Download (Students with questions should see the appropriate Professor)

(Students with questions should see the appropriate Professor)
FINAL EXAMINATION – April 2004 – Biology 202
Prof. Schoen’s Questions – 7 points
1. (1 point) Humans with the genotypes DD and Dd show the Rh+ blood phenotype,
whereas those with the genotype dd show the Rh- blood phenotype. In a sample of
400 Basques from Spain, 230 people were Rh+ and 170 people were Rh-. Assuming
that this population is in Hardy-Weinberg proportions, what is the allele frequency of
the allele D?
(a)
(b)
(c)
(d)
(e)
(a)
(b)
(c)
(d)
(e)
0.348
0.652
0.425
0.575
0.288
2. (2 points) In the Basque population mentioned above, what proportion of the Rh+
individuals would be expected to be heterozygote?
(a)
(b)
(c)
(d)
(e)
(a)
(b)
(c)
(d)
(e)
0.454
0.789
0.516
0.250
0.500
3. (2 points) Suppose the genotypes AA, Aa, and aa have frequencies in the zygotes of
0.16, 0.48, and 0.36, respectively. Suppose the relative fitnesses of these genotypes
are 1.0, 0.8, and 0.6, respectively. Selection occurs after the zygote stage but before
mating, and mating is random with respect to this locus. What are the zygote
frequencies in the next generation?
(a)
(b)
(c)
(d)
(e)
(a)
(b)
(c)
(d)
(e)
0.16, 0.48, 0.36, respectively.
0.16, 0.384, 0.216, respectively.
0.2105, 0.5053, 0.2842, respectively.
0.2145, 0.4973, 0.2882, respectively.
0.25, 0.50, 0.25, respectively.
4. (1 point) How can we explain the fact that the frequency of a recessive deleterious
allele (e.g., such as one that causes a severe genetic disease) seems to attain some
finite equilibrium value in a large population. In other words, why has natural
selection not removed the deleterious allele from the population and driven its
frequency down to zero?
(a) (a) Genetic drift prevents this from happening, because in large populations, the
deleterious allele would behave as if it was selectively neutral.
(b) (b) Selection cannot act to remove a recessive deleterious allele, because all the
deleterious alleles become “hidden” from selection in the heterozygotes.
(c) (c) An equilibrium is reached between the removal of the deleterious allele by
selection and its re-introduction into the population by the process of mutation.
(d) (d) Only in populations undergoing drastic reductions in size is it possible for
genetic drift to bring about the removal of deleterious alleles.
(e) (e) Such alleles are in Hardy-Weinberg equilibrium, and so their frequency is
constant.
5. (1 point) How does the neutral theory of molecular evolution explain the observation
that nucleotide variation among individuals in a population tends to be lower for genes
that code for a very important protein, such as cytochrome c (a protein involved in
cellular respiration), compared to genes that code for a less important protein, such as
fibrinopeptide (involved in blood coagulation)?
(a) (a) The theory predicts that the cytochrome c gene should undergo fewer
mutations than the fibrinopeptide, and therefore, exhibit less variation than
fibrinopeptide.
(b) (b) The theory predicts there should be many more alleles of fibrinopeptide than
of cytochrome c.
(c) (c) The theory predicts that heterozygosity levels should be lower for cytochrome
c than for fibrinopeptide.
(d) (d) The theory predicts that there are more amino acid residues in cytochrome c,
which if altered by an underlying DNA mutation, would lead to impaired protein
function (compared with the number of such residues in fibrinopeptide).
(e) (e) The theory predicts that the mutation rate varies over time more so in
cytochrome c than in fibrinopeptide.
Prof. Lasko’s Questions – 32 points
6. (1 point) A nucleic acid with a base composition of 22% adenine, 20% cytosine, 30%
uracil, and 28% guanine is:
(a)
(b)
(c)
(d)
(e)
double-stranded DNA
double-stranded RNA
single-stranded DNA
single-stranded RNA
Z-DNA
7. (1 point) The pyrimidines found in RNA are:
(a)
(b)
(c)
(d)
(e)
adenine and guanine
adenine and uracil
cytosine and uracil
guanine and thymine
cytosine and thymine
8. (1 point) Which one of the following statements about eukaryotic DNA replication is
not true?
(a) DNA polymerase delta is involved in the synthesis of the leading strand.
