Quiz
BIO 150 Midterm 1 & 2
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Fragen in diesem Quiz (20)
1Which statement describes the relationship between Tolloid and Sog?
- AA) Sog is the ligand for the Tolloid receptor
- B(B) Tolloid is a protease that degrades Sog
- C(C) Tolloid is the ligand for the Toll receptor and Sog inhibits Tolloid activation of Toll
- D(D) Tolloid is an inhibitor of Dpp signaling and Sog is an activator of Dpp signaling
2What allows transcription factors to act as morphogens in early-stage Drosophila embryos?
- A(A) the transcription factors have different activities (activator or repressor) depending on their concentration
- B(B) the early activation of zygotic transcription factors establishes morphogen gradients
- C(C) the mRNA for these transcription factors is maternally-deposited
- D(D) the early-stage embryo is a syncytium: there are no cell membranes separating nuclei and
transcription factors can freely diffuse throughout the embryo
3Colinearity of Hox genes refers to:
- AA) The relative position of each Hox gene along the chromosome corresponds to the relative
position of each Hox gene's expression along the anterior-posterior body axis. - B(B) Anterior Hox genes activate transcription of more posterior Hox genes
- C(C) Hox genes are conserved in insect lineages, but different Hox genes are conserved in vertebrate lineages.
- D(D) Redundancy in Hox gene function: one Hox gene can be replaced by the next posterior Hox gene without any developmental defects.
- AA) The relative position of each Hox gene along the chromosome corresponds to the relative
4You identify a region of ventral ectoderm in the earthworm embryo that later develops into brain cells in the adult worm (using standard fate mapping). You perform a heterotopic transplantation experiment and move this region of ventral ectoderm into the dorsal ectoderm of a worm embryo. You find that the transplanted cells develop into skin cells instead of brain cells. If you take these same ventral ectoderm cells and grow them by themselves in tissue culture, the cells develop into brain cells.
What do these results tell you about the ventral ectoderm cells?- AA) The cells are specified and determined to become brain cells
- BB) The cells require Hox gene expression to develop into brain cells
- CC) The cells require a morphogen signal to develop into brain cells
- DD) The cells are specified but not determined to become brain cells
5A mutation that caused all the follicle cells surrounding a Drosophila oocyte to produce Trunk would result in what phenotype?
- A(A) The entire larvae would be made of acron
- B(B) The anterior half of the larvae would be all acron, the posterior half would be all telson
- C(C) The entire larvae would be made of telson
- D(D) The anterior half of the larvae would be normal (acron, head, thorax) but the abdominal region would be replaced with telson.
6You are studying embryonic development in a species of earthworm. The earthworm genome sequence is available, but it is not possible to introduce transgenes or delete genes in this species. You identify an earthworm Wnt gene and want to know where it is expressed. What method could you use?
- A(A) reporter gene: use the Wnt gene enhancer to direct lacZ expression
- B(B) In situ hybridization
- C(C) Dye labeling
- D(D) Cre / loxP method: use the Wnt gene enhancer to direct Cre expression and remove a "stop" sequence blocking constitutive lacZ expression
7 Loss of Eve (Even-skipped) function during Drosophila embryogenesis causes:
- A(A) The normal 14 segments are formed but segment polarity is lost, only anterior fates are made.
- B(B) Only 7 segments develop, instead of the normal 14 segments.
- C(C) Additional segments form in the abdomen, most likely giving 21 total segments.
- D(D) Only the anterior segments are lost (head and thorax segments).
8Driesch's experiment showing that a four cell sea urchin embryo can be broken up into individual cells and each cell develops into a complete sea urchin is an example of what approach?
- A(A) Transplantation
- B(B) Recombination
- C(C) Isolation
- D(D) Defect
9The phenotype of a nanos loss-of-function Drosophila embryo would be most similar to which of the following mutants?
- A(A) Bicoid gain-of-function (bicoid expressed throughout the embryo)
- B(B) Torso loss-of-function
- C(C) Maternal hunchback gain-of-function (maternal hunchback expressed throughout the
embryo) - D(D) Caudal loss-of-function
10Under what situation does Dorsal act as a transcription repressor?
- A(A) When Dorsal is present at high concentrations
- B(B) When Dorsal and Groucho both bind to the region controlling gene transcription
- C(C) When Dorsal and Rhomboid both bind to the region controlling gene transcription
- D(D) When Dorsal is present at low concentrations
11If the Xenopus genome encoded two different N-cadherin proteins (N-cad A and N-cad B), how could you use a "dominant negative" approach to identify the cadherin that is necessary for neural tube closure?
- A(A) Inject a morpholino that targets the mRNA of both N-cad A and N-cad B and see if this alters neural tube closure
- B(B) Make mRNA encoding a mutant cadherin (either A or B) that lacks the intracellular domain
required for cadherin - actin cytoskeleton interactions and inject a large amount of this mutant - C(C) Inject mRNA encoding the normal / wildtype version of another adhesion molecule, such as the Immunoglobulin superfamily protein N-CAM, into the neural plate
- D(D) Inject a large amount of mRNA encoding the normal / wildtype cadherin protein (either A or B) into the neural plate
12What would happen if a mutation in a cell migration gene caused all neural crest cells to migrate along an identical route (for example, all neural crest cells migrate through somites and never through any other part of the embryo)?
