Question 1

Suppose that a homeotic gene from Drosophila is introduced into an Arabidopsis embryo with a mutation in one of its homeotic flower genes.What effect would you expect and why?
A)rescue (or negating)of the phenotype of the mutant because the fly Hox gene supplies the missing protein
B)correction of the sequence in the mutant by recombination with the Drosophila gene sequence
C)expression of the Drosophila homeotic gene correcting the mRNA of the mutant plant
D)a mixture of roughly equal numbers of Drosophila and plant gene product,producing an intermediate phenotype
E)no effect,because Hox genes are different from MADS box genes

Accepted Answer

Hox genes in Drosophila and MADS box genes in plants are different types of homeotic genes, so introducing a Drosophila gene into Arabidopsis would not rescue the mutant phenotype.

Question 2

In early vertebrate embryonic development,a number of pathways,such as Wnt,Hedgehog,and Notch,regulate differentiation and organ formation.They operate by
A)formation of new transcription factors at each stage.
B)transcription activation due to interaction among transcription factors.
C)cell signaling mediated by receptor binding.
D)establishing anterior-posterior and dorsal-ventral symmetry.
E)stimulating MADS box genes.

Accepted Answer

The pathways mentioned in the question operate through cell signaling which involves the binding of growth factors or ligands to receptors on the surface of cells. This leads to activation of downstream signaling pathways that ultimately regulate gene expression and cellular behavior, leading to differentiation and organ formation.

Question 3

Since mice embryos with loss-of-function Hox mutations result in homeotic alterations,then introduction of such a mutation might cause
A)ribs on normally rib-less vertebrae.
B)pointed rather than rounded ears.
C)a change in fur coloration along the dorsal midline.
D)behavioral alterations causing limb shaking.
E)paralysis of hindlimbs but not forelimbs.

Accepted Answer

Homeotic alterations refer to changes in body part identity, and loss-of-function Hox mutations can result in the transformation of one body part into another. In the case of the mice embryos, a loss-of-function Hox mutation may result in ribs on normally rib-less vertebrae, which would be a homeotic alteration. The other options do not directly relate to the function of Hox genes and their effect on body part identity.

Question 4

In Drosophila,a developmental mutation called ssᵃ (spineless aristapedia)results in flies with mini leg parts on the antennae.What kind of gene is this mutation?
A)gap gene
B)pair-rule gene
C)homeotic gene
D)segment polarity gene
E)maternal effect gene

Accepted Answer

Homeotic genes are responsible for determining the identity of body segments and regulate the development of body parts. The mutation resulting in antennal leg parts indicates a change in the identity of the antennal segment, indicating a homeotic mutation. Gap genes, pair-rule genes, segment polarity genes, and maternal effect genes are not directly involved in specifying body segments or body part identity.

Question 5

Segmentation genes in Drosophila act in which of the following orders?
A)segmental,helical,spherical
B)pair-rule,transdeterminal,gap genes
C)gap,pair-rule,segment polarity
D)pair-rule,segment polarity,gap genes
E)pair-rule,segment polarity,Hox genes

Accepted Answer

The answer of Segmentation genes in Drosophila act in which...

Question 6

An Antennapedia (Antp)mutant in Drosophila has legs in place of antennae.This is caused by
A)activation of the Antp gene in the head region instead of in the thorax.
B)the product of an abdominal segment gene being expressed in the head.
C)a chromosomal error that shifts this Antp gene to the anterior end of the complex.
D)the interaction of an inhibitor gene of the bithorax complex with the Antp mutation.
E)loss of function of the Antp gene.

Accepted Answer

The Antennapedia (Antp) gene is normally expressed in the thorax of Drosophila and is responsible for specifying the formation of legs. However, in the Antp mutant, the gene is activated in the head region instead of in the thorax, causing legs to form in place of antennae. This is a classic example of homeotic transformation, where a gene normally expressed in one body part is expressed in another, resulting in morphological changes.

Question 7

Human Hox genes are in four clusters: HOXA,HOXB,HOXC,and HOXD,for a total of 39 genes.Besides conserving the clustering of genes,evolution has also conserved
A)their chromosomal locations.
B)the order of the genes within them.
C)their expression along the dorsal-ventral axis.
D)the mutations to which they give rise.
E)their bilateral symmetry.

