3K and Akt kinases, and inhibited by PTEN phosphatases) can lead to an increase in nutrient uptake and utilization in cancer cells.

8. Sort the following organisms from the least to the most degree of asymmetry in the unfertilized egg (i.e. pre-defined axes of polarization). Your answer would be a three-letter string composed of letters A to C only, e.g. BCA. (A) M. musculus (B) D. melanogaster (C) X. laevis

ACB The mouse egg is almost perfectly symmetrical. In contrast, fly eggs are highly polarized. Many other organisms, such as the frog, lie between these two extremes.

2. The simplified drawing below depicts early stages of animal development. Indicate which letter (A to E) in the drawing corresponds to each of the following terms. Your answer would be a five-letter string composed of letters A to E only, e.g. ECDBA. fertilized egg --cleavage--> A --gastrulation--> B CDE ( ) Blastula ( ) Ectoderm ( ) Endoderm ( ) Mesoderm ( ) Gastrula

ACEDB Figure 21-3*

47. Sort the following phases in the order that they take place in neural development in vertebrates. Your answer would be a four-letter string composed of letters A to D only, e.g. DACB. (A) Neurons are assigned specific characters according to the place and time of their birth. (B) Neurons form synapses with other neurons or with muscle cells. (C) Synaptic connections are refined and adjusted. (D) Neurons extend axons and dendrites toward their target cells.

ADBC First, the neurons are assigned specific characters according to the place and time of their birth. Next, newborn neurons extend projections along specific routes toward their target cells. In the third phase, neurons form synapses (with other neurons or muscle cells) that are then refined and adjusted in the fourth phase.

From an immortalized human HeLa cell line with wild-type p53 genes, you have derived a line that lacks both copies of the gene. You then treat the original and derived cells with the anticancer drug doxorubicin, which can activate the p53 pathway in the cell by stalling DNA replication forks and inducing double-strand breaks in DNA. You measure cell proliferation in the presence of different doses of the drug in each of the two cell lines, and plot the results as shown in the graph below. Which cell line (A or B) do you expect to be the original HeLa line? Write down A or B as your answer.

B Cells in the original HeLa line better respond to the drug since they can trigger apoptosis through the p53 pathway. Knocking out p53 makes the cells more resistant to this effect of the drug.

The Ames test is used to test the mutagenicity of a compound suspected to be a carcinogen. In a simple form of the test, the carcinogen is first mixed with a rat liver extract. A disc of filter paper is soaked with this mixture and placed on a culture of a strain of Salmonella typhimurium that is defective in a gene involved in the synthesis of histidine, an amino acid that is essential for cell growth and proliferation. The strain is thus normally unable to grow into visible colonies when the histidine in the culture medium is depleted. In the presence of a mutagen, however, mutations (often "reverse mutations" in the same gene) can enable the bacteria to produce histidine on their own, and therefore grow into colonies. The results of the Ames test for three compounds A, B, and C—each used at the same concentration—are shown in the schematic diagram below. Colonies are indicated with black dots, and the disc is indicated with a white circle at the center of each plate. Which compound (A to C) appears to be a stronger mutagen in this assay? Write down A, B, or C as your answer.

B More colonies (revertants) appear in the presence of a stronger mutagen, with the distribution of colonies reflecting the dose-response relationship for the mutagen (stronger mutagens will affect bacteria further away from the disc).

41. Which cell behavior depicted below is involved in the tube formation that underlies the development of lungs and trachea?

B. *diagram* Budding (B) generates the tubes of developing lungs and trachea.

39. The following schematic drawings show an epithelial sheet. Which example shows a mutant with planar-cell-polarity defects but no other defects?

B. *diagram* In planar cell polarity, the cells in an epithelium are all arranged as if they had an arrow written on them, pointing in a specific direction in the plane of the epithelium. In the examples shown, A represents a normal epithelium, whereas B shows planar-cellpolarity defects. Note that C and D show defects in apical-basal polarity as well.

Which of the molecules (A or B) in the following drawing is a more potent mutagen? Write down A or B as your answer.

B. Aflatoxin-2,3-epoxide Unfortunately, carcinogens such as aflatoxin B1 are rendered more potent by the P-450 detoxification system in the liver

36. The Steel factor/Kit signaling pathway plays an important role in the migration of many types of cells during development as well as in the adult animal. Not surprisingly, loss-offunction mutations in Steel factor result in cell migration defects. Selective inactivation in these cells of which of the following proteins might be expected to rescue (i.e. partially restore to normal) the defective phenotype of Steel loss-of-function mutants?

B. Bax, an essential apoptotic protein Steel factor and its receptor provide survival signals for many migrating cells. In the absence of Steel factor, therefore, the cells would undergo apoptosis. This would be prevented if Bax is also deleted.

28. Which of the following evolutionary changes better explains morphological differences between different animals despite many common molecular mechanisms governing their development?

B. Changes in regulatory DNA controlling expression of key developmental genes. Changes in regulatory DNA controlling expression of key developmental genes are largely responsible for morphological differences between animal species.

A benign neoplasia of cartilage is called a ...

B. Chondroma Benign and malignant tumors of cartilage are termed chondroma and chondrosarcoma, respectively.

44. After reaching sexual maturity, the nematode Caenorhabditis elegans normally doubles in size within about two weeks. This doubling is mainly due to ...

B. DNA replication without cell division After cell division in the somatic tissue is halted in the adult worm, endoreplication makes these cells polyploid, which is accompanied by a larger cell size and a larger overall body size.

27. The result of Notch-mediated competitive lateral inhibition in a patch of wild-type cells is depicted on the left in the following schematic drawing, in which the black cells have become specialized. Consider a genetic mutation that interferes with intracellular Notch signaling in such a way that Notch can no longer regulate Delta effectively. Which drawing (1 or 2) better represents the outcome with cells bearing this mutation? Do the specialized cells (black hexagons) have active or inactive Notch at their surface?

B. Drawing 1; inactive Notch In lateral inhibition, Notch signaling typically leads to inhibition of the differentiation program as well as to reduction in the cell's display of Delta. Interfering with Notch signaling would therefore impair lateral inhibition—generating patterns in which two neighboring cells undergo specialization (drawing 1)—since Notch signaling can be off in both cells.