(b) Nucleosomes disassociate into individual histone proteins in advance of the
replication fork, and reassemble only after the replication complex passes.
(c) Eukaryotic chromosomes contain multiple replication origins.
(d) Telomerase contains an RNA template.
(e) DNA polymerase alpha has primase activity.
9. (1 point) Which of the following statements about equilibrium density-gradient
centrifugation is/are true?
(a) This technique can separate DNA molecules of different lengths.
(b) This technique can separate DNA molecules from organisms grown in the
presence of different isotopes of nitrogen.
(c) This technique can separate small ribosomal subunits from large ribosomal
subunits.
(d) Both (b) and (c) are true.
(e) (a), (b) and (c) are all true.
10. (1 point) The eukaryotic enzyme complex that catalyzes the transcription of proteincoding genes is:
(a)
(b)
(c)
(d)
(e)
DNA polymerase alpha
RNA polymerase I
RNA polymerase II
Eukaryotic initiation factor 4F
Poly(A) binding protein
11. (1 point) The molecule that transmits information, stored in the DNA of the
nucleus, to the ribosomes in the cytoplasm where it is read is called:
(a)
(b)
(c)
(d)
(e)
tRNA
rRNA
mRNA
snRNA
hnRNA
12. (1 point) A single base pair change in a gene that results in a single amino acid
substitution in the corresponding protein is called:
(a)
(b)
(c)
(d)
(e)
a nonsense mutation
a missense mutation
a conditional mutation
a frameshift mutation
a knockout mutation
13. (1 point) If a human cannot repair thymine dimers, which disease will occur?
(a)
(b)
(c)
(d)
(e)
progeria
xeroderma pigmentosum
alkaptonuria
Tay-Sachs’ disease
Cockayne syndrome
14. (1 point) Which of the following statements about Drosophila development is true?
(a) Genes on the Y chromosome determine male sexual identity.
(b) Sex determination is largely regulated through sex-specific alternative splicing
of the primary transcripts of one or more regulatory genes.
(c) The products of the gap genes are RNA binding proteins involved in alternative
splicing.
(d) After fertilization, nuclear divisions are coupled with cell divisions, so that the
cytoplasm inherited from the egg is subdivided into smaller and smaller cells.
(e) Segment-polarity genes specify the dorsal-ventral axis of the embryo.
15. (1 point) Which is the correct sequence of activation of expression of segmentation
genes in Drosophila?
(a)
(b)
(c)
(d)
(e)
maternal -> gap genes -> pair-rule genes -> segment-polarity genes
maternal -> pair-rule genes -> segment-polarity genes –> gap genes
maternal -> gap genes -> segment-polarity genes -> pair-rule genes
maternal -> segment-polarity genes -> gap genes -> pair-rule genes
maternal -> pair-rule genes -> gap genes -> segment-polarity genes
16. (1 point) Acridines usually produce which type of mutations?
(a)
(b)
(c)
(d)
(e)
frameshift mutations
transversion mutations
transition mutations
missense mutations
dominant mutations
17. (1 point) A researcher is planning to collect mutations in Drosophila gap genes.
What phenotype should the researcher look for?
(a)
(b)
(c)
(d)
(e)
maternal-effect lethal, deletions of several segments
maternal-effect lethal, holes in the cuticle
recessive lethal, deletions of several segments
recessive lethal, holes in the cuticle
recessive lethal, loss of pole cells
18. (2 points) Complex behavioral traits in humans are determined by a combination of
environmental and genetic factors. Analysis of monozygotic (identical) and
dizygotic (fraternal) twin pairs is valuable for determining the relative contributions
of environmental and genetic factors. Concordance is measured as the percentage of
the time that a particular trait shown by one twin will also be shown by the other. In
this context, which of the following statements is/are true?(everyone given credit
for this question)
(a) If a trait has nearly 100% concordance among monozygotic twins, and nearly
50% concordance among dizygotic twins, its causes are almost entirely genetic.
(b) If a trait has nearly 100% concordance among both monozygotic and dizygotic
twins, its causes are almost entirely environmental.
(c) If for a given trait the concordance is much higher among monozygotic twins
than among dizygotic twins, then there is a substantial genetic influence on its
determination.
(d) (a) and (c) are both true.
(e) (a), (b), and (c) are all true.
