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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)
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)
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)
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) Genetic drift prevents this from happening, because in large populations, the
deleterious allele would behave as if it was selectively neutral.
(b) Selection cannot act to remove a recessive deleterious allele, because all the
deleterious alleles become “hidden” from selection in the heterozygotes.
(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) Only in populations undergoing drastic reductions in size is it possible for genetic
drift to bring about the removal of deleterious alleles.
(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) The theory predicts that the cytochrome c gene should undergo fewer mutations
than the fibrinopeptide, and therefore, exhibit less variation than fibrinopeptide.
(b) The theory predicts there should be many more alleles of fibrinopeptide than of
cytochrome c.
(c) The theory predicts that heterozygosity levels should be lower for cytochrome c
than for fibrinopeptide.
(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) 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) a nonsense mutation
(b) a missense mutation
(c) a conditional mutation
(d) a frameshift mutation
(e) 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) maternal-effect lethal, deletions of several segments
(b) maternal-effect lethal, holes in the cuticle
(c) recessive lethal, deletions of several segments
(d) recessive lethal, holes in the cuticle
(e) 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)
(b)
(c)
(d)
(e)
recruitment of the initiator tRNA to the ribosome
formation of the IF-2/initiator tRNA complex
formation of the 30S initiation complex
recruitment of the large ribosomal subunit to the mRNA
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)
3)
Upon separation on a cesium chloride gradient, the minisatellites containing CA
repeats will be separated from the rest of genomic DNA.
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
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.
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.
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.
Microsatellite polymorphisms are generated by misalignment and unequal
crossing-over.
Although they exist, long DNA deletions are rarely seen in normal human DNA.
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.
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)
3)
Although quite useful, a multiple cloning site (MCS) is not an essential part of a
cloning vector.
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.
Although cosmid vectors will be packaged inside the head of a lambda virus, once
inside a bacteria, these vectors will behave like plasmids.
4)
5)
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.
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
probe will each always show one fluorescent dot.
(b) Selection of cell hybrids in media containing HAT medium will allow the
survival of cells that have functional HPRT and TK genes.
(c) Two different Alu primers are needed to amplify inter-Alu fragments from
human DNA.
(d) 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.
(e) 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?
1)
(a) There is only one gene responsible for the induction of senescence and it
is
2)
3)
(b)
several
4)
chromosomal
5)
human
6)
7)
(c)
and
located on human chromosome 21.
These hybrids show that cellular senescence can be induced by a gene or
genes located on normal human chromosome(s). Although the
location of the gene(s) cannot be determined precisely, it cannot be on
chromosome 5.
There are only two genes responsible for the induction of cell senescence
8)
9)
10)
11)
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?
12)
(a)
senescence
13)
14)
(b)
senescence
15)
16)
(c)
they are
17)
18)
(d)
conclusion
19)
that answers
20)
21)
(e)
one is
22)
23)
24)
41. (2 points)
With these new hybrids, one can conclude that there is only one
inducing gene located on human chromosome 20.
There are only two human chromosome portions that can encode
inducing genes: those are located on human chromosomes 12 and 21.
There are two genes that are needed in combination to induce senescence,
located on human chromosomes 5 and 21.
These new hybrids just complicate the analysis and invalidate the
reached upon analysis of the previous 8 hybrids; moreover they show
a) to c) above are incompatible with the properties of these 16 hybrids.
There are at least two genes responsible for the induction of senescence;
located on human chromosome 12 and the other one is located on human
chromosome 20.
Please read carefully the following statements:
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) The absence of introns in viral oncogenes is responsible for their tumor-promoting
activity.
3) Inactivation of the TP53 gene is implicated in the progression of both colorectal and
prostate cancers.
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:
Absent
Present
Mortal
Present
Absent
Absent
Absent
Normal
in cancer
cells
Present
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.
Prof. Chevrette 20 questions 29 points
1. (1 point) Identify the incorrect statement:
a) Using either G (Giemsa) or R (reverse) banding, the 23 pairs of human chromosomes
can be identified in metaphase chromosome spreads obtained from normal human cells.
b)The majority of human chromosomes are submetacentric, while there is no telocentric
human chromosome.
c)Endomitosis occurs in human liver cells and leads to tetraploid cells in this organ.
d)If non-disjunction of chromosomes 18 occurs during the first meiotic division of a
human gamete cell, after meiosis there will be two cells containing one chromosome
18, one cell containing two chromosomes 18 and one cell without any human
chromosome 18.
e)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".
2. (2 points) A plant species A, which has 9 chromosomes in its gametes, was crossed
with a related species B, which has 11 chromosomes in its gametes. The hybrids were
sterile, and microscopic observation of their pollen mother cells showed chromosome
pairing on two 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) There will be 20 chromosomes in its somatic cells.
b) There will be 11 chromosomes in the gametes of this plant.
c) There will be 36 chromosomes in its somatic cells.
d) There will be 40 chromosomes in its somatic cells.
e) There will be 32 chromosomes in its somatic cells.
3. (1 point) The figure on the right represents a VNTR analysis of
five individuals including the mother (M), her child (C)
and the three putative fathers (1, 2 and 3).
Which one of the following statements best describes
these results:
a) This VNTR analysis is uninformative.
b) The mother (M) is not the biological mother of the child (C)
c) There must have been cross-contamination of samples.
d) Putative father 3 is the father of the child (C).
e) It is impossible to determine the real father, since putative fathers 1 and 3 are
likely twins.
4. (2 points) Please read carefully the following statements:
1)In contrast to prokaryotic DNA, eukaryotic DNA contains many repetitive elements.
2)PCR amplification of the VNTR sequence using primers complementary to the VNTR
sequence has replaced the need of using Southern blot to detect the VNTRs
3)Mouse satellite DNAs are located near the centromere of the mouse chromosomes.
4)During renaturation kinetics studies, the Cot value for a 300 bp fragment representing a
highly repetitive DNA sequence will be higher than the Cot value of a 300 bp fragment
derived from a single copy gene.
5)In human, the Alu sequences are an example of moderately repetitive DNA.
Based on the previous statements, which one of the following analysis is RIGHT:
a)Statements 1) , 4) and 5) are right, while statements 2) and 3) are false.
b)Only statement 1) is right, all others are false.
c)All statements are right.
d)Statements 1) , 3) and 5) are right, while statements 2) and 4) are false.
e) Statements 2) , 3) and 4) are right, while statements 1) and 5) are false.
