Ch 8 - DNA: The Chemical Nature of the Gene
Which relation will be found in the percentages of bases of a double-stranded DNA molecule? (A + G)/(C + T) = 1.0 A/T = G/C A/G = C/T A + C = G + T A + T = G + C
(A + G)/(C + T) = 1.0 A/T = G/C A + C = G + T To determine if these equations are true, assign number values to the variable representing each nucleotide. For example, if there is a piece of double-stranded DNA five nucleotides long, there are ten nucleotides in that piece of DNA. If two of them are adenine, then two of them are thymine. This means three nucleotides will be cytosine and three nucleotides will be guanine. With numbers assigned to each nucleotide, you can test if the equations are true. A/T=C/G is true because (2/2)=1 and (3/3)=1. (A+G)/(C +T)=1.0 is true because (2 + 3) / (2 + 3) = 1. A+C=G+T is true because (2+3)=(2+3). A+T=G+C is not true because 2+2 does not equal 3+3. A/G=C/T is not true because (2/3) does not equal (3/2).
An analysis is performed to determine the proportions of each of the four nucleotide bases in the DNA of several tissue samples from various species. The results appear in the table. % Adenine(A) % Guanine(G) % Thymine(T) % Cytosine(C) Human. 31. 20. 30. 19 Chicken. 28. 21. 29. 22 Yeast. 32. 18. 33. 17 E. coli. 26. 25. 24. 25 Which conclusions can be drawn from this data? The proportion of A-T base pairs is roughly equal to the proportion of G-C base pairs.
- The proportion of pyrimidines is roughly equal to the proportion of purines. - The proportion of A is roughly equal to the proportion of T. - The proportion of G is roughly equal to the proportion of C. The biochemist Erwin Chargaff studied the chemical composition of DNA. His most important finding was that in any sample of double‑stranded DNA, the amount of adenine (A) always equaled the amount of thymine (T), and the amount of guanine (G) always equaled the amount of cytosine (C). More generally, Chargaff found that the total purine content (A + G) always equaled the total pyrimidine content (C + T) in double‑stranded DNA. This finding implied that the bases were half purine and half pyrimidine. These relationships between the proportions of the four bases in double‑stranded DNA have come to be known as Chargaff's rules. This insight was later incorporated into the Watson-Crick double helix model of DNA, in which every A is hydrogen bonded to a T, and every G to a C.
According to Chargaff's rules, what is the likely ratio for G/C nucleotides in a given genome? 0.25 0.75 1 0.50
1 The base composition of guanine is equal to cytosine so the ratio would be 1.
Which of the eukaryotic DNA sequences would likely be in a gene family together? Choose the most complete answer. (1) 5' ATCGTAAGCTA 3' (2) 5' GCTCCCCA 3' (3) 5' ATCGTATCGTA 3' (4) 5' ATTTTTTTGGGGG 3' (5) 5' ACCGTAAGGTT 3' 2, 5 1, 2, 3 1, 3, 4 1, 3, 5
1, 3, 5 Genes within unique-sequence DNA are present in several similar, but not identical, copies and are collectively referred to as a gene family.
What is the width, or diameter, of B-form DNA?
2 nm
What is the approximate average length of DNA from the beginning of one chromosome to the beginning of the next chromosome?
200 bp
A diploid human cell contains approximately 6.4 billion base pairs of DNA. How many histones are present in such a cell? Assume that the linker DNA encompasses 40 base pairs.
288 million Each nucleosome contains two of each of the four types of histones, and a nucleosome plus one molecule of histone H1 constitute the chromatosome.
Enter the complementary sequence to the DNA strand shown. 5′- TGACGTGAT -3′
3'-ACTGCACTA-5' To determine the complementary sequence to the DNA strand 5′- TGACGTGAT -3′, match each base in the original DNA strand with its complementary base. In DNA base pairing, adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C). The original DNA strand and its complementary sequence are shown. 5′- TGACGTGAT -3′ 3′- ACTGCACTA -5′
What would the sequence of the complementary DNA strand be for the strand 5'-ATGCTAGGC-3'? 5'-TACGATCCG-3' 5'-ATGCTAGGC-3' 5'-AUGCUAGGC-3' 3'-TACGATCCG-5'
3'-TACGATCCG-5' This is an accurate depiction of the complementary strand.
