Exam 3: 33 Genome and DNA Packaging (complete?) Flashcards
1) (AGCGCTAATA)n
- Which of the following nucleotide sequences is most likely compatible with the formation of nucleosome?
(Note: n means that the sequence gets repeated numerous times)
#2) (ACGCTGCATC)n
#3) (UATGCATUCG)n
#4) (GCCTATAAGC)n
A) #1; #2
B) #2; #4
C) #2; #3
D) #1; #4
D) #1; #4
Analyzing Nucleotide Sequences and Nucleosome Formation
You’re asking which of the given nucleotide sequences is most likely compatible with nucleosome formation. To effectively address this, let’s revisit the key factors influencing nucleosome formation, as highlighted in the sources:
- Bending Propensity: Nucleosomes have a preference for DNA sequences that possess an inherent tendency to bend. This preference stems from the requirement for DNA to bend significantly as it wraps around the histone octamer.
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Alternating A:T and G:C Rich Sequences: Sequences that exhibit an alternating pattern of A:T-rich and G:C-rich segments, with a periodicity of roughly 5 base pairs, are favored for nucleosome formation. This preference is rooted in the structural properties of DNA base pairs:
- A:T base pairs tend to bend towards the minor groove, facilitating favorable interactions with the histone core.
- G:C base pairs exhibit a bending tendency towards the major groove.
This alternating pattern creates a favorable landscape for DNA to wrap around the histone octamer, with A:T-rich segments bending towards the histone core and G:C-rich segments bending outwards.
Now, let’s analyze the provided sequences, keeping these criteria in mind:
- (AGCGCTAATA)n: This sequence presents a relatively balanced distribution of A:T and G:C base pairs, lacking a distinct pattern of alternating A:T and G:C rich segments. Consequently, it’s less likely to be highly compatible with nucleosome formation.
- (ACGCTGCATC)n: Similar to sequence #1, this sequence also displays a relatively even distribution of A:T and G:C base pairs without a clear pattern of alternating A:T and G:C rich regions. Therefore, its compatibility with nucleosome formation is likely to be moderate.
- (UATGCATUCG)n: This sequence incorporates uracil (U), a nucleotide base characteristic of RNA, not DNA. Consequently, it’s not a relevant candidate for nucleosome formation, as nucleosomes are composed of DNA wrapped around histone proteins.
- (GCCTATAAGC)n: This sequence stands out with a distinct pattern of alternating A:T and G:C rich regions. It commences with a G:C rich segment (GCCT), succeeded by an A:T rich segment (ATA), and concludes with another G:C rich segment (AGC). This pattern repeats throughout the sequence, enhancing its compatibility with nucleosome formation. The alternating pattern promotes the bending and wrapping of DNA around the histone core.
Concluding Remarks
Based on the criteria established by the sources, sequences #1 and #4 (AGCGCTAATA)n and (GCCTATAAGC)n are the most likely to be compatible with nucleosome formation. They exhibit the alternating pattern of A:T and G:C rich segments that favors the bending and wrapping of DNA around the histone core. Therefore, the correct answer is D) #1; #4.
- Nucleosome sliding can be facilitated by which of the following proteins?
A) DNA primase
B) Histone methyltransferase
C) Chromatin remodeling complex
D) Histone kinase
C) Chromatin remodeling complex
The question asks which protein facilitates nucleosome sliding. Nucleosome sliding is the process of repositioning histone octamers along the DNA molecule, which can alter the accessibility of DNA sequences and regulate gene expression.
The sources provide information about chromatin remodeling complexes and their role in modulating chromatin structure. These complexes are large protein assemblies that utilize the energy derived from ATP hydrolysis to modify the structure of nucleosomes.
- The sources specifically mention that chromatin remodeling complexes can facilitate nucleosome sliding by shifting the position of histone octamers along the DNA strand. This movement can either expose previously inaccessible DNA regions or occlude previously accessible regions, influencing the binding of regulatory proteins and the transcription of genes.
Let’s consider the other options:
- DNA primase is an enzyme involved in DNA replication. It synthesizes short RNA primers that serve as starting points for DNA polymerase. DNA primase does not play a direct role in nucleosome sliding.
- Histone methyltransferases are enzymes that catalyze the addition of methyl groups to histone proteins. While histone methylation can influence chromatin structure and gene expression, it does not directly facilitate nucleosome sliding.
- Histone kinases are enzymes that catalyze the addition of phosphate groups to histone proteins. Like methylation, histone phosphorylation can affect chromatin structure and gene expression but is not directly involved in nucleosome sliding.
Conclusion
Based on the information provided in the sources, the protein that facilitates nucleosome sliding is the chromatin remodeling complex (C).
Clicker:
Comparing the gene density of the genomes from the following organisms, which one has the LEAST dense genome?
A.Escherichia coli K-12
B.Saccharomyces cerevisiae
C.Drosophila melanogaster
D.Homo sapiens
D. Homo sapiens
More complex organisms have less dense genome.
- In a nucleosome core particle, a 146 bp long DNA fragment wraps around (1.7 turns) on a histone octamer in a left-hand manner.
Assume the DNA fragment (146 bp) adopts B-form DNA (or close to B-form), if the histone octamer is removed and two ends of the DNA are ligated to form a circular DNA. The resulting circular DNA is allowed to adopt its energetically relaxed condition (i.e. close to B-form).
What is the Wr (Writhe number) of this circular DNA?
A) +1.7
B) –1.7
C) +14.6
D –14.6
B) –1.7
Clicker:
Which of the following is true about DNA in our body?
