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to live to reproduce. Those with sickle cell disease will most likely not live long enough to
reproduce. Because the heterozygotes will more successful at reproduction, sickle cell
trait will be selected for in the population and its frequency will increase.
101. The answer is (B).
Individuals with PML may have bleeding secondary to disseminated
intravascular coagulation and are at high risk for strokes. A rapid cytogenetic diagnosis
can definitively diagnose PML so that appropriate preventative treatment can be admin-
istered.
102. The answer is (A).
Meiosis is suspended in prophase I during prenatal life and does not
resume until ovulation. At the end of meiosis in females, there is usually only one mature
ovum. Two polar bodies contain the genetic material from meiosis I and II cell divisions
and these usually degenerate. Crossing over occurs during meiosis in both sexes.
103. The answer is (B).
Because both non-disjoined homologues of chromosome 21 went to
the polar body, there are no copies in the ovum. When a normal sperm joins with the
ovum, it will contribute its one copy of chromosome 21, so there will only be one copy of
chromosome 21 in the conceptus.
104. The answer is (A).
Carriers of balanced reciprocal translocations have an overall risk of 5
to 15% to have a child with an unbalanced derivative chromosome depending on the sex
of the carrier and features of the translocation itself. The 11;22 translocation described
here is not a Robertsonian translocation because it does not involve acrocentric chromo-
somes.
105. The answer is (A).
Adjacent segregation of translocation chromosomes in a carrier parent
will give rise to unbalanced derivative chromosomes in the gametes. The sister may be a
balanced translocation carrier since it is virtually certain that one of the parents is a car-
rier.
106. The answer is (D).
The alleles (genes) present at any locus determine the genotype. Most
of the rest of the genome is not transcribed and its function is not very well understood at
present.
107. The answer is (B).
Exons are the functional portion of the gene that codes for proteins.
Introns may be transcribed but are cut out of the final messenger RNA. Promoters and
enhancers influence gene function but are not part of the functional portion of the gene.
108. The answer is (C).
The risk for the male fetus to be affected is 50%. If the male fetus has
inherited the X-linked dominant mutation, then there will be a miscarriage because the
condition is lethal prenatally in males. The patient’s healthy daughter has no risk to have
a child with incontinentia pigmenti as she is not affected with this autosomal dominant
condition.
109. The answer is (C).
The possible outcomes are as follows: there is a 50% risk that a child
will be a carrier and a 50% risk that a child will be affected.
110. The answer is (D).
Given the family history, it is obvious that this is a mitochondrial dis-
ease. It is passed on in a matrilineal fashion, so all of the daughter’s children will be
affected.
111. The answer is (A).
Because males do not pass on mitochondria via the sperm, there is no
risk of passing on a mitochondrial disease to offspring.
112. The answer is (C).
The number of mitochondria passed on with mutations will vary and
that will influence the severity of the disease in some cases. This condition is called het-
eroplasmy.
113. The answer is (B).
PKU is an autosomal recessive disease and the possible outcomes are
as follows: 25% will have PKU, 50% will be carriers, and 25% will be normal.
114. The answer is (C).
The possible outcomes are that 50% of the children will be normal, and
50% will be carriers.
Answers and Explanations
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115. The answer is (D).
NF-1 is an autosomal dominant disease and the risk of having a child
with the disease is 50%.
116. The answer is (C).
NF-1 exhibits variable expressivity, where some individuals in a family
have more severe disease than others.
117. The answer is (A).
The frequency of the mutated gene is “q” or 0.045. Because the gene is
relatively common, rounding off would introduce some error. So “p” in this case is 1.000
0.045 or 0.955. The carrier frequency, or 2pq, would be 2(0.955)(0.045) which would be
equal to 0.086 or 8.6%.
118. The answer is (D).
Heterozygotes for sickle cell trait are resistant to contracting malaria.
This confers a selective advantage on the sickle cell trait because heterozygotes are more
likely to survive to reproduce. Those who are not carriers are at risk for contracting the
disease and those with sickle cell disease are not as likely to survive to reproduce as those
who are normal or are carriers of sickle cell trait. Because heterozygotes have the best
chance to leave surviving offspring, the frequency of heterozygote carriers (AS) would be
expected to increase.
119. The answer is (D).
In any population, for two alleles, the population is made up of those
with “p” alleles and those with “q” alleles, in any combination. Therefore the population
can be described as p
q 1, where the population equals “1”. So, if “p” is found at a fre-
quency of 0.00005 in the population then “q” is 1
p or 1 0.00005 0.99995.
