Bio 1350 Prelim 4

A gene mutation in flies changes the normal expression pattern of the pair-rule gene even-skipped so that instead of being expressed in 7 stripes, it now is expressed only in 4 stripes. What kind of gene is this mutation likely in? A mutation in a Hox gene. A mutation in a segment polarity gene. A mutation in a gap gene. A mutation in a pair-rule gene.

A mutation in a gap gene.

What is the best rationale for the use of therapeutic cloning to replace defective cells or tissues in a patient? A. The genotype of the transplanted cells is identical to that of the recipient who donated the adult nucleus, so complications of immune rejection are avoided. B. Embryos do not have to be destroyed, so ethically there is little controversy. C. It is a very efficient process for generating the tissues that need to be replaced. D. All of the above.

A. the genotype of the transplanted cells is identical to that of the recipient who donated the adult nucleus, so complications of immune rejection are avoided.

Which statement is true about stem cells?Choose one: A. They can divide for the lifetime of the organism. B. They always divide asymmetrically, giving rise to two different daughter cells. C. They are usually present in large numbers, especially in tissues that undergo rapid renewal. D. They are terminally differentiated.

A. they can divide for the lifetime of the organism

What is meant by 'stem-cell niche'? What is the role of signaling? What types of cells are found there? If it helps, try providing specific examples from the intestinal stem cell niche discussed in lecture.

Adult stem cells reside in special regulatory environments called the stem cell niche (e.g., crypt = intestine stem cell niche) where signals regulate stem cell proliferation and differentiation (e.g., Wnt signaling in intestinal stem cell niche). Niche cells, cells that are not themselves stem cells but which surround stem cells, provide some of the necessary proteins and molecules to maintain a stem cell population (eg., Paneth cells in intestinal stem cell niche).

What is an organoid? Describe how they may be used in basic research and therapy.

An organoid is a mini-organ (mini-gut, mini-kidney, etc.) created by growing iPS cells in culture and then induced to form a three-dimensional structure that resembles some properties of the in vivo organ. Uses: Basic research: study organogenesis of humans Diagnostic and therapeutic tool to understand the underlying mechanisms of a disease and test therapeutic tools in-vitro

Choose ALL of the correct statements about cell senescence (you can select multiple answer options). A. refers to programmed cell death B. results from telomere attrition C. is enhanced by high rates of telomerase activity D. is a form of tumor suppression E. is a form of cellular aging

B, D, and E are correct. Cell senescence is a form of cellular aging characterized by permanent cell cycle arrest. When telomeres reach a certain critical short length, a signal is sent to initiate a program of cell senescence. Shutting down telomerase (thus enabling telomere attrition) in adult tissue is a mechanism to suppress cancer. A. is incorrect. Senescent cells do not die but they permanently withdraw from the cell cycle B. is incorrect. High rates of telomerase activity would reduce telomere attrition and thus reduce senescence.

Which of the following statements about iPS cells is FALSE? A. iPS cells can be made by adding a combination of transcription regulators to a fibroblast. B. iPS cells made from mouse cells can differentiate into almost any human cell type. C. Stimulation by extracellular signal molecules causes iPS cells to differentiate. D. A cell that is reprogrammed (dedifferentiated) to become an iPS cell will undergo changes to its gene expression profile.

B. is the FALSE statement. When iPS cells differentiate, the cells follow a program dictated by their genome. Thus, mouse cells can only differentiate into mouse cells and cannot become human cells.

When animal cells from an early blastula are placed in a culture with cells from the vegetal region _________. A. the animal cells become endoderm B. the animal cells become mesoderm C. the vegetal cells become ectoderm D. the vegetal cells become mesoderm.

B. the animal cells become mesoderm Nodal signals produced by vegetal (endoderm) cells induce mesoderm fate in animal cells (this also explains why D. is incorrect). A. and C. are incorrect. VegT in vegetal cells autonomously specifies endoderm fate.

Overproduction of cadherins, such as E-cadherin, ____________. A. is often found in cancers originating from epithelia. B. induces epithelial-mesenchymal transition. C. leads to stronger cell-cell adhesion. D. all of the above.

C is correct A higher level of cadherin typically leads to stronger cadherin-dependent adhestions

Before answering this question, work with your group members to actively recall what you learned about telomeres and their importance to aging. Without looking at your lecture notes, create a list of important facts and concepts that stood out to you in lecture. Next, check your understanding by selecting ALL TRUE statements about telomeres (you can select multiple answer options). If a statement is false, make sure that you can explain why and correct the statement to be true. A. They contain genes related to aging B. They are regions of decondensed DNA at the ends of chromosomes C. Their length decreases with each cell division in regular (non-stem) somatic cells (progenitors and differentiated cells). D. Their length can be maintained through cells divisions in pluripotent stem cells (ESCs, NTSCs, and iPS) and germline stem cells that express high levels of telomerase. E. They contains hundreds to thousands of repetitive DNA segments.

C, D, and E are TRUE. A. is FALSE. Telomeres are devoid of genes and thus do not contain genes related to aging. However, telomeres do play an important role in aging because the inability to fully replicate and hence maintain telomere length in tissue stem cells is an important contributor to aging (along with several other causes). B. is FALSE. Telomeres are heterochromatic regions at the ends of chromosomes. In Lecture 17: Regulation of Gene Expression II, you learned that heterochromatin is condensed chromatin and euchromatin is decondensed chromatin.

he activation of receptor tyrosine kinases is characterized by ___. A. GTP hydrolysis B. a phosphorylation cascade C. dimerization and phosphorylation D. dimerization and IP3 binding

C. dimerization and phosphorylation

Cells isolated from the developing neural plate (labeled green) and from the epidermis (labeled blue) were dissociated as single cells. They were then mixed up and placed in two culturing conditions: (1) with Ca2+; and (2) without Ca2+ in the growth medium. Which results do you expect to be true? Cells only sort based on color in condition (2). Cells aggregate together first, and then they sort out in green and blue cells in both conditions. Cells do not aggregate in any of the conditions and remain dispersed due to epithelial-mesenchymal transition (EMT). Blue and green cells will end up in one aggregated ball of mixed cells in both conditions. Cells only sort out in green and blue cells in condition (1).

Cells only sort out in green and blue cells in condition (1).

A basal lamina is ____________. A. is a thin layer of extracellular matrix underlying an epithelium. B. is attached to the apical surface of an epithelium. C. separates epithelial cells from each other. D. none of the above.

Correct Answer A. is a thin layer of extracellular matrix underlying an epithelium.

What class of patterning genes encodes components of cell signaling pathways? A. Pair-rule genes B. Segment polarity genes C. Gap genes D. Egg polarity genes

Correct! B. Segment polarity genes

DNA replication is considered semiconservative because ____. A. after many rounds of DNA replication, the original DNA double helix is still intact. B. each daughter DNA molecule consists of two new strands copied from the parent DNA molecule. C. each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand. D. new DNA strands must be copied from a DNA template.

Correct! C. each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand.

What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized? A. The origins of replication occur only at the 5′ end. B. DNA ligase only works in the 3' -> 5' direction C. DNA polymerase can join new nucleotides only to the 5′ end of a pre-existing strand, and the strands are antiparallel. D. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand, and the strands are antiparallel.

D. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand, and the strands are antiparallel.

Which answer best describes the role of telomerase in replicating the ends of linear chromosomes? A. It adds a 5' cap structure to the chromosome ends that resists degradation by nucleases. B. It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity. C. It cleaves short portions of telomeric sequences with each round of cell division D. It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.

D. It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.

Telomeres serve as caps protecting the ends of linear chromosomes. Which of the following is FALSE regarding the replication of telomeric sequences? A. The lagging-strand telomeres are not completely replicated by DNA polymerase. B. Telomeres are made of repeating sequences. C. Additional repeated sequences are added to the template strand. D. The leading strand doubles back on itself to form a primer for the lagging strand.

D. The leading strand doubles back on itself to form a primer for the lagging strand.

Which of the following statements correctly explains what it means for DNA replication to be bidirectional? Hint: how many replication forks are formed when an origin of replication is opened? A. The replication fork can open or close, depending on the conditions. B. The DNA replication machinery can move in either direction on the template strand. C. Replication-fork movement can switch directions when the fork converges on another replication fork. D. The replication forks formed at the origin move in opposite directions.

D. The replication forks formed at the origin move in opposite directions.

Depletion of VegT will lead to loss of ____________. A. endoderm only B. mesoderm only C. ectoderm only D. endoderm and mesoderm E. mesoderm and ectoderm F. endoderm, mesoderm, and ectoderm

D. endoderm and mesoderm

Which of the following statements about DNA polymerase is TRUE? Important: what are the implications for this for DNA synthesis given the directionality of DNA strands at a replication fork? A. DNA polymerase can start a new DNA strand from scratch and add nucleotides to both the 3' and 5' ends of a growing DNA strand B. DNA polymerase can start a new DNA strand from scratch and can only add nucleotides to the 3' end of a growing DNA strand C. DNA polymerase can start a new DNA strand from scratch and can only add nucleotides to the 5' end of a growing DNA strand D. DNA polymerase can only add nucleotides to the 3' end of an existing polynucleotide chain E. DNA polymerase can only add nucleotides to the 5' end of an existing polynucleotide chain F. Certain DNA polymerases can add nucleotides to the 3' end of an existing polynucleotide chain and other DNA polymerases can add to the 5' end.

D. is correct. DNA polymerase can only add nucleotides to the 3' end of an existing polynucleotide chain. DNA polymerase synthesizes DNA from 5' to 3'. Important implications: 1. DNA polymerase can only add nucleotides to the 3' end of an existing polynucleotide chain and cannot start a DNA strand from scratch. Thus, to begin the synthesis of a new DNA strand, the enzyme primase must synthesize RNA primers - a short length of RNA nucleotides base-paired to the template DNA - which provides a base-paired 3' end for DNA polymerase. 2. The two strands in a DNA double helix are antiparallel. Thus, at a replication fork, one new DNA strand is being made on a template that runs 3' to 5' and the other on a template that runs 5' to 3'. Because DNA polymerase only synthesizes DNA 5' to 3', the strand made on the template that runs 3' to 5' can be synthesized continuously (leading strand) while the strand made on the template that runs 5' to 3' must be synthesized discontinuously (lagging strand).