A diploid human cell contains approximately 6.4 billion base pairs of DNA. How many nucleosomes are present in such a cell? Assume that the linker DNA encompasses 40 base pairs. 32 billion 32 million 288 million 200
32 million Each nucleosome contains 140 bp associated with the core histone octamer, another 20 bp for histone H1, and 40 bp in the linker region.
What is the average length of linker DNA?
45 bp
Each sequence indicates one of two strands of a DNA molecule. When heated, which molecule would separate first? 5'-GGGGGGGGGG-3' 5'-ACACAACAAA-3' 5'-GACGAACCCC-3' 5'-ATAATAACAT-3'
5'-ATAATAACAT-3' The more A-T bonds, the easier it is to separate due to their two bond base activity.
Which of these sequences will most likely yield a Z-DNA structure configuration? 5'-GCGAATAACGCTTATG-3' 5'-GGGCCCGGTACCCGGG-3' 5'-ATTAATAGGCTAGCAT-3' 5'-GCGTTAAATGCTAGAT-3'
5'-GGGCCCGGTACCCGGG-3' This sequence has a high concentration of C and G sequences, so it would be the most likely to yield a Z-DNA configuration.
Each sequence indicates one of two strands of a DNA molecule. The two strands of which molecule would be held together most tightly? 5'-GAGAGAGAGA-3' 5'-GGGGGGGGGG-3' 5'-ATATATATAT-3' 5'-AAAAAAAAAA-3'
5'-GGGGGGGGGG-3' This molecule is held together by 30 hydrogen bonds.
Which form of DNA can occur when less water is present and is shorter and wider than other forms? X Z B A
A A-DNA is a structure forming a right-handed helix when less water is present.
The bacteria Streptococcus pneumoniae has a virulent S strain and a nonvirulent R strain. The S strain is lethal to mice. The S strain contains a chemical factor that can transform the R strain to be virulent. The diagram shows a series of experiments conducted by injecting combinations of these strains into mice to identify the transforming factor. A \ R strain (nonvirulent) B \ S strain (virulent) C \ Heat-treated S strain D \ R strain and heat-treated S strain E \ R strain and heat-treated S
A = alive B = dead C = alive D = dead E = dead F = alive This series of experiments demonstrates that DNA contains biological information that can be passed from one organism to another. Griffith discovered a transforming factor that held information to change the nonvirulent Streptococcus pneumoniae R strain into the virulent S strain. Avery, MacLeod, and McCarty built on the early work of Griffith to identify DNA as the molecule that stores biological information. Mice injected with the R strain are not infected and will survive, whereas mice injected with the S strain are infected and will die. Heat treatment kills live bacteria by breaking them open, rupturing, and releasing the cellular components. The heat‑treated S strain alone is not infectious, and mice will survive because the injected bacteria are all dead. When live R strain is combined with the heat‑treated S strain, the infected mice will die. An undetermined transforming factor in the cellular components of the hea
Which statement is true? All of the statements are true. Chromosomes never exist as naked, B-form DNA in a cell The length of a mitotic chromosome is less than the width of the average cell nucleus The chromosome is made up of DNA and protein The total length of naked DNA in any given chromosome far exceeds the width of an average cell nucleus.
All of the statements are true.
How does bacterial DNA differ from eukaryotic DNA? Eukaryotic DNA is complexed with histone proteins and is linear. Eukaryotic DNA is not complexed with histone proteins and is circular. Bacterial DNA is not complexed with histone proteins and is circular. Bacterial DNA is complexed with histone proteins and is linear.
Bacterial DNA is not complexed with histone proteins and is circular Unlike eukaryotic DNA, bacterial DNA is not attached to histone proteins and consists of a single circular DNA molecule.
Which base pair has the greatest number of hydrogen bonds? C-G C-T A-G A-T
C-G This base pairing has three hydrogen bonds.
