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BIO380 Multiple Choice Questions.pdf University of Toronto, Mississauga BIO 380

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1. What developmental mutation would most likely lead to a Cyclops (one eye) phenotype? a. Shh GOF mutation b. Ptc2 LOF mutation c. Rx LOF mutation d. Pax6 LOF mutation 2. Notch functions during somitogenesis to: a. Antagonize the Fgf8 morphogen gradient b. Generate somite boundaries c. Prevent the premature end of axis elongation d. Negatively regulate retinoic acid expression 3. In a frog post-gastrulation animal cap assay, the addition of a BMP antibody will result in: a. A change in the cell type normally observed, from neurons to epidermal cells b. A change in the cell type normally observed, from epidermal cells to neurons c. No change – epidermal cells will still be generated d. No change – neurons will still be generated 4. Why does the expression of the stripe 5 enhancer lacZ transgene extend posteriorly in giant mutant embryos? a. The stripes 6 and 7 enhancers are activated when giant is absent b. Giant protein binds to the stripe 5 enhancer as a repressor c. Giant is expressed in the future stripes 5-7 of wild-type embryos d. Giant binding is required for the repression of stripe 6 and 7 enhancers Giant is a negative regulator of stripe 5 and a positive regulator of stripe 1.In the wild-type embryo you can see with giant stripe 5 stays as a normal stripe. But in the mutant embryo without giant, stripe 5 loses definition and extends posteriorly. This means that giant must be negatively regulating stripe 5 so that it doesn't extend like that. The transgene only contains stripes 1,5 enhancer, so answers A and D do not make sense C is wrong because in the WT, stripe 1 and 5 is stained. giant can’t be expressed in stripe 5 otherwise it should be repressed and not stain in the giant mutant. Also, mutual repression of TFs is needed to define segments with sharp boundaries. 5. Which of the following techniques would allow you to identify new genes that are involved in a developmental process? a. In situ hybridization b. Immunohistochemistry c. Enhancer-reporter analysis d. Forward genetics screen e. Reverse genetics screen

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BIO380 Multiple Choice Questions


Class Poll Questions (2020)
1. What developmental mutation would most likely lead to a Cyclops (one eye) phenotype?
a. Shh GOF mutation
b. Ptc2 LOF mutation
c. Rx LOF mutation
d. Pax6 LOF mutation
Eyes close together Eyes further apart
Cyclops one eye phenotype Loss of eyes, no eyes

Shh LOF mutation Shh GOF mutation
Ptc2 and Pax2 LOF mutation Ptc2 and Pax2 GOF mutation
Rx LOF mutation
Pax6 LOF mutation


2. Notch functions during somitogenesis to:
a. Antagonize the Fgf8 morphogen gradient
b. Generate somite boundaries
c. Prevent the premature end of axis elongation
d. Negatively regulate retinoic acid expression
3. In a frog ​post-gastrulation​ animal cap assay, the addition of a BMP antibody will result in:
a. A change in the cell type normally observed, from neurons to epidermal cells
b. A change in the cell type normally observed, from epidermal cells to neurons
c. No change – epidermal cells will still be generated
d. No change – neurons will still be generated
4. Why does the expression of the stripe 5 enhancer lacZ transgene extend posteriorly in giant mutant
embryos?
a. The stripes 6 and 7 enhancers are activated when giant is absent
b. Giant protein binds to the stripe 5 enhancer as a repressor
c. Giant is expressed in the future stripes 5-7 of wild-type embryos
d. Giant binding is required for the repression of stripe 6 and 7 enhancers
Giant is a negative regulator of stripe 5 and a positive regulator of stripe 1.In the wild-type embryo you can
see with giant stripe 5 stays as a normal stripe. But in the mutant embryo without giant, stripe 5 loses
definition and extends posteriorly. This means that giant must be negatively regulating stripe 5 so that it
doesn't extend like that.
The transgene only contains stripes 1,5 enhancer, so answers A and D do not make sense
C is wrong because in the WT, stripe 1 and 5 is stained. giant can’t be expressed in stripe 5 otherwise it
should be repressed and not stain in the giant mutant.
Also, mutual repression of TFs is needed to define segments with sharp boundaries.
5. Which of the following techniques would allow you to identify new genes that are involved in a
developmental process?
a. In situ hybridization
b. Immunohistochemistry
c. Enhancer-reporter analysis
d. Forward genetics screen
e. Reverse genetics screen