19. (2 points) Srb and Horowitz isolated several Neurospora mutants that were
auxotrophic for arginine. The mutants were individually tested for their ability to
grow in the presence of compounds thought to be intermediate in the biochemical
pathway leading to arginine. Mutant 1 grew if ornithine, citrulline, or arginine was
added to the media. Mutant 2 grew if citrulline or arginine, but not ornithine was
added to the media. Mutant 3 grew if arginine was added to the media, but did not
grow if either citrulline or ornithine was added. Enzyme 1 is produced by the gene
mutated in Mutant 1, enzyme 2 is produced by the gene mutated in Mutant 2, and
enzyme 3 is produced by the gene mutated in Mutant 3.
From analysis of the above data what is the metabolic pathway leading to arginine,
and which enzymes catalyze each reaction?
(a) precursor –-enzyme 1--> citrulline –enzyme 2 --> ornithine –enzyme 3-->
arginine
(b) precursor –enzyme 1--> ornithine –enzyme 2--> citrulline –enzyme 3-->
arginine
(c) precursor –-enzyme 2--> citrulline –enzyme 3 --> ornithine –enzyme 1-->
arginine
(d) precursor –enzyme 2--> ornithine –enzyme 1--> citrulline –enzyme 3-->
arginine
(e) precursor –enzyme 3--> ornithine –enzyme 2--> citrulline –enzyme 1-->
arginine
20. (2 points) Which of the following statements about anterior-posterior patterning in
Drosophila is/are true?
(a) nanos RNA is localized to the posterior pole of the oocyte, dependent on
sequences in its 3’ UTR.
(b) Translation of nanos RNA is repressed in positions other than the posterior pole
of the oocyte, dependent on sequences in its 3’ UTR.
(c) Nanos protein is involved in translational repression of specific mRNAs.
(d) (a) and (b) are both true.
(e) (a), (b), and (c) are all true.
21. (2 points) Which of the following statements about dosage compensation is/are not
true?
(a) Dosage compensation serves to equalize X-linked gene expression in males and
females.
(b) Dosage compensation always involves reducing X-linked gene expression in
XX individuals.
(c) Noncoding X-chromosome binding RNAs are involved in dosage compensation
in mammals and Drosophila.
(d) The Drosophila roX1 and roX2 RNAs are expressed only in males, and are
involved in dosage compensation.
(e) (b) and (d) are both not true.
22. (2 points) Temperature-sensitive mutations are often valuable for geneticists. You
wish to isolate a random collection of temperature-sensitive mutants in yeast to
study metabolic pathways. Which would be the best mutagenic agent to use to do
this?
(a)
(b)
(c)
(d)
(e)
gamma-irradiation
proflavin
ethyl ethane sulfonate
X-irradiation
Kool-Aid
23. (2 points) Of the steps of prokaryotic translation initiation listed below, which
occurs earliest?
(a) recruitment of the initiator tRNA to the ribosome
(b) formation of the IF-2/initiator tRNA complex
(c) formation of the 30S initiation complex
(d) recruitment of the large ribosomal subunit to the mRNA
(e) positioning of the initiator tRNA at the start codon of the mRNA
24. (2 points) What would be the phenotype of a gain-of-function Sex-lethal mutation
in Drosophila?
(a)
(b)
(c)
(d)
(e)
XX flies would develop as males.
XX flies would die, XY flies would develop as normal males.
XY flies would die, XX flies would develop as normal females.
XY flies would develop as females.
None of the above.
25. (2 points) Secondary structure of a protein involves…
(a)
(b)
(c)
(d)
(e)
the spatial interrelationships of the amino acids in segments of the polypeptide
structures such as alpha-helices and beta-sheets
the association of two or more polypeptides in a multimeric protein
(a) and (b)
(a), (b), and (c).
26. (2 points) Which of the following statements about genes is/are not true?
(a) Enhancer elements are sometimes located within an intron.
(b) Enhancer elements are sometimes located 3’ to the transcribed region of the
gene.
(c) Promoter elements are sometimes located 3’ to the transcribed region of the
gene.
(d) Enhancer elements confer tissue-specificity to gene expression.
(e) (a) and (c).
27. (2 points) Which of the following statements about eukaryotic transcription and
RNA processing is/are true?
(a) The 5’-most residue of the primary transcript is always a G, transcribed from a
C on the template strand of the DNA, and modified with a methyl group added
enzymatically to its 7-nitrogen.
