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Use the information below to answer the next two questions:
Assume that high blood pressure is inherited as an autosomal dominant trait. You
genotype the individuals in your pedigrees with many markers and successfully
localize the trait to a region on chromosome 3. A1 and A2 are two codominant
alleles of the A marker gene on chromosome 3. Our representative family and their
genotypes at the A marker are shown below. Note: Within a given generation,
individuals are numbered consecutively from left to right (e.g., I-1, I-2 for the A1/A1
and A2/A2 parents in generation I).
1. (1 point) The individual(s) in the pedigree that has (have) produced informative
meioses for mapping of the A marker and high blood pressure locus is (are):
a. I-1 and I-2
b. II-3
c. III-2, III-4, and III-5
d. III-1 and III-3
e. None of the above
Answer b. In order for an individual to produce informative meioses for mapping
the disease and marker loci, it must be doubly heterozygous and produce progeny
whose genotypes allow inference of the gamete type that was contributed by the
individual in question (e.g., whether it was a parental or recombinant type of gamete).
Given the information provided in the pedigree, we see that only II-3 fits these criteria.
Note: Genotypes of individuals in generation III provide information for mapping, but
it is not because of their meioses (i.e., we are not looking at the progeny of
generation III!).
2. (2 points) Calculate a LOD score for linkage between the high blood pressure
locus and marker A at a distance of 10 cM.
a. 0.16384
b. -0.78558
c. -2.5406
d. -0.18352
e. None of the above
Answer c. We must inspect the 5 testcross progeny present in generation III and
consider the gamete types produced by II-3. Note that parental gametes of II-3 are
either A1d or A2D (where D denotes the dominant disease allele), and recombinant
gametes are A1D or A2d. Among the 5 testcross progeny, there is one that is the
product of a parental gamete of II-3 (individual III-3), whereas the remaining 4
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progeny are each the product of recombinant gametes of II-3. Thus, the appropriate
calculation of the LOD for 10 cM is:
LOD = Log [(0.054)*(0.451)/.255] = -2.5406 (where Log denotes the base 10 log of the
quantity in square brackets).
3. (1 point) In a complementation test the number of complementation groups
indicates
a.
b.
c.
d.
e.
the number of genes required for a specific phenotype.
the penetrance of the phenotype.
the number of phenotypes for a gene.
the number of chromosomes in an organism.
the quantity of gene product required for a phenotype.
Answer a.
Use the information below to answer the next two questions:
In a population, two different rare genetic diseases (extra finger and an eye disease)
are observed. In the pedigree below (taken from a family in the population), a cross
represents the occurrence of an extra finger and a black square represents the
occurrence of the eye disease:
4. (1 point) In the above pedigree, what is the most likely pattern of inheritance for
the occurrence of an extra finger?
a. autosomal dominant
b. autosomal recessive
c. X-linked dominant
d. X- linked recessive
e. Y-linked inheritance.
Answer a.
5. (1 point) In the above pedigree, what is the most likely pattern of inheritance for
the occurrence of eye disease?
a. autosomal dominant
b. autosomal recessive
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c. X-linked dominant
d. X- linked recessive
e. Y-linked inheritance.
Answer d.
Use the information below to answer the next two questions:
The following three recessive markers are known in lab mice: h, hotfoot; o, obese;
and wa, waved. A trihybrid of unknown origin is testcrossed, producing the
following offspring:
hotfoot, obese, waved
357
hotfoot, obese
74
waved
66
obese
79
wildtype
343
hotfoot, waved
61
obese, waved
11
hotfoot
9
6. (2 points) The map distance between the gene pairs o and h, and h and w are
respectively:
a. 14 cM, 16 cM
b. 16 cM, 14 cM
c. 16 cM, 16 cM
d. 14 cM, 14 cM
e. None of the above
Answer c. Note that there are excesses of the triple mutant and triple wild type
progeny, suggesting linkage of the three genes and a triple heterozygote parent in
which all mutant alleles are on one of the homologs, and all wild type on the other.
Looking at the pairs of genes (o, h and h, w individually) we see that there are
79+61+11+9=160 recombinants for o and h. And there are 74+66+11+9=160
recombinants for h and w. Since there are 1000 progeny in total, this gives map
distances of 16 cM for each pair.
7. (1 point) The linear order of the genes is:
a. o,h,w
b. o,w,h
c. h,w,o
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d. w,o,h
e. Both b. and c. above.
Answer a. As inferred from the low frequency of double crossovers and maps
distances above.
