RNA and Protein: Transcription, Translation, and the Genetic ...
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RNA and Protein: Transcription, Translation, and the Genetic ... RNA and Protein: Transcription, Translation, and the Genetic ... Document Transcript

  • DNA Structure, Function, and Replication 1. The backbone of each strand in a DNA molecule consists of: a. base-phosphate linkages b. sugar-base linkages c. sugar-phosphate linkages d. sugar-base-phosphate linkages e. none of the above 2. At a replication point on DNA, a. Okazaki fragments are made on both strands of the DNA. b. RNA primers are involved. c. only one protein is involved. d. the process is referred to as "conservative" because the old strands remain together. e. None of the above are correct. 3. What parts of adjacent nucleotides must the ligase proteins be linking together? 4. As one strand of a DNA molecule is replicating, the sequence of nucleotide bases that are added is ATC. Can you tell from this sequence what will be the next base to be added to the strand? 5. In a Meselson-Stahl experiment, what will be the expected density distribution and abundance of DNA after three generations? 6. In organisms living today, a. DNA self-replicates, without help from other molecules. b. DNA replication requires only one enzyme. c. DNA replication requires a number of enzymes. d. none of the above is correct. 7. Which of the following requires hydrogen bonds to maintain normal structure? a. DNA b. protein c. water d. all of the above are correct e. none of the above are correct RNA and Protein: Transcription, Translation, and the Genetic Code True/False 8. Base-pair substitutions are examples of point mutations. 9. Mutations can produce frame-shifts. 10. Point mutations cannot produce large changes in an organism. Multiple Choice 11. Point mutations or deletions in DNA: a. can result in the change of a single amino acid in a protein. b. can result in the premature termination of a protein. c. can result in a number of amino acids being changed. d. can produce no change in the amino acid sequence of a protein. e. All of the above are correct.
  • 12. Use the following information about the genetic code to answer the question below: UUU = phenylalanine; CUU = leucine; AAA = lycine. Assume that a normal mRNA creates the following amino acid sequence: ...-phenylalanine-phenylalanine-lycine-... A mutation substitutes leucine for the second phenylalanine. Using the above part of the genetic code, determine the sequence present in the coding (template) strand of the mutated DNA. a. AAAGAATTT b. AATAAATTT c. GGGTGGGGG d. TTTGTTAAA e. TTTATTGGG Short Answer 13. Distinguish between missense and nonsense mutations. 14. Can there be mutations that produce changes in organisms without changing the amino acid sequence in a particular protein? Cell Cycle, Mitosis, and Meiosis 15. When during the cell cycle does the shift to G0 usually occur? a. during G1 b. during G2 c. during S d. in the middle of mitosis e. during cytokinesis 16. Comparing a cell in the G1 phase before meiosis with the cells at the end of meiosis II, a. a diploid cell becomes four haploid cells. b. a diploid cell becomes two haploid cells. c. a diploid cell becomes two diploid cells. d. a diploid cell becomes four diploid cells. e. a haploid cell becomes two diploid cells. 17. Crossing over is: a. the shift of whole chromosomes from one side of the metaphase plate to the other side during meiosis. b. what the chicken did to get to the other side of the road. c. a form of genetic recombination where pieces of homologous chromosomes get exchanged. d. an event that occurs about 3 times per chromosome during meiosis I in humans. e. Both c and d, above, are correct. Mendelian Genetics 18. Consider the following cross: AaBbCc x AabbCC. What percent of the offspring will be homozygous dominant at the C-gene locus? a. 0% b. 25%
  • c. 50% d. 75% e. 100% 19. An individual with a dominant phenotype is crossed with an individual that exhibits a recessive phenotype for the same character. If the dominant individual is: a. heterozygous, all of the offspring will exhibit the dominant trait. b. homozygous, all of the offspring will exhibit the dominant trait. c. heterozygous, all of the offspring will exhibit the recessive trait. d. homozygous, all of the offspring will exhibit the recessive trait. e. homozygous, half of the offspring will exhibit the recessive trait. 20. Grandparents, neither of whom has a recessive disorder have a daughter who does not have the disorder. Nevertheless, two of their daughter's children (their grandchildren) have the disorder. Ignoring the possibility of new mutations, this suggests that: a. Both of the two grandparents could be homozygous dominant, since the "bad" gene could have come from the son-in-law. b. At least one of the grandparents must have been heterozygous. c. Both of the grandparents must have been heterozygous. d. the daughter must be homozygous recessive. 