Molecular Biology Principles And Practice 2nd ed by Micheal Cox Test Bank

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Molecular Biology Principles And Practice 2nd ed by Micheal Cox Test Bank

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WITH ANSWERS
Molecular Biology Principles And Practice 2nd ed by Micheal Cox

Molecular Biology- Principles and Practice 2e

Cox et. al

Chapter 2

 

  1. Mendels success in formulating his fundamental principles of inheritance can be attributed to which of the following?

 

  1. Mendel focused on the overall appearance of the plant rather than on individual traits.
  2. Mendel focused on individual traits of the plant rather than on the overall appearance.
  3. Mendel chose to study complex traits that result from interactions between multiple genes.
  4. Mendel used an organism that grew slowly, therefore having long generation times.
  5. Mendel used fruit flies to study inheritance.

 

Answer: B

Section: 2.1

Level: Easy

Blooms: Comprehension

 

  1. If a pea plant strain is true-breeding for a particular trait, then:

 

  1. crosses between two true-breeding plants with the same trait will have offspring identical to the parents.
  2. it will be homozygous for all genes.
  3. it must be a recessive trait.
  4. it is not useful for studying inheritance.
  5. it only self-fertilizes.

 

Answer: A

Section: 2.1

Level: Easy

Blooms: Comprehension

 

  1. True-breeding, purple-flowered pea plants were crossed with true-breeding, white-flowered pea plants. The F1 progeny all had purple flowers.  If the F1 offspring are self-fertilized, what progeny phenotypes would be expected?

 

  1. 1 purple : 2 pink : 1 white
  2. 3 white : 1 purple
  3. 3 purple : 1 white
  4. all purple
  5. all white

 

Answer: C

Section: 2.1

Level: Easy

Blooms: Application

 

  1. A cross of pea plants, both heterozygous for seed shape and color traits, resulted in progeny in the ratio 9/16 round yellow (R_Y_), 3/16 round green (R_yy), 3/16 wrinkled yellow (rrY_), and 1/16 wrinkled green (rryy). This result demonstrates that the:

 

  1. alleles responsible for seed color are segregating but those responsible for shape are not.
  2. alleles of seed shape and color are segregating independently.
  3. phenotypes corresponding to seed shape are segregating 1:2:1.
  4. alleles governing seed color display codominance.
  5. alleles for seed shape and color are linked.

 

Answer: B

Section: 2.1

Level: Medium

Blooms: Comprehension

 

  1. True-breeding pea plants with inflated, yellow pods were crossed with true-breeding plants with constricted, green pods. The F1 progeny all had inflated, green pods.  When the F1 is allowed to self-fertilize, what fraction of the F2 progeny will have constricted, yellow pods?

 

  1. 1/4
  2. 3/4
  3. 1/16
  4. 3/16
  5. 9/16

 

Answer: C

Section: 2.1

Level: Medium

Blooms: Application

 

  1. True-breeding tall snapdragons with red flowers were crossed with short snapdragons with white flowers. The resulting F1 progeny were all of medium height and had pink flowers.  When the F1 progeny are crossed with each other, what proportion of the F2 progeny would be of medium height and have pink flowers?

 

  1. 1/4
  2. 1/2
  3. 1/8
  4. 1/16
  5. 3/4

 

Answer: A

Section: 2.1

Level: Medium

Blooms: Application

 

  1. Which of the following traits will show Mendelian behavior?

 

  1. dominance
  2. semidominance
  3. codominance
  4. epistasis
  5. linkage

 

Answer: A

Section: 2.1

Level: Easy

Blooms: Knowledge

 

  1. The child of a mother with IAIA blood and father with IAIB blood  will have:

 

  1. 25% chance of being IAIA.
  2. 50% chance of being IAIA.
  3. 75% chance of being IAIA.
  4. 25% chance of being IAIB.
  5. 50% chance of being IBIB.

 

Answer: B

Section: 2.1

Level: Medium

Blooms: Application

 

  1. A fruit fly homozygous for wild-type eyes and straight wings was crossed with a fly homozygous for rosy eyes and wrinkled wings. The progeny all showed wild-type eyes and straight wings.  When crossing the F1, the resulting offspring showed 3 wild-type eyes/straight wings : 1 rosy eyes/wrinkled wings.   This data suggest:

 

  1. rosy eyes is dominant.
  2. straight wings is recessive.
  3. these two genes are linked.
  4. these two genes are sex-linked.
  5. these two genes are codominant.

 

Answer: C

Section: 2.1

Level: Hard

Blooms: Analysis

 

  1. Using the great lubber grasshopper as a model, Walter Sutton observed:

 

  1. separation of maternal and paternal chromosomes during mitosis.
  2. replication of chromosomes during meiosis.
  3. crossing over of homologous chromosomes.
  4. pairing of homologous chromosomes during mitosis.
  5. pairing of homologous chromosomes during meiosis.

 

Answer: E

Section: 2.1

Level: Medium

Blooms: Comprehension

 

  1. Photographs of cells from onion root tips are often used to show chromosomes primarily because these cells are:

 

  1. diploid.
  2. rapidly undergoing mitosis.
  3. rapidly undergoing meiosis.
  4. in stationary phase.
  5. lacking organelles.

 

Answer: B

Section: 2.2

Level: Medium

Blooms: Application

 

  1. Tetrads are:

 

  1. composed of a pair of nonhomologous chromosomes.
  2. composed of two pairs of sister chromatids.
  3. formed in prophase II of meiosis.
  4. lined up along the cell axis during metaphase II.
  5. formed during DNA synthesis.

 

Answer: B

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. The pair of sister-chromatids, formed by replication of a chromosome, are held together at a specific region of the chromosome called the:

 

  1. replication fork.
  2. telomere.
  3. centromere.
  4. centriole.
  5. origin.

 

Answer: C

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. Differentiated cells, such as fat cells, have acquired their specialized function and remain in:

 

  1. S
  2. G1
  3. G2
  4. M
  5. G0

 

Answer: E

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. The mitotic stage in which the chromosomes condense, nuclei disappear, and the mitotic spindle begins to form is known as:

 

 

Answer: A

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. What technical advance allowed Walther Flemming and other scientists of his generation to better understand the stages of mitosis and meiosis?

