Essential Cell Biology 4th Edition By Bruce Alberts - Test Bank

Essential Cell Biology 4th Edition By Bruce Alberts – Test Bank

 

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Essential Cell Biology, Fourth Edition

Chapter 5: DNA and Chromosomes

The Structure of DNA

 

5-1       Using terms from the list below, fill in the blanks in the following brief description of the experiment with Streptococcus pneumoniae that identified which biological molecule carries heritable genetic information. Some terms may be used more than once.

 

Cell-free extracts from S-strain cells of S. pneumoniae were fractionated to __________________ DNA, RNA, protein, and other cell components. Each fraction was then mixed with __________________ cells of S. pneumoniae. Its ability to change these into cells with __________________ properties resembling the __________________ cells was tested by injecting the mixture into mice. Only the fraction containing __________________ was able to __________________ the __________________ cells to __________________ (or __________________ ) cells that could kill mice.

 

carbohydrate               lipid                                         R-strain

DNA                           nonpathogenic                         RNA

identify                       pathogenic                               S-strain

label                             purify                                      transform

5-2       Many of the breakthroughs in modern biology came after Watson and Crick published their model of DNA in 1953. However, chromosomes were identified earlier. In what decade did scientists first identify chromosomes?

(a)        1880s

(b)        1920s

(c)        1940s

(d)       1780s

 

5-3       Mitotic chromosomes were first visualized with the use of very simple tools: a basic light microscope and some dyes. Which of the following characteristics of mitotic chromosomes reflects how they were named?

(a)        motion

(b)        color

(c)        shape

(d)       location

 

5-4       In the 1940s, proteins were thought to be the more likely molecules to house genetic information. What was the primary reason that DNA was not originally believed to be the genetic material?

(a)        DNA has a high density of negative charges.

(b)        Nucleotides were known to be a source of chemical energy for the cell.

(c)        Both protein and nucleic acids were found to be components of chromosomes.

(d)       DNA was found to contain only four different chemical building blocks.

 

5-5       In a DNA double helix, _____________________.

(a)        the two DNA strands are identical.

(b)        purines pair with purines.

(c)        thymine pairs with cytosine.

(d)       the two DNA strands run antiparallel.

 

5-6       You are a virologist interested in studying the evolution of viral genomes. You are studying two newly isolated viral strains and have sequenced their genomes. You find that the genome of strain 1 contains 25% A, 55% G, 20% C, and 10% T. You report that you have isolated a virus with a single-stranded DNA genome. Based on what evidence can you make this conclusion?

(a)        single-stranded genomes always have a large percentage of purines

(b)        using the formula: G – A = C + T

(c)        Double-stranded genomes have equal amounts of A and T

(d)       Single-stranded genomes have a higher rate of mutation

 

5-7       Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

1. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.
2. The polarity of a DNA strand results from the polarity of the nucleotide subunits.
3. There are five different nucleotides that become incorporated into a DNA strand.
4. Hydrogen bonds between each nucleotide hold individual DNA strands together.

 

5-8       Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists first demonstrated that cells contain some component that can be transferred to a new population of cells and permanently cause changes in the new cells?

(a)        Griffith

(b)        Watson and Crick

(c)        Avery, MacLeod, and McCarty

(d)       Hershey and Chase

 

5-9       Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists obtained definitive results demonstrating that DNA is the genetic molecule?

(a)        Griffith

(b)        Watson

(c)        Crick

(d)       Hershey and Chase

 

5-10     Fred Griffith studied two strains of Streptococcus pneumonia, one that causes a lethal infection when injected into mice, and a second that is harmless. He observed that pathogenic bacteria that have been killed by heating can no longer cause an infection. But when these heat-killed bacteria are mixed with live, harmless bacteria, this mixture is capable of infecting and killing a mouse. What did Griffith conclude from this experiment?

(a)        The infectious strain cannot killed by heating.

(b)        The heat-killed pathogenic bacteria “transformed” the harmless strain into a lethal one.

(c)        The harmless strain somehow revived the heat-killed pathogenic bacteria.

(d)       The mice had lost their immunity to infection with S. pneumoniae.

 

5-11     Avery, MacLeod, and McCarty carried out experiments to identify the class of biological molecule that carries heritable information. Explain how they identified the “transforming principle” that could convert a harmless strain of bacteria to a pathogenic one.

 

5-12     Hershey and Chase used radiolabeled macromolecules to identify the material that contains heritable information. What radioactive material was used to track DNA during this experiment?

(a)        ³H

(b)        ¹⁴C

(c)        ³⁵S

(d)       ³²P

 

5-13     Which of the following chemical groups is not used to construct a DNA molecule?

(a)        five-carbon sugar

(b)        phosphate

(c)        nitrogen-containing base

(d)       six-carbon sugar

 

5-14     Which of the following structural characteristics is not normally observed in a DNA duplex?