Accepted Answer

The answer of Human Hox genes are in four clusters:...

Question 8

Holoprosencephaly is a highly variable condition in mammals,including humans.Its effects range from an absence of brain areas,mid-face disorders,and cyclopia to slight dental abnormalities.One form,HPE3,is caused by Shh mutations.Familial cases are inherited as dominants.Which of the following do you expect?
A)HPE3 is actually a multigene complex.
B)Each type of malformation is due to a different HPE3 allele.
C)HPE3 expression is modified by genetic background.
D)HPE3 individuals also share a common environmental exposure.
E)All mutations in Shh that result in HPE3 are due to frameshifts.

Accepted Answer

Since Holoprosencephaly is highly variable and its effects range from an absence of brain areas, mid-face disorders, and cyclopia to slight dental abnormalities, we can expect that HPE3 expression is modified by genetic background. This means that the same mutation in different individuals may have different effects due to differences in their genetic background.

Question 9

Polycomb group (PcG)mutants cause gain-of-function of Antennapedia (Antp)and other homeotic genes by causing some of these genes to be expressed in tissues where they are usually repressed.PcG has a domain that binds to methylated histone H3.This finding supports which of the following hypotheses?
A)that PcG binds to Antp genes and serves as a transcription activator
B)that PcG mutants have phenotypes like those of Antp mutants
C)that PcG regulates genes by altering chromatin remodeling
D)that PcG initiates repression of Antp genes in tissues
E)that PcG is a repressor molecule like those found in bacteria

Accepted Answer

The fact that PcG binds to methylated histone H3 suggests that it plays a role in regulating gene expression by altering chromatin remodeling, which is consistent with choice C. The other choices are not directly supported by the information provided.

Question 10

A pair-rule gene in Drosophila called runt controls some of sex and nervous system formation.In the mouse,runt controls blood cell formation,osteogenesis,and genital structures.In humans,the analog of runt,when mutated,causes children to lack collar bones,and the opening at the top of the skull fails to close.What kind of experiment could tell you whether these genes are conserved in function?
A)sequencing of the runt genes from each of the three species of organisms
B)in vivo experiments inserting the copy of the moue gene into a segmented fly embryo
C)in vitro experiment that inserts mouse runt gene into cells from an unsegmented Drosophila embryo
D)sequencing the proteins encoded for by each of the species' runt genes
E)use of immunological methods to identify the target cells of the species' runt transcription factors

Accepted Answer

The most direct way to test whether the runt genes are conserved in function is to perform an experiment that inserts the mouse runt gene into cells from a fly embryo. This will allow for the evaluation of whether or not the mouse gene can carry out the same function as the fly gene. Sequencing or protein analysis will not directly test gene function, while in vivo experiments with mouse genes in fly embryos will not test conservation of function between the three species. Immunological methods may identify target cells but will not directly test gene function.

Question 11

The abdominal segments of Drosophila have no appendages.Loss of function of bithorax genes results in appendages forming in these abdominal segments.Therefore,
A)bithorax mutations are necessarily recessive.
B)bithorax genes normally activate appendage-forming genes.
C)ultrabithorax usually acts to oppose bithorax genes.
D)bithorax normally activates a gene that negatively regulates appendages in the abdomen.
E)expression of bithorax wild-type genes suppresses all appendage formation.

Accepted Answer

The answer of The abdominal segments of Drosophila have no...

Question 12

There is much controversy in the newsboth scientific and popularabout the use of embryonic stem cells for research.These cells are grown as cell lines in culture,created by taking the inner cell mass from 5-day-old embryos.What is the center of the controversy?
A)the fact that the embryos are mostly those unused by in vitro clinics
B)the knowledge that these cell lines have reverted spinal cord injuries in mice
C)the unproven usefulness of stem cells to treat human disorders
D)the realization that these cells are pluripotent rather than totipotent
E)the difference of opinion about whether these embryos are persons

Accepted Answer

The controversy surrounding embryonic stem cell research centers largely around the ethical question of whether these embryos are considered to be persons with rights. This leads to debates about the moral implications of using them for scientific research, especially since the embryos are destroyed in the process of harvesting the stem cells.