Which of the following sequential barriers to metastasis is the easiest to overcome for cancer cells in general?

B. Exit from the blood into a remote tissue or organ Traveling through the body's circulation, which includes survival in the circulation, arrest in capillaries or small vessels, and extravasation into remote tissues, is generally more efficient than the prior phase of escape from the parent tissue and the later phase of colonization of a remote site.

30. The gene encoding Hes7 in mouse contains three introns. Any or all of these introns can be deleted to alter the delay associated with transcription and splicing of Hes7 mRNA. As a result of such deletions, either the oscillation frequency of this gene's expression can change (result 1) or the oscillatory behavior can disappear altogether (result 2). One of these two results is obtained when only one intron is deleted from the gene, while the other result is obtained when all three introns are removed. Would you expect the oscillation frequency to increase or decrease if result 1 is obtained? Which result (1 or 2) would you expect to observe when all three introns are deleted?

B. Increase; result 2 Deleting only one intron shortens the gestation delay for the mRNA only slightly, leading to an overall shorter feedback delay and therefore higher oscillation frequency. Deleting all three introns shortens the delay to such an extent that a sustained oscillation is no longer supported.

16. In Drosophila melanogaster, the expression of genes Ultrabithorax and Antennapedia can normally be observed in the third thoracic segment which bears a pair of legs as well as a pair of halteres. Homeotic mutations associated with these two genes can give rise to remarkable disturbances in the organization of the adult fly: two pairs of wings in the case of Ultrabithorax, and legs in the place of antennae in the case of Antennapedia. Would you expect these to be gainof-function or loss-of-function mutations?

B. Loss-of-function for Ultrabithorax and gain-of-function for Antennapedia Loss of Ultrabithorax transforms the segment into a more anterior identity and gives rise to a pair of wings in this segment. Gain of Antennapedia transforms a head segment such that it bears a leg in the place of an antenna.

26. Which of the following is NOT true regarding cell-type specification in developing animals?

B. Members of the Achaete/Scute family of transcription regulators drive the cells to form endothelial layers. Members of the Achaete/Scute family of transcription regulators drive the cells to become neural, not endothelial, progenitors.

33. TSH is a pituitary hormone that stimulates the production of thyroid hormone by the thyroid gland. This production can be blocked by sodium perchlorate (SP), which inhibits the cellular import of iodine necessary for thyroid hormone synthesis. What would be the effect of TSH or SP exposure, respectively, on the timing of metamorphosis in frog larvae?

B. Premature metamorphosis; delayed metamorphosis Metamorphosis in amphibians is triggered by the secretion of thyroid hormone from the thyroid gland, which coordinates the various processes that are involved in this important developmental transition.

The immortalized non-malignant mouse cell line NIH-3T3 was derived from normal mouse fibroblasts in the early 1960s. These cells are able to readily take up exogenous DNA and are prone to transformation by cancer-causing agents, including some retroviruses. DNA extracted from a human bladder carcinoma line is able to transform these cells, as judged by a significant increase in the number of foci (cell clumps) in the cell-culture plates when the DNA is added. The malignant cells contain human DNA, and the DNA can be shown by sequence analysis to contain a single mutant gene that is present in the original bladder carcinoma cell line. The gene codes for a monomeric G protein and was one of the first cancer-critical genes to be identified in this way. The protein encoded by this gene is ...

B. Ras Ras was also found earlier to be the oncogene carried by some sarcoma viruses, and the identification of the same gene (with similar mutations) in both cases was an important discovery that advanced our understanding of the molecular biology of cancer.

Retinoblastoma is an early-onset cancer of the retina with a rapid progression, and is mostly diagnosed in children. In its hereditary form, multiple eye tumors usually arise in both eyes, while the nonhereditary form usually causes fewer tumors in only one eye. Treatment may involve a combination of chemotherapy, radiotherapy, and other therapies and the majority of patients can be cured if given the right treatment. However, survivors of one form of retinoblastoma (and not the other form) have a markedly increased frequency of subsequent neoplasms that can lead to other cancers later in life, especially soft-tissue sarcomas. These patients should therefore be closely monitored throughout their lives. Which gene is affected by the primary driver mutation in this cancer as well as the later sarcomas? Which form of retinoblastoma do you think is associated with a higher risk of subsequent neoplasms?

B. Rb; hereditary Due to a germ-line mutation in the Rb gene, people with hereditary retinoblastoma are at a higher risk of cancers in older age. Note that this is different from "relapse" of partially treated retinoblastomas, which can be accounted for by the survival of a few original cancer cells after the initial treatment.

43. You have obtained leaf samples from three strawberry varieties A, B, and C. You isolate the cells, extract their nuclei, and stain them with propidium iodide, a fluorescent dye that binds quantitatively to DNA. You then use a fluorescence-activated cell sorter (FACS) machine to sort the nuclei based on their fluorescence. Guessing from the results, presented in the following histograms, which variety would you expect to yield larger strawberries?

B. Variety B In plants, as in animals, cell size correlates with ploidy. Additionally, body and organ size in plants correlates well with ploidy. The fluorescent signal in this experiment corresponds to the nuclear DNA content. Varieties A, B, and C appear to be mostly diploid, octaploid, and tetraploid, respectively.

The genotypes of 400 colorectal cancer tumors are tabulated below, where the number of tumors with or without mutations in each of the two cancer-critical genes, β-catenin and Apc, are indicated. Which row (a or b) corresponds to those with a mutant Apc gene? Which column (1 or 2) corresponds to those with a mutant β-catenin gene?

B. a; 2 The majority of colorectal carcinoma cells have a loss-of-function mutation in Apc. Among those that do not, the majority have a gain-of-function mutation in β-catenin. Having both mutations is not common, implying that the activation of the Wnt pathway, and not the individual mutations, confers a selective advantage to the tumor cells.

According to the cancer stem-cell model for tumor growth and propagation, ...

B. transit amplifying cells constitute the great majority of the cells in the tumor. Cancer stem cells can be responsible for tumor growth and maintenance, yet remain only a small part of the tumor cell population.

