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Unformatted text preview: CBB Genetics
Immo E. Scheffler Fall 2000
FINAL EXAMINATION All answers are to be written into the Blue Book. Leave the first page blank for scoring.
The number of points for each question is equivalent to the number of minutes recommended. Please
note the question number clearly at the top of each page of the Blue Book.
"A fact is a contestable component of a theoretically constituted order of things".
M.E. Hawksworth, 1989
(Postmodernist Rhetoric is not acceptable on this examination)
Multiple Choice Questions
For each of the following questions, select the one best answer. QUESTION 1 (3 minutes)
A boy with Duchenne muscular dystrophy was born to a man and woman with no familial history of
the disease. The most likely explanation for this occurrence is:
B. a point mutation in the dystrophin gene.
C. a recombination event in the dystrophin gene gave rise to a frameshift mutation leading to an
D. a CGG expansion resulted in the disruption of the promoter of the dystrophin gene.
E. a translocation resulted in the disruption of the dystrophin gene.
Answer: C Duchenne muscular dystrophy (DMD) is a lethal, X-linked recessive disease affecting approximately 1
in 3300 live male births. The disease becomes symptomatic in early childhood. Inability to walk
occurs by the end of the first decade, and death usually occurs by the second decade. Nearly all
patients show the complete absence of the protein dystrophin, which is abundantly found in skeletal
and cardiac muscle. The DMD gene, located at Xp21, codes for this protein. It is approximately 2300
kb in size, making it one of the largest genes known of any species. It contains 24 regions of 109
amino acids, that are similar but not identical repeats of each other. In women, the similarity of these
sequences can lead to the misalignment of homologous material at meiotic synapsis. In combination
with a recombination event, this misalignment gives rise to frameshift mutations, leading to an
untranslatable mRNA. This series of events occurs at an extremely high rate of about 1 in 10,000. In
fact, one third of the DMD cases in each generation arise from this mechanism.
Infidelity is not a likely explanation for the child because the trait is an X-linked recessive condition.
A male child must receive the DMD trait on the maternal X chromosome. The father of the child only
contributes a Y chromosome. Many genetic diseases, such as cystic fibrosis, are due to a single point mutation within the coding region of the gene. Diseases such as Fragile X syndrome and Huntington
disease, are caused by the expansion of a trinucleotide repeat. Finally, rare cases of DMD in females
have been caused by an X chromosome-autosome translocation with the breakpoint on the X
chromosome within the DMD gene. After X inactivation occurs in the female zygote, the only cells
which survive have the normal X chromosome inactivated. This gives rise to a female who is
heterozygous for the DMD, but phenotypically expresses the disease. QUESTION 2 (3 minutes)
Klinefelter's syndrome can arise from all of the following events EXCEPT:
E. nondisjunction in meiosis I of the male
nondisjunction in meiosis II of the male
nondisjunction in meiosis I of the female
nondisjunction in meiosis II of the female
nondisjunction in the first mitotic division in the zygote Answer: B Klinefelter's syndrome, XXY, is characterized by poorly developed secondary sexual characteristics,
small testes, hyalinization of the seminiferous tubules and sterility. Many affected males are tall and
eunuchoid. Klinefelter's syndrome can arise from nondisjunction in any division in a female and from
nondisjunction in meiosis I in the male. It could also arise from a failure in an early mitotic division in
which XX and Y went to one pole and Y alone went to the other. Cells from the XXY line would be
viable, while cells containing only the Y would die. Klinefelter's syndrome cannot arise from
nondisjunction in meiosis II of the male because the possible gametes arising from a failure in this
division are either XX or YY. Since a female can only contribute an X chromosome, possible children
from a meiosis II failure in the male would be either XXX or XYY. The chart below illustrates the
possible offspring from failures during different meiotic divisions in the different sexes.
QUESTION 3 (3 minutes)
The autosomal recessive Tay Sach disease is caused by a deficiency for hexosaminidase A and is lethal
in childhood. In a population of Ashkenazi Jews, blood testing showed the frequency of heterozygotes
to be 0.1. What is the probability that the first child of two individuals from this population with no
family history of the disease will be affected with Tay Sach disease?