(b) The 5’ most residue of an intron is always a G, transcribed from a C on the
template strand of the DNA.
(c) Poly(A) tracts are present at the 3’ end of eukaryotic mRNAs because they are
added enzymatically after the primary transcript has been cleaved.
(d) (b) and (c)
(e) (a), (b), and (c).
Prof. Chevrette’s Questions – 26 points
28. (2 points) Please read carefully the following statements:
1) Using either G (Giemsa) or R (reverse) banding, the 23 pairs of human
chromosomes can be identified in interphase cells obtained from normal human
cells.
2) In human, only the trisomy of either human chromosome 13, 18 and 21 can
produce viable individuals.
3) Endomitosis has never been detected in human cells.
4) If non-disjunction of chromosomes 18 occurs during the first meiotic division of
a human gamete cell, after meiosis is fully completed, there will be two cells
containing one chromosome 18, one cell containing two chromosomes 18 and
one cell without any human chromosome 18.
5) Human males affected by the Klinefelter syndrome will have additional “X”
chromosome(s), while human females affected by the Turner syndrome will lack
one chromosome “X”.
Based on the previous statements, which one of the following analysis is RIGHT?
(a) Statements 1) , 2) and 5) are right, while statements 3) and 4) are false.
(b) Statements 2) and 5) are right, while statements 1), 3) and 4) are false.
(c) Only statement 5) is right, all others are false.
(d) Statements 1) , 3) and 5) are right, while statements 2) and 4) are false.
(e) All statements are false.
29. (2 points) A plant species A, which has 7 chromosomes in its gametes, was crossed
with a related species B, which has 13 chromosomes in its gametes. The hybrids
were sterile, and microscopic observation of their pollen mother cells showed
chromosome pairing of three individual pairs of chromosomes. A section from one
of the hybrids that grew vigorously was propagated vegetatively, producing a
healthy plant.
Which one of the following answers is more likely to describe the cells of this new
healthy plant?
(a)
(b)
(c)
(d)
(e)
There will be 20 chromosomes in its somatic cells.
There will be 11 chromosomes in the gametes of this plant.
There will be 38 chromosomes in its somatic cells.
There will be 34 chromosome in its somatic cells.
There will be 32 chromosomes in its somatic cells.
30. (2 points) Please read carefully the following statements:
1)
2)
1)
Upon separation on a cesium chloride gradient, the minisatellites
containing CA repeats will be separated from the rest of genomic DNA.
2)
PCR amplification and analysis of the VNTR sequence using primers
complementary to the VNTR sequence has replaced the need of using Southern
blot to detect the VNTRs
3)
3)
Mouse satellite DNAs are located near the centromere of the mouse
chromosomes.
4)
During renaturation kinetics studies, the Cot value for a 400 bp fragment
representing a highly repetitive DNA sequence will be lower than the Cot value of
a 400 bp fragment derived from a single copy gene.
5)
In humans, the DNA fingerprinting technique is based on the analysis of
RFLPs of particular single copy genes, and thus allows identification of different
individuals.
4)
5)
Based on the previous statements, which one of the following analysis is RIGHT?
(a)
(b)
(c)
(d)
(e)
Statements 1), 3) and 4) are right, while statements 2) and 5) are false.
Statements 3) and 4) are right, while statements 1), 2) and 5) are false..
All statements are right.
Statements 1), 3) and 5) are right, while statements 2) and 4) are false.
Only statement 2) is false; all others are right.
31. (2 points) Please read carefully the following statements:
1)
2)
3)
4)
5)
DNA polymerase pausing and slippage during DNA replication is the main cause
of minisatellite polymorphism.
2)
Microsatellite polymorphisms are generated by misalignment and
unequal crossing-over.
3)
Although they exist, long DNA deletions are rarely seen in normal
human DNA.
4)
Due to their small length (8 to 10 nucleotides), allele specific
oligonucleotides (ASO) can be used to identify alleles that differ by only one
nucleotide.
5)
RAPD is a PCR technique performed using a pair of two small different
primers and is a rapid way of analyzing polymorphic sequences.
Based on the previous statements, which one of the following analysis is RIGHT?
(a)
(b)
(c)
(d)
(e)
Statements 1), 2) and 5) are right, while statements 3) and 4) are false.
All statement are right.
Statements 1), 2) and 3) are right, while statements 4) and 5) are false.
