STUDY HINTS
Nucleic acids play a central role in both the coding and the decoding of genetic information. For this reason it is important to have a fairly detailed understanding of their structure and replication, and of the differences between deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). If you spend a little time looking closely at the types and arrangement of atoms in each molecular component of a nucleotide, you will probably find that the structures are not as complex as they may first appear (for example, T = 5 methyl uracil). Indeed, some of the most important are quite similar and are therefore fairly easy to learn and compare. The diagrams that follow can be used to prepare flash cards or other study aids. You can then arrange them to make your own nucleotides or chains.
Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are both composed of chains of subunits called nucleotides. Each nucleotide is, in turn, made up of a 5-carbon sugar, a phosphoric acid group, and a nucleotide base. DNA and RNA differ structurally in three main ways, as shown in the table of comparisons. First, the 5-carbon sugar in DNA is deoxyribose and is missing an oxygen (noted by the arrow) that is present in ribose, the 5-carbon sugar in RNA. Second, the nucleotide bases in DNA are of four types, two purines and two pyrimidines. The same is true of RNA, but there is a difference in one of the pyrimidines. DNA contains thymine, whereas RNA contains uracil. The third difference is not shown in this table. DNA is a double-stranded molecule containing two paired nucleotide chains. RNA, on the other hand, is typically a single-stranded molecule.
As we mentioned earlier, finding an appropriate mnemonic may help you remember important relationships. For instance, you might try “Agents are pure in heart” to help you remember the nucleotide base compositions. Agents will remind you of A (adenine) and G (guanine). Pure is for purines, and heart reminds you that purines have two attached rings (heartlike, if you have a good imagination). All other nucleotide bases are single-ring pyrimidines. In order to test your understanding of these genetic building blocks, you might set a hypothetical sequence to diagram. Use simplified representations of nucleotide components. Several of the problems in the following Problem Set are designed in this way. An almost endless variety of similar problems can also be written. Just remember, if you can set such a problem for yourself and solve it correctly, you can do the same thing on an examination.
Three classical experiments played a central role in confirming the role of DNA as the genetic material. These experiments were (1) the demonstration of a transforming factor by Griffith (1928); (2) the establishment that this transforming factor was DNA, as shown by Avery, MacLeod, and McCarty (1944); and (3) the confirmation that DNA was not only necessary but sufficient to code for the growth of a new organism, as reported by Hershey and Chase (1952). This sequence of experiments provides an outstanding example of the development and confirmation of an idea and the application of experimental methods. You should look at their designs and conclusions carefully.
In 1953, Watson and Crick proposed a model of DNA as a pair of helically coiled nucleotide chains or strands held together by hydrogen bonding between the nucleotide bases. Within a strand, one nucleotide is bound to the next by a phosphodiester bond between the 3′ carbon of one nucleotide and the 5′ carbon of the next. This creates an important asymmetry in the molecule, with a 3′ end and a 5′ end to each strand.
Two nucleotide strands are linked by hydrogen bonds between specific pairs of nucleotide bases. The purine adenine pairs with the pyrimidine thymine by two hydrogen bonds, and the purine guanine pairs with the pyrimidine cytosine by three hydrogen bonds. Furthermore, the orientation of the nucleotide bases in one chain is inverted, when compared to the other chain, so one side has a 3′ to 5′ orientation and the other has a 5′ to 3′ orientation. This is called antiparallel since the two strands are parallel but “backward” to each other. In the following diagram, we have exploded a small segment of DNA to show the components and their relationships.
DNA replication is summarized in the diagram. There are several key points to remember. First, the pairing of A and T, and of G and C, gives rise to Chargaff’s rule equating the members of each pair [A = T and G = C; A + G = C + T; or (A + G)/(C + T) = 1]. If Chargaff’s rule holds for a given set of nucleotides, the molecules must be double-stranded.
Second, remember that DNA polymerases synthesize new DNA strands by adding new nucleotides at the 3′ end of a chain. In other words, DNA elongates in the 5′ to 3′ direction. Since the two strands of the double helix are antiparallel, one new strand will be formed as a continuous chain, and the other is formed in short sections as unwinding proceeds. An RNA primer is first laid down, as indicated in the figure by the circles. The short segments, called Okazaki fragments, are then joined when the RNA is replaced by DNA nucleotides and the DNA is linked by ligase (see problem 15).