A probability cannot be calculated from the information given. Answer: A Since Tay Sach is a lethal in childhood, the only individuals carrying the gene are heterozygotes. The
only mating which produces progeny with the disease will be the mating between two heterozygotes.
Using the blood test data, the probability that a random individual is a heterozygote is 0.01. The probability that two heterozygotes will produce a child with Tay Sach disease is 0.25. So the overall
probability that two individuals will produce a child with Tay Sach is (0.1)(0.1)(0.25) or 0.0025.
What about the other choices?
Choice B, 0.0625 is a garbage answer which looks like it would be a reasonable guess for this kind of
probability. It is the product of (0.25)(0.25). Aside from that, it is difficult to come up with reasonable
probabilities in this problem which multiply out to that number
Choice C, incorrectly assumes that the parents had a sibling with Tay Sachs. This meant that the
grandparents carried the disease, and their normal children had a 0.67 chance of carrying the allele.
This gave the probability that the parent's child would have Tay Sach as (0.67)(0.67)(0.25) or 0.1122.
Choice D, 0.25 incorrectly assumed that the parents are carriers, so the probability of a Tay Sach child
is equal to the probability that two heterozygotes will produce a child with the disease or 0.25.
Choice E, incorrectly assumes that because q2, the frequency of the disease cannot be calculated from
the problem, and hence the allele frequencies, that an estimate of the frequency of heterozygotes
cannot be made. Without the probability of a heterozygous individual, the problem cannot be done. QUESTION 4 (3 minutes)
A 28 year old woman and a 25 year old man present for genetic counseling. Both are Caucasian and
have one sibling affected with cystic fibrosis. The most appropriate method to assess the risk of
transmitting cystic fibrosis to a potential child would be:
E. Biochemical testing
Fluorescence in situ hybridization (FISH)
Polymerase Chain Reaction (PCR)
Western blot analysis Answer: D
Cystic fibrosis is an autosomal recessive disease caused by point mutations or small deletions in an
integral membrane protein which functions as a chlorine transporter. Although many mutations within
the gene cause the phenotype, it is not practical or cost effective to sequence the entire region to screen
for a carrier. However, in the Caucasian population, the most frequent mutant allele causing cystic
fibrosis is due to a small deletion at phenylalanine 508 in exon 10. It accounts for greater than 50% of
mutant cystic fibrosis alleles in Caucasian populations. An amplification of this region using the
polymerase chain reaction (PCR) can be done, and the PCR products sequenced and compared against
the normal sequence for this region. If comparison reveals the deletion, the donor of that template
DNA would be classified as a carrier. It the mutation is not present, the probability that the donor of
that DNA is a carrier of cystic fibrosis is greatly reduced, but still exists. Most diagnostic laboratories
will use this method to screen for between 4 to 10 of the most common mutations. Note, that if a
person belongs to a different ethnic group, PCR must be used to amplify the exons which contain the
majority of the CF mutations within that particular ethnic group. Answer A, biochemical testing, is usually used to detect a defective enzyme or a reduced amount of
the normal enzyme. It is more economical than DNA testing for detecting carriers or affected
individuals. It is employed in the detection of carriers of autosomal recessive diseases such as TaySachs disease, sickle cell anemia and the thalassemias.
Answer B, karyotype analysis is used to determine whether an individual’s chromosomes are grossly
normal in their number and structure. It can be used to detect trisomies, monosomies, and
translocations as well as large inversions and deletions.
Answer C, Fluorescence in situ hybridization (FISH), employs a DNA probe that has been labeled with
a fluorescent compound to visualize locations on the chromosomes that have homology to the probe.
It can be used to detect microdeletions in a gene of interest.
Answer E, Western blot analysis is a technique that is used to detect the presence of antibody against
specific proteins in serum. It is most used in laboratory medicine as a confirmatory test for HIV
QUESTION 5 (3 minutes)
The Rh blood group has two alleles, the Rh-positive allele [R] and the Rh-negative allele [r].