 

  1. the ability to obtain single cells
  2. the ability to slice cells thinly
  3. microscopes with two eyepieces
  4. microscopes with high-quality oil immersion lens
  5. electron microscopes

 

Answer: D

Section: 2.2

Level: Medium

Blooms: Comprehension

 

  1. The mitotic stage in which the chromosomes decondense and the nuclear membrane reforms is:

 

 

Answer: D

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. The centromeres of sister chromatids uncouple and the chromatids separate in what meiotic phase?

 

  1. prophase I
  2. metaphase II
  3. anaphase I
  4. telophase II
  5. anaphase II

 

Answer: E

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. The primary result of meiosis is:

 

  1. gamete formation.
  2. two daughter cells from one parental cell.
  3. four daughter cells from one parental cell.
  4. gamete formation, two from one parental cell.
  5. gamete formation, four from one parental cell.

 

Answer: E

Section: 2.2

Level: Easy

Blooms: Knowledge

 

  1. In meiosis, the complexity of DNA (amount of information encoded) per cell is essentially halved as compared to the original cell. This reduction in information occurs immediately after:

 

  1. metaphase I.
  2. metaphase II.
  3. telophase I.
  4. telophase

 

Answer: C

Section: 2.2

Level: Medium

Blooms: Analysis

 

  1. In the ZW system of sex determination (as in birds), a male is _____________ while a female is ______________.

 

  1. ZZ : ZW
  2. ZW : ZZ
  3. Z0 : ZW
  4. Z : ZW

 

Answer: A

Section: 2.2

Level: Medium

Blooms: Knowledge

 

  1. The wild-type allele of a gene is:

 

  1. dominant to other alleles.
  2. recessive to other alleles.
  3. found at the lowest frequency in a population.
  4. found at the greatest frequency in a population.

 

Answer: D

Section: 2.3

Level: Easy

Blooms: Knowledge

 

  1. Genes are said to be linked when:

 

  1. they lie on the same chromosome.
  2. they lie on the same chromosome and are greater than 50 map units apart.
  3. they lie on the same chromosome and are less than 50 map units apart.
  4. they are alleles of the same gene.
  5. None of these choices are correct.

 

Answer: C

Section: 2.3

Level: Medium

Blooms: Comprehension

 

  1. What observation by Thomas Hunt Morgan led to the understanding that genes were on chromosomes?

 

  1. The white-eyed trait was recessive and disappeared in the F1
  2. In a cross of a red-eyed female and a white-eyed male, only females in the F2 generation were white-eyed.
  3. The inheritance of the white-eyed trait in fruit flies followed the same pattern of inheritance as that of the X chromosome.
  4. The inheritance of the white-eyed trait in fruit flies followed the same pattern of inheritance as that of the Y chromosome.
  5. Inheritance of the white-eyed trait was the same for female and male flies.

 

Answer: C

Section: 2.3

Level: Hard

Blooms: Comprehension

 

  1. Calvin Bridges, an associate of Thomas Hunt Morgans, found further evidence that genes were located on chromosomes. Which of the following is not a component of his study?

 

  1. He crossed white-eyed females (XwXw) with red-eyed males (XWY).
  2. The F1 progeny were mostly the red-eyed males and white-eyed females expected.
  3. A few rare white-eyed females and red-eyed males were observed, which he called primary exceptionals.
  4. He proposed that the rare white-eyed females were the result of abnormal chromosome number.
  5. Abnormal chromosome number was a result of nondisjunction in the female fly.

 

Answer: B

Section: 2.3

Level: Medium

Blooms: Comprehension

 

  1. The chromosomal abnormality Calvin Bridges observed as a characteristic of primary exceptionals is a result of:

 

  1. mutation.
  2. nondisjunction.
  3. chiasmata.
  4. recombination.
  5. linkage.

 

Answer: B

Section: 2.3

Level: Medium

Blooms: Comprehension

 

  1. The test cross ab/ab AB/ab is performed. The following number of progeny of each genotype are obtained: 38% AaBb, 12% Aabb, 12% aaBb, 38% aabb. What is the distance (in map units) between the two genes in question?

 

  1. 38 centimorgans (cM)
  2. 24 centimorgans (cM)
  3. 19 centimorgans (cM)
  4. 12 centimorgans (cM)
  5. 6 centimorgans (cM)

 

Answer: D

Section: 2.3

Level: Hard

Blooms: Analysis

 

  1. Five two-point crosses involving the four linked genes A, B, C, and D were performed in yeast with the following results:

 

AB ab:   41% AB 39% ab 9% Ab 11% aB

AC ac:   45% AC 48% ac  4% Ac  4% aC

BC bc:   44% BC 44% bc  6% Bc  6% bC

AD ad:   48% AD 46% ad 3% Ad  3% aD

BD bd:   37% BD 37% bd 12% bD 13% Bd

 

Which is the best map for these four genes?

 

  1. B 6mu        A             4mu        C              3mu        D
  2. B 12mu      C              8mu        A             6mu        D
  3. D 3mu        A             4mu        C              6mu        B
  4. A 4mu        C              6mu        B              12mu      D
  5. The data suggest that the genes are not linked.

 

Answer: C

Section: 2.3

Level: Hard

Blooms: Analysis

 

  1. Oswald Avery found that DNA was necessary and sufficient for bacterial transformation. Which of the following experimental results support this statement (select all that apply)?

 

  1. Removal of DNA from virulent Streptococcus pneumoniae cell extracts eliminated transformation.
  2. Removal of RNA from virulent Streptococcus pneumoniae cell extracts eliminated transformation.
  3. Protein alone will cause bacterial transformation to occur.
  4. DNA alone will cause bacterial transformation to occur.

 

Answer: A, D

Section: 2.4

Level: Medium

Blooms: Comprehension

 

  1. In the following pathway, mutation of enzyme II would result in the accumulation of ______________ and the mutant would require _____________ in the media in order to grow.

