Biology General Genetics Exams 1998

Biology General Genetics Exams 1998 - CURRENT SEMESTER'S...

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Unformatted text preview: CURRENT SEMESTER'S EXAMS (Fall, 1998) ------------------------------------------------------------------------------------------------- Final Exam, Part I (New Material, since last Mid-Term) [Answers at bottom] There are 40 questions (1-40) in Part I (New Material), and 45 questions (51-95) in Part II (Review Material). Note that there are no questions numbered 41-50. Part I and Part II will be graded separately. 1. A negatively regulated genetic system…. A. is more common in eukaryotes. B. requires an effector molecule acting at the promoter. C. is derepressed by an inducer. D. none of the above. E. all of the above. 2. Constitutive gene expression refers to… A. constant, unregulated expression. B. polycistronic mRNA. C. coupled transcription/translation. D. a situation where mutant repressor no longer responds to inducer. E. gene expression following addition of inducer. 3. An operator mutation may give rise to constitutive gene expression. A. true B. false 4. The product of the lacI gene in E. coli… A. is cis-dominant. B. cannot be complemented. C. induces lac operon transcription. D. binds to the lac operator. E. transports lactose into the cell. 5. The tryptophan operon is… A. inducible B. repressible. 6. Which one of the following features of gene regulation is present only in prokaryotes? A. gene dosage. B. activators. C. enhancers. D. alternative splicing. E. attenuators. 7. Genes which when mutated can cause cancer are called… A. proto-oncogenes. B. tumor-suppressor genes. C. regulator genes. D. more than one of the above. E. none of the above. 8. During embryonic development, cells all have the same set of genes (the result of mitotic divisions). Nevertheless, different cells in the embryo come to function differently because of… A. different chemical messages stored in different parts of the egg cytoplasm that are parceled out to different cells during the early mitotic division of the embryo. B. effects of different micro-environments on the cells of the early embryo. C. both of the above. D. neither of the above. 9. A mutation in which (on one DNA strand) one purine is substituted for another, or one pyrimidine is substituted for another, is called… A. a transition mutation B. a transversion mutation. 10. A mutation in which a G-C pair in DNA is substituted by a C-G pair is… A. a transition mutation B. a transversion mutation. C. not a mutation. 11. In sickle-cell anemia, the glutamic acid at position 6 in normal hemoglobin is replaced by valine. This is… A. a transition mutation B. a transversion mutation. 12. If a point mutation occurred just upstream from the coding region of a gene, the result could be… A. failure to produce mRNA. B. production of mRNA, but failure of the mRNA to be translated. C. both of the above. D. neither of the above. 13. Frameshift mutations are caused by insertions or deletions. A. true B. false 14. Which one(s) of the following kinds of mutation can give premature termination of the translation product (protein)? A. splice-site mutation B. frameshift mutation C. nucleotide substitution D. all of the above E. none of the above. 15. Which one of the following is a typical mutation rate for a typical gene? A. 1 x 10-1 B. 1 x 10-3 C. 1 x 10-5 16. Most spontaneous mutations, after they have initially occurred in a cell, are corrected by DNA repair mechanisms in the cell. A. true B. false 17. Hermann J. Muller was awarded the Nobel Prize for the discovery that x-rays could be used to induce new mutations. A. true B. false 18. Mutations caused by base substitutions undergo reversion readily, compared to large deletions, which almost never revert. A. true B. false 19. In the coding regions of genes coding for proteins, a base substitution in the first position of a codon always results in an amino acid substitution. A. true B. false 20. In the gene for the A protein of tryptophan synthetase in E. coli, fewer than 30 of the 268 amino-acid positions have been found to be altered, among several hundred mutations isolated. This result has probably been obtained because… A. many of the other amino-acid positions are in fact probably altered too, but most of