BIMM110 Human Molecular Genetics Spring 2004
Immo E. Scheffler
There are 15 questions. All answers are to be written into the Blue Book. Please leave the first
page blank for scoring. The number of points for each question is proportional to the number of
minutes recommended. Please make sure that the question number is clearly visible at the
beginning of each answer.
"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)
QUESTION 1 (6 minutes)
A 1.5 kb piece of DNA has been cloned from a region very close to or even within a suspected
gene. What are some of the criteria (experiments, data) that one would use to confirm that it is
indeed part of a coding sequence for a protein?
- look for an open reading frame (ORF), a series of codons uninterrupted by a stop codon;
the length is a subjective judgement
- do a “Zoo blot” with genomic DNA from different species; exon sequences tend to be
conserved, and one will observe hybridizations of the human probe with DNA from
closely and even not so closely related species
- use it as a probe on Northern blots with RNA from various tissues; if it hybridizes to a
specific mRNA it is from an exon of a gene
- if the mRNA is found in a specific tissue but not in other tissues one may already have a
clue about the role/involvement of this gene/mRNA in the expression of the disease
QUESTION 2 (10 minutes)
The following terms are associated with the phenomenon of X-inactivation in females:
dosage compensation, XIST RNA, Barr body, late replication, and hypermethylation
a) Give a brief description of each of these, indicating how these terms are mechanistically
Females have two X chromosomes. It is now clear that most of the genes on one X chromosome
are silenced to achieve dosage compensation, i.e. transcripts (mRNA) from X-linked genes
should be equivalent in XX females and XY males. In eutherians the paternal or maternal X
chromosome are inactivated randomly at an early stage in embryonic development. The inactive
X chromosome is visible in interphase cells as the Barr body, a condensed and highly stained
chromosome located near the nuclear periphery. There is no transcription from the majority of
the genes on the inactive X, but one exception is the XIST gene which is transcribed into XIST
RNA. This XIST RNA is not exported from the nucleus, it does not encode any protein, and it is
associated with the inactive X/Barr body, i.e. it acts in cis on the chromosome from which it was
transcribed. In an unknown way it initiates hypermethylation and inactive (hetero-) chromatin
QUESTION 3 (10 minutes)
Normal females have the karyotype 46, XX. However, individuals have been discovered with the
46, XX karyotype who nevertheless have some of the external physical characteristics of males
(pseudohermaphrodites). Without going into all the technical details, explain how these
observations can be explained, and how the findings and interpretations have contributed
significantly to the mapping and cloning of the gene responsible for testis determination
(TDF/SRY) in early embryonic development.
The X and Y chromosomes in male meiosis become paired and form a synaptonemal complex
between two short regions (the pseudo-autosomal regions) at the tips of the short arms of the X
and Y chromosomes. Recombination and chiasma formation takes place within the PARs, and
this mechanism is necessary for proper alignment of the two non-homologous chromosomes at
the metaphase plate of meiosis I. The alignment of the two chromosomes extends beyond the
PARs in the direction of the centromeres, and occasionally an illegitimate recombination event
occurs between the non-homologous regions of the X and Y chromosome. As a result, the tip of
the Y is translocated to the X and the tip of the X is translocated to the Y. Cytologically the
chromosomes are not distinguishable, but one can find XX males in which the tip of one “X” is
from the Y chromosome including the gene that encodes the testis-determining factor
(TDF/SRY). Conversely, one can find XY females in which the Y chromosome is missing the
These observations gave a strong hint that the TDF/SRY gene was located on the short arm of
the Y very close to the PAR. Focusing on this short region (defined ultimately by the XX male
with the shortest piece of the Y) the cloning of the TDF/SRY gene was accomplished
QUESTION 4 (14 minutes)
a. What is the distinction between the genetic linkage map and the physical map? What are
the units of measurement? 
A genetic map measures distances between genes in centimorgans, where 1 cM represents a
recombination frequency of 1% between the two genes
A physical map measures the distance between genes in base pairs or kilo base pairs (kb).
There is a rough proportionality between physical distance and recombination frequency; 1
cM corresponds roughly to 1 Mb (106 basepairs), but while the physical distance is the same
in males and females, recombination frequencies typically differ in oogenesis and
b. What is a speculative explanation that the genetic map is apparently longer when measured in
females compared to males? 
