Chapter Twenty-One: Advanced Topics in Genetics: Developmental
Genetics, Immunogenetics, and Cancer Genetics
COMPREHENSION QUESTIONS
*1. What experiments suggested that genes are not lost or permanently altered in
development?
The ability to clone plants and animals from differentiated cells showed that the
nuclei of these differentiated cells still retained all of the genetic information
required for the development of a whole organism.
2.
Briefly explain how the Dorsal protein is redistributed in the formation of the
Drosophila embryo and how this redistribution helps to establish the dorsal–ventral
axis of the early embryo.
Dorsal protein is distributed uniformly in the cytoplasm of the unfertilized egg.
After fertilization, at the stage that nuclei migrate to the periphery of the egg,
Dorsal protein migrates into the nucleus along the future ventral side of the egg,
but not along the future dorsal side of the egg. This gradient of nuclear-localized
Dorsal protein activates transcription of mesodermal genes along the ventral
surface of the embryo. Nuclei that lack Dorsal protein transcribe dorsal-specific
genes such as decapentaplegic.
*3. Briefly describe how the bicoid and nanos genes help to determine the anterior–
posterior axis of the fruit fly.
Maternally transcribed bicoid and nanos mRNAs are localized to the anterior and
posterior ends of the egg, respectively. After fertilization, these mRNAs are
translated, and the proteins diffuse to form opposing gradients: Bicoid protein
concentrations are highest at the anterior, whereas Nanos protein concentrations
are highest at the posterior. Bicoid protein at the anterior acts as a transcription
factor to activate transcription of hunchback, a gene required for the formation of
head and thoracic structures. Nanos protein at the posterior end inhibits translation
of hunchback mRNA, thereby preventing the formation of anterior structures in the
posterior regions.
*4. List the three major classes of segmentation genes and outline the function of each.
(1) Gap genes specify broad regions (multiple adjacent segments) along the
anterior–posterior axis of the embryo. Interactions among the gap genes
regulate transcription of the pair-rule genes.
(2) Pair-rule genes compartmentalize the embryo into segments and regulate
expression of the segment polarity genes. Each pair rule gene is expressed in
alternating segments.
(3) Segment polarity genes specify the anterior and posterior compartments within
each segment.
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5.
What role do homeotic genes play in the development of fruit flies?
Homeotic genes specify segment identity—expression of the homeotic genes informs
cells of their location or address along the anterior-posterior axis.
*6. What is apoptosis and how is it regulated?
Apoptosis is programmed cell death, characterized by nuclear DNA fragmentation,
shrinkage of the cytoplasm and nucleus, and phagocytosis of the remnants of the
dead cell. Apoptosis is regulated by internal and external signals that regulate the
activation of procaspases, cysteine proteases that are activated by proteolytic
cleavage. Once activated, these caspases activate other caspases in a cascade and
degrade key cellular proteins.
*7. Explain how each of the following processes contributes to antibody diversity.
(a) Somatic recombination: Recombination produces many combinations of
variable domains with junction segments and diversity segments.
(b) Junctional diversity: During recombination, the V, J, and D joining events are
imprecise, resulting in small deletions or insertions and frameshifts.
(c) Hypermutation: The V gene segments are subject to somatic hypermutation—
accelerated random mutation—that further diversifies antibodies.
8.
What is the function of the MHC antigens? Why are the genes that encode these
antigens so variable?
MHC proteins present fragments of foreign or self antigens to T cells. Variability of
the MHC proteins has the effect of making individuals virtually unique in terms of
their MHC genotype. This variability, or high degree of polymorphism, may have
evolved in response to selective pressure to present antigens from many different
pathogenic organisms for recognition by T cells.
*9. Outline Knudson’s multistage theory of cancer and describe how it helps to explain
unilateral and bilateral cases of retinoblastoma.
The multistage theory of cancer states that more than one mutation is required for
most cancers to develop. Most retinoblastomas are unilateral because the
likelihood of any cell acquiring two rare mutations is very low, and thus
retinoblastomas develop in only one eye. Bilateral cases of retinoblastoma occur in
people born with a predisposing mutation, so that only one additional mutational
event will result in cancer. Thus the probability of retinoblastoma is higher and
likely to occur in both eyes. Because the predisposing mutation is inherited, people
with bilateral retinoblastoma have relatives with retinoblastoma.
