Study Questions Set 2 Flashcards

1
Q

Thinking question re DNA/RNA structure: Certain chemical agents acting on DNA could convert cytosine to uracil through the process of deamination (chopping off the amino group). This mutation is routinely repaired by the existing repair mechanism (uracil is removed and it gets replaced by cytosine). Knowing this, how would you explain why DNA contains thymine and NOT uracil.

A

-it is important that DNA uses thymine because if uracil was used, the repair mechanism wouldn’t be able to differentiate between mutated cytosine and actual uracil that is important to the sequence of bases

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2
Q

How does complexity of bacterial genome differ from that of eukaryotic (calf) genome?

A
  • E. coli genome has no repetitive sequences, whole genome is one unique sequence
  • calf genome has lots of repetitive sequences, its unique sequences are more complex than E. coli however
  • A bacterial genome tends to have more complexity. This is because there are few non-repeating areas of the genome, while eukaryotic genome has many repeating parts.
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3
Q

Explain C value paradox.

A
  • no correlation between the amount of DNA and the apparent complexity of organisms -> prokaryotic genomes contain only non-repetitive DNA
  • The C value paradox (where C = DNA/haploid) is the idea that there is no correlation between amount of DNA and complexity of an organism.
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4
Q

List and briefly explain factors that influence DNA renaturation kinetics.

A
  • DNA concentration: greater [DNA], greater chance of faster renaturation
  • salt concentration: ionic conditions mask the repulsion forces of phosphate backbone
  • temperature: melting temperature = temp. at which 50% of DNA is denatured  high temperatures denature DNA
  • time
  • size of DNA fragment: bigger the DNA fragment, more difficult is renaturation
  • complexity: simple sequences renature faster than complex sequences
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5
Q

You have found a new species of insects. To evaluate the complexity of the genome of this species, you isolate genomic DNA from, fragment the DNA to uniform 500 base pair pieces, denature the DNA and measure the rate of reassociation. Your data is represented in the curve below (sorry for the bad drawing): a) How many classes of DNA are found here? b) What can you say about the relative complexity?

A

a) 3 classes
b)
- Highly repetitive/fast -> not complex at all, very repetitive (25%)
- moderately repetitive/medium -> somewhat complex (25%)
- Unique/slow -> highly complex

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6
Q

List 3 differences between prokaryotic topoisomerase I and gyrase

A
  • Topoisomerase I relaxes negative supercoils; gyrase introduces negative supercoils
  • Gyrase is a topoisomerase II, cuts double stranded DNA; topoisomerase I cuts one strand
  • Topoisomerase I changes L in steps of 1, gyrase changes L in steps of 2
  • Topo I makes a nick in one strand, topo II (gyrase) makes nick in both strands
  • Topo I uses ATP, Topo II does not
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7
Q

What are topological isomers of DNA?

A

DNA strands differing only in their states of supercoiling; a circular DNA without any superhelical turns is known as relaxed

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8
Q

Explain the role of DNA supercoiling in cell survival?

A

i. important for packaging of DNA, it allows DNA to be more condensed. one use of this is in heterochromatin.
ii. RNA polymerase will transcribe anything it will get its lil hands on, so it’s important to hide the DNA so it can’t be transcribed
iii. coiled structure protects from damage
iv. negative supercoils release tension when DNA is coiled/uncoiled
The DNA must supercoil in order to allow for cell survival because there is tension created as the DNA unwinds. The DNA must be able to unwind to allow it to express genes, as well as replicate. In gene expression, where portions of the DNA are fixed, it allows for the relaxation.
There is also supercoiling of bacterial DNA, and this is slightly different, where there is supercoiling in “threads” around the a protein core, and a nick allows the supercoils to become relaxed

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9
Q

What is the difference between primary, secondary, and tertiary structures of DNA/RNA?

A
Primary = the linear bonding. Primary is just the formation of phosphodiester bonds with the nitrogenous bases. Primary RNA vs DNA:
•	RNA = CGAU
•	DNA = CGAT
Secondary for RNA:
•	Triple bond
•	Folding back on self
•	GU = non canonical BP
•	Helices, stem-loop structure
2o for DNA:
•	Helical structure, H bonding between AT and GC, stacking interactions, salts
•	Triple helix
•	Cruciform, slipped structures
3o RNA vs DNA
•	Interactions of 2o structures
•	Lack of constraint through long-range structures means there is dynamic 3o structure in RNA
•	RNA has pseudoknots, where there are loops that are looped. 2 stem-loop structures are intercalated between the halves of another stem.
•	DNA has supercoiling.
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10
Q

You have a small 4800 bp long circular DNA. It has a linking number of 450 (L=450). What are the twist (T) and the writhe (W) of this DNA? What assumptions about the structure of the DNA have you made in your answer? (Hint: what is the definition of the twist #?)