Hershey and Chase demonstrated that _____ of a bacteriophage enters a host bacterial cell. protein lipid DNA RNA
DNA Hershey and Chase confirmed the utility of DNA as source material versus proteins in their experiment with phages.
The diagram depicts the molecular structure of DNA. Label the diagram with the names of the components of the nucleic acid.
DNA is a polymer composed of nucleotide monomers and is known as a polynucleotide. Each nucleotide consists of a five‑carbon sugar called deoxyribose, a phosphate group, and one of the nitrogen‑containing bases: adenine (A), guanine (G), thymine (T), or cytosine (C). Deoxyribose is the pentose sugar found in DNA. Its name indicates that it is a deoxy‑sugar, meaning that it is derived from the sugar ribose but has lost an oxygen atom. The absence of the oxygen atom in deoxyribose is responsible for the increased mechanical flexibility of DNA compared to RNA, which allows DNA to assume the double helix conformation. The phosphate group is part of the DNA backbone and bonds the deoxyribose groups together. The T and C nucleotides are single‑ring structures called pyrimidines, and the A and G nucleotides are larger, double‑ring structures called purines. In a DNA double helix, purines form hydrogen bonds to pyrimidines so that A bonds only to T with two hydrogen bonds, and C bo
The diagram depicts the general structure of DNA, with a single nucleotide circled. Label the diagram with the names of the three components of a nucleotide. Deoxyribose Phosphate group Nitrogenous base
DNA is a polymer of nucleotide monomers, and the polymer is known as a polynucleotide. Each nucleotide consists of a five‑carbon sugar called deoxyribose, a phosphate group, and one of the nitrogen‑containing bases: adenine (A), guanine (G), thymine (T), or cytosine (C). Deoxyribose is the pentameric sugar group found in DNA. As its name indicates, it is a deoxy‑ sugar, meaning that it is derived from the sugar ribose but has lost an oxygen atom. The absence of the oxygen atom in deoxyribose is responsible for the increased mechanical flexibility of DNA compared to RNA, which allows DNA to assume the double‑helix conformation. The phosphate group is part of the DNA backbone and bonds the deoxyribose groups together. The T and C nucleotides are single‑ring structures called pyrimidines, and the A and G are larger, double‑ring structures called purines. In a DNA double helix, purines form hydrogen bonds with pyrimidines, where A bonds only to T with two hydrogen bonds, and
Which of these is generally NOT true about epigenetic effects? Epigenetic effects are more easily reversed than genetic changes. Epigenetic effects are rare and do not affect very many traits. Epigenetic effects are often influenced by environmental factors. Epigenetic effects often involve changes in methylation of DNA bases.
Epigenetic effects are rare and do not affect very many traits. Scientists have discovered a wide variety of epigenetic effects in a large number of organisms.
Classify each description as applying to either heterochromatin or euchromatin. The form chromatin takes most often when transcription is not occurring. The expanded form of chromatin The highly compressed form of chromatin The form chromatin takes most often during transcription
Heterochromatin: - The form chromatin takes most often when transcription is not occurring. - The highly compressed form of chromatin Euchromatin: - The form chromatin takes most often during transcription - The expanded form of chromatin In eukaryotes, chromosomes exist as a complex of DNA and associated proteins. This complex is called chromatin. Chromatin may take on two different forms: heterochromatin and euchromatin. Euchromatin, or "true" chromatin, is the less compressed or less dense form of chromatin. Transcription of genes usually takes place when chromatin exists as euchromatin because the DNA is more available for transcription. Conversely, heterochromatin is a highly compressed or dense form of chromatin. Transcription of genes does not usually take place when chromatin exists as heterochromatin because the DNA is not readily available for transcription. Euchromatin and heterochromatin may be distinguished using staining. Euchromatin appears lighter because it is m
What is the function of histone H1?