34. The qualitative graph below shows the molecular changes accompanying vernalization in the flowering plant Arabidopsis thaliana. Indicate which curve (A to D) in the graph better represents the temporal changes in each of the following variables. Your answer would be a fourletter string composed of letters A to D only, with each letter used once, e.g. BACD. ( ) Level of Coolair noncoding RNA ( ) Level of Flowering locus C (FlC) gene product (transcriptional repressor) ( ) Level of Flowering locus T (Ft) gene product (transcriptional activator) ( ) Level of repressive chromatin marks at the FlC locus

BADC As Coolair levels (curve B) peak during vernalization, closed chromatin structures (curve C) are induced at the FlC locus, gradually shutting off FlC expression (curve A). This persisting change in turn allows flowering genes of the Ft locus to be expressed under the right conditions in the spring (curve D). Figure*

The following simplified diagram shows the typical sequence of genetic changes in a developing colorectal carcinoma. Indicate which event (A to C) corresponds to the following changes. Your answer would be a three-letter string composed of letters A to C only, e.g. CAB. normal epithelium --(A)--> early adenoma --(B)--> intermediate and late adenomas --(C)--> adenocarcinoma and metastases ( ) Activation of K-Ras ( ) Loss of p53 ( ) Loss of Apc

BCA This oversimplified diagram provides a general correspondence between mutations and the stages of cancer progression. In most cases, inactivating Apc mutations appear to occur early, followed by later activation of K-Ras and inactivation of p53. Many other mutations are generally involved. Note that genetic and epigenetic instability rises during the process.

Three fundamental controls seem to have been subverted in essentially every type of cancer. Choose these three among the following regulatory axes. Your answer would be a three-letter string composed of letters A to F only, in alphabetical order, e.g. BDF.

BCD Rb pathway - RTK/Ras/PI3K Pathway - p53 pathway Genes in each of these pathways are mutated in most cancer cells that have been analyzed.

25. The role of Chordin and Noggin in patterning in developing vertebrates is equivalent to that of ...(1) in Drosophila. They are secreted from the ...(2) pole and antagonize the activity of bone morphogenetic factors that are secreted throughout the embryo.

C. 1: Short gastrulation; 2: dorsal Like their Drosophila counterpart Sog, Noggin and Chordin antagonize the proteins of the bone morphogenetic protein (BMP) family (Decapentaplegic is a family member in the fruit fly). However, unlike Sog, Noggin and Chordin limit BMP activity to the ventral (rather than the dorsal) side of the embryo. This results in the inversion of the D-V axis in vertebrates compared to that of insects.

23. The following graph qualitatively represents the gradients of Nodal and Lefty gene products in an early frog embryo. The position along which primary axis is defined by these gradients? Which curves correspond to these two proteins?

C. A-V axis; curves 1 and 2 The cells at the vegetal pole secrete two opposing morphogens, Nodal (1) and Lefty (2), the latter being more rapidly diffusing. The result is the domination of Nodal near the vegetal end, and domination of Lefty near the animal end.

Mutation in which of the following genes is most prevalent in human colorectal cancer cells?

C. Apc The Apc tumor suppressor gene is mutated (and is defective) in over 80% of colorectal cancers.

22. Assuming that the only function of Short gastrulation (Sog) in fruit flies is to regulate Decapentaplegic (Dpp), which of the following genetic interactions would you NOT expect to observe?

C. Gain of Dpp suppresses the phenotype of partial loss of Sog. Given the antagonist role of Sog in regulating Dpp function, gain of Dpp is expected to enhance (rather than suppress) the phenotype of Sog loss-of-function mutations.

Once the molecular aberrations in a cancer are understood, drugs can be designed with a rational approach to treat the cancer. Which of the following is NOT true regarding such drugs?

C. Imatinib is most effective in treating chronic myelogenous leukemia (CML) in its acute blast-crisis phase. Imatinib is not very successful in treating CML once the cancer has advanced to a terminal phase, blast crisis, and no longer behaves like a "chronic" leukemia.

Which of the following is estimated to be the leading cause of death from cancer in the United States?

C. Smoking

In medical oncology, PET (positron emission tomography) is used to selectively image tumors in the body and to monitor cancer progression and response to treatment. Before performing a PET scan, the patient should fast for at least several hours for blood glucose to be sufficiently low. At the time of the scan, the positron-emitting glucose analog fluorodeoxyglucose (FDG) is injected into the bloodstream and the patient is asked to wait for up to an hour while avoiding physical activity. Finally, the scanner moves slowly over the body to reveal the location of possible tumors. Why do you think the patient should avoid physical activity before the scan?

C. To prevent the absorption of the radioactive tracer by healthy tissues When the blood glucose level is low, the added radioactive glucose analog is preferentially and rapidly taken up by tumor cells as a result of the Warburg effect. Physical activity stimulates uptake by muscle cells also (including heart muscle), which interferes with the imaging.

37. In classical experiments done half a century ago, the cells of early frog embryos were disaggregated and later reaggregated in desired combinations. The cells managed to rearrange and sort themselves out into an overall arrangement similar to that of a normal embryo. This effect is mainly due to ...

C. cell adhesion Cell adhesion molecules such as cadherins play a major role in forcing the cells into their natural arrangement.

Carcinoma cells that have acquired malignancy and started local invasiveness to begin metastasis ...

C. decrease the expression of E-cadherin and undergo epithelial-mesenchymal transition. The transition in cancer cells to become invasive involves shifting to a less adhesive and more motile character, resembling the epithelial-mesenchymal transition in normal development. A key part in this process involves switching off the expression of Ecadherin, which is a cell adhesion molecule.

45. Mutations in certain components of the cell-cycle machinery in Drosophila melanogaster can be used to slow down the rate of progression through the cell cycle in the wing imaginal discs of the fly larvae, without a major effect on the rate of cell growth. As a result, compared to wild-type flies, the wing in the mutant flies would ...

C. have the same size, but with a smaller number of relatively larger cells. The size of the disc is not set to contain a certain number of cells; instead, it is regulated by the total disc size.

17. The gene clusters known as the Bithorax complex and the Antennapedia complex contain...

C. homeobox-containing genes Hox genes lie in one or the other of two gene clusters, the Bithorax complex and the Antennapedia complex.

Most DNA tumor viruses inhibit the products of ...

C. p53 and Rb Small DNA tumor viruses encode proteins that interfere with the Rb and p53 pathways, allowing the replication of the viral genome.

