BL414 Genetics Spring 2006
Test 3
page 1
150 pts. total, 15% of final grade
April 24, 2006
Your name:____________________Key_________________________
1. (2.5pts) The human karyotype consists of how many chromosomes in
total? ____46____
(2.5pts) How many of these are autosomes? ____44___
2. (5pts) Chromosomes seen in a karyotype are in what stage of
mitosis?____________metaphase___________
3. (5pts) Do the dark bands seen in karyotype chromosomes contain
heterochromatin or euchromatin? _____heterochromatin___________
4. (5pts) Methylation of DNA is associatiated with what effect on the activity
of a nearby gene? __________silencing or imprinting_______________
5. (5pts) What effect does a shortening of polyA tail length have on mRNA
stability? ___________decreases stability_____________
6. (5pts) Sequence-specific double-stranded RNA introduced into cells
causes the down-regulation of the activity of a gene – this process targets
the degradation of what molecule? ________mRNA_________________
7. (5pts) The key tumor suppressor protein involved in control of DNA
damage, which is mutated in many cancers is ____p53________.
8. (5pts) Some of the proteins involved in cell-cycle control go through cycles
of high ad low levels of expression during the cell cycle. These proteins
are called ______________cyclins__________.
9. (5pts) When a normal cell is damaged, it will stop at certain points in the
cell-cycle, such as the G1 S transition. These are called
____________checkpoints_________.
10. (5pts) If a gene involved in cell cycle regulation undergoes a gain-offunction mutation that promotes cell proliferation, it becomes what type
of gene (or mutation)? _________oncogene________
11. (5pts) According to estimates in your textbook, what percentage of cancer
comes from familial cancer syndromes? _____1%_________
BL414 Genetics Spring 2006
Test 3
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150 pts. total, 15% of final grade
12. (5 pts) A group of several genes located in tandem and regulated by a
single promoter is called a(n) ___________operon_________.
13. (5 pts) If two chromosome arms are joined by fusing the centromeres of
two acrocentric chromosomes, this is called a _Robertsonian tranlocation.
14. (5 pts) What is the role of the GAL4 protein in yeast?
_________transcriptional activator______________
15. (5 pts) Name a protein that represses GAL4.______GAL80_________
16. (5 pts) Name a structural motif found in GAL4 and describe its function.
Helix-turn-helix or zinc finger motif – both involved in DNA binding
17. (5pts) The gene order on chromosome 3 of C. elegans is usually:
____sma-1_______mab-5_________unc-32_________lin-12_______
Using DNA from a mutant worm, you perform PCR on sequences in
this region of chromosome 3 and find the following arrangement.
Name the type of chromosome abnormality:_inverted duplication__
____sma-1_____mab-5_____unc-32_____lin-12______ lin-12_____unc-32____
18. (10pts) What does the term “13p42.1” indicate about the location of a
gene?
The gene is located on chromosome 13, on the p arm (shorter arm), on the first
band in subregion 42.
19. (10pts) Draw the following types of chromosomes, including the position
of the centromere:
a. (2.5pts) Metacentric chromosome – centromere is at the center –
arms are of equal length
b. (2.5pts) Submetacentric chromosome – centromere is slightly offcenter, one arm is slightly longer than the other one
BL414 Genetics Spring 2006
Test 3
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150 pts. total, 15% of final grade
c. (2.5pts) Acentric chromosome – no centromere
d. (2.5pts) Dicentric chromosome – two centromeres
20. (10pts) Diagram the mechanism proposed for expansion of triplet
nucleotide repeats. Give one example of this type of nucleotide expansion
in humans.
In humans, this type of mutation in the gene for FMR1 protein causes Fragile
X syndrome. There are other triplet expansion diseases, such as Friedreich’s
ataxia, spinal and bulbar muscular atrophy, Huntington disease, and
spinocerebellar ataxia.
21. (10pts) Can transposons cause mutations? What type of mutation could
they cause?
Yes, transposons can cause mutations. A transposon could insert into a region
of a chromosome containing a gene – causing an insertion. The insertion could
disrupt the promoter function, causing a loss of function of that gene, or could
BL414 Genetics Spring 2006
Test 3
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150 pts. total, 15% of final grade
insert into the coding region of the gene, causing a frameshift mutation.