Histone H1 clamps the DNA joining and leaving the nucleosome to the core protein octomer
Label the indicated components of the DNA double helix
In 1953, James Watson and Francis Crick proposed the double helix structure of DNA. This model incorporated the X‑ray diffraction data obtained by Rosalind Franklin as well as the nitrogenous base composition data obtained by Erwin Chargaff. The Watson-Crick model of DNA has two strands that wrap around each other, forming a pair of spirals, or helices. The two DNA helices are slightly offset from each other so that the two gaps between the strands are not exactly the same width. The wider of the two gaps is called the major groove, whereas the narrower gap is called the minor groove. The major groove exposes between 6 to 10 bases of DNA and is thus an important location for enzymes and other proteins to recognize DNA sequences without opening the double helix. Each strand is a polymer of nucleotides, which are chemical units composed of a nitrogenous base, the sugar deoxyribose, and a phosphate group. Covalent phosphodiester bonds link the nucleotides. The alternating deoxyribos
Label the components on the diagram of DNA structure by using all four of the provided terms. DNA molecule Chromosome Protein Chromatin
In humans, each cell contains approximately five feet of DNA. Because DNA is such a long molecule, it has to be compressed in order to fit into the tiny nucleus of a cell and for cell division to occur properly. The blue thread running through the diagram represents the long DNA molecule. A portion of the thread has been enlarged to show the double‑helix structure of the molecule in the right portion of the diagram. The level of DNA compression varies between a cell that is about to divide and a cell that is not about to divide. In cells, the DNA double helix is wrapped tightly around small proteins called histones. Individual histone proteins are represented by the pink globes in the diagram. This association of DNA and proteins is called chromatin. When viewed under a microscope, chromatin resembles beads on a string, as shown in the center of the diagram. Within the cell, chromatin becomes tightly wound into a thicker, helical fiber. This fiber then loops and folds in on itself,
Injection by Griffith of which sample of Streptococcus pneumonia yielded surprising results that indicated that transformation had occurred? living IIR + dead IIIS dead IIIS living IIIS (disease-causing) living IIR (not disease-causing)
Living IIR + dead IIIS This injection produced pneumonia due to transformation of IIR cells into IIIS cells by taking up DNA from the dead IIIS cells.
Which statement best describes chromatin associated with actively transcribed genes as compared to chromatin associated with genes which are not being actively transcribed? more compacted and less acetylated more compacted and more acetylated more relaxed and more acetylated more relaxed and less acetylated
More relaxed and more acetylated Acetylation of histone tails causes the histones to loosen their grip on DNA so that the genetic information is more accessible to the transcriptional machinery.
If Avery and his colleagues found that RNA was the transforming material, what would the results of their experiment look like? Only type IIR bacteria would be isolated from the RNase-treated filtrate. Only type IIR bacteria would be isolated from the DNase-treated filtrate. Only type IIIS bacteria would be isolated from the RNase-treated filtrate. Only type IIR bacteria would be isolated from the protease-treated filtrate.
Only type IIR bacteria would be isolated from the RNase-treated filtrate. Once IIR is isolated, it can be seen that the strain is now virulent with RNA as the transforming material.
Boris Magasanik collected data on the proportion of each base in RNA from different species. To compare the relative amount of each base among the sources, he standardized the value of adenine to 10. Relative amounts of the other bases were calculated based on this value for adenine. The data are shown: Organism & tissue: A : G : C : U Rat liver nuclei 10: 14.8: 14.3: 12.9 Rabbit liver nuclei. 10: 13.6: 13.1: 14.0 Cat brain 10: 14.7: 12.0: 9.5 Carp muscle. 10: 21.0
Organism and tissue : A+G/C+U Rat live nuclei : 10+14.8/14.3+12.9 = 0.911 Rabbit liver nuclei: 10+13.6/13.1+14.0 = 0.870 Cat brain: 10+14.7/12.0+19.5 = 1.14 Carp muscle: 10+21.0/19.0+11.0 = 1.03 Yeast: 10+12.0/8.0+9.8 = 1.235 The ratio of purines to pyrimidines is dependent of the species Nucleotides are divided into two categories based upon their chemical structure. Adenine and guanine are purines, whereas pyrimidines include thymine, cytosine, and uracil. Begin by calculating the ratios of purines to pyrimidines in each of the tissues. For RNA, use the equation ratio=A+G/C+U For rat liver nuclei, add the 10 adenine to the 14.8 guanine and divide the total by the sum of the 14.3 cytosine plus the 12.9 uracil. Thus, rat liver nuclei has a ratio of purines to pyrimidines of 0.91. For rabbit liver nuclei, add the 10 adenine to the 13.6 guanine and divide the total by the sum of the 13.1 cytosine plus the 14.0 uracil. Thus, rabbit liver nuclei has a ratio of purines to pyrimidines
What is the difference between positive and negative supercoiling?