One hypothesis is that in females meiosis I is arrested for years, and completed only at the time
of ovulation, one oocyte at a time. Thus there appears to be a longer time in females for
chromosomes to become aligned to form the synaptonemal complex and to undergo
recombination. A higher recombination frequency is interpreted as a longer genetic distance
c. The genetic distance (θ) between two loci at the extremes of the pseudoautosomal region of
sex chromosomes appears to be very large when measured in male spermatogenesis. Can you
explain briefly? 
There is one obligatory chiasma/recombination event within the PAR during spermatogenesis
(meiosis I). Thus, the loci at the extreme ends of the PAR appear to segregate randomly, or with
a recombination frequency of 50 cM.
In females chiasmata can form along the entire X chromosomes, and chiasma formation in the
PAR occurs at a significantly lower grequency (genetic size ~ 5 cM)
d. What is the estimated approximate size of a human chromosome in which on average three
chiasmata are observed per male meiosis? 
One chiasma represents a distance of 50 cM. Hence, three chiasmata observed suggest a genetic
length of 150 cM, and we can estimate the physical length to be ~ 150 Mb.
QUESTION 5 (10 minutes)
The following table contains a series of Lod scores for families in which linkage between two
loci is suspected.
Recombination fraction (θ)
0.00 0.05 0.10 0.15 0.20 0.25 0.30
Family I 0.602 0.558 0.511 0.460 0.408 0.352 0.292
Family II 1.505 1.395 1.275 1.150 1.020 0.880 0.730
Family III 1.204 1.116 1.020 0.920 0.816 0.704 0.584
Family IV -∞ - 0.116 0.321 0.397 0.418 0.403 0.362
a) Assuming that initially only the first three families (I – III) were studied, what would you
conclude about the linkage or non-linkage of these two loci; explain your answer briefly.
The highest value of z (lod score) in the first three families is for θ = 0. One would therefore
conclude that the two loci are in the same position or very close to each other (for example an
RFLP can be within a gene defining a genetic disease)
b) Eventually, a fourth family is found, and the lod scores for this family are given in the
last row. How would that change your interpretation?
When the fourth family is found, the lod score for θ = 0 is negative and very large (∞); therefore,
θ must be greater than zero. With the fourth family alone one would make a best estimate of q to
be ~ 0.2, but the maximum z is less than one, i.e. you would not want to bet on this. If the lod
scores for the four families are combined, one would come up with a lod score of z = 3.127 for
θ = 0.1. The chances are ~1000:1 that θ is in this neighborhood.
QUESTION 6 (20 minutes)
Somatic cell genetics is an approach that treats mammalian cells in tissue culture like
microorganisms. Certain mutant phenotypes can be selected for and para-sexual experiments can
be performed. Two cell lines are available to you:
a mouse cell line with the phenotype TK (BudRR) [TK - thymidine kinase]
a human cell line from a Lesh-Nyhan patient with the phenotype HPRT (TGR)
[HPRT - hypoxanthanthine phosphoribosyl transferase]
a) The two enzymes are referred to as salvage enzymes; explain briefly with a reaction or at least
in words 
Thymidine kinase converts thymidine (the nucleoside) to TMP (a nucleotide) using ATP as the
HPRT catalyzes the reaction between hypoxanthine (a base) and phophoribosyl pyrophosphate to
form a nucleotide.
These salvage enzymes can salve nucleosides and bases and they can sometimes substitute for
the de novo formation of these nucleotides
b) The two cells can be fused and hybrid cells can be selected. What is the selective medium
used in this case? (spell out the abbreviations normally used) 
The selective medium is HAT medium: it contains hypoxanthine, aminopterin and thymidine.
The aminopterin is a folate analogue that inhibits the de novo reactions; hence both salvage
enzymes and the substrates are required for cells to survive in this medium.
c) What chromosomes would you expect to find in the hybrid cells after some time of selection?
Which human gene would be mapped specifically in this case? 
