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10
BRS Genetics
8.
Genetic variability in an organism
(including humans) is significantly affected
by which one of the following?
(A)
microsatellite DNA
(B)
satellite DNA
(C)
transposons
(D)
heterochromatin
9.
The most abundant sequence in the
human genome is which one of the following?
(A)
rRNA tandem repeats
(B)
microsatellite DNA
(C)
satellite DNA
(D)
Alu repeats
10.
The modification of DNA that can make
transcription of a DNA segment unlikely and
thus “silence” a gene containing that seg-
ment is which one of the following?
(A)
methylation of cytosine nucleotides
(B)
acetylation of histones
(C)
retrotransposition
(D)
transcription
11.
Which noncoding DNA is found near the
telomeres of the chromosomes?
(A)
microsatellite DNA
(B)
hypervariable minisatellite DNA
(C)
satellite 1 DNA
(D)
alpha satellite DNA
12.
Which one of the following is the mecha-
nism responsible for genomic imprinting?
(A)
acetylation
(B)
phosphorylation
(C)
methylation
(D)
transposition
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1. The answer is (A).
Noncoding DNA such as introns, pseudogenes, and repetitive elements
(such as satellite DNA) make up the rest of the genome.
2. The answer is (E).
The main component of ribosomes is ribosomal RNA or rRNA. The other
RNAs participate in the processes of transcription and translation but are not components
of the ribosomes.
3. The answer is (B).
Because of genomic imprinting, both maternal and paternal haploid sets
of chromosomes are required for normal development. When there are two paternal hap-
loid sets of chromosomes in a conceptus, a placenta will develop but not an embryo.
4. The answer is (E).
Imprinting does not change the DNA sequence of a gene. The maternal
and paternal copies of genes are mostly active or silent at the same time. The end result of
a deletion of chromosome 15 or UPD is that there are no paternal copies of the gene(s)
involved in the syndrome, so there is no difference in phenotype.
5. The answer is (A).
If the normal gene is inactivated in a large enough number of cells, then
there would be more defective gene products present than normal gene products and dis-
ease symptoms will be the result.
6. The answer is (C).
Although protein-coding genes account for much of what are recognized
as heritable traits, RNA-coding genes and epigenetic control are also important in gene
expression, both normal and abnormal.
7. The answer is (A).
What has been determined so far is that the amount of noncoding DNA
corresponds with the biological complexity of an organism. The human genome is com-
posed of
98% noncoding DNA and no other organism studied to date has this amount of
noncoding DNA. Humans have
30,000 genes compared to 19,100 for the roundworm
Caenorhabditis elegans. The number of chromosomes has no correspondence with biologi-
cal complexity. For example, carp (a fish) have 100 chromosomes and humans have 46.
8. The answer is (C).
Transposons can cause mutations in their former site when they relo-
cate, alter gene expression at sites where they integrate, inactivate a gene by integrating
somewhere in its sequence, and move pieces of nontransposon DNA to a new location in
the genome. Although much of the time this is a detrimental event, sometimes the changes
are beneficial and spread through the population.
9. The answer is (D).
Alu repeats are located in the GC rich, R-band positive areas of the chro-
mosome that contain many genes. There are many copies of the other sequences in the
human genome, but they are not as abundant as the Alu sequences.
10. The answer is (A).
Methylation of DNA is one of the primary ways that a gene can be
“turned off”. Methylation plays a crucial role in genomic imprinting.
11. The answer is (B).
Hypervariable minisatellite DNA is found near the telomere and at other
chromosome locations. Satellite 1 and alpha satellite DNA are found at the centromeres.
Microsatellite DNA is dispersed throughout all the chromosomes.
12. The answer is (C).
DNA is chemically modified by methylation and is less likely to be tran-
scribed into RNA.
11
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12
I. THE BIOCHEMISTRY OF NUCLEIC ACIDS
A nucleoside consists of a nitrogenous base and a sugar. A nucleotide
consists of a nitrogenous
base, a sugar, and a phosphate group. DNA and RNA consist of a chain of nucleotides, which are
composed of the following components:
A. Nitrogenous Bases
1. Purines
a.
Adenine (A)
b.
Guanine (G)
2. Pyrimidines
a.
Cytosine (C)
b.
Thymine (T)
c.
Uracil (U), which is found in RNA
3. Base Pairing.
Adenine pairs with thymine or uracil (
A-T or A-U
). Cytosine pairs with gua-
nine (
C-G
).
