Practice Questions and Answers

9 Pages
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Department
Molecular and Cellular Biology
Course Code
MCB 2050
Professor
Jim Kirkland

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Chapter 20 20.1 Operons are common in bacteria but not in eukaryotes. Suggest a reason why. ANS: In multicellular eukaryotes, the environment of an individual cell is relatively stable. There is no need to respond quickly to changes in the external environment. In addition, the development of a multicellular organism involves complex regulatory hierarchies composed of hundreds of different genes. The expression of these genes is regulated spatially and temporally, often through intricate intercellular signaling processes. FEEDBACK: 20.6 DIFFICULTY: easy 20.2. In bacteria, translation of an mRNA begins before the synthesis of that mRNA is completed. Why is this “coupling” of transcription and translation not possible in eukaryotes? ANS: Coupling of transcription and translation is not possible in eukaryotes because these two processes take place in different cellular compartments—transcription in the nucleus and translation in the cytoplasm. FEEDBACK: 20.1 DIFFICULTY: easy 20.3 Muscular dystrophy in humans is caused by mutations in an X-linked gene that encodes a protein called dystrophin. What techniques could you use to determine if this gene is active in different types of cells, say skin cells, nerve cells, and muscle cells? ANS: Activity of the dystrophin gene could be assessed by blotting RNA extracted from the different types of cells and hybridizing it with a probe from the gene (northern blotting); or the RNA could be reverse transcribed into cDNA using one or more primers specific to the dystrophin gene and the resulting cDNA could be amplified by the polymerase chain reaction (RT-PCR). Another technique would be to hybridize dystrophin RNA in situ—that is, in the cells themselves—with a probe from the gene. It would also be possible to check each cell type for production of dystrophin protein by using anti-Dystrophin antibodies to analyze proteins from the different cell types on western blots, or to analyze the proteins in the cells themselves—that is, in situ. FEEDBACK: 20.1 DIFFICULTY: medium 20.4. Why do steroid hormones interact with receptors inside the cell, whereas peptide hormones interact with receptors on the cell surface? ANS: Steroid hormones are small, lipid-soluble molecules that have little difficulty passing through the cell membrane. Peptide hormones are typically too large to pass through the cell membrane freely; rather, their influence must be mediated by a signaling system that begins with a membrane-bound receptor protein that binds the hormone. FEEDBACK: 20.1 DIFFICULTY: easy 20.5 How could you use the polytene chromosomes of Dipteran insects to study the regulation of transcription? ANS: By monitoring puffs in response to environmental signals, such as heat shock, or to hormonal signals. FEEDBACK: 20.1 DIFFICULTY: easy 20.6. How would you distinguish between an enhancer and a promoter? ANS: An enhancer can be located upstream, downstream, or within a gene and it functions independently of its orientation. A promoter is almost always immediately upstream of a gene and it functions only in one direction with respect to the gene. FEEDBACK: 20.2 DIFFICULTY: easy 20.7 Tropomyosins are proteins that mediate the interaction of actin and troponin, two proteins involved in muscle contractions. In higher animals, tropomyosins exist as a family of closely related proteins that share some amino acid sequences but differ in others. Explain how these proteins could be created from the transcript of a single gene. ANS: By alternate splicing of the transcript. FEEDBACK: 20.2 DIFFICULTY: easy 20.8. A polypeptide consists of three separate segments of amino acids, A—B—C. Another polypeptide contains segments A and C, but not segment B. How might you determine if these two polypeptides are produced by translating alternately spliced versions of RNA from a single gene, or by translating mRNA from two different genes? ANS: Southern blotting of genomic DNA digested with an appropriate restriction enzyme, followed by hybridization of the blot with a probe containing the DNA encoding segments A and B, or B and C, or at least parts of these adjacent segments. If one DNA fragment is detected on the blot, the two polypeptides are encoded by a single gene whose RNA is alternately spliced to produce two mRNAs. If two DNA fragments are detected, the two polypeptides are encoded by two different genes. FEEDBACK: 20.2 DIFFICULTY: medium 20.9 What techniques could be used to show that a plant gene is transcribed when the plant is illuminated with light? ANS: Northern blotting of RNA extracted from plants grown with and without light, or PCR amplification of cDNA made by reverse transcribing these same RNA extracts. FEEDBACK: 20.2 DIFFICULTY: medium 20.10. When introns were first discovered, they were thought to be genetic “junk”—that is, sequences without any useful function. In fact, they appeared to be worse than junk because they actually interrupted the coding sequences of genes. However, among eukaryotes, introns are pervasive and anything that is pervasive in biology usually has a function. What function might introns have? What benefit might they confer on an organism? ANS: Introns make it possible for genes to encode different—but related—polypeptides by alternate splicing of their RNA transcripts. FEEDBACK: 20.2 DIFFICULTY: medium 20.11 Why does it make sense for the photosynthetic enzyme ribulose 1,5-bisphosphate carboxylase (RBC) to be synthesized specifically when plants are exposed to light? ANS: This enzyme plays an important role in photosynthesis, a light-dependent process. Thus, it makes sense that its production should be triggered by exposure to light. FEEDBACK: 20.2 DIFFICULTY: medium 20.12. Using the techniques of genetic engineering, a researcher has constructed a fusion gene containing the heat-shock response elements from a Drosophila hsp70 gene and the coding region of a jellyfish gene ( gfp) for green fluorescent protein. This fusion gene has been inserted into the chromosomes of living Drosophila by the technique of transposon- mediated transformation (Chapter 18). Under what conditions will the green fluorescent protein be synthesized in these genetically transformed flies? Explain. ANS: The green fluorescent protein will be made after the flies are heat shocked. FEEDBACK: 20.3 DIFFICULTY: medium 20.13 Suppose that the segment of the hsp70 gene that was used to make the hsp70/gfp fusion in the preceding problem had mutations in each of its heat-shock response elements. Would the green fluorescent protein encoded by this fusion gene be synthesized in genetically transformed flies? ANS: Probably not unless the promoter of the gfp gene is recognized and transcribed by the Drosophila RNA polymerase independently of the heat shock response elements. FEEDBACK: 20.3 DIFFICULTY: medium 20.14. The polypeptide products of two different genes, A and B, each function as transcription factors. These polypeptides interact to form dimers: AA homodimers, BB homodimers, and AB heterodimers. If the A and B polypeptides are equally abundant in cells, and if dimer formation is random, what is the expected ratio of homodimers to heterodimers in these cells? ANS: With equal abundance of the A and B polypeptides, AA homodimers should constitute 1/4 of the total dimers formed, BB homodimers should constitute 1/4 of the total, and AB heterodimers should constitute 1/2 of the total. The expected ratio of homodimers to heterodimers is therefore (1/4 + 1/4):(1/2) = 1:1. FEEDBACK: 20.3 DIFFICULTY: hard 20.15 A particular transcription factor binds to enhancers in 40 different genes. Predict the phenotype of individuals homozygous for a frameshift mutation in the coding sequence of the gene that specifies this transcription factor. ANS: The mutation is likely to be lethal in homozygous condition because the transcription factor controls so many different genes and a frameshift mutation in the coding sequence will almost certainly destroy the transcription factor’s function. FEEDBACK: 20.3 DIFFICULTY: medium 20.16. The alternately spliced forms of the RNA from the Drosophila doublesex gene encode proteins that are needed to block the development of one or the other set of sexual characteristics. The protein that is made in female animals blocks the development of male characteristics, and the protein that is made in male animals blocks the development of female characteristics. Predict the phenotype of XX and XY animals homozygous for a nul
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