Genetically knocking out both copies of the p53 gene in rats ...

C. results in a higher rate of cancer onset, but the rats are otherwise seemingly normal. Most rats (or mice) with a homozygous knockout of the gene encoding p53 develop cancer within a few months after birth but appear normal otherwise, suggesting that p53 function is required only in special circumstances.

9. Sort the following primary axes in the order that they are established during the development of Xenopus laevis. Your answer would be a three-letter string composed of letters A to C only, e.g. ACB. (A) A-P (B) D-V (C) A-V

CBA The animal-vegetal (A-V) asymmetry is established before fertilization. The point of sperm entry then helps determine the dorsoventral (D-V) axis. The anteroposterior (A-P) axis is established later.

Indicate whether each of the following cancers can be best classified as a carcinoma (C), sarcoma (S), or neither of the two (N). Your answer would be a four-letter string composed of letters C, S, and N only, e.g. SSNC. ( ) Breast cancer ( ) Lung cancer ( ) Colorectal cancer ( ) Myeloma

CCCN It is not surprising that many cancers are derived from epithelial cells (i.e. are carcinomas) as these cells are more proliferative and also more exposed to the environment.

48. A cross section of a developing spinal cord in a vertebrate embryo is shown in the schematic drawing below. Indicate which feature in the drawing (labeled A to E) better matches each of the following descriptions. Your answer would be a four-letter string composed of letters A to E only, e.g. AAEB. ( ) It contains motor neurons. ( ) It is the floor plate. ( ) It secretes BMP and Wnt signals. ( ) It contains sensory neurons.

CDAB A represents the roof plate which secretes BMP and Wnt signals; D represents the floor plate which secretes Sonic hedgehog protein. B and C represent the dorsal and ventral regions of the tube, containing sensory and motor neurons, respectively.

Indicate whether each of the following descriptions better applies to a cancer cell (C) or a normal adult cell (N). Your answer would be a four-letter string composed of letters C and N only, e.g. CCNC.

CNNC Transformed cells display an altered growth control (e.g. lack of contact inhibition or anchorage dependence) and an altered sugar metabolism (e.g. de-emphasized oxidative phosphorylation, increased glucose uptake, and increased lactic acid fermentation).

PARP inhibitors can efficiently kill many breast cancer cells that lack functional Brca1 or Brca2 genes. How do these drugs accomplish this?

D. By inhibiting a DNA repair pathway PARP inhibitors interfere with a DNA repair pathway that becomes essential for cancers that lack the alternative Brca-dependent pathway.

1. Plants and animals use different developmental strategies and have very different ways of life. In which of the following fundamental cellular processes during development are plants most different to animals?

D. Cell movement Plant cells do not actively migrate through the embryo.

29. Which of the following is NOT true regarding vertebrate segmentation? A. The presomitic mesoderm retreats tailward as new somites are generated. B. The period of gene-expression oscillations in the presomitic mesoderm determines the size of each somite. C. If feedback delays in the gene-expression oscillations are increased, somite size will consequently increase as well. D. Notch-mediated lateral inhibition ensures that the segmentation clock is different in neighboring cells in the presomitic mesoderm.

D. Notch-mediated lateral inhibition ensures that the segmentation clock is different in neighboring cells in the presomitic mesoderm. A gene-expression oscillator acts as a clock to control segmentation in vertebrates. During this process, Notch signaling ensures that the oscillations are in synchrony in neighboring cells in the presomitic mesoderm (i.e. the segmentation clock is the same). Note that this is opposite to the role of Notch signaling in lateral inhibition.

40. During branching morphogenesis in lung development, ...

D. Shh is produced by epithelial cells at the tip of the growing epithelial tubes. Fibroblast growth factor FGF10 is secreted by the mesenchymal cells, inviting the nearby epithelial cells at the tip of the growing tube to invade the mesenchyme. The Sonic hedgehog (Shh) signal is sent in the opposite direction—from the epithelial cells back to the mesenchymal cells—where it likely inhibits FGF10 production, leading to branching of the growing bud.

35. Stromal cell-derived factor 1 (SDF1) is a secreted protein that plays a major role in guiding the migration of various cells during development. It binds to a G-protein-coupled receptor on the surface of the migrating cell, which in turn activates a trimeric G protein containing a Gi subunit. Pertussis toxin (PTX) modifies and inactivates the Gi subunit. What would you expect to observe if migrating cells are treated with PTX?

D. They move in small steps in random directions. SDF1 biases the site of membrane protrusions (blebs) toward its source, but the cells still display blebbing even when SDF1 signaling is perturbed.

The effect of the deletion of one copy of the gene encoding p53 is different from the effects caused by other p53 mutations. For example, some loss-of-function mutations in the DNAbinding domain of p53 cripple its function as a transcription regulator. Such a mutation in only one copy of the p53 gene can be enough to confer a p53 loss-of-function phenotype, even when the other copy of the gene on the homologous chromosome is wild type. This is because ...

D. p53 forms a tetramer. The dominant negative effect of p53 mutations is due to the fact that defective p53 monomers can block the function of the tetramers in which they participate.

21. Cactus is a maternal-effect gene coding for an inhibitory protein in the Drosophila Toll signaling pathway that binds to Dorsal in the cytosol and keeps it from nuclear entry. Would you expect maternal loss-of-function mutations in Cactus to give rise to dorsalized or ventralized embryos? What about loss-of-function mutations in Decapentaplegic (Dpp)?

D. ventralized; ventralized Both of these mutations would result in a ventralized phenotype (similar to when Dorsal is activated in the entire embryo).

