Genetics Ch. 9
C2) After the DNA from type S bacteria is exposed to type R bacteria, list all of the steps that you think must occur for the bacteria to start making a capsule.
1. A fragment of DNA binds to the cell surface. 2. It penetrates the cell wall/cell membrane. 3. It enters the cytoplasm. 4. It recombines with the chromosome. 5. The genes within the DNA are expressed (i.e., transcription and translation). 6. The gene products create a capsule. That is, they are enzymes that synthesize a capsule using cellular molecules as building blocks.
Describe the previous experimental evidence that led Watson and Crick to the discovery of the DNA double helix.
1. The chemical structure of single nucleotides was understood by the 1950s. 2. Watson and Crick assumed DNA is composed of nucleotides linked together in a linear fashion to form a strand. They also assumed the chemical linkage between two nucleotides is always the same. 3. Franklin's diffraction patterns suggested several structural features. First, it was consistent with a helical structure. Second, the diameter of the helical structure was too wide to be only a single-stranded helix. Finally, the line spacing on the diffraction pattern indicated the helix contains about 10 bp per complete turn. 4. In the chemical analysis of the DNA from different species, the work of Chargaff indicated that the amount of adenine equaled the amount of thymine, and the amount of cytosine equaled that of guanine. 5. In the early 1950s, Linus Pauling proposed that regions of proteins can fold into a secondary structure known as an α helix. To discover this, Pauling built large models by linking together simple ball-and-stick units. In this way, he could determine if atoms fit together properly in a complicated three-dimensional structure. A similar approach was used by Watson and Crick to solve the structure of the DNA double helix.
You identify a new species of gecko on a Spring Break trip to Cancun. You decide to check and see if Chargraff's rules govern the composition of the DNA in this gecko. You find that it contains 18% adenine, but the other three bands are indecipherable. How much each of thymine, guanine, and cytosine would you expect to find?
18% thymine; 32% guanine; 32% cytosine.
C8) If one DNA strand is 5ʹ-GGCATTACACTAGGCCT-3ʹ, what is the sequence of the complementary strand?
3'-CCGTAATGTGATCCGGA-5'
Deoxycytidine refers to the nucleoside which includes the base cytosine and deoxyribose sugar. A) True B) False
A
Eukaryotic DNA is packaged around proteins to form compact loops. A) True B) False
A
T2 bacteriophage rely on the lytic life cycle to generate an ongoing infection. A) True B) False
A
The advent of molecular genetics has greatly improved our ability to perform research in other areas of genetics. A) True B) False
A
The backbones of the two DNA strands in a double helix run in opposite directions (one 5'-3' left to right, the other 5'-3' right to left). A) True B) False
A
The genetic material must be able to contain an infinite amount of information. A) True B) False
A
What functional role is the three dimensional structure of DNA likely to accomplish? A) Very long molecules of DNA can be packed into a small space. B) Information within DNA is made available to drive cellular functions. C) Information within DNA is modified from generation to generation. D) All of these. E) None of these.
A
What are the key differences between A, B, and Z DNA?
A and B are right handed helices, while Z DNA is a left handed helix. The numbers of base pairs per turn differ slightly (11.0 for A, 10.0 for B, and 12.0 for Z). B DNA forms major and minor grooves, while A and Z DNA have different external composition (for instance, Z DNA forms a single external groove). Other answers are possible.
One explanation for the amount of 32P in the solution (35%) is that not all the phage were able to infect bacterial cells. What would be likely to happen if a longer incubation period were allowed for infection?
A few more phage may be able to infect bacterial cells. However, with long incubation times, it becomes more likely that lysis of the bacteria may begin, releasing the originally labeled DNA into the solution again.
C28) What chemical group (phosphate group, hydroxyl group, or a nitrogenous base) is found at the 3ʹ end of a DNA strand? What group is found at the 5ʹ end?
A hydroxyl group is at the 3' end, and a phosphate group is at the 5' end.
E2) With regard to the experiment described in Figure 9.2, answer the following: A. List several possible reasons why only a small percentage of the type R bacteria was converted to type S. B. Explain why an antibody must be used to remove the bacteria that are not transformed. What would the results look like, in all five cases, if the antibody/centrifugation step had not been included in the experimental procedure? C. The DNA extract was treated with DNase, RNase, or protease. Why was this done? (In other words, what were the researchers trying to demonstrate?)