21. Which of the following is NOT correct? a. In a test cross, one is determining an organism's genotype by crossing it with a homozygous recessive individual. b. In incomplete dominance, an intermediate phenotype, between the phenotypes of the two alleles (when homozygous) is observed. c. a monohybrid cross follows the inheritance of one character. d. polygenic inheritance refers to characters depending on more than one gene. e. in a dihybrid cross the F2 generation always exhibit a ratio of four phenotypes of 9:3:3:1. Recombination, Linkage, and Mapping 22. A cross involving two genes (doubly heterozygous individual crossed with a doubly homozygous-recessive) gives 600 parental types and 602 recombinant types. What is the relationship between the two genes? a. they are separated on the same chromosome by about 50 map units. b. they are separated on the same chromosome by more than 50 map units. c. they are found on different chromosomes. d. all of the above are possibilities. 23. A woman has a disorder, and her husband does not. The disorder is inherited by half of her offspring, both male and female. Her children marry spouses who are normal. Of her daughters who have the disorder, half of their children also have the disorder. Of her sons who have the disorder, all of their daughters have the disorder, but none of their sons inherit it. The most likely explanation is: a. sex-linked recessive inheritance b. sex-linked dominant inheritance
  • c. autosomal-recessive inheritance d. autosomal-dominant inheritance 24. A female has a genetic disorder. Her husband does not. Among her sixteen (!) offspring, all of her sons inherit the disorder, but none of her daughters do. The most likely explanation is that this is: a. sex-linked recessive inheritance b. extrachromosomal inheritance c. autosomal inheritance 25. In the case of linked genes, recombinant phenotypes are produced by: a. crossing over b. sex-linked inheritance c. presence of the Barr body in females d. incomplete dominance e. codominance Genetics of Bacteria and Viruses 26. A bacteriophage can produce over 100 progeny in: a. 30 min. b. 6 hrs. c. a day d. a week e. a month 27. How could restriction enzymes be useful? 28. Using the numbers given in the textbook, calculate how many base pairs there are in E. coli for each gene. We know that there are no introns in prokaryotic DNA. Does your number suggest that there is no room for such introns? 29. In bacterial cells, the regulation of gene expression is: a. not reliable because bacteria are too small and events inside are too statistical in nature. b. not present because regulation of gene expression occurs only in eukaryotic cells. c. mostly at the level of translation. d. mostly at the level of transcription. e. both a and c, above, are correct. 30. Transduction a. is the taking up of bare DNA by a bacterium. b. occurs when a virus takes a piece of DNA from one bacterium to another. c. occurs when a plasmid is transferred between bacteria. d. is the synthesis of RNA on a DNA template. e. is the synthesis of DNA on an RNA template. 31. In the tryptophan (Trp) operon, a. the repressor is active unless it is bound to tryptophan. b. the repressor is inactive unless it is bound to tryptophan. c. there is only one structural gene.
  • d. there is a pre-mRNA which contains introns as well as exons. e. Both c and d, above, are correct. Gene Expression, Gene Technology, and Human Genetics 32. Both parents of a child have freckles, a dominant trait. The odds of the child having freckles are equal to or more than: a. 0 b. 1/4 c. 2/4 d. 3/4 e. 1 33. RFLPs a. are generated from DNA by restriction enzymes. b. are present because of differences in DNA sequences in samples of DNA from different individuals. c. are able to be seen by cutting DNA at particular sequences and separating the resulting DNA fragments. d. all of the above are correct. e. none of the above is correct. 34. RFLPs are: a. differences in lengths of restriction fragments. b. produced by differences in gene sequences that are recognized by restriction enzymes. c. made by the polymerase chain reaction. d. made by cloning genes in plasmids. e. both a and b, above, are correct. 35. The regulation of gene expression in eukaryotic cells: a. only occurs during transcription. b. only occurs in the cytosol, or cytoplasm. c. only occurs in the nucleus. d. can occur at many levels, including transcription, mRNA masking, mRNA degradation, etc. e. does not occur, because the "readout" of genes is not changed with time in such cells. 36. What do the alleles that cause sickle-cell and cystic fibrosis have in common? a. they both are recessive b. both are relatively common in some populations, suggesting that they confer a selective advantage when heterozygous c. they tend to occur together in the same people d. both a and b, above, are correct e. all of the above are correct 37. Gene expression in eukaryotic cells can be regulated by: a. regulation of mRNA degradation. b. DNA modification. c. transcription factors.