 

 

  1. Citrulline : arginine
  2. Ornithine : citrulline
  3. Precursor : ornithine
  4. Ornithine : precursor
  5. Arginine : citrulline

 

Answer: B

Section: 2.4

Level: Easy

Blooms: Comprehension

 

  1. Which of the following has the information flow of the central dogma in the correct order?

 

  1. DNA, translation, RNA, replication, protein
  2. DNA, transcription, RNA, replication, protein
  3. RNA, translation, DNA, transcription, protein
  4. DNA, transcription, RNA, translation, protein
  5. RNA, transcription, DNA, translation, protein

 

Answer: D

Section: 2.4

Level: Easy

Blooms: Knowledge

 

  1. When considering alleles of a gene, DNA is to genotype as ______________ is/are to phenotype.

 

  1. RNA
  2. mutation
  3. proteins
  4. lipids
  5. nucleotides

 

Answer: C

Section: 2.4

Level: Easy

Blooms: Knowledge

 

 

 

  1. In the early 1900s, determining the chemical basis of heredity was difficult because chromosomes are composed of:

 

  1. DNA.
  2. RNA.
  3. protein.
  4. DNA and protein.
  5. DNA, RNA, and protein.

 

Answer: D

Section: 2.4

Level: Medium

Blooms: Comprehension

 

  1. If protein had been the genetic material, then what would Oswald Avery have observed in his experiments?

 

  1. Bacterial extracts treated with proteinase would transform nonvirulent bacteria into the virulent strain.
  2. Bacterial extracts treated with proteinase would not transform nonvirulent bacteria into the virulent strain.
  3. Bacterial extracts treated with DNAse would transform nonvirulent bacteria into the virulent strain.
  4. Bacterial extracts treated with DNAse would not transform nonvirulent bacteria into the virulent strain.

 

  1. Statement 1
  2. Statement 2
  3. Statements 1 and 3
  4. Statements 2 and 3
  5. Statements 2 and 4

 

Answer. D

Section: 2.4

Level: Hard

Blooms: Synthesis

 

  1. Archibald Garrods evaluations of alkaptonuria (black urine in patients) led him to conclude:

 

  1. a mutation can result in a nonfunctional enzyme.
  2. enzymes are required for biochemical pathways.
  3. this defect is sex-linked and recessive.
  4. disjunction can result in a nonfunctional enzyme.

 

Answer:. A

Section: 2.4

Level: Medium

Blooms: Comprehension

 

 

  1. George Beadle and Edward Tatum used _____________as a model organisms and created mutants using ______________.

 

  1. Drosophila melanogaster : irradiation
  2. B. Escherichia coli : irradiation
  3. Neurospora crassa : irradiation
  4. Drosophila melanogaster : carcinogenic chemicals
  5. Neurospora crassa : carcinogenic chemicals

 

Answer. C

Section: 2.4

Level: Medium

Blooms: Knowledge

 

  1. George Beadle and Edward Tatum used auxotrophs to determine their one geneone enzyme hypothesis. An auxotroph is best described as:

 

  1. a defective enzyme.
  2. a result of nondisjunction.
  3. a mutant unable to grow on minimal media.
  4. a wild-type version of a model organism.
  5. an organism with a dominant lethal mutation.

 

Answer. C

Section: 2.4

Level: Medium

Blooms: Knowledge

 

 

  1. The one geneone enzyme hypothesis of George Beadle and Edward Tatum was an oversimplification because:

 

  1. mutations may not completely inactivate an enzyme.
  2. some enzymes contain more than one polypeptide.
  3. some enzymes can be phosphorylated.
  4. auxotrophs can be corrected.
  5. some RNAs are independently catalytic.

 

Answer. B

Section: 2.4

Level: Hard

Blooms: Comprehension

 

  1. George Beadle and Edward Tatum created a collection of mutants to study the following pathway.

 

 

 

A  mutant shows no growth on minimal media; however it shows growth if citrulline is provided.  These results indicate the mutation would most likely be in the gene for:

 

  1. Enzyme I.
  2. Enzyme II.
  3. Enzyme III
  4. either Enzyme I or Enzyme II.
  5. either Enzyme II or Enzyme III.

 

Answer. D

Section: 2.4

Level: Hard

Blooms: Analysis

 

  1. rRNA is (select all that apply):

 

  1. associated with ribosomes.
  2. the least abundant RNA in the cell.
  3. the most abundant RNA in the cell.
  4. relatively consistent in sequence between organisms.
  5. considerably variable in sequence between organisms.

 

Answer: A, C, D

Section: 2.4

Level: Hard

Blooms: Synthesis

 

  1. Cricks proposed adaptor molecule is now known to be:

 

  1. mRNA.
  2. rRNA.
    C. tRNA.
  3. imRNA.

 

Answer: C

Section: 2.4

Level: Medium

Blooms: Synthesis

 

  1. A mutation in a tRNA gene could potentially decrease the ability of that tRNA to (select all that apply):

 

  1. complex with ribosomes.
  2. bind to an amino acid.
  3. bind multiple amino acids simultaneously.
  4. control DNA replication.
  5. code for protein.

 

Answer: A, B

Section: 2.4

Level: Hard

Blooms: Analysis

 

  1. Epigenetic inheritance may depend on all the following except:

 

  1. the DNA sequence from a parent.
  2. inherited chemical modification of parental nucleotides.
  3. inherited nucleosome structure of chromosome.
  4. inherited chemical modification of parental proteins.

 

Answer: A

Section: 2.4

Level: Hard

Blooms: Analysis

 

  1. Hemophilia in humans results from an Xlinked recessive gene. Which of these statements regarding the inheritance of this trait is true?
  2. Heterozygous females will exhibit a milder form of hemophilia.
  3. All daughters of an affected mother will have hemophilia.
  4. Carrier females will always pass the hemophilia allele to their sons.
  5. Males who inherit the recessive allele from their fathers will exhibit hemophilia.
  6. All sons of an affected mother will have hemophilia.

 

Answer: E

Section: 2.4

Level: Medium

Blooms: Analysis

 

 

  1. A woman with apparently normal health (mother) and a colorblind man (father) had a son with hemophilia and a normal daughter. If the genes for both traits are on the X chromosome, which statement about this family is true?

 

  1. The maternal grandfather is colorblind.
  2. The daughter cannot be a carrier for colorblindness.
  3. The maternal grandfather is not hemophilic.
  4. There is a 50% chance that the maternal grandfather is hemophilic.
  5. The father is a carrier of hemophilia.