120. The answer is (A).
A balanced reciprocal translocation between a portion of the long
arms of chromosomes 9 and 22 produces the derivative chromosome 22 that was named
the “Philadelphia” chromosome because of the city where it was discovered. The translo-
cation is the leukemogenic event in chronic myeloid leukemia (CML) and juxtaposes the
ABL and BCR proto-oncogenes to form a fusion gene.
121. The answer is (C).
Most human cells have a limited lifespan due to the fact that with each
cell division, the telomeric repeats at the ends of the chromosomes get shorter. The
telomere repeat sequences at the ends of the chromosomes are not synthesized with the
rest of the DNA and eventually they disappear, causing cell death. In stem cells, telom-
erase is activated to synthesize new telomere sequences with each cell division, confer-
ring a degree of immortality upon them. This seems to happen in most cancers with the
tumor cells becoming “immortal” because they no longer have a finite lifespan.
122. The answer is (B).
Only the acrocentric chromosomes 13, 14, 15, 21, and 22 are involved
in Robertsonian translocations. In a Robertsonian translocation, the small short arms,
which contain heterochromatin and the highly repeated ribosomal gene arrays are lost
and the chromosomes fuse at the centromeres. This makes one chromosome out of the
fused long arms of the two acrocentric chromosomes involved in the translocation. The
acrocentric chromosomes and the other chromosomes in the genome can be involved in
reciprocal translocations, in which chromosomes exchange pieces.
123. The answer is (A).
Meiosis I is the “reduction division” in which the 2n number of chro-
mosomes, 46 in humans, is reduced by half to the “n” number, which is 23 in humans.
124. The answer is (C).
In mitosis, there are two copies of each chromosome on the metaphase
plate and as the cell starts to divide, the copies separate with one copy going into each of the
two new cells. If the two copies do not disjoin and go to the same cell, the other cell will not
get a copy. The other homologue separates normally and one copy joins the two copies in
one daughter cell while the other homologue becomes the sole copy of that chromosome in
the other daughter cell. In this case, the two copies of the X chromosome going to one
daughter cell along with one copy from the homologous X chromosome gives one cell three
copies for a 47,XXX constitution while the other daughter cell with no copies of the X chro-
mosome receives the other copy of the homologous X chromosome for a 45,X constitution.
125. The answer is (B).
Because there is increased chromosome breakage in Bloom syndrome,
there are also increased chromosome structural rearrangements. The increased chromosome
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rearrangements raise the risk that proto-oncogenes will be activated by being juxtaposed
through translocation or that tumor suppressor genes will be mutated through deletions
or breakage within the genes.
126. The answer is (B).
Because of the family history, the best course of action is to find out
if Miss Z has the mutation. If she does, then sigmoidoscopy will be useful in determin-
ing how many polyps are present, and then colectomy may be required to remove
affected sections of the colon. If she does not have the mutation, then she has the gen-
eral population risk for colon cancer and nothing needs to be done except routine
medical checkups.
127. The answer is (C).
Because FAP is an autosomal dominant disease, each child has a 50%
risk of inheriting the mutation.
128. The answer is (B).
PKU is a relatively common disease in Caucasians with an incidence of
1 in 10,000. The treatment consists of limiting phenylalanine in the diet and has to be
done throughout an affected individual’s lifetime. Failure to treat PKU results in progres-
sive mental retardation due to the buildup of phenylalanine in the brain.
129. The answer is (C).
With 50 of 200 or 25% of the cells having a yellow signal, this indicates
that a bcr/abl fusion has taken place in those cells and that there is residual disease in
this patient.
130. The answer is (C).
A mutation is anything that changes the nucleotide sequence of a gene.
Genomic imprinting, transcription, and synthesis do not change the nucleotide
sequence of genes.
131. The answer is (B).
In autosomal dominant disorders, the risk of having a child who is
homozygous for the disorder is 1/4 or 25%, the risk of having an affected heterozygote is
1/2 or 50%, and the chance for a normal, unaffected child is 1/4 or 25%. In the case of
achondroplasia, however, there is no risk for having a homozygous child because that
condition is lethal, so of the three remaining possible outcomes, the chance that the child
will be an affected heterozygote is 2 out of 3, or approximately 67% and the chance of
having a child of normal height is 1 of 3 or approximately 33%.
132. The answer is (D).
Because it is unlikely that a reproductive partner would be the carrier
of a cystic fibrosis mutation, he would only be able to contribute a normal allele to a con-
ception. Abby can only contribute a mutated allele since both of her alleles have the
mutation. So the only possible combination in any conception would be a normal allele
and a mutated allele, thus any child born to Abby and her normal partner would be a car-
rier.
