Exam 3 Cell Biology

Specific regions of eukaryotic chromosomes contain sequence elements that are absolutely required for the proper transmission of genetic information from a mother cell to each daughter cell. Which of the following is not known to be one of these required elements in eukaryotes? (a)terminators of replication (b)origins of replication (c)telomeres (d)centromeres

A

The chromatin remodeling complex uses the energy from ATP hydrolysis to catalyze _____________________________. (a)the sliding of nucleosomes along DNA. (b)acetylation of a histone tail. (c)the binding of histone H1 to the linker DNA. (d)the initiation of replication.

A

Transcription in bacteria differs from transcription in a eukaryotic cell because __________________________. (a)RNA polymerase (along with its sigma subunit) can initiate transcription on its own. (b)RNA polymerase (along with its sigma subunit) requires the general transcription factors to assemble at the promoter before polymerase can begin transcription. (c)the sigma subunit must associate with the appropriate type of RNA polymerase to produce mRNAs. (d)RNA polymerase must be phosphorylated at its C-terminal tail for transcription to proceed.

A

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking helicase were used to make the cell extracts? (a)initiation of DNA synthesis (b)Okazaki fragment synthesis (c)leading-strand elongation (d)lagging-strand completion

A

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking helicase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

A

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking primase were used to make the cell extracts? (a)initiation of DNA synthesis (b)Okazaki fragment synthesis (c)leading-strand elongation (d)lagging-strand completion

A

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking primase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

A

Which of the following statements about transcriptional regulators is false? (a)Transcriptional regulators usually interact with the sugar-phosphate backbone on the outside of the double helix to determine which DNA sequence to bind. (b)Transcriptional regulators will form hydrogen bonds, ionic bonds, and hydrophobic interactions with DNA. (c)The DNA-binding motifs of transcriptional regulators usually bind in the major groove of the DNA helix. (d)The binding of transcriptional regulators generally does not disrupt the hydrogen bonds that hold the double helix together.

A

How many possible nucleotide sequences are there for a stretch of DNA that is N nucleotides long, if it is (a) single stranded or (b) double-stranded?

A single-stranded DNA molecule that is N nucleotides long can have any one of 4N possible sequences, but the number of possible double-stranded DNA molecules is more difficult to calculate. Many of the 4N single stranded sequences will be the complement of another possible sequence in the list; for example, 5ʹ-AGTCC-3ʹ and 5ʹ-GGACT-3ʹ form the same double-stranded DNA molecule and therefore count as a single, double stranded possibility. If N is an odd number, then every single-stranded sequence will complement another sequence in the list so that the number of double stranded sequences will be 0.5 × 4N. If N is an even number, then there will be slightly more than this, since some sequences will be self-complementary (such as 5ʹ-ACTAGT-3ʹ) and the actual value can be calculated to be 0.5 × 4N + 0.5 × 4N/2.

Which of the following statements are correct? Explain your answers. A. A bacterial replication fork is asymmetrical because it contains two DNA polymerase molecules that are structurally distinct. B. Okazaki fragments are removed by a nuclease that degrades RNA. C. The error rate of DNA replication is reduced both by proofreading by DNA polymerase and by DNA mismatch repair. D. In the absence of DNA repair, genes are unstable. E. None of the aberrant bases formed by deamination occur naturally in DNA. F. Cancer can result from the accumulation of mutations in somatic cells.

A. False. Identical DNA polymerase molecules catalyze DNA synthesis on the leading and lagging strands of a bacterial replication fork. The replication fork is asymmetrical because the lagging strand is synthesized in pieces that are then stitched together. B. False. Only the RNA primers are removed by an RNA nuclease; Okazaki fragments are pieces of newly synthesized DNA on the lagging strand that are eventually joined together by DNA ligase. C. True. With proofreading, DNA polymerase has an error rate of one mistake in 107 nucleotides polymerized; 99% of its errors are corrected by DNA mismatch repair enzymes, bringing the final error rate to one in 109. D. True. Mutations would accumulate rapidly, inactivating many genes. E. True. If a damaged nucleotide also occurred naturally in DNA, the repair enzyme would have no way of identifying the damage. It would therefore have only a 50% chance of fixing the right strand. F. True. Usually, multiple mutations of specific types need to accumulate in a somatic cell lineage to produce a cancer. A mutation in a gene that codes for a DNA repair enzyme can make a cell more liable to accumulate further mutations, thereby accelerating the onset of cancer.