Positive supercoiling is overrotation and negative supercoiling is underrotation Supercoiling is a tertiary structure of DNA that forms when the DNA is overrotated or underrotated. The change in rotation occurs during the normal cellular processes of DNA replication and repair as proteins interact to separate the strands of the double helix. Even linear DNA becomes overrotated or underrotated due to secondary and tertiary structures that cause it to act more similar to a circular chromosome without free ends. The end result is a tertiary structure that loops back onto itself in a figure eight pattern. Topoisomerase relieves the stress of supercoiling and returns the DNA helix to a relaxed state. The DNA double helix has a relaxed state of about 10 bp per turn. When the helix is overrotated tighter and has fewer base pairs per turn, the DNA has positive supercoiling. Conversely, when the helix is underrotated and has more base pairs per turn, the DNA has negative supercoiling. Negat
In Hershey and Chase's experiment, they noticed that they could recover little to no radioactive sulfur from the infected bacteria. What did this suggest? Protein from the phage did not enter the cell and, therefore, could not be the genetic material. The phage did not infect the bacteria at all. DNA from the phage entered the cell and was therefore the genetic material. DNA from the phage did not enter the cell, so DNA could not be the genetic material.
Protein from the phage did not enter the cell and, therefore, could not be the genetic material. Sulfur is indicative of protein so, if there was little to none recovered, it can be assumed that protein did not enter the cell.
Which of the statements describes purines and pyrimidines in DNA molecules Pyrimidines form covalent bonds with purines Pyrimidines consist of a one-ring structure. Pyrimidines form hydrogen bonds with purines. Adenine and guanine are pyrimidines. Purines form hydrogen bonds with purines.
Pyrimidines consist of a one-ring structure. Pyrimidines form hydrogen bonds with purines. DNA molecules are composed of monomer subunits known as nucleotides. Nucleotides in DNA are composed of a phosphate group, a deoxyribose sugar, and a nitrogenous base. There are four bases found in DNA: adenine (A), guanine (G), thymine (T), and cytosine (C). Adenine and guanine are known as purines. Purines are composed of a double‑ring structure and are larger than pyrimidines, which are composed of a single‑ring structure. Thymine and cytosine are the pyrimidines in DNA. A DNA double helix is composed of two antiparallel strands of DNA. The bases in each DNA molecule are able to form hydrogen bonds to a complementary base, which stabilizes the double helix structure. The complement of a purine is always a pyrimidine. For example, adenine pairs with thymine, and guanine pairs with cytosine. The number of hydrogen bonds that form depend on which bases are interacting. Adenine and thymi
In Hershey and Chase's experiment, they found that progeny phages contained radioactive phosphorous. What did this suggest? Radioactive phage protein entered the bacterial cell but did not become incorporated into the progeny. Radioactive phosphorous was the genetic material. Radioactive phage DNA not only entered the bacterial cell but was also passed on to the phage progeny. Phage progeny pick up material from the environment.
Radioactive phage DNA not only entered the bacterial cell, but was also passed on to the phage progeny. The presence of genetic traits in the progeny is the result of inheritance from one generation to the next.
Which of these statements is TRUE concerning DNA's capacity for stacking? The degree of stacking doesn't affect the secondary or tertiary structure of the DNA molecule. Stacking assesses the binding forces of opposite bases parallel to one another. Stacking can create several types of secondary structures varying in prevalence from one organism to the next. The form of stacking is not affected by the concentration of bases in the DNA molecule.