10. Briefly explain how cancer arises through clonal evolution.
A mutation that relaxes growth control in a cell will cause it to divide and form a
clone of cells that are growing or dividing more rapidly than their neighbors.
Successive mutations that cause even more rapid growth, or the ability to invade
and spread, each produce progeny cells with more aggressive, malignant properties
that take over the original clone.
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*11. What is the difference between an oncogene and a tumor-suppressor gene? Give
some examples of functions of proto-oncogenes and tumor suppressors in normal
cells.
An oncogene stimulates cell division, whereas a tumor-suppressor gene puts the
brakes on cell growth. Proto-oncogenes are normal cellular genes that function in
cell growth and regulation of the cell cycle: from growth factors such as sis to
receptors such as ErbA and ErbB, protein kinases such as Src, and nuclear
transcription factors such as Myc. Tumor suppressors inhibit cell cycle
progression: RB and P53 are transcription factors and NF1 is a GTPase activator.
12. Why do mutations in genes that encode DNA repair enzymes and chromosome
segregation often produce a predisposition to cancer?
Mutations that affect DNA repair cause high rates of mutation that may convert
proto-oncogenes into oncogenes or inactivate tumor-suppressor genes. Similarly,
errors in chromosome segregation cause aneuploidy and chromosomal aberrations
that cause loss of tumor-suppressor genes or add extra gene doses of protooncogenes.
*13. What role do telomeres and telomerase play in cancer progression?
DNA polymerases are unable to replicate the ends of linear DNA molecules.
Therefore, the ends of eukaryotic chromosomes shorten with every round of DNA
replication, unless telomerase adds back special nontemplated telomeric DNA
sequences. Normally, somatic cells do not express telomerase; their telomeres
progressively shorten with each cell division until vital genes are lost and the cells
undergo apoptosis. Transformed cells (cancerous cells) induce the expression of the
telomerase gene to keep proliferating.
APPLICATION QUESTIONS AND PROBLEMS
14. If telomeres are normally shortened after each round of replication in somatic cells,
what prediction would you make about the length of telomeres in Dolly, the first
cloned sheep?
The chromosomes in the mammary cell nucleus that provided the nuclear DNA for
Dolly would have had shortened telomeres, compared to chromosomes in germ-line
cells (sperm and egg). If telomerase expression was not induced in the process of
cloning and subsequent embryonic development, we would predict that Dolly’s
somatic cells will have undergone further loss of telomeric DNA, so that Dolly
would have shorter telomeres than other sheep of the same age.
*15. Give examples of genes that affect development in fruit flies by regulating gene
expression at the level of (a) transcription and (b) translation.
(a) The products of bicoid and dorsal both affect embryonic polarity by regulating
transcription of target genes.
(b) The nanos gene regulates translation of hunchback mRNA.
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16. What would be the most likely effect on development of puncturing the posterior
end of a Drosophila egg, allowing a small amount of cytoplasm to leak out, and
then injecting that cytoplasm into the anterior end of another egg?
The posterior cytoplasm contains posterior determinants, such as maternal nanos
mRNA. Injecting cytoplasm containing nanos mRNA into the anterior end of an egg
would interfere with determination of anterior (head and thoracic) structures by
hunchback and result in an embryo lacking head structures.
*17. What would be the most likely result of injecting bicoid mRNA into the posterior
end of a Drosophila embryo and inhibiting the translation of nanos mRNA?
Bicoid mRNA in the posterior end of the embryo would cause transcription of
hunchback in the posterior regions. Without Nanos protein, the hunchback mRNA
would be translated to create high levels of Hunchback protein in the posterior as
well as in the anterior. The result would be an embryo with anterior structures on
both ends.
18. What would be the most likely effect of inhibiting the translation of hunchback
mRNA throughout the embryo?
If the translation of hunchback throughout the embryo were inhibited, no anterior
structures would form. The embryo would be entirely posteriorized, perhaps
forming all abdominal structures.