8. (1 point) In cocker spaniels, black color (B) is dominant over red (b), and solid
color (S) is dominant over spotted (s). The genes for these two characters are
located on different chromosomes. Suppose that a dog with genotype B/b s/s is
mated to a dog with genotype b/b S/s. What fraction of their offspring are expected
to be black and spotted?
a. 1/16
b. ¼
c. 9/16
d. 3/4
e. None of the above.
Answer b.
9. (1 point) In a cross between a female with genotype A/a B/b c/c d/d and a male
with genotype A/a b/b C/c D/d, what proportion of the progeny will be
phenotypically identical to the female parent? (Assume independent assortment of
all genes and complete dominance).
a. 1/8
b. 3/32
c. 1/64
d. 9/16
e. None of the above
Answer b.
10. (1 point) A common way to separate DNA molecules and fragments is gel
electrophoresis. When the DNA fragments are run on a gel for a Southern blot, what
does the separation depend upon?
a.
b.
c.
d.
e.
Radioactivity and fluorescence
Number and size of fragment
Size and radioactivity
Size and charge
Charge and radioactivity
11. (1 point) In Mendelian genetics, null mutations of different genes may interact
to produce deviations from the classic dihybrid ratio. Which of the following is most
likely due to interactions between genes in the same linear biosynthetic pathway
(i.e., PrecursorIntermediateFinal product).
a.
b.
c.
d.
e.
9:3:3:1
9:7
12:3:4
15:1
None of the above
Answer b. Interactions between null mutations of two genes in this pathway will
produce 9:7 ratio of wildtype:mutant.
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Use the information below to answer the next two questions:
The recessive mutations vg (vestigial wings) and br (brown eyes) identify two
autosomal genes on the second chromosome of Drosophila melanogaster. When
females heterozygous for these genes were crossed with vestigial winged, brown
eyed males, the following classes and numbers of progeny (out of 1000) were
obtained:
Phenotype
_
wildtype wings, wildtype eyes
vestigial wings, wildtype eyes
wildtype wings, brown eyes
vestigial wings, brown eyes
Number
180
317
308
195
12. (1 point) Based upon these results, the map distance between the vg and br
genes is estimated to be:
a. 18.8 map units.
b. 31.2 map units.
c. 37.5 map units.
d. greater than 50 units because all four classes of offspring were observed.
e. None of the above.
Answer c. Note that the double heterozygote must be of the type + br / vg + , as the
excess types of progeny are not either double wild type or double mutant.
13. (2 points) Which homozygous parent stocks could be used to produce the
heterozygous females used in the cross described above?
a. wildtype wings, wildtype eyes X vestigial wings, brown eyes
b. wildtype wings, brown eyes X vestigial wings, wildtype eyes
c. wildtype wings, wildtype eyes X wildtype wings, wildtype eyes
d. vestigial wings, brown eyes X vestigial wings, brown eye
e. None of the above
Answer b. Only this pair of parental stocks could give rise to the double heterozgote
parent noted in the above answer.
Use the information below to answer the next question:
A fruit fly of genotype a+/a b+/b is crossed to another fruit fly of genotype a/a b/b.
The progeny of this cross were:
Genotype
a+/a b+/b
a/a
b/b
a+/a b/b
a/a b+/b
Number of Individuals
39
33
10
18
14. (2 point) The Chi-Square associated with the test for independent assortment of
the two genes is:
a. 20.007 (with 3 degrees of freedom)
b. 20.007 (with 1 degrees of freedom)
c. 21.360 (with 1 degrees of freedom)
d. 21.360 (with 3 degrees of freedom)
e. Cannot be calculated from the information given.
Answer b or d. As stated, the null hypothesis is independent assortment. By rejecting
the null, we might be inclined to accept the alternative, namely that the genes are
linked. But rejection of the null could also be due to differential viability. Answer “b”
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takes into account the possibility of differential viability of the mutation-bearing
gametes, whereas answer “d” does not. However, I noted that there has been confusion
over this due to inconsistency in the way in which the text and MegaManual solves
these types of problems. Technically, the method that takes viability into account is
more accurate, but I will also accept the method that uses a 1:1:1:1 ratio (no viability
differences) as the null hypothesis.
15. (1 point) Non-homologous chromosomes:
a. are genetically identical.
b. are similar, but there are a few minor genetic differences between them.
c. pair with each other during meiosis I.
d. are found in mitochondria.
e. None of the above.
Answer e.