  • d. mRNA masking. e. all of the above. 38. The polymerase chain reaction: a. is used to make many copies of polymerase molecules. b. is used to make chains of polymerase molecules. c. is used to make many copies of particular DNA molecules. d. is called cloning. Evolution and Natural Selection 39. The inheritance of acquired characteristics: a. is a currently accepted theory of how evolution occurs. b. would suggest that, if you spend hours learning how to swim, your child will be born a better swimmer. c. explains how genetic recombination can occur. d. all of the above are correct. 40. Are humans the most advanced, the most evolved, and at the end of evolution? 41. In the study of English peppered moths, it was presumed or found that: a. the light and dark coloration are heritable traits in these moths. b. the coloration gives a selective advantage depending on the environment. c. if the environment changes, so can what is selected for. d. all of the above. 42. In an experiment on fruit flies, offspring produced when parent flies are old are chosen to produce the next generation. This is continued for many generations. It is found that the population of flies lives longer, compared to the flies that one started with. What is this an example of? 43. The presumptions that are involved in the theory of natural selection do NOT include which of the following? a. heritable variations b. an actual physical battle between individuals, with survival of the physically fittest individual c. limited resources d. the potential for overpopulation through reproduction Microevolution and Population Genetics 44. Gene flow occurs: a. after a bottleneck. b. after the first, few members of a species populate an island. c. when mutations occur. d. when members of one population migrate to another population. e. Both a and b, above, are correct. 45. A population in Hardy-Weinberg equilibrium has two alleles, B and b, at a given gene locus. Given that the frequency of the homozygous recessive organisms in the population is 25%, a)
  • what is the frequency of the dominant allele in the gene pool? b) What is the frequency of heterozygous individuals in the population? 46. Members of a population originally exhibit a range of gray color. Over a number of generations, the average color darkens as a result of natural selection. This is most likely an example of: a. non-random mating b. diversifying selection c. sexual selection d. directional selection e. stabilizing selection 47. What is the difference between non-random mating and sexual selection? 48. Natural selection: a. is a process whereby genes are selected randomly for preservation in the next generation. b. contributes to Hardy-Weinberg equilibrium. c. makes perfect creatures. d. produces organisms that are more "fit" to their environments. e. is a theory that opposes evolution. 49. Which of the following CANNOT occur while significant natural selection is occurring? a. adaptive evolution b. increased fitness c. Hardy-Weinberg equilibrium d. sexual selection e. a reduction in frequency of deleterious genes 50. Which of the following is an example of balanced polymorphism? a. heterozygote advantage b. non-random mating c. sexual selection d. directional selection e. all of the above 51. A population is NOT in Hardy-Weinberg equilibrium. At one gene locus there are two alleles at a gene locus, one dominant and one recessive. The fraction of homozygous dominant individuals in the population is 0.5; the fraction of homozygous recessive individuals is 0.3. What fraction of the gene pool consists of dominant alleles at this gene locus? a. 0.3 b. 0.6 c. 0.2 d. 0.1 e. 0.4 52. Hardy-Weinberg equilibrium requires: a. natural selection b. random mating
  • c. small population sizes d. significant migration between populations e. genetic drift 53. Which of the following is a form of natural selection? a. genetic drift b. non-random mating c. sexual selection d. isolation in a population e. no net mutations in a population Answers 1. c. The backbone is sugar-phosphate, repeating over and over as the nucleotides are linked to one another to form each chain. 2. b. The RNA primers are used by DNA polymerase to initiate the making of DNA. All of the other statements are incorrect. 3. Since the ligase enzymes are combining parts of what will become one strand of the DNA, they are linking a phosphate group on the nucleotide at the end of one chain with the ribose sugar group on the end-nucleotide of the next chain, thereby forming a single, longer strand. 4. No, you cannot tell which base will be next. You can only tell what the complimentary strand has as a sequence opposite the ATC. 5. After two replications, the distribution was half intermediate, half light. Since there will be one more semi-conservative replication with light isotopes, the distribution after three generations will be one-third intermediate and two-thirds light. 6. c. 7. d. 8. True. Base pair substitutions change the DNA at one point. 9. True. Examples of such frame-shifting mutations include deletions and additions of single base pairs. 10. False. Point mutations can cause an enzyme to be inactive, and if it is an essential enzyme, death of the organism can result. We will consider such a case in the next section. 11. e. Mutations that do not change amino acid sequence would include those that happen to generate a redundant codon and those occurring outside of reading frames in the DNA, such as in introns.
