The Molecular Biology of Eukaryotes Practice Test (page 3)
Review the following concepts if needed:
- Genome Size and Complexity for Genetics
- Gene Expression for Genetics
- Regulation of Gene Expression for Genetics
- Development and the Molecular Biology of Eukaryotes for Genetics
- Somatic Nuclear Transfer and Cloning for Genetics
- Organelles and the Molecular Biology of Eukaryotes for Genetics
The Molecular Biology of Eukaryotes Practice Test
For each of the following definitions, give the appropriate term and spell it correctly. Terms are single words unless indicated otherwise.
- The analogue in eukaryotes of the bacterial Pribnow box. (Two words.)
- Genes involved in metabolism that tend to be active at all times in all nucleated cells. (Two words.)
- The total complement of proteins in a cell.
- A DNA sequence in cis position with a structural gene that potentiates the transcriptional activity of a gene on that same DNA molecule even though it may be far distant upstream or downstream from the gene it influences.
- A genelike DNA sequence bearing close resemblance to a functional gene at a different locus, but rendered nonfunctional by additions or deletions in its structure that prevent its transcription and/or translation.
- A small protein that, when covalently attached to target proteins, marks them for destruction by proteases.
- The removal of intron sequences from mRNA and coupling of flanking exons.
- Descriptive of mutant genes that cause a normal body part to develop in an abnormal location.
- Programmed cell death.
- The name of a theory that explains the origin of mitochondria and chloroplasts in eukaryotic cells.
Choose the one best answer.
- Nucleolar organizing regions (NORs) are most closely associated with (a) protein-coding genes (b) rRNA genes (c) replication forks (d) cellular differentiation (e) homeoboxes
- Which of the following statements about mitochondria is incorrect? (a) Mitochondrial DNA has a higher mutation rate than nuclear DNA. (b) Replication of mitochondrial DNA is synchronized with that of chromosomal DNA. (c) The size of mitochondrial DNA varies considerably from one species to another. (d) Mitochondria contain a protein-synthesizing system of their own. (e) Mitochondria usually follow a maternal inheritance pattern in species with equalsized gametes.
- Differentiation of most somatic cells does not appear to involve the loss of genes or recombination of DNA segments. The most striking exception to this rule is found in (a) histone genes (b) rDNA (c) mitochondrial DNA (d) immunoglobulin genes (e) hemoglobin genes.
- Which of the following differentiates eukaryotic DNA replication from prokaryotic replication? (a) multiple origins of replication (b) bidirectional replication fork (c) no use of an RNA primer (d) use of only one DNA polymerase (e) none of the above.
- Hormones are thought to regulate gene activity primarily at the level of (a) transcription (b) mRNA processing (c) transport of RNA from nucleus to cytoplasm (d) translation (e) post-translation processing of protein
- Which of the following is not characteristic of most mRNA processing in eukaryotes? (a) addition of a poly-A tail at the 30 end (b) addition of an unusual guanine to the 50 end (c) removal of exons and splicing together of introns (d) removal of leader and trailer sequences (e) more than one of the above
- There are three kinds of RNA polymerases (I, II, III) in eukaryotic cells, each specific for one class of RNA molecule (mRNA, tRNA, rRNA). Which of the following is a correct match? (a) I = rRNA; II = tRNA (b) II = mRNA; III = rRNA (c) I = tRNA; III = rRNA (d) I = rRNA; II = mRNA (e) none of the above
- Apoptosis involves which of the following: (a) chromatin condensation (b) membrane blebbing (c) DNA fragmentation (d) cell dissolution (e) all of the above
- A cell in its final functional state and form is said to be (a) determined (b) transdetermined (c) differentiated (d) activated (e) stabilized
- Which of the following chain combinations could be found in a functional antibody combining site? (a) Llambda HMu (b) Llambda Lkappa (c) Hgamma Halpha (d) more than one of the above (e) none of the above
The Molecular Biology of Eukaryotes Questions
- The haploid genome of Drosophila melanogaster contains approximately 1.8 × 108 nucleotide pairs. The chromosomes of a polytene nucleus collectively have about 5000 bands. Assuming that 95% of the DNA is located in these bands and 5%is in the interband regions (a) determine the average number of nucleotide pairs in each band and interband region and (b) estimate the average number of genes per band and interband if an average gene contains 103 nucleotide pairs.
- The puffing pattern of Drosophila polytene chromosomes seems to change in a predictable pattern during larva development. (a) It has been suggested that these puffs are the sites of genes transcription. How could this hypothesis be tested experimentally? (b) The synthesis of a particular protein coincides with the appearance of a specific puff in one of the polytene chromosomes. What inference can be made from this observation?