Statements 1), 2), 3) and 4) are right, while statement 5) is false.
Statement 3) is right, all others are false
32. (1 point) Which of the following statements is incorrect?
(two accepted answers)
(a) When analyzing DNA with RFLPs, the deletion of a 1 kb fragment in the
middle of a five kb EcoRI fragment the detectable with the probe you are
using
will always be seen as a 4 kb fragment on the autoradiogram.
(b) A point mutation occurring in a DNA molecule will be seen as an RFLP only
if this mutation affects the restriction site of a given enzyme.
(c) A complete analysis of the human DNA sequence derived from two
individuals will reveal millions of single base polymorphisms.
(d) The DNA sequence flanking microsatellite DNA must be known in order to be
able to analyze this type of DNA polymorphism by PCR.
(e) The EcoRI digestion of a plasmid containing 3 restriction sites for this enzyme
will generate 4 fragments.
33. (1 point) Which of the following statements is incorrect?
(a) If we could analyze the following DNA sequence by SSCP;
GCATATGC
CGTATACG
after electrophoresis, we will detect two bands.
(b) PCR analysis using ASOs does not necessitate gel electrophoresis.
(c) RFLP analysis can be used even if we do not know the sequence of the
fragment to be analyzed.
(d) Expansion of microsatellite sequences (CAG) is the cause of the Huntington
disease.
(e) RFLP analysis could be used to detect DNA insertion or deletion
34. (1 point) Which of the following statements is incorrect?
(a) A yeast origin of replication, two yeast telomeric sequences, and one selectable
marker (such as URA3+) are needed in a vector to be grown in yeast.
(b) Genomic libraries can be made by generating genomic DNA fragments without
the use of restriction endonucleases.
(c) A cosmid vector is a plasmid containing the cos sites of the bacteriophage ,
and
it can be used to generate cDNA libraries.
(d) Cloning a DNA fragment into the sequence of the -galactosidase gene of a
plasmid containing also the gene for ampicillin resistance and plating the
bacteria on media containing ampicillin and X-Gal will allow selection for the
bacteria that have been transformed with plasmids that contain a foreign piece
of DNA.
(e) The number of clones to be screened to identify a particular gene in a genomic
library depends not only on the size of the genomic DNA that can be inserted in
the cloning vector, but also on the size of the genome of the species.
35. (2 points) Please read carefully the following statements:
1)
2)
1)
Although quite useful, a multiple cloning site (MCS) is not an essential
part of a cloning vector.
2)
A BAC vector can contain large fragments (100 kb to 500 kb) of human
genomic DNA and is a shuttle vector that can grow in bacteria and yeast.
3)
4)
5)
3)
Although cosmid vectors will be packaged inside the head of a lambda
virus, once inside a bacteria, these vectors will behave like plasmids.
4)
A 700 bp radioactive probe can be used in colony hybridization to
identify bacterial colonies that have integrated plasmids containing 2 kb inserts
which have a 500 bp portion that is homologous to the probe.
5)
Because it is derived from the single-stranded genome of the filamentous
M13 phage, phagemid (such as pUC118) will produce only single-stranded DNA
molecules.
Based on the previous statements, which one of the following analysis is RIGHT?
(a)
(b)
(c)
(d)
(e)
Statement 2) is false, all others are right.
All statement are right.
Statements 1), 2) and 3) are right, while statements 4) and 5) are false.
Statements 1), 3) and 5) are right, while statement 2) and 4) are false.
Statements 1), 3) and 4) are right, while statements 2) and 5) are false.
36. (1 point) Which of the following statements is incorrect.
(a) During chromosome walking, a fragment of a clone is used to find clones
containing adjacent and overlapping genomic DNA fragments.
(b) A genetic map of an organism will be expressed in megabases, and will be
obtained with different molecular markers and DNA sequences.
(c) Chromosome jumping cannot be used to detect DNA fragments located on two
different chromosomes.
(d) In microchip hybridization experiments, the oligonucleotides present on the
microchips are similar to ASO molecules.
(e) The generation of single-stranded DNA molecules from a phagemid will
necessitate the collaboration of a helper phage.
37. (1 point) The karyotype of female blood cells from a new rodent exhibiting
similar sex determination to humans, has just been revealed. Analysis of its
chromosomes shows that these cells contain only 8 pairs of chromosomes. Based on
this data, the minimum number of contig that a male from this organism will have
upon full sequence and analysis of its genome will be:
(a)
(b)
(c)
(d)
(e)
8
16
4
9
5
38. (1 point) Which one of the following statements is correct?