Which of the following statements are correct? Explain your answers. A. Each eukaryotic chromosome must contain the following DNA sequence elements: multiple origins of replication, two telomeres, and one centromere. B. Nucleosome core particles are 30 nm in diameter.

A. True. B. False. Nucleosome core particles are approximately 11 nm in diameter

What do you predict would happen if you replace the Lac operator DNA from the Lac operon with the DNA from the operator region from the tryptophan operon? (a)The presence of lactose will not cause allosteric changes to the Lac repressor. (b)The Lac operon will not be transcribed when tryptophan levels are high. (c)The lack of glucose will no longer allow CAP binding to the DNA.(d)RNA polymerase will only bind to the Lac promoter when lactose is present.

B

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking single-strand binding protein were used to make the cell extracts? (a)initiation of DNA synthesis (b)Okazaki fragment synthesis (c)leading-strand elongation (d)lagging-strand completion

B

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking single-strand binding protein were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

B

Which of the following statements is true with respect to this in vitro replication system? (a) There will be only one leading strand and one lagging strand produced using this template. (b) The leading and lagging strands compose one half of each newly synthesized DNA strand. (c) The DNA replication machinery can assemble at multiple places on this plasmid. (d) One daughter DNA molecule will be slightly shorter than the other.

B

Which posttranslational modification does not occur on histone tails?(a)Phosphorylation (b)Germination (c)Acetylation (d)Methylation

B

On average, how often would the nucleotide sequence CGATTG be expected to occur in a DNA strand 4000 bases long? Show your work and explain your answer.

Because 4^6 (= 4096) different sequences of six nucleotides can occur in DNA, any given sequence of six nucleotides would be expected to occur on average once in a DNA strand 4000 bases long, assuming a random distribution of sequences.

ATP powered enzyme that mechanically unwinds double stranded DNA. (a)DNA ligase (b)DNA polymerase (c)DNA helicase (d)Telomerase

C

Enzyme that catalyzes phosphodiester bond formation between a nucleoside triphosphate and the 3' nucleotide in a newly synthesized complementary strand of DNA. (a)Primase (b)Topoisomerase (c)DNA polymerase (d)DNA helicase

C

The Messelson-Stahl experiment showed that DNA replication is semiconservative because ____________________________. (a)after many rounds of DNA replication, the original DNA double helix is still intact. (b)each daughter DNA molecule consists of two new strands copied from the parent DNA molecule. (c)each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand. (d)new DNA strands must be copied from a DNA template.

C

The classic "beads-on-a-string" structure is the most decondensed chromatin structure possible. Which chromatin components are not retained when this structure is generated? (a)linker DNA (b)nucleosome core particles (c)linker histones (d)core histones

C

What happens to DNA ahead of the replication fork when topoisomerase activity is inhibited? (a)It remains double stranded and linear as the helicase approaches.(b)It unwinds and becomes single stranded before the arrives. (c)It becomes overwound and supercoiled, which eventually stops the helicase. (d)It becomes coated with replication protein A (RPA).

C

Transcription is similar to DNA replication in that ___________________. (a)an RNA transcript is synthesized discontinuously and the pieces are then joined together. (b)it uses the same enzyme as that used to synthesize RNA primers during DNA replication. (c)the newly synthesized RNA remains paired to the template DNA.(d)nucleotide polymerization occurs only in the 5′-to-3′ direction.

D

Unlike DNA, which typically forms a helical structure, different molecules of RNA can fold into a variety of three-dimensional shapes. This is largely because ___________________. (a)RNA contains uracil and uses ribose as the sugar. (b)RNA bases cannot form hydrogen bonds with each other. (c)RNA nucleotides use a different chemical linkage between nucleotides compared to DNA. (d)RNA is single-stranded.