B. Sugars.
1.
Deoxyribose, which is found in DNA
2.
Ribose, which is found in RNA
C. Phosphate (PO
4
3
).
A. Double Helix DNA.
1.
The DNA molecule is two complementary polynucleotide chains (or DNA strands)
arranged as a double helix, which are held together by
hydrogen bonding
between laterally
opposed base pairs (bps).
2.
DNA can adopt different helical structures, which include:
a. A-DNA:
a right-handed helix with 11 bp/turn
b. B-DNA:
a right-handed helix with10 bp/turn
c. Z-DNA:
a left-handed helix with 12 bp/turn
3.
In humans, most of the DNA is in the B-DNA form under physiological conditions.
B. Nucleosome.
1.
The most fundamental unit of packaging of DNA is the nucleosome. A nucleosome con-
sists of a histone protein octamer (two each of
H2A, H2B, H3, and H4 histone proteins
)
around which 146 bp of DNA is coiled in 1.75 turns.
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2.
The nucleosomes are connected by spacer DNA, which results in a 10nm diameter fiber
that resembles a “beads on a string” appearance by electron microscopy.
3.
Histones are small proteins containing a high proportion of
lysine
and
arginine
that
impart a positive charge to the proteins, which enhances its binding to negatively charged
DNA. Histones bind to DNA in A-T rich regions.
4.
Histone proteins have exposed N-terminal amino acid tails that are subject to modifica-
tion and are crucial in regulating nucleosome structure.
Histone acetylation
of lysine by
histone acetyltransferases (HATs)
and
histone deacetylation
by
histone deacetylases (HDACs)
are the most investigated histone modifications.
5.
Histone acetylation reduces the affinity between histones and DNA. An increased acetyla-
tion 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
acetylation of histones
expressed genes
).
C. 30 nm Chromatin Fiber.
1.
The 10 nm nucleosome fiber is joined by
H1 histone protein
to form a
30 nm chromatin
fiber.
2.
During interphase of mitosis, chromosomes exist as 30 nm chromatin fibers organized as
extended chromatin
(Note: when the general term “chromatin” is used, it refers specifically
to the 30 nm chromatin fiber organized as extended chromatin).
3.
The extended chromatin can also form
secondary loops.
4.
During metaphase of mitosis, chromatin undergoes
compaction.
A.
A centromere is a specialized nucleotide DNA sequence that binds to the mitotic spindle
during cell division.
B.
A major component of centromeric DNA is
-satellite DNA,
which consists of 171 bp repeat
unit.
C.
A centromere is associated with a number of centromeric proteins, which include:
CENP-A,
CENP-B, CENP-C,
and
CENP-G.
D.
Chromosomes have a single centromere that is observed microscopically as a
primary con-
striction,
which is the region where sister chromatids are joined.
E.
During prometaphase, a pair of protein complexes called
kinetochores
forms at the cen-
tromere and one kinetochore is attached to each sister chromatid.
F.
Microtubules produced the by
centrosome
of the cell attach to the kinetochore (called
kine-
tochore microtubules
) and pull the two sister chromatids toward opposite poles of the mitotic
cell.
IV. HETEROCHROMATIN AND EUCHROMATIN
A. Heterochromatin
is condensed chromatin and is
transcriptionally inactive.
In electron micro-
graphs, heterochromatin is electron dense (i.e., very black). An example of heterochromatin
is the
Barr body,
which can be seen in interphase cells from females, which is the inactive X
chromosome. Heterochromatin comprises
10% of the total chromatin.
1. Constitutive heterochromatin
is always condensed (i.e., transcriptionally inactive) and con-
sists of repetitive DNA found near the centromere and other regions.
Chapter 2
DNA Packaging
13
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2. Facultative heterochromatin
can be either condensed (i.e., transcriptionally inactive) or
dispersed (i.e., transcriptionally active). An example of facultative heterochromatin is the
XY body,
which forms when both the X and Y chromosome are inactivated for
15 days
during male meiosis and the inactivated X chromosome in females.
B. Euchromatin
is dispersed chromatin and comprises
90% of the total chromatin. Of this 90%,
10% is transcriptionally active and 80% is transcriptionally inactive. When chromatin is tran-
scriptionally active, there is weak binding to the H1 histone protein and
acetylation
of the
H2A, H2B, H3, and H4 histone proteins.
14
BRS Genetics
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