Test Bank 5th-Ed Molecular Biology of the Cell by Bruce Alberts

Test Bank 5th-Ed Molecular Biology of the Cell by Bruce Alberts


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Test Bank 5th-Ed Molecular Biology of the Cell by Bruce Alberts

Sample Chapter No 1                                               


Cells and Genomes

Molecular Biology of the Cell, Fifth Edition

Ó 2008 Garland Science Publishing






1-1       Your friend studies the SARS virus and tells you that the SARS virus packages its genetic material as a positive-strand single-stranded RNA. A positive-strand RNA is an RNA molecule that can serve as a translatable template to produce protein. Given what you know about the production of nucleic acids in cells, would you expect that the positive-strand RNA molecule could be used as a template to produce other positive-strand RNA molecules, or would you expect that another nucleic acid intermediary would have to be produced first? Explain.


1-2       You are interested in studying HIV. On reading the literature, you learn that the genetic information for the HIV virus is a single-stranded RNA molecule. Given what you know about the properties of nucleic acids, would you expect the ratios of purines to pyrimidine bases in the HIV viral RNA to be the same as the ratio found in the hereditary information of all living cells? Explain.


1-3       You have two computer files, one containing the complete Escherichia coli (bacterial) genomic sequence and the other containing the complete Drosophila melanogaster (fruit fly) genomic sequence. A new student in the lab made a mistake and renamed the files “A” and “B.” However, you do know that file A is 1125 kilobytes while file B is 30.5 megabytes in size. Without opening the file and examining the sequence, can you predict which file is more likely to contain the genomic sequence for E. coli and which file contains the genomic sequence for Drosophila? Explain.


1-4       You have two tubes. Tube 1 contains the DNA encoding the genome of a human being and tube 2 contains the DNA encoding the genome of a lithotrophic archaean that lives adjacent to a deep-sea hydrothermal vent that has an average temperature of 115°C (note that normal human body temperature is 37°C). You lend your tubes to a chemist friend, who loses the label on your tubes. However, he was able to determine the percentage molar ratios of As and Ts in these tubes, and finds that the DNA in tube 1 is composed of 59% As and Ts while the DNA in tube 2 is composed of 48% As and Ts. Which tube contains the human DNA and which tube contains the archaean DNA? Explain.


1-5       The adult human intestine contains hundreds of species of bacteria, and these bacteria often have specialized niches within the intestine. Members of the Bacteriodetes genus are common members of the intestinal flora, and the genomes of several species within this genus have been sequenced, including a species (B. dulcis) that is particularly capable of digesting plant polysaccharides. When the sequences of the Bacteriodetes species were compared, it was found that orthologous genes had between 60 to 80% amino acid identity between species. B. dulcis was found to have a set of genes encoding enzymes important for polysaccharide degradation that did not exist in the other species. Interestingly, sequence comparisons of these polysaccharide-degrading genes suggested that the closest relatives of these genes were in archaea and not in bacteria. Provide two possible explanations for how this set of genes could have arisen.


1-6       It is now generally accepted that mitochondria and chloroplasts originated as free-living bacteria that were engulfed by an ancestral eucaryotic cell. However, it is also known that the mitochondrial DNA and the chloroplast DNA lack genes for many essential functions. This means that, were the mitochondria or the chloroplast to “escape” the modern day eucaryotic cell, the mitochondria or the chloroplasts could not survive on their own. Explain what has happened that allows the mitochondria and chloroplasts to survive in the host cell, even though they lack the DNA encoding essential genes in their DNA.


1-7       Your friend obtained the sequence of an mRNA that encodes a gene important for tooth formation in cheetahs. Since she knows that the Us in an mRNA sequence were encoded by corresponding Ts in the DNA, she changes the Us in the mRNA to Ts and uses this nucleotide sequence to search the published genomic sequence of humans for a homologous gene. Her database search identifies a region of human DNA that has 35% identity with the cheetah nucleotide sequence (meaning, if you line up the two nucleotide sequences, 35% of the corresponding bases are exactly the same). However, when she showed this human DNA region to her boss, her boss took a quick look and told her that her search had not identified a homologous gene and that there was a better way to conduct her search. She has not yet taken cell biology, and turns to you for help. Given your knowledge of how a cell encodes and uses genetic information, devise a better strategy for using this nucleotide sequence to identify the homologous gene in humans.