A. Most human cells have 3.2 billion base-pairs of DNA
B. Most human cells have 6.4 billion base-pairs of DNA
C. Human has a total of 100,000 genes
D. Human has a total of 300,000 genes
B. Most human cells have 6.4 billion base-pairs of DNA
Most human cells are diploid; thus, they have 2 x 3.2 billion
base-pairs.
Clicker:
Which of the following describes the correct order
of events in nucleosome assembly?
A. Two H2A/H2B dimers form a tetramer; DNA wraps
around the tetramer; two H3/H4 dimers join in
B. Two H3/H4 dimers form a tetramer; DNA wraps around
the tetramer; two H2A/H2B dimers join in
C. One H2A/H2B dimer and one H3/H4 dimer form a
tetramer; DNA wraps around the tetramer; another
H2A/H2B dimer joins in
D. DNA wraps around one H2A/H2B dimer; H3/H4
tetramer joins in, followed by another H2A/H2B dimer
B. Two H3/H4 dimers form a tetramer; DNA wraps around
the tetramer; two H2A/H2B dimers join in
Histone octamer = one H3-H4 tetramer plus two H2A-H2B dimers
The histone octamer at the heart of a nucleosome is formed in a specific order:
- Two H3-H4 dimers associate to form a stable H3-H4 tetramer.
- The H3-H4 tetramer binds to DNA, initiating nucleosome formation.
- Two H2A-H2B dimers join the H3-H4 tetramer to complete the histone octamer.
- The DNA wraps around the histone octamer in about 1.7 turns.
Clicker:
Which of the following post-translational modifications preserves the positive charges of histones?
A: Acetylation of lysine.
B: Methylation of lysine.
C: Phosphorylation of serine.
B: Methylation of lysine.
The positive charge on lysine can be removed by acetylation. However, once lysine is methylated, it can not be acetylated without removing methylation. Thus, methylation can preserve the positive charge on the side chain of lysine.
- In the structure shown in Question 28, which amino acids are more frequently observed on the tails of each histone subunit, i.e., H2A, H2B, H3, H4?
A) Lysine, Glutamine
B) Lysine, Arginine
C) Glutamine, Asparagine
D) Leucine, Isoleucine
B) Lysine, Arginine
Based on the information from the sources, the amino acids most frequently observed on the tails of histone subunits H2A, H2B, H3, and H4 are lysine and arginine (B). The abundance of these positively charged amino acids is essential for the interaction of histones with DNA and the packaging of DNA into chromatin.
- Posttranslational modifications (PTMs) modulate DNA-histone interactions in nucleosome. Which of the following statement is correct?
A) Methylation of lysine residue occurs on the amide bond
B) Acetylation of lysine preserves the positive charge of lysine and consequently tend to enhance DNA and histone interactions
C) Phosphorylation of serine residues adds positive charge to histone and weakens DNA and histone interactions
D) Phosphorylation of serine residues adds negative charge to histone and weakens the DNA and histone interactions
D) Phosphorylation of serine residues adds negative charge to histone and weakens the DNA and histone interactions
Which of the following PTMs would MOST LIKELY lead to increased accessibility of DNA within a nucleosome?
A) Methylation of a lysine residue known to enhance DNA-histone interactions.
B) Acetylation of a lysine residue in the histone tail.
C) Phosphorylation of a serine residue in the histone tail.
D) Removal of a ubiquitin molecule from a histone protein.
B) Acetylation of a lysine residue in the histone tail.
A) Methylation of a lysine residue known to enhance DNA-histone interactions. This modification would likely lead to tighter DNA-histone interactions and therefore decreased DNA accessibility. The sources explain that methylation can have different effects depending on the specific lysine residue targeted, but in this case, the question specifies that the methylation enhances DNA-histone interactions.
B) Acetylation of a lysine residue in the histone tail. This modification is a strong contender for the correct answer. As the sources explain, acetylation neutralizes the positive charge of lysine, weakening its interaction with the negatively charged DNA backbone. This often leads to a more relaxed chromatin structure and increased DNA accessibility.
C) Phosphorylation of a serine residue in the histone tail. This option is also likely to increase DNA accessibility. The sources state that phosphorylation adds a negative charge to the histone tail, further repelling the negatively charged DNA and weakening their interaction.
D) Removal of a ubiquitin molecule from a histone protein. This option requires information that is not explicitly present in the sources. Ubiquitination is another type of PTM, but the sources don’t discuss its effects on DNA-histone interactions. Therefore, it’s difficult to assess whether this modification would increase or decrease DNA accessibility based solely on the information provided.
Which of the following describes Acetylation of Lysine?
A) Neutralizes (+) Lysine; weakening DNA-histone interactions, promoting gene expression
B) Preserves or enhances (+) Lysine; can either activate or repress gene expression
C) Adds (-) charge to histones; weakening DNA- histone interactions;, related to DNA damage response and chromosome condensation during cell division.
A) Neutralizes (+) Lysine; weakening DNA-histone interactions, promoting gene expression
Which of the following describes Methylation of Lysine?
A) Neutralizes (+) Lysine; weakening DNA-histone interactions, promoting gene expression
B) Preserves or enhances (+) Lysine; can either activate or repress gene expression
C) Adds (-) charge to histones; weakening DNA- histone interactions;, related to DNA damage response and chromosome condensation during cell division.
B) Preserves or enhances (+) Lysine; can either activate or repress gene expression
Which of the following describes Phosphorylation of Serine?
A) Neutralizes (+) Lysine; weakening DNA-histone interactions, promoting gene expression
B) Preserves or enhances (+) Lysine; can either activate or repress gene expression
C) Adds (-) charge to histones; weakening DNA- histone interactions;, related to DNA damage response and chromosome condensation during cell division.
C) Adds (-) charge to histones; weakening DNA- histone interactions;, related to DNA damage response and chromosome condensation during cell division.