Two types of epithelial cells from the lining of the stomach and the esophagus are held together at a region called a 'transitional zone' via cell-cell adhesion. Among the lists of molecules below, which pair of molecules do you think would be able to mediate this adhesive interaction between two types of epithelial cells? Integrin and Fibronectin Delta and Notch E-cadherin and P-cadherin Wnt and the Frizzled receptor E-cadherin and E-cadherin

E-cadherin and E-cadherin

If embryos are disaggregated and the individual cells are mixed together in media lacking calcium, what will likely happen? Choose ALL correct answers (you can select multiple answer options). Important follow-up: what will happen if you add Ca2+ to the media? A. Cells expressing the same type of integrin will be together. B. Cells will likely sort themselves so that cells expressing the same type of cadherin will be together. C. Cells expressing different amounts of the same type of cadherin will sort out, with those with the greatest adhesion found near the inside. D. Cadherins will bind to RGD motifs on the same type of cadherin. E. The cells will remain dissociated.

E. cadherins remain dissociated

Cortical rotation following fertilization in frog embryos places the future _______ side at the point of sperm entry, while Wnt11 mRNA is transported to the future _______ side. A. anterior; posterior B. posterior; anterior C. animal; vegetal D. dorsal; ventral E. ventral; dorsal

E. ventral; dorsal

Which of the following statements is TRUE? Enzyme-coupled receptors have a catalytic domain on the extracellular side of the plasma membrane. G-protein-coupled receptors are GTP-binding proteins. Each extracellular signal molecule interacts with a single class of cell-surface receptor. Many ion-channel-coupled receptors have an intrinsic catalytic domain on the cytosolic side of the plasma membrane. GTP-binding proteins that act as molecular switches inside cells are inactive when GDP is bound.

GTP-binding proteins that act as molecular switches inside cells are inactive when GDP is bound.

1.product of an egg-polarity gene (e.g., Bicoid) 2.product of a gap gene (e.g., Hunchback) 3. product of a pair rule 4. product of a segment-polarity gene (e/g. Engrailed)

In a developing Drosophila embryo, a hierarchy of gene regulatory interactions subdivides the embryo to regulate progressively finer details of patterning. Match expression patterns 1 - 4 in the images below to the most appropriate protein. 1. 2. 3. 4.

Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and is therefore not expressed. An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression. Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.

In the absence of telomerase, the life-span of a cell and its progeny cells is limited. With each round of DNA replication, the length of telomeric DNA will shrink, until finally all the telomeric DNA has disappeared. Without telomeres capping the chromosome ends, the ends might be treated like breaks arising from DNA damage, or crucial genetic information might be lost. Cells whose DNA lacks telomeres will stop dividing or die. However, if telomerase is provided to cells, they may be able to divide indefinitely because their telomeres will remain a constant length despite repeated rounds of DNA replication.

Ras promotes cell proliferation and acts as a 'molecular switch' that functions in a manner similar to the alpha-subunit of the trimeric G protein. Which mutation is most likely to contribute to the uncontrolled cell proliferation of cancer cells? One in which the Ras GTPase function is inhibited. One that prevents Ras from being made. One that decreases the affinity of Ras for GTP. One that increases the affinity of Ras for GDP.

One in which the Ras GTPase function is inhibited.

In terms of its function, is telomerase more like DNA polymerase or Primase? Explain your reasoning. Your Answer:

Primase is a type of RNA polymerase, making RNA primers necessary for strand elongation by DNA polymerase. Telomerase is more like DNA polymerase and is highly specialized to help complete DNA replication on the lagging strand. Telomerase carries its own RNA primer, which pairs to repeated sequences in the telomeric region of the chromosome. It then synthesizes new DNA to extend the repeats.

Which protein or proteins function(s) downstream of receptor tyrosine kinases? Protein kinase A (PKA) MAP kinase Ras and protein kinase A (PKA) Ras and MAP kinase Ras

RAS and mak

If Ras contains a mutation that leads to a defect in GTP hydrolysis, this could fuel uncontrolled proliferation in cancer becauseChoose one: Ras is unable to bind to the GEF for activation. Ras is unable to send signals to downstream pathways. Ras is able to migrate to the nucleus and deactivate transcription of proliferation genes. Ras is able to signal to downstream pathways inappropriately.

Ras is able to signal to downstream pathways inappropriately.

The Drosophila pair-rule gene odd-skipped is expressed in seven stripes. Three transcription regulators, TR1, TR2 and TR3, control expression of odd-skipped in one stripe, shown as a dotted box in the figure below. The figure shows the expression patterns of these three transcription regulators as shaded areas. From this data, which of the following is most likely? TR3 is an activator and TR1 and TR2 are inhibitors. TR2 is an activator and TR1 and TR3 are inhibitors. There is not enough information to make any conclusions. TR2 and TR3 are activators and TR1 is an inhibitor. TR1, TR2, and TR3 are all activators.

TR2 and TR3 are activators and TR1 is an inhibitor.

Create a diagram that illustrates how telomerase prevents linear eukaryotic chromosomes from shortening with each cell division.

Telomerase prevents chromosome shortening: to complete the replication of the lagging strand at the ends of a chromosome, the template strand is first extended beyond the DNA that is to be copied. To achieve this, the enzyme telomerase adds to the telomere repeat sequences at the 3ʹ end of the template strand, which then allows the newly synthesized lagging strand to be lengthened by DNA polymerase. The telomerase enzyme itself is made of a protein component known as TERT and an RNA component known as TERC with a sequence that is complementary to the DNA repeat sequence; this RNA acts as the template for telomere DNA synthesis by the TERT, which acts as a reverse transcriptase (makes DNA on an RNA template). In this way, the newly synthesized lagging strand contains all the original DNA information.

Create a diagram and/or in your own words, explain why chromosomes would shorten with each replication cycle in the absence of telomerase (the end replication problem).

The end replication problem: DNA synthesis begins at origins of replication and continues until the replication machinery reaches the ends of the chromosome. The leading strand is synthesized in its entirety. The synthesis of the lagging strand involves the continual synthesis of RNA primers, which are then elongated by DNA polymerase, as the replication fork moves along a DNA molecule. But the ends of the lagging strand can't be completed, because once the final RNA primer has been removed, there is no mechanism for replacing it with DNA. Complete replication of the lagging strand requires a special mechanism to prevent the loss of this DNA during each round of cell division. This problem is solved by telomerase, which extends the template strand by a repeat unit, thus providing a template for primase and, consequently, replication of the lagging strand to its full length.

What problem with replication of linear chromosomes does telomerase address? The leading strand stops short of the 5' end during replication, so chromosomes would shorten in each replication cycle without telomerase. The lagging strand stops short of the 3' end during replication, so chromosomes would shorten in each replication cycle without telomerase. The lagging strand continues farther during replication, so chromosomes would lengthen in each replication cycle without telomerase. The leading strand continues farther during replication, so chromosomes would lengthen in each replication cycle without telomerase.

The lagging strand stops short of the 3' end during replication, so chromosomes would shorten in each replication cycle without telomerase.

The procedure for therapeutic cloning is as follows. The nucleus of an individual is transplanted into an enucleated donor egg. The egg is then activated, and an embryo develops into a blastocyst in vitro. ES cells are isolated and induced to form the desired cell type by application of the appropriate signaling factors. These differentiated cells are then transplanted into the individual who provided the nucleus. There are several problems with this approach, but which of the following is NOT a problem? The new tissue is likely to be rejected because it is not immune-compatible. The differentiated cells suffer stress during the procedure, which may affect their survival. Nuclear transplantation is technically difficult, and it is inefficient to do in humans. Differentiation pathways needed to produce desired cell types are still poorly understood. It requires a large source of human eggs.

The new tissue is likely to be rejected because it is not immune-compatible.

A stem cell divides into two daughter cells. One of the daughter cells goes on to become a terminally differentiated cell. What is the typical fate of the other daughter cell?

The other daughter cell typically remains a stem cell. Stem cells are cells that are capable of extensive self-renewal. Stem cells create (1) more stem cells and (2) more differentiated progeny.

Telomerase was first described in the ciliate Tetrahymena thermophila by Elizabeth Blackburn and her student Carol Greider. They, along with Jack Szostak, subsequently won a Nobel Prize for this discovery. The template RNA sequence in Tetrahymena is 3'-ACCCCAAC-5'. The telomerase protein and RNA template together extend the 3' end of the chromosome by adding 5'-GGGTTG-3' repeats to the chromosome. The complementary strand is then synthesized by DNA polymerase α. Blackburn's lab altered the sequence of the telomerase RNA. Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated to 3'-ACCCCGAC-5'. The telomere sequence would be altered to 5'-GGGCTG-3'. There would be no changes in telomeres compared to normal cells. The telomere sequence would be altered to 5'-GTCGGG-3'. Telomeres would become shorter every generation compared to normal cells.

The telomere sequence would be altered to 5'-GGGCTG-3'.

Which statement is TRUE about adult stem cells? They play essential roles in tissue renewal (homeostasis) and injury repair throughout life. They are pluripotent cells that can differentiate to all cell types in a body. They reside in a stem cell niche that promotes their differentiation. They are abundantly present in many tissues and therefore can be easily used in cell therapy. They exclusively divide asymmetrically, and their two daughter cells generated at each division always adopt fates autonomously.

They play essential roles in tissue renewal (homeostasis) and injury repair throughout life.

A biological problem contributing to organismal aging is: Tissue stem cells undergoing replicative senescence. Terminally differentiated cells not maintaining their telomeres. The RNA template promotes telomere elongation in the absence of the telomerase protein. Telomerase expression is randomly turned on in aging tissues.

Tissue stem cells undergoing replicative senescence.

Spemann-Mangold separated the two blastomeres formed by the first cleavage of the frog embryo and let each grow separately. What were the results? When both cells received the gray crescent, each of them would develop into an embryo belly piece. When one cell receives the gray crescent, it forms an embryo with two heads and the other cell does not develop further. When both cells received the gray crescent, each of them would develop into a normal tadpole. When only one cell receives the gray crescent, it becomes a belly piece and the other becomes a co-joined twin sharing the ventral side.

When both cells received the gray crescent, each of them would develop into a normal tadpole.

