Chapter 10 Problem Set (Solutions)
List the functions of the 1', 2', 3', and 5' carbons in the pentose sugar found in DNA and RNA.
1' carbon - links sugar to the base 2' carbon - covalently bonded to an OH group in RNA and H in DNA 3' and 5' carbons - the two carbons used to create phosphodiester linkages between nucleotides
What four 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 have the capacity to vary or mutate to generate diversity. 4) The genetic material must encode the phenotype or have the ability to code for traits.
One nucleotide strand of a double stranded DNA molecule has the base sequence illustrated below. 5' ATTGCTACGG 3' Give the base sequence and label the 5'and 3'ends of the complementary DNA nucleotide strand.
3' TAACGATGCC 5'
Which bases are capable of forming hydrogen bonds with each other?
Adenine is capable of forming two hydrogen bonds with thymine. Guanine is capable of forming three hydrogen bonds with cytosine.
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.
What experiments demonstrated that DNA is the genetic material?
Experiments by Hershey and Chase in the 1950s using the bacteriophage T2 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.
In DNA and RNA 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.
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.
Heinz Shuster collected the following data on the base composition of ribgrass virus. On the basis of this information, is the hereditary information of the ribgrass virus RNA or DNA? Is it likely to be single stranded or double stranded? Percent AGCTU Ribgrass virus 29.3 25.8 18.0 0.0 27.0
Most likely, the ribgrass viral genome is a single-stranded RNA. The presence of uracil indicates that the viral genome is RNA. For the molecule to be double-stranded RNA, we would predict equal percentages of adenine and uracil bases and equal percentages of guanine and cytosine bases. Neither the percentages of adenine and uracil bases nor the percentages of guanine and cytosine bases are equal, indicating that the viral genome is likely single stranded.
DNA has polarity. What does this mean? What is the polarity of each strand of DNA in a double helix relative to the other?
One end of a polynucleotide chain of has a free phosphate group covalently linked to the 5' C of the first nucleotide. This is the 5' end of the DNA molecule. The other end has a free OH group covalently linked to the 3' C of the last nucleotide. This is the 3' end. The two DNA strands in a double helix are arranged with opposite polarity (i.e. they are antiparallel).
How does a purine differ from a pyrimidine? What purines and pyrimidines are found in DNA and RNA?
Pyrimidines have a single C-N ring structure. Purines have a double C-N ring structure. 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.
Explain how heat-killed type IIIS bacteria in Griffith's experiment genetically altered the live type IIR bacteria.
The IIR strain of Streptococcus pneumonia must have been naturally competent or, in other words, was capable of taking up DNA from the environment. The heat-killed IIIS bacteria lysed releasing their DNA into the environment allowing for IIIS chromosomal DNA fragments to come in contact with IIR cells. The IIIS DNA responsible for the virulence of the IIIS strain was taken up by an IIR cell and integrated into the IIR cell's chromosome, thus "transforming" the IIR cell into a virulent IIIS cell.
What different types of chemical bonds are found in DNA and where are they found?
The deoxyribonucleotides in a single chain or strand of DNA are held by covalent bonds called phosphodiester linkages between the 3' carbon of one deoxyribose sugar of a nucleotide and the 5' carbon of the deoxyribose sugar of the next nucleotide in the chain. Two chains of deoxyribonucleotides are held together by hydrogen bonds between the complementary nitrogenous bases of the nucleotides in each chain.
Predict what would happen if Griffith had mixed some heat-killed type IIIS bacteria and some heat-killed type IIR bacteria and injected these into a mouse. Would the mouse have contracted pneumonia and died? Explain why or why not.
The mouse would not have contracted pneumonia and died. Although the mouse would have received IIIS DNA, which codes for virulent Streptococcus pneumoniae, there are no live bacteria for this DNA to transform. Live bacteria are required for pneumonia to develop.
If a double-stranded DNA molecule is 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%.
What results would you expect if the bacteriophage that Hershey and Chase used in their experiment had contained RNA instead of DNA?
The results would be the same. RNA, like DNA, contains phosphorus but not sulfur, so the 32P would have been incorporated into the RNA and entered the cell hosts during the course of the infection. Ultimately the 32P would be incorporated into the RNA of progeny phage during the course of phage reproduction. Protein contains sulfur but not phosphorus, so 35S would have been assimilated into the phage protein coat; it would have remained with the viral coats that do not enter the cells and not transferred to progeny phage. The 35S- containing protein coats would still be recovered in the fluid recovered following centrifugation.
Identify the three parts of a DNA nucleotide.
The three parts of a DNA nucleotide are a phosphate group, deoxyribose sugar, and a nitrogenous base.
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 RNA 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.
Imagine you are a student in Alfred Hershey and Martha Chase's lab in the late 1940s. You are given five test tubes containing E. coli bacteria that were infected with T2 bacteriophage that have been labeled with either 32P or 35S. Unfortunately, you forgot to mark the tubes and are now uncertain which were labeled with 32P and which with 35S. You place the contents of each tube in a blender and turn it on for a few seconds to shear off the protein coats. You then centrifuge the contents to separate the protein coats and the cells. You check for the presence of radioactivity and obtain the following results. Which tubes contained E. coli infected with 32P- labeled phage? Explain your answer. Tube number 1: Presence of radioactivity in 2: cells 3: protein coats protein coats 4: cells 5: cells
Tubes 1, 4, and 5. The DNA of the bacteriophage contains phosphorous and the protein contains sulfur. When the bacteriophages infect the cell, they inject their DNA into the cell, but the protein coats stay on the surface of the cell. The protein coats are sheared off in the blender, while the cells with the DNA pellet at the bottom of the tube. Thus, cells infected with 35S -labeled bacteriophage will have radioactivity associated with the protein coats, whereas those cells infected with 32P-labeled bacteriophage will have radioactivity associated with the cells.
The relative amounts of each nucleotide base are tabulated here for four different viruses. For each virus listed in the following table, indicate whether its genetic material is DNA or RNA and whether it is single stranded or double stranded. Explain your reasoning. Virus TCUGA I 0 12 9 12 9 II 23 16 0 16 23 III 34 42 0 18 39 IV 0 24 35 27 17
Virus I is a double-stranded RNA virus. Uracil is present indicating an RNA genome and we see equal percentages of adenine and uracil and equal percentages of guanine and cytosine, which we would expect if it is a double-stranded genome. Virus II is a double-stranded DNA virus. The presence of thymine indicates that the viral genome is DNA. As expected for a double-stranded DNA molecule, we see equal percentages of adenine and thymine bases and equal percentages of guanine and cytosine bases. Virus III is a single-stranded DNA virus. Thymine is present suggesting a DNA genome. However, we see unequal percentages of thymine and adenine and unequal percentages of guanine and cytosine, which suggest a single-stranded DNA molecule. Virus IV is a single-stranded RNA virus. Uracil is present indicating an RNA genome. However, the percentage of adenine does not equal the percentage of uracil, and the percentage of guanine does not equal the percentage of cytosine. These unequal amounts suggest a single-stranded genome.
A sample of normal double helical DNA was found to have a guanine content of 18%. a. What is the expected percentage of adenine, cytosine and thymine in this DNA? b. What is the expected percentage of pyrimidines and purines in this DNA?
a. G+C will be 36% therefore A+T will be 64% so %A = %T = 32%; %C = 18% b. 50% pyrimidines and 50% purines. Purines base pair with pyrimidines so they will each make up half of the bases in double stranded DNA.