Erythroblastosis fetalis (EF) is a hemolytic anemia of newborn infant caused by the transplacental
transmission of maternally antibody against Rh-positive blood. Rh-positive babies born to Rhnegative mothers (rr) are at risk for the condition. If the frequency of the Rh-negative allele is 0.3, the
frequency of matings at risk for producing a child with EF is:
The frequency cannot be calculated from the information given. Answer: B P(mating at risk for erythroblastosis )
=P(Rh+ male x rh- female) = P(RR male x rr female) + P(Rr male x rr female)
= p2q2 + (2pq3)
This table shows the possible matings relative to the Rh blood groups. The highlighted matings are between Rh positive males and Rh
negative females. ♂ p2 (RR)
q2 (rr) ♀ p2 (RR)
p2q2 2pq (Rr)
2pq3 q2 (rr)
q4 In this case, the frequency of Rh- = r= 0.3; The frequency of Rh+ = R = 0.7
Plugging these numbers into the first formula gives us:
(0.7)2(0.3)2 + 2(0.7)(0.3)3 = 0.082
What about the other answers?? Answer A, 0.063, is the probability of a child having erythroblastosis or
=P(Rh+ male x rh- female) P(Rh+ child)
= P(RR male x rr female)P(Rh+ child) + P(Rr male x rr female)P(Rh+ child)
= p2q2 + (2pq3)1/2 = p2q2 + pq3
Answers C and D, 0.126 and 0.164, result when the answer to either A or B is multiplied by two.
Remember,if the sex of the parent didn’t matter, there were often 2 ways of fulfilling the question.
(See problem set XVII, numbers 1 and 5 for more practice.)
QUESTION 6 (3 minutes)
An RFLP located next to the centromere of chromosome 21 has two haplotypes, A and B. An AA
woman and an AB man produce a child with Trisomy 21 that is AAB. The child’s trisomy arose from
a failure in which of the following:
E. nondisjunction in meiosis I of the male
nondisjunction in meiosis II of the male
nondisjunction in meiosis I of the female
nondisjunction in meiosis II of the female
The failure cannot be determined from the information provided. Answer: E The RFLP detects a region near the centromere of chromosome 21. The region around the centromere
exhibits a phenomenon called crossover suppression. Since genetic exchange cannot happen in this
area, the probe is a reliable marker for the individual chromosomes. The male is AB and the woman is
AA. The child is AAB. If chromosomes failed to disjoin in Meiosis I of the mother, eggs containing
AA would be produced. If a B sperm fertilized this egg, an AAB child would be produced. If sister
chromatids failed to disjoin in Meiosis II of the mother, an AA egg would be produced. If a B sperm
fertilized this egg, an AAB child would also be produced. If chromosomes failed to disjoin in Meiosis
I of the father, sperm containing AB would be produced. If that sperm fertilized any egg from the
mother, an AAB child would be produced. In fact, the only meiotic failure that would be unable to
produce an AAB child would be a failure in Meiosis II in the male. If sister chromatids failed to
disjoin, AA or BB sperm would be produced. Once one of these sperm fertilized one of the mother’s
eggs, either an AAA or an ABB child would be produced. QUESTION 7 (3 minutes)
A 26 year old woman presents to your clinic during a spontaneous abortion from which she recovers
uneventfully. She and her 29 year old husband have been trying to have a child for the last 3 years,
during which time the woman has had 5 spontaneous abortions. The karyotype of the aborted fetus
contained 46 chromosomes, and all pairs were normal in except for the pairs shown below.