 

Answer: D

Section: 2.4

Level: Hard

Blooms: Analysis

 

  1. A patient in his forties is diagnosed with Huntington disease. The patients mother is healthy and a genetic evaluation shows 25 and 26 CAG repeats within her HTT alleles.  The patients father died at age 24 when the patient was 2 and no genetic evaluation can be done.  It is most likely that:

 

  1. the father was heterozygous for the defective HTT allele.
  2. the father was homozygous for the defective HTT allele.
  3. the mother is a carrier for the defective HTT allele.
  4. the HTT gene acquired additional CAG repeats during the 40 years of the patients life.

 

Answer: A

Section: 2.4

Level: Hard

Blooms: Synthesis

 

  1. A newborn is identified as having cystic fibrosis. While both parents are healthy, a genetic screening of both parents will most likely reveal:

 

  1. the father is a carrier of the mutant CF allele.
  2. the mother is a carrier of the mutant CF allele.
  3. both mother and father are carriers of a mutant CF allele.
  4. neither mother nor father are carriers; spontaneous mutation caused the babys disease.

 

Answer: C

Section: 2.4

Level: Hard

Blooms: Evaluation

 

  1. Cystic fibrosis is most often caused by the CFTRF508 mutation. This mutation:
  2. reduces the function of a chloride ion channel.
  3. increases ATP binding to chloride.
  4. is autosomal dominant.
  5. causes symptoms in carriers.

 

Answer: A

Section: 2.4

Level: Easy

Blooms: Knowledge

 

 

  1. A newborn is identified as having sickle-cell anemia. While both parents are healthy, a screening of both parents will most likely reveal:

 

  1. both mother and father are carriers of a mutant chain gene.
  2. only one parent has a mutant chain gene since mutation is dominant.
  3. both mother and father are carriers of a mutant chain gene.
  4. only one parent has a mutant chain gene since mutation is dominant.
  5. both parents carry a mutant hemoglobin gene and are highly susceptible to malaria.

 

Answer: C

Section: 2.4

Level: Medium

Blooms: Application

 

 

  1. In order to prove genetic recombination occurred through material (DNA) exchange between homologous maize (corn) chromosomes, Harriet Creighton and Barbara McClintock:

 

  1. stained individual chromosomes to see banding, and then observed segregation of bands.
  2. observed specific morphologic features of one chromosome being traded with its homolog.
  3. created mutations of chromosomes via irradiation and observed transfer of damaged DNA between homologs.
  4. labeled one chromosome with 32P and observed the transfer of radioactivity to the homolog.

 

Answer: B

Section: 2.4

Level: Hard

Blooms: Comprehension

 

Molecular Biology- Principles and Practice 2e

Cox et. al

Chapter 10

 

  1. Chromatin is composed of:

 

  1. DNA.
  2. protein.
  3. DNA and RNA.
  4. DNA and protein.
  5. RNA and protein.

 

Answer: D

Section: 10.1

Level: Easy

Blooms: Knowledge

 

 

  1. Of the proteins associated with chromatin, which of the following represents the largest component?

 

  1. histones
  2. topoisomerases
  3. SMC proteins
  4. transcriptional regulators
  5. ribosomal proteins

 

Answer: A

Section: 10.1

Level: Easy

Blooms: Knowledge

 

 

  1. The first evidence for nucleosome formation came from digesting chromosomal DNA with a nonspecific nuclease. Gel electrophoresis of the DNA after the reaction revealed:

 

  1. a ladder of protected fragments about 1,000 bp apart.
  2. a ladder of protected fragments about 200 bp apart.
  3. a large band corresponding to 200 bp.
  4. random distribution of bands corresponding to different sizes.
  5. four bands, each composed of a different histone.

 

Answer: B

Section: 10.1

Level: Easy

Blooms: Comprehension

 

 

  1. Each nucleosome contains:

 

  1. four proteins: H2A, H2B, H3, and H4.
  2. a histone octamer composed of two copies of each of the following proteins: H1, H2, H3, and H4.
  3. a histone octamer composed of two copies of each of the following proteins: H2A, H2B, H3, and H4.
  4. four proteins: SMC1, SMC2, SMC3, and SMC4.
  5. a mixture of histone and SMC proteins.

 

Answer: C

Section: 10.1

Level: Easy

Blooms: Knowledge

 

 

  1. Which of the following is true of nucleosome structure?

 

  1. Nucleosomes are composed of about 1,000 bp of DNA wrapped around a histone core.
  2. Nucleosomes are composed of about 200 bp of DNA wrapped around a histone core.
  3. The DNA wraps around the core nearly two times, forming a right-handed solenoidal supercoil.
  4. The DNA wraps around the core nearly three times, forming a left-handed solenoidal supercoil.
  5. None of these choices is correct.

 

Answer: B

Section: 10.1

Level: Easy

Blooms: Knowledge

 

 

  1. Histones:

 

  1. have highly conserved sequences among eukaryotes.
  2. are rich in basic amino acids; about 25% of the total amino acids are lysine and arginine.
  3. are small proteins, ranging in molecular weight from 11,000 to 21,000.
  4. contain a histone-fold motif comprised of three -helices linked by two short loops.
  5. All of these choices are correct.

 

Answer: E

Section: 10.1

Level: Medium

Blooms: Comprehension

 

 

  1. The contacts between the DNA and the histones of the nucleosome are:

 

  1. mainly between the R groups of the histones and the phosphate backbone of the DNA.
  2. mainly between the conserved histone folds and the bases that are exposed in the major groove of the DNA.
  3. mainly between the conserved histone folds and the bases that are exposed in the minor groove of the DNA.
  4. more prevalent in regions of the DNA that have tracts of GC base pairs.
  5. primarily between the DNA and the histone peptide backbone.

 

Answer: E

Section: 10.1

Level: Medium

Blooms: Comprehension

 

 

  1. Which of the following is not true of histone tails?

 

  1. They protrude between the two DNA strands that supercoil around the nucleosome.
  2. They are the target of numerous chemical modifications.
  3. They experience changes in net charge, shape, and other properties of histones in response to modifications.
  4. They are recognized by acetylases and methylases.
  5. They have enzymatic properties that can covalently modify DNA.