5. ( 2 points) Please read carefully the following statements:
1) Slippage during DNA replication is the main cause of minisatellite polymorphism.
2)Microsatellite polymorphisms are generated by misalignment and unequal crossingover.
3)The RAPD protocol will be used to analyze genome of new organism where sequence
information is not available
4)Due to their small length (10 nucleotides), allele specific oligonucleotides (ASO) can
be used to identify alleles that differ by only one nucleotide.
5)RAPD (random amplification of polymorphic DNA) is performed using 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)Statement 3) is right, all others are false.
b)All statements are right.
c)Statements 1), 2) and 3) are right while statements 4) and 5) are false.
d)Statements 1), 2), 3) and 4) are right while statement 5) is false.
e)Statements 3), 4) and 5) are right while statements 1) and 2) are false.
6. ( 1 point) Which of the following statements is incorrect.
a)When analyzing DNA with RFLPs, the insertion of a 1 kb fragment in the middle
of a five kb EcoRI fragment detectable with the probe you are using will always be
seen as a 6 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)RFLP analysis could be used to detect DNA insertion or deletion.
d)Restriction enzymes are part of a bacterial protective mechanism against foreign DNA.
e)SSCP analysis can be used to detect DNA polymorphism, even if the sequence of this
polymorphism is not known.
7. (1 point) Which of the following is not required in the construction of a shuttle vector
to be grown in bacteria and yeast:
a)A yeast origin of replication
b)A selectable marker to be used in bacteria (like Ampr)
c)Two yeast telomere sequences (Tel)
d)A bacterial origin of replication (like Ori)
e)A selectable marker to be used in yeast (like URA3+)
8. ( 1 point) Which of the following statements is incorrect.
a) A phagemid is a vector containing sequence derived from a virus and can be converted
in a plasmid. It can be used to generate single-stranded DNA molecule.
b)Genomic libraries can be made by generating genomic DNA fragments without the use
of restriction endonucleases.
c)Cloning a DNA fragment into the sequence of the  -galactosidase gene and
plating the bacteria on media containing ampicillin will allow selection for the
bacteria that have been transformed with plasmids that contain a foreign piece of
DNA.
d)A cosmid vector is a plasmid containing the cos sites of the bacteriophage  and can
thus be packaged in virus particles.
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.
9. (2 points) Please read carefully the following statements:
1)Although quite useful, a multiple cloning site (MCS) is not an essential part of a
cloning vector.
2)Since about one third of the genome of the lambda phage is not essential for its
replication, it can be replaced by exogenous DNA.
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)Using a radioactive probe in colony hybridization will permit to identify only bacterial
colonies that have integrated a plasmid containing an insert that is identical to the probe
used in the hybridization.
5)Because it is derived from the single-stranded genome of the filamentous M13 phage,
phagemid (such as pUC118) will produced only single-stranded DNA molecules.
Based on the previous statements, which one of the following analysis is RIGHT:
a)Statement 3) is right, all others are false.
b)All statements are right.
c)Statements 1), 2) and 3) are right while statements 4) and 5) are false.
d)Statements 1), 2), 3) and 5) are right while statement 4) is false.
e)Statements 1) and 3) are right while statements 2), 4) and 5) are false.
10. ( 1 point) Which of the following statements is incorrect.
a)Chromosome walking and restriction fragment analysis can be used to generate a contig
map of a chromosomal region.
b) Although sequencing the human genome has pinpointed the chromosomal location of
most if not all human genes, we still do not know the function of many of their encoded
proteins.
c)A genetic map of an organism will be expressed in centiMorgans.
d)During chromosome jumping, a fragment of a clone (the probe) is used to find
adjacent genomic fragments.
e) In microchip hybridization experiments, the oligonucleotides present on the
microchips are similar to ASO molecules.
11. ( 2 points) The karyotype of a female blood cells from a new primate exhibiting
similar sex determination than human, has just been revealed. Analysis of its
chromosomes shows that these cells contain 16 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:
a08
b)16
c)32
d)17
e)33
12. ( 1 point) Which of the following statements is incorrect.
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 probe will
each show two fluorescent dots corresponding to the position of the probe on the two
sister chromatids of each chromosome.
b)On chromosome spreads obtained from the cells cultured from a patient with Down
syndrome, a human chromosome 21 specific painting probe will reveal three
chromosomes.
c)During fluorescent activated chromosome sorting (FACS), the "X" chromosomes can
be isolated from the other chromosomes using a combination of fluorescent dyes.
d)Microcell fusion is used to generate hybrids that contain different small portions
of the genome of one of the parental cell.
e)By combining the results of FISH and Giemsa staining, the BAC clones can be
localized on the cytogenetic map.
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 want 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 (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; 21; Y
1p; 3; 5; 8; 15; 16q; 20; X; Y
3; 7; 9; 11; 12; 16; 19; X
3; 6; 8; 9; 10; 13; 15; 16; 22
2; 3; 4; 7q; t(8p;16q); 10; 18; 20; X
2; 3; 4; 5; 14; 16; 17; 21
1; 3; 4; 5; 8; 10; 16; 17; 18; 20; X; Y
3; 7; 11; 19; 21; 22; Y
13. ( 2 points) Which one of the following statements is in complete agreement with
these results:
a) There is no selectable marker on human chromosomes in these hybrids.
b)There is only one gene responsible for the induction of senescence and it is located on
human chromosome X
c)A selectable marker could be present on human chromosome 3, while a gene
responsible for the induction of cellular senescence is located on human chromosome 7q.
d)These hybrids show that cellular senescence can be induced by gene or 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 3.
e) There are only two genes responsible for the induction of cell senescence and they are
located on human chromosomes 12 and 20.
14. ( 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
SOS9
SOS10
SOS11
SOS12
SOS13
SOS14
SOS15
SOS16
Phenotype
no senescence
senescence
no senescence
no senescence
senescence
no senescence
senescence
no senescence
Human chromosome content
1: 3: 7; 10; 12; 15; 18; Y
1 ; 2; 3; 5; 1 6; 16; 18; 19; 21; X
1p; 3; 5; 8; 15; 16q; 20; Y
1; 2: 7; 9; 11; 12; 16; 19; 20; X
3; 6; 8; 9; 10; 13; 15; 16; X
2; 4; 7q; t(8p;16q); 10; 18; X
2; 3; 4; 5; 14; 16; 17; 21; X
1; 4; 5; 8; 10; 12: 16; 17; 18; 20; X; Y
From the analysis of all the 16 hybrids obtained by these scientists, you can conclude
that:
a)There are only two human chromosome portions that can encode senescence inducing
genes: those are located on human chromosomes 12 and X.
b)With these new hybrids, one can conclude that there is only one senescence inducing
gene located on human chromosome 20.
c) 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.
d) There are two genes that are needed in combination to induce senescence, they
are located on human chromosomes 3 and X.
e)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 d) above
are incompatible with the properties of these 16 hybrids.
