HW #5
What are the specialized DNA sequences that are at the ends of most eukaryotic chromosomes called? Choose one: A. telomeres B. centrosomes C. nucleosomes D. centromeres E. histones
A. telomeres Explanation: Telomeres contain repeated nucleotide sequences that are required for the ends of chromosomes to be fully replicated.
The Encyclopedia of DNA Elements (ENCODE) project aims to discover functional elements in DNA. One technique employed is DNase-seq. This technique employs a DNase enzyme that digests accessible regions of chromatin, while inaccessible regions remain undigested. The undigested DNA is then sequenced. Which of the following is likely true of the DNase-seq results? Choose one or more: A.Unsequenced DNA is likely part of euchromatin. B.Centromeric DNA is likely to be sequenced in all samples. C.Unsequenced DNA regions were likely bound to nucleosomes. D.The sequenced DNA will be the same from cell type to cell type.
A.Unsequenced DNA is likely part of euchromatin. B.Centromeric DNA is likely to be sequenced in all samples Explanation: DNA bound by nucleosomes or tightly packed into heterochromatin will not be digested. The digested sequences are linker DNA between nucleosomes and reflect euchromatin regions of DNA that may be actively transcribed. Since different cell types transcribe different genes, the regions of the genome that are euchromatin vs. heterochromatin may differ between cell types.
The Hershey and Chase experiment. Identify where viral protein and viral DNA would be found by dragging the labels to the correct targets.
When the researchers measured the radioactivity, they found that much of the 32P-labeled DNA had entered the bacterial cells, while the vast majority of the 35S-labeled proteins remained in solution with the spent viral particles. So the DNA is at the bottom. RNA is extracellular at the top.
Which option correctly describes the two strands of DNA in a double helix? Choose one: held together by covalent bonds between bases antiparallel in orientation identical parallel in orientation
antiparallel in orientation explanation: The two strands of DNA are considered to be antiparallel in orientation. This means that one strand runs 5' to 3' against the 3'-to-5' orientation of the other strand.
Which of the following represents the specialized DNA sequence that attaches to microtubules and allows duplicated eukaryotic chromosomes to be separated during M phase? Choose one: telomere mitotic spindle centrosome centromere nucleosome histone
centromere. Explanation: During mitosis, DNA condenses, adopting a more compact structure and ultimately forming mitotic chromosomes. Once the chromosomes have condensed, the centromere allows the mitotic spindle to attach to each duplicated chromosome in a way that directs one copy of each chromosome to be segregated to each of the two daughter cells.
What is the term that describes the complex of DNA and proteins that makes up a eukaryotic chromosome? Choose one: chromatin centromere centrosome centriole
chromatin. Explanation: DNA is wrapped around histone octamers, creating nucleosomes, which are the fundamental unit of chromatin.
What does each eukaryotic chromosome contain? Choose one: one long ribonucleotide sequence one long double-stranded DNA molecule one long sequence of amino acids one single long gene one long DNA strand
one long double-stranded DNA molecule.
Prokaryotes have chromosomes that are circular in structure. Which of the following would such chromosomes lack? Choose one: A. DNA double helix B. telomeres C. replication origin D. sugar-phosphate backbone E. complementary base pairs
B. telomeres Explanation: Telomeres are a structure of linear chromosomes that helps prevent the shortening of chromosomes over time due to the replication machinery being unable to attach to the lagging strand of DNA near the end of the chromosome.
Which of the following statements is not true? Choose one: A. A cell will temporarily decondense its chromatin to silence genes during differentiation. B. A cell will temporarily decondense its chromatin to give proteins rapid, localized access to specific DNA sequences. C. A cell can permanently condense and silence an entire chromosome during development. D. When a cell divides, its chromatin structures will typically be inherited by its daughter cells. E. A cell will temporarily decondense its chromatin to allow access to specific DNA sequences for replication, repair, or gene expression.
