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Chapter Ten: DNA: The Chemical Nature of the Gene
Chapter Ten: DNA: The Chemical Nature of the Gene
COMPREHENSION QUESTIONS
*1.What three general characteristics must the genetic material possess?
(1) The genetic material must contain complex information.
(2) The genetic material must replicate or be replicated faithfully.
(3) The genetic material must encode the phenotype.
2.Briefly outline the history of our knowledge of the structure of DNA until the time of Watson and Crick. Which do you think were the principle contributions and developments?
1869: Johann Friedrich Miescher isolates nuclei from white blood cells and extracts a substance that was slightly acidic and rich in phosphorous. He calls it nuclein.
Late 1800s: Albrecht Kossel determines that DNA contains the four nitrogenous bases: adenine, guanine, cytosine, and thymine.
1920s: Phoebus Aaron Levine discovers that DNA consists of repeating units, each consisting of a sugar, a phosphate, and a nitrogenous base.
1950: Erwin Chargaff formulates Chargaff’s rules (A=T and G=C).
1947: William Ashbury begins studying DNA structure using X-ray diffraction.
1951–1953: Rosalind Franklin, working in Maurice Wilkins’ lab, obtains higher resolution pictures of DNA structure using X-ray diffraction techniques.
1953: Watson and Crick propose the model of DNA structure.
All of these scientists contributed information that helped Watson and Crick determine the structure of the DNA double helix. Erwin Chargaff and Rosalind Franklin made two important contributions that directly led to the discoveries by Watson and Crick. By combining Chargaff’s rules with Rosalind Franklin’s X-ray diffraction data, Watson and Crick were able to predict accurately the structure of the DNA double helix.
*3.What experiments demonstrated that DNA is the genetic material?
Experiments by Hershey and Chase in the 1950s using the bacteriophage T2 and E. coli cells demonstrated that DNA is the genetic material of the bacteriophage. Also, the experiments by Avery, Macleod, and McCarty demonstrated that the transforming material initially identified by Griffiths was DNA.
4.What is transformation? How did Avery and his colleagues demonstrate that the transforming principle is DNA?
Transformation occurs when a transforming material (or DNA) genetically alters the bacterium that absorbs the transforming material. Avery and his colleagues demonstrated that DNA is the transforming material by using enzymes that destroyed the different classes of biological molecules. Enzymes that destroyed proteins or nucleic acids had no effect on the activity of the transforming material. However, enzymes that destroyed DNA eliminated the biological activity of the transforming material. Avery and his colleagues were also able to isolate the transforming material and demonstrate that it had chemical properties similar to DNA.
*5.How did Hershey and Chase show that DNA is passed to new phages in phage reproduction?
Hershey and Chase used the radioactive isotope 32P to demonstrate that DNA is passed to new phage particles during phage reproduction. The progeny phage released from bacteria infected with 32P-labeled phages emitted radioactivity from 32P. The presence of the 32P in the progeny phage indicated that the infecting phage had passed DNA on to the progeny phage.
6.Why was Watson and Crick’s discovery so important?
By deciphering the structure of the DNA molecule, Watson and Crick provided the foundation for molecular studies of the genetic material or DNA, allowing scientists to discern how genes function to produce phenotypes. Their model also suggested a possible mechanism for the replication of DNA that would ensure the fidelity of the replicated copies.
*7.Draw and label the three parts of a DNA nucleotide.
The three parts of a DNA nucleotide are phosphate, deoxyribose sugar, and a nitrogenous base.
8.How does an RNA nucleotide differ from a DNA nucleotide?
DNA nucleotides, or deoxyribonucleotides, have a deoxyribose sugar that lacks an oxygen molecule at the 2' carbon of the sugar molecule. Ribonucleotides, or RNA nucleotides, have a ribose sugar with an oxygen linked to the 2' carbon of the sugar molecule. Ribonucleotides may contain the nitrogenous base uracil, but not thymine. DNA nucleotides contain thymine, but not uracil.
9.How does a purine differ from a pyrimidine? What purines and pyrimidines are found in DNA and RNA?