  • 12. a. Splitting the codons gives AAA/GAA/TTT, which is transcribed to: UUU/CUU/AAA in the mRNA, which translates to: phenylalanine/leucine/lysine in the protein. 13. Missense mutations substitute one amino acid for another. Nonsense mutations produce a stop codon which prematurely terminates the polypeptide chain. 14. Yes. Consider the possibility of a mutation in the region of DNA coding for a tRNA or rRNA. Furthermore, changes in promoter regions or binding regions for RNA polymerase could change the abundance of a gene product without changing the product itself. 15. a. The shift occurs before DNA is duplicated. 16. a. The cell starts as a diploid cell, duplicates its DNA once, then forms a total of four daughter cells, so each must have half the initial amount of DNA, or a haploid amount. 17. e. (although someone might wish to argue for answer b, as well) 18. c. No, we did not consider three-factor crosses, but you only need to realize that the question only asks about one of the gene loci. You can essentially ignore the information about the other two gene loci, A and B. Then it is merely a matter of noticing that we are crossing Cc with CC. Clearly we have a simple cross, considering the gametes that can form gives us: C C CC c cC It should be evident that 50% of the offspring are homozygous dominant (CC). 19. b. This is just a test cross, as described in the book. 20. b. Since the grandchildren are "aa," they must have inherited one "a" from each side of the family, so at least one grandparent must be a carrier (Aa). 21. e. This answer is incorrect because there are a number of different genotypes that can be used in a dihybrid cross. 22. d. With a recombination frequency of about 50%, these two genes appear to be unlinked. 23. b. 24. a. 25. a. 26. a. The process is remarkably rapid. 27. Restriction enzymes cut DNA at particular sequences of nucleotides. Such restriction enzymes can be used to detect such sequences in a DNA, by cutting any DNA containing the
  • sequence into fragments. In later chapters it will be shown how such fragments can be useful in several different ways. 28. 4 million base pairs divided by 3,000 genes gives about 1,300 nucleotide base pairs/gene. Since codons are three nucleotides long, this would provide for an average protein size of 430-440 amino acids, which is about typical for known proteins. This suggests that there is very little "extra" DNA that could be used for things like introns, especially when one realizes that information for making rRNAs and tRNAs also must be coded by the bacterial genome. 29. d. In prokaryotic cells transcription is the major site for regulation of gene expression. 30. b. 31. b. 32. d. Each parent must be either Ff or FF (where F = freckles). Since the question asks for the lowest odds, then we can presume each parent to be Ff, so the odds of the child being Ff or FF follows from the possible gametes (bold) from each parent: F f F FF Ff f fF ff Thus there are four, equally probable offspring. 3/4 of them have dimples. If either parent is FF, the odds become 100%, so answer d is the minimum chance. 33. d. The polymorphisms in RFLPs are present because of the differences in DNA nucleotide sequences between the DNAs from different sources. 34. e. 35. d. 36. b. 37. e. 38. c. With each cycle, PCR doubles the copies of the DNA. 39. b. This is a hypothesis that was shown to be false. 40. We are the most intelligent (hopefully not too smart--consider atomic weapons), but we cannot climb a tree like our close cousin the chimp, nor are we as numerous as say, our friends the cockroaches. We grow as many bacteria in each of us each day as there are humans on earth. All organisms have evolved to fit niches, and there are many which fit very well into niches quite different from ours. There also is no sense in which we are the end of evolution, unless it be because, by mistake, we were to destroy all life on earth, an unlikely occurrence. 41. d.
  • 42. This is an example of artificial selection, and is an experimental example of how natural selection can work. In this case selection for one trait, the having of late offspring, influences another trait, the average life span of members of the population. 43. b. 44. d. Answers a and b are examples of genetic drift, not gene flow. 45. a) Since the frequency of homozygous recessive individuals is 0.25, the frequency of the recessive allele in the population, q, is the square root of 0.25 = 0.5, or 50%. So, the frequency of the dominant allele, p = 1 - q = 0.5, or 50%. b) The frequency of heterozygous individuals is 2pq = 2 (0.5)(0.5) = 0.5, or 50%.. 46. d. Directional selection would describe this shift of color. Although less likely, answer c, sexual selection, could not be ruled out on the basis of the information given since the darker gray color could be a secondary sex characteristic.. 47. Non-random mating is a violation of Hardy-Weinberg equilibrium. It is not the same as sexual selection because it does not entail a change in the frequency of alleles in gene pools. Non-random mating includes inbreeding and assortative mating. Sexual selection is a form of natural selection and so does influence the frequency of alleles in gene pools. Sexual selection involves secondary sex characteristics. You should not confuse the two.. 48. d. 49. c. 50 a. 51. b. The fraction of heterozygous individuals must be .2 (the remainder after summing both homozygous groups). Thus the frequency of homozygous dominant genes in the gene pool must be: (.5+.5+.2)/2 = .6 52. b. 53. c. The other answers might influence microevolution, but are not examples of natural selection.