A. A cell will temporarily decondense its chromatin to silence genes during differentiation. Explanation: DNA carries a vast amount of information that cells utilize to perform their various and specific functions. Regulation of the genetic information on chromosomes can occur at many steps, with regulation of chromatin structure being fundamental among them. Interphase chromosomes contain chromatin in its most relaxed state, yet much of it remains tightly condensed in a form termed heterochromatin. Genes found within heterochromatin are inaccessible and not expressed, which is most notably observed in X chromosome inactivation in mammalian females. On the other hand, relaxed chromatin, or euchromatin, is accessible to transcription, replication, and repair machinery and genes in this type of chromatin are able to be expressed.
In their 1953 paper on the double-helical structure of DNA, Watson and Crick famously wrote: "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material." What did they mean?Choose one: A. Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand. B. When a cell divides, each DNA helix is split between the daughter cells. C. The sugar-phosphate backbone of DNA holds the helix together in a way that allows the genetic information to be copied. D. Sexually reproducing organisms swap their DNA.
A. Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand Explanation: Because each DNA strand in a double helix contains a sequence of nucleotides that is complementary to the sequence of its partner strand, each strand can serve as a template to direct the synthesis of a new strand identical in sequence to its former partner. Watson and Crick were hinting at the "semiconservative" mechanism of DNA replication that we now know occurs.
Another step in PCR requires small single-stranded DNA primers to anneal to a target sequence on denatured DNA. Two primers are used. Ideally these primers will have similar melting temperatures. The melting temperature is defined as the temperature at which 50% of the DNA is in the single-stranded form. Which of the following primers will have the highest melting temperature? Choose one: A. G G G G A A A T T T C C C C B. A A A A G G C C T T T T C. A A A A G G G C C C T T T T D. G G G A A A T T T C C C
A. G G G G A A A T T T C C C C Explanation: Since G-C base pairs contain one more hydrogen bond than A-T base pairs, it takes more heat energy to separate G-C base pairs than A-T base pairs. This primer, due to its length and G-C content, has the highest melting temperature of the listed primers.
What histone protein is thought to act as a linker that pulls nucleosomes together into a regular repeating array, resulting in a 30-nm fiber? (Hint: Nucleosomes contain DNA wrapped around a protein core of these eight histone molecules, which one listed is not a part of the histone octamer?) Choose one: A. H1 B. H3 C. H4 D. H2A E. H2B
A. H1 Histone 1 (H1) protein is thought to act as a linker that pulls nucleosomes together into a regular repeating array, resulting in a 30-nm fiber. Both histone H1 and a set of specialized non-histone chromosomal proteins are known to help drive these condensations.
Which statement is true about the association of histone proteins and DNA? Choose one: A. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone. B. Histone proteins have a high proportion of negatively charged amino acids, which bind tightly to the positively charged DNA backbone. C. Each histone protein has a deep groove into which a DNA double helix tightly fits. D. Histone proteins insert themselves into the major groove of DNA.E. Histone proteins form hydrogen bonds with the nucleotide bases of DNA.
A. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone. Explanation: Peripheral arginine and lysine side chains, which are positively charged, interact with the phosphate groups of DNA.
What is the general name given to the most highly condensed form of chromatin? Choose one: A. heterochromatin B. X chromatin C. nucleosome D. 30-nm chromatin fiber E. euchromatin
A. heterochromatin Explanation: "Heterochromatin" is the general name given to the most highly condensed form of chromatin, which can be observed under the light microscope.
What type of bond connects base pairs? Choose one: A. hydrogen B. van der Waals C. covalent D. ionic
A. hydrogen Explanation: A double-stranded DNA molecule is composed of two polynucleotide chains (DNA strands) held together by hydrogen bonds between the paired bases.
In the early part of the twentieth century, scientists thought that proteins were the most likely carriers of genetic information. What characteristic of proteins led scientists to this belief? Choose one: their well-defined, three-dimensional structure their variety of shapes their abundance in cells their large size their chemical diversity
Answer: their chemical diversity Explanation: Proteins, which are built from a set of 20 different amino acids, were thought to have the chemical diversity necessary to encode the information needed to build cells. DNA, which is built from a set of only four nucleotides, seemed too chemically simple for the task.