 

Answer: E

Section: 10.1

Level: Medium

Blooms: Comprehension

 

 

  1. What best describes the composition of a nucleosome complex (with no H1)?
  2. H2A, H2B, H3, H4, along with 147 bp of DNA
  3. H2A, H2B, H3, H4, along with 168 bp of DNA
  4. 2 molecules each of H2A, H2B, H3, and H4, along with 147 bp of DNA
  5. 2 molecules each of H2A, H2B, H3, and H4, along with 168 bp of DNA

 

Answer: D

Section: 10.1

Level: Medium

Blooms: Comprehension

 

 

  1. Which of the following is not a way nucleosome formation affects DNA?

 

  1. compacts the DNA in the nucleus
  2. alters access to DNA by enzymes
  3. regulates the rate of DNA replication in mitochondria
  4. inhibits transcription by the internucleosome connections mediated by histone tails
  5. competes with transcription factor Sp1 to maintain transcriptional repression

 

Answer: C

Section: 10.1

Level: Hard

Blooms: Analysis

 

 

  1. Denaturation of protein X yields two subunits, A (MW 8,000) and B (MW 20,000). In order to understand the quaternary structure, a crosslinking experiment was performed with cell extracts.  Analysis of the crosslinking products by SDS-PAGE showed four bands of the following molecular weights: 7,800, 19,500, 39,000, and 48,000.  This data suggests that protein X functions in vivo as:
  2. a dimer.
  3. a trimer.
  4. a tetramer.

D  The results are inconclusive.

 

Answer: B

Section: 10.1

Level: Hard

Blooms: Analysis

 

 

  1. The histone octamer forms a tightly packed particle, but the N-termini that protrude from the octamer are less ordered. Which is the best explanation for this observation?
  2. The charges on the amino acid residues in the N-terminal tails are primarily acidic and repel each other.
  3. The charges on the amino acid residues in the N-terminal tails are primarily basic and flex to accommodate the DNA.
  4. The N-terminal tails must be flexible because they bind to and then release DNA during transcription.
  5. The N-terminal tails must be more flexible because they form intermolecular contacts with other nucleosomes.
  6. The N-terminal tails must be more flexible because they interact within the minor groove of DNA.

 

Answer: D

Section: 10.1

Level: Hard

Blooms: Synthesis

 

 

  1. When complexing with histones, the DNA phosphodiester backbone is in close proximity to:
  2. the histone fold, two helices connected by two short links, this motif being found within H1, H2A, H2B, H3, and H4.
  3. the histone fold, two helices connected by two short links, this motif being found within H2A, H2B, H3, and H4.
  4. the histone fold, three helices connected by two short links, this motif being found within H1, H2A, H2B, H3, and H4.
  5. the histone fold, three helices connected by two short links, this motif being found within H2A, H2B, H3, and H4.

 

Answer: D

Section: 10.1

Level: Hard

Blooms: Application

 

 

  1. In the presence of DNA-binding protein Sp1, transcription is:
  2. increased if H1 is at low levels.
  3. decreased if H1 is at low levels.
  4. increased if H1 is at high levels.
  5. not affected by H1.

 

Answer: A

Section: 10.1

Level: Medium

Blooms: Comprehension

 

 

  1. A temperature sensitive mutant of Sp1 is isolated. When cells are grown at the permissive temperature, cell function appears normal.  However, at high temperatures, the phenotypic effect of this mutation would most likely be:
  2. A. reduced transcription.
  3. increased transcription.
  4. reduced levels of H1.
  5. increased levels of H1.
  6. problems with DNA packaging.

 

Answer: A

Section: 10.1

Level: Hard

Blooms: Synthesis

 

 

Short Answer

  1. Why is eukaryotic DNA underwound even though eukaryotes lack topoisomerases that can introduce supercoils?

 

Answer:  Formation of a left-handed, solenoidal supercoil requires the removal of about one turn from the DNA.   This effect combined with the removal of unbound positive supercoils by topoisomerases leads to a net underwound state.

 

Section: 10.1

Level: Hard

Blooms: Synthesis

 

 

  1. H1 binding to DNA is different than that of the other four histones because H1 binds to:

 

  1. the major groove of the DNA only.

B   the DNA as it comes off both sides of the nucleosome.

  1. as a component of the octamer; however, unlike H2A, H3, and H4, H1 lacks the histone tail.
  2. one strand of the linker DNA as it comes off the nucleosome and binds a second site in the central region of the DNA supercoil.
  3. the DNA as a homotetramer.

 

Answer: D

Section: 10.2

Level: Hard

Blooms: Synthesis

 

 

  1. Which of the following is not a function of histone H1?

 

  1. stabilizing nucleosomes
  2. promoting higher order chromosome structure
  3. enhancing the repression of transcription by nucleosomes
  4. binding to regions undergoing active RNA synthesis
  5. All of these options are functions of histone H1.

 

Answer: D

Section: 10.2

Level: Medium

Blooms: Synthesis

 

 

  1. In different regions of the chromosome, the ratio of histone H1 to histone H2A may vary, but the ratio of H2A to histone H2B is generally the same. If the amount of H1 increases in a region of chromatin, what will be its effect on compaction of the DNA and transcription in that region?

 

  1. Transcription in that region will increase.
  2. Transcription in that region will decrease.
  3. The DNA in that region will become less compact.
  4. DNA compaction in that region will be unchanged.
  5. Transcription of the DNA in that region will remain the same.

 

Answer: B

Section: 10.2

Level: Medium

Blooms: Comprehension

 

 

  1. The beads-on-a-string form of DNA condenses into a compact fiber called the 30 nm filament. Which of the following is true about its formation?

 

  1. Histone H1 is essential for formation of the 30 nm filament.
  2. The histone tails of the octamer histones are absolutely required for filament formation.
  3. The zigzag model describes how the nucleosome array adopts a spiral shape when forming the filament.
  4. The 30 nm filament is also called a tetranucleosome.
  5. The linker histone is essential for formation of the 30 nm filament.

 

Answer: B

Section: 10.2

Level: Medium

Blooms: Comprehension

 

 

  1. Higher chromosome structure depends on:

 

  1. SMC proteins, one of the major components of the chromosome scaffold.
  2. the formation of a chromosome scaffold by topoisomerases.
  3. the presence of the H1 protein, which promotes formation of the nucleoid.
  4. the formation of rosettes, which are stabilized by histone proteins.
  5. the formation of tetranucleosomes.