Recombination during meiosis could occur between two homologous
transposon sequences and could lead to a duplication or deletion of the genetic
material in between the two transposons.
22. (10pts) Laboratory mice used for a genetic screen were exposed to the
mutagen EMS. The parental generation exhibited a shortened life span
and were found to have a much higher than normal rate of tumor
formation. Explain these findings.
EMS causes single base transitions. The parental generation of mice were
exposed to a high dosage of EMS and presumably had many mutations in their
somatic tissue and their germ line tissue. A high number of tumors formed in
the somatic tissue of these mice because two or more genes involved in cancer
were mutated by EMS. These genes were either tumor-suppressor genes like
p53, which undergo a loss-of-function mutation in order to increase the
likelihood of tumor formation, or proto-oncogenes which would undergo a
gain of function mutation in order to increase the likelihood of cancer.
23. (10pts) One of the offspring from this screen was found to have a specific
developmental defect and PCR revealed that a transcription factor gene
hox-1 was affected. The gel below shows the restriction fragments for
DNA containing this gene. Explain the mutation and the potential effect
on the animal’s development.
Normal mouse
__________
Mutant offspring
__________
__________
__________
__________
The mutant has one missing restriction fragment for the gene. EMS causes
single base changes, so it is unlikely that a large deletion or inversion
occurred. EMS probably caused a mutation that changed the restriction site so
that it was no longer recognized by the restriction enzyme. The mutation was
mostly likely a single base transition resulting in a missense or nonsense
mutation that disrupted the function of the hox-1 gene product.
24. (10pts) Laboratory mice were exposed to the mutagen proflavin, and the
first generation of offspring were found to be deficient in the levels of
collagen protein in their bodies. DNA analysis using restriction digests
BL414 Genetics Spring 2006
Test 3
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150 pts. total, 15% of final grade
did not reveal any difference in restriction fragment lengths from wild
type mice. Give one possible explanation for the protein deficiency.
Normal mouse
Mutant offspring
__________
__________
__________
__________
__________
__________
Proflavin is an intercalating agent that causes single base insertion or
deletions. The proflavin caused a mutation that was only one base pair
different in length from the normal DNA fragments, which would not be a
visible change in the DNA restriction fragment gel. A single base insertion or
deletion in the gene for collagen may have caused a nonsense mutation, or a
frameshift mutation. A nonsense mutation could cause the mRNA to be
degraded by mRNA surveillance machinery, or could result in translation of a
truncated protein, which may not function properly in the cell. A frameshift
mutation could also result in a protein of abnormal sequence and length. This
abnormal protein may not properly undergo translation termination and be
subsequently degraded by protein degradation machinery, or the protein
produced will be non-functional. Either a nonsense or a frameshift mutation
could explain deficient levels of collagen protein in the offspring.
Bonus questions:
25. (25 pts) What is the mechanism in humans for dosage compensation of the
X chromosome. In what sex does the mechanism occur? Describe the
process, mention the role of mosaicity, the X-inactivation center, Xist, and
Barr bodies.
X-inactivation is the mechanism for dosage compensation in humans. Xinactivation occurs in females, during early embryonic development, causing
one X chromosome to be inactivated in every cell of a female. The paternal or
maternal X chromosome is randomly inactivated. All descendants of the
embryonic cells keep the same X chromosome inactivated. X-inactivation
initiates on the chromosome at the XIC – “X-inactivation center” and the
chromosome condenses and becomes coated with an RNA transcript encoded
by the Xist gene. Heavy DNA methylation occurs along the inactivated
chromosome. In some cells, the inactivated X chromosome is visible as a
densely staining Barr body. Because of X-inactivation, human females are
mosaic for X-linked genes, because in different cells, either a maternal or
paternal X-linked gene is the active copy.
BL414 Genetics Spring 2006
Test 3
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150 pts. total, 15% of final grade
26. (15 pts) Explain the difference between autopolyploidy and
allopolyploidy in plants.
Polyploidy refers to multiple sets of homologous chromosomes. Higher ploidy
is often seen in plants. Autopolyploidy refers to a species in which all of the
chromosomes in the species derive from a single ancestral set. Allopolyploidy
is the case where complete sets of chromosomes have come from two or more
ancestral species.