(a)        purine–pyrimidine pairs

(b)        external sugar–phosphate backbone

(c)        uniform left-handed twist

(d)       antiparallel strands

 

5-15     Which of the following DNA strands can form a DNA duplex by pairing with itself at each position?

(a)        5′-AAGCCGAA-3′

(b)        5′-AAGCCGTT-3′

(c)        5′-AAGCGCAA-3′

(d)       5′-AAGCGCTT-3′

 

5-16     The DNA from two different species can often be distinguished by a difference in the ______________________.

(a)        ratio of A + T to G + C.

(b)        ratio of A + G to C + T.

(c)        ratio of sugar to phosphate.

(d)       presence of bases other than A, G, C, and T.

 

5-17     For a better understanding of DNA structure, it helps to be able to compare physical characteristics evident from a side view of double-stranded DNA with those of individual base pairs.

1. Use brackets to designate the major and minor grooves on Figure Q5-17A and shade in the surface that will be exposed in the major groove in Figure Q5-17B.
2. If base pairs were aligned and stacked directly on top of each other, the major and minor grooves would be linear depressions all along the DNA. Explain why this is not the actual conformation of a DNA molecule.

 

Figure Q5-17

 

5-18     Which DNA base pair is represented in Figure Q5-18?

(a)        A-T

(b)        T-A

(c)        G-C

(d)       C-G

Figure Q5-18

 

5-19     Use the terms listed to fill in the blanks in Figure Q5-19.

1. A-T base pair
2. G-C base pair
3. deoxyribose
4. phosphodiester bonds
5. purine base
6. pyrimidine base

 

Figure Q5-19

 

5-20     The structures of the four bases in DNA are given in Figure Q5-20.

 

 

­­­­Figure Q5-20

 

1. Which are purines and which are pyrimidines?
2. Which bases pair with each other in double-stranded DNA?

 

5-21     Using the structures in Figure Q5-20 as a guide, sketch the hydrogen bonds between the base pairs in DNA. Hint: The bases in the figure are all drawn with the –NH– that attaches to the sugar at the bottom of the structure.

 

5-22     Because hydrogen bonds hold the two strands of a DNA molecule together, the strands can be separated without breaking any covalent bonds. Every unique DNA molecule “melts” at a different temperature. In this context, T_m (melting temperature) is the point at which two strands separate, or become denatured. Order the DNA sequences listed below according to relative melting temperatures (from lowest T_m to highest T_m). Assume that they all begin as stable double-stranded DNA molecules. Explain your answer.

1. GGCGCACC
2. TATTGTCT
3. GACTCCTG
4. CTAACTGG

 

5-23     Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

1. Each strand of DNA contains all the information needed to create a new double-stranded DNA molecule with the same sequence information.
2. All functional DNA sequences inside a cell code for protein products.
3. Gene expression is the process of duplicating genes during DNA replication.
4. Gene sequences correspond exactly to the respective protein sequences produced from them.

 

5-24     The complete set of information found in a given organism’s DNA is called its ____________.

(a)        genetic code.

(b)        coding sequence.

(c)        gene.

(d)       genome.

 

5-25     The manner in which a gene sequence is related to its respective protein sequence is referred to as the _________ code.

(a)        protein

(b)        genetic

(c)        translational

(d)       expression

 

5-26     The information stored in the DNA sequences is used directly as a template to make ___________.

(a)        lipids.

(b)        RNA.

(c)        polypeptides.

(d)       carbohydrates.

 

5-27     Given the sequence of one strand of a DNA helix (below), provide the sequence of the complementary strand and label the 5′ and 3′ ends.

5′-GCATTCGTGGGTAG-3′

 

5-28     When double-stranded DNA is heated, the two strands separate into single strands in a process called melting or denaturation. The temperature at which half of the duplex DNA molecules are intact and half have melted is defined as the T_m.

1. Do you think T_m 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 T_m?
2. Under standard conditions, the expected melting temperature in degrees Celsius can be calculated from the equation T_m = 59.9 + [0.41 × %(G + C)] – (675/length of duplex). Does the T_m increase or decrease if there are more G + C (and thus fewer A + T) base pairs? Does the T_m increase or decrease as the length of DNA increases? Why?
3. Calculate the predicted T_m for a stretch of double helix that is 100 nucleotides long and contains 50% G + C content.

 

5-29     Consider the structure of the DNA double helix.

1. You and a friend want to split a double-stranded DNA molecule so you each have half. Is it better to cut the length of DNA in half so each person has a shorter length, or to separate the strands and each take one strand? Explain.
2. In the original 1953 publication describing the discovery of the structure of DNA, Watson and Crick wrote, “It has not escaped our notice that the specific pairings we have postulated immediately suggest a possible copying mechanism for the genetic material.” What did they mean?

 

5-30     In principle, what would be the minimum number of consecutive nucleotides necessary to correspond to a single amino acid to produce a workable genetic code (it can specifically code for all 20 amino acids)? Assume that each amino acid is encoded by the same number of nucleotides. Explain your reasoning.