Question 13

Aniridia is a human condition in which the eye has no iris.The protein encoded by the gene responsible for aniridia is very nearly identical to the protein product of the fly Eyeless protein.What experiment could provide evidence that the two genes are functionally equivalent?
A)Introduce the aniridia mutation into Drosophila embryos to look for iris formation.
B)Sequence the Eyeless and aniridia DNA sequences and regulators.
C)Use the Eyeless mRNA as a probe in other invertebrate and non-mammalian species.
D)Introduce the mouse wild-type sequence into the fly egg to see whether mouse eyes develop.
E)Introduce the mouse aniridia wild-type sequence into the fly to see whether fly eyes develop.

Accepted Answer

Introducing the mouse aniridia wild-type sequence into the fly to see whether fly eyes develop would provide evidence of functional equivalence, as it would demonstrate that the mouse gene can substitute for the fly gene in eye development.

Question 14

Mutations of regulatory elements of developmental genes may have either dominant or recessive effects.If a mutation results in a silencer element,that negatively regulates,which do you expect?
A)a recessive effect in which cis elements are up-regulated
B)a recessive effect in which cis elements are down-regulated
C)a dominant effect in which the silencer down-regulates gene activity
D)a dominant effect in which gene activity is up-regulated
E)a dominant effect in which the silencer is transactivating

Accepted Answer

The answer of Mutations of regulatory elements of developmental genes...

Question 15

As a Drosophila larva molts and becomes a pupa,most of the larval tissues die and adult structures are formed from cells in clusters known as imaginal discs that have remained unchanged throughout the larval stages.Therefore,the disc cells are
A)maternal.
B)zygotic.
C)embryonal.
D)differentiated.
E)transcriptionally activated.

Accepted Answer

The imaginal discs are clusters of embryonic cells that remain unchanged throughout the larval stages and give rise to adult structures during metamorphosis. Therefore, the disc cells are embryonal in nature.

Question 16

Drosophila mutations such as bcd,if homozygous in the mother,produce offspring with lethal defects,no matter what the genotype of the offspring.For some other genes,the mutant phenotype can be "rescued" by introducing a wild-type allele.Why the difference?
A)Non-rescuable mutations are of paternal origin.
B)Rescuable mutations act as dominants.
C)bcd is a maternal effect mutation.
D)Rescue requires recombination between the maternal and introduced alleles.
E)bcd is a Hox gene.

Accepted Answer

The fact that the mutant phenotype is only observed when the mutation is homozygous in the mother indicates that the mutation is a maternal effect mutation. This means that the genotype of the mother determines the phenotype of the offspring. In the case of bcd, the mutant phenotype cannot be rescued by introducing a wild-type allele because the mutation affects early developmental processes that are dependent on the maternal mRNA and protein deposits in the egg. On the other hand, mutations in other genes may be rescued by introducing a wild-type allele because the gene product is required later in development and can be supplied by the introduced allele.

Question 17

Use of double mutants has been invaluable in studying developmental pathways.In C.elegans,sex is determined by the ratio of sets of autosomes to X chromosomes.A diploid male has one X,and diploid hermaphrodites have two X's.The loss-of-function gene her-1 produces XO worms that are hermaphrodites,but XX worms are unaffected.Mutation of a second gene,tra,causes the XX to develop as male.Given this information,what would be the phenotype of a tra,her-1 double mutant?
A)a male nematode
B)a female nematode
C)a hermaphrodite
D)an intersex variant
E)lethality

Accepted Answer

The answer of Use of double mutants has been invaluable...

Question 18

A homeotic mutant flower has an arrangement (from outside in)of carpels,stamens,stamens,carpels instead of the usual sepal,petals,stamens,carpels.In general,Class A genes specify sepals,Class A and B genes specify petals,Class B and C genes specify stamens,and Class C specifies carpels.This mutant must therefore have a mutation in
A)a Class A gene.
B)a Class B gene.
C)a Class C gene.
D)both A and B.
E)both B and C.

Accepted Answer

The mutant flower has carpels where there should be sepals and petals, indicating a loss of Class A gene function, which is responsible for specifying sepals and, together with Class B, for specifying petals.

