Добавлен: 10.02.2019
Просмотров: 10167
Скачиваний: 3
7.
The mechanism of X chromosome inactivation also involves the
XIST gene,
which
encodes for
XIST RNA
that is the primary signal for spreading the inactivation along the X
chromosome from which it is expressed.
G. Micro RNA (miRNA) Genes.
The miRNA genes encode for
miRNAs
that block the expression of
other genes.
H. Antisense RNA Genes.
The antisense RNA genes encode for
antisense RNA
that binds to
mRNA and physically blocks translation.
I. Riboswitch Genes.
The riboswitch genes encode for
riboswitch RNA,
which binds to a target
molecule, changes shape, and then switches on protein synthesis.
A. Chemical Modification of DNA.
1.
DNA can be chemically modified by
methylation of cytosine nucleotides
using
methylating
enzymes
.
An increased methylation of a DNA segment will make that DNA segment less
likely to be transcribed into RNA and hence any genes in that DNA segment will be
silenced (i.e., cc
methylation of DNA
silenced genes
). The DNA nucleotide sequence is not
altered by these modifications.
2.
DNA methylation plays a crucial role in the epigenetic phenomenon called
genomic
imprinting
.
Genomic imprinting is the differential expression of alleles depending on
whether the allele is on the paternal chromosome or the maternal chromosome.
3.
We are generally accustomed to the normal situation where the allele on the paternal
chromosome and the allele on the maternal chromosome are expressed or silenced at the
same time. When a gene is imprinted, only the allele on the paternal chromosome is
expressed while the allele on the maternal chromosome is silenced (or visa versa). Hence,
there must be some mechanism that distinguishes between paternal and maternal alleles.
4.
During male and female gametogenesis, male and female chromosomes must acquire
some sort of
imprint
that signals the difference between paternal and maternal alleles. The
role of genomic imprinting is highlighted by several rare diseases like the Prader-
Willi/Angelman syndromes (see Chapter 11-II-B), complete hydatidiform moles, and
Beckwith-Wiedemann syndrome that show abnormal DNA methylation patterns.
B. Clinical Considerations.
1. Complete hydatidiform mole.
a.
A hydatidiform mole (complete or partial) represents an abnormal placenta character-
ized by marked enlargement of chorionic villi. A complete mole (no embryo present) is
distinguished from a partial mole (embryo present) by the amount of chorionic villous
involvement.
b.
A complete mole occurs when an “empty” ovum is fertilized by a haploid sperm, which
then duplicates. This results in a
46,XX karyotype
with all nuclear chromosomes of
pater-
nal
origin. A
46,YY karyotype
complete mole does not occur since this is a genetic lethal
condition.
c.
A complete mole may also occur when an “empty” ovum is fertilized by two sperm
(3%–13% of complete moles). This results in a
46,XY karyotype
with all nuclear chromo-
somes of
paternal
origin.
d. Clinical features include:
gross, generalized edema of chorionic villi forming grapelike,
transparent vesicles; hyperplastic proliferation of surrounding trophoblastic cells;
absence of an embryo/fetus; preeclampsia during the first trimester; elevated hCG lev-
els (
100,000 mIU/mL); and an enlarged uterus with bleeding; follow-up visits after a
mole are essential because 3%–5% of moles develop into gestational trophoblastic neo-
plasia.
Chapter 1
The Human Nuclear Genome
5
LWBK274-C01_01-11.qxd 06/02/2009 03:29 PM Page 5 Aptara
2. Beckwith-Wiedemann syndrome (BWS).
a.
BWS is caused by abnormal transcription and regulation of various genes located in the
imprinted domain on chromosome 11p15.5.
b.
The causes of BWS involve:
(i)
The
KCNQ1OT1 gene
on chromosome 11p15.5 which encodes for a
paternally
expressed K
+
voltage-gated ion channel
.
In
60% of BWS cases, KCNQ1OT1 gene
hypomethylation is detectable.
(ii)
The
H19 gene
on chromosome 11p15.5 which encodes for a
maternally expressed
H19 untranslated mRNA
that functions as a tumor suppressor. In
7% of BWS cases,
H19 gene hypermethylation is detectable.
(iii)
The
CDKN1C gene
on chromosome 11p15.5 which encodes for
cyclin-dependent
kinase inhibitor 1C
that functions as a tumor suppressor. In
40% of familial BWS
cases, CDKN1C gene mutations have been detected.
(iv)
In
20% of BWS cases,
paternal uniparental disomy
has been detected.
c. Prevalence.
The prevalence of BWS is 1/14,000 births.
d. Clinical features include:
macrosomia; macroglossia; visceromegaly; embryonal tumors
(e.g., Wilms tumor, hepatoblastoma, neuroblastoma, rhabdomyosarcoma); omphalo-
cele; neonatal hypoglycemia; ear creases/pits; adrenocortical cytomegaly; and renal
abnormalities.