 

5-31     A.        Explain the reason why the cell requires a mechanism for identifying specific sequences of DNA.

1. 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.
2. 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?

 

 

The Structure of Eukaryotic Chromosomes

 

5-32     For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

 

In eukaryotic __________________, DNA is complexed with proteins to form __________________. The paternal and maternal copies of human Chromosome 1 are __________________, whereas the paternal copy of Chromosome 1 and the maternal copy of Chromosome 3 are __________________. Cytogeneticists can determine large-scale chromosomal abnormalities by looking at a patient’s __________________. Fluorescent molecules can be used to paint a chromosome, by a technique that employs DNA __________________, and thereby to identify each chromosome by microscopy.

 

bands                           extended                     kinetochore

chromatin                    homologous                 nonhomologous

chromosomes              hybridization

condensation               karyotype

 

5-33     A.        Define a gene.

1. Consider two different species of yeast that have similar genome sizes. Is it likely that they contain the same number of genes? A similar number of chromosomes?

 

5-34     The human genome is divided into linear segments and packaged into structures called chromosomes. What is the total number of chromosomes found in each of the somatic cells in your body?

(a)        22

(b)        23

(c)        44

(d)       46

 

5-35     The human genome is a diploid genome. However, when germ-line cells produce gametes, these specialized cells are haploid. What is the total number of chromosomes found in each of the gametes (egg or sperm) in your body?

(a)        22

(b)        23

(c)        44

(d)       46

 

5-36     What type of macromolecule helps package DNA in eukaryotic chromosomes?

(a)        lipids

(b)        carbohydrates

(c)        proteins

(d)       RNA

 

5-37     The number of cells in an average-sized adult human is on the order of 10¹⁴. Use this information, and the estimate that the length of DNA contained in each cell is 2 m, to do the following calculations (look up the necessary distances and show your working):

1. Over how many miles would the total DNA from the average human stretch?
2. How many times would the total DNA from the average human wrap around the planet Earth at the Equator?
3. How many times would the total DNA from the average human stretch from Earth to the Sun and back?
4. How many times would the total DNA from the average human stretch from the Earth to Pluto and back?

 

5-38     The process of sorting human chromosome pairs by size and morphology is called karyotyping. A modern method employed for karyotyping is called chromosome painting. How are individual chromosomes “painted”?

(a)        with a laser

(b)        using fluorescent antibodies

(c)        using fluorescent DNA molecules

(d)       using green fluorescent protein

 

5-39     The human genome comprises 23 pairs of chromosomes found in nearly every cell in the body. Answer the quantitative questions below by choosing one of the numbers in the following list:

23        69        >200

46        92        >10⁹

 

1. How many centromeres are in each cell? What is the main function of the centromere?
2. How many telomeres are in each cell? What is their main function?
3. How many replication origins are in each cell? What is their main function?

 

5-40     Explain the differences between chromosome painting and the older, more traditional method of staining chromosomes being prepared for karyotyping. Highlight the way in which each method identifies chromosomes by the unique sequences they contain.

 

5-41     Which of the following questions would not be answered by using karyotyping?

(a)        Is the individual genetically female or male?

(b)        Do any of the chromosomes contain pieces that belong to other chromosomes?

(c)        Does the individual have an extra chromosome?

(d)       Do any chromosomes contain point mutations?

 

5-42     Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

1. Comparing the relative number of chromosome pairs is a good way to determine whether two species are closely related.
2. Chromosomes exist at different levels of condensation, depending on the stage of the cell cycle.
3. Eukaryotic chromosomes contain many different sites where DNA replication can be initiated.
4. The telomere is a specialized DNA sequence where microtubules from the mitotic spindle attach to the chromosome so that duplicate copies move to opposite ends of the dividing cell.

 

5-43     For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

 

Each chromosome is a single molecule of __________________ whose extraordinarily long length can be compacted by as much as __________________-fold during __________________ and tenfold more during __________________. This is accomplished by binding to __________________ that help package the DNA in an orderly manner so it can fit in the small space delimited by the __________________. The structure of the DNA–protein complex, called __________________, is highly __________________ over time.

 

10,000                         chromosome                mitosis

100                              different                      nuclear envelope

1000                            DNA                           nucleolus

cell cycle                     dynamic                      proteins

cell wall                       interphase                    similar

chromatin                    lipids                           static

 

5-44     The chromosomes we typically see in images are isolated from mitotic cells. These mitotic chromosomes are in the most highly condensed form. Interphase cells contain chromosomes that are less densely packed and __________________________.

(a)        occupy discrete territories in the nucleus.

(b)        share the same nuclear territory as their homolog.

(c)        are restricted to the nucleolus.

(d)       are completely tangled with other chromosomes.

 

5-45     Figure Q5-45 clearly depicts the nucleolus, a nuclear structure that looks like a large, dark region when stained. The other dark, speckled regions in this image are the locations of particularly compact chromosomal segments called ____________.