The vegetal side of the embryo in frogs will form _____. at the pigmented side of the oocyte on the same side where sperm fertilizes the egg at the unpigmented side of the oocyte at the side that will become ectoderm

at the unpigmented side of the oocyte

["continuous replication", "discontinuous replication", "both modes of replication"] leading strand lagging strand primase RNA primers Okazaki fragments DNA helicase DNA ligase

continuous replication discontinuous replication both modes of replication both modes of replication discontinuous replication both modes of replication discontinuous replication

The Yamanaka reprograming transcription factors are Oct4, Sox2 and Klf4. They induce the expression of transcription factors Nanog and Myc, and together these factors re-set the genome of a differentiated cell to the pluripotent state. What do you predict would happen if the expression of these transcription factors could not be turned off at subsequent stages after reprogramming? If Klf4 could not be turned off, then the embryonic stem cells genes will be repressed. f Myc could not be turned off, then the reprogrammed cells will divide uncontrollably resulting in cancers. If Nanog could not be turned off, then the differentiation genes will be expressed. If Oct4 could not be turned off, then the reprogrammed cells will differentiate prematurely without having a chance to proliferate.

f Myc could not be turned off, then the reprogrammed cells will divide uncontrollably resulting in cancers.

Following the binding of an extracellular signal molecule, receptor tyrosine kinases (RTKs) do which of the following?Choose one: become internalized so that they can phosphorylate and activate various intracellular signaling proteins form dimers in which each polypeptide chain cross-phosphorylates the cytoplasmic tail of its partner. rearrange their multiple membrane-spanning helices in a way that enables self-phosphorylation form dimers in which each polypeptide chain phosphorylates its own cytoplasmic tail activate (or inactivate) the α and βγ subunits of G proteins

form dimers in which each polypeptide chain cross-phosphorylates the cytoplasmic tail of its partner.

What attaches epithelial cells to the basal lamina?Choose one: tight junctions hemidesmosomes cadherins desmosomes adherens junctions

hemidesmosomes

The enzyme-coupled receptor signaling pathway, briefly summarized here, promotes cell survival: insulin growth factor (IGF) → RTK → PI3 kinase → AKT __| Bad __| Bcl2. (→ denotes 'activate' and __| denotes 'inhibit'). A mutation in PI3 kinase that kills its catalytic activity is found in a neurodegenerative disease where neurons die by apoptosis. To prevent this neurodegeneration a drug should ____. promote tight binding of Bad to Bcl2 in neurons make AKT constitutively active activate RTK in the absence of signal enhance the binding affinity of IGF for RTK

make AKT constitutively active

Mouse and human embryonic stem cells are ____. cells that self-renew and differentiate in the body for the entire animal's life pluripotent cells with limited self-renewal potential in cell culture pluripotent cells that can differentiate to all cell types found in the embryo body isolated from embryos during gastrula stage cells that can give rise to placenta, blood, neurons, and all the other cells found in an adult body

pluripotent cells that can differentiate to all cell types found in the embryo body

Which of the following are activated by many receptor tyrosine kinases (RTKs)?Choose one or more: the monomeric GTP-binding protein Ras phosphoinositide 3-kinase phospholipase C the MAP-kinase signaling module trimeric GTP-binding proteins

the monomeric GTP-binding protein Ras phosphoinositide 3-kinase phospholipase C the MAP-kinase signaling module

What type of junctions present in epithelial cells serves to seal neighboring cells, creating a barrier to solute diffusion? None of the above Tight junction Gap junction Adherens junction Hemidesmosome

tight junt

One of the two types of GTP-binding proteins, often called G-proteins, are membrane bound. These are theChoose one: monomeric GTP-binding proteins. trimeric GTP-binding proteins. cyclic GMPs. G-protein coupled receptors.

trimeric GTP-binding proteins.

Cortical rotation following fertilization in X. laevis places the _____ pole at the point of sperm entry, while Wnt11 mRNA is transported to the _____ pole. ventral; dorsal anterior; posterior animal; vegetal posterior; anterior dorsal; ventral

ventral; dorsal

Which of the following statements is FALSE regarding mesoderm migration during gastrulation in frog embryos? A. the blastocoel roof expresses integrin receptors B. mesoderm cells express integrin receptors C. fibronectin in the extracellular matrix coats the blastocoel cavity D. integrin receptors become active when bound to the RGD motif on fibronectin E. activation of integrin receptors leads to reorganization of the actin cytoskeleton facilitates migration

, C, D, and E are TRUE. Fibronectin-integrin interactions are essential for mesoderm migration during gastrulation in frog embryos. During gastrulation, mesoderm cells from the organizer express integrin receptors. These cells can then move as a sheet along the roof of the blastocoel cavity, which is coated with fibronectin. Integrin receptors bind to the actin cytoskeleton and become active when bound to RGD motif on ECM proteins, altering cell behavior.

Use the dropdown menus to answer the following questions. A. In which region would you expect to find the reporter protein put under the control of the regulatory module mentioned above? [ Select ] B. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Giant? It would fail to express efficiently in the stripe B region. C. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Bicoid? It would be expressed in all seven stripes.

1. B 2. it would expand to cover a broad anterior region of the embryo 3. It would fail to express efficiently in the stripe B region.

Many G proteins activate membrane-bound enzymes that produce small messenger molecules. Specifically, the two major targets of GPCRs are phospholipase C (PLC), which produces certain lipids, and adenylyl cyclase, which generates cAMP. Outline how GPCR activation of phospholipase C leads to the activation of protein kinase C (PKC).

1. Binding of a signal molecule activates the GPCR, which then activates its G protein. 2. The activated G protein activates the membrane-bound enzyme phospholipase C (PLC). 3. The activated phospholipase C hydrolyzes the membrane inositol phospholipid to produce inositol 1,4,5-trisphosphate (IP3) and diacylglycerol (DAG). 4. IP3 diffuses through the cytosol and binds to Ca2+ channels in the ER membrane (ion channel-coupled receptors). Ca2+ rushes out of the ER and into the cytosol. 5. DAG remains in the plasma membrane and, together with Ca2+, helps activate the enzyme protein kinase C (PKC). PKC then phosphorylates its own set of intracellular proteins, further propagating the signal. Second messengers: IP3, DAG, Ca2+

Receptor tyrosine kinases (RTKs) are a major type of enzyme-coupled receptors. RTKs commonly activate the small GTP-binding protein Ras. Write out a series of steps describing how activation of an RTK leads to the activation of Ras. Be sure to describe the effects of active Ras on the cell.

1. Binding of a signal molecule in the form of a dimer dimerizes the RTK, activating their tyrosine kinase activity. 2. The RTK cytoplasmic domains function as tyrosine kinases, phosphorylating a number of sites on their cytoplasmic domains which can then serve as binding sites for adaptor proteins. 3. Adaptor proteins then activate Ras-GEFs, which stimulate the exchange of GDP to GTP on Ras, resulting in active Ras-GTP. 4. Active Ras stimulates cell proliferation through the MAP kinase signaling cascade.

Use the dropdown menus to choose whether each of the following statements is TRUE or FALSE. If false, make sure that you can explain why it is incorrect. 1. Cadherins bind catenins on neighboring cells, mediating heterotypic interactions in adherens junctions. 2. Adhesion by cadherins is dependent on Ca2+ ions. 3. Cells can attach to integrins in the extracellular matrix using fibronectin, a transmembrane protein. 4. Cells in both epithelial sheets and epithelial tubes use adherens junctions to stay connected with each other. 5. Adherens junctions involve fibronectin-integrin interactions. 6. Integrin receptors are activated upon binding to RGD motifs in extracellular matrix proteins, altering cell behavior. 7. Fibronectin-integrin interactions can alter cell behavior by changing the organization of the actin cytoskeleton.

1. FALSE. The extracellular (EC) domain of a cadherin binds to the EC domain of a cadherin of the same type on an adjacent cell (cadherins mediate cell-cell attachments through homophilic interactions (cadherin molecules only bind to cadherin molecules of the same type; N cadherin binds N cadherin and not E cadherin). The intracellular domain of cadherin is connected with the actin cytoskeleton via the catenin complex. 2. TRUE 3. FALSE. Cells can attach to fibronectin in the extracellular matrix using integrins, which are transmembrane proteins. 4. TRUE 5. FALSE. Adherens junctions involve homotypic interactions between cadherin molecules on neighboring epithelial cells. The intracellular domains of cadherins connect indirectly to the actin cytoskeleton through the catenin complex. Integrin is a cell surface receptor that binds extracellular matrix proteins (like fibronectin) and indirectly connects to the cytoskeleton inside the cell. 6. TRUE. 7. TRUE. Previous

Working with your group, create a labeled diagram and step-wise description outlining how G protein-coupled receptors (GPCRs) function by activating their trimeric G protein. You can refer to the blank diagrams below to organize your thoughts,

1. The binding of a signal molecule activates the G protein-coupled receptor (GPCR) on its extracellular domain, causing a conformational change that is transmitted to the GPCR's cytoplasmic domain. 2. The active GPCR activates its trimeric G protein (α, β, γ subunits) by acting as a guanosine exchange factor (GEF), stimulating the dissociation of GDP from the α subunit of the G protein (α subunit). GTP can then bind to the α subunit, activating both the α subunit and the βγ complex. *Remember: G proteins are inactive in GDP-bound state and active in GTP-bound state* 3. Both the α subunit and the βγ complex can then each activate different effector proteins - many G proteins activate membrane-bound enzymes: phospholipase C (PLC) and adenylyl cyclase are the two main targets. 4. Normally, the α subunit then acts as a GTPase, hydrolyzing its bound GTP to GDP within seconds, inactivating the α subunit. 5. The inactive α subunit reassembles with the inactivated βγ complex to reform an inactive G protein. The two major targets of GPCRs are phospholipase C (PLC) and adenylyl cyclase.

Write out a series of steps and/or create a diagram to describe how changes in cadherin expression and changes in cell shape facilitate neurulation.

