Lodish 8e Ch08 Test Bank PDF

8 Section 8.1 1\. Which one of the following regarding pseudogenes is NOT true? a\. They are present in the eukaryotic genome. b\. They encode miRNAs. c\. They mark the region of gene duplications. d\. They always encode functional products. Ans: b 2\. Which of the following is a typical fea...

8 Section 8.1 1\. Which one of the following regarding pseudogenes is NOT true? a\. They are present in the eukaryotic genome. b\. They encode miRNAs. c\. They mark the region of gene duplications. d\. They always encode functional products. Ans: b 2\. Which of the following is a typical feature of prokaryotic genes? a\. polycistronic messenger RNAs b\. complex transcription units c\. introns d\. a and c Ans: a 3\. The chicken lysozyme gene is considered to be a solitary gene because a\. it contains no introns. b\. it is not present on a chromosome. c\. it is represented only once in the haploid genome. d\. none of the above Ans. c 4\. All the following statements about complex transcription units are true except: a\. They can have multiple poly(A) sites. b\. They can generate multiple mRNAs. c\. They can generate multiple polypeptides. d\. They are common in bacteria. Ans: d 5\. In eukaryotes, tandemly repeated genes encode a\. rRNAs. b\. cytoskeletal proteins. c\. β-globin. d\. all of the above Ans. a 6\. Short micro RNAs (miRNAs) a\. code for proteins. b\. are common in bacteria but not eukaryotes. c\. are involved in regulation of gene expression. d\. have no known function. Ans: c 7\. Give a functional definition of a gene. Ans: A gene consists of the entire DNA sequence required for synthesis of a functional protein or RNA molecule. In addition to the coding regions or exons, a gene includes transcription control regions, such as enhancers, and other critical noncoding regions such as poly(A) sites and splice sites. Sometimes essential control regions can even be located in introns. 8\. What is the advantage of complex transcription units over simple transcription units? Ans: A complex transcription unit can encode mRNAs processed in multiple ways to generate different proteins. A simple transcription unit can code for only one RNA and one protein. Complex transcription units allow for a greater diversity of proteins from the same number of genes. 9\. After a diagnostic sequencing analysis of an individual's DNA, you find that this person has a number of microsatellite triplet repeats within a region of their huntingtin gene. Specifically, these CAG repeats code for long stretches of: a\. glycines. b\. prolines. c\. glutamines. d\. stop codons. Ans: c 10\. Blood samples were retrieved from a crime scene, and three suspects were arrested on suspicion of committing the crime. Which of the following techniques could be used to identify the suspect(s) responsible for crime? a\. DNA fingerprinting b\. polymerase chain reaction c\. in situ hybridization d\. DNA fingerprinting and polymerase chain reaction Ans: d 11\. Which of the following organisms has the greatest amount of DNA per cell? a\. chicken b\. fruit fly c\. tulip d\. human Ans: c 12\. All the following statements about microsatellite DNA are true except: a\. It consists of a repeat length of 1--13 base pairs. b\. It can cause neurological diseases such as myotonic dystrophy. c\. It can occur within transcription units. d\. all of the above Ans: d 13\. Which of the following classes of repetitive DNA is most abundant in the human genome? a\. simple-sequence DNA b\. non-LTR transposons c\. LTR transposons d\. DNA transposons Ans: b Difficulty: Moderate 14\. Describe the general organization of protein coding genes in the yeast and human genomes. Ans: In yeast, the protein coding regions are closely spaced along the DNA sequence. In contrast, in the human genome, only a small fraction of the DNA encodes for protein. Thus, the density of protein coding genes per length of DNA is higher in yeast than it is in humans. Put another way, the human genome contains a much higher proportion of noncoding to coding sequences than does the yeast genome. 15\. Describe the proposed mechanism discussed in this chapter for the origin of gene families. Ans: A gene family consists of a set of duplicated genes that encode proteins with similar but not identical amino acid sequences. An example of a gene family is the genes encoding the β-like globins. The different genes in the gene family probably arose by duplication of an ancestral gene, most likely as a result of an unequal crossover during meiotic recombination. Over time, these duplicated genes accumulated random mutations. In some cases, a protein with a slightly different function emerged; in other cases, the mutations led to a nonfunctional gene known as a pseudogene. 