(a)        euchromatin.

(b)        heterochromatin.

(c)        nuclear pores.

(d)       nucleosomes.

 

Figure Q5-45

 

5-46     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

 

5-47     Mitotic chromosomes are _____ times more compact than a DNA molecule in its extended form.

(a)        10,000

(b)        100,000

(c)        1000

(d)       100

 

5-48     Interphase chromosomes are about______ times less compact than mitotic chromosomes, but still are about______ times more compact than a DNA molecule in its extended form.

(a)        10, 1000

(b)        20, 500

(c)        5, 2000

(d)       50, 200

 

5-49     For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement about nucleosomes.

1. Nucleosomes are present in [prokaryotic/eukaryotic] chromosomes, but not in [prokaryotic/eukaryotic] chromosomes.
2. A nucleosome contains two molecules each of histones [H1 and H2A/H2A and H2B] as well as of histones H3 and H4.
3. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately [twice/three times/four times].
4. Nucleosomes are aided in their formation by the high proportion of [acidic/basic/polar] amino acids in histone proteins.
5. Nucleosome formation compacts the DNA into approximately [one-third/one-hundredth/one-thousandth] of its original length.

 

5-50     The classic “beads-on-a-string” structure is the most decondensed chromatin structure possible and is produced experimentally. Which chromatin components are not retained when this structure is generated?

(a)        linker histones

(b)        linker DNA

(c)        nucleosome core particles

(d)       core histones

 

5-51     A.        How can nucleosome core particles be isolated from chromatin?

1. What molecular components were identified after this treatment was complete?
2. What portion of the nucleosome was destroyed/removed during this treatment and what function does it normally serve?

 

5-52     Nucleosomes are formed when DNA wraps _____ times around the histone octamer in a ______ coil.

(a)        2.0, right-handed

(b)        2.5, left-handed

(c)        1.7, left-handed

(d)       1.3, right-handed

 

5-53     For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

 

Interphase chromosomes contain both darkly staining __________________ and more lightly staining __________________. Genes that are being transcribed are thought to be packaged in a __________________ condensed type of euchromatin. Nucleosome core particles are separated from each other by stretches of __________________ DNA. A string of nucleosomes coils up with the help of __________________ to form the more compact structure of the __________________. A __________________ model describes the structure of the 30-nm fiber. The 30 nm chromatin fiber is further compacted by the formation of __________________ that emanate from a central __________________.

 

30-nm fiber                             heterochromatin          linker

active chromatin                      histone H1                   loops

axis                                          histone H3                   more

beads-on-a-string                    histone H4                   synaptic complex

euchromatin                            less                              zigzag

 

5-54     The octameric histone core is composed of four different histone proteins, assembled in a stepwise manner. Once the core octamer has been formed, DNA wraps around it to form a nucleosome core particle. Which of the following histone proteins does not form part of the octameric core?

(a)        H4

(b)        H2A

(c)        H3

(d)       H1

 

5-55     The core histones are small, basic proteins that have a globular domain at the C-terminus and a long, extended conformation at the N-terminus. Which of the following is not true of the N-terminal “tail” of these histones?

(a)        It is subject to covalent modifications.

(b)        It extends out of the nucleosome core.

(c)        It binds to DNA in a sequence-specific manner.

(d)       It helps DNA pack tightly.

 

5-56     Stepwise condensation of linear DNA happens in five different packing processes. Which of the following four processes has a direct requirement for histone H1?

(a)        formation of “beads-on-a-string”

(b)        formation of the 30-nm fiber

(c)        looping of the 30-nm fiber

(d)       packing of loops to form interphase chromosomes

 

5-57     Evidence suggests that the replication of DNA packaged into heterochromatin occurs later than the replication of other chromosomal DNA. What is the simplest possible explanation for this phenomenon?

 

5-58     Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

4. The histone proteins that constitute the core nucleosome include tetramers of histones H2A, H2B, H3, and H4.
5. Linker histones help compact genomic DNA by influencing the path of the DNA after it has wrapped about the nucleosome core.
6. Histone proteins have a lower-than-average number of lysines and arginines in their polypeptide chains.
7. Interphase chromosomes represent a physical state of the chromatin with the highest order of packaging.

 

The Regulation of Chromosome Structure

 

5-59     Although the chromatin structure of interphase and mitotic chromosomes is very compact, DNA-binding proteins and protein complexes must be able to gain access to the DNA molecule. Chromatin-remodeling complexes provide this access by __________________.

(a)        recruiting other enzymes.

(b)        modifying the N-terminal tails of core histones.

(c)        using the energy of ATP hydrolysis to move nucleosomes.

(d)       denaturing the DNA by interfering with hydrogen-bonding between base pairs.