1. The neural plate and epidermis are initially connected (all cells initially express E-cadherin) Remember: cadherins engage in homotypic interactions. 2. Neural tube cells switch from expressing E-cadherin to N-cadherin. The apical actin bundle contracts and invagination begins. N-cadherin to N-cadherin homotypic interactions lead to neural tube closure and separation from epidermis. 3. Epidermis reconnects due to homotypic interactions between cells expressing E-cadherins. 4. Neural crest cells stop expressing all cadherins, undergo epithelial-to-mesenchymal transition, and migrate away.

1. The unfertilized frog egg is asymmetric along the ______ 2. ____ and ____ mRNAs are maternal components deposited in the unfertilized egg by the mother. 3. VegT mRNA is located in the ____ ___ ____ 4. Wnt11 mRNA is located in the ____ ___ ____ ____ 5. Cortical rotation is triggered by ___ and leads to the formation of the gray crescent. 6. The gray crescent forms on the side opposite to sperm entry and marks the future ____ side. 7. Centrioles contributed by the sperm lead to the formation of organized microtubules bundles near the ________ 8. The motor protein ______ moves _______ to the future dorsal side.

1. The unfertilized frog egg is asymmetric along the animal-vegetal axis. 2. VegT and Wnt11 mRNAs are maternal components deposited in the unfertilized egg by the mother. 3. VegT mRNA is located in the vegetal pole cytoplasm. 4. Wnt11 mRNA is located in the vegetal pole cortical cytoplasm. 5. Cortical rotation is triggered by sperm entry and leads to the formation of the gray crescent. 6. The gray crescent forms on the side opposite to sperm entry and marks the future dorsal side. 7. Centrioles contributed by the sperm lead to the formation of organized microtubules bundles near the vegetal cortex. 8. The motor protein kinesin moves Wnt11 mRNA to the future dorsal side.

Use the dropdown menus to complete the following statements about the Spemann-Mangold organizer. The Spemann-Mangold organizer forms in the dorsal ectoderm where mesoderm cells induced by the TGF-β signaling molecule VegT coincide with nodal target cells. The organizer dorsalizes the embryo by secreting Chordin and Noggin. Chordin and Noggin stimulate the ventralizing activity of the TGF-β BMP4 signaling molecule expressed throughout the embryo. This results in the formation of corresponding ventral to dorsal structures in all three germ layers. 1. 2. 3. 4. 5. 6.

1. dorsal mesoderm 2. Veg T 3. Wnt11 4. dorsalizes 5. inhibit 6. ventralizing

Use the dropdown menus to complete the following statements.Cells signal to one another in various ways. Some use extracellular signal molecules that are [ Select ] , such as steroids, which can diffuse across the plasma membrane and bind to [ Select ] . In contrast, signals that cannot diffuse across the plasma membrane bind to [ Select ] , which include ion channel-coupled receptors, G protein-coupled receptors (GPCRs), and enzyme-coupled receptors.Ion channel-coupled receptors allow certain ions to pass across the membrane when stimulated by a signal, leading to [ Select ] and providing a [ Select ] response. G protein-coupled receptors (GPCRs) activate [ Select ] , many of which activate membrane-bound enzymes (phospholipase C, adenylyl cyclase) that lead to the production of [ Select ]. Phospholipase C (PLC) cleaves [ Select ] to generate the two small messenger molecules (which remains embedded in the plasma membrane) and [ Select ] (which diffuses into the cytosol). Activation of adenylyl cyclase leads to the production of [ Select ] .Enzyme-coupled receptors act as enzymes themselves or associate with enzymes that can be activated when the receptors are stimulated by the signal molecules. One of the largest classes of enzyme-coupled receptors is receptor tyrosine kinases (RTKs), which lead to the activation of [ Select ] and a MAP kinase signaling pathway that influences cell proliferation.

1. small hydrophobic molecules 2. nuclear hormone receptors 3. cell surface receptors 4. a change in membrane potential 5. fast 6. trimeric g proteins 7. second messneger molecules 8. inositol phospholipid (PIP2) 9. diglycerol (DAG) 10. inositol 1.4.5.triphosphate (IP3) 11. cycliic AMP 12. the small GTP-binding protein RAS

Identify the type of cell junction in #2-6 below. Follow-up: identify the apical and basal sides. 2. Seals the epithelial cells together, serving as a barrier to solute diffusion, preventing molecules from leaking through intercellular spaces from one side to the other side. 3. Binds epithelial cells to each other via the action of cell adhesion molecules and allow the actin filaments from two adjacent epithelial cells to "connect" (not directly but indirectly). 4. "Connects" (not directly but indirectly) the intermediate filaments from two adjacent epithelial cells. 5. Allows small molecules to pass from cell to cell. 6. Anchors the epithelial cells to the basal lamina.

2. tight juntion 3. adherens junction 4. desmosome junction 5. gap junction 6. hemidesmosome junction

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. B. C. D. E.

A Engrailed, Wingless, and Hedgehog are segment polarity genes expressed at the posterior end of each segment in neighboring cells. The expression of segment polarity genes needs to be maintained during development so that the segments (boundary, polarity and number) are maintained through gastrulation and metamorphosis and in adults. The auto-regulatory feedback loop involving mutually reinforcing signals between Wingless expressing cells and Hedgehog expressing cells maintains their narrow stripes of expression through development (cell memory). Cells that express the paracrine signal protein Wingless induce their neighboring cells to express the transcription regulator Engrailed, which activates the expression of the paracrine signal protein Hedgehog in these cells. Hedgehog signaling, in turn, activates the expression of Wingless in the first cell, maintaining the fine patterning within the segment.

Adult stem cells can be used in therapy, such as bone marrow transplantation to treat leukemia. Which of the following is a challenge associated with the therapeutic use of adult stem cells? Choose ALL correct statements (you can select multiple answer options). A. Need to overcome histoincompatibility (rejection of foreign substances by the immune system). B. Need different types of adult stem cells for different therapies (adult stem cells can differentiate into a limited number of cell types). C. Need to identify the niche cells and the regulatory signals for each type of adult stem cells D. Scarcity (adult stem cells are rare) E. Only transiently present in the early embryo F. The safety of reprogramming using three transcription regulators is unknown.

A, B, C, and D are all challenges of using adult stem cells in therapy. E. is incorrect. The statement refers to a challenge associated with the use of embryonic stem cells, which are only transiently present in early embryos. Adult stem cells are present throughout the life of an individual. F. is incorrect. The statement refers to a challenge associated with the use of induced pluripotential stem cells. Adult stem cell therapies, such as bone marrow transplants, do not require them to be reprogrammed back to pluripotency).

Which of the following acts as a second messenger molecule? Choose ALL that apply A. Ca2+ B. diacylglycerol (DAG) C. IP3 D. cyclic AMP (cAMP)

A,B,C,D

Which of the following sequences is correct? Choose ALL correct sequences (you can select multiple answer options). As you start practicing material from Lecture 19: Cell Signaling II, it is fine if you have trouble answering this question initially. Be sure to come back to this question after practicing to A. diffusion of a signaling molecule across the plasma membrane → binding of the signaling molecule to its receptor → movement of the signaling molecule-receptor complex into the nucleus → transcription B. binding of a signaling molecule to its receptor → G protein activation → adenylyl cyclase activation → cAMP production → protein kinase activation → cellular response C. binding of a signaling molecule to its receptor → G protein activation → phospholipase C activation → IP3 production → increase in cytoplasmic calcium concentration D. binding of a growth factor to its receptor → phosphorylation cascade → activation of transcription factor → transcription

A,B,C,D

Much attention has been paid to the process of segmentation in the Drosophila embryo. What are segments, and what is their significance to the strategy used by Drosophila to carry out pattern formation? Do humans have segments? If so, give examples. What have been some of the key developmental and cellular concepts that have been learned, perhaps unexpectedly, from the analysis of Drosophila segmentation?

A. Each segment has distinctive and characteristic structural features, which allowed investigators to determine which segment is which even if their formation was not entirely normal. Drosophila have 14 segments. B. The most easily observed evidence of segmentation in humans is our vertebrae. Cervical vertebrae (in the neck) are distinct from thoracic vertebrae, which are also distinct from lumbar vertebrae (the "tailbone"). C. Scientists studying development expected to use the analysis of Drosophila mutants to understand the process by which the segments form and acquire their distinct identities. This would also include how the anterior-posterior body axis is determined, which in turn affects which body parts form on which segment. They did learn that the acquisition of segment position and identity occurs progressively. That is a general principle of development in all organisms, although not always involving gap genes, pair-rule genes, and segment polarity genes. While the specific genes involved in the anterior-posterior axis (gap genes) and segment position (pair-rule genes) are generally not used for this process in other animals, the genes involved in segment identity (the Hox genes) are in fact highly conserved and are important for the overall body plan in all animals. In addition, many of the genes that are important in carrying out the specific functions of the segment—the segment polarity genes—are also highly conserved and important in many developmental processes in other animals.

Embryonic stem (ES) cells and induced pluripotent stem (iPS) cells are both pluripotent stem cells but are obtained in different ways. A. Describe how each is obtained. B. What are the potential advantages of iPS cells over ES cells for therapeutic purposes? C. What are the disadvantages of iPS cells for therapeutic purposes? D. How are ES and iPS cells different?

A. Embryonic stem (ES) cells are obtained from the inner cell mass of an embryo. Induced pluripotent stem (iPS) cells are stem cells made by inducing mature differentiated cells to "de-differentiate" with three key transcriptional regulators (Oct4, Sox2, and Nanog activate ES cell genes). B. Advantages of iPS cells over ES cells include: no histo-incompatibility issue no ethical issue regarding destroying human blastocysts to generate ES cells C. The disadvantages of iPS cells are that they are not identical to ES cells (may contain epigenetic marks) and the somatic cells used to make iPS cells could contain damaging mutations. There is always the question about whether the induction process has other effects (safety of reprogramming unknown). D. ES cells and iPS cells are not identical, as they may have differences in epigenetic marks (histone modifications, DNA methylation).

