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Biology 4285
Human Molecular Genetics
Midterm I
February 1, 2007
8:30 – 10:00 am
READ INSTRUCTIONS CAREFULLY:
Answers are to be written in this booklet only. THERE ARE 7 PAGES TO THIS EXAM,
INCLUDING THIS PAGE.
Please answer every question. YOU MAY USE POINT FORM!!!!!
Total marks on this exam are 50.
Maximum time is 1 hr 20 min
GOOD LUCK!
1
1. Your PhD project is to analyze igf2 and h19 knockout mice. A previous student has created
the heterozygotes and you must now cross them to investigate phenotype.
You cross a male h19-/+ heterozygote with a female igf2-/+ heterozygote. Use a Punnett
square to illustrate genotype and describe the phenotype of the expected progeny. (10)
Male
H19+
Igf2Female
Igf2+
H19-
A.
H19
Igf2 -
C.
H19Igf2-
B.
H19
Igf2
D.
H19Igf2+
A. Paternal h19, Maternal igf2- : Since igf2 is silenced on the maternal chromosome because of
h19, and igf2 is normal on the paternal chromosome, progeny would be normal.
B. Normal progeny.
C. Paternal h19-, Maternal igf2-: Normal igf2 expression since h19 is deleted on the paternal
chromosome, igf2 is deleted from the maternal chromosome, where it is normally silenced.
Normal progeny.
D. Paternal h19-, maternal igf2. Maternal igf2 locus will be silenced, and paternal igf2 will be
expressed because of h19-. Normal progeny.
2
2. Discuss the relationship between CDKN1C and IGF2 expression. Describe how the failure of
this regulation can cause human disease (10).
- CDKN1C and IGF2 are both expressed from the same chromosomal region on chromosome 11.
They are imprinted, with opposing parent of origin expression. IGF2 is a growth factor, while
CDKN1C is a cell cycle inhibitor. Imprinting of each regulates gene dosage during early
development.
- Misregulation of CDKN1C and/or IGF2 can lead to a condition called Beckwith-Wiedemann
Syndrome.
- Overexpression of IGF2, and/or underexpression of CDKN1C.
- Increased growth during fetal development and after birth.
- Large organs, large tongue, hypoglycemia.
- Increased risk of cancer such as Wilms tumor.
Epigenetic mistakes :
- loss of KvDMR methylation
- Maternal H19 methylation
- Errors in CDKN1C gene causing loss of gene expression
- UPD causing biallelic expression of IGF2 and/or suppression of CDKN1C
- Other chromosome abnormalities
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3. Rett syndrome is noted primarily in females. Provide two theories that attempt to explain this
gender bias, providing supporting arguments for each (10).
- Errors in the DNA methyl-binding protein, MeCP2
- 80% of Retts caused by mutations within MeCP2 gene.
- Neurological disorder
- Lower expression of MeCP2 in females results in reduced gene silencing.
Theory 1 for female bias:
Males are hemizygous for X chromosome gene expression, and a mutation of MeCP2 will lead
to a loss or partial inactivation of the MeCP2 function. The severe phenotype in males results in
early lethality.
Theory 2 for female bias:
Theory 1 does not explain why there are some males with Rett syndrome, if it should be lethal.
Kleinfelters syndrome (47 XXY) is possible for some of these males, but the alternate theory is
that the mutations to MeCP2 are more likely to occur in the male germ line. During
spermatogenesis, DNA methylation is higher than in females, and this might cause a mutation of
MeC –> T through deamination. Thus, the sporadic mutations that cause most of Rett’s
syndrome are more likely to occur in male germ line, and when he passes on this mutant X
chromosome it will always be to a daughter.
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4. Describe the process of X – chromosome inactivation and the involvement of Xist. Explain
why a female with a balanced, reciprocal translocation between one X chromosome and an
autosome, it is the intact X chromosome that is usually inactivated. (10)
-
X chromosome inactivation limits only one X-chromosome per cell.
Xist is a RNA molecule expressed cis to the X chromosome to be inactivated.
Xist “coats” the X chromosome and recruits Histone- and DNA- modifiying enzymes.
Histone deacetylation and DNA methylation lead to chromosome condensation and
inactivation
The mechanism of inactivating all but one X chromosome is called count, and the
mechanisms to decide which X chromosome inactivates is called choice.
Xist expression is controlled by Tsix, the anti-sense RNA molecule of Xist.
Tsix expression is controlled by Xcite, another regulatory protein which together control
both choice and count.
-
Balanced translocation would place some X chromosome onto an autosome. This
autosome cannot be inactivated.
-
Gene dosage of X chromosome and autosome would only be correct if the intact X
chromosome were inactivated.
-
Thus, there would be selection of early embryonic cells for the correct X and autosome
gene dosage.
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I1
II1
II2
I2
II3
II4
II5
III1
I1
I2
II1
II2
II3 II4
II5
III1
6.5
2.2
5. The Southern blot shown analyzed DNA from individuals in a family shown above. The DNA
was hybridized with a probe to the fmr1 gene. DNA samples were digested with EcoRI and EagI
(methylation sensitive). (10)
1. Shade in the persons in each generation who you think would have the disease phenotype.
2. Who might you think would be the most affected by this disease? Why? (use the next blank
page).
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III1 will be the most affected: The probe binds to both of her X chromosomes. One chromosome
is producing a very high molecular weight fragment, consistent with this chromosome being
hypermethylated on DNA surrounding the probe. The other chromosome is producing a
fragment that is also methylated, and represents the inactive chromosome. Thus, she has no fmr1
gene that is hypomethylated, and would likely not express any of this gene. Thus, she would be
very severe for Fragile X syndrome.
You would not be able to say if II3 has the disease phenotype. She has a hypermethylated fmr1
gene but that appears to be restricted to the X chromosome.
II2 might or might not have a very strong phenotype. She has some active X chromosome with
hypermethylated fmr1.
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