(a) Upon hybridization of a fluorescent probe corresponding to a single copy gene
to a normal human chromosome spread, the two chromosomes that contain this
(b)
(c)
(d)
(e)
probe will each always show one fluorescent dot.
Selection of cell hybrids in media containing HAT medium will allow the
survival of cells that have functional HPRT and TK genes.
Two different Alu primers are needed to amplify inter-Alu fragments from
human DNA.
Using a combination of fluorescent dyes, during fluorescent activated
chromosome sorting (FACS) of cells from a child, the human chromosome 21
inherited from the father can be distinguished from the chromosome 21
inherited from the mother.
Microcell fusion is used to generate hybrids that contain different small portions
of the genome of one of the parental cell.
The next two questions are derived from the following observations:
When human cells are cultured in vitro, they become senescent (they stop growing) after
a certain number of cell division. In contrast, mouse cancerous cells never become
senescent.
Using cell fusion a group of scientists lead by Dr. Sue R. Vhiving wants to identify the
gene responsible for the induction of senescence. They have thus made cell hybrids
between normal human cells (senescent) and a mouse melanoma cell line (non-senescent)
and characterized the hybrids they have obtained (these hybrids are called SOS for
Senesce Or Survive). All hybrids have retained most mouse chromosomes.
Hybrids
SOS1
SOS2
SOS3
SOS4
SOS5
SOS6
SOS7
SOS8
Phenotype
no senescence
senescence
senescence
no senescence
senescence
no senescence
senescence
no senescence
Human Chromosome content
1; 2; 3: 5; 6; 16; 18; 19; X; Y
1p; 3; 5; 8; 15; 16q; 20; 21; Y
5; 7; 9; 11; 12; 16; 19; 21
5; 6; 8; 9; 10; 13; 15; 16; 22
2; 4; 5; 7q; t(8p;16q); 10; 18; 20; 21
2; 3; 4; 5; 14; 16; 17; X
1; 3; 4; 5; 8; 10; 16; 17; 18; 20; 21; Y
5; 7; 11; 19; 22; X; Y
39. (2 points) Which one of the following statements is in complete agreement with
these results?
(a) There is only one gene responsible for the induction of senescence and it is
located on human chromosome 21.
(b) These hybrids show that cellular senescence can be induced by a gene or several
genes located on normal human chromosome(s). Although the chromosomal
location of the gene(s) cannot be determined precisely, it cannot be on human
chromosome 5.
(c) There are only two genes responsible for the induction of cell senescence and
they are located on human chromosomes 12 and 20.
(d) There is no selectable marker on human chromosomes in these hybrids.
(e) A selectable marker could be present on human chromosome 5, while a gene
responsible for the induction of cellular senescence is located on human
chromosome 7q.
40. (2 points) Following these results, the same group of scientists has generated more
hybrids. The properties of these new hybrids, all containing most mouse
chromosomes are reported below:
Hybrids
Phenotype
Human chromosome content
SOS9
SOS10
SOS11
SOS12
SOS13
SOS14
SOS15
SOS16
no senescence
senescence
no senescence
no senescence
senescence
no senescence
senescence
no senescence
1: 5: 7; 10; 12; 15; 18; Y
1; 2; 3; 5; 6; 16; 18; 19; 21; X
1p; 3; 5; 8; 15; 16q; 20; Y
1; 2: 7; 9; 11; 12; 16; 19; 20; 21
5; 6; 8; 9; 10; 13; 15; 16; 21
2; 4; 7q; t(8p;16q); 10; 18; 21
2; 3; 4; 5; 14; 16; 17; 21; X
1; 3; 4; 8; 10; 12: 16; 17; 18; 20; 21; Y
From the analysis of all the 16 hybrids obtained by these scientists, what are you able to
conclude?
(a) With these new hybrids, one can conclude that there is only one senescence
inducing gene located on human chromosome 20.
(b) There are only two human chromosome portions that can encode senescence
inducing genes: those are located on human chromosomes 12 and 21.
(c) There are two genes that are needed in combination to induce senescence, they
are
located on human chromosomes 5 and 21.