Use the dropdown menus to choose whether each of the following statements about the Spemann-Mangold organizer is TRUE or FALSE. Make sure that you are able to explain why any false statements are incorrect. A. The organizer forms where VegT signaling molecules and Wnt11 signaling molecules coincide. B. BMP4 is a TGF-β signaling molecule with ventralizing activity that is expressed throughout the frog embryo. C. Chordin and Noggin secreted by the organizer lead to a graded distribution of BMP4 activity, with BMP4 activity highest near the organizer and lowest on the side opposite the organizer. D. The organizer is responsible for patterning along the animal-vegetal axis. E. Nodal, Wnt11, BMP4 are all members of the transforming growth factor β (TGF-β) family of signaling molecules. T/F

A. FALSE. The organizer forms where Nodal signaling and Wnt11 signaling coincide. (VegT is a transcription factor that activates expression of the TGF-β signaling molecule Nodal). B. TRUE C. FALSE. Chordin and Noggin secreted by the organizer inhibit BMP4 activity. Thus, BMP4 activity is lowest near the organizer and greatest on the side opposite the organizer. D. FALSE. The organizer is responsible for patterning along the dorsal-ventral axis. E. FALSE. Nodal and BMP4 are both TGF-β signaling molecules. Wnt11 belongs to the family of Wnt signaling molecules.

EXTRA PRACTICE Your friend is a pioneer in ES cell research. In her research, she uses an ES cell line that originated from an inbred strain of laboratory mice called FG426. She has just figured out methods that allow her to grow an entire liver from an ES cell and has successfully grown 10 livers. She demonstrates that the newly grown livers are functional by successfully transplanting one of the new livers into a FG426 laboratory mouse. You are particularly excited about this because you have a sick pet mouse called Squeaky. You are very attached to Squeaky, as you found him when you were out camping in New Hampshire. Unfortunately, Squeaky has developed liver disease and will not live much longer without a liver transplant. After you see your friend on TV talking about her new method for growing mouse livers, you immediately grab your cell phone to ask her whether Squeaky could have one of the newly grown livers. Just as you are about to dial your friend, you remember something you learned in cell biology and realize that instead, you should ask your friend about possibly creating mouse iPS cells for Squeaky's benefit. A. Why do you think that one of the newly grown livers may not work in Squeaky? B. Explain how the use of iPS cells could solve this problem.

A. For organ transplantation to be successful, the donor and the recipient should be as close a genetic match as possible to minimize the risk of immunological rejection (i.e., histo-incompatibility). Because you found Squeaky in the fields of New Hampshire, Squeaky is likely to have genetic differences from the FG426 inbred laboratory mice that would cause the livers to be rejected. B. The use of iPS cells may help save Squeaky if the iPS cells were generated from Squeaky himself and therefore would be genetically identical to Squeaky. You hope that your friend will be successful in using iPS cells derived from Squeaky and can grow a functional liver for him.

The artificial introduction of the three key transcription factors __________ into an adult cell can convert the adult cell into a cell with the properties of embryonic stem (ES) cells. A. Sox2, Oct4, and Nanog B. Sox2, Oct4, and Hedgehog C. TGF-beta, Wnt, and Hedgehog D. Winken, Blinken, and Nod

A. Sox2, OCt4, Nanog

In wildtype Drosophila embryos, the bicoid mRNA is localized to the anterior end of the embryo (bicoid is an egg-polarity gene). If bicoid mRNA was injected into the embryo's posterior end as well, which of the following developmental events would occur? A. The embryo would show anterior structures at both ends of the embryo. B. The embryo would grow extra wings and legs. C. The embryo would develop normally. D. The embryo would develop normal segment but segmentaiton would not be maintained into adulthood.

A. The embryo would show anterior structures at both ends of the embryo.

Which of the following is the correct hierarchy of gene regulatory interactions leading to anterior-posterior patterning in Drosophila? A. egg-polarity, gap, pair-rule, segment-polarity B. egg-polarity, pair-rule, gap, segment-polarity C. segment-polarity, egg-polarity, gap, pair-rule D. segment-polarity, egg-polarity, pair-rule. gap

A. egg-polarity, gap, pair-rule, segment-polarity

Identify the fate that best corresponds to zones A-E in the fate map for the frog blastula below. Important follow-up: describe the developmental significance of the cell fate E. A. B. C. D. E.

A. epidermis B. neural tube C. blood, heart, kidney D. endoderm E. notochord

Which type of protein in a fibroblast's plasma membrane attaches to the extracellular matrix on the outside of the cell and (through adapter molecules) to actin inside the cell?Choose one: A. integrin B. fibronectin C. proteoglycan D. cadherin E. collagen

A. integrin

Which of the following mechanisms is NOT directly involved in inactivating an activated RTK? A. dephosphorylation by serine/threonine phosphatases B. dephosphorylation by protein tyrosine phosphatases C. removal of the RTK from the plasma membrane by endocytosis D. digestion of the RTK in lysosomes

A. is correct. RTKs are phosphorylated on tyrosines by their dimerization partner, which is also a tyrosine kinase, and thus the reversal of these phosphorylations involves protein tyrosine phosphatases, and not protein serine/threonine phosphatases. Endocytosis of the receptor and its ultimate digestion in the lysosome are other methods that the cell uses to down-regulate active receptors.

Binding of a growth factor to its receptor is most likely to immediately activate which of the following molecules? A. protein kinase B. Ca2+ C. cAMP D. adenylyl cyclase

A. is correct. RTKs, one of the largest class of enzyme-coupled receptors, are activated by growth factors (signals), such as FGF (fibroblast growth factor), EGF (epidermal growth factor), etc. They regulate growth, proliferation, differentiation and survival of cells

Cells of the inner cell mass in an early mammalian embryo can be isolated and grown in culture. Which of the following is FALSE regarding these cells? A. They are totipotent. B. They are stem cells that are only transiently present during embryogenesis. C. They can remain undifferentiated when cultured under the correct conditions. D. They can be manipulated in culture to give rise to almost any type of differentiated cell.

A. is the correct answer. Embryonic stem cells (ES cells) are not totipotent; they are pluripotent. This limitation is only minor, and requires that they are injected into preformed early embryos, as opposed to being directly implanted.

Which of the following statements regarding Hox genes in Drosophila is FALSE? Important follow-up: Do Hox genes regulate patterning along the A-P axis in both vertebrates and invertebrates? A. If all the Hox genes in an embryo are deleted, segmentation still occurs but distinct segment identities are lost (i.e., all of the segments are identical). B. Hox proteins are homeodomain transcription factors that regulate the expression of multiple genes. C. Hox proteins act as master regulators. D. A mutation in a Hox gene results in a homeotic mutation. E. The expression pattern of Hox genes mirrors their chromosomal location F. All of the above statements are TRUE.

All are true

How does diacylglycerol (DAG) function in the inositol phospholipid pathway?Choose one: Along with Ca2+, it recruits and activates PKC at the plasma membrane. It produces cyclic AMP.It activates phospholipase C. It produces IP3. It binds to and opens Ca2+ channels in the plasma membrane, allowing extracellular Ca2+ to enter the cytosol.

Along with Ca2+, it recruits and activates PKC at the plasma membrane.

Which of the following is a major distinction between cells in a mesenchymal arrangement and cells in an epithelial arrangement? A. Mesenchymal cells are joined together in an orderly side-by-side arrangement. B. Mesenchymal cells can migrate as individuals. C. Mesenchymal cells form tubes and sheets. D. Mesenchymal cells are tightly connected to each other.

B is correct. Mesenchymal cells are loosely aggregated and can migrate as individuals. A, C, and D are incorrect because they describe epithelial cells, not mesenchymal cells.

You conduct a series of cell sorting experiments with three types of cells: gray, white, and black. Experiment 1: you mix the gray and white cells and find that the gray cells sort to the inside. Experiment 2: you mix the black and gray cells and find that the black cells sort to the inside. Experiment 3: you mix the black and white cells. What do you predict will happen in experiment 3? A. The white cells will be inside B. The black cells will be inside C. The white and black cells will remain mixed (will not sort out) D. All of the cells will turn gray

B.

Which of the following epithelial cell junctions is not involved in providing mechanical strength to an epithelial sheet by connecting the cytoskeletons of adjacent cells?Choose one: A. desmosomes B. gap junctions C. hemidesmosomes D. adherens junctions

B. Gap junctions

Which of the following statements best describes an aspect of signal transduction? Important follow-up: synthesize your knowledge of cell signaling and explain why the incorrect answer options are incorrect. A. When signal molecules first bind to receptor tyrosine kinases, the receptors phosphorylate a number of nearby molecules. B. Protein kinase A activation is one possible result of signal molecules binding to G protein-coupled receptors. C. In response to some G protein-mediated signals, a special type of lipid molecule associated with the plasma membrane is cleaved to form IP3 and calcium. D. In most cases, signal molecules interact with the cell at the plasma membrane, enter the cell, and eventually enter the nucleus.

B. Protein kinase A activation is one possible result of signal molecules binding to G protein-coupled receptors.

Which type of developmental defect could result from a mutation in a Hox gene? A. Several adjacent segments will be missing from the resulting embryo. B. Segments will develop into a different type of segment than they would normally. C. The anterior portion of the embryo will not develop. D. All of the above.

B. Segments will develop into a different type of segment than they would normally.

Create a diagram describing the formation of the three germ layers in frog embryos. A. Draw a blastula-stage frog embryo and label the animal and vegetal poles and the regions that will become endoderm, mesoderm, and ectoderm. B. Show how endoderm fate is specified (name the molecule(s) and describe the mechanism(s)). C. Show how mesoderm fate is induced (name the molecule(s) and describe the mechanism(s)). D. At this stage in embryonic development, are cells near the animal pole specified or determined to become ectoderm? Explain.

B. VegT mRNA is deposited in the unfertilized egg by the mother. The VegT protein is a transcription factor that acts autonomously to specify endoderm fate in cells located near the vegetal pole. C. VegT activates the expression of the TGF-β signaling molecule Nodal in vegetal cells. Nodal diffuses away from the vegetal pole and induces signal transduction cascade in cells above vegetal cells, inducing mesoderm fate. D. Cells near the animal pole are specified, but not determined, to become ectoderm. Recall how isolation and transplantation experiments test for the two stages of cell commitment: specification and determination. When animal cap cells are isolated, they become ectoderm (animal cap cells are specified to be ectoderm). When isolated animal cap cells receive Nodal, they become mesoderm (animal cap cells are not determined to be ectoderm).

