Chapter 9: Molecular Structure of DNA and RNA

BIO 184 Laboratory

CSU, SacramentoFall 2007

Supplemental Problems for Exam 1

Chapter 9: Molecular Structure of DNA and RNA

Conceptual problems: C4, C7, C8, C9, C13, C16, C17, C18, C19, C20, C28, C33

Experimental problems: E1

C4.The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and phosphate. 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 a hydroxyl group at the 3 position on another nucleotide.

C7.The bases conform to the AT/GC rule of complementarity. There are two hydrogen bonds between A and T and three hydrogen bonds between G and C. The planar rings of the bases stack on top of each other within the helical structure to provide even more stability.

C8.3–CCGTAATGTGATCCGGA–5

C9.The sequence of nucleotide bases.

C13.DNA has deoxyribose as its sugar while RNA has ribose. DNA has the base thymine while RNA has uracil. DNA is a double helical structure. RNA is single stranded although parts of it may form double-stranded regions.

C16.Double-stranded RNA is more like A DNA than B DNA. See the text for a discussion of A-DNA structure.

C17.The sequence in part A would be more difficult to separate because it has a higher percentage of GC base pairs compared to the one in part B. GC base pairs have three hydrogen bonds compared with AT base pairs, which only have two.

C18.Its nucleotide base sequence.

C19.Complementarity is important in several ways. First, it is needed to copy genetic information. This occurs during replication, when new DNA strands are made, and during transcription, when RNA strands are made. Complementarity is also important during translation for codon/anticodon recognition. It also allows RNA molecules to form secondary structures and to recognize each other.

C20.G = 32%, C = 32%, A = 18%, T = 18%

C28.A hydroxyl group is at the 3 end and a phosphate group is at the 5 end.

C33.Yes, as long as there are sequences that are complementary and antiparallel to each other. It would be similar to the complementary double-stranded regions observed in RNA molecules (e.g., see Figures 9.23 and 9.24).

E1.A trait of pneumococci is the ability to synthesize a capsule. There needs to be a blueprint for this ability. The blueprint for capsule formation was being transferred from the type IIIS to the type IIR bacteria. (Note: At the molecular level, the blueprint is a group of genes that encode enzymes that can synthesize a capsule.)

Chapter 10: Chromosome Organization and Molecular Structure

Conceptual problems: C3, C5, C7 (but not the Z DNA question), C13, C14, C15, C16, C19, C24, C27

Experimental problems: E2, E3

C3.The bacterial nucleoid is a region in a bacterial cell that contains a compacted circular chromosome. Unlike eukaryotic nuclei, a nucleoid is not surrounded by a membrane.

C5.One mechanism is DNA looping. Loops of DNA are anchored to DNA-binding proteins. Secondly, the DNA double helix is twisted further to make it more compact, much like twisting a rubber band.

C7.DNA is a double helix. The helix is a coiled structure. Supercoiling involves additional coiling to a structure that is already a coil. Positive supercoiling is called overwinding because it adds additional twists in the same direction as the DNA double helix; it is in a right-handed direction. Negative supercoiling is in the opposite direction.

C13.Centromeres are found in eukaryotic chromosomes. They provide an attachment site for kinetochore proteins so that the chromosomes are sorted (i.e., segregated) during mitosis and meiosis. They are most important during M phase.

C14.Highly repetitive DNA, as its name suggests, is a DNA sequence that is repeated many times. It can be tandemly repeated or interspersed. Tandemly repeated DNA often has a base content that is significantly different from the rest of the chromosomal DNA so it sediments as a satellite band. In DNA renaturation studies, highly repetitive DNA renatures at a much faster rate because it is found at a higher concentration.

C15.A nucleosome is composed of double-stranded DNA wrapped 1.65 times around an octamer of histones. In the
30 nm fiber, histone H1 helps to compact the nucleosomes. The three-dimensional zigzag model is a current model that describes how this compaction occurs. It looks like a somewhat random (zigzagging) of the nucleosomes within the 30 nm fiber.

C16.During interphase (i.e., G1, S, and G2), the euchromatin is found primarily as a 30 nm fiber in a radial loop configuration. Most interphase chromosomes also have some heterochromatic regions where the radial loops are more highly compacted. During M phase, each chromosome becomes entirely heterochromatic. This is needed for the proper sorting of the chromosomes during nuclear division.

