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The Molecular Biology of Eukaryotes Practice Test (page 3)

By — McGraw-Hill Professional
Updated on Aug 23, 2011

Answers

Vocabulary

  1. Hogness box
  2. housekeeping genes
  3. proteome
  4. enhancer
  5. pseudogene
  6. ubiquitin
  7. splicing
  8. homeotic
  9. apoptosis
  10. endosymbiosis

Multiple-Choice

  1. b
  2. b
  3. d
  4. a
  5. a
  6. e (c and d)
  7. d
  8. e
  9. c
  10. a

The Molecular Biology of Eukaryotes

  1.  
    1. (1.8 × 108 nucleotide pairs × 0:95)/(5 × 103 bands) = 3.4 × 104 nucleotide pairs per band (1.8 × 108 nucleotide pairs × 0.05)/(5 × 103 bands) = 1.8 × 103 nucleotide pairs per interband
    2. (3 × 104 nucleotide pairs per band) / (103 nucleotide pairs per gene) = 30 genes per band (1 × 103 nucleotide pairs per interband)/(103 nucleotide pairs per gene) = 1 gene per interband
  2. (a) Affix the cells containing the polytene chromosomes to a slide and expose them to radioactive uracils. Active genes will synthesize RNAs containing the labeled uracils. Then wash the slide to remove any unincorporated label and cover it with photographic film sensitive to the radiation (autoradiography). If dots on the developed film are concentrated in the puffed areas, the hypothesis would be confirmed.   (b) It might be tempting to speculate that one or more genes in the puffed regions is actively synthesizing the protein. However, it is also possible that the protein itself might be responsible for the change in the puff pattern; or perhaps some other protein synthesized at the same time might be responsible. Thus, cause and effect cannot be established from this observation.
  3. In most female mammals, including humans, one of the X chromosomes is inactivated. In females that areheterozygous for the mutation, about half of the cells would be expected to have an inactive X chromosome containing the normal G6PD gene and an active X chromosome containing the mutant gene; these cells would have a deficiency of G6PD and would not be parasitized. In the remainder of these female cells, the other X chromosome bearing the normal gene would be active, would not be G6PD deficient, and therefore would be parasitized. Males do not inactivate their single X chromosome. Males hemizygous for G6PD deficiency would therefore not be parasitized.
  4. Expose cells to one or more radioactive ribonucleotides for a defined short period of time (a pulse) followed by large amounts of unlabeled ribonucleotides (the chase). After various lengths of chase, expose all of the labeledm RNAs to a single-stranded cDNA that is of a nucleotide sequence complementary to that of the mRNA species under consideration. Measure the amount of radioactivity in the mRNA-cDNA hybrids trapped on a nitrocellulose filter. For any given chase time, the relative stability of an mRNA should be directly related to the amount of radioactivity detected in the hybrids.
  5. Proteins that are destined to cross the endoplasmic reticulum (ER; which contributes to the Golgi apparatus) contain an N-terminal leader sequence called the signal peptide. This peptide is cleaved after it has performed its job of aiding the polypeptide in passing through the ER membrane.
  6. Offspring bearing the normal gene gs+ are fertile and can produce a second generation. F1 females of genotype gs/gs produce offspring (regardless of the genotype of their mates) that will be sterile; hence, those F1 females cannot have "grandchildren." The name of this mutation is "grandchildless."
  7. Antennapedia is a homeotic gene in which mutation causes transformation of one body part into another. In other words, it makes the right structure in the wrong place. Homeotic genes contain a regulatory sequence that responds to signals from other control genes. In embryonic cells of the imaginal (imago = adult insect) leg disks, the wild-type antennapedia gene is normally active; in cells of the imaginal antennae disks, the gene is normally inactive (silenced). A mutant antennapedia gene might fail to respond to the signals that normally turn off its normal allele in the antennae disks, and thus it is somehow able to direct development of legs instead of antennae.
  8. (a) The female will be sterile, i.e., not able to produce viable offspring.   (b) The F1 offspring will be viable; however, 1/4 of F2 females will be sterile.   (c) The mother of this fly must be heterozygous in order to produce viable offspring, so the male parent must be homozygous recessive.
  9. (a) The offspring will be viable: 1/2 z+ /z : 1/2 z+/=z+.   (b) 1/4 of the offspring (z/z) will not produce viable embryos.   (c) It is not expected that a homozygous recessive zygotic gene mutation would be able to develop properly, so no flies of this genotype are expected. However, it may depend on the particular gene and its action.
  10. (a) The bicoid protein will not be as concentrated in the anterior portion of the oocyte (i.e., more of it will be located in the posterior than normal); thus, anterior structures may develop in the posterior end.   (b) The effect would be exaggerated since nanos helps suppress anterior development in the posterior end.
  11. (a) Initial crosses should reveal females that are sterile, because a dominant gene will have an effect even in a heterozygous state. (b) The F2 of mutagenized flies would have to be examined for embryonic lethals. The F2 is necessary to generate homozygous genotypes from the initial heterozygous mutant (a new mutation is expected in only one of two alleles).
  12. When more bicoid protein is present, there are higher levels of hunchback protein. If three of five bicoid binding sites are mutated, the effect would be similar to having less bicoid protein and so less hunchback gene activation would occur. This would result in a reduced concentration of hunchback protein and a less dramatic gradient throughout the embryo. Formation of head, thorax, and other anterior structures would be altered. Posterior structure formation would most likely be unaffected.
  13. All eukaryotic cytoplasmic mRNAs are monocistronic, whereas many prokaryotic mRNAs are polycistronic. Thus, within a bacterial operon, if the product of one structural gene is needed and another is not, then gene expression must be controlled at the translation level (or perhaps posttranslationally).
  14. (a) Loss of ced-3 would result in survival of cells that would normally die.   (b) Gain of ced-3 function would perhaps result in premature death of cells programmed to die.   (c) Yes, only loss of function mutations will have a cancerous effect.
  15. (a) Hydrophobic amino acids are expected in the signal peptide so that it can easily be inserted into the hydrophobic lipid membrane bilayer to initiate transport of the attached protein.   (b) This is because these organelles are surrounded by a double membrane system, whereas the endoplasmic reticulum membrane consists of a single membrane.   (c) There are pores in the nuclear membrane.
  16. Transcriptional control may be operative. However, posttranscription regulation may be solely responsible or in conjunction with transcription control. Primary eukaryotic mRNA transcripts are subject to 5' capping, 3' polyadenylation, intron removal and splicing of exons, RNA degradation, and transport across the nuclear envelope; each operation presents a potential control point.
  17. If cytoplasmic mRNA could be copied into DNA, it would have the characteristics described for processed pseudogenes. The retroviral enzyme reverse transcriptase can make DNA from viral RNA, and this cDNA can then become integrated into the host's genome, thus providing a model for the origin of processed pseudogenes.
  18. (g) No, because ribosomal RNAs and ribosomal proteins are assembled into ribosomes in the cytoplasm, not in the nucleus where Miller trees are formed.   (h) Yes, but there will be ribosomes present on the transcripts.