these mutations do not affect the function of the protein significantly, and so the cells carrying them are never isolated as "mutant" cells. B. mutations are decidedly non-random, affecting only certain regions of the gene -- a common finding for mutations in protein-coding genes in general. C. such mutation studies are not extensive enough to uncover all the mutations that could theoretically be detected; to have a reasonable chance of uncovering mutations at all or most of the 268 amino-acid positions would require the isolation of thousands, not merely hundreds, of mutant cells. D. most mutations in this gene would result in death of the mutant cells, and so would never be recovered. E. most mutations do not lead to amino-acid substitutions. 21. In humans, a polymorphism exists at the D17S213 locus, which can be detected by Southern blotting using EcoRI digestion and a suitable probe: some chromosomes have a restriction fragment that is broken into two pieces by the enzyme (allele "2"), while other chromosomes carry at that locus a base substitution that abolishes the target site (allele "1"). In a survey of 500 people in Atlanta, Georgia, 289 had only allele 1, 29 had only allele 2, and 182 had both alleles. What is the gene frequency for allele 2? (Choose closest answer.) A. 0.60 B. 0.51 C. 0.42 D. 0.33 E. 0.24 22. Ataxia-telangiectasia (A-T) is an autosomal recessive, progressive, degenerative disease characterized by cerebellar degeneration, immunodeficiency, sensitivity to x-rays, and predisposition to cancer. A-T children appear normal at birth, with the first signs of disease usually appearing during the second year of life, commonly a "wobbly" lack of balance and slurred speech caused by ataxia (lack of muscle control). Epidemiologists estimate that the frequency of A-T births is about 1 in 40,000. What is the estimated frequency of carriers? (Choose closest answer.) A. 1 in 100,000 B. 1 in 10,000 C. 1 in 1000 D. 1 in 100 E. 1 in 10 23. If the frequency of a homozygous dominant genotype in a randomly mating population is 10%, what is the frequency of the dominant allele? (Choose closest answer.) A. 16% B. 32% C. 48% D. 60% E. 72% 24. (Refer to above question.) What is the frequency of the dominant phenotype? (Choose closest answer.) A. 40% B. 44% C. 48% D. 52% E. 56% 25. Adrenoleukodystrophy (ALD) is a rare metabolic disorder inherited as an X-linked recessive. (ALD was the illness afflicting young Lorenzo Odone of Maryland, whose story was the basis of the 1993 film, "Lorenzo's Oil," starring Susan Sarandon and Nick Nolte.) In ALD the fatty covering (myelin sheath) on nerve fibers in the brain is lost, and the adrenal gland degenerates, leading to progressive neurological disability and death. It occurs at a frequency of about 2 in every 100,000 male births. What is the allele frequency for (the defective form of) this gene? A. less than 0.00002 B. 0.00002 C. 0.0002 D. 0.002 E. 0.02 26. (Refer to previous question.) What is the frequency of female carries in the population? A. less than 1/25,000 B. 1/25,000 C. 1/2500 D. 1/250 E. more than 1/250 27. In a Pygmy group in Central Africa, the frequencies of the alleles for the ABO blood group are estimated at 0.74 for IO, 0.16 for IA, and 0.10 for IB. What is the expected frequency of type A blood in this population? (Choose closest answer.) A. 21% B. 23% C. 26% D. 29% E. 32% 28. (Refer to previous question.) Did you need to rely on the Hardy-Weinberg law to answer this question? A. yes B. no 29. Which one of the following populations is most likely to be engaging in inbreeding? (The numbers are the fractions of the genotypes AA, Aa, and aa, respectively, in the populations.) A. 0.38 0.53 0.09 B. 0.46 0.38 0.16 C. 0.42 0.46 0.12 30. The "normal distribution" is also known as… A. the standard deviation. B. the standard statistical limit theorem. C. the bell curve. 