15. (1 point) Which of the following statements is incorrect.
a)Combined with cell fusion, chromosome translocation can facilitate the mapping of
genes.
b)Selection of cell hybrids in media containing hypoxanthine, aminopterin and
thymidine (HAT medium) will allow the survival of cells that have functional HPRT
and APH genes.
c)RFLPs, microsatellites, ESTs and VNTRs can be used to distinguish between human
and rodent genes.
d)A single Alu primer can be used to amplify inter-Alu fragments from human DNA.
e)Using a radiation hybrid panel, the location of new genes could easily be assigned
to specific (few megabases) chromosomal regions.
16. ( 2 points) Please read carefully the following statements:
1)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.
2)During the cell cycle, the passage through the "Start" checkpoint in G1 is controlled by
the association between a kinase and a lamin.
3)The hMSH2 tumor suppressor gene is implicated in the control of the cell cycle.
4)In human , mutations that activate the Ras oncogene will keep the ras protein in an
active, GDP-binding form.
5)According to Knudson’s two hit hypothesis, two oncogenes will have to be activated to
produce a cancerous cell.
Based on the previous statements, which one of the following analysis is RIGHT:
a) Statement 5) is right, all others are false.
b) Statements 2), 3) and 4) are false, while statements 1) and 5) are right
c) Statement 4) is right while statements 1), 2), 3) and 5) are false.
d) Statements 1), 3) and 4) are false while statements 2) and 5) are right.
e) All statements are false.
17. (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:
Absent
Present
Mortal
Present
Absent
Absent
Absent
Normal
in cancer
cells
Present
Absent
Immortal
Greatly increased
Present
Present
Present
Many abnormalities
Which properties are incorrectly described for normal or cancer cells.
a)Contact inhibition and telomerase activity
b)Gap junction and contact inhibition
c)Telomerase activity and angiogenesis
d)Angiogenesis and irradiation induced cell death
e0Irradiation induced cell death and contact inhibition.
18. (1 point) Which of the following statements is incorrect.
a)The bcr-abl oncogene is not found in individual having a normal karyotype.
b)Mutations of the p53 gene will have an effect on both the cell-cycle arrest and the
apoptotic pathways of a cell.
c)Inactivation of the TP53 gene is implicated in the progression of both colorectal and
prostate cancers.
d)The sporadic 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.
19. ( 2 points) Please read carefully the following statements:
1)Viral oncogenes (v-onc) are derived from cellular genes, but are often mutated and
constitutively active.
2)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.
3)The absence of introns in viral oncogene is responsible for their tumor-promoting
activity,
4)Phosphorylation of pRB results in the release of the E2F transcription.
5) In human cancer, ras activation often involved 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)Statements 1) and 5) are right, while statements 2), 3) and 4) are false.
b)Statements 1), 2) and 3) are right, while statements 4) and 5) are false.
c)Statements 1) and 4) are right while statements 2), 3) and 5) are false.
d)Statements 1), 2) and 4) are right while statements 3) and 5) are false.
e) Statements 2) and 4) are right, while statements 1), 3) and 5) are false
20. ( 1 point) Which of the following statements is incorrect.
a)The 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 able to divide.
b)The long terminal repeats (LTR) of a retroviral vector are needed for the integration of
such vector in the genome of the infected cells.
c)Using an antisense oligonucleotide will interfere with the transcription of a gene
and thus inhibit the production of the encoded protein.
d)One of the advantages of using an adenovirus-derived vector is the fact that nondividing cells will be infected by such vector.
e)Even with all the studies that have been done so far, one cannot eliminate that there
could still be some danger in using retroviral vectors in human.
Prof. Schoen 5 questions 7 points
21.(2 points) In human populations, the widow’s peak phenotype (in which the hairline
comes to a V-shaped point on the forehead) is coded for by a dominant allele. If the
genotypes underlying the presence/absence of the widow’s peak are in Hardy-Weinberg
proportions, and the frequencies of individuals with and without the widow’s peak
phenotype are 0.36 and 0.64, respectively, what is frequency of the widow’s peak allele?
a)0.8
b)0.6
c)0.2
d)0.32
e) 0.36
22. (2 points) In the population referred to above, a strange disease appears that reduces
the survival prospects of all individuals who lack the widow’s peak, such that they are
25% LESS likely to survive to reproduce compared with individuals who possess the
widow’s peak. What is the expected frequency of the widow’s peak allele at the start of
the NEXT generation?
a).238
b).244
c).385
d).445
e).571
23.(1 point) Which evolutionary force, acting over time, reduces the diversity of alleles
per locus in populations?
a)Non-random mating.
b)Genetic Drift.
c)Inbreeding.
d)Random mutation.
e)None of the above.
24.(1 point) Which evolutionary force (or combination of forces) can result in the
maintenance of more than one allele per locus in a population?
a)Directional selection in which separate subsets of the
population undergo selection for different alleles of the
same locus.
b)Selection favouring heterozygotes.
c)Genetic drift and mutation acting together.
d)Mutation and genetic drift acting together.
e)All of the above.