The growth factor RGF stimulates proliferation of cultured rat cells. The receptor that binds RGF is a receptor tyrosine kinase called RGFR. Which of the following types of alteration would be most likely to prevent receptor dimerization? A. a mutation that increases the affinity of RGFR for RGF B. a mutation that prevents RGFR from binding to RGF C. changing the tyrosines that are normally phosphorylated on RGFR to alanines D. changing the tyrosines that are normally phosphorylated on RGFR to glutamic acid

B. a mutation that prevents RGFR from binding to RGF A. is incorrect. A mutation that increases the affinity of RGFR for RGF will increase dimerization in the presence of ligand. C. is incorrect. RTKs dimerize upon ligand binding, and subsequently become phosphorylated. However, changing the relevant tyrosines to alanine will block receptor activation but should not cause or prevent dimerization. D. is incorrect. Because glutamic acid is negatively charged, it can mimic the addition of a phosphate to an amino acid; thus, changing the relevant tyrosines to glutamic acid may mimic receptor activation, but it should not cause or prevent receptor dimerization.

Adult stem cells __________. A. are terminally differentiated. B. can divide for the entire lifetime of the organism. C. reside in a stem cell niche that produces signals that promote differentiation. D. always divide asymmetrically to give rise to two different daughter cells. E. can only remain undifferentiated under certain culture conditions.

B. can divide for the entire lifetime of the organism. B. is correct. For the most part, adult stem cells are present throughout the life of an individual and can self-renew by unlimited divisions (at least for the lifetime of the animal). A. is incorrect. Adult stem cells are not terminally differentiated and can give rise to different cell types in the tissue in which they reside. C. is incorrect. Adult stem cells do reside in a stem cell niche, but the niche produces signals that maintain the proliferative and undifferentiated state of cells, not differentiation. D. is incorrect. Adult stem cells can divide symmetrically (producing daughter cells that both remain as stem cells) or asymmetrically (producing a daughter cell that remains as a stem cell and a daughter cell that differentiates). D. is incorrect. Cells of the inner cell mass (ICM), not adult stem cells, can only remain undifferentiated in culture as embryonic stem cell lines (ES cell lines). Otherwise, ICM cells are only transiently present during embryogenesis and will differentiate.

A pluripotent cell ________. Important follow-up: explain the difference between totipotent cells, pluripotent stem cells, adult stem cells, committed progenitor cells, and terminally differentiated cells. A. can only be produced in the laboratory. B. can give rise to all cell types in an adult body. C. can only give rise to stem cells. D. is considered to be terminally differentiated. E. None of the above.

B. can give rise to all cell types in an adult body.

Which type of cell-surface receptor(s), when activated, catalyze(s) a reaction inside the cell?Choose one: A. ion-channel-coupled receptors B. enzyme-coupled receptors C. G-protein-coupled receptors D. enzyme-coupled receptors AND G-protein-coupled receptors

B. enzyme coupled receptors

Match each protein with the correct label in the figure below. Note: A is collagen, a major component of extracellular matrix. B. C. D.

B. fibronectin C. integrin D. actin

Which of the following statements is FALSE regarding embryonic stem (ES) cells? Choose ALL false statements (you can select multiple answer options). A. They are cells of the inner cell mass in a blastocyst stage embryo B. They are present throughout the lifetime of an individual. C. They can remain undifferentiated when cultured with the appropriate signaling molecules and conditions. D. They can be manipulated in culture to give rise to all differentiated cell types in the body. E. They are identical to induced pluripotent stem cells (iPS)

B. is FALSE. Embryonic stem cells are only transiently present during early embryogenesis. E. is FALSE. Induced pluripotent stem cells (iPS) are not identical to ES cells (eg. differences with some epigenetic marks - histone modifications, DNA methylation)

Which of the following statements is TRUE? A. MAP kinase kinase kinase is important for phosphorylating Ras. B. a change in membrane potential resulting from activation of an ion channel-coupled receptor serves as a fast block to polyspermy C. Ras becomes activated when an RTK phosphorylates its bound GDP to create GTP. D. Dimerization of GPCRs leads to Gα activation.

B. is TRUE. The entry of the first sperm sends a signal that activates an ion channel-coupled receptor (Na+ channel). Opening of the Na+ channel allows Na+ to rush into the cell, altering the membrane potential, making the egg membrane impermeable to additional sperm. Very fast response to prevent polyspermy. A. is false. Ras is activated when Ras-GEF stimulates the exchange of GDP for GTP. The active form of Ras, Ras-GTP then activates the MAP kinase pathway and thus cell proliferation (MAP kinase kinase kinase phosphorylates MAP kinase kinase which phosphorylates MAP kinase (MAPKKK > MAPKK > MAPK). C. is false. Ras exchanges its GDP for GTP when activated. D. is false. Receptor tyrosine kinases, not GPCRs, dimerize upon ligand binding.

You are interested in further understanding the signal transduction pathway that controls the production of Pig1, a protein important for regulating cell size. Activation of the TRK receptor leads to activation of the GTP-binding protein, Ras, which then activates a protein kinase that phosphorylates the SZE transcription factor. SZE only interacts with the nuclear transport receptor when it is phosphorylated. SZE is a gene activator for the Pig1 gene. This pathway is diagrammed in the figure below. Normal cells grown under standard conditions (without ligand) are 14 μm in diameter while normal cells exposed to TRK ligand are 10.5 μm in diameter. Given this situation, which of the following conditions do you predict will more likely lead to smaller cells? A. addition of TRK ligand and a drug that stimulates the GTPase activity of Ras B. addition of TRK ligand and a drug that inhibits the activity of the phosphatase that acts on SZE C. addition of TRK ligand and a drug that stimulates the degradation of Pig1 D. addition of TRK ligand and a drug that inhibits Pig1 binding to DNA

B. is correct. The activation of the TRK receptor and its downstream signaling pathway leads to smaller cells. Thus, by inhibiting the dephosphorylation of SZE, SZE will likely activate the expression of Pig1 longer than it would in normal cells, leading to smaller cells. All other scenarios interfere with TRK receptor signaling, and should lead to cells that are not as small.

The schematic drawing below shows the cross-section of the stem cell niche in a mammalian intestine. Choose whether each of the following descriptions best applies to the areas labeled A, B, or C. Fig. 22-1A MBC6 Contains stem cells interspersed with Paneth cells (niche cells) that produce Wnt signaling Wnt signaling pathway is active and cells are dividing Wnt signaling pathway is inactive, cells no longer proliferate and have differentiated into different intestinal cell types.

C, B,A Area C: in the intestinal stem cell niche, stem cells are located near the base of the crypt, interspersed between Paneth cells (niche cells), which produce signals to maintain the proliferative and undifferentiated potential of stem cells (Wnt signaling). Area B: cells receiving Wnt signaling (Wnt pathway active) divide rapidly (proliferate). Area A: cells no longer receive Wnt signaling (Wnt pathway inactive). Cells have differentiated and no longer divide.

Which of the following junctions binds a cell to another cell and is linked to keratin intermediate filaments?Choose one: A. hemidesmosome B. adherens junction C. desmosome D. tight junction

C. Desmosome

Earlier in the course, you learned about different approaches to study development. Which approach was used to elucidate the mechanisms underlying anterior-posterior patterning in Drosophila in the Nobel-prize winning research by Ed Lewis, Christine Nusslein-Volhard, and Eric Weischaus? Important follow-up: work with your group to create a diagram or description that explains how this approach allows scientists to determine regulatory relationships between two gene products. A. Descriptive embryology B. Experimental embryology C. Developmental genetics D. Comparative embryology

C. Developmental genetics Developmental genetics (the study of mutants). The regulatory logic was deciphered by genetic screens (looking for mutants with defects). In this approach, one can also examine the expression of gene A in embryos mutant for gene B to infer whether and how the product of gene B regulates the expression of gene A.

EXTRA PRACTICE The length of time a G protein will signal is determined by the ____________ . A. activity of phosphatases that turn off G proteins by dephosphorylating G-protein α-subunit. B. activity of phosphatases that turn GTP into GDP. C. GTPase activity of G protein α-subunit. D. degradation of the G protein after the α-subunit separates from βγ-subunit.

C. GTPase activity of G protein α-subunit.

Which of the statements below regarding the pattern of cell proliferation in the epithelium that forms the lining of the mammalian intestine is FALSE? A. Wnt signaling is important for maintaining the proliferative and undifferentiated potential of intestinal stem cells. B. When an intestinal stem cell divides, daughter cells that stay in the crypt and do not move away will continue to receive Wnt signaling and will not differentiate. C. Stem cell progeny that move away from the crypt continue to receive Wnt signaling and to proliferate. D. Constitutively active Wnt signaling could lead to the formation of a cancerous tumor

C. Stem cell progeny that move away from the crypt continue to receive Wnt signaling and to proliferate.

Which of the following happens when a G protein-coupled receptor (GPCR) activates its trimeric G protein? A. The active GPCR acts as a GTPase activating protein (GAP) for its G protein, stimulating the α subunit to hydrolyze GTP to GDP.. B. The GDP bound to the α subunit is phosphorylated to form bound GTP. C. The α subunit exchanges its bound GDP for GTP. D. It activates the α subunit and inactivates the βγ subunit.