C19.Heterochromatin is more tightly packed. This is due to a greater compaction of the radial loop domains. Functionally, euchromatin can be transcribed into RNA, while heterochromatin is inactive.

C24.The role of the core histones is to form the nucleosomes. In a nucleosome, the DNA is wrapped 1.65 times around the core histones. Histone H1 binds to the linker region. It may play a role in compacting the DNA into a 30 nm fiber.

C27.During interphase, much of the chromosomal DNA is in the form of the 30 nm fiber, and some of it is more highly compacted heterochromatin. During metaphase, all of the DNA is highly compacted, as shown in Figure 10.21d. A high level of compaction prevents gene transcription and DNA replication from taking place. Therefore, these events occur during interphase.

E2.This type of experiment gives the relative proportions of highly repetitive, moderately repetitive, and unique DNA sequences within the genome. The highly repetitive sequences renature at a fast rate, the moderately repetitive sequences renature at an intermediate rate, and the unique sequences renature at a slow rate.

E3.It affects only the rate of renaturation. Denaturation occurs because the heat breaks the hydrogen bonds between the two strands. The rate of denaturation depends on the hydrogen bonding, not on the number of copies of a sequence. The rate of renaturation, however, depends on the two complementary strands “finding” each other. The rate at which two complementary strands find each other will be faster if the concentration of the two strands is higher. That is why highly repetitive sequences renature at a faster rate.

Chapter 11: DNA Replication

Conceptual problems: C1, C2, C3, C7, C8, C15, C16, C20, C21, C22 (only first part of question), C23, C28, C29, C30, C32

Experimental problems: E1, E4

C1.It is a double-stranded structure that follows the AT/GC rule.

C2.Bidirectionality refers to the idea that two replication forks emanate from one origin of replication. There are four DNA strands being made in two directions radiating outward from the origin.

C3.Statement C is not true. A new strand is always made from a preexisting template strand. Therefore, a double helix always contains one strand that is older than the other.

C7.

5—————————DNA Polymerase——>3

3————————————————————————————————————5

Template strand

C8.DNA polymerase would slide from right to left. The new strand would be

3–CTAGGGCTAGGCGTATGTAAATGGTCTAGTGGTGG–5

C15.A.The removal of RNA primers occurs in the 5 to 3 direction, while the proofreading function occurs in the 3 to 5 direction.

B.No. The removal of RNA primers occurs from the 5 end of the strand.

C16.A.The right Okazaki fragment was made first. It is farthest away from the replication fork. The fork (not seen in this diagram) would be to the left of the three Okazaki fragments, and moving from right to left.

B.The RNA primer in the right Okazaki fragment would be removed first. DNA polymerase would begin by elongating the DNA strand of the middle Okazaki fragment and remove the right RNA primer with its 5 to 3 exonuclease activity. DNA polymerase I would use the 3 end of the DNA of the middle Okazaki fragment as a primer to synthesize DNA in the region where the right RNA primer is removed. If the middle fragment was not present, DNA polymerase could not fill in this DNA (because it needs a primer).

C.You only need DNA ligase at the right arrow. DNA polymerase I begins at the end of the left Okazaki fragment and synthesizes DNA to fill in the region as it removes the middle RNA primer. At the left arrow, DNA polymerase I is simply extending the length of the left Okazaki fragment. No ligase is needed here. When DNA polymerase I has extended the left Okazaki fragment through the entire region where the RNA primer has been removed, it hits the DNA of the middle Okazaki fragment. This occurs at the right arrow. At this point, the DNA of the middle Okazaki fragment has a 5 end that is a monophosphate. DNA ligase is needed to connect this monophosphate with the 3 end of the region where the middle RNA primer has been removed.