IMPORTANT TERMS
Endonuclease
Eukaryote
Exonuclease
Gyrase
Helicase
Ligase
Nucleoside
Nucleotide
Phosphodiester bond
Polymerase
Primase
Primer
Primosome
Prokaryote
Purine
Pyrimidine
Repetitive DNA
Replication
Semiconservative replication
Single-strand binding proteins
Template DNA
Topoisomerase
Transcription
Translation
PROBLEM SET 3
1. A single strand of DNA contains the base sequence 5′AGCTG3′.
(a) What is the complementary strand sequence?
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(b) If the complementary strand is used during transcription, what is the base sequence of the resulting RNA?
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2. What kind of evidence indicates that nucleotides of most organisms occur in matched pairs in the DNA molecules?
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3. If it is known that about 22 percent of the double-stranded DNA of an organism consists of thymine, can the other base percentages be determined? If so, what are they?
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4. The gene for a polypeptide (double-stranded DNA with 300 nucleotide pairs) has a base composition of A = 0.32, G = 0.18, C = 0.18, and T = 0.32. Assume that a single 300-base strand of RNA is transcribed from this gene. Can you determine, from the information given, the base composition of the RNA? If so, what is it?
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5. A certain DNA virus has a base ratio of (A + G)/(C + T) = 0.85. Is this single- or double-stranded DNA? Please explain your answer.
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6. Label the nucleosides in the following figure as RNA or DNA components, and give the reason for your answers.
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7. Multiple choice (in this and later multiple choice questions): Select the best answer and be certain you can explain the reason for your choice. There is a reasonably good correlation between the amount of DNA and the degree of complexity of the organism in which of the following?
(a) viruses and bacteria;
(b) viruses, bacteria, and eukaryotes;
(c) only eukaryotes.
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8. Using the following symbols for nucleotide components, construct a single strand of DNA with this sequence: 5′ATGCGT 3′. Now, using your newly constructed DNA strand as a template, construct the complementary strand as if this DNA were replicating. Indicate the direction of growth.
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9. Why does DNA polymerase III require an RNA primer?
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10. In double-stranded nucleic acid, the proper alignment of bases needed for the hydrogen bonding of complementary bases is possible only when the two poly-nucleotide chains are oriented in which of the following:
(a) the same direction?
(b) opposite directions?
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11. The Hershey and Chase experiment used bacteriophage and a chemical difference between proteins and nucleic acids to demonstrate the role of proteins and DNA in genetic transmission. Explain the chemical differences the researchers utilized, and describe their experiment.
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12. The genetic material in a hypothetical organism has the base ratio (U + C)/(A + G) of approximately 1.02 as determined experimentally by chemical analysis. This organism is also known to contain 20 percent adenine. Can the other base percentages be determined? If so, what are they?
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13. From the following figure, indicate whether transcription is occurring in a prokaryotic or a eukaryotic cell. Please explain.
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14. The Meselson and Stahl experiment used Escherichia coli with 15N-labeled DNA transferred to a 14N medium. The isotope constitution of its replicated DNA was studied by density-gradient separation. Since DNA replication is semiconservative, the Meselson–Stahl experiment would be expected to show which of the following density patterns, after the third replication in 14N?
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15. The following figure shows a replication fork during the replication of double-stranded DNA. Briefly identify the numbered proteins and describe their functions in DNA replication.
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ANSWERS TO PROBLEM SET 3
1. Since adenine pairs with thymine and cytosine pairs with guanine,
(a) RNA polymerase transcribes 5′ to 3′; therefore, the complementary strand is TCGAC.
(b) The base sequence of the resulting RNA is the same as the original base sequence, except that thymine is replaced by uracil: AGCUG.
2. The ratios of adenine to thymine and of cytosine to guanine are 1:1. This relationship can be represented in many different forms (some of which are illustrated in the following answers) and is commonly known as Chargaff’s rule.
3. Yes, if 22 percent of the molecule is thymine, then 22 percent is also adenine. The remaining 56 percent is divided equally between cytosine and guanine (28 percent each).
4. No, the base composition of the RNA molecule would be determined by the base sequence of only one of the DNA strands. There is no way to determine the base ratios from the data given.
5. The DNA in this virus is single-stranded, because the 1:1 ratio of A to T and of G to C typical of a double-stranded molecule is not found here. Note also that the equality pairs one purine (e.g., adenine) with one pyrimidine (e.g., thymine). In the formula given in the problem, (A + G)/(C + T) should equal 1 in a double-stranded molecule, because the number of purines equals the number of pyrimidines in such a molecule. Be sure to look at a given ratio carefully, however, because a test maker could easily make alterations that would change the expectation; for instance, (A + T)/(C + G) would not be predictable in an organism.
6.
(a) RNA nucleoside, because the sugar is ribose.
(b) DNA nucleoside, because the sugar is deoxyribose (literally, “missing an oxygen”).