A. There are different possible reasons why most of the cells were not transformed. 1. Most of the cells did not take up any of the type S DNA. 2. The type S DNA was usually degraded after it entered the type R bacteria. 3. The type S DNA was usually not expressed in the type R bacteria. B. The antibody/centrifugation steps were used to remove the bacteria that had not been transformed. It enabled the researchers to determine the phenotype of the bacteria that had been transformed. If this step was omitted, there would have been so many colonies on the plate it would have been difficult to identify any transformed bacterial colonies, because they would have represented a very small proportion of the total number of bacterial colonies. C. They were trying to demonstrate that it was really the DNA in their DNA extract that was the genetic material. It was possible that the extract was not entirely pure and could contain contaminating RNA or protein. However, treatment with RNase and protease did not prevent transformation, indicating that RNA and protein were not the genetic material. In contrast, treatment with DNase blocked transformation, confirming that DNA is the genetic material.
A nucleoside consists of a nitrogenous base, a sugar, and up to three phosphate groups. A) True B) False
B
Avery, MacLeod, and McCarty were able to demonstrate that their biochemical extracts were perfectly pure. A) True B) False
B
B DNA is the only biologically relevant form of DNA. A) True B) False
B
DNA is the genetic material for all organisms. A) True B) False
B
Hershey and Chase demonstrated that DNA is the genetic material for eukaryotic organisms. A) True B) False
B
Hydrogen bonding is the primary chemical bond holding a strand of DNA together. A) True B) False
B
RNA molecules follow the same base pairing rules as DNA. A) True B) False
B
Rosalind Franklin was given proper recognition for her critical role in the discovery of the structure of DNA. A) True B) False
B
What key characteristic of T2 bacteriophage allowed Hershey and Chase to use it in their studies of the genetic material? A) Its genes encode proteins that assemble to produce the viral coat. B) It injects its genetic material into a bacterial cell. C) It can undergo either the lytic or lysogenic life cycle. D) It enters the bacterial cell to cause infection.
B
How did Hershey and Chase's results support the idea that DNA is the genetic material?
Bacteriophage were known to inject their genetic material into the infected bacterium. When the bacteriphage DNA was radioactively labeled, the majority of it remained with bacterial cells rather than in the supernatant. When bacteriophage protein was labeled, the opposite result was obtained. This indicated that DNA rather than protein was likely to be functioning as genetic material for the virus.
What factors contribute to RNA tertiary structure?
Base pairing to form double stranded helices, stacking between bases, hydrogen bonding between bases and backbone regions, interactions with ions, small molecules, and large proteins.
List three possible secondary structures for RNA.
Bulge loop, internal loop, multibranched loop, stem-loop.
What is the difference between purines and pyrimidines? A) Purines are found in RNA while pyrimidines are found in DNA. B) Pyrimidines are found in RNA while purines are found in DNA. C) Purines contain a double ring backbone, while pyrimidines contain a single ring backbone. D) Pyrimidines contain a double ring backbone while purines contain a single ring backbone.
C
Which of the following is a line of evidence that indicates triplex DNA may exist in vivo? A) Triplex DNA can be synthesized in the lab. B) Triplex DNA follows certain base pairing rules. C) Triplex DNA is recognized by certain proteins present in cells. D) A and B. E) None of these.
C
Which of the following statements about Griffith's type IIR strain of bacteria is incorrect? A) It cannot make a polysaccharide capsule. B) It must be competent to pick up DNA from the environment. C) It contains a mutation in a gene that would allow synthesis of a type III capsule. D) All of these are correct statements.
C
Which of the following statements about the structure of DNA is incorrect? A) One complete turn requires 3.4nm and 10 base pairs. B) The backbones of each strand run in opposite directions relative to each other. C) Each pair of nucleotides is held together by three hydrogen bonds. D) The width of the molecule is a constant 2nm.
C
Which statement is false? A) RNA and DNA are both composed of nucleotides joined by phosphodiester bonds. B) RNA and DNA both have secondary and tertiary structure. C) RNA is always single stranded while DNA is always double stranded. D) A and B are both false. E) None of these (A, B and C are all true statements).