In such a hybrid we would expect to find all mouse chromosomes and some small subset of
human chromosomes, with different hybrids containing a different subset. The one human
chromosome that would always be present when the HAT selection is made is the chromosome
with the TK gene (chromosome 17).
d) What is a hybrid panel, and how can it be used to map genes (very general description) 
A hybrid panel is a collection of different hybrids from human and rodent cells. Each hybrid cell
line will have a distinct subset of a small number of human chromosomes; the complete panel
will have each human chromosome represented at least once in the collection, and no single
human chromosome should be present in all the hybrids.
Each hybrid cell line can be examined for the presence of the human gene of interest (or the
human protein, if it can be distinguished from the rodent version), and a correlation will be made
between the presence of the gene and the presence of a particular human chromosome.
QUESTION 7 (6 minutes)
Describe in outline the procedure referred to as FISH (fluorescence in situ hybridization) and
indicate what type of information one can obtain from it.
A metaphase spread of chromosomes is made on a microscope slide. They are denatured by a
brief exposure to alkaline pH, but remain stuck to the slide. A specific probe is made
representing a unique DNA sequence using deoxy nucleotide triphosphates of which one has a
base modified with a chromophore (fluorescent molecule) attached; this chromophore does not
interfere with the incorporation of the nucleotide into the DNA probe. The slide may be pre-
hybridized with highly repeated human DNA to react with all the repeated sequences before the
slide is hybridized with the fluorescent probe. The slide is then washed thoroughly to remove
excess probe and examined under a fluorescence microscope. Two fluorescent dots should be
seen on each of the two homologous chromosomes. Each dot represents the target gene on a
We learn about the location of the gene on a specific chromosome and roughly on its position on
the chromosome (which arm, which region, which band).
QUESTION 8 (18 minutes)
The fragile X syndrome has been a puzzle to geneticists for over two decades, because of
some unusual features in its pattern of inheritance, which did not fit the conventional pattern
for X-linked recessive or dominant mutations.
The above pedigree includes a non-penetrant transmitting male (NTM) in generation I, several
affected individuals in generation III (black (dark) symbols), normal individuals in every
generation (no fill), and some others in generations II and III (light shade) that show no
symptoms of the disease.
A PCR-based analysis is to be performed on selected individuals to establish their genotype.
a) describe the essential information and materials needed for the analysis by PCR (no need to
describe the PCR procedure in any detail) 
One would need to know the unique sequences of the DNA flanking the CGG repeat in the
FMR1 gene to design oligonucleotides for PCR.
PCR would be used to amplify the DNA segment including the CGG repeat using as template
DNA from each individual of interest.
The size of the PCR product will reflect the level of amplification of the CGG repeat. One
product would be seen from male genomic DNA, while two products might be expected from the
females (one from each allele on two X chromosomes; these alleles are frequently polymorphic
even in normal individuals)
A molecular weight (length) calibration of the gel electrophoresis of the PCR fragments would
be used to estimate the number of repeats in each amplified fragment.
b) What would you be looking for in interpreting the PCR results? Rationalize or explain. 
One would look at the repeat number for the CGG trinucleotide repeat on the X chromosomes of
the various members in the pedigree.
c) Discuss your findings for individuals I-1, I-2, II-2, III-1, III-4 and III-5. In your discussion
you should introduce and explain the terms “pre-mutation” and “full mutation” and how they
relate to your molecular data. 
Let n be the number of repeats in a given fragment. A normal allele has less than 50 repeats; the
pre-mutation allele has between 50 and 200 repeats; the full mutation allele has >>200 repeats.
Expectations are as follows:
I-1: a normal female with two alleles with n less than 50, but the fragments might be slightly
I-2: a normal transmitting male would be expected to have the pre-mutation with 50<n<200
II-2: a normal carrier female with two alleles; n < 50, and n between 50 and 200
III-1: an affected female with one normal allele (n < 50), and a second allele with n >> 200
III-4: a carrier female similar to II-2
III-5: a fragile X male with one allele with n >> 200
d) Explain the difference between females such as II-2 and III-1 with regard to phenotype, and
with regard to their risk of having affected sons. 
II-2 is an unaffected carrier with a high risk if having affected sons
III-3 is an affected female (mild symptoms) with a 50% chance of having affected sons.