24. In the following fate map for the Xenopus blastula, indicate which zone in the map (A to D) better corresponds to each of the following fates. Your answer would be a four-letter string composed of letters A to D only, e.g. DCAB. ( ) Lung ( ) Blood ( ) Brain ( ) Skin

DCBA Figure 21-28*

5. Imagine a morphogen gradient established from left to right in a field of cells in a developing tissue, as shown in the following schematic diagram. Below a first threshold of morphogen concentration, cells do not respond to the morphogen and express gene "red" by default. Cells exposed to morphogen concentrations above this threshold respond by expressing gene "white" instead, while those exposed to even higher concentrations, above a second threshold, express gene "blue." As shown, the initial pattern resembles a French flag with equally wide blue, white, and red expression domains. With no other change, if the diffusion rate of the morphogen is increased (by a modification that decreases its affinity for heparan sulfate proteoglycans, for example), the gradient profile changes from the gray curve to the black curve, as indicated. Under this new condition, indicate whether each of the following would be expected to increase (I), decrease (D), or remain unchanged (U) in its range. Your answer would be a three-letter string composed of letters I, D, and U only, e.g. UUI. ( ) Blue expression domain ( ) White expression domain ( ) Red expression domain

DID Changes in parameters such as diffusion rate can affect the patterning of cells by a morphogen. In this example, the more shallow gradient widens the gap between positions corresponding to the first and second thresholds—i.e. the "white" expression domain—at the expense of the other two domains.

20. Indicate whether each of the following proteins is most concentrated near the dorsal (D) or ventral (V) side of the Drosophila embryo. Your answer would be a four-letter string composed of letters D and V only, e.g. DDDV. ( ) Decapentaplegic ( ) Dorsal (nuclear fraction) ( ) Toll (active form) ( ) Twist

DVVV Activated Toll receptors on the ventral side of the embryo result in the nuclear localization of Dorsal at this side. The highest nuclear Dorsal concentration activates the expression of Twist, whereas the lowest concentration de-represses Decapentaplegic (Dpp). At intermediate concentrations, Short gastrulation (Sog) expression is induced.

The requirement for accumulation of multiple mutations in cancer progression is manifested in the normalized percentage of new cases of cancer diagnosed in different age groups. Which of the curves A to E in the following graph better represents the incidence of human cancers as a function of age?

E The incidence of cancers grows exponentially as a function of age in adulthood, although it is thought to reach a plateau or even decline after the age of 80.

You are studying the rising incidence of a certain subtype of cervical cancer in Oceania, and are curious to know whether environmental factors are the dominant cause of the disease. You collect the incidence statistics from indigenous populations as well as from two different immigrant populations in three different countries, as shown in the following table. Do these data appear to be consistent with a dominant role of environmental risk factors (E) or a genetic background (G) for this type of cancer? Write down E or G as your answer.

E The incidence rates for this type of cancer in each immigrant population mirrors that of the host country, suggesting an overall dominance of environmental factors.

10. Cortical rotation following fertilization in X. laevis places the ...(1) pole at the point of sperm entry, while Wnt11 mRNA is transported to the ...(2) pole. A. 1: anterior; 2: posterior B. 1: posterior; 2: anterior C. 1: animal; 2: vegetal D. 1: dorsal; 2: ventral E. 1: ventral; 2: dorsal

E. 1: ventral; 2: dorsal Fertilization in Xenopus laevis, through a reorganization of the microtubule cytoskeleton, triggers a rotation of the egg cortex. This places the ventral pole at the point of sperm entry. The reorganization also leads to the transport of several cytoplasmic components, including the Wnt 11 mRNA, which becomes concentrated at the dorsal side.

The Rb gene in retinoblastomas is similar to the Apc gene in polyposis colon carcinomas in that both genes ... A. are tumor suppressors. B. are mutated in one copy in all cells of patients with a hereditary form of the cancer. C. are in a locus that shows loss of heterozygosity in the hereditary form of the cancer. D. should be inactivated in both copies to cause the nonhereditary form of the cancer. E. All of the above.

E. All of the above. Apc and Rb are tumor suppressors, and inactivation of both copies of each gene occurs in colorectal cancers and retinoblastomas, respectively. The hereditary forms, however, are easier to develop since the person receives only one functional copy to begin with, which can become lost or defective.

11. Which of the following is true regarding maternal-effect genes? A. Bicoid and Nanos are maternal-effect genes. B. A female homozygous for a loss-of-function maternal-effect gene mutation can be fully normal, but her offspring will show the phenotype. C. The offspring of a female homozygous for a loss-of-function maternal-effect gene mutation will show the phenotype regardless of the paternal genotype. D. The second-generation offspring of a male homozygous for a loss-of-function maternal-effect gene mutation can show the phenotype. E. All of the above.

E. All of the above. For maternal-effect genes, it is the mother's genome rather than the zygotic genome that is critical.

19. Which of the following is correct regarding the regulation and maintenance of Hox gene expression in Drosophila? A. Proteins of the Polycomb and Trithorax groups maintain inactive and active states of Hox gene expression, respectively. B. If Polycomb or Trithorax group genes are defective, Hox gene expression patterns are still initiated, but cannot be correctly maintained. C. The remodeled chromatin at the Hox complex is heritable through cell generations. D. If all the Hox genes in an embryo are deleted, segmentation still occurs but distinct segment identities are lost. E. All of the above.

E. All of the above. Hox proteins give each segment its individuality; without them, segments will lose their individual characters. Trithorax and Polycomb group proteins work together, in opposite ways, to enable the Hox complexes to maintain a permanent record of positional information. Without them, the Hox gene expression pattern is set up correctly, but cannot be maintained.

Which of the following proteins is NOT encoded by a proto-oncogene? A. Src B. Ras C. EGF receptor D. Myc E. E-cadherin

E. E-cadherin Loss-of-function mutations in the tumor suppressor E-cadherin promote the epithelial-mesenchymal transition and local invasiveness.

Which of the following can lead to p53 stabilization and activation? A. Hypoxia B. Overexpression of Myc C. DNA Damage D. Telomere Loss E. all of the above

E. all of the above Cells raise their concentration of p53 in response to a whole range of conditions.

Consider a healthy adult animal in which 10^15 cell divisions have taken place since birth. Spontaneous mutations can occur at a rate of approximately one nucleotide out of about ten billion nucleotides every time DNA is replicated. The animal has a diploid genome size of about 2 billion nucleotide pairs. Assuming that only about 5% of mutations occur within genes or gene regulatory sequences, and further assuming that about 0.1% of those may cause cancer, how many potential cancer-causing mutations has the animal been able to successfully suppress (i.e. has been able to survive) during its lifetime? A. ten thousand B. one hundred thousand C. ten million D. one billion E. ten billion