31. Which of the following is a meristic trait? A. Milk production in cows B. Yield in soybeans per year C. Growth rate of a line of turkeys D. Blood pressure E. Number of eggs laid by a hen per week 32. In a cross between two inbred lines of Drosophila, the F1 population had a variance of 0.87 for abdominal bristles, and the F2 generation had a variance of 3.2. What is VG (the variance due to genetic factors)? A. 0.87 B. 1.7 C. 2.3 D. 3.2 E. 3.7 33. In one study, the correlation coefficient of IQ between identical twins was 0.82, while that between fraternal twins was 0.55. Which of the following conclusions can be drawn from these observations? (Choose best answer.) A. The heritability of IQ is about 82%. B. The heritability of IQ is about 55%. C. The heritability of IQ is about 27%. D. IQ is determined mostly by genes. E. The environment has some influence on IQ. 34. Inbreeding depression results because… A. rare harmful dominant alleles become homozygous. B. rare harmful recessive alleles become heterozygous. C. rare harmful recessive alleles become homozygous. D. heterozygosity is increased. E. mutation rates are increased. 35. If the heritability of a trait is zero, genes are not involved in determining the trait. A. true B. false 36. In the tomato, the locations of many of the quantitative trait loci for fruit weight, acidity, and amount of soluble solids are approximately the same. The most likely explanation for this is… A. high heritability. B. pleiotropy. C. epistasis. D. polygenes. E. linkage. 37. In the tomato, variation in environmental factors accounts for about 13% of the total (phenotypic) variance in fruit weight. What is the (broad-sense) heritability of this trait? (Choose best answer.) A. 0.13 B. 0.50 C. 0.87 38. In one study of maze-learning ability in rats, the mean number of trials necessary to learn the maze was 10.8, with a variance of 4.0. Rats with an ability to learn the maze in 6 trials or less (average, 5.6 trials) were interbred. Their offspring required, on average, 8.2 trials to learn the maze. What is the heritability of maze-learning ability? (Choose closest answer.) A. 10% B. 20% C. 30% D. 40% E. 50% 39. "Neuroticism" (anxiety and emotional stability) has a heritability that is probably in the 30%50% range. A. true B. false 40. Part of the variation from person to person in "neuroticism" (anxiety and emotional stability) can be related to variation in a serotonin-transporter gene. A. true B. false Answers to Final Exam, Part I. 1. C 2. A 3. A 4. D 5. B 6. E 7. D 8. C 9. A 10. B 11. B 12. C 13. A 14. D 15. C 16. A 17. A 18. A 19. B 20. A 21. E 22. D 23. B 24. D 25. B 26. B 27. C 28. A 29. B 30. C 31. E 32. C 33. E 34. C 35. B 36. B 37. C 38. E 39. A 40. A ------------------------------------------------------------------------------------------------- Final Exam, Part II (Review Material) There are 45 questions in this part of the exam. Choose the best answer for each question. [Answers at bottom] You may want to use some or all of the following information. Avogadro's number is 6.023 x 1023. The average "molecular weight" of a nucleotide pair is about 650. The distance between successive base pairs along the axis of the DNA double helix is 0.34 nm. One picogram = 10-12 g Protein molecules have about 350 amino acids, on average. Code table: U C A G U phe ser tyr cys U/C leu ser * * A leu ser * trp G C leu pro his arg U/C leu pro gln arg A/G A ile thr asn ser U/C ile thr lys arg A met thr lys arg G G val ala asp gly U/C val ala glu gly A/G 51. Genetic maps in E. coli are sometimes represented in units of time, rather than units of recombination frequency (%). A. true B. false 52. A person's DNA contains information about how to build… A. nucleic acids. B. proteins. C. ribosomal RNA. D. all of the above. E. none of the above. 53. In double-stranded DNA, the number of phosphate groups is equal to the number of bases. A. true B. false 54. In double-stranded DNA, the two purines are paired (by hydrogen-bonding) with one another, as are the two pyrimidines. A. true B. false 55. The haploid genome of baker's yeast, Saccharomyces cerevisiae, consists of 16 chromosomes containing a total of 0.0131 pg of DNA. If all the DNA molecules were stretched out end-to-end, how long would all this DNA be? A. 4 nanometers B. 4 micrometers C. 4 millimeters D. 4 meters E. 4 kilometers 56. (Refer to above question.) About half of the DNA of baker's yeast consists of DNA between genes; the other half is genic. Estimate the total number of genes this yeast cell has. A. 600 B. 6000 C. 60,000 D. 600,000 E. 6,000,000 57. To make the calculation in the question 55, it is necessary first to determine the total number of base pairs in yeast DNA. A. true B. false 58. Which one(s) of the following kinds of observation indicate that the physical nature of the gene is the DNA molecule? A. transformation in the bacterium Streptococcus. B. transfer of radioactive material from parent bacteriophage T2 to progeny bacteriophage T2. C. the correlation between base changes in DNA and amino acid changes in proteins. D. two of the above. E. all three of the above. 