16. (1 point) An individual who is heterozygous has brittle bone disease disease (in
the mutant heterozygote an abnormal copy of the protein wraps around one or two
copies of the other and distorts the conformation of the functional trimeric
molecule). This type of mutation is called a:
a. a leaky mutation
b. a dominant-negative mutation
c. recessive mutation
d. haplo-insufficient mutation
e. haplo-sufficient mutation
Answer b. Note: “Haplo-insufficient mutation” is an incorrect answer. While this is a
dominant mutation, the phenotypic basis does NOT arise because two functional doses
of an enzyme are required for completion of a biosynthetic pathway. Rather, the
mechanism is that of “poisoning” of one protein unit by another (a dominant
negative), as discussed in the text and class (and contrasted with haplo-insufficient
mutations) under the topic of dominant mutations (see Lecture 8).
17. (1 point) A man with blood type "B" marries a woman with blood type "B".
Which of the following blood types might you expect to see in their children?
a. Type A and B would be possible.
b. Only type O would be possible.
c. Only type B would be possible.
d. Types B and O would be possible.
e. Only Type AB would be possible
Answer d.
18. (1 point) After finding a new species of rabbit in the deep forest of Lac-St-Jean,
scientists decided to analyze its karyotype. Such analysis, performed on the white
blood cells of a male, revealed 12 chromosomes (you can assume that the sex
determination system in these rabbits is equivalent to the one found in humans).
Based on these data, the minimum number of contigs that a female from this animal
species will have upon full sequence and analysis of its genome will be:
a. 7
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b. 6
c. 24
d. 12
e. 13
Answer b.
19. (2 points) Your thesis supervisor, who is working on a new rabbit species
recently found in the deep forest of Lac-St-Jean, has isolated a 1 kilobase cDNA
fragment from the liver of this mammal. The mRNA (3 kb in length) from which the
cDNA was derived seems to be expressed only during the (very cold) winter and
could protect the rabbit against freezing temperature. Your supervisor asks you to
isolate the promoter from this gene. Which combination of the 15 statements
mentioned below could be successful?
1)
Isolate genomic DNA from the blood of this mammal, shear it and produce
cohesive ends using appropriate enzymes.
2)
Isolate RNA from the liver cells of this mammal.
3)
Isolate and analyze the cDNA clones.
4)
Generate cDNA molecules from the sample(s) you have isolated.
5)
After sequencing the 1 kb cDNA fragment, generate primers and use PCR to
amplify the whole gene.
6)
Make a genomic library from the DNA you have isolated.
7)
Introduce the amplified DNA in a cosmid vector.
8)
Label (with radioactive nucleotides) your 1 kilobase cDNA fragment and
screen the appropriate library that you have generated.
9)
Isolate RNA from the blood of this rabbit.
10)
Repeat the process until you clone the promoter of this gene.
11)
Use chromosome jumping to isolate genomic DNA fragments.
12)
Make a library by introducing your cDNAs into a plasmid vector.
13)
Use chromosome walking to isolate all the needed clones from the
appropriate library, and generate a map using restriction enzymes.
14)
Introduce the genomic DNA into a BAC vector having compatible ends.
15)
Use the longer DNA clone you have identified and use it as a probe to screen
a cDNA library made from RNA isolated from the liver of this mammal.
a. 2); 4); 12); 8); 15); 10).
b. 2); 5); 7); 8); 15); 10).
c. 9); 4); 5); 7); 8); 10).
d. 1); 14); 8); 13); 10).
e. 1); 6); 8); 11); 15); 10).
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Answer d.
20. (1 point) You have just cloned a DNA fragment that you have isolated from a
genomic library that your supervisor has generated from the DNA of a new species
of moose found in Lac-St-Jean. You know that this DNA fragment contains at least
three large exons and two small introns, so you perform a Northern blot analysis
using as a probe either your genomic fragment or an actin cDNA (from this moose
species). You obtained the illustrated results from the RNA isolated from five
different tissues from the moose:
M: muscle
A: adrenals
R: retina
I: intestines
O: ovary
M
A
R
I
O
3 kb
2 kb
Your genomic
1 kb
fragment
Actin cDNA
Please read carefully the following putative conclusions from these results:
1) The three bands seen in the RNA from the intestines (I) correspond to the three
exons present in your genomic fragment.
2) The muscle cells do not express the gene encoded by the DNA fragment that you
have used as a probe.
3) The fragments of 1 kb and 2 kb seen in the intestine sample (I) may represent a
partial restriction enzyme digest of this sample.