D

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking DNA ligase were used to make the cell extracts? (a)initiation of DNA synthesis (b)Okazaki fragment synthesis (c)leading-strand elongation (d)lagging-strand completion

D

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking DNA ligase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

D

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking the exonuclease activity of DNA polymerase were used to make the cell extracts? (a) initiation of DNA synthesis (b) Okazaki fragment synthesis (c) leading-strand elongation (d) lagging-strand completion

D

You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit. What is the most likely explanation for this? (a)Coding sequences are repaired more efficiently. (b)Coding sequences are replicated more accurately. (c)Coding sequences are packaged more tightly in the chromosomes to protect them from DNA damage. (d)Mutations in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

D

____________________ bonds hold DNA base pairs together and ______________________ bonds form the DNA backbone. (a)phosphodiester, hydrogen (b)van der Waals, hydrogen (c)ionic, phosphodiester (d)hydrogen, phosphodiester

D

Define the following terms and their relationships to one another: A. Interphase chromosome B. Mitotic chromosome C. Chromatin D. Heterochromatin E. Histones F. Nucleosome

DNA assembles with specialized proteins to form chromatin. At a first level of packing, histones form the core of nucleosomes. In a nucleosome, the DNA is wrapped almost twice around this core. Between nuclear divisions—that is, in interphase—the chromatin of the interphase chromosomes is in a relatively extended form in the nucleus, although some regions of it, the heterochromatin, remain densely packed and are transcriptionally inactive. During nuclear division—that is, in mitosis—replicated chromosomes become condensed into mitotic chromosomes, which are transcriptionally inactive and are designed to be readily distributed between the two daughter cells.

Know that interphase chromosomes are not a tangled mess.

Different chromosomes occupy different regions of the nucleus during interphase. The DNA remains relatively condensed during interphase. This is due to post-translational modifications of histones.

The sliding clamp is loaded once on each DNA strand, where it remains associated until replication is complete. True or False

False. Although the sliding clamp is only loaded once on the leading strand, the lagging strand needs to unload the clamp once the polymerase reaches the RNA primer from the previous segment and then reload it where a new primer has been synthesized.

Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication. True or False

False. Primase does not have a proofreading function, nor does it need one because the RNA primers are not a permanent part of the DNA. The primers are removed, and a DNA polymerase that does have a proofreading function fills in the remaining gaps

Know why the structure of interphase and metaphase chromosomes is different.

Gene expression and chromosome duplication occur during interphase. Chromosome alignment and separation occur during mitosis. Also requires breakdown of the nuclear envelope. Histone proteins are responsible for the differences in chromosome structure.

Assuming that the histone octamer (shown in Figure 5-21) forms a cylinder 9 nm in diameter and 5 nm in height and that the human genome forms 32 million nucleosomes, what volume of the nucleus (6 μm in diameter) is occupied by histone octamers? (Volume of a cylinder is πr2h; volume of a sphere is 4/3 πr3.) What fraction of the total volume of the nucleus do the histone octamers occupy? How does this compare with the volume of the nucleus occupied by human DNA?

Histone octamers occupy about 9% of the volume of the nucleus. The volume of the nucleus is V = 4/3 × 3.14 × (3 × 103 nm)3 V = 1.13 × 1011 nm3 The volume of the histone octamers is V = 3.14 × (4.5 nm)2 × (5 nm) × (32 × 106) V = 1.02 × 1010 nm3 The ratio of the volume of histone octamers to the nuclear volume is 0.09; thus, histone octamers occupy about 9% of the nuclear volume. Because the DNA also occupies about 9% of the nuclear volume, together they occupy about 18% of the volume of the nucleus

DNA mismatch repair enzymes preferentially repair bases on the newly synthesized DNA strand, using the old DNA strand as a template. If mismatches were simply repaired without regard for which strand served as template, would this reduce replication errors? Explain your answer.