Came from the normal parent
1 3 2
5 Chromosome 7
Mother's Karyotype Father's Karyotype Karyotype of
Aborted Fetus Came from the parent with the translocation. Note both
of the centromeres are from the 7 chromosome, so
Adjacent II disjunction occurred. The event which lead to the child's abnormal karyotype was:
5 Chromosome 12 Note-In the adjacent I segregation pattern, the
cell correctly places homologous centromeres
in different cells. In the adjacent II
segregation pattern, a breakpoint of the
translocation is so close to a centromere that
the cell can’t tell which centromere it is
looking at. To our eyes, it looks like
homologous centromeres went to the same
pole at the end of Meiosis I. By looking at the
centromeres, you can quickly determine which
type of segregation occurred. adjacent I segregation
adjacent II segregation
a single recombination event within a pericentric inversion
a single recombination event within a paracentric inversion Answer B
The couple is experiencing infertility and a high rate of spontaneous abortions because the father has a
reciprocal translocation between chromosome 7 and chromosome 12. His karyotype is 46, XY, t(7;12)
(12qter→12q23::7p13→7qter; 12pter→12q23::7p13→7pter). During meiosis I, the four chromosomes
involved in the translocation will pair together in a structure called a quadrivalent, which is pictured
below. 12qter 12qter 12q23
7qter 7p13 12pter 12q23 12pter 7p13 7qter 7pter 7pter There are three ways that the quadrivalent can separate during meiosis I. They are called adjacent I,
adjacent II and alternate segregation. Adjacent II segregation only occurs when the breakpoint of the
translocation is so close to the centromere of a chromosome that the cell cannot distinguish between
the centromeres. In this specific case, the fetus received a normal chromosome 7 and a normal
chromosome 12 from his mother. From his father, the fetus received a normal chromosome 7, and a
chromosome composed of 12qter→12q23:: 7p13→7qter. This gives the fetus a total of three copies of
the material from 7p13→7qter, and only one copy of material from 12pter→12q23. Note that the
zygote has three copies of the centromere of chromosome 7, the hallmark of adjacent II segregation.
Since the zygote is unbalanced, it aborts.
What about the other choices?
Choice A, Adjacent I segregation produces cells which contain different centromeres but duplication
and deletions which lead to spontaneous abortion. In this case, the products of Adjacent I segregation
would be 12qter→12q23::7p13→7qter and the normal chromosome 12, or 12pter
→ 12q23::7p13→7pter and the normal chromosome 7.
Choice C, Alternate segregation gives the only outcomes that produce viable progeny. The two
products are produced. One contains a normal chromosome 7 and a normal chromosome 12, and the
second product the balanced translocation. This cell has all the information, it is just in a unique order.
In this case, 12qter→12q23::7p13→7qter and 12pter→12q23::7p13→7pter.
Choice D, a recombination event within a pericentric inversion leads to the formation of an acentric
fragment and a dicentric bridge. This cell will not complete meiosis.
Choice E, a recombination event within a paracentric inversion leads to the chromatids that are
duplicated for material on one of the arms and deleted for material on the other arm. QUESTION 8 (3 minutes)
Lesch Nyhan syndrome is caused by a lack of the enzyme hypoxanthine-guanine phosphoribosyl
transferase. In a certain population, if the frequency of affected males is 0.01, what is the probability
that a woman selected at random will carry the trait?
The probability cannot be determined from the information given.
B Lesch Nyhan syndrome is an X-linked recessive trait. A male is hemizygous for the X chromosome,
and thus has only one copy of each trait. The frequency of an X-linked recessive in males is thus equal
to the frequency of the allele in the population. From this, we know that q= 0.01 and p = .99. A
woman has two copies of each gene on the X chromosome, so the equation for Hardy Weinberg
equilibrium is the same as for the autosomal traits. In this case, a heterozygous female would occur at
a frequency of 2pq or 0.0198.
What about the other choices?
Choice A, 0.0099, is obtained if the incorrect formula, pq, is used for the frequency of female
Choice C, 0.09, makes the assumption that the trait is autosomal, and so the frequency of affected
males and affected females are equal, and also uses the incorrect formula, pq.
Choice D, 0.18, makes the assumption that the trait is autosomal, and so the frequency of affected
males and affected females are equal and uses the correct formula, 2pq.
Choice E, The probability cannot be determined from the information given. If a person is guessing,
s/he might find this attractive, however, if you are given the frequency of one of the alleles, you should
be able to calculate the frequency of any of the genotypes. QUESTION 9 (3 minutes)
In a certain population, the frequency of the alleles of the ABO blood group are A= 0.1,
B = 0.3 and O = 0.6. In the following pedigree from that population, the ABO phenotypes of the
family are listed beneath their symbols. What is the probability that III-1 will have type O blood?