Stacking can create several types of secondary structures varying in prevalence from one organism to the next. Stacking allows for different configurations to be sustained especially with the variation in base composition in DNA.
DNA had unique properties that allow it to accurately retain genetic information, even after multiple rounds of replication. One aspect of DNA that allows it to accurately store genetic information is the base pairing from Chargaff's first rule of the four nucleotide bases. If the A content of a DNA molecule is 22%, what are the percentages of the remaining bases?
T = 22 G = 28 C = 28 Chargaff's first rule states that in DNA, the amount of purines will always equal the amount of pyrimidines. Specifically, the amount of the purine adenine (A) will always equal the amount of the pyrimidine thymine (T), and the amount of the purine guanine (G) will equal that of the pyrimidine cytosine (C). Chargaff's first rule holds true because DNA has complementary base pairing. This means that A always pairs with T and G always pairs with C. For example, if A is found at 22%, then T will also be found at 22%. The total amount of all four bases must equal 100%; therefore, since the total percentage of A and T is 44%, there is 56% remaining. The 56% is split evenly between the complementary bases G and C to comply with Chargaff's first rule. The nucleotides G and C are each found at 28%.
What are telomeres? Enzymes that add new nucleotides to template DNA strands during replication Sections of repetitive DNA sequences located anywhere within a eukaryotic DNA strand The region of sister chromatids where the sister chromatids are bound together. The DNA located at the ends of eukaryotic DNA molecules
The DNA located at the ends of eukaryotic DNA molecules Telomeres are the DNA located at the ends of eukaryotic DNA molecules. Telomeres, which do not contain genes, consist of repetitive nucleotide sequences. Telomeres prevent degradation of the coding ends of linear chromosomes that can occur through the process of DNA replication. During replication, DNA polymerase completes the synthesis of the leading strand of DNA. However, because DNA polymerase can only add nucleotides to the 3′ end of a DNA molecule, the synthesis of the lagging strand is incomplete. Once DNA polymerase synthesizes the last Okazaki fragment on this strand, a 100−200 base‑long region of the strand remains unreplicated, shortening the chromosome. The presence of telomeres at the ends of eukaryotic chromosomes protects the coding regions of the DNA molecules by preventing the coding regions from being shortened during replication; rather, the non‑coding DNA of the telomeres is removed. In some types
If Levene's tetranucleotide hypothesis was correct, what nucleotide ratios would Chargaff find? The amount of guanine was only equal to the amount of cytosine. The amount of all the nucleotides would have been the same. The amount of adenine was only equal to the amount of thymine. The combined amount of adenine and cytosine was not equal to the combined amount of thymine and guanine.
The amount of all the nucleotides would have been the same. This would demonstrate the lack of variability that was proposed by the tetranucleotide hypothesis.
Which of these is NOT a key feature of a centromere? The centromere is located at the ends of a chromosome. Most centromeres are defined by epigenetic modifications. There are no specific sequences that are found in all centromeres. Most of the centromere is made up of heterochromatin.
The centromere is located at the ends of a chromosome. The centromere is located in the middle of a chromosome.
Suppose a chemist develops a new drug that neutralizes the positive charges on the tails of histone proteins. What would be the most likely effect of this new drug on chromatin structure? It would cause the formation of epigenetic changes. The histones and DNA would not be tightly associated. The histones and DNA would be tightly associated. It would cause the formation of chromosome puffs.
The histones and DNA would not be tightly associated. The positive tails of histones are attracted to the negatively charged DNA molecule.
DNA molecules of different sizes are often separated with the use of a technique called electrophoresis. With this technique, DNA molecules are placed in a gel, an electrical current is applied to the gel, and the DNA molecules migrate toward the positive (+) pole of the current. What aspect of its structure causes a DNA molecule to migrate toward the positive pole? The deoxyribose sugar had a negative charge The phosphate groups of the backbone have a positive charge The phosphate groups o
The phosphate groups of the backbone have a negative charge Gel electrophoresis relies upon the charge of DNA to cause it to migrate through the gel in response to an electric current. DNA contains phosphate groups that form the DNA backbone and have a negative charge. When an electric current is introduced, these negatively charged phosphate groups cause the DNA to migrate towards the positive pole. DNA also contains deoxyribose sugars but deoxyribose is not charged and is not affected by an electric current. Like charges repel each other, so a positively charged structure would not migrate towards the positive pole but would instead migrate away from it.