 

Answer: A

Section: 10.2

Level: Medium

Blooms: Comprehension

 

 

  1. Condensation of DNA yields the 30 nm filament. This filament:
  2. is the most compact form of eukaryotic DNA.
  3. is moderately condensed DNA, as facilitated by H1 and the N-terminal tails.
  4. is also called the chromosomal scaffold.
  5. includes tetranucleosomes, each 30 nm in diameter.
  6. would not include histone variants.

 

Answer: B

Section: 10.2

Level: Medium

Blooms: Synthesis

 

 

  1. Through an electron microscope, chromatin will have the appearance of beads on a string. What protein(s) will tighten the beads to create a pronounced zigzag appearance?
  2. scaffolding proteins
  3. NAP-1
  4. H1 histone
  5. Sp1
  6. bromodomain proteins

 

Answer: C

Section: 10.2

Level: Medium

Blooms: Analysis

 

 

  1. Which of the following is a difference between the solenoid and zigzag models for the formation of the 30 nm filament?

 

  1. Most of the evidence to date supports the solenoid model, while the zigzag model is unlikely to occur in cells.
  2. Linker DNA is at the center of the filament in the zigzag model, and on the outside of the filament in the solenoid model.
  3. The solenoid model is a one-start helix and the zigzag model is a two-start helix.
  4. The helical pitch in the solenoid model is higher than the helical pitch in the zigzag model.
  5. The zigzag model occurs in higher eukaryotes such as humans, while the solenoid model occurs in lower eukaryotes such as yeast.

 

Answer: C

Section: 10.2

Level: Hard

Blooms: Synthesis

 

 

  1. Which of the following may be a reason for bacteria not having nucleosomelike structures?

 

  1. Bacteria need to respond quickly to the environment and therefore need ready access to the genome.
  2. Cell division in bacteria occurs in as little as 15 minutes.
  3. A much larger proportion of the bacterial chromosome codes for proteins.
  4. Higher rates of metabolism in bacteria mean that a much larger proportion of the DNA is being transcribed or replicated at a given time.
  5. All of these choices are correct.

 

Answer: E

Section: 10.2

Level: Hard

Blooms: Synthesis

 

 

  1. Which of the following features of DNA organization do eukaryotes and bacteria have in common (select all correct choices)?
  2. A. DNA is associated with basic proteins.
  3. DNA is found within a 30 nm filament.
  4. DNA breakage in one loop will not affect DNA in other loops.
  5. A change of supercoiling in a cleaved loop would not affect other loops.
  6. Under electron microscope, DNA will have a beads-on-a-string appearance.

 

Answer: A, C, D

Section: 10.2

Level: Hard

Blooms: Synthesis

 

 

  1. The higher order structure of chromatin contains topologically constrained loops. What would happen if a single loop was cleaved?

 

  1. The entire chromosome would unravel.
  2. The genes of that loop could not be transcribed.
  3. Only that loop would relax because each loop is separately constrained by proteins.
  4. The genes in the cleaved loop could not be replicated.
  5. Supercoiling in the cleaved loop would not be affected.

 

Answer: C

Section: 10.2

Level: Medium

Blooms: Synthesis

 

 

  1. If histones are extracted from DNA, what would remain?

 

  1. DNA only
  2. DNA and tetranucleosomes
  3. DNA and chromosomal scaffolding proteins
  4. degraded segments of DNA

 

Answer: C

Section: 10.2

Level: Hard

Blooms: Application

 

 

Short Answer

  1. Why is histone H1 referred to as the linker histone?

 

Answer:  When chromatin is treated with nucleases, initially a larger portion of the DNA is protected, but when more extensive digestion is carried out and H1 is released, the DNA fragment protected is smaller. This shows that H1 binds to linker DNA unlike the histones of the nucleosome.

 

Section: 10.2

Level: Hard

Blooms: Synthesis

 

 

Section 10.3

  1. Transcriptionally active genes are characterized by:

 

  1. an increase in H1.
  2. the presence of bound nucleosomes at the promoter regions.
  3. the presence of specialized chromosome remodeling complexes.
  4. the absence of histone variants such as H2AZ and H3.3.
  5. All of these choices are correct.

 

Answer: C

Section: 10.3

Level: Medium

Blooms: Comprehension

 

 

  1. In SDSpolyacrylamide gel electrophoresis proteins are partially denatured and separated entirely as a function of their mass. Histones are an exception; they migrate more slowly than they should, as though they are much larger than they actually are.  Why does this occur?

 

  1. Histones form dimers and tetramers, making them appear larger.
  2. Histones are heavily glycosylated, greatly adding to their mass.
  3. Histones have a large number of positively charged amino acid residues. The binding of SDS to the histones is not sufficient to mask these charges, resulting in slower migration.
  4. Histones have an affinity for polyacrylamide.
  5. Histones are somewhat insoluble, which slows their passage through the gel.

 

Answer: C

Section: 10.3

Level: Medium

Blooms: Application

 

 

  1. Which of the following is not a way chromatin remodeling complexes affect nucleosomes?

 

  1. They can slide a nucleosome to a different location.
  2. They can modify the N-terminal tails of histones.
  3. They can eject a nucleosome from the DNA.
  4. They can replace the nucleosome with a new nucleosome that contains a variant histone.
  5. All of these choices are correct.

 

Answer: B

Section: 10.3

Level: Medium

Blooms: Comprehension

 

 

  1. Which of the following is not typically a modification of histones?

 

  1. adenylation
  2. phosphorylation
  3. methylation
  4. acetylation
  5. adenylation and phosphorylation

 

Answer: A

Section: 10.3

Level: Medium

Blooms: Comprehension

 

 

  1. A histone modification that attracts other proteins such as a transcription factor is said to be acting:

 

  1. in cis.
  2. in trans.
  3. epigenetically.
  4. as a remodeling protein.
  5. as a chaperone.

 

Answer: B

Section: 10.3

Level: Medium

Blooms: Comprehension

 

 

  1. Which of the following is not true about histone modification and the enzymes that create them?

 

  1. The enzymes responsible are often part of multiprotein complexes that include transcription factors.
  2. The enzymes responsible can work with chromatin remodeling complexes to alter specific regions of chromatin in response to environmental signals.
  3. The modifications can be passed on by epigenetic inheritance.
  4. The enzymes responsible can create histone variants.
  5. All of these choices are correct.