- A sex-linked mutation in humans results in deficiency of the enzyme glucose-6-phosphate dehydrogenase (G6PD). Some individuals with this enzyme defect are more resistant to malaria than are those without this enzyme defect. Among those parasitized, approximately half of the blood cells of the resistant females contain the causative parasite; the cells of G6PD-deficient males are not parasitized. How can these observations be explained?
- Different mRNA molecules have characteristic half-lives. Propose a method for estimating the half-life of a specific mRNA.
- If a certain protein is found in the Golgi apparatus, how can you explain the fact that its cytoplasmic mRNA transcript contains 24 codons at its 5' end that are not represented by corresponding amino acids at the amino terminus of the protein?
- The gonads of Drosophila develop from material in the posterior end of the oocyte cortex (outer layer) that contains densely staining polar granules. An autosomal recessive mutation gs, when homozygous, causes adult females to produce oocytes without polar granules; progeny that develop from such eggs do not develop gonads. If parents are of genotype gs+/gs, predict the results for the next two generations.
- A mutant gene in Drosophila called antennapedia causes legs to develop on the head where antennae normally appear. To what class of developmental control genes does antennapedia belong? Offer an explanation as to how this mutation might cause abnormal development.
- A female fruit fly is homozygous recessive for a mutation in a maternal effect gene important for embryonic development (m/m). (a) What kind of offspring will this fly produce when mated to a wild type male (m+ /m+)? (b) What kind of offspring will be produced by a male that is homozygous recessive for the same mutation when mated to a wild-type female? (c) How is it possible to generate this female fly (m/m) from a cross involving a homozygous recessive parent?
- A female fruit fly is heterozygous recessive for a mutation in a zygotic gene important for embryonic development (z+/z). (a) What kind of offspring will this fly produce when mated to a wild-type male (z+/z+)? (b) What kind of offspring will this fly produce when mated to a male heterozygous for the same mutation? (c) Is it possible to isolate a mature fly that is homozygous recessive for a zygotic gene mutation? Explain.
- The bicoid protein (see Example 13.20) is localized to the anterior portion of the Drosophila oocyte by the action of two proteins swallow and exuperantia. (a) What effect would a mutation in the swallow gene have on embryonic development? (b) How would this effect be altered in a fly containing a mutation in both the swallow and the nanos genes?
- (a) Design a mutant screen to identify dominant (gain-of-function) mutations in maternal effect genes in Drosophila. (b) Design a screen to identify recessive zygotic gene mutations.
- The Drosophila gene hunchback (see Example 13.20) contains five repeats of 5'-TCTAATCCCC-3' in its promoter. Bicoid protein is known to bind to each of these sites. When all five sites are bound with bicoid, maximal transcriptional activation occurs. What would be the phenotype of a fly that contains a deletion of three of the five repeats?
- Why are translation controls less critical for eukaryotic genes than for many prokaryotic genes?
- The ced-3 gene (see Example 13.22) is required for apoptosis in C. elegans. (a) How would a loss-of-function mutation in this gene effect C. elegans development? (b) What about a gain-of-function mutation? (c) Is there likely to be a human oncogene counterpart to the ced-3 gene?
- The N terminus of a polypeptide destined to cross a membrane contains a signal sequence or signal peptide that is removed by a signal peptidase enzyme sometime during the passage of the rest of the polypeptide through the membrane. (a) What kinds of amino acids would be expected to predominate in the signal peptide? (b)Why is protein translocation through the mitochondrial or chloroplast membranes potentially more complex than that through the endoplasmic reticulum? (c) In what major respect does the nuclear membrane differ from the membranes of other organelles?
- The concentration of cytoplasmic mRNA is observed to be higher under one condition when compared with another. Does this observation indicate that transcription control of that gene is operative? Explain.
- Pseudogenes are nontranscribed DNA sequences that are highly homologous in nucleotide sequence to functional genes found elsewhere in the same genome. One class of pseudogenes, known as "processed pseudogenes," is characterized by absence of introns and upstream promoter sequences and presence of 3' terminal poly-A tracts. Propose a mechanism that might account for the origin of processed pseudogenes.
- The diagram below represents a spread of nucleolar chromatin, showing gene transcription on a segment of DNA containing a tandemly arranged series of nucleolar rRNA genes. They give the appearance of a linear series of Christmas-tree-like structures, first identified in electron micrographs by O. L. Miller and B. R. Beatty (1969), and have been subsequently referred to as "Miller trees." Identify the following structures or regions. (a) The limits of an rRNA gene for the 38S rRNA precursor molecule. (b) A nontranscribed spacer DNA region between the rDNA repeats. (c) Promoter or initiator region. (d) Terminator of an rDNA gene. (e) RNA polymerase molecules. (f ) 5' end of an rRNA transcript. (g) Do you expect to find ribosomes? (h) Is a similar phenomenon expected in E. coli? If so, explain any predicted differences.