C. The α subunit exchanges its bound GDP for GTP.

You are interested in how cyclic-AMP-dependent protein kinase A (PKA) functions to affect learning and memory, and you decide to study its function in the brain. It is known that, in the cells you are studying, PKA works via a signal transduction pathway like the one depicted in the figure below. Furthermore, it is also known that activated PKA phosphorylates the transcriptional regulator called Nerd that then activates transcription of the gene Brainy. Which situation described below will lead to an increase in Brainy transcription? A. a mutation in the Nerd gene that produces a protein that cannot be phosphorylated by PKA B. a mutation in the nuclear import sequence of PKA from PPKKKRKV to PPAAAAAV C. a mutation in the gene that encodes cAMP phosphodiesterase that makes the enzyme inactive D. a mutation in the gene that encodes adenylyl cyclase that renders the enzyme unable to interact with the α subunit of the G protein

C. a mutation in the gene that encodes cAMP phosphodiesterase that makes the enzyme inactive

The growth factor Superchick stimulates the proliferation of cultured chicken cells. The receptor that binds Superchick is a receptor tyrosine kinase (RTK), and many chicken tumor cell lines have mutations in the gene that encodes this receptor. Which of the following types of mutation would be expected to promote uncontrolled cell proliferation? A. a mutation that prevents the catalytic domains on the RTK cytosolic tails from getting close to each other. B. a mutation that destroys the kinase activity of the receptor C. a mutation that inactivates the protein tyrosine phosphatase that normally removes the phosphates from tyrosines on the activated receptor D. a mutation that prevents the binding of the normal extracellular signal to the receptor

C. a mutation that inactivates the protein tyrosine phosphatase that normally removes the phosphates from tyrosines on the activated receptor

Adherens junctions have an important role in development; they are composed of ___________. A. cadherins linked to the same type of cadherin extracellularly in the absence of Ca2+, and linked to actin cytoskeleton intracellularly via the catenin complex B. cadherins linked to the same type of cadherin extracellularly in the presence of Ca2+, and to intermediate filaments intracellularly C. cadherins linked to the same type of cadherin extracellularly in the presence of Ca2+, and linked to the actin cytoskeleton intracellularly via the catenin complex D. cadherins linked to the same type of cadherin extracellularly, and bound solely to Ca2+ intracellularly E. integrin receptors bound to the RGD motif on extracellular matrix proteins, such as fibronectin

C. is correct. Adherens junctions are composed of cadherins linked to the same type of cadherins extracellularly in the presence of Ca2+, and to actin filaments indirectly via catenin complexes intracellularly. (Cadherins are Ca2+-dependent cell adhesion molecules). A. is incorrect. Cadherin-cadherin interactions require Ca2+. B. is incorrect. Cadherins are not linked to intermediate filaments intracellularly. D. are incorrect. The intracellular domain of cadherins binds to the actin cytoskeleton indirectly via the catenin complex. E. is incorrect. This does not describe an adherens junction. However, integrin receptors do bind to the RGD motif on extracellular matrix proteins, such as fibronectin and integrin-fibronectin interactions play an important role in mesoderm migration during frog gastrulation

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 _____. A. differential integrin-fibronectin interactions B. epithelial-to-mesenchymal transition C. differential cell adhesion D. differences in their basal lamina

C. is correct. Cadherins are a major type of cell adhesion molecules. Cells sort out due to differential adhesion which can be due to (1) cells expressing different types of adhesion molecules or (2) cells expressing different amounts of the same adhesion molecules.

The figure below shows how normal signaling works with a Ras protein acting downstream of a receptor tyrosine kinase (RTK). You examine a tumor cell line with a constitutively active Ras protein that is always signaling. Which of the following conditions will turn OFF signaling in this cell line? A. addition of a drug that prevents protein X from activating Ras B. addition of a drug that prevents the RTK from being activated C. addition of a drug that blocks protein Y from interacting with its target D. addition of a drug that increases the activity of protein Y

C. is correct. Signaling in this cell line will be turned off by addition of a drug that blocks protein Y from interacting with its target. If protein Y cannot interact with its target, signaling will not occur. A, B, and D are incorrect. Increasing the activity of protein Y would not turn off signaling. Preventing the RTK from being activated or preventing protein X from activating Ras would have no effect in a cell line with a constitutively active Ras protein, because Ras is already active.

The first cleavage division in a fertilized frog egg normally bisects the gray crescent. If the two resulting blastomeres are separated, they can each develop into a normal embryo. However, if the plane of the first cleavage is altered so that only one of the resulting blastomeres receives the gray crescent, the blastomere that receives the gray crescent will develop into a normal embryo while the other blastomere will develop into __________, indicating that the gray crescent contains _________. A. endoderm; endoderm determinants B. mesoderm; mesoderm determinants C. a ventral piece; dorsal determinants D. a dorsal piece; ventral determinants

C. is correct. The gray crescent marks the future dorsal side and is necessary for dorsal formation. Wnt11, which moves to the area of the gray crescent (future dorsal side) during cortical rotation, is a dorsal fate determinant. Thus, in this experiment, the blastomere that does not receive the gray crescent will develop into a ventral piece, indicating the gray crescent is the source of dorsal fate determinants.

Which of the following transplantation experiments conducted with frog embryos would result in an embryo with a secondary dorsal side? Important follow-up: explain how a second cortical rotation would also produce an embryo with a secondary dorsal side. A. Transplantation of cells fated to become ventral mesoderm onto the dorsal side of a host embryo. B. Transplantation of cells fated to become ventral mesoderm onto the ventral side of a host embryo. C. Transplantation of cells fated to become dorsal mesoderm onto the dorsal side of a host embryo. D. Transplantation of cells fated to become dorsal mesoderm onto the ventral side of a host embryo.

D. Transplantation of cells fated to become dorsal mesoderm onto the ventral side of a host embryo.

Which of the following statements about the role of cell surface receptors during fertilization is TRUE? A. activation of an ion channel-coupled receptor leads to a rapid change in membrane potential, resulting in a fast block to polyspermy B. activation of phospholipase C can lead to an increase in cytosolic Ca2+ concentrations, resulting in physical changes in the egg shell that prevent polyspermy pathway, C. DAG, in combination with Ca2+, can activate protein kinase C (PKC), leading to new protein synthesis D. all of the above statements are true

D. all of the above statements are true

Which of the statements is FALSE regarding the two major types of stem cells: adult stem cells and embryonic stem (ES) cells? A. Adult stem cells can give rise to a more limited set of cells than ES cells. B. ES cells are only transiently present during early embryogenesis, while adult stem cells are present throughout the life of an individual. C. Both can be maintained in culture with the appropriate signals and conditions. D. Adult stem cells are cells that have been induced back to an undifferentiated state using three key transcription factors.

D. is FALSE. Adult stem cells are committed stem cells that give rise to a limited set of cells and tissue types (multipotent). Induced pluripotent stem cells (iPS cells) are stem cells derived from terminally differentiated cells that have been induced back to pluripotency using three key transcription factors (Oct4, Sox2, and Nanog).

Which of the following statements regarding the creation of nuclear transfer stem cells (NTSCs) is FALSE? Choose ALL false statements (you can select multiple answer options). A. It can be used to create patient-specific ES cells, overcoming issues of histo-incompatibility. B. A nucleus from a fully differentiated cell in an adult's body can be used. C. It requires the use of many donor eggs D. The resulting embryo is genetically identical to the egg donor. E. They are not associated with ethical problems because they do not require the generation of a blastocyst.

D. is FALSE. Nuclear Transfer Stem Cells (NTSCs) are generated through therapeutic cloning, a process in which one takes the nucleus from a differentiated cell (of a patient, for example) and transplants it to an enucleated oocyte, generating a blastocyst-stage embryo whose entire genome is identical to the nuclear donor, not the egg donor. The egg has had its nucleus removed prior to the transplantation. E. is FALSE. The generation of NTSCs through therapeutic cloning still requires the generation of a blastocyst. A. and B. are TRUE. Personalized ES cells can be generated through therapeutic cloning, a technique in which one takes the nucleus from a differentiated cell in the patient and transplants it to an enucleated oocyte, generating a blastocyst-stage embryo whose entire genome is from the nuclear donor, thus overcoming issues of histo-incompatibility. The inner cell mass from the blastocyst can be used to make ES cell culture for therapeutic cloning. C. is TRUE. There is a very low success rate, so the technique requires the use of many donor eggs.

What is the subcellular event that provides the force required for puckering the epithelium during invagination? A. Inhibition of myosin motor activity B. Inhibition of inductive signals which inform the expression of genes that instruct changes in cell identity C. Endocytosis of E-cadherins D. Asymmetric actin-based contraction

D. is correct. Contractile actin filaments composed of actin and its motor protein myosin contract at the outer edge of a subset of cells within an epithelium. This contractile force makes them wedge-shaped, thus causing the epithelium in that region to pucker and invaginate.

Nodal inhibitors would block formation of _____. A. endoderm and mesoderm B. mesoderm and ectoderm C. mesoderm and the gray crescent D. mesoderm and organizer formation

D. mesoderm and organizer formation A. is incorrect. VegT specifies endoderm fate. B. is incorrect. Cells located near the animal pole that do not receive Nodal signaling become ectoderm. C. is incorrect. Sperm fertilization triggers cortical rotation, resulting in the formation of the gray crescent opposite the side of sperm entry.

The expression of which group(s) of genes must be maintained throughout development and in adults? Choose ALL correct answers (you can select multiple answer options). Follow-up: briefly describe the general mechanism that maintains the expression of this/these group(s) of genes throughout development. A. egg-polarity genes B. gap genes C. pair-rule genes D. segment polarity genes

D. segment polarity genes D. is correct. Both larvae and adults share the 14 segmented body plan. Thus, expression of the segment-polarity genes (which control the boundary, number, and polarity of segments) needs to be maintained during gastrulation, through metamorphosis and in adults. A. is incorrect. Egg polarity genes are maternal genes, whose mRNAs are deposited in the egg by the mother before fertilization (maternal gene products). Egg polarity gene products regulate expression of gap genes. B. and C. are incorrect. Gap genes and pair-rule genes are transiently expressed in the syncytial blastoderm. Gap gene and pair-rule gene products regulate the expression of pair-rule genes and segment polarity genes, respectively. Follow-up: Segment polarity gene expression is maintained throughout development through an auto-regulatory feedback loop acting through two cell signaling pathways: the wingless pathway and the hedgehog pathway.

Which group(s) of genes are expressed in cellularized blastoderm? Choose ALL correct answers (you can select multiple answer options). A. egg-polarity genes B. gap genes C. pair-rule genes D. segment-polarity genes E. Hox genes

D. segment-polarity genes E. Hox genes

Passive leakage of partially digested material from the small intestine into the human abdominal cavity may result from defects in which of the following type of intercellular junction? A. desmosomes B. gap junctions C. adherens junctions D. tight junctions

D. tight junctions

What is the function of receptor tyrosine kinases? A. binding to nonpolar signaling molecules such as nitric oxide or steroid hormones B. enzymatic degrading of GTP to GDP C. allowing specific ions to enter the cell after ligand binding D. enzymatic hydrolysis of the signaling molecule shortly after its arrival E. enzymatic phosphorylation of tyrosine in the receptor protein

E. is correct. Phosphorylated receptor tyrosine kinases then interact with relay proteins within the cell.

What is unique about the cleavage divisions in early Drosophila embryos? How does this unique aspect of Drosophila cleavage divisions influence embryonic development?