E. ten billion

50. In the following schematic drawing, a growing axon (in red) is extended to a target neuron to form three synapses. In its journey, it is guided by a variety of mechanisms. Sort the following mechanisms to reflect the order in which they guide the navigation of the growth cone toward the target neuron in this example. Ignore the final chemoattraction step toward the target neuron itself. Your answer would be a six-letter string composed of letters A to F only, e.g. FACBDE. (A) Chemoattraction (B) Chemorepulsion (C) Cell surface adhesion (D) Contact inhibition (E) Extracellular matrix adhesion (F) Guidance by pioneer neuron

ECFADB Growth cones use various cues to navigate toward their targets. Figure 21-73*

12. Indicate whether each of the following descriptions better applies to egg-polarity genes (E), gap genes (G), pair-rule genes (P), or segment-polarity genes (S). Your answer would be a four-letter string composed of letters E, G, P, and S only, e.g. SSGG. ( ) Mutations in these genes show a maternal effect. ( ) Mutations in these genes leave the embryo with only half the number of normal segments. ( ) Mutations in these genes produce a normal number of segments but with part of each segment replaced by a mirror-image duplicate of other parts of the segment. ( ) Mutations in these genes can eliminate one or two groups of adjacent segments altogether.

EPSG Egg-polarity genes in Drosophila are maternal-effect genes: their mRNA transcript is deposited by the mother into the egg, and hence it is the mother's genome (rather than the zygote's genome) that is critical. Mutations in gap genes result in the elimination of one or more adjacent segments altogether. Mutations in pair-rule genes cause a series of deletions affecting alternate segments, leaving the embryo with only half as many segments as usual. Finally, mutations in segment-polarity genes produce the right number of segments but with a part of each segment replaced by a mirror-image duplicate of another part of the segment.

Indicate true (T) and false (F) statements below regarding cancer. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancers become less and less heterogeneous as they progress. ( ) Knocking out Ras or Myc genes individually leads to a higher incidence of cancers in mice, and knocking out both genes simultaneously has an even stronger phenotype. ( ) Wnt signaling is important in colon epithelial cells and, correspondingly, mutations in genes in the Wnt pathway are present in most colorectal cancers. ( ) Genome destabilization in a subset of colorectal cancers that have defects in DNA mismatch repair takes the form of chromosome breaks, translocations, and deletions.

FFTF Cancer risk is increased by gain-of-function mutations in Ras and Myc, not by their knockout. It is not surprising that in cancer cells with defects in DNA mismatch repair, many point mutations are found throughout the genome. Cancer progression coincides with increasing heterogeneity of the tumor cell population

Indicate true (T) and false (F) statements below regarding cell proliferation in human somatic cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer cells show replicative cell senescence. ( ) Cancer cells maintain their telomeres by inhibiting the enzyme telomerase. ( ) Some cancer cells do not rely on telomerase for telomere lengthening. ( ) Most cancer cells lack telomeres.

FFTF Mammalian cancer cells avoid replicative cell senescence (which generally depends on telomere shortening) in two ways: they can either maintain telomerase activity to prevent telomere shortening, or evolve an alternative mechanism based on homologous recombination to lengthen their telomeres.

Indicate true (T) and false (F) statements below regarding the mutational landscape of cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF.

FFTT The number of driver mutations in most cancers is estimated to be on the order of 10. About 300 genes (less than 2% of our genes) are strongly suspected to be cancercritical. They encode proteins of various functions. In addition to DNA sequence changes, chromosomal breaks and translocations are common in cancer cells.

31. Indicate true (T) and false (F) statements below regarding developmental timing in animals. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Cell-division cycles generally serve as an intracellular timer that controls the timing of cellular differentiation. ( ) Developmental transitions are often regulated by microRNAs. ( ) Intracellular developmental programs are often followed even if the cells are taken from the developing embryo and maintained in culture. ( ) Timing of developmental transitions can be coordinated by cell-cell interactions as well as globally by hormones.

FTTT Cells rarely count cell divisions to time their development. In many cases, intracellular developmental programs help determine the time course of a cell's development, allowing it to step through more or less the same program whether it is within the embryo or in culture. The timing of developmental transitions can also be coordinated by cell-cell interactions and a global coordinating signal. The transitions are often regulated and sharpened by microRNAs in the cell.

Indicate true (T) and false (F) statements below regarding colorectal cancers. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Observation of polyps in the colon epithelium of a patient is an indication of a malignant carcinoma. ( ) Progression of colorectal cancer is very slow and normally takes over 10 years to turn into malignancy. ( ) Colorectal cancers are usually diagnosed later in life. ( ) Invasive colorectal cancer cells usually metastasize to lymph nodes via lymphatic vessels and then into the bloodstream.

FTTT The adenomatous polyps are believed to be the precursors of most colorectal cancers, but progression into malignancy is slow and the average time from detection of the polyp to cancer diagnosis is about 12 years. This usually happens later in life. Colorectal cancers tend to metastasize through lymphatic nodes after spreading through the layers of tissues lining the gut.

7. Indicate whether each of the following descriptions better applies to these poles established in an early embryo: anterior (A), posterior (P), dorsal (D), ventral (V), animal (N), or vegetal (G). Your answer would be a six-letter string composed of letters A, D, G, N, P, and V only, e.g. ADGPVN. ( ) It defines the part to become internalized in gastrulation. ( ) It defines the parts to remain external. ( ) It defines the location of the future head. ( ) It defines the location of the future tail. ( ) It defines the location of the future belly. ( ) It defines the location of the future back.

GNAPVD Three axes generally have to be established in the early embryo. In most animals, the animal-vegetal (A-V) axis defines which parts are to remain external in gastrulation and which are to become internalized. The anteroposterior (A-P) axis specifies the locations of the future head and tail. The dorsoventral (D-V) axis specifies the future back and belly.

32. Does the nuclear-to-cytoplasmic ratio increase (I) or decrease (D) during cleavage in early vertebrate development? Would you expect tetraploid embryos to undergo the maternalzygotic transition earlier (E) or later (L) compared to diploids? If the availability of a transcription regulatory protein determines the timing of this transition, should the protein be a transcriptional activator (A) or repressor (R)? Write down your answer as a three-letter string using the letters in the parentheses above, e.g. IEA.