 

5-60     You are studying a newly identified chromatin-remodeling complex, which you call NICRC. You decide to run an in vitro experiment to characterize the activity of the purified complex. Your molecular toolbox includes: (1) a 400-base-pair DNA molecule that has a single recognition site for the restriction endonuclease EcoRI, an enzyme that cleaves internal sites on double-stranded DNA (dsDNA); (2) purified EcoRI enzyme; (3) purified DNase I, a DNA endonuclease that will cleave dsDNA at nonspecific sites if they are exposed; and (4) core octamer histones. You are able to assemble core nucleosomes on this DNA template and test for NICRC activity. Figure Q5-60A illustrates the DNA template used and indicates both the location of the EcoRI cleavage site and the size of the DNA fragments that are produced when it cuts. Figure Q5-60B illustrates how the DNA molecules in your experiment looked after separation according to size by using gel electrophoresis. Your experiment had a total of six samples, each of which was treated according to the legend below the gel. The sizes of the DNA fragments observed are indicated on the left side of the gel.

 

Figure Q5-60

 

1. Explain the results in lanes 1–4 and why it is important to have this information before you begin to test your remodeling complex.
2. What can you conclude about your purified remodeling complex from the results in lanes 5 and 6?

 

5-61     The N-terminal tail of histone H3 can be extensively modified, and depending on the number, location, and combination of these modifications, these changes may promote the formation of heterochromatin. What is the result of heterochromatin formation?

(a)        increase in gene expression

(b)        gene silencing

(c)        recruitment of remodeling complexes

(d)       displacement of histone H1

 

5-62     Methylation and acetylation are common changes made to histone H3, and the specific combination of these changes is sometimes referred to as the “histone code.” Which of the following patterns will probably lead to gene silencing?

(a)        lysine 9 methylation

(b)        lysine 4 methylation and lysine 9 acetylation

(c)        lysine 14 acetylation

(d)       lysine 9 acetylation and lysine 14 acetylation

 

5-63     When there is a well-established segment of heterochromatin on an interphase chromosome, there is usually a special barrier sequence that prevents the heterochromatin from expanding along the entire chromosome. Gene A, which is normally expressed, has been moved by DNA recombination near an area of heterochromatin. None of the daughter cells produced after this recombination event express gene A, even though its DNA sequence is unchanged. What is the best way to describe what has happened to the function of gene A in these cells?

(a)        barrier destruction

(b)        heterochromatization

(c)        epigenetic inheritance

(d)       euchromatin depletion

 

5-64     Which of the following best describes the mechanism by which chromatin-remodeling complexes “loosen” the DNA wrapped around the core histones?

(a)        They use energy derived from ATP hydrolysis to change the relative position of the DNA and the core histone octamer.

(b)        They chemically modify the DNA, changing the affinity between the histone octamer and the DNA.

(c)        They remove histone H1 from the linker DNA adjacent to the core histone octamer.

(d)       They chemically modify core histones to alter the affinity between the histone octamer and the DNA.

 

5-65     Which of the following is not a chemical modification commonly found on core histone N-terminal tails?

(a)        methylation

(b)        hydroxylation

(c)        phosphorylation

(d)       acetylation

 

5-66     How do changes in histone modifications lead to changes in chromatin structure?

(a)        They directly lead to changes in the positions of the core histones.

(b)        They change the affinity between the histone octamer and the DNA.

(c)        They help recruit other proteins to the chromatin.

(d)       They cause the histone N-terminal tails to become hyperextended.

 

5-67     Most eukaryotic cells only express 20–30% of the genes they possess. The formation of heterochromatin maintains the other genes in a transcriptionally silent (unexpressed) state. Which histone modification directs the formation of the most common type of heterochromatin?

(a)        H3 lysine 4 methylation

(b)        H3 lysine 9 methylation

(c)        H3 lysine 14 methylation

(d)       H3 lysine 27 methylation

 

5-68     Describe the mechanism by which heterochromatin can spread, once it has been established in one region of the chromosome.

 

5-69     The inactivation of one X chromosome is established by the directed spreading of heterochromatin. The silent state of this chromosome is _______________ in the subsequent cell divisions.

(a)        completed

(b)        switched

(c)        erased

(d)       maintained

 

5-70     Your friend is working in a lab to study how yeast cells adapt to growth on different carbon sources. He grew half of his cells in the presence of glucose and the other half in the presence of galactose. Then he harvested the cells and isolated their DNA with a gentle procedure that leaves nucleosomes and some higher-order chromatin structures intact. He treated the DNA briefly with a low concentration of M-nuclease, a special enzyme that easily degrades protein-free stretches of DNA. After removing all the proteins, he separated the resulting DNA on the basis of length. Finally, he used a procedure to visualize only those DNA fragments from a region near a particular gene called Sweetie or another gene called Salty. The separated DNA fragments are shown in Figure Q5-70. Each vertical column, called a lane, is from a different sample. DNA spots near the top of the figure represent DNA molecules that are longer than those near the bottom. Darker spots contain more DNA than fainter spots. The lanes are as follows:

 

1. “marker” containing known DNA fragments of indicated lengths
2. cells grown in glucose, DNA visualized near Sweetie gene
3. cells grown in galactose, DNA visualized near Sweetie gene
4. cells grown in glucose, DNA visualized near Salty gene
5. cells grown in galactose, DNA visualized near Salty gene

 

Figure Q5-70

 

1. The lowest spot (as observed in lanes 2, 4, and 5) has a length of about 150 nucleotides. Can you propose what it is and how it arose?
2. What are the spots representing longer lengths of DNA? Why is there a ladder of spots?
3. Notice the faint spots and extensive smearing in lane 3, suggesting the DNA could be cut almost anywhere near the Sweetie gene after growth of the cells in galactose. This was not observed in the other lanes. What probably happened to the DNA to change the pattern between lanes 2 and 3?
4. What kinds of enzymes might have been involved in changing the chromatin structure between lanes 2 and 3?
5. Do you think that gene expression of Sweetie is higher, lower, or the same in galactose compared to glucose? What about Salty?

 

 

answers

 

5-1       Cell-free extracts from S-strain cells of S. pneumoniae were fractionated to purify DNA, RNA, protein, and other cell components. Each fraction was then mixed with R-strain cells of S. pneumoniae. Its ability to change these into cells with pathogenic properties resembling the S-strain cells was tested by injecting the mixture into mice. Only the fraction containing DNA was able to transform the R-strain cells to pathogenic (or S-strain) cells that could kill mice.

 

5-2       (a)

 

5-3       (b)

 

5-4       (d)

 

5-5       (d)

 

5-6       (c)

 

5-7       A.        False. DNA is double-stranded. It is actually made of two polymers that are complementary in sequence.

1. True.
2. False. There are four different nucleotides that are used to make a DNA polymer: adenine, thymine, guanine, and cytosine. A fifth nucleotide, uracil, is found exclusively in RNA molecules, replacing thymine nucleotides in the DNA sequence.
3. False. Nucleotides are linked covalently through phosphodiester bonds. Hydrogen-bonding between nucleotides from opposite strands holds the DNA molecule together.

 

5-8       (a)

 

5-9       (d)

 

5-10     (b)

 

5-11     They prepared extracts from the infectious, pathogenic bacterial strain, and separated the different types of macromolecules (RNA, DNA, protein, lipids, and carbohydrates). Each of these materials was incubated separately with the noninfectious strain. The researchers were able to conclude that DNA was the “transforming principle” because it was the only macromolecule isolated from the pathogenic strain that was able to convert the noninfectious strain into an infectious one.

 

5-12     (d)

 

5-13     (d)

 

5-14     (c)

 

5-15     (d)

 

5-16     (a)

 

5-17     A.        See Figure A5-17 below.

Figure A5-17

 

1. The DNA base pairs are rotated with respect to each other. For a double-stranded DNA molecule with 10 base pairs, a full 360° rotation has occurred. This is referred to as one turn of the helix, and it can be seen in Figure A5-17A in the alignment of the A-T base pair at the bottom with the G-C base pair at the top.

 

5-18     (c)

 

5-19     See Figure A5-19 below.

Figure A5-19

 

 

5-20     A.        Adenine and guanine are purines; cytosine and thymine are pyrimidines.

1. Cytosine pairs with guanine, and adenine with thymine.

 

5-21     See Figure A5-21 below.

 

Figure A5-21

 

5-22     The order in which the DNA molecules would denature as the temperature increases is:

 

1—B; 2—D; 3—C; 4—A

 

All the DNA molecules are the same length, so only the A + T and G + C content determines their relative T_m. Molecules with higher G + C content will be more stable than molecules with a high A + T content. This is because there are three hydrogen bonds between each G-C base pair but only two between each A-T base pair. More energy (heat) is required to disrupt a larger number of hydrogen bonds.

 

5-23     A.        True.

1. False. Some sequences encode only RNA molecules, some bind to specific regulatory proteins, and others are sites where specific chromosomal protein structures are built (for example, centromeric and telomeric DNA).
2. False. Gene expression is the process of going from gene sequence to RNA sequence, to protein sequence.
3. False. This statement is false for two reasons. First, genes often contain intron sequences. Second, genes always contain nucleotides flanking the protein-coding sequences that are required for the regulation of transcription and translation.

 

5-24     (d)

 

5-25     (b)

 

5-26     (b)

 

5-27     5′-CTACCCACGAATGC-3′

 

5-28     A.        T_m 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.

1. The T_m 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.
2. Inserting values into the equation in part B gives T_m = 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.

 

5-29     A.        It is better to separate the strands and each take a single strand, because all of the information found in the original molecule is preserved in a full-length single strand but not in a half-length double-stranded molecule.

1. Watson and Crick meant that the complementary base-pairing of the strands allows a single strand to contain all of the information necessary to direct the synthesis of a new complementary strand.