Which statement about C-values is TRUE? There is not a strong correlation between C-value and organismal complexity among eukaryotes. Organisms with very high C-values do not have very many highly repeated DNA sequences. Typically, bacteria have a higher C-value than eukaryotes. Humans have a higher C-value than salamanders.
There is not a strong correlation between C-value and organismal complexity among eukaryotes. Most of the variation in C-value among eukaryotes is due to vast differences in the amount of highly repetitive DNA sequences that do not seem to contribute to structural complexity.
In Fraenkel-Conrat and Singer's experiments using the tobacco mosaic virus (TMV), what result led them to believe that RNA was the virus's genomic material? They were able to make hybrid TMV with type A RNA with type B protein and vice versa. They were able to separate TMV RNA from protein. There were two different types of TMV. Hybrid TMV infected tobacco leaves.
There were two different types of TMV. Since the virulent strain was reflective of one type of TMV, it can be assumed that strain carries virulence.
Which of these is NOT a characteristic of DNA sequences at the telomeres? They are located at the center of a chromosome. They consist of repeated sequences. One strand consists of guanine and adenine (or thymine) nucleotides. One strand protrudes beyond the other, creating some single-stranded DNA at the end.
They are located at the center of a chromosome. Telomeres are located at the ends of a chromosome.
Which of these statements about highly repetitive DNA sequences is incorrect? They tend to be present in hundreds of thousands to millions of copies that repeat and cluster in certain regions of the chromosomes. They are sometimes referred to as satellite DNA due to the behavior during centrifugation. They are rarely transcribed into RNA molecules. They tend to contain protein coding sequences in tandem repeats.
They tend to contain protein coding sequences in tandem repeats. Protein coding sequences are found in unique-sequence DNA.
Which of these descriptions of the arrangement of DNA in eukaryotic cells is most accurate? tightly packed and static not tightly packed and static not tightly packed and dynamic tightly packed and dynamic
Tightly packed and dynamic DNA must be tightly packed to fit within a cell, but the packing must be reduced sometimes to allow for access to the genetic information contained within the DNA.
Stanley is sequencing a new species in the genetics laboratory. He determines to note his findings in order to uncover the structure that he is observing. What type of DNA configuration can be assumed based on the data? DNA parameters Observation Type of helix left-handed Helical diameter 1.6 nm Base orientation angled toward from main axis A-DNA Z-DNA Y-DNA B-DNA
Z-DNA Z-DNA is the only left-handed helix to note and meets the criteria for the data presented.
Comparing the C-values of which two organisms best illustrates the C-value paradox? human and Drosophila melanogaster (insect) Zea Mays corn and humans Arabidopsis thaliana (plant) and Amphiuma (salamander) E. coli (bacterium) and yeast
Zea Mays corn and humans The genome of corn is actually larger than the genome of humans.
What would be the most likely unit of measurement for a C-value? grams base pairs meters milliliters
base pairs The size of a genome is measured in terms of base pairs, which is considered the unit of measurement in DNA.
What is the primary level of DNA structure?
double-stranded B-form DNA
What drives the formation of the 30 nm fiber of chromatin structure?
histone protein interactions
Heating a double-stranded DNA molecule will cause chemical bonds to break. Which bonds will break first, resulting in separation of the DNA molecule? covalent bonds formed between base pairs hydrogen bonds between bases on opposite strands hydrogen bonds between the sugar and phosphate of the adjoining nucleotides on the same strand phosphodiester bonds of the sugar phosphate backbone
hydrogen bonds between bases on opposite strands Hydrogen bonds are weaker bonds especially between bases of distance.