A purine consists of a six-sided ring attached to a five-sided ring. A pyrimidine consists of only a six-sided ring. In both DNA and RNA, the purines found are adenine and guanine. DNA and RNA differ in their pyrimidine content. The pyrimidine cytosine is found in both RNA and DNA. However, DNA contains the pyrimidine thymine, whereas RNA contains the pyrimidine uracil but not thymine.
*10.Draw a short segment of a single polynucleotide strand, including at least three nucleotides. Indicate the polarity of the strand by labeling the 5' end and the 3' end.
11.Which bases are capable of forming hydrogen bonds with each another?
Adenine is capable of forming two hydrogen bonds with thymine. Guanine is capable of forming three hydrogen bonds with cytosine.
*12.What is local variation in DNA structure and what causes it?
Since DNA is not a static, rigid structure that is invariant, the local variation in DNA structure refers to the actual variations that exist in a DNA molecule. For instance, B-DNA is described as having an average of 10 bases per turn. However, the actual values may be less than or greater than 10, depending on the environmental conditions.
13.What are some of the important genetic implications of the DNA structure?
Referring back to question 1, the structure of DNA gives insight into the three fundamental genetic processes. The Watson and Crick model suggests that the genetic information or instructions are encoded in the nucleotide sequences. The complementary polynucleotide strands indicate how faithful replication of the genetic material is possible. Finally, the arrangement of the nucleotides is such that they specify the primary structure or amino acid sequence of protein molecules.
*14.What are the major transfers of genetic information?
The major transfers of genetic information are replication, transcription, and translation. These are the components of the central dogma of molecular biology.
15.What are hairpins and how do they form?
Hairpins are a type of secondary structure found in single strands of nucleotides. The formation of hairpins occurs when sequences of nucleotides on the single strand are inverted complementary repeats of one another.
16. What is DNA methylation?
DNA methylation is the addition of methyl groups (–CH3) to certain positions on the nitrogenous bases on the nucleotide.
APPLICATION questions and problems
17.A student mixes some heat-killed type IIS Streptococcus pneumonia bacteria with live type IIR bacteria and injects the mixture into a mouse. The mouse develops pneumonia and dies. The student recovers some type IIS bacteria from the dead mouse. It is the only experiment conducted by the student. Has the student demonstrated that transformation has taken place? What other explanations might explain the presence of the type IIS bacteria in the dead mouse?
No, the student has not demonstrated that transformation has taken place. Unlike Griffiths, who used strains IIR and IIIS to demonstrate transformation, the student is using strains IIR and IIS. A mutation in the IIR strain injected into the mouse could be sufficient to convert the IIR strain into the virulent IIS strain. By not conducting the appropriate control of injecting IIR bacteria only, the student cannot determine whether the conversion from IIR to IIS is due to transformation or to a mutation.
Although heat may have killed all the IIS bacteria, the student has not demonstrated that the heat was sufficient to kill all the IIS bacteria. A second useful control experiment would have been to inject the heat-killed IIS into mice and see if any of the IIS bacteria survived the heat treatment.
*18. (a) Why did Hershey and Chase choose 32P and 35S for use in their experiment?
Proteins contain sulfur in the amino acids cysteine and methionine. However, proteins do not typically contain phosphorous (or have very limited amounts due to the phosphorylation of certain proteins by protein kinases). DNA contains much phosphorous due to its sugar-phosphate backbone but no sulfur. Hershey and Chase chose the isotopes 32P and 35S because these radioactive elements would allow them to distinguish between proteins and DNA molecules. Only DNA would contain the isotope 32P, and only proteins would contain the isotope 35S.
(b) Could they have used radioactive isotopes of carbon (C) and oxygen (O) instead? Why or why not?
No, they could not have used radioactive isotopes of oxygen and carbon. Both DNA and proteins contain significant amounts of carbon and oxygen. The molecules were labeled with the radioactive isotopes in vivo by using phage to infect E. coli cells grown in media containing the radioactive isotopes. Since both proteins and DNA contain carbon and oxygen, both molecules in the phage progeny would have received the radioactive isotopes. Hershey and Chase would have been unable to isolate only DNA molecules or proteins that contain radioactive isotopes of these elements.