16\. What is the underlying mechanism behind why gene mutations that lead to Huntington's disease act as dominant mutations? Ans: The mutations that lead to Huntington's disease are examples of expanded microsatellite repeats. In the case of the gene responsible for Huntington's disease, there is a triplet CAG repeat in the first exon. Expansion of this repeat results in synthesis of long stretches of polyglutamine. Over time, the protein products that contain long stretches of polyglutamine aggregate. Protein aggregation leads to neuronal cell death, which in turn gives rise to the symptoms of Huntington's disease. These microsatellite mutations are dominant because the presence of aggregated proteins causes symptoms, even though some normal proteins are produced from the normal allele. *Section 8.3* 17\. *Drosophila* is considered a model system because it is quite easy to create transgenic lines harboring a variety of different genes. As a *Drosophila* geneticist, which one of the following is a DNA transposon that you would exploit to create a transgenic line? a\. copia element b\. N element c\. P element d\. Ty element Ans: c Difficulty: Moderate 18\. All the following steps are performed by the enzyme transposase during transposition of bacterial insertion sequences except a\. excision of the IS element from the donor DNA molecule. b\. introduction of staggered cuts into the target DNA molecule. c\. ligation of the IS element to the target DNA. d\. synthesis of DNA to fill in the single-stranded gaps. Ans: d Difficulty: Difficult 19\. Which of the following is not a mobile DNA element? a\. transposon b\. long terminal repeats (LTR) c\. long interspersed elements (LINES) d\. insertion sequence (IS) elements Ans: b Difficulty: Moderate 20\. Which of the following mobile elements is a retrotransposon? a\. yeast Ty element b\. bacterial IS sequence c\. *Drosophila* P element d\. maize activator (Ac) element Ans: a Difficulty: Moderate 21\. SINES (short interspersed elements) a\. are approximately 300 base pairs long. b\. are LTRs containing retrotransposons. c\. are present in over 1 million copies in the human genome. d\. a and c Ans: d Difficulty: Moderate 22\. Mobile DNA elements likely contributed to the evolution of higher organisms by the a\. generation of gene families by gene duplication. b\. creation of new genes by exon shuffling. c\. formation of more complex regulatory regions. d\. all of the above Ans: d Difficulty: Difficult *Section 8.4* 23\. Which of the following is an algorithm designed to compare the sequence of a newly identified protein with sequences already stored in the GenBank database? a\. LINES b\. BLAST c\. HATs d\. 3C Ans: b Difficulty: Easy 24\. The DNA and protein sequences of the α-tubulin genes in humans and in fish are similar, and because each arose due to speciation, these genes would be considered: a\. homologous. b\. orthologous. c\. paralogous. d\. autologous. Ans: b Difficulty: Moderate 25\. Describe the two major pathways for transposition of mobile elements. Ans: Mobile elements fall into two major classes. Insertion sequences and transposons move via a DNA intermediate, whereas retrotransposons transpose via an RNA intermediate. DNA elements encode a transposase enzyme, which catalyzes the transposition event. A retrotransposon is first transcribed into RNA, which is then used as a template for synthesis of double-stranded DNA by the action of the retrotransposon-encoded enzyme, reverse transcriptase. The resulting double-stranded DNA is then integrated into the host genome. *Section 8.5* 26\. To examine the folding and compaction of chromatin during mitosis, you will need to isolate and stain chromosomes at a particular stage using a special spreading preparation technique. For the best analysis, the chromosomes must be at which one of the following stages? a\. metaphase b\. interphase c\. telophase d\. anaphase Ans: a Difficulty: Easy 27\. There are five major types of histone proteins, but only four of them are considered as core histones. Which one of the following is NOT considered a core histone protein? a\. H1 b\. H2B c\. H3 d\. H4 Ans: a Difficulty: Easy 28\. Histone modifications play integral roles in chromatin condensation and function. Which of the following is NOT considered to be a histone modification? a\. acetylation b\. methylation c\. phosphorylation d\. prenylation Ans: d Difficulty: Easy 29\. You are studying the regulation of a group of genes and have determined that the full activation of transcription of these genes occurs when histone acetyl transferases have made post-translational modifications specifically to which one of the following amino acids? a\. glycine b\. glutamine c\. lysine d\. proline Ans: c Difficulty: Moderate 30\. "3C" or chromosome conformation capture methods, used to determine the three-dimensional spatial organization of chromatin within nuclei of interphase cells, rely on a series of steps where the end result is the sequence analysis of purified DNA fragments. Which one of the following presents the correct order of steps you as an investigator need to follow in a 3C method strategy? a\. shear DNA to 200--600 bp; cross-link proteins and DNA with formaldehyde b\. ligate linkers marked with biotin onto DNA fragments; dilute and ligate the fragments c\. cross-link streptavidin to DNA; purify and shear biotin-labeled fragments d\. none of the above Ans: b Difficulty: Difficult 31\. Which of the following pairs of proteins are considered to be paralogous? a\. yeast α-tubulin and yeast β-tubulin b\. yeast α-tubulin and worm α-tubulin c\. fly β-tubulin and human β-tubulin d\. worm β-tubulin and human α-tubulin Ans: a Difficulty: Easy 32\. How many genes are estimated to be in the human genome? a\. 