D.As mentioned in the answer to part C, the 5 end of the DNA in the middle Okazaki fragment is a monophosphate. It is a monophosphate because it was previously connected to the RNA primer by a phosphoester bond. At the location of the right arrow, there was only one phosphate connecting this deoxyribonucleotide to the last ribonucleotide in the RNA primer. For DNA polymerase to function, the energy to connect two nucleotides comes from the hydrolysis of the incoming triphosphate. In this location shown at the right arrow, however, the nucleotide is already present at the 5 end of the DNA, and it is a monophosphate. DNA ligase needs energy to connect this nucleotide with the left Okazaki fragment. It obtains energy from the hydrolysis of ATP. C17. DNA methylation is the covalent attachment of methyl groups to bases in the DNA. Immediately after replication, there has not been sufficient time to attach methyl groups to the bases in the newly made daughter strand. The time delay of DNA methylation helps to prevent premature DNA replication immediately after cell division.

C20.The picture would depict a ring of helicase proteins traveling along a DNA strand and breaking the hydrogen bonding between the two helices, as shown in Figure 11.6.

C21.An Okazaki fragment is a short segment of newly made DNA in the lagging strand. It is necessary to make short fragments because the fork is exposing the lagging strand in a 5 to 3 direction but DNA polymerase can slide along a template strand in a 3 to 5 direction. Therefore, the newly made lagging strand is synthesized in short pieces in the direction away from the replication fork.

C22.The leading strand is primed once, at the origin, and then DNA polymerase III makes it continuously in the direction of the replication fork. In the lagging strand, many short pieces of DNA are made. This requires many RNA primers and DNA polIII. The primers are removed by polI, which then fills in the gaps with DNA. DNA ligase then covalently connects the Okazaki fragments together.

C23.The active site of DNA polymerase has the ability to recognize a distortion in the newly made strand and remove it. This occurs by a 3 to 5 exonuclease activity. After the mistake is removed, DNA polymerase resumes DNA synthesis.

C28.The opposite strand is made in the conventional way by DNA polymerase using the “telomerase added strand” as a template.

C29.Fifty, because two replication forks emanate from each origin of replication. DNA replication is bidirectional.

C30.The ends labeled B and C could not be replicated by DNA polymerase. DNA polymerase makes a strand in the 5 to 3 direction using a template strand that is running in the 3 to 5 direction. Also, DNA polymerase requires a primer. At the ends labeled B and C, there is no place (upstream) for a primer to be made.

C32.As shown in Figure 11.23, the first step involves a binding of telomerase to the telomere. The 3 overhang binds to the complementary RNA in telomerase. For this reason, a 3 overhang is necessary for telomerase to replicate the telomere.

E1.A.Four generations: 7/8 light, 1/8 half-heavy

Five generations: 15/16 light, 1/16 half-heavy

B.All of the DNA double helices would be 1/8 heavy.

C.The CsCl gradient separates molecules according to their densities. 14N-containing compounds have a lighter density compared to 15N-containing compounds. The bases of DNA contain nitrogen. If the bases contain only 15N, the DNA will be heavy; it will sediment at a higher density. If the bases contain only 14N, the DNA will be light; it will sediment at a lower density. If the bases in one DNA strand contain 14N and the bases in the opposite strand contain 15N, the DNA will be half-heavy; it will sediment at an intermediate density.

E4.You would need to add a primer (or primase), dNTPs, and DNA polymerase. If the DNA were double stranded, you would also need helicase. Adding single-strand binding protein and topoisomerase may also help.

Chapter 12: Gene Transcription and RNA Modification

Conceptual problems: C1, C2, C3, C8, C9, C12, C14, C15, C16, C17, C20, C25, C29, C31, C33

Experimental problems:

C1.A.tRNA genes encode tRNA molecules, and rRNA genes encode the rRNAs found in ribosomes. There are also genes for the RNAs found in snRNPs, etc.

B.The term template strand is still appropriate because one of the DNA strands is used as a template to make the RNA. The term coding strand is not appropriate because the RNA made from nonstructural genes does not code for a polypeptide sequence.

C.Yes

C2.The formation of the open complex, and the release of sigma factor.

C3.A consensus sequence is the most common nucleotide sequence that is found within a group of related sequences. An example is the –35 and –10 consensus sequence found in bacterial promoters. At –35, it is TTGACA, but it can differ by one or two nucleotides and still function efficiently as a promoter. In the consensus sequences within bacterial promoters, the –35 site is primarily for recognition by sigma factor. The –10 site, also known as the Pribnow box, is the site where the DNA will begin to unwind to allow transcription to occur.

C8.This will not affect transcription. However, it will affect translation by preventing the initiation of polypeptide synthesis.