(d) These new hybrids just complicate the analysis and invalidate the conclusion
reached upon analysis of the previous 8 hybrids; moreover they show that
answers
a) to c) above are incompatible with the properties of these 16 hybrids.
(e) There are at least two genes responsible for the induction of senescence; one is
located on human chromosome 12 and the other one is located on human
chromosome 20.
41. (2 points) Please read carefully the following statements:
1) 1) Mutations of the TP53 gene (encoding the p53 protein) will have an effect on
both the cell-cycle arrest and the apoptotic pathways of a cell.
2) 2) The absence of introns in viral oncogenes is responsible for their tumorpromoting activity.
3) 3) Inactivation of the TP53 gene is implicated in the progression of both colorectal
and prostate cancers.
4) 4) Since the incidence of the different types of cancer varies between countries, it
argues that most of the time, cancer is not due to DNA mutations.
5) In human cancer, ras activation often involves a mutation of its promoter and results
in constitutive expression of this oncogene.
Based on the previous statements, which one of the following analysis is RIGHT?
(a)
(b)
(c)
(d)
(e)
Statement 3) is right, all others are false.
Statements 1), 2) and 3) are right, while statements 4) and 5) are false.
Statement 1) is right, while, statements 2), 3), 4) and 5) are false.
Statements 1), 3) and 5) are right, while statements 2) and 4) are false.
Statements 1) and 3) are right, while statements 2), 4) and 5) are false.
42. (1 point) The following table compares the properties of normal and cancer cells.
Properties
in normal
cells
Angiogenesis
Irradiation induced cell death:
Life span:
Gap junction:
Contact inhibition :
Telomerase activity
Metastatic potential
Karyotype:
in cancer
cells
Absent
Present
Present
Mortal
Present
Absent
Absent
Absent
Normal
Absent
Immortal
Greatly increased
Present
Present
Present
Many abnormalities
Which properties are incorrectly described for normal or cancer cells?
(a)
(b)
(c)
(d)
(e)
Contact inhibition and telomerase activity
Irradiation induced cell death and contact inhibition.
Telomerase activity and angiogenesis
Angiogenesis and irradiation induced cell death
Gap junction and contact inhibition.
43. (1 point) Which of the following statements is incorrect?
(a) The bcr-abl oncogene is not found in individuals having a normal karyotype.
(b) According to Knudson’s two hit hypothesis, two oncogenes will have to be
activated to produce a cancerous cell.
(c) In human, mutations that activate the Ras oncogene will keep the ras protein in
an
active, GTP-binding form.
(d) The inherited cases of retinoblastomas usually occur at early ages and these
individuals are often affected in both of their eyes.
(e) The growth of a cell population is due to a delicate balance between the action
of
growth-stimulating and apoptotic factors.
44. (1 point) Which of the following statements is correct?
(all answers accepted)
(a) Cellular oncogenes (c-onc) are derived from, but have lost the oncogenic
properties, of viral genes.
(b) During the cell cycle, the passage through the “Start” checkpoint in G1 is
controlled by the association between a kinase and a lamin.
(c) In colorectal cancer, mutations of the APC gene will increase the ability of the
pAPC protein to bind to the beta-catenin transcription factor, and will thus result
in the activation (transcription) of target genes.
(d) Phosphorylation of pRB results in the release and activation of the E2F
transcription.
(e) The hMSH2 tumor suppressor gene is implicated in the control of the cell cycle.
45. (1 point) Which of the following statements is right?
(a) A gene of interest (like the ADA gene) present in a retroviral vector used in
gene
therapy experiments will be integrated in the genome of the infected cells as
long
as these cells are not able to divide.
(b) The long terminal repeats (LTR) of a retroviral vector are used as selectable
markers upon integration of such a vector in the genome of the infected cells.
(c) With all the studies that have been done so far, one can affirm that there is
absolutely no danger in using retroviral vectors in human.
(d) Using an antisense oligonucleotide will interfere with the translation of a gene
and
thus inhibit the production of the encoded protein.
(e) One of the advantages of using an adenovirus-derived vector is the fact that only
dividing cells will be infected by such vector.
Answer key
a, b, d, c ,d , d, c, b, b, c, c, b, b, b, a ,a , c, all accepted, b, e, b, c, b, c, d, c, d ,c, d, b, e, a
& e ,a ,c , e, b, d, b, b, c, e, e, b, all accepted, d.