  19. It appears that the chromosomal gene ij, when homozygous, induces irreversible changes in normal plastids. The plastids exhibit autonomy in subsequent generations, being insensitive to the presence of Ij in single or double dose. Random distribution of plastids to daughter cells could give all normal plastids to some, all defective plastids to others, and a mixture of normal and defective plastids to still others. All plastids are not rendered defective in the presence of ij/ij as this would produce only white (lethal) seedlings.
  20. It is important to distinguish between congenital defects (recognizable at birth) that are acquired from the environment during embryonic development and genetic defects that are produced in response to the baby's own genotype. The former may be produced by infective agents such as the virus of German measles, which is not really a part of the baby's genotype but is acquired through agents external to the developing individual. An active case of this disease usually produces immunity in the mother so that subsequent children of this mother should not be susceptible to the crippling influences of this virus. A hereditary disease is one produced in response to instructions of an abnormal gene belonging to the diseased individual and that can be transmitted in Mendelian fashion from generation to generation.
  21. Parents: s+ s female × s/? male; F1: ss
  22. Cross ss mt (male) × sd mt+ (female); if chromosomal, 25% of the sexual progeny should be ss mt ; 25% ss mt+, 25% sd mt, 25% sd mt+; if cytoplasmic, almost all of the progeny should follow the maternal line (streptomycin-dependent) as in Example 13.25, while mating type segregates 1 mt: 1 mt+.
  23. From a cross of minute × normal, a nuclear gene will segregate in the spores in a 1 : 1 ratio (e.g., segregational petite in Problem 13.3). If an extranuclear gene is involved, segregation will not be evident and all spores will be normal (e. g., neutral petite in Example 13.26).
  24. (a) All phenotypically normal; 1/2F' (normal cytoplasm) : 1/2 F (normal cytoplasm).   (b) and   (c) 1/2 fast-poky; F (poky cytoplasm): 1/2 poky; F' (poky cytoplasm)
  25. One strain may have an abnormal cytochrome a but a normal cytochrome b. The other strain might have an abnormal cytochrome b but a normal cytochrome a. The normal cytochromes in the heteroplasmon complement each other to produce rapid growth.
  26. (a) A plant in which a dominant genic male sterile gene arose by mutation of a normal gene would be unable to fertilize itself and would be lost unless cross-pollinated by a fertile plant. The gene would be rapidly eliminated from heterozygotes within a few generations by continuous backcrossing to normal pollen parents.   (b) F1: +/ms, fertile; F2: = + / +; = + /ms; 1/4 ms/ms; 3/4 fertile, 1/4 male sterile.   (c) Male sterile cytoplasm is transmitted to all F1 progeny; a selfed F2 cannot be produced because none of the F1 plants can make fertile pollen.
  27. Plant the seeds and pollinate the resulting plants with normal pollen from a strain devoid of male sterility. If the F1 is sterile, then it is cytoplasmic; if the F1 is fertile, it is genic.
  28. (a) If interspecific crosses can transmit the spirochaete, it is probably relatively insensitive to the chromosomal gene complement. Backcrossing would cause no change in the SR trait; indeed, this is how the culture is maintained.   (b) The sex ratio would probably be normal. It is unlikely that the spirochaete would be included in the minute amount of cytoplasm that surrounds the sperm nucleus.
  29. Previously, geneticists did not have knowledge of the location or sequence of genes. Gene function was discovered through the isolation of mutants with particular mutant phenotypes. With the knowledge of gene location and sequence, a different approach can be taken. For example, in yeast, strains containing deletions of genes whose function is not known can be created and their phenotype examined. In addition, gene sequence knowledge allows inference of amino acid sequence and thus allows comparisons of protein sequence with other known genes. This can help geneticists make better educated guesses regarding gene function. Experiments can be tailored to answer specific questions based on these sequence comparisons. For example, if a particular motif is known to be important in a previously identified gene and a newly sequenced gene is found to contain the same motif, mutations can be directed to that region to determine if they have similar functions. This approach is more efficient that screening through thousands of mutant organisms looking for a particular mutant phenotype.
  30. Africa was probably the place where humans first evolved, and our ancestors emigrated from there to populate the world. Mitochondria are maternally inherited. From this fact and the single family tree for mitochondrial DNA, it has been inferred that there probably was a single ancestral woman (our "mitochondrial Eve") from which all humans derived their mitochondrial DNA.
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