133. The answer is (C).
In autosomal recessive disorders, there is a 1/4 or 25% chance that a
child will not be a carrier or affected a 1/2 or 50% of being a carrier, and a 1/4 or 25%
chance of being affected with the disorder. In this family, the brothers are not affected so
they only have a risk to be carriers. Of the remaining possibilities then, there is a 1/3 or
33% chance for each of them to be a homozygote and not a carrier, and a 2/3 or 67% risk
for each of them to be a carrier.
134. The answer is (A).
A disorder appearing in every generation that can be passed on from
father to son is most likely autosomal dominant. Males can only pass on the Y chromo-
some to sons, so X-linked inheritance is not a possibility in this case. Autosomal recessive
and multifactorial inheritance are generally not multi-generational and have low recur-
rence risks, so they would not be expected to show up in every generation.
135. The answer is (C).
In small populations, rare genes may be present in proportions greater
than in the general population. If even one surviving male had the mutation, that would
constitute more than 10% of the male “population” at that time. It is likely that this gene
would remain at a high frequency as the population grew, depending on the reproductive
success of those possessing it.
Answers and Explanations
227
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228
From Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore: Lippincott
Williams & Wilkins; 2007:84.
From Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore: Lippincott
Williams & Wilkins; 2007:91.
From Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore: Lippincott
Williams & Wilkins; 2007:49, 50.
From Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore: Lippincott
Williams & Wilkins; 2007:51. (B) From Dudek RW. HY Cell and Molecular Biology, 2nd ed.
Baltimore: Lippincott Williams & Wilkins; 2007:51. (C) From Dudek RW. HY Cell and
Molecular Biology, 2nd ed. Baltimore: Lippincott Williams & Wilkins; 2007:51.
Modified from Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore:
Lippincott Williams & Wilkins; 2007:76.
Modified from Dudek and Fix. BRS Embryology, 3rd ed. Baltimore: Lippinctott
Williams & Wilkins; 2005:231.
From McMillan JA et al. Oski’s Pediatrics: Principles and Practice, 3rd ed.
Baltimore: Lippincott Williams & Wilkins; 1999:2243. (B) Courtesy of Dr. Ron Dudek. (C)
From Dudek RW and Louis TM. High Yield Gross Anatomy, 3rd ed. Baltimore: Lippincott
Williams & Wilkins; 2008:76. (D) From Eisenberg RL. Clinical Imaging: An Atlas of
Differential Diagnosis, 4th ed. Baltimore: Lippincott Williams & Wilkins; 2003:193. (E)
McMillan JA et al. Oski’s Pediatrics: Principles and Practice, 3rd ed. Baltimore: Lippincott
Williams & Wilkins; 1999:1828. (F) From Swischuk LE. Imaging of the Newborn, Infant, and
Young Child, 5th ed. Baltimore: Lippincott Williams & Wilkins; 2004:853. (G) Damjanov I.
Histopathology: A Color Atlas and Textbook. Baltimore: Lippincott Williams & Wilkins;
1996:431. (H) From McMillan JA et al. Oski’s Pediatrics: Principles and Practice, 3rd ed.
Baltimore: Lippincott Williams & Wilkins; 1999:2004. (I) From McMillan JA et al. Oski’s
Pediatrics: Principles and Practice, 3rd ed. Baltimore: Lippincott Williams & Wilkins;
1999:248. (J) From Damjanov I. Histopathology: A Color Atlas and Textbook. Baltimore:
Lippincott Williams & Wilkins; 1996:453. (K) From: Damjanov I. Histopathology: A Color Atlas
and Textbook. Baltimore: Lippincott Williams & Wilkins; 1996:453. (L) From Dudek R and Fix
J. BRS Embryology, 3rd ed. Baltimore: Lippincott Williams & Wilkins; 2005:188. (M) From
Brandt WE, Helms CA. Fundamentals of diagnostic radiology, 2nd ed. Baltimore: Lippincott
Williams & Wilkins; 1999:555.
From Dudek RW. HY Genetics. Baltimore: Lippincott Williams & Wilkins; 2008.
From McMillian JA, DeAngelis CD, Feigin RD, Warshaw JB. Oski’s Pediatrics:
Principles and Practice, 3rd ed. Baltimore: Lippincott Williams & Wilkins; 1999:2232; no fig-
ure number. (B) From Rubin R, Strayer DS. Rubin’s Pathology: Clinicopathologic Foundations
of Medicine, 5th ed. Baltimore: Lippincott Williams & Wilkins; 2008:1225. (C, D, E) From
Westman JA. Medical Genetics for the Modern Clinician, 1st ed. Baltimore: Lippincott
LWBK274-FCRE_228-232.qxd 06/02/2009 05:45 PM Page 228 Aptara
Williams & Wilkins; 2006:68. (F) From Rubin R and Strayer DS. Rubin’s Pathology, 5th ed.