1-8       Explain what an amphipathic molecule is and why in an aqueous solution a bilayer of amphipathic molecules self-aggregates to form a vesicle that contains water, whereas a monolayer of amphipathic molecules self-aggregates into a sphere that excludes water.







1-1       Another nucleic acid intermediary would have to be produced first. According to base-pairing rules, a single-stranded RNA molecule could not directly replicate itself. However, if either a complementary DNA or RNA molecule were produced as an intermediary, that intermediary could produce more of the positive-strand RNA molecule. (A positive-strand virus has the same polarity as the viral mRNA and can be used as a direct template for translation but not for viral replication. A positive-strand virus must produce small amounts of a negative-strand RNA intermediate or a reverse-transcribed DNA intermediate, which is then used to generate more positive-strand viral RNA that is then encapsulated into new virus particles.)


1-2       No, you would not expect the molar ratio of purine to pyrimidine bases in the HIV RNA to be the same as that found in double-stranded DNA, the hereditary information of all living cells. In double-stranded DNA, the molar ratio of purine to pyrimidine bases is equal to 1 because As pair with Ts, and Cs pair with Gs. Because the genetic material of HIV is single-stranded RNA, base-pairing rules do not apply. Although the single-stranded RNA may have some base-pairing, which contributes to its secondary structure, the nucleotides in the single-stranded RNA molecule do not have to be paired with another strand.


1-3       File A is likely to contain the E. coli genomic sequence, and file B is likely to contain the Drosophila genomic sequence. In general, procaryotic genomes are smaller than eucaryotic genomes. (The E. coli genome is made of 4.6 million nucleotide pairs, while the Drosophila genome is made of 122 million nucleotide pairs, making the Drosophila genome more than 20 times the size of the E. coli genome.)


1-4       Tube 1 probably contains the human genomic DNA, and tube 2 contains the DNA from the deep-sea lithotropic archaea. The DNA in tube 1 has a greater AT content than the DNA in tube 2. Because a CG base pair is stronger than an AT base pair, the melting temperature of the DNA in tube 1 will be lower than that for the DNA in tube 2. Because humans exist at lower temperatures than temperatures at the deep-sea hydrothermal vents, it is likely that human genomic DNA will have a lower melting temperature than the genomic DNA of organisms that live at high temperatures.


1-5       The genes involved in polysaccharide degradation could have arisen by

(A) horizontal gene transfer from another species, or

(B) differential gene loss such that the ancestral Bacteriodetes had genes encoding the polysaccharide-degrading enzymes, but these genes have been lost from all other members of the genus that have been sequenced.



1-6       Although the many essential genes needed for survival of the mitochondria and chloroplast are no longer encoded by their respective DNAs, these genes have not been lost. Instead, these “missing genes” are now in the DNA of the eucaryotic host-cell nucleus. Thus, the mitochondria and the chloroplasts are now symbionts that require their host cells for survival.


1-7       Ultimately, protein function is what is conserved over evolutionary time, and thus homologous proteins are most similar at the amino acid level. Because mRNAs are translated by tRNAs that recognize three nucleotides at a time and because more than one three-nucleotide combination can code for a particular amino acid, your friend should translate her mRNA into the protein that it encodes and use the amino acid sequence to search against a database that contains the potential translated proteins from the human genome. By searching using the nucleotide sequence, your friend will obtain sequences that may seem to weakly match the cheetah gene at the nucleotide level but cannot code for the protein she is interested in studying.


1-8       An amphipathic molecule has one part that is water insoluble (hydrophobic) and another part that is water soluble (hydrophilic). A monolayer of amphipathic molecules in an aqueous solution will self-aggregate to form a sphere that excludes water because all the hydrophilic regions will want to come in contact with each other and will avoid interacting with water. These interactions will form a sphere with the hydrophilic regions on the outside of the sphere and the hydrophobic regions enclosed inside the sphere, avoiding interaction with water (Fig. A1-8A). On the other hand, a bilayer of amphipathic molecules in an aqueous solution could start out arranged in a sheet with their hydrophilic regions exposed to the surface and their hydrophobic regions contacting one another, thus avoiding interaction with water. However, hydrophobic regions will be exposed to water at the edges of the sheet. Thus, the sheet will close up into a vesicle. Water will be enclosed in the center of the vesicle in contact with the hydrophilic regions of the molecule (Fig. A1-8B).


Figure A1-8

Test Bank 5th-Ed Molecular Biology of the Cell by Bruce Alberts

Test Bank 5th-Ed Molecular Biology of the Cell by Bruce Alberts