C
Why might heat-killed bacteria be useful as a vaccine? A) It can cause a lethal infection. B) Heat degradation of proteins changes their shape. C) Molecules from the cell surface are still intact and can provoke an immune response. D) DNA molecules can transform other strains of bacteria.
C
Why was the DNase treatment used by Avery, MacLeod, and McCarty an important step? A) This allowed them to isolate pure DNA samples. B) This allowed them to isolate pure protein samples. C) This allowed them to demonstrate that removing the DNA prevents transformation. D) This allowed them to demonstrate that mixing rough cells with DNA prevents transformation.
C
The chromosomal theory of inheritance was easily accepted, and the chemical composition of chromosomes was determined, but this did not provide researchers with information about the nature of the genetic material. Why?
Chromosomes are made up of both DNA and protein.
In studying a virus, you find the following proportions of nitrogenous bases present: adenine 23%, guanine 31%, cytosine 23%, unexpected base 17%. What statement(s) can you make about this virus? A) It probably uses RNA as its genetic material. B) The genetic material of this virus is probably single stranded. C) Base pairing rules in this virus probably include adenine:cytosine. D) A and B. E) None of the above.
D
The genetic material must show variation. Which of the following is NOT an example of genetic variation? A) New mutations can occur during meiosis. B) Different species have different numbers of chromosomes. C) Different individuals within a species have different phenotypes. D) Individuals of different species may have similar phenotypes.
D
What is the key difference between 'rough' and 'smooth' S. pneumoniae? A) Rough bacteria can evade the immune system of a mouse. B) Rough bacteria have more genes than smooth bacteria. C) Smooth bacteria have more genes than rough bacteria. D) Smooth bacteria are able to synthesize a polysaccharide capsule.
D
What occurred in the transformation observed by Griffith? A) A rough strain underwent mutation to convert it to a smooth strain. B) A smooth strain underwent mutation to convert it to a rough strain. C) A rough strain passed genetic information to a smooth strain. D) A smooth strain passed genetic information to a rough strain.
D
What type of bond connects nucleotides? A) Phosphodiester. B) Hydrogen. C) Ionic. D) A and B. E) None of these.
D
Which of the following is a key component of a DNA nucleotide? A) Nitrogenous base. B) Deoxyribose sugar. C) Phosphate group. D) All of these. E) None of these.
D
Within living cells, many different proteins play important functional roles by binding to DNA and RNA. As described throughout your textbook, the dynamic interactions between nucleic acids and proteins lie at the heart of molecular genetics. Some proteins bind to DNA (or RNA) but not in a sequencespecific manner. For example, histones are proteins important in the formation of chromosome structure. In this case, the positively charged histone proteins actually bind to the negatively charged phosphate groups in DNA. In addition, several other proteins interact with DNA but do not require a specific nucleotide sequence to carry out their function. For example, DNA polymerase, which catalyzes the synthesis of new DNA strands, does not bind to DNA in a sequence-dependent manner. By comparison, many other proteins do interact with nucleic acids in a sequence-dependent fashion. This means that a specific sequence of bases can provide a structure that is recognized by a particular protein. Throughout the textbook, the functions of many of these proteins will be described. Some examples include transcription factors that affect the rate of transcription, proteins that bind to centromeres, and proteins that bind to origins of replication. With regard to the three-dimensional structure of DNA, where would you expect DNA-binding proteins to bind if they recognize a specific base sequence? What about DNA-binding proteins that do not recognize a base sequence?
DNA-binding proteins that recognize a base sequence must bind into a major or minor groove of the DNA, which is where the bases are accessible to a DNA-binding protein. Most DNA-binding proteins, which recognize a base sequence, fit into the major groove. By comparison, other DNA-binding proteins, such as histones, which do not recognize a base sequence, bind to the DNA backbone.
C16) Compare the structural features of a double-stranded RNA structure with those of a DNA double helix.
Double-stranded RNA is more like A DNA than B DNA. See the text for a discussion of A-DNA structure
C20) An organism has a G + C content of 64% in its DNA. What are the percentages of A, T, G, and C?
G = 32%, C = 32%, A = 18%, T = 18%.
What four characteristics are essential for genetic material?
It must contain information, be transmitted from parent to offspring, be replicated reliably, and allow for variation.
C18) What structural feature allows DNA to store information?
Its nucleotide base sequence
Describe several influences on Watson and Crick in their quest to elucidate the structure of DNA.