25.(1 point) The frequencies of the blood type alleles, A, B, and O in a certain human
population are 0.3, 0.4, and 0.3. In a sample of 1000 individuals from this population,
how many individuals would you expect to be homozygous at this locus? (Assume the
population is in Hardy-Weinberg proportions at this locus)
a)340
b)580
c)300
d)548
e)90
Prof. Lasko 20 questions 29 points
26.(2 points) You are studying a gene that encodes a metabolic enzyme in E. coli,
and you wish to isolate a temperature-sensitive mutant allele of it. Which would
be the best mutagenic agent to use to do this?
a)acridine orange
b)X-irradiation
c)transposable elements
d)ethyl methane sulfonate
e)Cheez-Whiz
27.(1 point) What is the Ames test?
a)a method to determine the frequency of tautomeric shifts
at a particular base pair
b)a method that uses special strains of Salmonella to
determine the mutagenicity of chemicals
c)a method used to diagnose inherited disorders like
xeroderma pigmentosum
d)a method to determine whether two mutations affect the
same gene
e)a method to determine if a gene produces multiple
alternatively-spliced transcripts
28.(2 points) Which of the following statements about genes is not true?
a)RNA can be transcribed from all genes.
b)The initiator codon of an mRNA encodes the most Cterminal amino acid in the protein that is translated from it.
c)All genes include information for the production of one
or more polypeptides.
d)(a) and (c)
e)(b) and (c)
29.(2 points) Which of the following statements about DNA replication is not
true?
(two accepted answers)
a)DNA replication is semiconservative.
b)DNA replication is continuous on one strand, but
discontinuous on the other strand.
c)DNA replication in prokaryotic cells initiates at
multiple sites.
d)Replication of telomeres poses a complex problem
because there is no 3’-hydroxyl group to prime the
polymerization of the lagging strand after the RNA primer
of the terminal Okazaki fragment has been excised.
e)(c) and (d).
30.(1 point) Which of the following techniques was used to investigate the
question of whether DNA replication is conservative, semiconservative, or
dispersive?
a)electron microscopy
b)autoradiography
c)molecular cloning
d)(a) and (b)
e)(a), (b), and (c)
31.(1 point) Different organisms use different mechanisms to equalize gene
expression from the sex chromosomes in males and females. Which of the
following statements is not true?
a)One of the X chromosomes is heterochromatic in
mammalian female cells.
b)Heterochromatin is depleted for acetylated histone H4.
c)MSL proteins bind to specific sites on the Drosophila
X chromosomes in females and increase transcription
from them.
d)(a) and (b).
e)(a), (b), and (c)
32.(2 points) Which of the following statements about RNA processing is not
true?
a)different cells and tissues, to produce different proteins.
b) RNA processing occurs in the cytoplasm.
c) Poly(A) tracts are present at the 3’ end of eukaryotic mRNAs because
they are transcribed from the chromosomal DNA.
d) (a) and (b)
e) (b) and (c)
33.(1 point) bumless is a recessive maternal-effect mutation of Drosophila that
causes embryonic lethality. What percentage of progeny embryos would survive
from a cross between bumless heterozygotes (that is, bumless/+ males x
bumless/+ females)?
a)0%
b)25%
c)50%
d)75%
e)100%
34.(1 point) Which of the following statements about pair-rule genes is not true?
a)Pair-rule genes encode transcription factors.
b)Pair-rule genes are expressed in a series of transverse
bands along the anterior-posterior axis of the Drosophila
embryo.
c)Pair-rule genes begin to be expressed before gap genes
begin to be expressed.
d)Pair-rule genes begin to be expressed before segment
polarity genes begin to be expressed.
e)(c) and (d).
35.(2 points) Which of the following statements about Drosophila embryonic
patterning is not true?
a)Embryos mutant for a homeotic gene have fewer segments than wildtype embryos.
b)The gradients of Hunchback and Caudal proteins required for anterior- posterior
patterning are established entirely by transcriptional control.
c)The Dorsal protein is a transcription factor.
d)(a) and (b)
e)(a) and (c)
36.(1 point) Which of the following statements about circadian rhythms in
Drosophila is not true?
a)PER and TIM proteins are synthesized with the onset of
night.
b)CLK and CYC proteins are active as a heterodimer, and
activate transcription of the per and tim genes.
c)PER and TIM bind to CLK and CYC and increase their
activity.
d)PER and TIM are active in the cytoplasm.
e)(c) and (d)
37. (2 points) Among monozygotic (MZ) twin pairs with one member identified
as alcoholic, the co-twin is alcoholic 55% of the time. Among dizygotic (DZ)
twin pairs with one member identified as alcoholic, the co-twin is alcoholic 28%
of the time. What conclusions can be drawn from this? (QUESTION
REMOVED)
a)Genetic factors influence the trait of alcoholism.
b)Non-genetic factors influence the trait of alcoholism.
c)Genetic factors have a relatively less important role with
regard to the trait of alcoholism than they do with regard to
intelligence as measured by IQ (intelligence quotient).
d)(b) and (c)
e)(a), (b), and (c)
38. (1 point)What are pole cells?
a)the cells in the C. elegans embryo that give rise to muscle
b)the cells in the Drosophila embryo that give rise to eggs or sperm
c)the cells that result from the asymmetric cleavages that occur during female meiosis
d) the cells in the Drosophila embryo that concentrate Dorsal protein in their nuclei
e) the cells in the C. elegans embryo that give rise to intestine
39. (2 points) Which of the following statements about Drosophila P-elements is
not true?
a)P elements only transpose in germ line cells because only
those cells have transposase activity.
b)A dysgenic hybrid results when a male of P cytotype is
crossed with a female of M cytotype.
c)P elements isolated from natural strains of flies often
carry antibiotic-resistant genes.
c)P elements can be used to introduce foreign genes into
the Drosophila germ line, allowing them to be stably
inherited.
d)(b) and (c)
40. (2 points) Which of the following statements about temperature-sensitive
mutants is not true?
a)Temperature sensitivity usually results from increased
heat lability of the mutant gene product.
b)Temperature-sensitive mutants are phenotypically normal
at low temperature and phenotypically mutant at a higher
temperature.
c)There are very few temperature-sensitive mutants in
mice, and it is extremely difficult to screen for them.
d)Temperature sensitivity often results from misregulation of the heat shock response genes.
e)(c) and (d)
41.(1 point) What is meant by the tertiary structure of a protein?
a) the association of two or more polypeptides in a multimeric protein
b) the overall folding of a polypeptide in three-dimensional space
c) the spatial interrelationships of the amino acids in segments of the polypeptide
d) the amino acid sequence of the protein
e) the series of third bases of each of the codons in the mRNA that encodes the
protein
42.(1 point) Which is the final step of prokaryotic translation initiation?
a)recruitment of the initiator tRNA to the ribosome
b)recruitment of the small ribosomal subunit to the mRNA
c)recruitment of the large ribosomal subunit to the
mRNA
d)recruitment of IF-2 to the initiator tRNA
e)recruitment of IF-3 to the small ribosomal subunit
43. (1 point) Which of the following events of embryonic development is the earliest?
a)completion of the blastula
b)neurulation
c)gastrulation
d) cleavage divisions
e) cellular differentiation
44.(1 point) Which of the following statements about embryonic stem cells (ES cells) is
not true?
a)ES cells are derived from the inner cell mass of mouse embryos.
b) DNA injected into ES cells can inactivate the normal allele on the
chromosomes.
c) ES cells can be transferred to other mouse embryos and can
contribute to adult tissues.
d)ES cells provide a means of creating mutant mice in any gene of
interest.
e)None of the above.