59. Which of the following is not a base found in DNA? A. adenine B. guanine C. uracil D. thymine E. 5-methyl-cytosine 60. The sequence of one strand of DNA is 5' AGCTAG 3'. The sequence of the complementary strand would be A. 5'-AGCTAG-3' B. 5'-TCGATC-3' C. 5'-CTAGCT-3' D. 5'-GCTAGC-3' E. 5'-GATCGA-3' 61. Which type(s) of RNA is produced by the process of transcription? A. mRNA B. tRNA C. rRNA D. mRNA and tRNA only E. mRNA, tRNA and rRNA 62. The process of making an RNA strand from a DNA template is called… A. translation. B. transcription. C. reverse transcription. D. RNA processing. E. replication. 63. What amino acid does the codon UAG code for? A. cysteine B. tryptophan C. leucine D. isoleucine E. It does not code for any amino acid. 64. What amino acid does the codon AAU code for? A. isoleucine B. leucine C. asparagine D. aspartic acid E. It does not code for any amino acid. 65. How many codons correspond to the amino acid serine? A. two B. three C. four D. five E. six 66. For each amino acid, there is one and only one tRNA molecule. A. true B. false 67. In human DNA, the molar content of cytosine is 20%. What is the adenine content? A. 20% B. 30% C. 40% D. 50% 68. A DNA sequence of unknown function that could potentially be translated into protein is a(n)… A. cDNA. B. STS. C. cistron. D. contig. E. ORF. 69. Overlapping clones used in gene mapping are called… A. contigs. B. STSs. C. ORFs. D. T-DNAs. E. cDNAs. 70. A restriction enzyme with which one of the following target sites would be expected to produce the largest DNA fragments from human DNA? (The parentheses indicate that either of the enclosed two bases could be at that position in the target sequence.) A. AGCT B. TC(A/T)GA C. GCATGC 71. The approach to understanding the molecular basis of human genetic disease that involves first finding the gene and then determining the amino-acid sequence of the gene-product in order to get information about its molecular function, is called… A. insertional mutagenesis. B. fine-structure genetic mapping. C. transgene analysis. D. reverse genetics. E. saturation mutagenesis. 72. The structure called a "YAC" is… A. a plasmid. B. a short base sequence consisting of a purine, an adenine, and a cytosine. C. a recombinant-DNA vector that can be cloned in yeast cells. D. the forward three-fifths of a large sailboat. E. a large member of the cattle family used as a beast of burden in Tibet. 73. Sister centromeres separate at anaphase of mitosis, but in meiosis they remain associated through anaphase I and don't separate from each other until anaphase II. A. true B. false 74. In meiosis, crossing-over occurs… A. after the last mitotic division, but before pre-meiotic DNA synthesis. B. after pre-meiotic DNA synthesis, but before the end of prophase I. C. between the end of prophase I and the beginning of prophase II. D. during prophase II. E. after the end of prophase II. 75. How many individual chromatids are present in human spermatocytes in mid-prophase I of meiosis? A. 22 B. 23 C. 46 D. 92 E. some other number. 76. An inversion heterozygote is partially sterile (relative to the homozygote for the standard chromosome arrangement) because of the production of gametes carrying abnormal chromosomes. A. true B. false 77. An inversion homozygote shows an even greater degree of sterility (i.e., produces fewer offspring) than the inversion heterozygote. A. true B. false 78. Down syndrome is caused by… A. monosomy for one of the sex chromosomes. B. trisomy for one of the autosomes. C. trisomy for the sex chromosomes. D. a partial deletion of one of the sex chromosomes. E. a partial deletion of one of the autosomes. 79. Turner syndrome is caused by… A. monosomy for one of the sex chromosomes. B. trisomy for one of the autosomes. C. trisomy for the sex chromosomes. D. a partial deletion of one of the sex chromosomes. E. a partial deletion of one of the autosomes. 