Linus Pauling's use of ball-and-stick models, Rosalind Franklin's X-ray diffraction patterns of DNA fibers, Chargaff's discover that amounts of A=T and G=C in DNA. Others are possible.
C24) Within a protein, certain amino acids are positively charged (e.g., lysine and arginine), some are negatively charged (e.g., glutamate and aspartate), some are polar but uncharged, and some are nonpolar. If you knew that a DNA-binding protein was recognizing the DNA backbone rather than base sequences, which amino acids in the protein would be good candidates for interacting with the DNA?
Lysines and arginines, and also polar amino acids
C30) The genetic material found within some viruses is single-stranded DNA. Would this genetic material contain equal amounts of A and T and equal amounts of G and C?
Not necessarily. The AT/GC rule is required only of double-stranded DNA molecules.
The formation of a double-stranded structure must obey the rule that adenine hydrogen bonds to thymine (or uracil) and cytosine hydrogen bonds to guanine. Based on your previous understanding of genetics (from this course or a general biology course), discuss reasons why complementarity is an important feature of DNA and RNA structure and function.
Note: Many of the topics described next are discussed in Chapters 10 through 13 . One way that complementarity underlies function is that it provides the basis for the synthesis of new strands of DNA and RNA. During replication, the synthesis of the new DNA strands occurs in such a way that adenine hydrogen bonds to thymine, and cytosine hydrogen bonds to guanine. In other words, the molecular feature of a complementary double-stranded structure makes it possible to produce exact copies of DNA. Likewise, the ability to transcribe DNA into RNA is based on complementarity. During transcription, one strand of DNA is used as a template to make a complementary strand of RNA. In addition to the synthesis of new strands of DNA and RNA, complementarity is important in other ways. As mentioned in this chapter, the folding of RNA into a particular structure is driven by the hydrogen bonding of complementary regions. This event is necessary to produce functionally active tRNA molecules. Likewise, stem-loop structures also occur in other types of RNA. For example, the rapid formation of stem-loop structures is known to occur as RNA is being transcribed and to affect the termination of transcription. A third way that complementarity can be functionally important is that it can promote the interaction of two separate RNA molecules. During translation, codons in mRNA bind to the anticodons in tRNA (see Chapter 13 ). This binding is due to complementarity. For example, if a codon is 5ʹ-AGG-3ʹ, the anticodon is 3ʹ-UCC-5ʹ. This type of specific interaction between codons and anticodons is an important step that enables the nucleotide sequence in mRNA to code for an amino acid sequence within a protein. In addition, many other examples of RNA-RNA interactions are known and will be described throughout this textbook.
List the four levels of complexity of DNA structure.
Nucletides (1) are assembled into a linear strand (2) which interacts with a second strand to form a double helix (3). This double helix is then packaged by wrapping around proteins or otherwise assumes three dimensional structure (4).
C22) On further analysis of the DNA described in conceptual question C21, you discover that the triplex DNA in this alien organism is composed of a double helix, with the third helix wound within the major groove (just like the DNA in Figure 9.19). How would you propose that this DNA is able to replicate itself? In your answer, be specific about the base pairing rules within the double helix and which part of the triplex DNA would be replicated first.
One possibility is a sequential mechanism. First, the double helix could unwind and replicate itself as described in Chapter 11. This would produce two double helices. Next, the third strand (bound in the major groove) could replicate itself via a semiconservative mechanism. This new strand could be copied to make a copy that is identical to the strand that lies in the major groove. At this point, you would have two double helices and two strands that could lie in the major groove. These could assemble to make two triple helices.
Why were 35S and 32P used to differentiate between protein and DNA?
Proteins contain sulphur but DNA does not, while DNA contains phosphorus while proteins do not. Both these isotopes are radioactive, so they can be tracked within the experiment.
C4) What are the building blocks of a nucleotide? With regard to the 5ʹ and 3ʹ positions on a sugar molecule, how are nucleotides linked together to form a strand of DNA?
The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and a phosphate group. In a nucleotide, the phosphate is already linked to the 5' position on the sugar. When two nucleotides are hooked together, a phosphate on one nucleotide forms a covalent bond with the 3' hyrdroxyl group on another nucleotide.