- A(A) The neural crest cells would fail to migrate posteriorly and no posterior nervous system would form but brain / anterior nervous system would be normal
- B(B) The neural crest cells would fail to migrate laterally and no lateral / peripheral nervous system would form but the spinal cord / central nervous system would be normal
- C(C) Since the neural crest cells are multipotent they would still make all their expected progeny but the progeny cells would be in the wrong place
- D(D) The neural crest cells would all make the same progeny cell type and fail to make the full diversity of cell types normally produced based on different migration routes
13There is a Xenopus mutant in which the DNA replication machinery makes extra copies of all chromosomes at each round of replication, starting from the one cell stage. This mutation would likely cause:
- A(A) Failure to induce the neural plate due to excess DNA blocking neural gene expression.
- B(B) The expression of zygotic genes (the "mid-blastula transition") to occur earlier than usual.
- C(C) The chromosome looping required for Xlim1 expression to occur earlier than usual.
- D(D) Muscle differentiation to occur late due to increased cell cycle timing.
141. The equivalent of Spemann's organizer in a zebrafish embryo is:
- A(A) The Shield
- B(B) Koller's sickle
- C(C) The posterior marginal zone
- D(D) Hensen's node
15Which of the following statements properly describes the source of lateral plate mesoderm (blood, kidney) and notochord during chicken development?
- A(A) The lateral mesoderm is specified during migration of epiblast cells through the primitive
- B(B) The lateral mesoderm is formed from epiblast cells in the ventral marginal zone while the notochord is formed from epiblast cells in the dorsal marginal zone
- C(C) The notochord forms as Hensen's node migrates anteriorly and the lateral mesoderm forms later during regression of Hensen's node
- D(D) Lateral mesoderm and notochord are both derived from pre-somitic mesoderm: the earliest forming pre-somites become lateral mesoderm and later forming pre-somites become notochord
16The Nieuwkoop center and Spemann organizer are both specified by high levels of Wnt / - catenin activity. Why don't cells in the Nieuwkoop center end up having the same organizing function as the Spemann organizer cells?
- A(A) Cells in the Nieuwkoop center never produce Nodal and Nodal is required for the organizer fate.
- B(B) The Nieuwkoop center forms on the future ventral side of the embryo and therefore produces the opposite signals from the dorsal organizer.
- C(C) Cells in the Nieuwkoop center are the first to migrate through the blastopore during gastrulation, so they are not in the proper position to have an organizer function.
- D(D) Cells in the Nieuwkoop center have higher levels of VegT and this induces endoderm fate.
The zygotic gene expression required for organizer function only occurs in the dorsal
mesoderm, not the dorsal endoderm, and so the Nieuwkoop cells never express organizer
genes.
17How does Ectodermin inhibit all TGF-Beta family signaling in Xenopus animal cap cells?
- A(A) It is a transmembrane protein that acts as a dominant negative inhibitor of Type I and Type II receptors because it lacks a kinase domain.
- B(B) It is a transcription factor that represses expression of all SMAD target genes.
- C(C) It binds to all TGF- signaling proteins before they bind their receptors.
- D(D) It is a ubiquitin ligase that causes destruction of Co-SMAD. Co-SMAD is necessary for all
TGF- signaling pathways.
18What would happen if the endoblast cells of an early blastula chicken embryo expressed Cerberus?
- A(A) Anterior mesoderm fates would be normal but posterior mesoderm fates would be missing
- B(B) The ectoderm would not be divided into neural plate and epidermis
- C(C) The primitive streak would never form
- D(D) Primitive streak formation would occur throughout the marginal zone
19The "phylotypic stage", when vertebrate embryos appear most similar to each other (mainly
due to the presence of the notochord, somites, and head versus tail structures) is the:- A(A) Neurula stage
- B(B) Early gastrula stage
- C(C) Late blastula stage
- D(D) Mid blastula stage
20If the number of pre-somitic cells produced in a chicken embryo and a snake embryo is the same, what could explain the higher number of somites formed in a snake?
- A(A) The "clock" of cyclical gene expression is faster in the pre-somites of a snake
- B(B) There are more Hox genes in a snake and this expanded diversity of Hox genes can specify more somites
- C(C) There is no difference between anterior and posterior somites in a snake, so there is no need for a "clock" mechanism and all pre-somites become somites
- D(D) The "clock" of cyclical gene expression is slower in the pre-somites of a snake
Informationen
- Beschreibung
- Biology Quiz am BIO 150 Midterm 1 & 2, erstellt von Aimee Armas am 23/04/2018.
- Autor
- Aimee Armas
- Sprache
- English
- Aufrufe
- 3
- Schlagwörter
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