45. (2 points) What is the chemical nature of Griffith’s "transforming principle"?
a)RNA
b)DNA
c)protein
d)polysaccharides
e)None of the above.
Biology 177-202B Final Examination April 24th 2002
Note that some of the questions have been corrected in response to the relevant
challenges ( they appear in bold type). You may contact Kathy (W3/23A), to view your
final exam once the marks have been posted. Due to the ambiguity of one of Dr. Lasko's
questions all students have received 2 points.
Prof. Schoen's Questions- 30 points
A
B
1. (1 point) Precursor -------> Intermediate -------> Red pigment
Consider the above biochemical pathway. A and B are dominant forms of two different
genes that code for functional enzymes. In addition to the wildtype alleles A and B, there
exist, as well, mutations a and b that are loss of function mutations (i.e., the enzyme
product they code for is not functional). Which of the following genotypes is capable of
coding for the red pigment?
a) aaBB
b) AAbb
c) Aabb
d) A-Be) All of the above
2. (2 points) Suppose you conduct a cross involving one parent with purple flowers and
another parent with white flowers. You then intercross the F1 hybrids and discover a 15:1
ratio of purple:white flowered progeny in the F2 generation. If there are two separate
genes that code for flower colour (each gene with dominant and recessive alleles), what is
the genotype of the white-flowered plant in the F2?
a) AaBb
b) aaBb
c) aabb
d) Aabb
e) AABB
3. (1 point) Which of the following is (NOT) a feature of continuous traits?
a) The environment influences the phenotype.
b) The effect of environment can vary with genotype.
c) Mendel's laws (do not apply to these traits)
d) Two or more genes are often involved.
e) All of the above
4. (2 points) The distance between two points on a genetic map of a chromosome can be
determined by:
a) the average number of crossover events between the two points.
b) determining the frequency of recombinant chromosomes.
c) mating individuals with two mutations on the same chromosome with wild-type
individuals, test crossing the F1 and counting the progeny types in the F2.
d) a three-point testcross.
e) all of the above.
5. (2 points) A P1 with blue-flowered, short-stalked plants and white-flowered, longstalked plants is crossed as is the resulting F1. You find: 400 blue, short: 400 white, long:
100 blue, long: 100 white, short. What is the recombination frequency?
a) 0.25
b) 0.2
c) 0.8
d) 0.1
e) 0.05
6. (1 point) A mating between a true breeding red long-stamen plant and a true breeding
white short-stamen plant yields an F1. This F1 is self-fertilized to produce an F2 that has
300 red long-stamen plants and 100 white short-stamen plants. What can you conclude?
a) The white-short stamen plants are the result of recombination.
b) A chiasma has been formed.
c) Mendel's law of segregation is supported by the evidence.
d) Mendel’s law of segregation is NOT supported by the evidence.
e) The genes are tightly linked.
7. (1 point) A pedigree reveals that genes coding for two recessive traits blue eyes and
obesity, are linked in humans. What would be true about the pedigree?
a) Few people in the pedigree will be both blue-eyed and obese, due to crossing over.
b) Many people in the pedigree will be both blue-eyed and obese.
c) The lod score for these genes will be near 0.
d) The lod score for these genes will be 0.5.
e) a and d.
8. (1 point) Crossing yeast with two different mutations yields almost all asci with a
parental ditype pattern. We can conclude that:
a) the genes are loosely linked.
b) the genes are on two different chromosomes.
c) the genes are closely linked.
d) neither of the two genotypes are like the parental genotype.
e) we need more data to determine the linkage.
9. (2 points) Nonparental ditype asci in a cross with two linked genes are the result of:
a) Tightly linked genes.
b) No crossing over events between tightly linked genes.
c) A two-strand crossover between loosely linked genes.
d) A four-strand double crossover event.
e) A tetratype event.
10. (1 point) Inbreeding results in:
a) An increase in the frequencies of rare alleles.
b) A decrease in the frequencies of rare alleles.
c) Increased frequency of heterozygotes.
d) Increased frequency of homozygotes.
e) a and d.
11. (1 point) Proteins evolve at different rates because:
a) Introns are not under selection.
b) Some base substitutions do not cause an amino acid substitution.
c) Some amino acid substitutions occur in non-functional parts of a protein.
d) Some mutations are more likely than others.
e) Some proteins are more functionally constrained than others.
12. (2 points) Non-Mendelian inheritance of organelles may be characterized by:
a) Preferential transmission of organelles through gametes of one sex.
b) Maternal inheritance.
c) Irregular segregation of phenotypes.
d) Paternal inheritance.
e)All of the above.
13. (2 points) According to the theory that argues that much of the DNA level variation
in populations is selectively neutral (i.e., has no impact on organismal fitness), which of
the following conditions would lead to the highest level of molecular level variation at
the population level?
a) Random mating and high mutation rate.
b) Random mating
c) Large population size.
d) High mutation rate and large population size
e) Outbreeding and heterozygote advantage.
14. (2 points) If a woman who is red-green colour-blind marries a man with normal
vision, what phenotypes would you expect their children to have?
a) All their children will be colour-blind.
b) All their daughters will be colour-blind, but all their sons will have normal vision.
c) All their daughters will be carriers and all their sons will be colour-blind.
d) All their daughters will have normal vision and will not be carriers, but all their sons
will be colour blind.
e) Half their daughters will be carriers and the other half will be fully normal; half their
sons will be colour-blind and the other half will have normal vision.
15. (2 points) Human albinism is an autosomal recessive trait. Suppose that you find a
village in the Andes where 1/4 of the population is albino. If the population size is 1000
and the population is in Hardy-Weinberg proportions with respect to this trait, how many
individuals are expected to be heterozygotes (i.e., carriers)?
a) 50
b) 250
c) 750
d) 500
e) 25
16. (2 points) Broad sense heritability is defined as the proportion of phenotypic variance
that is attributable to genetic differences among individuals in a specific target population
and test environment. Corn height is a continuous trait. Two different inbred (truebreeding) lines of corn are crossed. The variance of height in the F1 generation is 1.5.