19.What results would you expect if the Hershey and Chase experiment were conducted on tobacco mosaic virus?
Infection by TMV results in both the protein coat and the RNA genome entering the host cell. Inside the plant cell, the TMV protein coat unwinds, releasing the viral genome, which initiates infection. If Hershey and Chase had used 32P and 35S to label TMV particles, the RNA molecules would have been labeled with the 32P and the viral proteins would have been labeled with 35S. However, both the protein coat and the RNA genome would have entered the cell, so “radioactive ghost proteins” would not have been located outside the cell. Newly synthesized viral RNAs would have contained measurable levels of 32P.
*20. Each nucleotide pair of a DNA double helix weighs about 1 × 10–21 g. The human
body contains approximately 0.5 g of DNA. How many nucleotide pairs of DNA are in the human body? If you assume that all the DNA in human cells is in the B-DNA form, how far would the DNA reach if stretched end to end?
If each nucleotide pair of a DNA double helix weighs approximately 1 × 10–21 g, and the human body contains 0.5 grams of DNA, then the number of nucleotide pairs can be estimated as: (0.5 g DNA/human)/(1 × 10–21 g/ nucleotide) = 5 × 1020 nucleotides pairs/human.
DNA that is in B form has an average distance of 3.4 nm between each nucleotide pair. If a human possesses 5 × 1020 nucleotide pairs, then that DNA stretched end to end would reach: (5 × 1020 nucleotides/ human) × (3.4 nm/nucleotide pair) = 1.7 × 1021 nm or 1.7 × 109 km.
21.What aspects of its structure contribute to the stability of the DNA molecule? Why is RNA less stable than DNA?
Several aspects contribute to the stability of the DNA molecule. The relatively strong phosphodiester linkages connect the nucleotides of a given strand of DNA. The helical nature of the double-stranded DNA molecule results in the negatively charged phosphates of each strand being arranged to the outside and away from each other. The complementary nature of the nitrogenous bases of the nucleotides helps hold the two strands of polynucleotides together. The stacking interactions of the bases, which allow for any base to follow another in a given strand, also play a major role in holding the two strands together. Finally, the ability of DNA to have local variations in its secondary structure contributes to its stability.
RNA nucleotides or ribonucleotides contain an extra oxygen at the 2' carbon. This extra oxygen at each nucleotide makes RNA a less stable molecule.
*22.Which of the following relations will be found in the percentages of bases of a double-stranded DNA molecule?
A double-stranded DNA molecule will contain equal percentages of A and T nucleotides and equal percentages of G and C nucleotides. The combined percentage of A and T bases added to the combined percentage of the G and C bases should equal 100.
(a) A + T = G + CNo.
(b) A + G = T + C Yes.
(c) A + C = G + TYes.
(d) No.
(e) Yes.
(f) No.
(g) Yes.
(h) No.
*23.If a double-stranded DNA molecule has 15% thymine, what are the percentages of all the other bases?
The percentage of thymine (15%) should be approximately equal to the percentage of adenine (15%). The remaining percentage of DNA bases will consist of cytosine and guanine bases (100% – 15% – 15% = 70%); these should be in equal amounts (70%/2 = 35%). Therefore the percentages of each of the other bases if the thymine content is 15% are adenine = 15%; guanine = 35%; and cytosine = 35%.
24.A virus contains 10% adenine, 24% thymine, 30% guanine, and 36% cytosine. Is the genetic material in this virus double-stranded DNA, single-stranded DNA, double-stranded RNA, or single-stranded RNA? Support your answer.
Most likely the viral genome is single-stranded DNA. The presence of thymine indicates that the viral genome is DNA. For the molecule to be double-stranded DNA, we would predict equal percentages of adenine and thymine bases and equal percentages of guanine and cytosine bases. Neither the percentages of adenine and guanine bases nor the percentages of guanine and cytosine bases are equal, indicating that the viral genome is single-stranded.
*25.A B-DNA molecule has 1 million nucleotide pairs.
(a) How many complete turns are there in this molecule?
B-form DNA contains approximately 10 nucleotides per turn of the helix. A B-DNA molecule of 1 million nucleotide pairs will have about the following number of complete turns: (1,000,000 nucleotides)/ 10 nucleotides/turn) = 100,000 complete turns.