Question 19

In Drosophila,germ cells develop from pole nuclei isolated from the rest of the embryo at the posterior end.How are they "protected" from the effects of posterior concentrations of zygote-produced mRNAs and proteins?
A)Their genes have unresponsive promoters.
B)The cytoplasm at the pole is separated by a compartmentalizing membrane.
C)The pair-rule,gap,and segmentation genes are not expressed in these cells.
D)Homeotic genes in these cells are inhibited by zygotic genes from the mother.
E)Paternal effects operate on the embryo only in the pole cells.

Accepted Answer

Pole cells are protected by a compartmentalizing membrane that separates them from the rest of the embryonic cytoplasm, including the posterior concentrations of zygote-produced mRNAs and proteins. This allows them to develop independently and unaffected by the patterning signals present in the rest of the embryo.

Question 20

In an organism such as a sea urchin,administration of actinomycin D prevents RNA synthesis.However,early development up to gastrula formation can take place in the presence of actinomycin D but gastrulation cannot.This implies that
A)development beyond gastrulation requires actD.
B)actD is part of the normal environment of a sea urchin and serves as a developmental signal molecule.
C)actD receptors begin to form only on cell surfaces of the gastrula.
D)maternal effect genes regulate development until gastrulation.
E)immature sea urchin embryos produce actD to regulate further development.

Accepted Answer

The fact that early development up to gastrula formation can take place in the presence of actinomycin D but gastrulation cannot suggests that there are maternal effect genes regulating development up to gastrulation. Once gastrulation begins, the zygotic genes become active and are necessary for further development, but they require RNA synthesis which is prevented by actinomycin D. Therefore, this supports the idea that maternal effect genes are responsible for regulation of development up to gastrulation.

Quiz 20: Developmental Genetics

Suppose that a homeotic gene from Drosophila is introduced into an Arabidopsis embryo with a mutation in one of its homeotic flower genes.What effect would you expect and why?

In early vertebrate embryonic development,a number of pathways,such as Wnt,Hedgehog,and Notch,regulate differentiation and organ formation.They operate by

Since mice embryos with loss-of-function Hox mutations result in homeotic alterations,then introduction of such a mutation might cause

In Drosophila,a developmental mutation called ssᵃ (spineless aristapedia) results in flies with mini leg parts on the antennae.What kind of gene is this mutation?

Segmentation genes in Drosophila act in which of the following orders?

An Antennapedia (Antp) mutant in Drosophila has legs in place of antennae.This is caused by

Human Hox genes are in four clusters: HOXA,HOXB,HOXC,and HOXD,for a total of 39 genes.Besides conserving the clustering of genes,evolution has also conserved

Polycomb group (PcG) mutants cause gain-of-function of Antennapedia (Antp) and other homeotic genes by causing some of these genes to be expressed in tissues where they are usually repressed.PcG has a domain that binds to methylated histone H3.This finding supports which of the following hypotheses?

The abdominal segments of Drosophila have no appendages.Loss of function of bithorax genes results in appendages forming in these abdominal segments.Therefore,

There is much controversy in the newsboth scientific and popularabout the use of embryonic stem cells for research.These cells are grown as cell lines in culture,created by taking the inner cell mass from 5-day-old embryos.What is the center of the controversy?

Aniridia is a human condition in which the eye has no iris.The protein encoded by the gene responsible for aniridia is very nearly identical to the protein product of the fly Eyeless protein.What experiment could provide evidence that the two genes are functionally equivalent?

Mutations of regulatory elements of developmental genes may have either dominant or recessive effects.If a mutation results in a silencer element,that negatively regulates,which do you expect?

As a Drosophila larva molts and becomes a pupa,most of the larval tissues die and adult structures are formed from cells in clusters known as imaginal discs that have remained unchanged throughout the larval stages.Therefore,the disc cells are

Drosophila mutations such as bcd,if homozygous in the mother,produce offspring with lethal defects,no matter what the genotype of the offspring.For some other genes,the mutant phenotype can be "rescued" by introducing a wild-type allele.Why the difference?

In Drosophila,germ cells develop from pole nuclei isolated from the rest of the embryo at the posterior end.How are they "protected" from the effects of posterior concentrations of zygote-produced mRNAs and proteins?

In an organism such as a sea urchin,administration of actinomycin D prevents RNA synthesis.However,early development up to gastrula formation can take place in the presence of actinomycin D but gastrulation cannot.This implies that