How do chromatin-remodeling complexes work? Choose one: A. They bind to nucleosomes in the 30-nm fiber and induce another level of packing, obscuring DNA from binding by other proteins. B. They use the energy from ATP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins. C. They use the energy from GTP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins. D. They remove acetyl groups from the tails of histones, rendering the DNA more accessible to other proteins. E. They add methyl groups to the tails of histones in order to attract other proteins.
B. They use the energy from ATP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins. explanation: Changing the conformation of chromatin is an energy-intensive process and can result in making a segment of DNA more or less accessible to proteins involved, for example, in transcription.
In 1952, Alfred Hershey and Martha Chase, working with a virus called T2, conducted what is now considered a landmark experiment to determine whether genes are made of DNA or protein. When this virus, which is made entirely of DNA and protein, infects E. coli, it injects its genetic material into the bacterial cell, leaving the empty virus head stuck to the cell surface. To determine whether it was DNA or protein that enters the infected bacterial cell, the researchers radioactively labeled one batch of T2 with the isotope 35S, which resulted in radiolabeled viral proteins, and a second batch of T2 with 32P, which resulted in radiolabeled DNA. They then incubated the radioactive viruses with E. coli. After allowing a few minutes for the viruses to transfer their genetic material to the bacterial cells, the researchers used a blender to shear the empty virus heads from the bacterial cell surface. They then used a centrifuge to separate the infected bacteria from the empty virus heads: spinning the sample at high speed caused the heavier, infected bacteria to pellet at the bottom of the centrifuge tube, while the lighter, empty virus heads remained in solution. Using this protocol, what should the researchers have seen in terms of the distribution of radioactivity in the centrifuged sample? A. 35S in the pellet, 32P in the solution B. 32P in the pellet, 35S in the solution C. no radioactivity in the pellet D. an equal amount of 35S and 32P in the pellet and the solution E. no radioactivity in the solution
B. 32P in the pellet, 35S in the solution Explanation: Their hypothesis was that either DNA or protein was injected during infection. Because DNA is the genetic material and the viruses inject their genetic material into the bacterial cells, after infection most of the DNA should have been in the bacterial cells in the pellet. Thus, the 32P that was used to label the DNA should have been in the pellet. The head of the virus is composed mainly of protein, so the 35S should have remained with the empty virus heads in the solution.
Which chemical group is at the 3' end of a DNA strand? Choose one: A. a carboxyl group B. a hydroxyl group C. a nitrogenous base D. a phosphate group
B. a hydroxyl group Explanation: A hydroxyl group (-OH) is attached to the 3' carbon of the pentose sugar and (because of the polarity of DNA) is also found at the free, 3' end of a DNA strand.
In the 1920s, bacteriologist Fred Griffith demonstrated that a heat-killed, infectious pneumococcus produced a substance that could convert a harmless form of the bacterium into a lethal one. Fifteen years later, researchers prepared an extract from the disease-causing S strain of pneumococci and showed that this material could transform the harmless R-strain pneumococci cells into the infectious S-strain form. This change to the bacteria was both permanent and heritable, suggesting that this "transforming principle" represents the elusive genetic material of the cells. The researchers subjected their extract to a variety of tests to determine the chemical identity of the "transforming principle." In one experiment, they treated the material with enzymes that destroy all proteins. This treatment did not affect the ability of the extract to transform harmless bacteria into an infectious form.From this result, what could the researchers conclude? Choose one: A. Proteins act as the genetic material. B. The transforming principle is not DNA. C. The genetic material is not protein. D. The transforming principle is not genetic material. E. DNA acts as the genetic material.
C. The genetic material is not protein. From the results, the researchers could conclude that the genetic material is not protein. The results ruled out protein as the genetic material because experimentally destroying the proteins had no effect on the ability of the extract to transform harmless bacteria into an infectious form.