Plants from the F1 are then intercrossed, and the F2 generation progeny are grown out in
the same environment as the F1. The variance of height in the F2 is 6.0. What is the
broad sense heritability of height?
a) 0.25
b) 0.5
c) 0.75
d) 0
e) 1.0
17. (1 point) Which of the following forces leads to the loss of genetic variation in
populations?
a) Directional selection favouring one phenotype in all environments.
b) Genetic drift.
c) Inbreeding.
d) Migration between populations.
e) a and b.
18. (2 points) For many generations, the following percentages of genotypes were
observed in a large, random mating population of birds: 4 percent AA, 32 percent Aa, and
64 percent aa. Suppose a change in climate brings about a sudden decrease in fitness such
that individuals with genotype aa can no longer survive or reproduce. Individuals with
genotypes AA or Aa, however, experience no change in their mortality or reproductive
rates. Which of the following is a TRUE statement?
a) There will be no a alleles in the population in the next generation.
b) The frequency of the a allele will decline by one half in the next generation.
c) The frequency of homozygote AA individuals will increase.
d)There will be no a alleles in the population in the next generation, except for those
that are reintroduced by mutation or migration from other populations.
e) a and c
19. (2 points) Phenotypic resemblance among siblings may arise from:
a) Shared environmental effects.
b) Shared genes.
c) The norm or reaction.
d) Genotype x environment interaction.
e) a and b
Prof. Suter's Questions -11 points
20. (2 points) You are interested in identifying the genes that are essential for anterior
development of the Drosophila embryo. Recently you received a collection of 600 lines
of single gene mutations (mut1-600, each line has only one mutation mut) and you
performed a series of tests on them. Which result indicates that the mutant has nothing to
do with your project? (Note: if a result is non-informative, the mutation in question is still
a potentially interesting candidate).
All crosses shown are female crossed with male (female X male)
a) a cross of mut / + X mut / + reveals 25% embryos with posterior defects.
b) a cross + / + X mut / mut reveals 0% embryos with defects
c) a cross mut / mut X mut / + reveals 100% embryos with anterior defects.
d) a cross mut / mut X +/ + reveals 100% embryos with anterior defects.
e) a cross mut / + X mut / + reveals 25% embryos with anterior defects.
21. (2 points) The hat (head-and-tail-less) mutation causes the formation of embryos
without head and without tail. You conclude that this mutation disrupts an important
patterning gene and decide to study it genetically. Following are the crosses that were set
up to characterize hat with the results they yielded.
All crosses shown are female crossed with male (female X male)
1) hat /+ X hat /+ ==> phenotype of all embryos is wild type
2) hat / + X hat / hat ==> phenotype of all embryos is wild type
3) hat / hat X hat / + ==> phenotype of all embryos is hat
4) hat / hat X hat / hat ==> phenotype of all embryos is hat
5) + / + X hat / hat ==> phenotype of all embryos is wild type
6) hat / hat X + / +==> phenotype of all embryos is hat
From these results you draw one conclusion:
a) hat is partially penetrant
b) hat is a dominant zygotic mutation
c) hat is a recessive zygotic mutation
d) hat is a dominant maternal effect mutation
e) hat is a recessive maternal effect mutation
22. (2 points) From what you know about Drosophila Sex determination, which of the
following procedures would fail to cause sexual transformation?
a) expressing tra female cDNA constitutively in both sexes.
b) expressing Sxl male cDNA constitutively in both sexes
c) expressing dsx female cDNA constitutively in both sexes
d) expressing dsx male cDNA constitutively in both sexes
e) preventing expression of tra-2 in both sexes
23. (1 point) Identify the wrong answer:
The per gene:/Per polypeptide
a) is found in Drosophila.
b) is functionally equivalent to a molecular oscillator.
c) controls the periodicity of the courtship song.
d) is phosphorylated by Cry.
e) forms a complex with Tim.
24. (2 points) Over-expression of shaggy
1) causes Tim to enter the nucleus prematurely
2) directs Per to degradation
3) lengthens the periodicity of the circadian clock
4) shortens the periodicity of the circadian clock
5) does not affect the periodicity of the circadian clock
a) 1) and 3) are true.
b) 2) and 4) are true
c) 1) and 4) are true
d) 2) and 3) are true
e) 5) is true
25. (1 point) Which of the following is true of either per or tim, but not both?
a) involved in control of circadian rhythms in Drosophila
b) mRNA levels declining during the night
c) protein product destroyed by light
d) transcription occurring during the day
e) translation occurring during the night
26. (1 point) Which of the following does Huntington's disease have in common with
both PKU and Lesch-Nyhan syndrome?
a) The defective protein appears to influence energy metabolism.
b) The phenotype is due to detrimental effects on nerve cells.
c) It is a metabolic disorder.
d) It encodes a transcription factor
e) It alters the periodicity of the circadian clock
Prof. Chevrette's Questions -24 points
27. (1 point) Identify the incorrect statement:
a) In contrast to prokaryotic DNA, eukaryotic DNA contains many repetitive
elements.
b) Human DNA is composed of only two types of DNA : the unique
sequences and the highly repetitive sequences.
c) Although the exact function of repetitive DNA is not known, it could represent
a repository of unessential DNA for future evolution.
d) Mouse satellite DNAs are located near the centromere of the mouse
chromosomes.
e) In human, the Alu sequences are an example of moderately repetitive DNA.
28. (2 points) Which of the following statements regarding VNTRs is incorrect.
a) In an individual the number of repeats at a given locus is not always the
same.
b) They can be used in DNA fingerprinting.
c) PCR amplification of the VNTR sequence using primers complementary to
the VNTR sequence has replaced the need of using Southern blot to detect
the VNTRs.
d) In DNA fingerprinting, the restriction enzyme will digest the DNA on each
side of a VNTR sequence.
e) Since VNTRs varies from one individual to another, they can be used to
identify a specific individual.
29. (2 points) Which of the following statements is incorrect.
a) Restriction enzymes are part of a bacterial protective mechanism against
foreign DNA.
b) RFLPs can be used in genome mapping and they will segregate as other
markers.
c) A mutation occurring in a DNA molecule will be seen as an RFLP only if
this mutation affect the restriction site of a given enzyme.
d) In absence of sequence information, analyzing the human genome for RFLPs
will require the use of Southern blot.
e) RFLP analysis could be used to detect DNA insertion or deletion.