QUESTION 9 (12 minutes)
Duchenne muscular dystrophy (DMD) and cystic fibrosis (CF) are very common genetic
diseases due to single gene defects. Genetic counseling is strongly indicated, especially when
there is already an affected child in the family or among close relatives. The relevant genes have
been cloned, and carriers can in principle be identified by molecular-genetic analyses.
For each of these diseases discuss some of the potential technical problems/obstacles that make
carrier identification less than a standard routine.
(Hint: size of the genes, the different nature of the mutations encountered in each gene; cost
The dystrophin gene is exceptionally large (~ 2.5 Mb) and it has a large number of exons as well
as a very large transcript (mRNA) of ~ 15 kb. Sequencing all the exons would be tedious and
expensive. Many mutations in this gene involve deletions or insertions due to unequal cross
overs between imperfect repeats (encoding amino acid sequence repeats of 109 aa). Thus, a
Southern analysis with a series of cDNA probes would be a good start to analyze the gene. With
a given restriction enzyme one would establish a pattern of bands from a normal individual, and
this would then be compared with the bands obtained from the suspected carrier; she would have
one normal X and one with the mutation.
One should be aware that ~ 1/3 of all DMD males result from a spontaneous new mutation
during oogenesis in their mothers, and a female with an affected son may not necessarily be a
The gene for the CFTR is also large (~250 kb) with many introns and a transcript of ~ 6.5 kb.
Most of the mutations that have been found in the human population are point mutations (e.g.
premature stop codons) or very small deletions, and a Southern analysis would not be likely to
reveal any mutations, unless one was very lucky and the mutation was precisely within the
restriction site for the chosen restriction enzyme.
The only option is sequencing. If one knows something about the various mutations and
mutation frequencies in different ethnic groups, for example the ∆Phe508 mutation found in 70%
of Caucasian carriers, one could focus the sequencing on a few selected exons, but there is
always the risk of missing a mutation. This risk, and the associated fear of litigation ahs made
carrier testing for CF problematical at the beginning.
More recently very promising approaches for rapid (and cheap?) sequencing of specific genes
with micro-array technology have been developed.
QUESTION 10…. (15 minutes)
"Mitochondrial diseases" have become recognized in the past decade, originating from
mutations in the mitochondrial genome. They are heritable.
a. Explain the term "non-mendelian inheritance" with the help of a diagram
(pedigree) covering several generations. 
The mutation is passed only through the maternal lineage, since all mtDNA is derived from
There should be a three generation pedigree shown with maternal inheritance, i.e. any
mutation would be passed through the maternal lineage.
b. Explain the terms homoplasmy and heteroplasmy, and relate such findings to
an explanation for the variable severity of mitochondrial diseases. 
In most individuals all mtDNA are identical in sequence - homoplasmy; in individuals with
mitochondrial diseases we often find two populations of mtDNA in a cell: one with the
normal wild type sequence and another with a specific mutation/deletion - heteroplasmy.
It is found that the proportion of mutated mtDNA is roughly correlated with the symptoms
and the severity of the disease
c. Genetic counseling of families with children suffering from mitochondrial
disease is difficult. Even if a mutation in the mtDNA of the unaffected mother has been
established, predictions about the outcome of another pregnancy are highly uncertain.
Can you explain briefly? 
When a female is found to be heteroplasmic for a mitochondrial DNA mutation, she may
have only a low proportion of mutated mtDNAs and have no symptoms. She would
have been identified because she had a child with a mitochondrial disease and
heteroplasmy with a relatively high proportion of mutated mtDNA. The outcome of
another pregnancy will be difficult to predict precisely, since the heteroplasmy in the
child is difficult to predict. This is due to the bottleneck effect.
QUESTION 11 (10 minutes)
We have encountered several examples of epigenetic phenomena in this course. Give a definition
of this term, and cite four examples and distinguish between normal and potentially pathological
Epigenetic changes are responsible for alterations in gene expression that are not due to changes
in the DNA sequence. Examples:
1. X-inactivation: a normal process of silencing most of the second X chromosome in females for
2. Imprinting of selected autosomal genes: differential silencing of a gene on either the maternal
or paternal chromosome, - a normal and essential process.
3. Abnormal hypermethylation of a tumor suppressor gene that silences it and is one mechanism
in the progression of tumorigenesis.