- A recessive chromosomal gene produces green and white stripes in the leaves of maize, a condition called "japonica." This gene behaves normally in monohybrid crosses giving a 3 green : 1 striped ratio. Another striped phenotype was discovered in Iowa, named "iojap" (a contraction of Iowa and japonica), which is produced by a recessive gene ij when homozygous. If a plant with iojap striping serves as the seed parent, then the progeny will segregate green, striped, and white in irregular ratios regardless of the genotype of the pollen parent. Backcrossing striped progeny of genotype Ij/ij to a green pollinator of genotype Ij/Ij produces progeny that continue to segregate green, striped, and white in irregular ratios. White plants die due to lack of functional chloroplasts. Green plants produce only green progeny except when the genotype of the progeny is ij/ij; striping then reappears. Interpret this information to explain the inheritance of iojap.
- If a woman contracts German measles during the first trimester of pregnancy, the child may be seriously affected even though the mother herself suffers no permanent physical effects. Such anomalies as heart and liver defects, deafness, cataracts, and blindness often occur in the affected children at birth. Can these phenotypic results be considered hereditary abnormalities?
- A snail produced by a cross between two individuals has a shell with right-hand twist (dextral). This snail produces only left-hand (sinistral) progeny by selfing. Determine the genotype of this snail and its parents. See Solved Problem 13.2.
- Most strains of Chlamydomonas (Fig. 5-1) are sensitive to streptomycin (see Example 13.25). A strain is found that requires streptomycin in the culture medium for its survival. How could it be determined whether streptomycin-dependence is due to a chromosomal gene or to a cytoplasmic element?
- A yeast (Fig. 6-4) culture, when grown on medium containing acriflavine, produces numerous minute cells that grow very slowly. How could it be determined whether the slow growth was due to a cytoplasmic factor or to a nuclear gene?
- Determine the genotypes and phenotypes of sexual progeny in Neurospora from the following crosses: (a) fast-poky male × normal female of genotype F', (b) poky female × fast-poky male, (c) fast-poky female × poky male (see Solved Problem 13.4).
- The cells of a Neurospora mycelium are usually multinucleate. Fusion of hyphae from different strains results in the exchange of nuclei. A mycelium that has genetically different nuclei in a common cytoplasm is called a heterokaryon. Moreover, the union results in a mixture of two different cytoplasmic systems called a heteroplasmon or a heterocytosome. The mycelia of two slow-growing strains, each with an aberrant cytochrome spectrum, fuse to form a heteroplasmon that exhibits normal growth. Abnormal cytochromes a and b are still produced by the heteroplasmon. Offer an explanation for this phenomenon.
- Male sterile plants (Solved Problem 13.5) in corn may be produced either by a chromosomal gene or by a cytoplasmic factor. (a) At least 20 different male-sterile genes are known in maize, all of which are recessive. Why? Predict the F1 and F2 results of pollinating (b) a genetic male sterile by a normal, and (c) a cytoplasmic male sterile by a normal.
- Given seed from a male sterile line of corn (see Solved Problem 13.5), how would you determine if the sterility was genic or cytoplasmic?
- A bacterial spirochaete that is passed to the progeny only from the maternal parent has been found in Drosophila willistoni. This microorganism usually kills males during embryonic development but not females. The trait is called "sex ratio" (SR) for obvious reasons. Occasionally, a son of an SR female will survive. This allows reciprocal crosses to be made. The SR condition can be transferred between D. equinoxialis and D. willistoni. The spirochaete is sensitive to high temperatures, which inactivates them, forming "cured" strains with a normal sex ratio. (a) What would you anticipate to be the consequence of repeated backcrossing of SR females to normal males? (b) A "cured" female is crossed to a rare male from an SR culture. Would the sex ratio be normal? Explain.
- How will the sequencing of whole genomes change the way that geneticists approach the understanding of gene structure and function?
- The evolution of mitochondrial DNA occurs at a rate much faster than that of nuclear DNA. Hence, there is much greater variation from one person to another in mitochondrial DNA sequences than in nuclear DNA sequences. Of all the existing human populations, there appears to be greater variation in those of Africa than any other place on Earth. Furthermore, all the human mitochondrial DNA sequences can be arranged into a single phylogenetic tree. Assuming that the mitochondrial DNA of our most ancient ancestors had the same amount of individual variation as that of modern mitochondrial DNA, what are the evolutionary implications of these facts?
- Hogness box
- housekeeping genes
- e (c and d)
The Molecular Biology of Eukaryotes
- (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
- (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
- (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.
- 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.
- 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.
- 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.
- 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."
- 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.
- (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.
- (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.
- (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.
- (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).
- 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.
- 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).
- (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.
- (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.
- 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.
- 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.
- 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.
- 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.
- Parents: s+ s female × s/? male; F1: ss
- 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+.
- 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).
- (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)
- 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.
- (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.
- 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.
- (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.
- 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.
- 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.
(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.
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