(C) Chemical Modification of Histones.
Histone proteins can be chemically modified by
acetylation,
methylation, phosphorylation
,
or
addition of ubiquitin
(
sometimes called
epigenetic marks
or
epige-
netic tags
). An increased acetylation of histone proteins will make a DNA segment more likely to
be transcribed into RNA and hence any genes in that DNA segment will be expressed (i.e., cc
acety-
lation of histones
expressed genes
). The mechanism that determines the location and combi-
nation of epigenetic tags is unknown. This is another part of the
epigenetic code
that must be
deciphered.
A. Satellite DNA.
Satellite DNA is composed of very large-sized blocks (100 kb
→
several Mb) of
tandemly repeated noncoding DNA
.
Large-scale variable number tandem repeat (VNTR) polymor-
phisms
are typically found in satellite DNA. The function of satellite DNA is not known.
B. Minisatellite DNA.
Minisatellite DNA is composed of moderately-sized blocks (0.1 kb
→
20 kb) of
tandemly repeated noncoding DNA
.
Simple VNTR polymorphisms
are typically found in
minisatellite DNA.
Telomeric DNA
(a type of minisatellite DNA) allows for the replication of
DNA ends in the lagging strand during chromosome replication.
C. Microsatellite DNA (Simple Sequence Repeat; SSR).
Microsatellite DNA is composed of small-
sized blocks (
100 bp) of
tandemly repeated noncoding DNA
.
Simple VNTR polymorphisms
are
typically found in microsatellite DNA. The function of microsatellite DNA is not known.
D. Transposons (Transposable Elements; “Jumping Genes”).
Transposons are composed of
inter-
spersed repetitive noncoding DNA,
which that make up an incredible
45% of the human nuclear
genome
.
Transposons are mobile DNA sequences that jump from one place in the genome to
another (called
transposition
).
1. Types of transposons.
a. Short interspersed nuclear elements (SINEs).
The
Alu repeat
(280 bp) is a SINE that is the
most abundant sequence in the human genome.
When Alu repeats are located within
genes, they are confined to introns and other untranslated regions.
b. Long interspersed nuclear elements (LINEs).
LINE 1 (
6.1 kb) is the
most important human
transposon
in that it is still actively transposing (jumping) and occasionally causes dis-
ease by disrupting important functioning genes.
c. Long terminal repeat (LTR) transposons.
6
BRS Genetics
LWBK274-C01_01-11.qxd 06/02/2009 03:29 PM Page 6 Aptara
d. DNA transposons.
Most DNA transposons in humans are no longer active (i.e., they do
not jump) and therefore are considered
transposon fossils
.
2. Mechanism of transposition (Figure 1-2A,B).
Transposable elements jump either as double-
stranded DNA using
conservative transposition
(a “cut-and-paste” method) or through a
RNA intermediate using
retrotransposition.
a. Conservative transposition.
In conservative transposition, the transposon jumps as dou-
ble-stranded DNA.
Transposase
(a recombination enzyme similar to an integrase) cuts the
transposable element at a site marked by
inverted repeat DNA sequences
(about 20 base
pairs long). Transposase is encoded in the DNA of the transposable element. The transpo-
son is inserted at a new location, perhaps on another chromosome. This mechanism is
similar to the mechanism that a
DNA virus
uses in its life cycle to transform host DNA.
Chapter 1
The Human Nuclear Genome
7
A
C
E
F
D
B
Transposase
DNA repair
T
Inverted
repeat
DNA
sequences
Host chromosome
+
Target chromosome
T
T
Host chromosome
Target chromosome
Integrase
RNA
copy
RT
Transcription
Reverse
transcriptase
Transposase
DNA repair
T
T
T
T
T
Target DNA
Active
gene
Decreased
gene
expression
Mutation
Active
gene
Target DNA
Inactive
gene
Tet
R
Tet
R
Tet
R
Bacterial DNA
Phage DNA
Phage DNA with Tet
gene infects other
bacteria and confers
tetracycline resistance
R
Cut or nick site
+
FIGURE 1-2. Transposition. (A,B) Mechanisms of transposition. (A)
Conservative Transposition. (B) Retrotransposition.
(C-F) Transposons and genetic variability. (C)
Mutation at the former site of the transposon (D) Level of gene expression
(E)
Gene Inactivation. (F) Gene Transfer. T
transposon; RT: RNA code for reverse transcriptase (^): cut sites Tet
R
: gene
for tetracycline resistance.
LWBK274-C01_01-11.qxd 06/02/2009 03:29 PM Page 7 Aptara
b. Retrotransposition.
In retrotransposition, the transposon jumps through a RNA inter-
mediate. The transposon undergoes transcription, which produces a RNA copy that
encodes a reverse transcriptase enzyme.