Early Drosophila embryos undergo rapid and synchronous nuclear divisions without cell divisions (no cytokinesis), forming a syncytium, a collection of nuclei that share the whole cytoplasm of the egg (one giant cell with many nuclei; the nuclei are not in separate cells). Because the syncytial blastoderm is one giant cell with many nuclei, transcriptional regulatory proteins can diffuse and establish concentration gradients across the embryo. These transcriptional regulators can act as morphogens, with concentration-dependent effects on gene expression of nuclei throughout the syncytium. Egg-polarity genes, gap genes and pair-rule genes are expressed when the embryo is a syncytial blastoderm. (Segment-polarity genes and Hox genes are expressed when the embryo is a cellularized blastoderm). For example, the Bicoid protein, a transcription regulatory protein, can acts as a morphogen, diffusing away from its source and creating a concentration gradient in the syncytial cytoplasm. The effects of Bicoid on gene expression in nuclei along the anterior-posterior axis will differ depending on its concentration.

A. Create a labeled diagram of the unfertilized frog egg that includes the following: animal pole, vegetal pole, cortical cytoplasm, pigmented cytoplasm, oocyte nucleus, yolk, VegT mRNA, Wnt11 mRNA, region of future sperm entry. B. Create a labeled diagram of the fertilized frog egg immediately after cortical rotation that includes the following: animal pole, vegetal pole, location of sperm entry, pigmented cytoplasm, gray crescent, cortical cytoplasm, VegT mRNA, Wnt11 mRNA, future dorsal side, future ventral side C. What triggers cortical rotation? Describe the mechanism underlying cortical rotation.

Sperm could enter anywhere in the animal half. Sperm entry triggers cortical rotation (sperm contribute centrioles, aka, microtubule-organizing centers) Following cortical rotation, the gray crescent will form on the side opposite to sperm entry. Cortical rotation (rotation of the cortical cytoplasm) moves maternal components in the vegetal cortical cytoplasm (such as Wnt11 mRNA) to the future dorsal side. VegT mRNA is located in the vegetal pole cytoplasm and does NOT move with cortical rotation. C. Sperm entry during fertilization triggers cortical rotation. Cortical rotation results from organized microtubule bundle formation near the vegetal cortex. The motor protein kinesin can then move maternal components in the vegetal cortical cytoplasm (Wnt11 mRNA) to the future dorsal side.

Apply your knowledge of stem cells to explain how stem cell research can enhance our understanding of cancer. Your Answer:

Stem cells are likely important targets of cancer: Stem cells are long-lived and can therefore accumulate mutations. Stem cells have proliferative abilities, so fewer mutations could induce cancer. Evidence of cancer stem cells: stem cell populations that produce cancerous cells Links between inappropriate activation of signaling systems that produce cancer and signaling systems that promote stem cell maintenance.

explaining how G protein-coupled receptors (GPCRs) function by activating their trimeric G proteins. As a group, discuss ways to highlight important concepts and details. You can refer to the blank figures below the table to organize your thoughts. Your group will then share your creation with the class!

Story Spine Prompts Fill in your story here! Once upon a time... There was a conscientious GPCR that cared about its cell's ability to communicate with other cells. Indeed, the GPCR was so intimately tied to its cell that it passed across the cell's plasma membrane seven times as alpha helices. And every day (or/ and they loved) The GPCR wished it could help transduce an extracellular signal to the cell's intracellular environment, activating target proteins. Until one day A specific hydrophilic signaling molecule, diffusing along on its merry way in the extracellular environment, was a perfect fit and became bound to the GPCR. Excitedly, the GPCR so enthusiastically changed its conformation that the conformational change was transmitted to its cytoplasmic domain! And because of that In this new signal-bound conformation, the enthusiastic GPCR acted as a GEF (guanosine exchange factor) for many of its trimeric G-proteins. For each G-protein, the GPCR decreased the α subunit's affinity for GDP, enabling the exchange of GDP for GTP. And because of that Clutching their GTP, the activated α subunits detached from the activated βγ complexes. And because of that The activated α subunits and βγ complexes then each interacted directly with their targets, which in turn relayed the signal to other destinations in the cell. Until finally Within seconds, the α subunits switched themselves off by hydrolyzing their bound GTP to GDP. They then reassembled with the βγ complexes, switching them off as well. And, when the initial signaling molecule dissociated from the GPCR, the GPCR changed back to its initial conformation, no longer activating any more of its trimeric G proteins. And ever since then The GPCR has waited for its specific type of signaling molecule to bind, enabling it to activate many more of its trimeric G-proteins and produce signaling cascades!

A. Create a diagram of the early gastrula-stage embryo showing the location of the organizer, the locations of Chordin, Noggin, the TGF-β signaling molecule BMP4, and BMP4 activity. Be sure to label the animal and vegetal poles. B. Describe the mechanism by which BMP4, the organizer, Chordin, Noggin contribute to dorsal-ventral patterning.

The TGF-β signaling molecule BMP4 has ventralizing activity and is found throughout the embryo. The organizer (dorsal mesoderm) is where mesoderm and Wnt11 target cells coincide. The organizer secretes the BMP4 inhibitors Chordin and Noggin which bind to and sequester BMP4, thus preventing BMP4 from binding to its receptors to activate downstream signaling. This results in a graded distribution of BMP4 activity, with BMP4 activity lowest near the organizer (future dorsal side) and highest on the side opposite the organizer (future ventral side). As a result, the organizer side is dorsalized (low BMP4 activity) and the side opposite the organizer is ventralized (high BMP4 activity).

Egg-polarity genes encode maternal gene products. What are maternal gene products? What does this tell us about the initial establishment of the anterior-posterior organization in Drosophila?

The mRNAs of egg-polarity genes are deposited asymmetrically in the unfertilized egg by the mother. The localization of mRNAs of different egg-polarity genes sets up the initial anterior-posterior patterning before fertilization. After fertilization, the mRNAs are translated to form proteins that act as transcription regulatory proteins influencing the expression of gap genes in a concentration-dependent manner (morphogens). Example: anterior egg-polarity gene Bicoid and posterior egg-polarity gene Nanos set up opposing concentration gradients.

Which is true of cadherin proteins?Choose one: They create an electrical and metabolic coupling between epithelial cells. They allow ions and small molecules to pass from one plant cell to another. They link epithelial cells together by binding to similar cadherins in adjacent epithelial cells. They anchor epithelial cells to the basal lamina.

They link epithelial cells together by binding to similar cadherins in adjacent epithelial cells.

Match the class of cell surface receptor (1-3) with the best description of its function (A-E). Active practice ideas: Create a table that compares and contrasts the three main classes of cell surface receptors. Describe their general structure and function and relevant signal transduction pathways. Include specific examples during early embryonic development. Create diagrams or flow charts for each class of cell surface receptor that illustrate the signal transduction events following activation of the receptor by a signal. *Note: some of the following questions in this practice quiz will walk you through a few of these. As you continue to study this material, practice writing out this table and drawing these diagrams from memory. activate trimeric G proteins, many of which activate membrane-bound enzymes that produce second messenger molecules _________ !alter the membrane potential directly by allowing certain ions to pass through the plasma membrane __________ act as enzymes themselves or associate with enzymes that are activated when the receptors are stimulated by signal molecules ___________

a. G protein-coupled receptor b. ion channel-coupled receptor c. enzyme-coupled receptor

The enzyme cyclic AMP phosphodiesterase helps terminate a response mediated by an increase in cyclic AMP by doing what? activating an inhibitory G protein, Gi converting ATP to cyclic AMP converting cyclic AMP to AMPin activating adenylyl cyclase converting cyclic AMP to ATP

converting cyclic AMP to AMPin

Match each description to the appropriate type of stem cell. (Embryonic stem cell, inducedpluripotent stem cell, adult stem cell, nuclear transfer stem cell) An undifferentiated cell type derived from the inner cell mass of an early mammalian embryo and capable of differentiating to give rise to any of the specialized cell types in the adult body. Relatively undifferentiated, self-renewing cell that produces daughter cells that can either differentiate into more specialized cell types of a certain class (for example: various types of neurons, various types of blood cells) or can retain the developmental potential of the parent cell. Cells that resemble and behave like pluripotent embryonic stem cells through the introduction of a set of genes encoding particular transcription regulators to differentiated somatic cells. Personalized embryonic stem cells created by transplanting the nucleus from a differentiated cell into an enucleated oocyte, generating a blastocyte embryo with a genome from the nuclear donor (therapeutic cloning)

embryonic stem cell adult stem cell indued pluripotent stem cell nuclear transfer stem cell

Which of the following would be a plausible strategy for reducing the activity of the Ras protein?Choose one: introducing a mutation that enhances the protein's affinity for GDP introducing a mutation that inactivates the protein's GTPase activity inhibiting the GTPase-activating protein (GAP) that promotes GTP hydrolysis introducing a mutation that reduces the protein's affinity for GDP introducing a non-hydrolyzable form of GTP

introducing a mutation that enhances the protein's affinity for GDP

Egg-polarity gene products form two opposing [ Select ] and act as transcription factors regulating the expression of gap genes. Gap genes are expressed in [ Select ] and encode transcription factors that regulate the expression of pair-rule genes. Pair-rule genes are expressed in [ Select ] affecting alternating segments of the embryo. Pair-rule gene expression in each stripe is regulated by a specific [ Select ] which is bound by a unique set of transcription factors. Pair-rule genes encode [ Select ] and regulate the expression of segment polarity genes. Segment polarity genes encode [ Select ] and are expressed in a pattern of 14 stripes . Hox genes are active at the same time as segment polarity genes.

morphogen gradients, large domains, 7 stripes, enhancer element, transcription fators, transcription factors and signaling pathways, 14 stripes, at the same time

Totipotent, multipotent, pluripotent A cell capable of producing cell types with restricted specificity for the tissue in which it resides. _____ A cell capable of producing all the cell types of the embryonic lineage that form the body and germ cells. ______ A cell capable of producing all the cell types of the embryonic lineage that form the body and germ cells and the extraembryonic lineages that form the placenta, amnion, and yolk sac. ______

multipotent pluripotent totipotent