SINEs and LINEs are examples of which type of eukaryotic DNA sequence? unique-sequence DNA highly repetitive DNA tandem repeat sequences interspersed repeat sequences
interspersed repeat sequences Interspersed repeat sequences are a type of moderately repetitive DNA that are dispersed throughout the genome and are nonadjacent.
When Griffith injected mice with living IIR (not disease-causing) and heat-killed IIIS (disease-causing) bacteria, he found that the mice died of pneumonia. What kind of bacteria did he then isolate from the dead mice that suggested transformation had occurred? dead IIR living IIR dead IIIS living IIIS
living IIIS The use of IIIS in combination with IIR showed a transformation of IIR into IIIS strains that caused death in the mice.
Would you expect a chromosome that has a large amount of repeated DNA sequences to have a high or low density of genes? high; the repeated sequences each encode a protein, so there would be many copies of that gene/protein present high; since there is a high difficulty of repeated DNA, the chromosome itself would be very large and have a high density of genes low; repeated DNA sequences do not typically encode genes, so there would be a low density of genes in the chromosome relative to the r
low; repeated DNA sequences do not typically encode genes, so there would be a low density of genes in the chromosome relative to the repetitive DNA sequences Typically the DNA that codes for proteins is considered unique-sequence DNA, which is not repeated.
A DNA molecule 300 base pairs long has 20 complete rotations. This means that the DNA molecule is
negatively supercoiled. Under rotated molecules exhibit negative supercoiling.
All histones have a high percentage of arginine and lysine, charged amino acids that give the histones a _____ charge. These charges attract the _____ charges on the phosphates of DNA; this attraction holds the DNA in contact with the histones. negative; negative negative; positive positive; positive positive; negative
positive; negative The amino acids arginine and lysine have positive charged residues while the phosphate of DNA carries a negative charge.
A stretch of 100 base pairs of DNA contains 15 complete rotations. This DNA molecule is relaxed. negatively supercoiled. positively supercoiled. denatured.
positively supercoiled. This DNA molecule has more rotations per 100 bp than relaxed DNA.
A graduate student hoping to specialize in virology is studying the process of viral DNA integration within host cells by bacteriophages. If they desire to configure a means of mitigating that integration, which of these steps would not help in decreasing bacteriophage replication? modifying replication components to shut down at the entry site of viral DNA introducing cell markers to recognize foreign DNA removal of phage head after host attachment cell signaling cascade for cell death at e
removal of phage head after host attachment Removing the phage head would not be beneficial as the DNA has already been injected into the cell.
How does DNA supercoiling arise?
rotation during DNA replication and transcription
What level of chromatin structure is most common in mitotic chromosomes?
spiraled 250 nm fibers creating a coil of about 700 nm across
What mechanism prevents the ends of chromosomes from being degraded or repaired? Choose the most complete answer. t- loops and shelterins t-loops, shelterins, and proteins that bind to G-rich single-stranded sequences t-loops shelterin proteins
t-loops, shelterins, and proteins that bind to G-rich single-stranded sequences All three of these structures serve to protect the ends of chromosomes.
What level of chromatin structure is most common in interphase DNA?
tethered 30 nm loops on the nuclear scaffold creating a 250 nm fiber
Which of these does not affect the three-dimensional shape of a DNA molecule? the sequence of the DNA strands amount of water surrounding the DNA molecule the type of RNA molecule it encodes what type of cell the DNA molecule is in
the type of RNA molecule it encodes DNA conformation is not dependent on RNA.
Neutralizing their positive charges would have which effect on the histone proteins? It would have no effect on its association with DNA. They would bind more tightly to DNA. They would bind less tightly to DNA. It would cause the protein to denature.
they would bind less tightly to the DNA The positive charge of histones is attracted to the negatively charged DNA, so a reduction in the positive charge will weaken the bond between them.
What is the protein composition of the nucleosome?
two copies each of H2A, H2B, H3, and H4
A gene that encodes a protein is most likely to be found in . . . tandem repeats. unique-sequence DNA. highly repetitive DNA. moderately repetitive DNA.
unique-sequence DNA Genes that encode proteins are typically found in one or a few copies.