80. If a normal man and a normal woman each carrying a defective cystic fibrosis allele marry, what is the probability that their first child will have cystic fibrosis? A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 81. (Refer to previous question.) If this couple has two children, what is the probability that neither one of them will have cystic fibrosis? A. 1/16 B. 4/16 C. 9/16 D. 13/16 E. 15/16 82. The most common defective allele causing cystic fibrosis is one in which three adjacent A-T pairs are deleted. This is… A. a base substitution. B. a frameshift mutation. C. a nonsense mutation. D. a missense mutation. E. none of the above. 83. At the PGM1 locus on human chromosome no. 1, five different alleles have been identified by electrophoresis. How many different genotypes are expected to be encountered at this locus in human populations? A. 5 B. 8 C. 10 D. 15 E. 20 84. How many different kinds of gamete could be formed by an individual with the genotype AaBbCcDd? A. 2 B. 4 C. 8 D. 16 E. 32 85. How many different phenotypes are possible among the progeny of a cross AaBbCc x AaBbCc, where complete dominance occurs at the A and B loci, but partial dominance occurs at the C locus? A. four B. six C. eight D. ten E. twelve 86. A woman with blood type A has a child with blood type O. Which one of the following best represents the possible blood type(s) of the father? A. O B. A C. A or B D. A, B, or O E. A, B, AB, or O 87. In humans, a mild autosomal recessive condition (cc) affects the formation of collagen in connective tissue, and another mild autosomal recessive condition (pp) affects purine metabolism in the liver, leading to occasional gout-like symptoms. Both of these conditions are relatively rare, but over the years many three-generation pedigrees have been found in which, in the first generation, a man with one condition married a woman with the other condition. In what arrangement do the two genes occur in the offspring of such marriages? A. cis B. trans 88. (Refer to above question.) Among all these pedigrees there are several cases of first-cousin marriages in which two double heterozygotes married. If the two genes assort independently, what is the probability than a given child of any of these first-cousin marriages will suffer the symptoms of both diseases? (Choose closest answer.) A. zero B. 5% C. 10% D. 15% E. 20% 89. (Refer to above question.) The two genes in question are in fact known to be located on chromosome 1, about 20 map units apart. With this additional information, is the probability than a given child of any of these first-cousin marriages will suffer the symptoms of both diseases higher, or lower, than your estimate in the previous question? A. higher B. lower 90. (Refer to above question.) The C gene is located at 1p36, and the P gene at 1p32. This means that the genes are located on… A. the short arm of chromosome 1. B. the long arm of chromosome 1. 91. Recombination frequencies increase with increasing distance of the genes on a chromosome. A. true B. false 92. In the domestic cat (Felis catus), three genes (C, A, and T) on chromosome 3 have been the subject of recombination studies. The recombination frequencies are: 23% for C and A; 16% for A and T; and 9% for C and T. What is the order of these three genes on the chromosome? A. CAT B. CTA C. TCA 93. In tetrad analysis of the progeny from a cross between two strains, osm NIT and OSM nit, in the green alga Chlamydomonas reinhardtii, the following numbers of different kinds of tetrads were obtained: 30 parental ditypes (PD), 2 nonparental ditypes (NPD), and 18 tetratypes (T). Are the two genes linked? A. yes B. no 94. (Refer to above question.) Spores with the genotype OSM NIT would be found in… A. PD tetrads. B. NPD tetrads. C. T tetrads. D. PD and T tetrads. E. NPD and T tetrads. 95. (Refer to question 93.) What is the map distance between the two genes? (Choose best answer.) A. The question cannot be answered, because the two genes don't lie on the same chromosome. B. 10 map units C. 20 map units D. 30 map units E. 40 map units Answers to Final Exam, Part II. 51. 61. 71. 81. 