,December Exam (2018)
1. The clock-wavefront model for somitogenesis proposes that:
a. The differential expression of Hox genes along the A/P axis drives the wave of somitogenesis
b. A moving boundary between gradients of FGF and retinoic acid regulates somitogenesis
c. A segmentation clock involving Fgf8 signaling converts mesoderm into somites
d. A wave of Shh from the notochord drives somitogenesis

2. Which of the following statements regarding somites is false?
a. The newly formed somites contain 2 main compartments
b. Somites form in a cranial-caudal direction
c. Somites develop from the paraxial mesoderm
d. The scleretome generates rib cartilage
e. The myotome contributes to limb muscles

3. What do the transcription factors Id and E12 share in common?
a. Both are required for muscle-specific gene activation
b. Both work better when they dimerize with E12
c. Both can form heterodimers with MyoD
d. Both are activated by myostatin signaling

4. Notch functions during somitogenesis to:
A. Antagonize the Fgf8 morphogen gradient
B. Generate somite boundaries
C. Prevent the premature end of axis elongation
D. Negatively regulate retinoic acid expression

5. What gene’s loss-of-function produces an overly muscular dog?
A. Myostatin
B. MyoD
C. Myf-5
D. Myf-6
E. Myogenin

6. The rib arises from which mesodermal tissue?
A. The intermediate mesoderm
B. The dermatome
C. The lateral plate mesoderm
D. The sclerotome
E. None of the above



7. Which of the following organs develops from the intermediate mesoderm?
A. Brain
B. Skin
C. Kidney

, D. Heart
E. Lung

8. Differentiation, as opposed to determination, of muscle cells is dependent on:
A. Myf-5
B. Myf-6
C. Myogenin
D. Myostatin

9. The majority of transcription factors that regulate myogenesis are of this class:
A. Homeodomain
B. Basic Helix-Loop-Helix
C. Zinc-finger
D. Repressor

10. Which dimer of interacting proteins results in the strongest activation of muscle-specific genes?
A. MyoD-E12
B. MyoD-MyoD
C. MyoD-Id
D. Id-Id

11. The loss of Fgf8 expression in the tailbud of the embryo would lead to which of the following
phenotypes?
a. An increase in somite number
b. No somite formation
c. A decrease in RA activity
d. A decrease in the rate of the clock’s oscillation


12. Which of the following associations of tissues contributing to the limb is incorrect?
A. Somite – muscle
B. Ectoderm – pigment cells
C. Neural tube – motor neurons
D. Lateral plate mesoderm – bone



13. What is a principal function of the apical ectodermal ridge?
A. Stimulating outgrowth of the limb bud
B. Setting up the anterior-posterior axis of the limb bud
C. Determining the characteristics of the ectodermal appendages of the limb
D. Determining the pattern of neural ingrowth in the limb

14. Which transcription factor is required for the development of the ​hindlimb​?
A. Tbx5
B. Tbx4
C. Fgf10

, D. Hox11a

15. The transplantation experiment that tested the role of the ZPA on limb bud patterning demonstrated
that:
A. The ZPA is required for dorsal-ventral limb patterning
B. The ZPA is sufficient to induce digit duplications
C. The ZPA secretes Shh
D. The ZPA is sufficient for the limb bud to regulate for lost or added parts

16. Which of the following statements regarding limb development is not true?
A. In early limb development, mesoderm is the primary bearer of the limb blueprint
B. Feedback loops involving Tbx-5/FGF-10 and FGF-10/FGF-8 drive forelimb growth
C. Proximal-distal specification requires retinoic acid signaling
D. The loss of Hox genes results in the transformation of one segment into another

17. The presence of two AERs on the limb bud leads to:
A. Diplopodia
B. Multilimbed tree frogs
C. Loss of zeugopod bones
D. The failure to form a limb

18. Inserting a bead soaked with an anti-Fgf8 antibody below the AER of a chick forelimb bud would lead to:
A. The formation of a chimeric wing/leg structure
B. Normal wing development
C. A duplication of the wing
D. The generation of a truncated wing

19. Which of the following limb field experiments would result in the production of two limbs?
A. Combining two identical limb disc halves
B. Removing part of the limb disc
C. Superimposing two complete limb discs
D. Splitting the limb disc into two halves

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