4. Silencing of the FMR1 gene in fragile X syndrome.
QUESTION 12 (20 minutes)
The XIST RNA is not expressed from the single active X chromosome in a normal male. The
XIST“gene” is postulated to be silenced by hypermethylation in its promoter region.
We now have the entire sequence for this region, and hence we can predict restriction sites and
design oligonucleotide primers as required. The necessary probes are also easily made.
a) Describe how you would use restriction enzymes and Southern blotting to establish that a
relevant segment of the gene is hypermethylated. (Note: MspI and HpaII recognize the
CCGG sequence, but HpaII is inactive when cytosines in this sequemce are methylated).
Draw a highly schematic diagram indicating the appearance of Southern blots for various
combinations of restriction enzymes. The diagram should be accompanied with a brief
explanation and be clearly labeled. 
- genomic DNA from an XY male is isolated
- the hypermethylated promoter region is identified and a restriction site for MspI/HpaII is
- restriction sites for one or more restriction enzymes flanking the promoter region are selected;
these enzymes cut outside of the methylated region; for example the enzyme EcoRI is found to
- two aliquots of DNA would be digested with EcoRI, and either with HpaII or with MspI
- the digest is fragmented by electrophoresis and the fragments are transferred to paper by
- radioactive probe from the region of interest is used to detect the fragments
In the simplest case one would observe two fragments with the EcoRI/MspI combination, and
one EcoRI fragment with the EcoRI/HpaII combination of restriction enzymes
b) An alternative approach would be to use PCR in the analysis. Primers flanking the
hypermethylated segment are available for PCR. One student uses genomic DNA from a
male, carries out the PCR amplification and then tests whether his PCR product can be
cut with HpaII or MspI. Another student takes the genomic DNA, divides it into two
samples, cuts one with MspI and the other with HpaII, and then carries out PCR
amplification with each separate sample as template.
Which of the two students would you expect to obtain a result that could confirm
hypermethylation of the site in question? Explain briefly why one student is wrong in his/her
The second student would find that he would obtain the expected PCR product after digesting the
genomic DNA with HpaII but before carrying out the PCR; if the template DNA is first cut with
MspI it would be cut between the locations for the two primers, and no PCR product would be
Testing the PCR product with the MspI/HpaII combination of ezymes would be meaningless,
since the PCR does not utilize methylated dCTPs and all the products are unmethylated,
regardless of the status of the template DNA.
QUESTION 13 (12 minutes)
a) Explain the difference between an oncogene and the corresponding proto-oncogene.Why
is the oncogene dominant in a cell? Give one example.
the proto-oncogene is a normal and essential gene for the normal functioning of a cell; an
oncogene can be derived from such a gene by various paths
- a mutation that prevents down-regulation of the activity, e.g a mutation that prevents GTP
hydrolysis in the RAS gene
- a translocation that places the coding sequence of a proto-oncogene next to a strong promoter
or enhancer; such translocations may also yield chimeric proteins in which a truncated form of a
protein has abnormally high activity, e.g. the tyosine kinase of c-abl.
b) In contrast, mutations in tumor suppressor genes are recessive. Explain briefly and give
- an example of a tumor suppressor gene is the retinoblastoma gene (RB).
- in heterozygous individuals/cells 50% of a tumor suppressor activity is generally sufficient
QUESTION 14 (12 minutes)
Familial cancers, for example retinoblastoma, are frequently due to recessive mutations that
cause a “predisposition towards cancer”. Explain with the help of schematic diagrams three
mechanism leading to "loss of heterozygosity" and a step forward in tumorigenesis.
1. mitotic nondisjunction: loss of the chromosome carrying the wild type allele
2. mitotic crossing over: abnormal crossing over between non-sister chromatids of homologous
chromosomes between the centromere and the RB gene
3. loss of the wild type allele by an independent mutation or small interstitial deletion.
QUESTION 15 (4 minutes)
Your grandmother asks you to explain to her the molecular and genetic basis of "maple syrup
urine disease" and you don't have a clue. However, you have learned in this course about some
authoritative sources for information on genetic diseases. Where would you look (other than
Google, although Google would probably lead you there, too) ?
the best source is a website OMIM, Online Mendelian Inheritance in Man, now operated by the
National Library of Medicine