91. A E D C A 52. D 62. B 72. C 82. E 92. B 53. A 63. E 73. A 83. D 93. A 54. B 64. C 74. B 84. D 94. E 55. C 65. E 75. D 85. E 95. C 56. B 57. A 58. E 59. C 60. C 66. B 67. B 68. E 69. A 70. C 76. A 77. B 78. B 79. A 80. A 86. D 87. B 88. B 89. B 90. A ------------------------------------------------------------------------------------------------ Biology 250 (General Genetics) November 9, 1998 (50 questions) Answers are at end of exam Fall, 1998 Exam III 1. What is the diploid number of the common chimpanzee (Pan troglodytes)? A. 40 B. 42 C. 44 D. 46 E. 48 2. The Y chromosome of Homo sapiens is a small acrocentric chromosome. The Y chromosome of the common chimpanzee is a small ___________ ____________. A. metacentric chromosome. B. submetacentric chromosome. C. subtelocentric chromosome. D. telocentric chromosome. E. acrocentric chromosome also. 3. Crossing over within the loop of an inversion heterozygote containing the chromosomes ABCDEFGH and ABCFEDGH could produce a gamete carrying which of the following chromatids? (The sequence of letters indicates the sequence of chromosome segments.) A. ABCFEDCBA B. ABCFEEFGH C. ABCDEFHG D. ABCDFEDHG E. ABCDEEDCGH 4. A translocation heterozygote is partially sterile (relative to the homozygote for the standard chromosome arrangement) because of the production of gametes carrying duplications and deficiencies of certain chromosome segments. A. true B. false 5. A translocation homozygote shows an even greater degree of sterility (i.e., produces fewer offspring) than the translocation heterozygote. A. true B. false 6. Down syndrome, besides being caused by non-disjunction, can also be the indirect result of a Robertsonian translocation in one of the parents. A. true B. false 7. (This question is taken directly, without change, from the "Analysis and Applications? section at the end of Chapter 7.) Curly wings (Cy) is a dominant mutation in the second chromosome of Drosophila. A Cy/+ male was irradiated with x rays and crossed with +/+ females, and the Cy/+ sons were mated individually with +/+ females. From one cross, the progeny were curly males 146 wildytpe males 0 curly females 0 wildtype females 163 What abnormality in chromosome structure in the most likely explanation for these results? A. an inversion affecting the X chromosome B. an inversion affecting the Y chromosome C. an inversion affecting the second chromosome D. a translocation between the Cy-chromosome and the X E. a translocation between the Cy-chromosome and the Y 8. The temporal genetic map of Escherichia coli is circular, and so is the E. coli chromosome. A. true B. false 9. The recombination map of the bacteriophage T4 is circular, and so is the T4 chromosome. A. true B. false 10. A recombination map of the l prophage would be expected to be… A. linear B. circular 11. Bacterial transformation is a process in which an F+ bacterium transfers a sex-factor to an Fbacterium, thus transforming the latter into another F+ bacterium. A. true B. false 12. The first demonstration of a kind of sexual process in E. coli was based on finding a few wildtype cells (requiring no nutritional supplements) in mixed cultures of two parental cell strains each of which had several genetic deficiencies requiring several nutritional supplements. A. true B. false 13. Sex in E. coli involves cell fusion. A. true B. false 14. Which of the following statements is true regarding temporal (time-of-entry) mapping in E. coli? A. The sex factor always integrates at the same location to form an Hfr strain. B. Transfer begins at the origin of replication on the E. coli chromosome, regardless of the site of sex-factor integration. C. Genes are transferred to the recipient cell in a linear order. D. All donor cells initiate DNA transfer simultaneously. E. None of the above are true. 15. The process by which phage DNA is integrated into the host chromosome is called… A. sexduction. B. transduction. C. lysogeny. D. lysis. E. attenuation. 16. In the lytic cycle of a virulent bacteriophage… A. Phage DNA recombines with the host chromosome B. Recombination may occur between different infecting phage particles C. The host bacterium transmits the bacteriophage to daughter cells D. Two of the above E. All of the above 17. Restriction enzymes… A. produce uniform DNA fragments. B. produce a random collection of DNA fragments. C. produce DNA fragments the chemical nature of whose ends depends on the source of DNA. D. produce DNA fragments the chemical nature of whose ends is independent of the source of DNA. E. more than one of the above. 