A B 0.045
0.133 Answer E
P(OO child/B male from the population x A female with pedigree data mating)=
This expression reads What is the probability of an OO child given a mating between a male
from the population that is phenotype B and a female who has pedigree data that is phenotype
P(AO/A male) P(AO female/A phenotype) P(OO child)=
(p2 + 2pr ) (2/3)(1/4)=
(0.09 + 0.36 ) (2/3)(1/4)=
(0.8) (0.67) (0.25)=
=0.133 Matching Questions
QUESTIONS 10-12 (3 minutes each)
A. B. C. D. For each of the following patients described, choose the most likely mode of inheritance for the
10. A 20 year old man myoclonic epilepsy with ragged-red muscle fibers (MERRF)
MERRF exhibits a mitochondrial inheritance pattern. The hallmark of this pattern is matrilineal
inheritance. All of the children of an affected woman will be affected since they receive mitochondrial
genes only from the female parent. Affected males do not contribute mitochondria to progeny, so their
children will not receive the trait.
11. A 12 year old boy with hemophilia
Hemophilia is an X-linked recessive disorder that affects the blood clotting pathway. The hallmark of
X-linked recessive inheritance is an abundance of affected males and an absence of affected females.
Males are hemizygous for the X chromosome, so the phenotype is expressed with only one dose of the
gene. Females have two copies of the X chromosome, so they appear phenotypically normal although
they may carry the recessive allele. Since a male inherits his X chromosome from his mother, if he is
affected, she must carry the trait. Other X-linked recessives include glucose-6-phosphate
dehydrogenase deficiency, colorblindness, and Duchenne’s muscular dystrophy.
12. A 28 year old man with Marfan’s syndrome.
Marfan’s syndrome shows an autosomal dominant inheritance pattern. Since the trait is a dominant,
the phenotype can be expressed if the allele is present in one dose. Deleterious autosomal dominants
occur at very low frequencies, so affected individuals are almost always heterozygous for the trait.
Since the gene is located on an autosome, both male and female progeny can be affected. Common
autosomal dominants include Ehlers-Danlos syndrome, neurofibromatosis, Niemann-Pick disease and
osteogenesis imperfecta. Answer C shows an autosomal recessive pattern. For the phenotype of the autosomal recessive to be
expressed, the recessive allele must be present in two doses. Both male and female children can be
affected. The hallmark feature is that unaffected parents have affected children of both sexes.
Common autosomal recessives include phenylketonuria, sickle cell anemia, and Tay-Sachs disease.
Short Answer Questions
QUESTION 13 (12 minutes)
The polymerase chain reaction (PCR) has become one of the most powerful means of genetic analysis.
a. Explain briefly how it is useful in diagnosing fragile X syndrome (a schematic diagram with
appropriate labels will be very helpful)
With PCR we can use primers flanking the CGG repeat and amplify the corresponding genomic DNA;
from a comparison of the PCR product from normal males with that from males with fragile X
syndrome one can see the degree of amplification of the CGG repeat (>> 200 repeats in affected
b. Describe with a schematic diagram what you would expect to see in a nonpenetrant
transmitting male and his daughters.
A nonpenetrant transmitting male would yield a PCR product of intermediate size, or closer in size, but
larger than that from normal males (in the premutation the number of repeats is in the range of 60 –
200); carrier females would yield two diffuse bands, one from their normal X chromosome and the
second showing the premutation
c. What do we mean when we speak of the phenomenon of anticipation in case of fragile X,
Huntington's disease, and similar diseases?
Anticipation means that the severity of the disease gets worse in progressive generations of a pedigree,
and the age of onset may also become reduced progressively.
d. The Dutch-Belgian Fragile X Consortium has published the following paper :
"Fmr1 knockout mice: a model to study fragile X mental retardation"
Explain why such a mouse knockout model would not be useful to study the Sherman
paradox of the genetic of Fragile X syndrome in humans.
The mouse with the fmr1 point mutation may show the symptoms of fragile X (i.e the consequences of
missing the FMR1 protein, but this would be due to a “classical” mutation (deletion of an exon), and
one would not see the stochastic phenomena associated with trinucleotide expansion.
QUESTION 14 (12 minutes)
a. Define the phenomenon of imprinting in two or three sentences.
Imprinting is the phenomenon that describes the differential expression of certain loci (autosomal)
depending on whether the chromosome passed through spermatogenesis or oogenesis. An imprinted gene is expressed only from either the paternal or the maternal (never both) chromosomes. Example:
Igf2 is always expressed from the paternal chromosome, never from the maternal chromosome.
b. Give a prominent example from the human genetics literature and illustrate diagrammatically how
the different symptoms are interpreted.