If the old strand were "repaired" using the new strand that contains a replication error as the template, then the error would become a permanent mutation in the genome. The old information would be erased in the process. Therefore, if repair enzymes did not distinguish between the two strands, there would be only a 50% chance that any given replication error would be corrected.

Histone proteins are among the most highly conserved proteins in eukaryotes. Histone H4 proteins from a pea and a cow, for example, differ in only 2 of 102 amino acids. Comparison of the gene sequences shows many more differences, but only two change the amino acid sequence. These observations indicate that mutations that change amino acids must have been selected against during evolution. Why do you suppose that amino acid-altering mutations in histone genes are deleterious?

In contrast to most proteins, which accumulate amino acid changes over evolutionary time, the functions of histone proteins must involve nearly all of their amino acids, so that a change in any position would be deleterious to the cell.

Calculate the predicted Tm for a stretch of double helix that is 100 nucleotides long and contains 50% G + C content.

Inserting values into the equation in part B gives Tm = 59.9 + (0.41 × 50) - (675/100) = 73.65°C, which is about twice the normal temperature of the human body and nearly too hot to touch.

Explain the reason why the cell requires a mechanism for identifying specific sequences of DNA.

Sequence information contains indicators important for the regulation of gene expression and DNA packaging. Examples include sequence indicators for where a gene starts and ends, where transcription begins, and where to assemble specific protein complexes at specialized sequences such as those found in telomeric or centromeric DNA

Molecular processes depend upon sequence-specific interactions of proteins with DNA. Recognition sequences can be 4, 5, 6, 7, or even 8 base pairs in length for a single protein. What might be the advantages of a short recognition sequence? What might be the advantage of a longer recognition sequence?

Short recognition sequences do not have as many sequence-specific contacts (which means they don't bind as tightly to the binding site in question), and they are more likely to be found randomly throughout the genome. Using the same type of calculation from part B, there are 256 possible combinations for a 4-base-pair recognition sequence, which could be found 15-16 times over a 4000-base-pair segment by random chance. This could be useful for proteins that need to bind to a large number of sites with low affinity. If we take the case of the 8-base-pair sequence, there are 65,536 different possible sequences. So, not only do they represent high-affinity binding sites, they are much less likely to be found by random chance

In the DNA of certain bacterial cells, 13% of the nucleotides are adenine. What are the percentages of the other nucleotides?

The DNA is made of four nucleotides (100% = 13% A + x% T + y% G + z% C). Because A pairs with T, the two nucleotides are represented in equimolar proportions in DNA. Therefore, the bacterial DNA in question contains 13% thymidine. This leaves 74% [= 100% - (13% + 13%)] for G and C, which also form base pairs and hence are equimolar. Thus y = z = 74/2 = 37.

Under standard conditions, the expected melting temperature in degrees Celsius can be calculated from the equation Tm = 59.9 + [0.41 × %(G + C)] - (675/length of duplex). Does the Tm increase or decrease if there are more G + C (and thus fewer A + T) base pairs? Does the Tm increase or decrease as the length of DNA increases? Why?

The Tm increases as the proportion of G + C bases increases and as the length increases. The thermal energy required for melting depends on how many hydrogen bonds between the strands must be broken. Each G-C base pair contributes three hydrogen bonds, whereas an A-T base pair contributes only two.

The nucleotide sequence of one DNA strand of a DNA double helix is 5'-GGATTTTTGTCCACAATCA-3'. What is the sequence of the complementary strand?

The complementary strand reads 5ʹ-TGATTGTGGACAAAAATCC-3ʹ

What are the steps required to pack genomic DNA into a mitotic chromosome?

The next state of condensation creates a fiber or packed nucleosomes 30 nm thick. This is controlled by a linker histone (H1). Chromatin fibers get folded into large loops that create a cylindrical shape 700 nm in diameter. A mitotic chromosome is about 10,000 times shorter than the extended sequence of DNA.

What are the components of the nucleosome core particle and how does it form?