4) The RNA from your gene is likely alternatively spliced in the different tissues
analyzed.
5) This Northern analysis allows you to definitely conclude that from the five
tissues analyzed, the adrenals (A) are expressing your gene the most, while the
ovary (O) is expressing it the least.
Based on the previous statements, which one of the following analyses is RIGHT:
a. Only statement 4) is true, all others are false.
b. Statements 1), 4) and 5) are true, while statements 2) and 3) are false.
c. All statements are true.
d. Statements 2) and 4) are true, while statements 1), 3) and 5) are false.
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e. Statements 4) and 5) are true, while statements 1), 2) and 3) are false.
Answer a.
21. (1 point) Which one of the following statements is incorrect?
a. The C-value paradox defining the lack of correlation between the size of the
genome and the biological complexity of an organism can be explained
largely by the presence of repetitive DNA.
b. It is possible to generate genomic DNA fragments without the use of
restriction endonucleases, and to eventually make genomic libraries from
such fragments.
c. Cesium chloride gradients can be used to isolate satellite DNA from
eukaryotes.
d. A cosmid vector is a plasmid containing the cos sites of the bacteriophage 
and can thus be packaged in virus particles.
e. Most RFLPs are multi-allelic and can thus be used to distinguish many
different individuals in a given population.
Answer e
22. (1 point) Which one of the following statements is true?
a. The presence of an origin of replication (ori) in a plasmid will allow its
replication as an extrachromosomal entity in both prokaryote and eukaryote
cells.
b. Because highly repetitive DNA sequences have been amplified during
evolution, they are present at more than 100,000 copies/genome and will
thus be seen on an ethidium bromide stained agarose gel upon digestion of
human DNA.
c. In a plasmid containing an ampicillin resistance gene, cloning a DNA
fragment into the sequence of the lacZ gene and plating the transformed
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.
d. Slipped mispairing during DNA replication is the main cause of minisatellite
polymorphism.
e. 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. Answer c.
23. (1 point) The following figure represents a VNTR analysis of seven different
individuals including the mother (M), her child (C) and five other individuals (1, 2, 3,
4 and 5). Which one of the following statements best describes these results:
M C 1 2 3
4
5
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a. Individual 3 is the real biological father of the child (C).
b. Based on this analysis, it is impossible to determine who is the real biological
father from the 5 individuals (1 to 5).
c. Individual 5 is the real biological father of the child (C).
d. Individual 1 is the real biological father of the child (C).
e. None of these individuals could be the real biological father of this child (C).
Answer b.
24. (1 point) Using ASO hybridization (comparing hybridization results obtained
with normal and disease ASOs), your supervisor asked you to analyze the DNA from
the individuals represented in the following pedigree. Assuming that the disease
allele is dominant and that it represents the only mutation leading to the disease,
which one of the following statements is true?
A
C
D
E
B
F
G
H
I
J
K
L
M
N
O
P
a. DNA from individuals E and M will show hybridization with the disease allele
only.
b. Not being from the same family, DNA from individuals D, F, H, L, N, and P will
show no hybridization with the ASOs used in this analysis.
c. DNA from individual B will show hybridization with both ASOs.
d. Although individuals C, G, I, J, K, and O do not have the disease, their DNA could
still hybridize with the disease allele.
e. Assuming that individuals M and N will have five children, all of them will show
hybridization with both ASOs when analyzed later on.
Answer c.
25. (1 point) Please read carefully the following statements:
1.
2.
3.
4.
5.
A mutation occurring in a DNA molecule will be seen as an RFLP only if this
mutation is located directly in the restriction site of a given enzyme.
Using a labeled 10 kb genomic fragment as a probe on a microchip will never
produce a hybridization signal since the microchips contain oligonucleotides
specific for RNA molecules.
A cosmid is a smaller vector containing sequences derived from a virus and
can thus be used to make cDNA libraries.
If one wants to construct a representative genomic human DNA library,
complete digestion with the EcoR1 restriction enzyme cannot be used to
generate human genomic DNA fragments to be inserted into the
bacteriophage  vector.
During chromosome jumping, a fragment of a clone (the probe) is used to
find the next adjacent genomic fragments.
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Based on the previous statements, which one of the following analyses is RIGHT:
a.
b.
c.
d.
e.
Statements 2), 3) and 5) are true, while statements 1) and 4) are false.
Only statement 1) is false, all others are true.
All statements are true.
Statements 2) and 4) are true, while statements 1), 3) and 5) are false.
Only statement 4) is true, all others are false.
Answer e.