21,000 b\. 35,000 c\. 75,000 d\. 100,000 Ans: a Difficulty: Easy 33\. Open reading frame (ORF) analysis is not effective in identifying genes in higher eukaryotes because of the presence of a\. promoters. b\. enhancers. c\. introns. d\. repetitious DNA. Ans: c Difficulty: Moderate 34\. Which of the following lines of evidence is indicative of the presence of a gene in an unknown DNA sequence? a\. alignment to a partial cDNA sequence b\. sequence similarity to genes of other organisms c\. ORF consistent with the rules for exon and intron sequences d\. all of the above Ans: d Difficulty: Moderate *Section 8.6* 35\. Which of the following terms describes the phenomenon of genes occurring in the same order on a chromosome in two different species? a\. heterochrony b\. neoteny c\. synteny d\. phylogeny Ans: c Difficulty: Easy 36\. Which of the following terms describes when a chromosome is replicated everywhere except the telomeres and centromere, but the daughter chromosomes do not separate? a\. hybridization b\. polytenization c\. polymerization d\. heterochromatization Ans: b Difficulty: Moderate 37\. Which of the following is NOT a functional element required for any eukaryotic chromosome to replicate and segregate correctly? a\. replication origin b\. centromere c\. kinetochore d\. telomeres Ans: c Difficulty: Easy 38\. All the following statements are true about a nucleosome except: a\. It contains an octamer core of histones. b It is about 10 nm in diameter. c\. It is the "string" of the "beads-on-a-string" appearance. d\. It contains approximately 150 base pairs of DNA. Ans: c 39\. DNA that is transcriptionally active a\. is more susceptible to DNase I digestion. b\. is tightly packed into a solenoid arrangement. c\. contains nonacetylated histones. d\. is more condensed than nontranscribed DNA. Ans: a 40\. All of the following can be found in chromatin except a\. DNA. b\. histones. c\. RNA. d\. transcription factors. Ans: c 41\. Which of the following statement(s) is (are) true of a eukaryotic chromosome? a\. It is a linear structure. b\. It consists of a single DNA molecule. c\. It can contain greater than a billion base pairs of DNA. d\. all of the above Ans: d 42\. In mammals, X-chromosome inactivation a\. occurs in half the diploid cells of the adult female. b\. results from the ionization of the X-chromosome. c\. is considered an epigenetic event. d\. b and c Ans. c 43\. Describe how modification of histone tails can control chromatin condensation. Ans: The amino termini of histones, which are known as histone tails, extend from the structure of the nucleosome. Positively charged lysine side chains present in the histone tails may interact with linker DNA or other nucleosomes. Acetylation of the lysine side chains neutralizes the positive charges, thereby eliminating the potential interaction with the negatively charged DNA phosphate groups. Thus, acetylation of histones makes the chromatin less likely to form a condensed structure. Deacetylation of the histones once again allows the positively charged lysines to interact with the DNA phosphate groups, leading to chromatin condensation. *Section 8.7* 44\. The karyotype for any particular species is characterized by a\. the number of metaphase chromosomes. b\. the size and shape of the metaphase chromosomes. c\. the banding pattern of the metaphase chromosomes. d\. all of the above Ans: d 45\. Chromosome painting involves a\. staining chromosomes with Giemsa reagent. b\. hybridizing fluorescent probes to chromosomes. c\. hybridizing radioactive probes to chromosomes. d\. a and b Ans: b 46\. All the following statements about heterochromatin except: a\. It is a dark-staining area of a chromosome. b\. It is usually transcriptionally active. c\. It is often simple sequences of DNA. d\. It is a region of condensed chromatin. Ans: b 47\. Telomeres a\. consist of repetitive sequences with high G content. b\. are a few hundred base-pairs long in vertebrates. c\. have specific proteins bound at the DNA ends. d\. a and c Ans: d 48\. Why is there a need for a specialized structure at the ends of eukaryotic chromosomes and for the enzyme telomerase? Ans: Because all known DNA polymerases elongate DNA in the 5´ to 3´ direction, all require a RNA or DNA primer to initiate synthesis. As the replication fork approaches the end of the chromosome, DNA synthesis on the leading strand continues to the end of the chromosome without a problem. However, because the lagging strand is synthesized discontinuously, it cannot be replicated in its entirety. When the RNA primer is removed, a short segment of DNA remains single-stranded with no way to make this region double-stranded. If there were no specialized mechanism for replicating DNA at the ends, then the chromosome would shorten with each round of replication. Telomerase is the enzyme that completes DNA synthesis at the telomeres.