The DNA in eukaryotic chromosomes is folded into a compact form by interactions with which of the following? Choose one: A. centrioles B. microtubules C. histones D. RNA
C. histones Explanation: Histones are responsible for the first and most fundamental level of chromatin packing: the formation of the nucleosome.
Which of the following is true for most genes? Choose one: A. A gene is a segment of DNA that contains the instructions for making a particular protein. B. A gene is a segment of DNA that contains the instructions for making a particular RNA. C. A gene is a unit of heredity that contains instructions that dictate the characteristics of an organism. D. All of the above are true regarding genes. E. None of the above are true regarding genes.
D. All of the above are true regarding genes. Explanation: These are all interrelated by the central dogma of molecular biology.
One of the first steps in obtaining a karyotype (such as that shown below of a cancer cell) is treating cells with a drug that stalls cells in mitosis. Why must cells arrest in mitosis for karyotype analysis? Choose one: A. Only mitotic chromosome DNA is separated into single strands, allowing for staining by dyes. B. Only in mitosis are homologous chromosomes paired up. C. Only mitotic cells contain homologous chromosomes, because mitosis happens after DNA replication. D. Only mitotic chromosomes are highly condensed and visible with a light microscope.
D. Only mitotic chromosomes are highly condensed and visible with a light microscope. Explanation: During interphase, individual chromosomes are not visible with a light microscope. During mitosis, chromosomes become very tightly packed to assist with proper segregation of the chromosomes into the two daughter cells. (Karyotype analysis can detect chromosomal abnormalities such as extra or missing chromosomes. Some of these abnormalities are inherited and others arise during carcinogenesis. Mitotic chromosomes are much more highly condensed than interphase chromosomes, so cells are treated with a compound to arrest cells in mitosis and then painted with tags that recognize specific chromosomes. The resulting image is manipulated to place homologous chromosomes together in descending size order.)
PART 1:Some applications in biology, such as polymerase chain reaction (PCR), require melting the DNA double helix into single strands of DNA. This can be accomplished by heating the DNA. As DNA is heated, why does the double helix structure denature into single strands of DNA but not into individual nucleotides? In other words, why do the single strands remain intact even though the double helix does not? Choose one: A. The double helix structure is important for DNA replication, while the single strands carry the information. B. The single strands are wrapped around histones, which protects them from denaturing. C. The double helix is held together with phosphodiester bonds, while the single strands are linked by hydrogen bonds. D. The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds.
D. The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds. explanation: Hydrogen bonds are weaker than covalent phosphodiester bonds. Hence heating can separate the strands from each other without breaking up the individual strands.
The tails of the core histone proteins can be chemically modified by the covalent addition of what type of chemical group? Choose one: A. acetyl B. methyl C. phosphate D. all of these E. none of these
D. all of these Explanation: The tails of the core histone proteins can be chemically modified by the covalent addition of a methyl group, an acetyl group, or a phosphate group. Each modification alters the physiology of the histone, with some promoting or maintaining euchromatin and others promoting heterochromatin.
Which structure is normally on the 5' end of a DNA strand? Choose one: A. sulfur group B. nitrogenous base C. hydroxyl group D. phosphate group
D. phosphate group Explanation: On the 5' end of a DNA strand, a phosphate group is attached to the ribose sugar. DNA is polar, so the chemistry at the 5' end differs from the chemistry at the 3' end. The hydroxyl is attached to the 3' carbon.
Which of the following statements about nucleosomes is FALSE?Choose one: A. Nucleosomes convert a DNA molecule into a chromatin thread about one-third the length of the initial DNA. B. A nucleosome consists of DNA wrapped around eight histone proteins, plus a short segment of linker DNA. C. Nucleosomes represent the first and most fundamental level of chromatin packing. D. Nucleosomes can be seen in the electron microscope. E. Nucleosomes are found only in mitotic chromosomes.
E. Nucleosomes are found only in mitotic chromosomes. Explanation: Generally speaking, nucleosomes are found within chromatin regardless of the cell cycle stage.
