cancer and dna replication Flashcards
Draw and label a pie chart to show the relative amount of time spent in each phase of the cell cycle including the stages of interphase and mitosis, as well as cytokinesis.
Explain how cyclins affect control the progression of a cell through the cell cycle.
1) Cells cannot progress to the next stage of the cell cycle unless the specific cyclin reaches it threshold
2) Cyclins bind to enzymes called cyclin-dependent kinases
3) These kinases then become active and attach phosphate groups to other proteins in the cell
4) The attachment of phosphate triggers the other proteins to become active & carry out tasks (specific to one of the phases of the cell cycle)
Outline the roles of the four cyclines involved in control of the cell cycle.
Cyclin A: activates DNA replication inside the nucleus in S phase
Cyclin B: promotes the assembly of the mitotic spindle and other tasks in the cytoplasm to prepare for mitosis
Cyclin D: Triggers cells to move from the G0 to G1 and from G1 into S-phase
Cyclin E: prepares the cell for DNA replication in S phase
Explain what is meant by ‘semi-conservative’ in terms of DNA replication
DNA replication is a semi-conservative process, because when a new double-stranded DNA molecule is formed, it contains one original strand and one newly synthesized strand
- the two strands of the double helix separate, each strand serving as a guide
- new strands are creating by adding nucleotides and linking them together
- because of complementary base pairing, the base sequence on the template strand determines the base sequence of the new strand
- the result is two DNA molecules, both composed of an original strand and a newly synthesized strand, and both identical to the original DNA molecule
Outline the three hypotheses that had been proposed for the replication of DNA
- Conservative Model – An entirely new molecule is synthesised from a DNA template
- Dispersive Model – New molecules are made of segments of new and old DNA
- Semi-Conservative Model – Each new molecule consists of one newly synthesised strand and one template strand (this is the currently accepted model)
How did Meselson and Stahl’s results support the theory of semi-conservative replication of DNA?
- Meselson and Stahl were able to experimentally test the validity of these three models using radioactive isotopes of nitrogen
- Nitrogen is a key component of DNA and can exist as a heavier 15N or a lighter 14N
- DNA molecules were prepared using the heavier 15N and then induced to replicate in the presence of the lighter 14N
- DNA samples were then separated via centrifugation to determine the composition of DNA in the replicated molecules
- The results after two divisions supported the semi-conservative model of DNA replication
- After one division, DNA molecules were found to contain a mix of 15N and 14N, disproving the conservative model
- After two divisions, some molecules of DNA were found to consist solely of 14N, disproving the dispersive model
Differentiate between the leading strand and the lagging strand
When replication begins, the two parent DNA strands are separated. One of these is called the leading strand, and it is replicated continuously in the 3’ to 5’ direction. The other strand is the lagging strand, and it is replicated discontinuously in short sections.
Outline the role of helicase
- Helicase unwinds and separates the double-stranded DNA by breaking the hydrogen bonds between base pairs
- This occurs at specific regions (origins of replication), creating a replication fork of two strands running in antiparallel directions
- The two separated polynucleotide strands will act as templates for the synthesis of new complementary strands
Outline the role of DNA gyrase
- DNA gyrase reduces the torsional strain created by the unwinding of DNA by helicase
- It does this by relaxing positive supercoils (via negative supercoiling) that would otherwise form during the unwinding of DNA
Outline the role of DNA primase
- DNA primase generates a short RNA primer (~10–15 nucleotides) on each of the template strands
- The RNA primer provides an initiation point for DNA polymerase III, which can extend a nucleotide chain but not start one
Outline the role of DNA polymerase III
- Free nucleotides align opposite their complementary base partners (A = T ; G = C)
- DNA pol III attaches to the 3’-end of the primer and covalently joins the free nucleotides together in a 5’ → 3’ direction
- As DNA strands are antiparallel, DNA pol III moves in opposite directions on the two strands
- On the leading strand, DNA pol III is moving towards the replication fork and can synthesise continuously
- On the lagging strand, DNA pol III is moving away from the replication fork and synthesises in pieces (Okazaki fragments)
Outline the role of DNA polymerase I
- As the lagging strand is synthesised in a series of short fragments, it has multiple RNA primers along its length
- DNA pol I removes the RNA primers from the lagging strand and replaces them with DNA nucleotides
Outline the role of single stranded binding proteins
- SSB proteins bind to the DNA strands after they have been separated and prevent the strands from re-annealing
- These proteins also help to prevent the single stranded DNA from being digested by nucleases
- SSB proteins will be dislodged from the strand when a new complementary strand is synthesised by DNA polymerase III
Outline the role of DNA primase
- DNA primase generates a short RNA primer (~10–15 nucleotides) on each of the template strands
- The RNA primer provides an initiation point for DNA polymerase III, which can extend a nucleotide chain but not start one
Define Okazaki fragment
Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging template strand during DNA replication.