IER Rapid cleavage increases total DNA without a concomitant increase in the total cytoplasmic volume, thereby increasing the ratio of DNA to diffusible repressors. This ratio reaches a threshold at the maternal-zygotic transition (MZT). This occurs after fewer rounds of cell division if the cells are polyploid.

Indicate whether each of the following descriptions better applies to trastuzumab (T) which targets Her2, imatinib (I) which targets Bcr-Abl, or ipilimumab (P) which targets the CTLA4 protein. Your answer would be a three-letter string composed of letters T, I, and P only, e.g. TTP. ( ) It is NOT an antibody. ( ) It counters the immunosuppressive microenvironment of tumors. ( ) It does not bind to a cancer cell component.

IPP By binding to an inhibitory receptor on T cells, the monoclonal antibody ipilimumab helps block immunosuppression in tumors. Trastuzumab is also a monoclonal antibody, but it directly binds to EGF receptors on some cancer cells to block their function.

Indicate whether each of the following viruses is mostly associated with cervical cancer (C), Kaposi's sarcoma (K), liver cancer (L), or stomach cancer (S). Your answer would be a fourletter string composed of letters C, K, L, and S, e.g. CKLS. ( ) Hepatitis-B virus (HBV) ( ) Human immunodeficiency virus (HIV) ( ) Human papillomavirus (HPV) ( ) Helicobacter pylori

LKCS A small but significant proportion of human cancers is thought to arise from infections.

X-chromosome inactivation in female mammals occurs mostly randomly early in development, resulting in a heterogeneous cell population, with each cell having inactivated one or the other of its X chromosomes and passing on the same X-inactivation choice to its offspring. The inactivated X chromosome is generally hypermethylated and transcriptionally inactive. You are studying a newly discovered type of colon tumor in women that has a morphology distinct from that of other colon adenomas. You extract chromosomal DNA from the tumor cells. You then either keep the DNA untreated, or digest the DNA with a methylation-sensitive restriction enzyme that only cleaves its recognition DNA sequence if the sequence is not methylated. Finally, you amplify by polymerase chain reaction (PCR) a locus on the X chromosome known to be polymorphic in length (i.e. it is expected to be of different sizes in different X chromosomes). The locus has a restriction site for the mentioned enzyme, such that cleavage would prevent PCR amplification. You quantify the amount of PCR products corresponding to shorter and longer versions of the locus, and obtain the results shown in the following table. Do these data appear to be in better agreement with a monoclonal (M) or a polyclonal (P) origin of cancer? A monoclonal origin would mean that all cells in the tumor are the clonal descendants of a single abnormal cell, while a polyclonal tumor is composed of cells from different lineages. Write M or P as your answer.

M The individual is heterozygous in the locus, represented in a ratio close to 1:1 for the PCR products of the longer to shorter alleles when an undigested DNA is used. When fully digested, however, only the methylated DNA is expected to be represented in the PCR products. Whereas the polyclonal healthy tissue sample still has a mostly unbiased 1:1 ratio, the monoclonal tumor cells (all with the same inactive X chromosome) have a skewed ratio.

3. Indicate whether each of the following organs or tissues arises from ectoderm (C), mesoderm (M), or endoderm (N). Your answer would be a four-letter string composed of letters C, M, and N only, e.g. MMCC. ( ) Blood ( ) Liver and pancreas ( ) Brain ( ) Bone and cartilage

MNCM Ectoderm gives rise to the epidermis and the nervous system; endoderm gives rise to the gut tube and its appendages, such as lung, pancreas, and liver; mesoderm gives rise to muscles, connective tissues (e.g. bone and cartilage), blood, kidney, and various other components.

Suppose you are studying tumor heterogeneity in a certain type of melanoma. You have used fluorescence-activated cell sorting (FACS) to specifically isolate those melanoma tumor cells that either do (first category) or do not (second category) express a specific marker present in normal stem cells in the tissue of origin (i.e. the melanocyte stem cells). You implant the same number of cells from each of these categories into severely immunodeficient mice and compare the tumor-formation efficiencies after several weeks, which turn out to be significantly higher for the first category. You then analyze the new tumors using FACS, and find out that the majority of the cells in the tumors that originated from the first category of cells harbor the stem-cell marker, whereas the majority of the cells in the tumors that originated from the second cell category lack the marker, just like their respective founder cells. Do these observations support the existence of cancer stem cells? Write down Yes or No as your answer.

No Cancer stem cells are expected to constitute a minority of the tumor cell population. Even though the transplantation is more efficient with the subpopulation of cells that expresses the stem-cell marker, this might simply reflect the heterogeneity in the original tumor, with these cells harboring a higher ability to found new tumors.

46. Indicate whether each of the following conditions favors a larger (L) or smaller (S) tissue or body size. Your answer would be a four-letter string composed of letters L and S only, e.g. SSSS. ( ) Hippo overexpression ( ) Insulin-like growth factor overexpression ( ) Myostatin deletion ( ) Growth hormone deficiency

SLLS Growth hormone stimulates growth, and its deficiency leads to a smaller body size. It acts by inducing the expression of insulin-like growth factor, overexpression of which may lead to a larger body size in an animal. Myostatin specifically inhibits the growth and proliferation of myoblasts; consequently, loss of myostatin results in greatly enlarged muscle tissue. The Hippo pathway inhibits organ and body growth in general; overexpression of Hippo, an upstream component of this pathway, inhibits growth.

6. Indicate whether each of the following descriptions better applies to patterning by lateral inhibition (L), reaction-diffusion systems (R), or sequential induction (S). Your answer would be a four-letter string composed of letters L, R, and S only, e.g. SSRR. ( ) It does NOT generate asymmetrical patterns from an initial noisy field. ( ) It is based on short-range activation and long-range inhibition. ( ) It can readily generate complex patterns resembling the spots of a leopard or stripes of a zebra. ( ) It is commonly mediated by Notch signaling.

SRRL Lateral inhibition, commonly mediated by Notch signaling, can generate asymmetry when combined with positive feedback. Reaction-diffusion systems, based on short-range activation and long-range inhibition, are also capable of spontaneous pattern generation through positive feedback, and give rise to complex patterns such as spots and stripes. Sequential induction, on the other hand, is a strategy to refine initial patterns into more complicated ones, and cannot create asymmetrical patterns by itself.