Reverse transcriptase
makes a double-
stranded DNA copy of the transposon from the RNA copy. The transposon is inserted at
a new location using the enzyme
integrase
.
This mechanism is similar to the mecha-
nism that a
RNA virus (retrovirus)
uses in its life cycle to transform host DNA.
3. Transposons and genetic variability (Figure 1-2 C-F).
The main effect of transposons is to affect
the genetic variability of the organism. Transposons can do this in several ways:
a. Mutation at the former site of the transposon.
After the transposon is cut out of its site in the
host chromosome by transposase, the host DNA must undergo DNA repair. A mutation
may arise at the repair site.
b. Level of gene expression.
If the transposon moves to the target DNA near an active gene,
the transposon may affect the level of expression of that gene. While most of these
changes in the level of gene expression would be detrimental to the organism, some of the
changes over time might be beneficial and then spread through the population.
c. Gene inactivation.
If the transposon moves to the target DNA in the middle of a gene
sequence, the gene will be mutated and may be inactivated.
d. Gene transfer.
If two transposons happen to be close to one another, the transposition
mechanism may cut the ends of two different transposons. This will move the DNA
between the two transposons to a new location. If that DNA contains a gene (or an exon
sequence), then the gene will be transferred to a new location. This mechanism is espe-
cially important in
development of antibiotic resistance
in bacteria. Transposons in bacter-
ial DNA can move to bacteriophage DNA, which can then spread to other bacteria. If the
bacterial DNA between to the two transposons contains the gene for tetracycline resist-
ance, then other bacteria will become tetracycline resistant.
8
BRS Genetics
LWBK274-C01_01-11.qxd 06/02/2009 03:29 PM Page 8 Aptara
1.
The human genome codes for
30,000
genes that make up
2% of the DNA in the
human nuclear genome. The remaining
nuclear genome consists of which of the fol-
lowing DNA elements?
(A)
noncoding DNA
(B)
repetitive DNA
(C)
intron DNA
(D)
pseudogenes
(E)
satellite DNA
2.
The central dogma of molecular biology
is that DNA is transcribed into RNA, which is
then translated into a protein. The translation
takes place on the ribosomes. Which of the
following RNAs are the main components of
the ribosomes?
(A)
tRNA
(B)
snoRNA
(C)
snRNA
(D)
mRNA
(E)
rRNA
3.
A 24-year-old woman is diagnosed as hav-
ing a complete molar pregnancy with
enlargement of the chorionic villi and
absence of an embryo. Cytogenetic analysis
of the products of conception revealed a
46,XX karyotype. The molar pregnancy was
caused by which one of the following?
(A)
preeclampsia
(B)
two haploid sets of paternal chromosomes
(C)
trophoblastic neoplasia
(D)
elevated hCG levels
(E)
enlarged uterus
4.
Which of the following is a characteristic
of genomic imprinting?
(A)
Most genes must bear the parent of ori-
gin imprint for proper expression.
(B)
The parent of origin copy to be imprinted
differs from gene to gene, and most
genes require an imprint.
(C)
The phenotype of a child with Prader
Willi syndrome is different depending on
whether the child has a deletion on chro-
mosome 15 or UPD for the chromosome.
(D)
During gamete formation, the imprint is
removed from the genes and replaced
with an imprint of the opposite sex.
(E)
Imprinting does not disturb the primary
DNA sequence.
5.
Some female carriers of hemophilia B (an
X-linked recessive disease) have symptoms
of the disease. Which of the following is the
most likely explanation for how this occurs?
(A)
The X chromosome for the normal gene
is inactivated in a majority of cells in the
body.
(B)
Triplet repeat expansion.
(C)
Incomplete penetrance.
(D)
Variable expressivity.
6.
Heritable traits, both normal and disease
producing, are determined by which of the
following?
(A)
introns and exons of protein-coding
genes with epigenetic control
(B)
RNA-coding genes under epigenetic con-
trol
(C)
protein-coding genes, RNA-coding
genes, and epigenetic control
(D)
protein-coding genes, processed pseudo-
genes and retrogenes, and epigenetic
control
7.
The genomes of a number of organisms,
including humans, have now been charac-
terized and compared. Which of the follow-
ing describes one of the findings of these
endeavors?
(A)
There is a correspondence between the
biological complexity of an organism and
the amount of noncoding DNA.
(B)
There is a correspondence between the
biological complexity of an organism and
the amount of coding DNA.
(C)
There is a correspondence between the
biological complexity of an organism and
the number of chromosomes.
(D)
There is no correspondence between the
biological complexity of an organism and
the amount of coding DNA, noncoding
DNA, or the number of chromosomes.
9
LWBK274-C01_01-11.qxd 06/02/2009 03:29 PM Page 9 Aptara