The octamer is made up of 4 pairs of histone proteins. It encapsulates 147 base pairs of double stranded DNA. There is another 50+ base pairs that separate two nucleosomes. It is 11 nm in diameter. The persistence length of double stranded DNA is 50 nm. This is the length the DNA chain will extend before it bends in a different direction. Because DNA is relatively stiff it would be difficult for it all to fit comfortably in the nucleus. Histones induce curvature in the DNA to promote more efficient compaction/organization.

A single nucleosome core particle is 11 nm in diameter and contains 147 bp of DNA (the DNA double helix measures 0.34 nm/bp). What packing ratio (ratio of DNA length to nucleosome diameter) has been achieved by wrapping DNA around the histone octamer? Assuming that there are an additional 54 bp of extended DNA in the linker between nucleosomes, how condensed is "beads-on-a-string" DNA relative to fully extended DNA? What fraction of the 10,000-fold condensation that occurs at mitosis does this first level of packing represent?

The packing ratio within a nucleosome core is 4.5 [(147 bp × 0.34 nm/bp)/(11 nm) = 4.5]. If there is an additional 54 bp of linker DNA, then the packing ratio for "beads-on-a-string" DNA is 2.3 [(201 bp × 0.34 nm/ bp)/(11 nm + {54 bp × 0.34 nm/bp}) = 2.3]. This first level of packing represents only 0.023% (2.3/10,000) of the total condensation that occurs at mitosis.

The two strands of a DNA double helix can be separated by heating. If you raised the temperature of a solution containing the following three DNA molecules, in what order do you suppose they would "melt"? Explain your answer. A. 5'-GCGGGCCAGCCCGAGTGGGTAGCCCAGG-3' 3'-CGCCCGGTCGGGCTCACCCATCGGGTCC-5' B. 5'-ATTATAAAATATTTAGATACTATATTTACAA-3' 3'-TAATATTTTATAAATCTATGATATAAATGTT-5' C. 5'-AGAGCTAGATCGAT-3' 3'-TCTCGATCTAGCTA-5'

Therefore, helix C (containing a total of 34 hydrogen bonds) would melt at the lowest temperature, helix B (containing a total of 65 hydrogen bonds) would melt next, and helix A (containing a total of 78 hydrogen bonds) would melt last.

What are the three essential sequence elements contained in eukaryotic chromosomes?

They need a replication origin. - This is where polymerases start making complementary strands of DNA during replication. They need a centromere. - This is where the mitotic spindle attaches to the chromosome. They need telomeres. -Repeated DNA sequences at the ends of chromosomes required for replication and suppression of DNA repair.

Do you think Tm is a constant, or can it depend on other small molecules in the solution? Do you think high salt concentrations increase, decrease, or have no effect on Tm?

Tm depends on the identity and concentration of other molecules in the solution. High salt concentrations are more effective at shielding the two negatively charged sugar-phosphate backbones in the double helix from each other, so the two strands repel each other less strongly. Thus, a high salt concentration stabilizes the duplex and increases the melting temperature

Suppose you had a method of cutting DNA at specific sequences of nucleotides. How many nucleotides long (on average) would such a sequence have to be in order to make just one cut in a bacterial genome of 3 × 106 nucleotide pairs? How would the answer differ for the genome of an animal cell that contains 3 × 109 nucleotide pairs?

To specify a unique sequence which is N nucleotides long, 4N has to be larger than 3 × 106. Thus, 4N > 3 × 106, solved for N, gives N > ln(3 × 106)/ ln(4) = 10.7. Thus, on average, a sequence of only 11 nucleotides in length is unique in the genome. Performing the same calculation for the genome size of an animal cell yields a minimal stretch of 16 nucleotides. This shows that a relatively short sequence can mark a unique position in the genome and is sufficient, for example, to serve as an identity tag for one specific gene

Primase is needed to initiate DNA replication on both the leading strand and the lagging strand. True or False

True

Telomerase is a DNA polymerase that carries its own RNA molecule to use as a primer at the end of the lagging strand. True or False

True

Why are histones a special class of DNA binding proteins?

Unlike transcription factors and initiation factors, histones can bind to almost any sequence of DNA. Their sequences contain a lot of positively charged lysine and arginine amino acids that interact primarily with the phosphate backbone of the double helix