C32) A double-stranded DNA molecule is 1 cm long, and the percentage of adenine is 15%. How many cytosines would be found in this DNA molecule?
The first thing we need to do is to determine how many base pairs are in this DNA molecule. The linear length of 1 base pair is 0.34 nm, which equals 0.34 × 10^−9 m. One centimeter equals 10^−2 meters. 10^-2/0.34 x 10^-9 = 2.9 × 10^7 bp There are approximately 2.9 × 10^7 bp in this DNA molecule, which equals 5.8 × 10^7 nucleotides. If 15% are adenine, then 15% must also be thymine. This leaves 70% for cytosine and guanine. Because cytosine and guanine bind to each other, there must be 35% cytosine and 35% guanine. If we multiply 5.8 × 10^7 times 0.35, we get (5.8 × 10^7)(0.35) = 2.0 × 10^7 cytosines, or about 20 million cytosines
Describe the concept of base stacking.
The nitrogenous bases in DNA are flattened structures which 'stack' together as they hydrogen bond to each other. This is a key structural feature that stabilizes the entire molecule.
C12) What parts of a nucleotide (namely, phosphate, sugar, and/or bases) occupy the major and minor grooves of double-stranded DNA, and what parts are found in the DNA backbone? If a DNA-binding protein does not recognize a specific nucleotide sequence, do you expect that it recognizes the major groove, the minor groove, or the DNA backbone? Explain.
The nucleotide bases occupy the major and minor grooves. Phosphate and sugar are found in the backbone. If a DNA-binding protein does not recognize a nucleotide sequence, it probably is not binding in the grooves but instead is binding to the DNA backbone (i.e., sugar- phosphate sequence). DNA-binding proteins that recognize a base sequence must bind into a major or minor groove of the DNA, which is where the bases would be accessible to a DNA-binding protein. Most DNA-binding proteins that recognize a base sequence fit into the major groove. By comparison, other DNA-binding proteins, such as histones, which do not recognize a base sequence, bind to the DNA backbone.
Give a definition for molecular genetics.
The study of DNA structure and function at the molecular level.
List key differences between DNA and RNA structure.
The sugar in RNA nucleotides is ribose, while that of DNA nucleotides is deoxyribose; RNA tends to be single stranded with small regions of base pairing, while DNA is usually double stranded; RNA contains uracil where DNA contains thymine. Other answers are possible.
C26) A double-stranded DNA molecule contains 560 nucleotides. How many complete turns would be found in this double helix?
This DNA molecule contains 280 bp. There are 10 base pairs per turn, so there are 28 complete turns.
Describe deoxyguanosine monophosphate.
This structure contains a guanine nucleotide attached to a deoxyribose sugar and a single phosphate group.
What process of transfer of genetic information was observed by Griffith?
Transformation, where a bacterium can take up free DNA from the environment.
An important feature of triplex DNA formation is that it is sequence-specific. The synthetic third strand incorporates itself into a triple helix, so a thymine in the synthetic DNA binds near an AT pair in the biological DNA, and a cytosine in the synthetic DNA binds near a GC pair. From a practical point of view, this opens the possibility of synthesizing a short strand of DNA that forms a triple helix at a particular target site. For example, if the sequence of a particular gene is known, researchers can make a synthetic piece of DNA that forms a triple helix somewhere within that gene according to the T to AT, and C to GC rule. Triplex DNA formation is known to inhibit gene transcription. In other words, when the synthetic DNA binds within the DNA of a gene, the formation of triplex DNA prevents that gene from being transcribed into RNA. Discuss how this observation might be used to combat diseases.
Triplex DNA formation opens the exciting possibility of designing synthetic pieces of DNA to inhibit the expression of particular genes. Theoretically, such a tool could be used to combat viral diseases or to inhibit the growth of cancer cells. To combat a viral disease, a synthetic DNA could be made that specifically binds to an essential viral gene, thereby preventing viral proliferation. To inhibit cancer, a synthetic DNA could be made to bind to an oncogene. (Note: As described in Chapter 22 , an oncogene is a gene that promotes cancerous growth.) Inhibition of an oncogene could prevent cancer. At this point, a primary obstacle in applying this approach is devising a method of getting the synthetic DNA into living cells.
Name the five nitrogenous bases found in nucleic acids.
adenine, guanine, thymine, cytosine, uracil.