A prominent example from human genetic are the Angelman syndrome and the Prader-Willi
syndrome. A small deletion on chromosome 15 can cause Angelman syndrome, if the deletion is on
the maternal chromosome, and Prader-Willi syndrome if the deletion is on the paternal chromosome.
QUESTION 15 (12 minutes)
"Mitochondrial diseases" have become recognized in the past decade, originating from mutations
in the mitochondrial genome. They are heritable, but inheritance is not mendelian.
a. What is the current explanation (one sentence) for the observation that heteroplasmy in the
offspring from a heteroplasmic female is highly variable?
There is an apparent bottleneck in the segregation of mtDNA during oogenesis; only a very small
number of mtDNAs are selected to be amplified to end up in the oocytes produced
b. How would this variability be reflected in the observed symptoms of these children?
The symptoms are generally related to the degree of heteroplasmy, that is, the higher the proportion of
mutant mtDNAs, the more severe the disease
c. How is it possible that an individual suffering from Leber's hereditary optic neuropathy can be
homoplasmic for the mutation in the mitochondrial ND4 gene?
If the mutation is in a structural gene and causes a missense amino acid substitution, the resulting
protein may be partially functional; thus, even when 100% of the protein is mutant, one may have
sufficient functional electron transport to support live individuals
QUESTION 16 (16 minutes)
The Worton team in Toronto identified a DNA fragment from the X chromosome that was also
very close to the breakpoint/junction of the X-autosome translocation in the female suffering from
Duchenne muscular dystrophy.
a. How can we be reasonably certain that this piece of DNA is from the relevant gene (DMD,
The hypothesis is that such females are suffering from DMD because the DMD gene is broken in the
translocation; the normal DMD gene on the normal X chromosome is inactive, because the entire
normal X chromosome is preferentially inactivated.
b. This piece of DNA is used in an analysis of RFLPs in several pedigrees of families in which
DMD is observed. With the restriction enzyme Taq1 RFLPs are observed in females with affected children. Draw a pedigree with three generations and several children in each that can
be used to illustrate linkage of the locus defined by the RFLP locus and the DMD locus. Also,
show a representative schematic Southern blot illustrating the RFLPs in several individuals.
c. A lod score analysis between the Taq1 RFLP and the DMD loci shows that θ = 0.0 with a lod
score of 3.4. What does that tell us?
A recombination frequency of θ = 0.0 indicates that the two loci are very closely linked, i.e.
inseparable by recombination; a lod score of 3.4 suggests that the odds are greater than 1000:1 that this
value of theta is correct.
QUESTION 17 (12 minutes)
The following molecules/structures are associated with the phenomenon of X-inactivation in females:
XIST RNA, Barr body
a. What is their relation to X-inactivation and to each other? What is very special about the XIST
XIST RNA is expressed from the inactive X chromosome and it acts in cis on the chromosome from
which it is transcribed to inactivate it; the Barr body represents the inactive X in a highly condensed
(inactive) form. It was found that the XIST RNA is tightly associated with the Barr body.
XIST RNA is never exported from the nucleus; it has no open reading frame; it is never found on
b. Give a few arguments rationalizing the observation that in females with balanced, reciprocal Xautosome translocations the intact X chromosome is generally inactivated.
Such females have a normal X, and an X chromosome carrying an autosomal fragment; if the
derivative X chromosome was inactivated, the inactivation may spread into the autosomal region, and
hence there would be a deficiency (imbalance) with regard to the autosomal genes.
The second derivative chromosome would have a piece of X chromosome on the autosomal fragment;
this piece of X chromosome might not be inactivated, since it is removed from the X inactivation
center. Hence one would not be achieving dosage compensation with regard to those X-linked genes. QUESTION 18 (8 minutes) Osteogenesis imperfecta type I (OI) is an autosomal dominant that causes a generalized decrease in bone
mass (osteopenia) and makes the bones brittle. The disease is often associated with blue sclerae, dental
abnormalities and progressive hearing loss.