 

Answer: D

Section: 10.3

Level: Medium

Blooms: Analysis

 

 

  1. So far, three classes of chromatin remodeling complexes have been identified. Which of the following pairings is correct?

 

  1. SWI/SNF complexes activation
  2. SMC complexes repression
  3. ISWI complexes activation
  4. Mi2/NURD activation
  5. SWI/SNF complexes repression

 

Answer: A

Section: 10.3

Level: Medium

Blooms: Knowledge

 

 

  1. How can repositioning nucleosomes affect transcription (select all correct choices)?

 

  1. Moving nucleosomes could expose a promoter resulting in repression of transcription.
  2. Moving nucleosomes could cover a promoter resulting in repression of transcription.
  3. Moving nucleosomes could expose a promoter resulting in activation of transcription.
  4. Transcription is not affected by repositioning of nucleosomes.

 

 

Answer: B, C

Section: 10.3

Level: Medium

Blooms: Application

 

 

  1. Which of the following methods can be used to determine the position of nucleosomes in the genome?

 

  1. RNA-Seq
  2. Western blotting
  3. ChIP-Seq
  4. epigenetic immunoprecipitation
  5. immunofluorescence

 

Answer: C

Section: 10.3

Level: Medium

Blooms: Application

 

 

  1. Which of the following is the third step of a ChIP-Seq protocol?

 

  1. digestion of chromatin with nuclease
  2. washing away of unbound DNA, reversal of crosslinks, and removal of free histones
  3. identification of DNA by sequencing
  4. immunoprecipitation of nucleosome-DNA complexes with histone antibody
  5. crosslinking of nucleosome histones to DNA in chromatin

 

Answer: D

Section: 10.3

Level: Medium

Blooms: Analysis

 

 

  1. Histone variants:

 

  1. have been identified for H2B and H4 but not for H2A and H3.
  2. differ in their C-terminal sequences but not in their N-terminal sequences.
  3. that have been identified are associated with transcriptional activation only.
  4. have variant patterns of modifications as a result of the structural differences in their terminal sequences.
  5. have been found to affect all cellular processes that involve DNA except DNA repair.

 

Answer: D

Section: 10.3

Level: Hard

Blooms: Synthesis

 

 

  1. Histone chaperones:

 

  1. are required for direct assembly of the 30 nm filament.
  2. are acidic proteins that bind individual histones and assemble them into dimer and tetramer forms.
  3. can also modify the histones covalently by phosphorylation.
  4. are acidic proteins that bind to H2A-H2B dimers and H3-H4 tetramers before assembly into nucleosomes.
  5. can also modify the histones covalently by acetylation.

 

Answer: D

Section: 10.3

Level: Hard

Blooms: Synthesis

 

 

  1. Which of the following are correct pairings of histone variants and their functions?

 

  1. H3.3 stabilizes the open state (transcriptionally active)
  2. CENPA maintains kinetochore attachment
  3. H2AX DNA repair
  4. macroH2A X chromosome inactivation
  5. All of these choices are correct.

 

Answer: E

Section: 10.3

Level: Hard

Blooms: Comprehension

 

 

  1. Proteins that bind to acetylated lysine residues in the histone tail contain:

 

  1. bromodomains.
  2. DNA-binding domains.
  3. chromodomains.
  4. kinase domains.
  5. histone-fold motifs.

 

Answer: A

Section: 10.3

Level: Medium

Blooms: Comprehension

 

 

  1. Which of the following is not affected by the presence of epigenetic marks?

 

  1. development
  2. imprinting
  3. X chromosome inactivation
  4. maturation of RNA
  5. unique expression patterns in different cells

 

Answer: D

Section: 10.3

Level: Medium

Blooms: Application

 

 

  1. Activation of the IFN- globin promoter is an example of a histone code. Of the steps given, which is the third in this process?

 

  1. The TATA box in the promoter is exposed.
  2. Complexes containing Gcn5 bind to the promoter and acetylate nearby nucleosomes.
  3. A chromatin remodeling complex such as SW1/SNF binds to and moves the nucleosome.
  4. Initiation of transcription occurs.
  5. A protein kinase binds to the Gcn5 complex and phosphorylates nearby nucleosomes.

 

Answer: C

Section: 10.3

Level: Hard

Blooms: Analysis

 

 

  1. Activation of the IFN- globin promoter is an example of a histone code. Of the steps listed, which would be the last step?
  2. A. exposure of the TATA box in promoter
  3. acetylation of nucleosomes
  4. transport of the nucleosome
  5. methylation of nucleosomes
  6. phosphorylation of nucleosomes

 

Answer: A

Section: 10.3

Level: Medium

Blooms: Analysis

 

 

  1. Chaperones mediate nucleosome assembly. From the list below, what must happen first?
  2. The CAF-1 chaperone binds an H3-H4 dimer and deposits it onto the DNA strand.
  3. H3 and H4 form heterotetramers.
  4. The NAP-1 chaperone assembles two H2A-H2B heterodimers with a H3-H4 tetramer.
  5. The CAF-1 chaperone binds an H2A-H2B dimer and deposits it onto the DNA strand.

 

Answer: B

Section: 10.3

Level: Medium

Blooms: Analysis

 

 

  1. Which of the following are ways in which histones may be modified (select all correct choices)?

 

  1. phosphorylation of serine residues
  2. methylation of arginine residues
  3. phosphorylation of threonine residues
  4. acetylation of lysine residues
  5. phosphorylation of glycine residues

 

: Answer A, B, C, D

Section: 10.3

Level: Medium

Blooms: Analysis

 

  1. Proteins with ______________ bind to acetylated Lys residues on the histone tail, recruiting other histone modifying complexes and spreading the pattern of acetylation to nearby histones.