C9.RNA polymerase holoenzyme consists of sigma factor plus the core enzyme, which is a tetramer, a2ßß. The role of sigma factor is to recognize the promoter sequence. The a subunits are necessary for the assembly of the core enzyme and for loose DNA binding. The ß and ß subunits are the portion that catalyze the covalent linkages between adjacent ribonucleotides.

C12.DNA-G/RNA-C

DNA-C/RNA-G

DNA-A/RNA-U

DNA-T/RNA-A

The template strand is 3–CCGTACGTAATGCCGTAGTGTGATCCCTAG–5 and the coding strand is 5–GGCATGCATTACGGCATCACACTAGGGATC–3. The promoter would be to the left (in the 3 direction) of the template strand.

C14.Transcriptional termination occurs when the hydrogen bonding is broken between the DNA and the part of the newly made RNA transcript that is located in the open complex.

C15.In ρ-dependent termination, the p protein binds to the RNA transcript after the psite has been transcribed. Eventually, RNA polymerase will transcribe a termination stem-loop that will cause it to pause in the transcription process. As it is pausing, the p protein, which functions as a helicase, will catch up to RNA polymerase and knock it off the DNA. In p-independent transcription, there is no p protein. Again, the RNA polymerase transcribes a termination hairpin that causes it to pause. However, when it pauses, an AU-rich region is left base pairing in the open complex. Since this is holding on by fewer hydrogen bonds, it is rather unstable. Therefore, it tends to dissociate from the open complex and thereby end transcription.

C16.Helicase and p protein bind to a nucleic acid strand and travel in the 5 to 3 direction. When they encounter a double-stranded region, they break the hydrogen bonds between complementary strands. p protein is different from DNA helicase in that it moves along an RNA strand, while DNA helicase moves along a DNA strand. The purpose of DNA helicase function is to promote DNA replication while the purpose of p protein function is to promote transcriptional termination.

C17.RNA and DNA polymerase are similar in the following ways:

1.They both use a template strand.

2.They both synthesize in the 5 to 3 direction.

3.The chemistry of synthesis is very similar in that they use incoming triphosphates and make a phosphoester bond between the previous nucleotide and the incoming nucleotide.

4.They are both processive enzymes that slide along a template strand of DNA.

RNA and DNA polymerase are different in the following ways:

1. RNA polymerase makes an RNA strand while DNA polymerase makes a DNA strand.

2. RNA polymerase uses only one DNA strand as a template.

3. RNA polymerase recognizes a particular DNA sequence (i.e., promoters) and synthesizes only a defined region of RNA from the promoter to the terminator.

4. RNA polymerase cannot proofread.

5. RNA polymerase does not need a primer.

C20.Eukaryotic promoters are somewhat variable with regard to the pattern of sequence elements that may be found. In the case of structural genes that are transcribed by RNA polymerase II, it is common to have a TATA box, which is about 25 bp upstream from a transcriptional start site. The TATA box is important in the identification of the transcriptional start site and the assembly of RNA polymerase and various transcription factors. The transcriptional start site defines where transcription actually begins.

C25.Only the first intron would be spliced out. The mature RNA would be: exon 1–exon 2–intron 2–exon 3.

C29.A gene is colinear when the sequence of bases in the sense strand of the DNA (i.e., the DNA strand that is complementary to the template strand for RNA synthesis) corresponds to the sequence of bases in the mRNA. Most prokaryotic genes and many eukaryotic genes are colinear. Therefore, you can look at the gene sequence in the DNA and predict the amino acid sequence in the polypeptide. Many eukaryotic genes, however, are not colinear. They contain introns that are spliced out of the pre-mRNA.

C31.In eukaryotes, pre-mRNA can be capped, tailed, and spliced and then exported out of the nucleus.

C33.Alternative splicing occurs when exons are spliced out or alternative splice sites are used at intron-exon boundaries. The biological significance is that two or more polypeptide sequences can be derived from a single gene. This is a more efficient use of the genetic material. In multicellular organisms, alternative splicing is often used in a cell-specific manner.

Chapter 13:

C1.The start codon begins at the fifth nucleotide. The amino acid sequence would be Met Gly Asn Lys Pro Gly Gln STOP.