A

T = twisting number. It is the crossing of one strand of DNA over the other. It is a measure of how tightly the helix is wound. It is calculated through:
T = total # of BP/ # of BP per turn.
T can be positive or negative. Positive if right handed. Negative if left handed.
W = Writhing number. This is the number of superhelical turns, so how many times the duplex DNA crosses over itself. Can be negative or positive, reflecting positive or negative supercoiling.
L = linking number. It is the total number of times a close molecule of dsDNA encircles the other. Reflects both the T and the W. it can only be changed by breaking one or both strands of DNA, winding it tighter or looser, and then rejoining the ends. It is a constant unbroken DNA, so any change in W must reflect a change in T, and vice versa.

Okay, on to the question:
L = 450. T and W = ?
T = 4800/# of turns/bp. In order to solve this, the number of turns /bp must be assumed as 10 for a normal DNA. 4800/10 = 480.
L = T + W.
W = L – T = 450 – 480 = -30.
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11
Q

Thinking question 1: Many different mutations have been observed in almost all genes. However, only few have been isolated in histones. How would you explain this finding?

A

This is likely because of the importance of the histones. Any mutation in histones could result in non-functioning histones, and then the DNA couldn’t condense properly. The cell would likely just not exist as opposed to having a mutated histone

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12
Q

What are some of the distinctive features of eukaryotic chromosomes? (note: I expect you to first define chromosomes and after that you have to briefly explain nucleosomes/histone proteins/octet +H1/wrapped DNA, different levels of chromosome condensation, centromere and telomere regions)

A

• Chromosome: a single strand of condensed double-helix DNA.
• The DNA is condensed, with the help of proteins, to form chromatin. This chromatin organizes into chromatin fibers, which further condense to form the chromosome
o Chromatin Organization: the basic unit of the chromatin fiber is a nucleosome (the DNA & it’s associated proteins). The associated proteins are called histones. The histone proteins involved in the nucleosome consist of an octet that bind together, and the DNA then wraps around the octet. H1 is the histone protein which helps bind the nucleosomes together into a strand, using ‘linker DNA’.
o Levels of Chromosomes: primary structure of DNA –winds around histones nucleosomes –nucleosomes condense into 30nm strands chromatin fibre –fibres condense chromosome
o Centromere: the specific nucleotide sequence which serves as an attachment point for sister chromatids & spindle fibres; a highly repetitive sequence
o Telomere: the ends of the chromosome; a highly repetitive sequence which helps protect the chromosome from degredation

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13
Q

What is unusual about the amino acid composition of histones? How is the function of histones related to their amino acid composition?

A

• A: histones are proteins that are rich in lysine and arginine – both basic amino acids. At normal pH, these basic amino acids accept protons at their 2nd amino group, and they become NH3+. These positive charges attract the DNA through electrostatic attractions.

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14
Q

Name few nonhistone proteins which are a part of chromatin structure and explain why you would expect them to be found there.

A
  • S/MARs (scaffold/matrix attachment regions) - AT rich, cis acting regions that regulate transcription, they are the binding site of TopoII
  • structural maintenance of chromosome (SMC) proteins
  • Chaperone proteins: DNA replication proteins, transcription factors, chaperone proteins, topoisomerases
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15
Q

What are heterochromatin and euchromatin? What is their importance in DNA replication and transcription?

A

heterochromatin: tightly packed DNA -> less susceptible to DNase digestion -> less susceptible to DNA replication and transcription
o Constitutive: always highly condensed chromatin, highly repetitive sequence, very few genes
o Facultative: forms under specific circumstances and/or in certain tissues, to silence gene expression
 X-chromosome inactivation
 Imprinting
 Cell type
-euchromatin: lightly packed DNA -> more susceptible to DNase digestion -> consists of transcriptionally active DNA -> more susceptible to DNA replication and transcription

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16
Q

Does the degree of chromosomal condensation play a role in controlling replication and transcription? How (explain briefly; use your own words)?

A

Yes, it plays a role. The more condensed the chromosome, the harder it is to access for transcription & replication. Therefore if a gene does not need to be transcribed anymore, it can be condensed further in order to prevent the transcription of it. If a gene is then needed for replication or transcription, the condensation can loosed to allow access to the gene of interest.

17
Q

Would you expect there to be more histones per kilobase in euchromatin or heterochromatin? Explain your reasoning.

A

You would expect more histones per kilobase in heterochromatin, as it is transcriptionally inactive, and thus can be very tightly condensed as it never needs to be accessible

18
Q

Thinking question: the sequences of a particular set of genes are found by in situ hybridization (note: methodology is irrelevant for this question, so do not think about it) to be heterochromatic in some cells and euchromatic in cells at different stages of development. How would these sequences be categorized (what would be your conclusion about these sequences in respect to gene expression)?

A
  • This would suggest that this set of genes plays a role in the cell development and commitment. The genes would be used (i.e. are euchromatic) to produce proteins that are needed for development. Once the cell has developed past that point, the gene would become inactive (i.e. heterochromatic) as the proteins it codes for are no longer needed.
19
Q

Thinking question 2: would it be an easy task to produce polyclonal antibodies to histones? Explain your reasoning

A
  • Polyclonal antibody: an antibody which can bind to multiple sites on the target molecule. Generally, they are cheaper and easier to produce than monoclonal antibodies
  • A: where histones rarely ever have mutations, it should be quite easy to produce an antibody for it since they are so highly conserved