Answer: bromodomains

Section: 10.3

Level: Medium

Blooms: Synthesis

 

  1. Which chromatin remodeling complexes increase transcription and include an ATPase subunit?
  2. bromodomain proteins
  3. chromodomain proteins
  4. chaperone proteins
  5. both bromodomain and chromodomain proteins

 

Answer: A

Section: 10.3

Level: Medium

Blooms: Synthesis

 

  1. How are the epigenetic marks of a particular chromatin state preserved during cell division?

 

  1. Parental, marked H2A-H2B dimers stay associated with strands after replication and recruit new H3-H4 tetramers.
  2. H2A-H2B dimers and H3-H4 tetramers come off during replication and assemble with unmarked histones. These new octamers then rebind to the newly replicated DNA to form nucleosomes.
  3. Parental, marked H3-H4 tetramers stay associated with strands after replication and recruit new H2A-H2B dimers.
  4. Histone chaperones make sure that at least one component of each new nucleosome is marked.
  5. None of these choices is correct.

 

Answer:  B

Section: 10.3

Level: Hard

Blooms: Analysis

 

 

  1. How do modified and unmodified histone subunits become distributed among nucleosomes after replication?

 

  1. The newly synthesized DNA strand contains unmodified histone subunits, while the old strand retains the modified histones.
  2. Both strands of DNA are repopulated first with a mixture of modified and unmodified H2A histones, followed by mixtures of H2B, H3, and, finally, H4.
  3. Both strands of DNA are repopulated first with a mixture of modified and unmodified H2A and H2B histones, followed by mixtures of H3 and finally mixtures of H4.
  4. Both strands of DNA retain H3-H4 modified pairs, and then are repopulated with unmodified H3-H4 pairs, followed by modified and unmodified H2A-H2B.
  5. Both strands of DNA retain H2A-H2B modified pairs, and are then repopulated with unmodified H2A-H2B pairs and then mixtures of modified and unmodified H3-H4 pairs.

 

Answer: D

Section: 10.3

Level: Hard

Blooms: Comprehension

 

 

Short Answer

  1. How can the disruption of epigenetic markers result in cancer?

 

Answer: The disruption of epigenetic pathways can activate genes that promote cell growth or repress genes that limit cell growth, leading to tumor formation.

 

Section: 10.3

Level: Hard

Blooms: Synthesis

 

 

  1. What observation first suggested that H2A and H2B formed heterodimers?

 

Answer:  A milder protein purification technique that did not denature the histones resulted in two peaks upon purification, one containing H2A and H2B and one containing H3 and H4.  This observation suggested that they were physically associated in some way and lead Kornberg to carry out his crosslinking experiments.

 

Section: 10.3 [How We Know]

Level: Hard

Blooms: Synthesis

 

 

  1. When the gene that encodes the histones acetylase (HAT) was isolated, it was shown to have homology to the yeast protein, Gcn5. Why was this an exciting find?

 

Answer:  Gcn5 is a transcriptional activator that implies that the acetylase activity may account for the activators regulatory function.  Ultimately Gcn5 was shown to have histone acetylase (HAT) activity.

 

Section: 10.3 [How We Know]

Level: Hard

Blooms: Synthesis

 

 

 

Molecular Biology- Principles and Practice 2e

Cox et. al

Chapter 22

 

  1. An RNA processing event that frequently leads to different protein products encoded by a single gene is:

 

  1. alternative splicing.
  2. nucleolytic proofreading.

 

Answer: B

Section: 22.1

Level: Easy

Blooms: Knowledge

 

  1. In Drosophila cells, the expression of the Sxl gene to form active protein product requires:

 

  1. that the Dpn (Deadpan) repressor be present in higher concentration than the SisA and SisB activators.
  2. transcription from the Pe promoter in male flies.
  3. the removal of exon L3 in the pre-mRNA.
  4. All of the choices are required for expression of the Sxl gene to form active protein product.

 

Answer: C

Section: 22.1

Level: Medium

Blooms: Comprehension

 

  1. Which of the following is not a function of the Drosophila Sxl protein?

 

  1. prevention of the splicing machinery from recognizing the splice junctions of the E1 or L3 exons of the Sxl mRNA
  2. mediation of alternative splicing that produces transcripts for functional Tra protein
  3. binding specifically to the regulated 3 splice site of the tra transcript
  4. mediation of the splice variant of the dsx transcript leading to repression of female-specific genes

 

Answer: D

Section: 22.1

Level: Medium

Blooms: Comprehension

 

  1. The 3 cleavage and polyadenylation of mRNA is partly defined by the sequence:

 

 

Answer: A

Section: 22.1

Level: Medium

Blooms: Knowledge

 

  1. In response to the presence of extracellular antigen, B cells rapidly proliferate and the immunoglobulin M (IgM) proteins produced are soluble rather than anchored in the membrane. This change in IgM protein state is due to:

 

  1. the cleavage of several hydrophobic amino acid residues from the IgM protein.
  2. the differential expression of two different IgM genes.
  3. the predominant use of the first 3 end cleavage site of the IgM pre-mRNA.
  4. inclusion of the M exon in mature IgM mRNA.

 

Answer: C

Section: 22.1

Level: Medium

Blooms: Comprehension

 

 

 

  1. The function of the Rev protein encoded by the HIV genome is to:

 

  1. reverse transcribe its RNA genome to DNA.
  2. ensure that HIV gene transcripts containing introns are transported to the cytoplasm.
  3. integrate the HIV genome into the host genome.
  4. None of the above is correct.

 

Answer: B

Section: 22.1

Level: Hard

Blooms: Knowledge

 

  1. Treatment of eukaryotic cells with bacterial polypeptide actinomycin D will cause:

 

  1. interference with the movement of tRNAs during polypeptide elongation on the ribosome.
  2. blockage of splicing as mediated by snURP2.
  3. no change; bacterial proteins will have no impact on the function of eukaryotic cells.
  4. interference with transcript elongation by RNA polymerase II.

 

Answer:  D

Section: 22.1

Level: Medium

Blooms: Knowledge

 

  1. Splicing silencing sites can be located in both introns and exons.

 

  1. True
  2. False

 

Answer:  A

Section: 22.1

Level: Easy

Blooms: Knowledge

 

  1. A B cell CstF-64 mutant will likely:

 

  1. constitutively secrete IgM antibodies.
  2. selectively remove the VDJ region from IgM antibodies.
  3. cut IgM pre-mRNA at every available 3-end cleavage site.
  4. undergo IgM pre-mRNA splicing between the heavy chain exons and the membrane anchoring exon.

 

Answer:  D

Section: 22.1

Level: Hard

Blooms: Application

 

  1. Sxl protein mediates

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