*19. Molecular geneticists have performed experiments in which they altered the number
of copies of the bicoid gene in flies, affecting the amount of Bicoid protein
produced.
(a) What would be the effect on development of an increased number of copies of
the bicoid gene?
Females with an increased number of copies of the bicoid gene would have
higher levels of bicoid maternal mRNA in the anterior cytoplasm of their eggs,
and thus higher levels of Bicoid protein in the embryo after fertilization. The
resulting Bicoid protein gradient would extend further to the posterior, resulting
in the enlargement of anterior and thoracic structures.
(b) What would be the effect of a decreased number of copies of bicoid?
Conversely, a decreased number of copies of the bicoid gene would ultimately
result in a reduced Bicoid protein gradient in the eggs. Thus, sufficient Bicoid
protein concentrations for head structures would be found in a smaller, more
anterior portion of the embryo, resulting in an embryo with smaller head
structures.
20. What would be the most likely effect on fruit-fly development of a deletion in the
nanos gene?
Female flies homozygous for deletions in the nanos gene would lay eggs lacking
nanos mRNA. Upon fertilization, the resulting embryos would form all anterior and
thoracic structures, with no posterior structures. If the egg does have maternal
nanos mRNA (because the mother was a heterozygote), and even if the zygote was
homozygous for a deletion in the nanos gene, the embryo would develop normally.
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21. Give an example of a gene found in each of the categories of genes (egg-polarity,
gap, pair-rule, and so forth) listed in Figure 21.12.
Egg-polarity (maternal effect): bicoid, nanos.
Gap: hunchback, Krüppel.
Pair-rule: even-skipped, fushi tarazu.
Segment-polarity: gooseberry.
*22. In a particular species, the gene for the kappa light chain has 200 V gene segments
and 4 J segments. In the gene for the lambda light chain, this species has 300 V
segments and 6 J segments. Considering only the variability arising from somatic
recombination, how many different types of light chains are possible?
Light chain genes undergo recombination to join one V gene segment to one J
segment, in any combination. The number of different possible V and J
combinations is given by the product of the number of V segments and the number
of J segments for each light chain.
kappa light chain: 200 × 4 = 800
lambda light chain: 300 × 6 = 1800
Total light chains = kappa + lambda = 800 + 1800 = 2600
23. In the fictional book Chromosome 6 by Robin Cook, a biotechnology company
genetically engineers individual bonobos (a type of chimpanzee) to serve as future
organ donors for clients. The genes of the bonobos are altered so that no tissue
rejection takes place when their organs are transplanted into the client. What genes
would need to be altered for this scenario to work? Explain your answer.
The MHC genes of the bonobos must be humanized. The T-cells of the immune
system recognize antigen complexed with the body’s own MHC molecules. Non-self
MHC molecules trigger an immune response and tissue rejection. Therefore, the
bonobo MHC molecules must be replaced with the human patient’s own MHC
molecules to prevent rejection.
*24. A couple has one child with bilateral retinoblastoma. The mother is free from
cancer, but the father had unilateral retinoblastoma and he has a brother who has
bilateral retinoblastoma.
(a) If the couple has another child, what is the probability that this next child will
have retinoblastoma?
First, we summarize the information with a pedigree. The shaded boxes
represent bilateral retinoblastoma; the striped box represents unilateral
retinoblastoma.
Familial retinoblastoma is caused by mutation of the RB tumor-suppressor
gene (see Table 21.9). Because the loss of a functional RB allele means that
only one additional mutation event will completely eliminate RB function and
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lead to retinoblastoma, loss-of-function RB mutations have dominant effects
with regard to retinoblastoma. If the father with unilateral retinoblastoma is
heterozygous for an RB mutation, then the chance of another child inheriting
the mutant RB allele is ½. Of course, if the father is homozygous for the RB
mutation, then the chance of the child having retinoblastoma is nearly 100%.
(b) If the next child has retinoblastoma, is it likely to be bilateral or unilateral?