30. (1 point) Which of the following statements is incorrect.
a)Polymerase pausing during replication is the main cause of microsatellite
polymorphism.
b)Although they exist, long DNA deletions are rarely seen in normal human DNA.
c)A rapid analysis of RFLPs among individuals within a population can be performed by
combining PCR amplification, restriction enzyme digestion and gel electrophoresis.
d)About 100,000 different minisatellite sequences are found in the human genome
and they result from unequal crossingovers.
e)PCR is the method of choice to detect microsatellite variation between individuals.
31.(2 points) Please read carefully the following statements:
1)Due to their small length (10 nucleotides), allele specific oligonucleotides (ASO) can
be used to identify alleles that differ by only one nucleotide.
2)SSCP analysis is based on the fact that single-stranded DNA molecules will adapt
different conformation even if there is only one nucleotide difference between two
sequences.
3)RAPD (random amplification of polymorphic DNA) is performed using two small
complementary primers and is a rapid way of analyzing polymorphic sequences.
4)A complete analysis of the human DNA sequence derived from two individuals will
reveal millions of single base polymorphisms.
5)The DNA sequence flanking microsatellite DNA must be known to be able to analyze
this type of DNA polymorphism.
Based on the previous statements, which one of the following analyses is RIGHT:
a) Statement 3) is false, all others are right.
b) Only statement 2) is right, all others are false.
c) Statements 1), 2) and 3) are false while statements 4) and 5) are right.
d) Statements 1), 3) and 4) are false while statements 2) and 5) are right.
e) Statements 1) and 3) are false while statements 2) , 4) and 5) are right.
32.(1 point) Which of the following statements is incorrect.
a) The presence of an origin og replication (ori) in a plasmid will allow its replication as
an extrachromosomal entity in bacteria.
b) The number of clones to be screened to identify a particular gene in a genomic
library depends only on the size of the genomic DNA that can be inserted in the
cloning vector.
c) Cloning a DNA fragment into the sequence of the  -galactosidase gene and plating
the bacteria on media containing IPTG will allow selection for the bacteria that have been
transformed with plasmids that contain a foreign piece of DNA.
d)A cosmid vector is a plasmid containing the cos sites of the bacteriophage  and can
thus be packaged in virus particles.
e)Genomic libraries can be generated by generating genomic DNA fragments without the
use of restriction endonucleases.
33.(1 point) Which of the following is not required in the construction of a shuttle vector
to be grown in bacteria and yeast:
a) A yeast origin of replication
b) A bacterial origin of replication (like Ori)
c) A selectable marker to be used in bacteria (like Ampr)
d) A selectable marker to be used in yeast (like Leu2+)
e) Two yeast telomere sequences (Tel)
34. (1 point) Which of the following statements is incorrect.
a)A genetic map of an organism will be expressed in megabases, and will be
obtained with different molecular markers and DNA sequences.
b)During chromosome walking, a fragment of a clone is used to find clones containing
adjacent genomic DNA fragments.
c)Colony hybridization is used to identify bacteria which have a plasmid that contains a
gene or a cDNA of interest.
d)Cloning a cDNA into a phagemid will allow the generation of single-stranded or
double-stranded DNA molecules that can be used as DNA probes.
e)The generation of single-stranded DNA molecules from a phagemid will necessitate the
collaboration of a helper phage.
35.(1 point) The karyotype of a male germ cell (sperm) from a new organism has just
been revealed. Analysis of its chromosomes shows that it contains 14 chromosomes plus
an X or a Y. 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) 9
b) 16
c) 30
d) 7
e) 2
36.(2 points) Please read carefully the following statements:
1) During chromosome jumping, self-ligation of a long genomic DNA fragment will
bring in close proximity, portions of genomic DNA that were originally separated by
large distances.
2) Microchips hybridization will be performed to identify genes that are expressed
differentially between two cell types.
3) There are two types of oligonucleotides present on a microchip: the one representing a
perfect match to the genes of interest ; and other unrelated oligonucleotides used as
controls.
4) Upon hybridization of a fluorescent probe corresponding to a single copy gene to a
normal human chromosome spread, the two chromosomes that contain this probe will
each show two fluorescent dots corresponding to the position of the probe on the two
sister chromatids of each chromosome.
5) On chromosome spreads obtained from normal human skin, a human chromosome 18
specific painting probe will reveal a single chromosome.
Based on the previous statements, which one of the following analyses is RIGHT:
a) Statement 3) is false, all others are right.
b) Only statement 2) is right, all others are false.
c) Statements 1), 2) and 3) are right while statements 4) and 5) are false.
d) Statements 1), 2) and 4) are right while statements 3) and 5) are false.
e) Statements 1) and 3) are false while statements 2) , 4) and 5) are right.
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 want to identify the
gene responsible for the induction of senescence. They thus made cell hybrids between
normal human cells (senescent) and a mouse melanoma cell line (non-senescent) and
characterized the hybrids they obtained (called SOS for Senesce Or Survive). All hybrids
retained most mouse chromosomes.
Hybrids
SOS1
SOS2
SOS3
SOS4
SOS5
SOS6
SOS7
SOS8
X
SOS9
Phenotype
Human Chromosome content
senescence
no senescence
senescence
no senescence
senescence
senescence
no senescence
senescence
2; 4; 6; 8; 12; 16; 20; X; Y
1; 3; 4; 7; 9; 13; 17; 18; 21
4; 5; 10; 11; 14; 22; X; Y
1; 2; 4p; 7; 13; 17; 19; 20; 21
2; 4; 10; 12; 14; 18; 22; X
2; t(4p;8q); 5; 6; 12; 16; 20; X
1; 3; 4; 9; 13; 17; 18; 20; 22; X; Y
1q; 2; 4; 6; 7p; 9; 13q; 14; 19; 20;
no senescence
1; 2; 3; t(4p;8q); 5; 7p; 18; 20; 21
37. (2 points) Which one of the following statements is in complete agreement with these
results:
a) The properties of these hybrids show that there is no gene that can induce senescence.
b) There is at least one gene responsible for the induction of senescence and it is located
on human chromosome 4p.
c)These hybrids show that cellular senescence can be induced by gene or genes
located on normal human chromosome(s); however, the chromosomal location of
such gene(s) cannot be determined precisely.
d)A selectable marker could be present on human chromosome 4, while a gene
responsible for the induction of cellular senescence is located on human chromosome X.
e) There is no selectable marker on human chromosomes in these hybrids.