Study the image depicting the Avery, McCarty, and MacLeod experiment, and then answer the question. Which of the statements below is supported by the Avery, McCarty, and MacLeod experiment and the data in the figure above? Choose one: A. Carbohydrates from one strain of bacteria can alter the phenotype of another strain if those carbohydrates are taken up by the other strain. B. Lipids from one strain of bacteria can alter the phenotype of another strain if those lipids are taken up by the other strain. C. RNA from one strain of bacteria can alter the phenotype of another strain if that RNA is taken up by the other strain. D. Protein from one strain of bacteria can alter the phenotype of another strain if that protein is taken up by the other strain. E. DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain.
E. DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain.
Why are individual DNA strands considered polar? Choose one: A. The strands twist preferentially into a right-handed helix, not a left-handed helix. B. One end contains the telomere, while the other contains the centromere. C. Each end contains a different, complementary nucleotide sequence. D. One end is positively charged, while the other is negative. E. One end terminates in a hydroxyl group, while the other terminates in a phosphate.
E. One end terminates in a hydroxyl group, while the other terminates in a phosphate. Explanation: The end with the free hydroxyl group (which is attached to the 3' carbon position of the sugar) is called the 3' end, while the end bearing the phosphate group (which is attached via the 5' hydroxyl group of the sugar) is called the 5' end.
In the late 1920s, bacteriologist Fred Griffith was studying Streptococcus pneumoniae. This bacterium comes in two forms: one that is highly infectious (called the "S strain" because it forms colonies that appear smooth when grown on a nutrient plate in the lab) and one that is relatively harmless (called the "R strain" because its colonies appear rough). When injected into mice, the S strain is lethal, whereas the R strain causes no ill effect. Griffith confirmed that when the S strain is killed by heating, it is no longer infectious. But he then discovered that if he injected mice with both the heat-killed S strain pneumococci and the live, harmless R strain bacteria, the animals died of pneumonia. Furthermore, their blood was swarming with live, S strain bacteria that, when grown in culture, remained infectious and lethal. Based on these results, what could Griffith conclude? A. The infectious S strain of bacteria cannot be killed by heating. B. The R strain of bacteria is more deadly than previously thought. C. DNA is conclusively confirmed as the genetic material of cells. D. S. pneumoniae is a poor choice for investigating the molecular basis of heredity, as most bacteria behave unpredictably in the laboratory. E. Some substance in the infectious S strain can change the harmless R strain into the more lethal form.
E. Some substance in the infectious S strain can change the harmless R strain into the more lethal form. Explanation: Based on his results, Griffith could conclude that some substance in the infectious S strain could change the harmless R strain into the more lethal form. The Griffith experiments were performed in the early 1900s, well before the structure of DNA was solved.
Where is heterochromatin not commonly located? Choose one: A. telomeres B. centromeres C. silenced X chromosomes D. gene-poor regions of chromosomes E. chromosomal regions carrying genes that encode ribosomal proteins
E. chromosomal regions carrying genes that encode ribosomal proteins Explanation: Chromosomal regions carrying genes that encode ribosomal proteins are active in gene transcription; therefore, you would not expect to see heterochromatin in these areas. Telomeric DNA and centromeric DNA do not encode any genes, and as such, these regions are commonly heterochromatic like an inactivated X chromosome.
Determine whether the following statement is true or false: Histones are an example of a sequence-specific DNA binding protein.
False. Explanation: As the proteins that help package DNA into chromosomes, histones must interact with DNA of many sequences. Thus, they do not bind only to specific nucleotide sequences.
Identify which statement(s) is/are true regarding telomere characteristics. Choose one: A. Telomeres allow duplicated chromosomes to become separated into daughter cells during M phase. B. Telomeres are found in all living cells. C. Telomeres cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair. D. Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes. E. All of the above are true regarding telomere characteristics. F. Only statements B and D are true regarding telomere characteristics. G. Only statements C and D are true regarding telomere characteristics.
G. Only statements C and D are true regarding telomere characteristics. Explanation: Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes. These telomeric repeated nucleotide sequences cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair.