Bio Factsheet. How Science Works: Meselson and Stahl s Classic Experiment. Number 207.

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1 Number 207 How Science Works: Meselson and Stahl s lassic Experiment n 1953 James Watson and Francis rick built their model of the structure of DNA, which is still accepted today: DNA is an anti-parallel double stranded helix It consists of a sugar-phosphate backbone and a sequence of bases omplementary base pairing takes place A cytosine () base on one strand always pairs with a guanine (G) base on the other strand and an adenine (A) base on one strand always pairs with a thymine (T) base on the other strand This led them to propose a model for DNA replication in which each strand serves as a template for a newly synthesised strand; however Watson and rick had no evidence to back up this proposal. In 1957, American geneticists and biologists, Matthew Meselson and Franklin Stahl provided evidence of the mechanism of DNA replication in their classic experiment. Their research was important in showing how DNA replicates, recombines and is repaired in cells. Fig. 1: Meselson (left) and Stahl (right) Fig 2: Semi-conservative replication Typical Exam Question DNA replication can be described as being semi-conservative. What does this mean? When DNA replicates the two strands separate and each original strand of the DNA molecule acts as a template and is copied; therefore each new DNA molecule produced consists of one original and one new DNA strand. This is known as semi-conservative replication (Fig. 2). original DNA Newly synthersised DNA If DNA replication was conservative, both strands would still be copied, however the two original strands would join back together and the two new strands would join together (Fig. 3). DNA does not replicate in this way, as Meselson and Stahl demonstrated. Fig 3: onservative replication 1

2 How does DNA replicate semi-conservatively? DNA replicates to preserve its information; this information can then be passed on to new cells. It occurs during interphase before mitosis and meiosis in the nucleus. A cell can spend up to 90% of its life replicating DNA, i.e. it occurs almost all of the time. DNA replication begins at a sequence of nucleotides called the origin of replication (Fig 4): Fig 4: Semi-conservative DNA replication 1 1. Helicase enzymes unwind the DNA double helix by breaking hydrogen bonds between the bases. 21 replication bubble RNA primer 2. The strands separate forming one of many replication bubbles. These normally form in regions rich in A-T base-pairs as there are only two hydrogen bonds between A and T, but three between and G. RNA primers (single polynucleotide strands approximately ten nucleotides long) are attached by primase, an RNA polymerase enzyme. They signal where DNA polymerase is to start adding new DNA nucleotides DNA polymerase then starts to add new complementary DNA nucleotides to both template strands (from the 3 end of the primer only). This occurs at the rate of approximately 80 nucleotides per second in humans. Primase adds more RNA primer in the gaps and DNA polymerase then fills in the gaps with new complementary nucleotides Exonuclease removes the RNA primers and ligase inserts any missing phosphates to join up the sugarphosphate backbone. new complementary DNA nucleotides 5' 3' 5 5. Each of the new DNA molecules consists of one new and one old strand. 2

3 How does the structure of DNA make it suited for replication? (In exam questions always relate structure to function, even if it is only for one mark). DNA is double stranded and each strand holds the same genetic information; therefore both strands serve as templates for the reproduction of the opposite strand. The template strand is preserved in its entirety and the new strand is assembled from new nucleotides. The hydrogen bonds holding the two strands together are weak and split easily allowing replication to take place (they are normally protected by being inside the double helix). omplementary base-pairing enables accurate copying, allowing the information stored in the base sequence to be preserved. How did Meselson and Stahl provide evidence for the semi-conservative mechanism of DNA replication? In their classic experiment Meselson and Stahl devised a strategy that would enable original and new strands of DNA to be distinguished. Nitrogen is a major constituent of DNA, it is found in all four of the possible bases. The normal isotope of nitrogen, otherwise known as light nitrogen, 14 N, is by far the most abundant isotope of nitrogen. This is the isotope of nitrogen normally found in DNA. DNA can also be made with a heavy nitrogen isotope, 15 N (see Fig. 5 for an example of the DNA base thymine labelled with 15 N) Fig 5: Thymine labelled with 15 N H 15 N H The Experiment 1. Meselson and Stahl grew the bacterium Escherichia coli for fourteen generations in a medium containing 15 NH l as the 4 only source of nitrogen. This resulted in all of the DNA being labelled with high density 15 N. This was called generation This DNA was centrifuged in caesium chloride. As expected, they observed a band of DNA at the bottom of the tube (these are the original strands): Generation 0: All heavy DNA 3. E. coli with only 15 N DNA were transferred to a 14 N medium (from this point on the only new DNA available for new strand synthesis consisted of the normal lighter 14 N nucleotides). Again, the bacteria were allowed to divide once only. This was called generation When centrifuged in caesium chloride the band in the tube was in between that of 15 N DNA and 14 N DNA, i.e. of intermediate density. What Meselson and Stahl learned from this is that each of the original 15 N DNA strands must have acted as a template to which new 14 N DNA was added. Hence replication must have been semi-conservative: O 15 N H O H 3 Generation 0: All heavy DNA The 15 N isotope is not radioactive, only denser than 14 N. When centrifuged in caesium chloride, DNA settles out as bands in the tube, the more dense the DNA, the further down the tube it will be. Dense 15 N DNA settles out as a band at the bottom of the tube and normal less dense 14 N DNA settles out as a band near the top of the tube (see Fig. 6) Fig N and 14 N DNA settled out as bands when centrifuged with caesium chloride 15 N 14 N Generation 1: All Inermediate DNA If DNA replication was conservative replication, Meselson and Stahl would have observed equal amounts of DNA of the higher and lower densities, i.e. a band at the top and bottom with no intermediate band. They did not observe this and so rejected the mechanism of conservative DNA replication. 5. To conclusively prove their findings Meselson and Stahl grew E. coli for several more generations in the 14 N medium only, each time DNA was centrifuged in caesium chloride. Interpretation of the DNA in the bands 3

4 6. In generation 2 they observed two bands in the centrifuge tube, the same intermediate band in the middle of the tube and a new low density band at the top of the tube with the size ratio of the low density band to the intermediate band being of 1:1. This meant that half of the generation 2 bacteria would have one strand of the original 15 N DNA along with one new 14 N DNA strand (accounting for the intermediate band), while the DNA in the other half of the bacteria would consist entirely of new 14 N DNA; one strand synthesised in generation 1, and the other in generation 2 (accounting for the low density band at the top): Generation 1: All Inermediate DNA Generation 2: DNA in the ratio 1:1 7. In the generations 3 and 4 the same bands were seen as generation 2; however the band size ratio of low density to intermediate density was different. In generation 3 it was 3:1 and in generation 4 it was 7:1. Each generation the bacteria reproduced, requiring new DNA. The only DNA available for new strand synthesis was 14 N DNA, therefore the proportion of 14 N DNA in each successive generation increased, i.e. the low density band size in the tube increased. Generation 2: DNA in the ratio 1:1 Generation 3: DNA in the ratio 3:1 Generation 4: DNA in the ratio 7:1 4

5 8. The Important thing to note is that the semi-conservative strands were still being produced in every generation. Meselson and Stahl could therefore conclude that DNA replication was semi-conservative (Exam questions often show the centrifuge tubes blank and ask for the appropriate bands to be put in each generation, it the test tubes on the question are graduated, remember to put in the appropriate size bands). Fig 7: Summary of Meselson and Stahl s results Generation entrifuge tube Interpretation of DNA in the bands Explanation 0 All heavy DNA 1 All intermediate DNA 2 DNA in the ratio 1:1 3 DNA in the ratio 3:1 4 DNA in the ratio 7:1 Questions 1. The following drawing shows part of a DNA molecule: G Samples of cells were removed from the radioactive solution after one cell division. hromosomes from the cells were tested for radioactivity. The results are shown in the table below: phosphate A T A hromosomes from cells dividing in unlabelled thymine No radioactivity Deoxyribose (a) Describe and explain three structural features shown in the diagram that account for the ability of DNA to replicate (3 marks). G G B hromosomes from cells that divided once in radioactive thymine All chromosomes radioactive Use your knowledge of semi-conservative DNA replication to explain why all the chromosomes in B were radioactive (4 marks). (b) Rapidly dividing cells were grown in a medium containing unlabelled thymine. Some cells were transferred to a second medium in which all of the thymine molecules were radioactively labelled. 5

6 2. There are two forms of nitrogen. 15 N is a heavier isotope than the normal isotope 14 N. In an investigation a culture of Escherichia oli was obtained in which only contained 15 N DNA. The bacteria were transferred to a medium containing only 14 N DNA and allowed to divide once. A sample of this first generation was removed. The DNA was extracted and centrifuged at high speed. This was repeated with samples of the second and third generation. The diagram below shows the results of this: Generation 0 Generation 1 Generation 2 Generation 3 E.coli with 15 N labelled DNA cells replicate once on 14 N medium cells replicate a second time on 14 N medium cells replicate a third time on 14 N medium (a) Which component of DNA would contain nitrogen (1 mark)? (b) Explain how centrifugation allows DNA containing different isotopes of nitrogen to be separated and distinguished (3 marks). (c) Explain why the DNA from generation 1 is found in the position shown in the diagram (2 marks). (d) omplete the diagram to show the positions of the DNA from generations 2 and 3 (2 marks). (e) Select the letter or letters from the diagram below representing the bacterial DNA in (3 marks): i. Generation 1; ii. Generation 2; iii. Generation 3. Answers 1. a. Two strands/double stranded, each strand acts as a template/ is copied; omplementary base pairs present, allows accurate copying/ replication; Weak hydrogen bonds, easily broken for replication; b. DNA strands separate/hydrogen bonds are broken (a labelled diagram could show this); Each strand forms a template/is copied/one new strand & one old (a labelled diagram could show this); omplementary base pairing; Radioactivity incorporated into (all) new strands; 2. (a) Base/any named base; (b) Use of sl solution; oncentration gradient in tube (set up by centrifugation); DNA molecules move in the concentration gradient (until they reach a place where their density equals that of the caesium chloride)/dna molecules separate according to density/weight/mass; DNA with 15 N will be further down the tube (accept converse) (c) Semi-conservative replication; ontains one heavy and one light strand/half 15 N and half 14 N; (d) Generation 2 - band at top of tube and band at middle of tube; Generation 3 - band at top of tube and band in middle (e) i. A ii. iii. and E (f) 28; 32, 28, 22. represents DNA with 15 N A B represents DNA with 14 N D E F f. In a further investigation, the DNA of the bacterium was isolated and separated into single strands. The percentage of each nitrogenous base in each strand was found. The table shows some of the results. DNA Sample Percentage of base present Adenine ytosine Guanine Thymine Strand Strand 2 18 Use your knowledge of base pairing to complete the table. (2 marks) 6 Acknowledgements: This Factsheet was researched and written by John Greenhalgh. urriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. s may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN

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