 

5-30     Because there are 20 amino acids used in proteins, each amino acid would have to be encoded by a minimum of three nucleotides. For example, a code of two consecutive nucleotides could specify a maximum of 16 (4²) different amino acids, excluding stop and start signals. A code of three consecutive nucleotides has 64 (4³) different members and thus can easily accommodate the 20 amino acids plus a signal to stop protein synthesis.

 

5-31     A.        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.

1. Because 4⁶ (= 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.
2. 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.

 

5-32     In eukaryotic chromosomes, DNA is complexed with proteins to form chromatin. The paternal and maternal copies of human Chromosome 1 are homologous, whereas the paternal copy of Chromosome 1 and the maternal copy of Chromosome 3 are nonhomologous. Cytogeneticists can determine large-scale chromosomal abnormalities by looking at a patient’s karyotype. Fluorescent molecules can be used to paint a chromosome, by a technique that employs DNA hybridization, and thereby to identify each chromosome by microscopy.

 

5-33     A.        A gene is a segment of DNA that stores the information required to specify the particular sequence found in a protein (or, in some cases, the sequence of a structural or catalytic RNA).

1. A similar genome size indicates relatively little about the number of genes and virtually nothing about the number of chromosomes. For example, the commonly studied yeasts Saccharomyces cerevisiae (Sc) and Schizosaccharomyces pombe (Sp) are separated by roughly 400 million years of evolution, and both have a genome of 14 million base pairs. Yet Sc has 6500 genes packaged into 16 chromosomes and Sp has 4800 genes in 3 chromosomes.

 

5-34     (d)

 

5-35     (b)

 

5-36     (c)

 

5-37     A.        2 × 10¹⁴ m = 124,274,238,447 miles.

1. The Earth’s circumference at the Equator is 24,902 miles. The length of DNA from the average human body could wrap around the Earth 4,990,532 times.
2. The average distance from the Earth to the Sun is 93,000,000 miles. So, the round-trip distance is 186,000,000 miles. The length of DNA from the average human body could stretch from the Earth to the Sun and back 668 times.
3. The distance from the Earth to Pluto is, on average, about 39 × 93,000,000 miles. So, the round trip distance is 78 × 93,000,000 miles. The length of DNA from the average human body could stretch from the Earth to Pluto and back 17 times.

 

5-38     (c)

 

5-39     A.        There are 46 centromeres per cell, one on each chromosome. The centromeres have a key role in the distribution of chromosomes to daughter cells during mitosis.

1. There are 92 telomeres per cell, two on each chromosome. Telomeres serve to protect the ends of chromosomes and to enable complete replication of the DNA of each chromosome all the way to its tips.
2. There are far more than 200 replication origins in a human cell, probably about 10,000. These DNA sequences direct the initiation of DNA synthesis needed to replicate chromosomes.

 

5-40     Chromosome painting relies on the specificity of DNA complementarity. Because unique sequences for each chromosome are known, short DNA molecules matching a set of these sites can be designed for each chromosome. Each set is labeled with a specific combination of fluorescent dyes and then allowed to hybridize (form base pairs) with the two homologous chromosomes that contain the unique sequences being targeted.

 

Giemsa stain is a nonfluorescent dye that has a high affinity for DNA, and specifically accumulates in regions that are rich in A-T nucleotide pairs. This dye produces a pattern of dark and light bands, which differ for each chromosome on the basis of the distribution of AT-rich regions.

 

5-41     (d)

 

5-42     A.        False. There are several examples of closely related species that have a drastically different number of chromosome pairs. Two related species of deer—Chinese and Indian muntjac—have 23 and 3, respectively.

1. True.
2. True.
3. False. The telomere is a specialized DNA sequence, but not for the attachment of spindle microtubules. Telomeres form special caps that stabilize the ends of linear chromosomes.

 

5-43     Each chromosome is a single molecule of DNA whose extraordinarily long length can be compacted by as much as 1000-fold during interphase and tenfold more during mitosis. This is accomplished by binding to proteins that help package the DNA in an orderly manner so it can fit in the small space delimited by the nuclear envelope. The structure of the DNA–protein complex, called chromatin, is highly dynamic over time.

 

5-44     (a)

 

5-45     (b)

 

5-46     (a)

 

5-47     (a)

 

5-48     (b)

 

5-49     A.        Nucleosomes are present in eukaryotic chromosomes, but not in prokaryotic chromosomes.

4. A nucleosome contains two molecules each of histones H2A and H2B as well as of histones H3 and H4.
5. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately twice.
6. Nucleosomes are aided in their formation by the high proportion of basic amino acids in histone proteins.
7. Nucleosome formation compacts DNA into approximately one-third of its original length.

 

5-50     (a)

 

5-51     A.        In a test tube, the nucleosome core particle can be released from chromatin by treatment with a nuclease that degrades the exposed, linker DNA, but not the DNA wrapped around the nucleosome core.