Baltimore: Lippincott Williams & Wilkins; 2008:1224. (G) From Eisenberg RL. Clinical
Imaging; An atlas of differential diagnosis, 4th ed. Baltimore: Lippincott Williams & Wilkins;
2003:1121. (H) From Swischuk LE. Imaging of the newborn, infant, and young child, 5th ed.
Baltimore: Lippincott Williams & Wilkins; 2004:1097.
From Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore: Lippincott
Williams & Wilkins; 2007:94. (B) From Dudek RW. HY Embryology, 3rd ed. Baltimore:
Lippincott Williams & Wilkins; 2007:168.
From Kirks DR. Practical Pediatric Imaging: Diagnostic Radiology of Infants and
Children, 3rd ed. Baltimore: Lippincott Williams & Wilkins; 1998:186. (B) From Kirks DR.
Practical Pediatric Imaging: Diagnostic Radiology of Infants and Children, 3rd ed. Baltimore:
Lippincott Williams & Wilkins; 1998:187. (C) From Rubin R and Strayer DS. Rubin’s
Pathology; Clinicopathologic Foundations of Medicine, 5th ed. Baltimore: Lippincott
Williams & Wilkins; 2008:1167. (D) From Rubin R and Strayer DS. Rubin’s Pathology;
Clinicopathologic Foundations of Medicine, 5th ed. Baltimore: Lippincott Williams &
Wilkins; 2008:1167. (E) From Rubin R and Strayer DS. Rubin’s Pathology; Clinicopathologic
Foundations of Medicine, 5th ed. Baltimore: Lippincott Williams & Wilkins; 2008:1167. (F)
From Rubin R and Strayer DS. Rubin’s Pathology; Clinicopathologic Foundations of
Medicine, 5th ed. Baltimore: Lippincott Williams & Wilkins; 2008:1167.
Modified from Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore:
Lippincott Williams & Wilkins; 2007:127.
Modified from Dudek RW. HY Cell and Molecular Biology, 2nd ed. Baltimore:
Lippincott Williams & Wilkins; 2007:80–81.
From Westman JA. Medical Genetics for the Modern Clinician, 1st ed. Baltimore:
Lippincott Williams & Wilkins; 2006:20. Courtesy of Dr. G. Wenger, Children’s Hospital,
Columbus, OH. (B) From Westman JA. Medical Genetics for the Modern Clinician, 1st ed.
Baltimore: Lippincott Williams & Wilkins; 2006:25. Courtesy of Dr. G. Wenger, Children’s
Hospital, Columbus, OH. (C) From Sadler TW. Langman’s Medical Embyology, 10th ed.
Baltimore: Lippincott Williams & Wilkins; 2006:21. Courtesy of Dr. Barbara duPont,
Greenwood Genetics Center, Greenwood, SC. (D) Westman JA. Medical Genetics for the
Modern Clinician, 1st ed. Baltimore: Lippincott Williams & Wilkins; 2006:20. Courtesy of Dr.
G. Wenger, Children’s Hospital, Columbus, OH. (E) From Sadler TW. Langman’s Medical
Embyology, 10th ed. Baltimore: Lippincott Williams & Wilkins; 2006:21. Courtesy of Dr.
Barbara duPont, Greenwood Genetics Center, Greenwood, SC. (F) Westman JA. Medical
Genetics for the Modern Clinician, 1st ed. Baltimore: Lippincott Williams & Wilkins; 2006:20.
Courtesy of Dr. Krzysztof Mrozek, The Ohio State University at Columbus, Columbus, OH.
(G)
From Rubin R and Strayer DS. Rubin’s Pathology; Clinicopathologic Foundations of
Medicine, 5th ed. Baltimore: Lippincott Williams & Wilkins; 2008:186. (H) From Rubin R and
Strayer DS. Rubin’s Pathology: Clinicopathologic Foundations of Medicine, 5th ed. Baltimore:
Lippincott Williams & Wilkins; 2008:186.
From Dudek RW. HY Embryology, 3rd ed. Baltimore: Lippincott Williams &
Wilkins; 2007:149.
Modified from Dudek RW. HY Embryology, 3rd ed. Baltimore: Lippincott
Williams & Wilkins; 2007:161.
From Dudek RW. HY Embryology, 3rd ed. Baltimore: Lippincott Williams &
Wilkins; 2007:150.
Figure Credits
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