18. Which of the following restriction enzymes would be expected to produce the largest DNA fragments from human DNA? A. AGCT B. GCGC C. GCCGGC D. ATATAT 19. DNA termini without single-stranded overhangs produced by endonuclease digestion are called… A. cohesive termini. B. sticky ends. C. blunt ends. D. oligonucleotides. E. none of the above. 20. The enzyme used to covalently join DNA segments to form recombinant DNA molecules is called… A. DNA ligase. B. reverse transcriptase. C. DNA polymerase holoenzyme. D. Klenow fragment. E. Okazaki fragment. 21. Which of the following is not used as a vector in DNA cloning? A. YAC B. BAC C. pBR322 D. lambda phage E. transposable element Alu 22. A human genomic library is constructed after digestion of the DNA with the restriction endonuclease Not1 (target sequence 5'-GCGGCCGC-3'). What is the average size of the DNA fragments expected to be produced with this enzyme? (Choose closest answer.) A. 400 kb B. 65 kb C. 4 kb D. 1.0 kb E. 0.25 kb 23. (Refer to above question.) Which of the following vectors would be best to use to construct this library? A. pUC plasmid B. lambda C. cosmid D. YAC E. all of the above could be used equally well. 24. (Refer to question 22.) Excluding the approximately 40% of human DNA that is made up of repetitious (and probably non-functional) sequences, how many different fragments would occur in the Not1 library? (Choose closest answer.) A. 1,000,000 B. 200,000 C. 60,000 D. 5000 E. 400 25. The approach to understanding the molecular basis of human genetic disease that involves first finding the gene and then, after figuring out what the amino-acid sequence of the gene product must be, determining the molecular function of the protein, is called… A. insertional mutagenesis. B. gene therapy. C. transgene analysis. D. reverse genetics. E. saturation mutagenesis. 26. Which one of the following depends on the availability of a genomic or cDNA library? A. reverse genetics B. genomic complementation C. gene tagging by insertional mutagenesis 27. A chromosomal site that can be used for mapping purposes does not always have to be a functional gene with different alleles that give different phenotypes. It can also be… A. a restriction fragment length polymorphism. B. a VNTR region that can be analyzed with the help of a restriction enzyme. C. a YAC. D. two of the above. E. all three of the above. 28. To prepare clones representing expressed genes from a eukaryotic cell, you would mix total cellular mRNA with… A. Reverse transcriptase B. RNA polymerase C. DNA polymerase D. Integrase E. RNAse 29. Salmon have been genetically engineered to grow much faster than normal by… A. inserting multiple copies of a growth hormone gene. B. altering the promoter region of a growth hormone gene so that the gene is transcribed at a higher than normal rate. C. using the regulatory region of a metallothionein-gene placed next to a growth hormone gene. D. inserting a gene than blocks the normal action of a growth-hormone inhibitor. 30. Which enzyme functions to polymerize DNA from an RNA template? A. DNA polymerase B. RNA polymerase C. Reverse transcriptase D. Ligase E. Endonuclease 31. What amino acid is specified by the codon AAU? A. isoleucine (ile) B. asparagine (asn) C. phenylalanine (phe) D. arginine (arg) E. lysine (lys) 32. The repeating polynucleotide ---ACACACAC---- is used as an artificial mRNA in an in vitro (in a test tube) protein-synthesizing system, in which the ribosomes start at random positions. What polypeptide would be synthesized? (No particular starting point is implied by the sequences as given.) A. ---thr-thr-thr-thr--B. ---his-his-his-his--C. ---thr-his-thr-his--D. more than one of the above E. none of the above 33. How many different hexanucleotides could code for the same dipeptide histidine-lysine (his-lys)? A. only one B. two C. three D. four E. more than four 34. The following sequence is taken from the middle part of an exon. -----C C A T G C T T C C A G G G C---------G G T A C G A A G G T C C C G---A portion of the protein corresponding to part of this exon is known to include the amino acid sequence (NH2)---threonine-phenylalanine-valine---(COOH). The top one of the two DNA strands shown above is in which orientation? A. with the 5' end toward the left B. with the 5' end toward the right 35. (Refer to previous