Since retinoblastoma in this family is most likely an inherited disorder, a child
with retinoblastoma will more likely have bilateral retinoblastoma. Unilateral
retinoblastomas are usually spontaneous in origin, requiring two independent
mutations in a single somatic retinal cell. Familial retinoblastomas occur in
family members that inherited one of the two mutations required for
retinoblastoma. As only one additional mutation is required in the somatic
retinal cells, retinoblastoma occurs in both eyes and at earlier ages than
spontaneous unilateral retinoblastomas.
(c) Propose an explanation for why the father’s case of retinoblastoma was
unilateral, while his sons and brother’s cases were bilateral.
The father may have unilateral retinoblastoma because of incomplete
penetrance of the mutation in the RB gene. Alleles at another locus or multiple
other loci may have contributed to resistance to retinoblastoma in the father, so
that he suffered retinoblastoma in only one eye. Alternatively, it may have been
just good fortune (random chance) that one of his eyes was spared the second
mutation event that led to retinoblastoma in his other eye.
25. Some cancers are consistently associated with the deletion of a particular part of a
chromosome. Does the deleted region contain an oncogene or a tumor-suppressor
gene? Explain why.
The deleted region contains a tumor-suppressor gene. Tumor suppressors act as
brakes on cell proliferation. The deletion of tumor-suppressor genes will therefore
permit the uncontrolled cell proliferation that is characteristic of cancer.
Oncogenes, on the other hand, function as stimulators of cell division. Deletion of
oncogenes will therefore prevent cell proliferation, and cannot usually cause
cancer.
26. Cells in a tumor contain mutated copies of a particular gene that promotes tumor
growth. Gene therapy can be used to introduce a normal copy of this gene into the
tumor cells. Would you expect this therapy to be effective if the mutated gene were
an oncogene? A tumor-suppressor gene? Explain your reasoning.
Gene therapy to introduce a normal copy of the gene into tumor cells will not work
for oncogenes because oncogenes are dominant, activating mutations of protooncogenes. Gene therapy may work if the tumor arises from a mutation that
inactivates a tumor-suppressor gene. Loss-of-function mutations are recessive;
therefore, a normal copy of the gene will be dominant and restore regulation of cell
proliferation in the tumor cells.
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CHALLENGE QUESTIONS
27. As we have learned in this chapter, the Nanos protein inhibits the translation of
hunchback mRNA, thus lowering the concentration of Hunchback protein at the
posterior end of a fruit-fly embryo and stimulating the differentiation of posterior
characteristics. The results of experiments have demonstrated that the action of
Nanos on hunchback mRNA depends on the presence of an 11-base sequence that is
located in the 3' untranslated region of hunchback mRNA. This sequence has been
termed the Nanos response element (NRE). There are two copies of NREs in the
trailer of hunchback mRNA. If a copy of NRE is added to the 3' untranslated region
of another mRNA produced by a different gene, the mRNA now becomes repressed
by Nanos. The repression is greater if several NREs are added. On the basis of these
observations, propose a mechanism for how Nanos inhibits Hunchback translation.
Nanos may bind to the NREs, either directly or indirectly as a complex with other
proteins. Multiple NRE binding elements may enhance the binding either by simply
providing a higher concentration of binding sites or through cooperativity (the
binding of protein to one NRE enhances the binding of protein to other NREs). The
complex of Nanos (and other proteins) bound to the NREs at the 3' untranslated
region may target the mRNA for rapid degradation. Alternatively, the NRE-bound
protein complexes may interfere with ribosome binding or translational initiation at
the 5' end of the mRNA, although this postulates that mRNAs assume a circular
topology where the 3' and 5' ends of the mRNA are in close proximity.
28. Offer a possible explanation for the widespread distribution of Hox genes among
animals.
One explanation is that the Hox genes arose in the common evolutionary ancestor
of the animals that have Hox genes. This ancestor used the ancestral Hox gene for
body patterning. The complex of Hox genes and the overall function was then
retained in the descendant lineages.
29. Many cancer cells are immortal (will divide indefinitely) because they have
mutations that allow telomerase to be expressed. How might this knowledge be
used to design anticancer drugs?
Since cancer cells depend on telomerase activity to preserve their telomeres, drugs
that target telomerase enzymatic activity may limit the ability of cancer cells to
divide indefinitely.