(b) If this same molecule were in the Z-DNA configuration, how many complete turns would it have?
If the same DNA molecule assumes a Z-DNA configuration then each turn would consist of about 12 nucleotides. The determination of the number of complete turns in the 1 million nucleotide molecule is:
(1,000,000 nucleotides)/(12 nucleotides / turn) = 83333.3 or 83333 complete turns.
26.For entertainment on a Friday night, a genetics professor proposed that his children diagram a polynucleotide strand of DNA. Having learned about DNA in preschool, his 5-year-old daughter was able to draw a polynucleotide strand, but she made few mistakes. The daughter’s diagram (represented here) contained at least 10 mistakes.
(a) Make a list of all the mistakes in the structure of this DNA polynucleotide strand.
(1)Neither 5' carbon of the two sugars is directly linked to phosphorous.
(2)Neither 5' carbon of the two sugars has an OH group attached.
(3)Neither sugar molecule has oxygen in its ring structure between the 1' and 4' carbons.
(4)In both sugars, the 2' carbon has an –OH group attached, which does not occur in deoxyribonucleotides.
(5)At the 3' position in both sugars, only hydrogen is attached, as opposed to an –OH group.
(6)The 1' carbon of both sugars has an –OH group, as opposed to just a hydrogen attached.
(b) Draw the correct structure for the polynucleotide strand.
27.Chapter 1 considered the theory of the inheritance of acquired characteristics and noted that this theory is no longer accepted. Is the central dogma consistent with the theory of the inheritance of acquired characteristics? Why or why not?
The central dogma of molecular biology is not consistent with the theory of inheritance of acquired characteristics. The flow of information predicted by the central dogma is:
DNA RNA Protein
One exception to the central dogma is reverse transcription, whereby RNA codes for DNA. However, biologists currently do not know of a process that will allow for the flow of information from proteins back to DNA. The theory of inheritance of acquired characteristics necessitates such a flow of information from proteins back to the DNA.
*28. Write a sequence of bases in an RNA molecule that would produce a hairpin structure.
For a hairpin structure to form in a RNA molecule, an inverted complementary RNA sequence separated by a region of noncomplementary sequence is necessary. The inverted complements form the stem structure, and the loop of the hairpin is formed by the noncomplementary sequences.
5'—UGCAU—3'…unpaired nucleotides…5'—AUGCA—3'
U A
A U
C G
G C
5'------U A------3'
29.The following sequence is present in one strand of a DNA molecule:
5'—CATTGACCGA—3'
Write out the sequence on the same strand that produces an inverted repeat and the sequence on the complementary strand.
The presence of inverted complementary sequences on the same strand of a DNA molecule will result in a double-stranded DNA molecule containing a palindrome and could result in the formation of a cruciform structure.
5'—CATTGACCGA—3'……………………..5'—TCGGTCAATG—3'
3'—GTAACTGGCT—5'……………………..3'—AGCCAGTTAC—5'
CHALLENGE QUESTIONS
30. Suppose that an automated, unmanned probe is sent into deep space to search for extraterrestrial life. After wandering for many light years among the far reaches of the universe, this probe arrives on a distance planet and detects life. The chemical composition of life on this planet is completely different from that of life on Earth, and its genetic material is not composed of nucleic acids. What predictions can you make about the chemical properties of the genetic material on this planet?
Although the chemical composition of the genetic material may be different DNA, it
more than likely will have similar properties to DNA. As discussed earlier in the
chapter, the genetic material must possess three general characteristics:
(1)It must contain complex information.
(2)It must replicate or be replicated faithfully.
(3)It must encode the phenotype.
Even if the material is not DNA, it must meet these criteria. For instance, if the material could not be replicated or duplicated faithfully, then life on that planet could not continue, since ultimately no offspring could be produced. A lack of fidelity would result in the loss of information. Genetic material from any lifeform has to store the information necessary for the survival of that organism. Also the genetic material will need to be stable. Unstable molecules will not allow for long-term storage of information, resulting in the loss of information and change in phenotype.