1. The core nucleosome was revealed to contain two molecules of the histones H2A, H2B, H3, and H4, as well as a 147-base-pair (bp) fragment of DNA.
2. Nuclease treatment degrades linker DNA, which can be up to 80 bp in length. This region of DNA is typically bound to linker histones (H1), which are involved in higher-level packing of the chromatin.

 

5-52     (c)

 

5-53     Interphase chromosomes contain both darkly staining heterochromatin and more lightly staining euchromatin. Genes that are being transcribed are thought to be packaged in a less condensed type of euchromatin. Nucleosome core particles are separated from each other by stretches of linker DNA. A string of nucleosomes coils up with the help of histone H1 to form the more compact structure of the 30-nm fiber. A zigzag model describes the structure of the 30-nm fiber. The 30 nm chromatin fiber is further compacted by the formation of loops that emanate from a central axis.

 

5-54     (d)

 

5-55     (c)

 

5-56     (b)

 

5-57     The DNA double helix in heterochromatin may be so tightly packed and condensed that it is inaccessible to the proteins that bind replication origins, including the DNA replication machinery. It may take extra time to remodel the chromatin to make it more accessible to the proteins required to initiate and perform DNA replication.

 

5-58     A.        False. When the core nucleosome is analyzed, it is revealed that there are H2A/H2B tetramers and H3/H4 tetramers in solution. Each of the tetramers has two subunits of the respective histone proteins.

1. True.
2. False. Histones have a higher number of lysines and arginines than most proteins. These amino acids are positively charged and help to increase the nonspecific affinity between the histones and the negatively charged phosphates in the DNA backbone.
3. False. When cells enter mitosis, the interphase chromosomes undergo at least one more level of packaging, which facilitates the segregation of sister chromatids.

 

5-59     (c)

 

5-60     A.        Sample 1 confirms the location of the EcoRI restriction site and shows what those fragments should look like when separated on the gel. The cleavage is the readout that will tell us whether the remodeling complex is working. Sample 2 demonstrates that DNA that is not assembled into nucleosomes can be cut into many small fragments by DNaseI. That is why we do not see discrete bands. Sample 3 demonstrates that when nucleosomes are assembled on the DNA, DNase I cuts only in one place, presumably in the linker region between two assembled nucleosomes. Sample 4 demonstrates that EcoRI cannot access its cleavage site when nucleosomes are assembled over it.

1. Sample 5 demonstrates that our purified complex is working. It must be moving the nucleosomes, providing access for EcoRI, such that it can now cleave at its restriction site, which was not possible in the absence of NICRC. Sample 6 shows that NICRC will function only if ATP is available as an energy source for the remodeling process.

 

5-61     (b)

 

5-62     (a)

 

5-63     (c)

 

5-64     (a)

 

5-65     (b)

 

5-66     (c)

 

5-67     (b)

 

5-68     Once the initial H3 lysine 9 methylation is established on core histone octamers in one region, the modification attracts a specific set of proteins and other histone-methylating enzymes. These enzymes create the same modification on adjacent histone octamers, which continue to recruit more heterochromatin-specific proteins and enzymes, creating a wave of heterochromatin spreading along the chromosome.

 

5-69     (d)

 

5-70     A.        The lowest spot represents DNA of a length similar to that of the segment of DNA found in a nucleosome core particle. Partial digestion with an enzyme such as M-nuclease causes breaks in the DNA backbone primarily within the linker DNA or other DNA segments not bound tightly to histones. Thus, this band probably comprises the DNA bound tightly to a single histone octamer and it arose by cutting the linker DNA outside a single nucleosome core particle.

1. The ladder of bands representing longer lengths of DNA probably corresponds to stretches of DNA associated with increasing numbers of nucleosomes (1, 2, 3, 4, 5, and so on). In support of this proposal, adjacent bands differ in size by roughly 200 nucleotides, which is the length of DNA found in a nucleosome core particle plus neighboring linker DNA. This interpretation must mean that the M-nuclease digestion did not go to completion, because if all non-nucleosomal DNA were digested completely, the samples would contain only the 150-base-pair fragment.
2. Given the ability of M-nuclease to cut anywhere near Sweetie after growth in galactose, it seems that the DNA is no longer protected from digestion by binding to histones. Perhaps the wrapping of DNA within the nucleosomes has been loosened considerably. This change in the nucleosomes must be specific to the Sweetie gene, because it is not seen at the Salty gene or throughout the genome.
3. The main candidates for enzymes that catalyzed the nucleosome alterations near Sweetie are chromatin-remodeling complexes and enzymes that covalently modify histone tails with methyl, acetyl, or phosphate groups.
4. As the chromatin seems to have been loosened near Sweetie, it seems likely that Sweetie gene expression is increased when cells are grown in galactose rather than glucose, whereas Salty gene expression is likely to be the same under the two conditions. Perhaps the Sweetie gene contains instructions for a protein that is required for cells to metabolize galactose but not glucose.