The gene for OI has been cloned and mapped to chromosome 22. However, each family usually has a
unique mutation, which would require the sequencing and analysis of more than 5000 base pairs in each of
the two genes to identify the exact mutation. For these reasons, the strategy of choice for genetic
diagnosis is to trace the inheritance of the disease allele using a RFLP close to the locus. RFLP SC3 is
known to map to the q arm of the chromosome 22. It has four haplotypes, A, B, C and D. It is
hypothesized that OI is linked to this RFLP. Use the data from this family to answer the following
BD CD II
DD Part a B
I BC AA AB AC C
AB AC C
i NR NR Part b B
i R B
R AC C
i NR AB B
I NR AC C
i NR a. What is the linkage phase of II-2? (2 pts)
b. Write the general Lod score expression for the family of II-2 x II-3. (2pts for determining
recombinants versus nonrecombinant progeny. 2pts for writing the general lod expression.)
z= log10 (1-θ)5(θ)2
c. The following table is the cumulative Lod values of all the families in the study. State explicitly
what the Lod scores indicate about the possibility of linkage between the RFLP and OI. Explain
0.02 0.05 0.07 0.10 0.12 0.15 0.20 0.30 0.40
Lod 3.126 4.000 3.868 3.487 2.673 2.262 1.764 1.566 1.223
The maximum lod score occurs at θ = 0.05; a lod score of 4.0 indicates that there is a 10,000:1 chance
of θ = 0.05 is the best estimate for the recombination frequency. By convention genes are considered
linked at the highest lod value greater than or equal to 3. QUESTION 19 (12 minutes) Wilson’s disease is a rare autosomal recessive disorder of copper metabolism. The impairment of the
normal excretion of copper over time results in toxic accumulations of the metal in liver, brain and
other organs. III-1 and III-2 present for genetic counseling.
Normal a. What is the probability that their first child will have Wilson’s disease?
W= normal; w = Wilson’s disease
III-1 is not affected with the disease so he received a normal allele from his mother and a w from his father who has the
disease. So, III-1 is Ww. III-2 is not affected with the disease so she received a normal allele from her father and a w from
her mother who has the disease. So III-2 is Ww. The probability that their first child will be affected is:
P(III-is Ww) x P(III-2 is Ww) x P( a ww child is produced)
(1) x x (1/4) = (1) ¼ (2 pts) Since many mutant forms of the normal gene have been identified, linkage analysis is the technique of
choice for calculating the probability of an affected child. The locus for Wilson’s disease has been
mapped to 13q14. An RFLP is located 4 cM from the locus causing Wilson’s disease. The RFLP
contains two variable EcoRI sites labeled E1 and E2. Invariant sites (restriction sites that are always
present) are labeled E. The following diagram shows the region of genomic DNA with homology to
the probe. The results of a Southern blot of the family’s genomic DNA are provided below.
Map of the RFLP located 4cM from Wilson's's Disease
10Kb 6 Kb 5 kb probeJ1
E E1 E2 E Haplotype
D E1 E2
- Band Size in KB
unique to that
haplotype. (+ = enzyme cuts; - = enzyme cannot cut) b. Using the restriction map of the RFLP and the data from the Southern blot, determine the RFLP
haplotypes for the individuals listed on the Southern blot. (3 points)
II-1 BC BC DD AC CD CD II-2 II-4 II-5 III-1 III-2 IV-1 D D C C DD
B C markers
6.0 Use the unique
bands to set each
haplotype. c. What is the probability that IV-1 will have the disease?
Wilson's Disease AA WW
A C W w III-2’s
gametes C 0.48
gametes C D w CD W D 0.48 W w Normal DD WW BC ww 0.02 C
W D 0.02 w A A C A D A C A D W W w W W W W W w C C C C D C C C D w 0.48 w w w W w W w w 0.48 0.02 C C C C D C C C D W 0.02 W w W W W W W w A A C A D A C A D w w w w W w W w w We know that IV-1 is CD,
so one of the yellow
squares happened. The
chance of having Wilson’s
Disease and being CD is
P(CD and ww)/ P(CD)=
w w CD
(0.48)(0.02) = 0.0384
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This note was uploaded on 04/19/2009 for the course BIMM BIMM 110 taught by Professor Mcginnis during the Spring '09 term at UCSD.
- Spring '09