38. (3 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
SOS10
SOS11
SOS12
SOS13
SOS14
SOS15
SOS16
senescence
no senescence
no senescence
senescence
no senescence
senescence
no senescence
Human chromosome content
2; 3; 5; 6; 16; 18; 19; 21; X
1p; 5; 8; 15; 16q; 20; Y
3; 7; 9; 11; 12; 16; 19
6; 8; 9; 10; 13; 15; 16
2; 4; 7q; t(8p;16q); 10; 18
2; 3; 4; 5; 14; 16; 17; 21
1; 5; 8; 10; 16; 17; 18; 20
From the analysis of all the 16 hybrids obtained by these scientists, you can conclude
that:
a) The only senescence inducing gene is located on human chromosome 6; in this new set
of hybrids (SOS 10 to SOS 16), the selectable marker could be located on human
chromosome 16q.
b) With these new hybrids, one can conclude that there is only one senescence inducing
gene located on human chromosome 12.
c) There are at least two genes responsible for the induction of senescence; one is
located on human chromosome 6 and the other one is located on human
chromosome 14.
d) There are only two human chromosome portions that can encode senescence inducing
genes: those are located on human chromosomes 12 and 14.
e) Although a senescence inducing gene is located on human chromosome 6, these 16
hybrids show that there is additional gene(s) able to induce senescence. However their
location cannot be linked to any specific chromosome or chromosome segments.
39. (2 points) Please read carefully the following statements:
1) The incidence of the different types of cancer varies between countries.
2) The viral oncogenes (v-onc) are derived from cellular genes, but are often
mutated and constitutively active.
3) In human , mutations that activate the Ras oncogene will keep the ras protein in
an active, GTP-binding form.
4) The oncogenic properties of the bcr-abl oncogene is due to a chromosomal
translocation.
5) During the cell cycle, the passage through the "Start" checkpoint in G1 is
controlled by the association between a kinase and a cyclin.
Based on the previous statements, which one of the following analyses is RIGHT:
a) Statement 5) is false, all others are right.
b) Statements 1), 2) and 3) are right, while statements 4) and 5) are false.
c) Statements 1), 3) and 4) are right while statements 2) and 5) are false.
d) Statements 1), 2) and 4) are right while statements 3) and 5) are false.
e) All statements are right.
40.(1 point) The following table compares the properties of normal and cancer cells.
Properties
in normal
in cancer
cells
cells
Contact inhibition :
Present
Irradiation induced cell death: Present
Life span:
Mortal
Karyotype:
Normal
Gap junction:
Present
Telomerase activity
Absent
Angiogenesis
Present
Metastatic potential
Absent
Absent
Absent
Immortal
Many abnormalities
Greatly reduced
Present
Absent
Present
Which property is incorrectly described for normal and cancer cells.
a)Contact inhibition
b)Gap junction
c)Telomerase activity
d)Angiogenesis
e)Irradiation induced cell death
41. (2 points) Please read carefully the following statements:
1) The 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 able to divide.
2) According to Knudson’s two hit hypothesis, two oncogenes will have to be activated
to produce a cancerous cell.
3) Antisense RNA could be used in gene therapy to decrease the expression of a
particular protein.
4) Mutations of the p53 gene will have an effect on both the cell-cycle arrest and the
apoptotic pathways of a cell.
5) In colorectal cancer, mutations of the APC gene will interfere with the ability of the
pAPC protein to bind to the E2F transcription factor, and will thus result in the activation
(transcription) of target genes.
Based on the previous statements, which one of the following analyses is RIGHT:
a) Statement 5) is false, all others are right.
b) Statements 1), 2) and 3) are right, while statements 4) and 5) are false.
c) Statements 1), 3) and 4) are right while statements 2) and 5) are false.
d) Statements 1), 2) and 4) are right while statements 3) and 5) are false.
e) All statements are right.
Prof. Lasko's Questions -5 points
42. (1 point) Which of the following statements is/are false?
i. In a complementation test, the cis heterozygote should have the mutant
phenotype
regardless of whether the mutations are in the same or in different genes.
ii.
Null mutations give the least ambiguous results in complementation tests.
iii. Intragenic complementation always occurs between mutations that affect a
gene
that encodes a protein that functions as a dimer.
a) Statement i only.
b) Statement ii only.
c) Statement iii only.
d) Statements i and iii.
e) Statements ii and iii.
43. (2 points) Thanks to Biology 202B, you have become a disciple of Thomas
Hunt Morgan, and you have decided to emulate his pioneering work on fruit flies
by developing the cockroach as a model genetic organism. To do this, you feed
large quantities of mutagenic chemicals to male cockroaches, and establish many
different mutant lines from their progeny. You are particularly interested in two
mutant lines that affect the structure of the exoskeleton, that you have named
crunchy and armor-plated. Both crunchy and armor-plated are recessive
mutations, and you want to know whether they affect the same or different genes.
Which of the following experiments will help you answer this question?
i Cross crunchy homozygotes to amor-plated homozygotes, and determine if
the progeny are wild-type in phenotype.
ii Cross a cockroach that is heterozygous for both crunchy and armor-plated to a
cockroach that is heterozygous for crunchy, and determine if any progeny are
crunchy or armor-plated in phenotype.
iii Cross a cockroach that is heterozygous for both crunchy and armor-plated to a
cockroach that is wild-type, and determine if any progeny are crunchy or armorplated in phenotype.
iv Cross a cockroach that is heterozygous for crunchy to a cockroach that is
heterozygous for armor-plated, and determine if any progeny are mutant.
a) Experiments i and iv
b) Experiments i and iii.
c) Experiments iii and iv.
d)Experiments i, ii, and iv.
e) All four experiments.
44.(2 points) Which of the following statements is/are true?
i Complementation tests tell you how far apart genes are located on a chromosome.
ii Complementation does not depend upon any direct interaction between the
chromosomes.
iii Complementation tests cannot be used to study dominant mutations.
a)
b)
c)
d)
e)
Statement i only.
Statement ii only.
Statement iii only.
Statements i and iii.
Statements ii and iii.