3.4 & 7.2 DNA Replication
06/11/2012 § Leave a comment
The process of DNA replication is semi-conservative – one strand of the original strand from the parent molecule is used to make the second (new) strand to complete the set. The code (a.k.a. the nitrogenous bases) from the parent strands provides the template to make the new strand – as we know, the complementary base pairing of the bases allows for DNA to create exact copies of itself during replication.
The first part of DNA replication involves the enzyme helicase which is in charge of unzipping (technically, unwinding and separating) the DNA molecule. Floating extra nitrogenous bases are linked to the parent strands with hydrogen bonds. The enzyme DNA polymerase is responsible for linking the new nucleotide to the sugar molecule of the original parent nucleotide to which it is complementary. It does this by using a covalent bond.
DNA replication (in both eukaryotes and prokaryotes) begin in sites called origins of replication. Eukaryotes have many of these sites while prokaryotes only have one. There’re a lot of chemical reactions involved in replication for multiple purposes so let’s just… go through them? But first – it’s vital to know that a nucleoside is a molecule made up of a sugar and a base. Done. Therefore, a nucleotide is equal to a nucleoside monophosphate – because of the additional phosphate group! Okay. Good. A nucleoside triphosphate is what binds to the new DNA strand and they use hydrogen bonds to link the bases.
Two molecules of phosphate are hydrolyzed so that the nucleoside triphosphate can become a nucleotide. Two other molecules of phosphate are hydrolyzed to provide energy to add the nucleotide to the growing link of nucleotides (a polymer, excuse me). The nucleoside triphosphates can only be added at the 3′ side of the adult nucleotide. The entire process of replication is catalyzed by enzyme DNA polymerase III. The enzyme primase is what makes the primer, the sequence made of RNA that starts the process. DNA polymerase I is what digests the RNA primer and replaces it with DNA. DNA is replicated in either a continuous pattern or in lagging strands, or fragments. These fragments (called Okazaki fragments) on the lagging strand are joined together by enzyme DNA ligase.
- Most of the DNA of a human cell is contained in the nucleus. Distinguish between unique and highly repetitive sequences in nuclear DNA. (5)
- Draw a labelled diagram to show four DNA nucleotides, each with a different base, linked together in two strands. (5)
- Explain the structure of the DNA double helix, including its subunits and the way in which they are bonded together. (8)
- Outline the structure of the nucleosomes in eukaryotic chromosomes. (4)
- State a role for each of four different named enzymes in DNA replication. (6)
- Explain the process of DNA replication. (8)
- Explain how the process of DNA replication depends on the structure of DNA. (9)
- Describe the genetic code. (6)
- Discuss the relationship between genes and polypeptides. (5)
- Explain briefly the advantages and disadvantages of the universality of the genetic code to humans. (4)
- Compare the processes of DNA replication and transcription. (9)
- Distinguish between RNA and DNA. (3)
- Describe the roles of mRNA, tRNA and ribosomes in translation. (6)
- Outline the structure of tRNA. (5)
- Outline the structure of a ribosome. (4)
- Explain the process of translation. (9)
- Compare DNA transcription with translation. (4)
DATA BASED QUESTIONS
Page 63, the Meselson-Stahl experiment
In order for cell division to occur, DNA must be duplicated to ensure that progeny cells have the same genetic information as the parent cells. The process of duplicating DNA is termed replication. The Meselson-Stahl experiment sought to understand the mechanism of replication. Did it occur in a conservative fashion, a semi-conservative fashion or in a dispersive fashion?
It occurred in a semi-conservative fashion, replication did. (c)
Explain why the data in Figure 14 suggests that replication is semi-conservative and why the other models are not supported.
Firstly, I feel as though I need to explain everything just to make sure I know what’s going on in this DBQ. The different nitrogen mediums will show us different concentrations in the DNA, and will show specifically the different densities. 15N is the heavier nitrogen (it’s stated that it’s the heavier medium), which will then give us the heavier DNA, or one type of DNA. The 14N is clearly the lighter medium, and we can think of it as the light nitrogen. We also know that these different nitrogen mediums are heavy and light because of the graphs. The first graph shows that the DNA strands in 15N are heavier (density: 1.724) while the second graph shows that the DNA strands in 14N are lighter (density: 1.710).
Now, in the third graph, we know that replication requires for one of the original adult strands (which, in this case will be the first medium, which is 15N) to help formulate the new strand (which, in this case, comes from the second medium, 14N). The combination of these two different strands of two different densities changes the overall density of the new (and improved!) DNA strand, which now has a density that’s balanced out from the original 14N and 15N densities: 1.717. The same thing happens in the fourth graph except there is one more generation of 14N strands, therefore, there’s an extra supply of light weight DNA strands, which is represented by the curve on the left. The balanced density of the combined DNA strands made up of both 14N and 15N strands represent the semi-conservativeness of DNA strands during replication. It shows the combination of an original strand and a new strand, shown in the third graph with the 15N strand (the first medium used) and the 14N strand (the second medium used). The other models, such as conservative or dispersive fashions because the different strands aren’t distributed equally in a way that creates the exact same DNA strand from the original adult strand. The semi-conservative fashion allows for the exact replication of the original adult strand because the new strand matches its bases to the adult strand’s bases.This is shown in the third graph’s results.
Page 64, Evidence for discontinuous synthesis
1. Compare the sample that was pulsed for 10 seconds with the sample that was pulsed for 30 seconds.
The sample that was pulsed for 10 seconds stays below 1000 cpm/0.1m^-1 while the sample that was pulsed for 30 seconds is higher than that, at above 3,000 cpm/0.1m^-1. Both samples start to decrease at 0 (distance from the top of the centrifuge tube).
2. Explain why the sample that was pulsed for 30 seconds proves evidence for the presence of both a leading strand and many lagging strands.
The first peak that we see in the 30-second sample graph (at around 0.8 units from the top of the centrifuge) represent radioactive nucleosides that are lighter than the nucleosides represented at the second peak (at around 1 unit from the top of the centrifuge). The fact that the second peak is further from the top of the centrifuge shows that those nucleosides are heavier than the first nucleosides. This means that the first – the lighter – nucleosides are the lagging strands because they are lighter and are made up in broken fragments. Then, the second – the heavier – nucleosides are therefore the leading strands because they are heavier and attached to something – other nucleosides and molecules that connect it to the forming strand.
3. Explain why the sample that was pulsed for 60 seconds provides evidence for the activity of DNA ligase.
The peak of the graph of the 60-second sample is located at around the same place as the second peak of the 30-second sample (around 1.1 units from the top of the centrifuge). The heavier sample here shows that ligase is taking the fragmented (lagging) strands from the first 10, 30, and so on seconds and has been joining them together, which makes the nucleosides in the samples heavier.