(c) RNA nucleoside, because the sugar is ribose and the base is uracil.
7.
(a) Viruses and bacteria. There is a tendency for an increase in the amount of DNA in eukaryotes. Some protozoa, club mosses, and algae, however, have more DNA per cell than do mammals. There are numerous examples among the lower vertebrates in which the amount of DNA is larger than in more advanced vertebrates. In fact, animals of similar morphologic characteristics may also have quite different DNA content. Depending upon the organism, as much as 90 percent of the DNA in eukaryotes may not code for unique polypeptides and therefore will not reflect morphological or physiological complexity.
8. The direction of growth is the 5′ end of one nucleotide attaching to the 3′ end of the growing strand.
9. DNA polymerase can add nucleotides only to an open 3′ OH end, whereas RNA polymerase does not require such a preexisting “primer.” The RNA polymerase produces a 3′ OH end for the DNA polymerase III to begin replication.
10.
(b) Opposite directions, an arrangement called antiparallel.
11. They used the T2 virus, which is composed of a strand of DNA contained within a protein capsule. The protein capsule has some sulfur-containing amino acids but no phosphorus, whereas DNA contains phosphorus but no sulfur. They marked two cultures of T2 viruses by growing one on E. coli containing radioactive sulfur and the other on E. coli containing radioactive phosphorus. Thus they obtained 32P-labeled DNA in one culture and 35S-labeled protein in the other. These were then used to infect nonradioactive E. coli. It was found that 30 percent of the radioactive phosphorus was transmitted to the next generation, but none of the radioactive sulfur was transmitted. This experiment, in agreement with the study by Avery, MacLeod, and McCarty, showed DNA to be the genetic material.
12. Yes, 1.02 is close enough to 1.00 to be the result of a minor experimental error in measuring the equal base ratios of a double-stranded RNA, and the other base percentages would be 20 percent uracil, 30 percent cytosine, and 30 percent guanine.
13. Transcription is occurring in a prokaryote, since the ribosomes have attached to the messenger RNA (mRNA) while it is still being transcribed by RNA polymerase. If this were a eukaryote, the mRNA would have to be processed and leave the nucleus before ribosomes could complex with it.
14.
(c) After each replication, the proportion of 14/14 will increase. At the first replication, all of the new DNA will be 14/15 hybrid molecules. At the next replication, the ratio of 14/14 to 14/15 will be 1:1. At the third replication, the ratio will be 3:1.
15.
(1) Helicase or helix unwinding protein, which separates the parental strands.
(2) Single-strand binding protein or helix destabilizing protein, which prevents separated strands from rejoining.
(3) RNA primase or primosome complex, which synthesizes the RNA primer.
(4) DNA polymerase III, which synthesizes DNA using the RNA primer.
(5) DNA polymerase I, which extends the DNA synthesized by DNA polymerase III and removes the RNA primer of the Okazaki fragment in front of it.
(6) Ligase, which joins the last nucleotide of an Okazaki fragment to the adjacent fragment, forming a phosphodiester bond.
(7) Gyrase, which relaxes the supercoils that are extra coils of the coiled molecule caused by unwinding DNA.
CROSSWORD PUZZLE 3
Nucleic Acids: DNA and RNA
Across
1. Bond between the 3′ end of one nucleotide and the 5′ end of another nucleotide
7. Purine base that does not contain a double-bond oxygen
9. RNA sequence that is required for DNA replication
11. Molecule that contains a nucleoside and a phosphate group
12. DNA’s 5-carbon sugar
15. DNA-unwinding enzyme
19. Enzyme responsible for producing the RNA primer of a replicating DNA molecule
20. Pyrimidine base found in RNA but not DNA
21. Protein that prevents the reannealing of DNA during replication
23. Term for the replicating fragments on the so-called lagging strand of DNA
24. Term for the single-ringed nitrogen-containing bases
Down
2. Complex of proteins involved in the primer formation
3. Pyrimidine that is found in DNA, but not found in RNA
4. Term for the synthesis of DNA
5. Double-ringed nitrogen-containing bases
6. Enzyme that joins nucleotides together
8. One of the men that demonstrated DNA was the genetic material
10. Enzyme that removes the terminal nucleotide of nucleic acids
13. Describes the way DNA replication occurs
14. Enzyme that cuts the sugar-phosphate backbone of DNA
16. Pyrimidine with only one double-bond oxygen
17. Term for a molecule containing a 5-carbon sugar and a nitrogen containing base
18. Man who first indicated the 1:1 ratio of A to T and G to C.
22. Enzyme that removes supercoils from DNA helix as it unwinds during replication