The homologous chromosome pairs in our cells do not carry identical sequences in all loci. This heterozygosity (difference between the two copies) can be altered in cancer: in fact, loss of heterozygosity at many loci is observed in cancer cells, through an increase in either homozygosity (two identical copies) or hemizygosity (i.e. loss of one copy). Researchers can take advantage of this loss of heterozygosity in cancer cells to identify genomic loci that contain cancer-critical genes. What type of gene would you expect to find in chromosomal regions with a loss of heterozygosity? Proto-oncogenes (P) or tumor suppressor genes (T)? Write down P or T as your answer.

T Once the first copy of a tumor suppressor gene is lost or inactivated, the remaining copy is commonly lost by a less specific mechanism, leading to loss of heterozygosity in the gene as well as in its neighboring loci. This represents a way of finding loci that contain tumor suppressor genes.

18. Indicate true (T) and false (F) statements below regarding Hox genes. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) The order of expression of Hox genes along the body corresponds to their order in the Hox complex. ( ) Generally, the more anterior of the Hox genes dominate (or suppress) posterior Hox genes. ( ) When a posterior Hox gene is artificially expressed in an anterior region of the embryo, the tissue maintains its anterior character. ( ) Hox genes control the A-P axis in both vertebrates and invertebrates.

TFFT The genomic location of Hox genes (which control the anteroposterior [A-P] axis in both vertebrates and invertebrates) corresponds well with their order of expression along the A-P axis. In other words, Hox genes are expressed according to their order in the Hox complex. Generally, the more posterior of the Hox genes dominate the more anterior ones, so that when both are expressed in a given segment, the segment will assume an identity dictated by the more posterior Hox genes.

Indicate true (T) and false (F) statements below regarding cancer incidence and cancer prevention. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Over half of all cancers are preventable by lifestyle changes. ( ) The age-adjusted cancer death rates have increased steadily since 1900, mostly due to the industrial way of life. ( ) Currently, more than half of all cancer patients survive at least five years postdiagnosis. ( ) About half of all cancers are thought to arise by infection with viruses, bacteria, or parasites.

TFTF Except for the increase in cancers caused by smoking, age-adjusted death rates for most common cancers have either stayed the same since half a century ago, or have even declined significantly. Survival rates have also improved: currently, more than twothirds of cancer patients live more than five years from the time of diagnosis. Fifty percent of cancers could be prevented by changes in lifestyle such as altered smoking, eating, and exercise habits. Only a small proportion of human cancers (perhaps ~15%) are thought to arise from infection.

Indicate true (T) and false (F) statements below regarding the properties of cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer cells invade and colonize territories that normally belong to other cells. ( ) Unlike in normal tissue, cell death is extremely rare in tumors. ( ) Cancer cells grow and proliferate in defiance of normal restraints. ( ) Malignant tumors are composed of cells that grow and proliferate, but still have not acquired invasiveness.

TFTF Unrestrained growth and division plus invasiveness are two key features of true malignant cancer cells. Cell within a tumor still die in large numbers.

Indicate whether each of the following descriptions better applies to proto-oncogenes (P) or tumor suppressor genes (T). Your answer would be a four-letter string composed of letters P and T only, e.g. PPPT. ( ) Cancer mutations in these genes are usually recessive. ( ) Cancer mutations in these genes include gene duplications. ( ) Cancer mutations in these genes are responsible for most hereditary cancers. ( ) Cancer mutations in these genes are commonly in the form of nonsense (truncating) mutations that abort protein synthesis.

TPTT Truncating mutations in tumor suppressor genes can create recessive, loss-offunction mutants associated with most hereditary cancers. A proto-oncogene, in contrast, can be activated by a limited set of dominant, gain-of-function mutations (e.g. missense point mutations or other accidents such as gene duplication, which can produce an abnormally large amount of gene product).

4. Indicate true (T) and false (F) statements below regarding animal development. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) A fertilized egg is totipotent. ( ) Differences in their regulatory DNA can largely explain the differences between animal species. ( ) Inductive signaling is mostly mediated through G-protein-coupled receptors. ( ) A cell's response to a signal depends on its exposure to other signals at that present time as well as in the past.

TTFT A fertilized egg is totipotent, able to give rise to all the different cell types in an organism. Changes in regulatory DNA seem to be mainly responsible for the dramatic differences between one class of animals and another, even though the coding DNA has been, for the most part, highly conserved between the classes. Most known inductive signaling events during animal development are mediated by the transforming growth factor-β (TGFβ), Wnt, Hedgehog, and receptor tyrosine kinase (RTK) pathways. During development and in an adult organism, the cellular response to a signal depends not only on the identity of the signal, but also on the other signals that the cell is receiving, as well as on the previous experiences of the cell.

14. Indicate whether each of the following groups of genes typically creates a transient pattern in the developing embryo (T) or a long-lived pattern that is preserved (L). Your answer would be a five-letter string composed of letters T and L only, e.g. TTTLT. ( ) Egg-polarity genes ( ) Pair-rule genes ( ) Hox genes ( ) Gap genes ( ) Segment-polarity genes

TTLTL Egg-polarity, gap, and pair-rule genes create transient patterns that are remembered by segment-polarity and Hox genes.

Indicate true (T) and false (F) statements below regarding cancer. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer can be induced by infectious agents such as viruses. ( ) The earlier a cancer is diagnosed, the better the chances are for a cure. ( ) Most cancers originate from a single aberrant cell. ( ) A single mutation is NOT enough to turn a normal cell into a cancer cell.

TTTT A single mutation is not enough to change a normal cell into a cancer cell: most cancers develop gradually from a single aberrant cell by the accumulation of a number of genetic and epigenetic changes over time. Treatment is generally easier if the cancer is diagnosed at earlier stages. Some cancers can be induced by infectious agents.

38. Fill in the blank in the following paragraph describing collective cell rearrangements. Do not use abbreviations. "Cells form lamellipodia and attempt to crawl over one another, essentially pulling their neighbors inward into a narrow zone. This is accompanied by elongation along the long axis of the narrow zone. This process of ... depends on Wnt signaling and is observed multiple times during the development of a vertebrate."

convergent extension In convergent extension, the cells in a sheet crawl over one another in a coordinated fashion, causing the sheet to narrow along one axis (converge) and elongate along another (extend).