question.) Which of the two DNA strands shown is the template strand for making the mRNA? A. the top strand B. the bottom strand 36. The following mRNA sequence includes somewhere in it the codon for the last amino acid of its protein. What is this last amino acid? 5'-CAUGUCAACCAUGCGAAUCCAAAGUCGAACCGUCUGACAGAACUCCCCACACC-3' A. leucine B. glycine C. valine D. alanine E. none of the above 37. The following DNA sequence (written in the usual, conventional, way) is known to be an internal part of a gene. Is it more likely to be part of an intron, or an exon? (Hint: What amino acid sequence might it code for?) A. intron B. exon ----CAGGGATAACCCAGGCTGAAACGGCGTAGCAGACCT---38. What amino acid could be represented by the "aa" in the following "charged" tRNA (whose anticodon is shown)? ("I" stands for inosine, and it can pair with A or C or U, but not G.) Remember that amino acids are attached to the 3'-end of tRNA. aa--------------------UAI-----------A. methionine (met) B. isoleucine (ile) C. tyrosine (tyr) D. cysteine (cys) E. more than one of the above 39. TATA boxes are found… A. in mRNA. B. at transcription-termination sites. C. at transcript polyadenylation sites. D. at exon-intron boundaries. E. in promoter sites. 40. A DNA sequence that is without stop codons for several dozen base pairs or more is known as a(n)… A. cDNA. B. STS. C. cistron. D. contig. E. ORF. 41. To describe the genetic code as "degenerate" means that … A. mRNA is rapidly degraded. B. the code is not universal among organisms. C. some amino acids have more than one codon. D. frameshift mutations are tolerated. E. stop codons may have corresponding tRNA molecules. 42. The 3' end of mRNA corresponds to the carboxyl terminus of the protein. A. true B. false 43. For a protein to function properly, each amino acid at each position has to be a certain one. A. true B. false 44. Transcript RNA processing involves removal of exons. A. true B. false 45. Newly synthesized RNA molecules, like newly synthesized DNA molecules, are made from the 5'-end to the 3'-end. A. true B. false 46. Termination of transcription depends on a consensus sequence in DNA being recognized by the RNA polymerase. A. true B. false 47. Polycistronic messenger RNA is found in prokaryotes, but not in eukaryotes. A. true B. false 48. The ribosome binding site lies at the 3' end of mRNA. A. true B. false 49. The anticodon of one tRNA can usually bind with several codons. A. true B. false 50. All messenger RNA's are polyadenylated. A. true B. false Answers to Exam III: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 EAAAB AE A B A BA BCC BDCCA E 26 27 28 29 30 BDAC C 21 22 23 24 25 ADD D 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 BCDB B C A B EE CA BBA B ABA A Answer to 34-35: middle part of an exon: 3'-----C C A T G C T T C C A G G G C-----5' -----G G T A C G A A G G T C C C G---Given the aa sequence thr-phe-val, the code table says that the mRNA must have the corresponding sequence 5'-ACNUUYGUN-3' (where N = any base, Y = either pyrimidine). This sequence is found, in DNA form and in reverse orientation, on the top strand above (shown in bold). The top strand must therefore be the "sense" or anti-template strand, the one that's just like the mRNA (except for T in place of U). Its orientation must be 3'---5', as shown, to make the same-sense mRNA; and the other strand must be the template strand. ------------------------------------------------------------------------------------------------EXAM II Biology 250 (General Genetics) Fall, 1998 SECOND MID-TERM EXAMINATION October 12, 1998 Some or all of the following information may be useful: Avogadro's number is 6.023 x 1023. The average "molecular weight" of a nucleotide pair is about 650. The distance between successive base pairs along the axis of the DNA double helix is 0.34 nm. 1. Differences in the density of DNA molecules can be detected by centrifugation in a solution of A. cesium chloride. B. potassium chloride. C. sucrose. D. magnesium phosphate. E. sodium phosphate. 2. The fundamental chemical units of DNA, the sum of all of which account for all the atoms in DNA, are… A. nucleotides. B. nitrogenous bases. C. 5-carbon sugars. D. nucleosides. E. phosphodiester bonds. 3. Which pair of bases in double-stranded DNA forms more hydrogen bonds? A. adenine-cytosine B. adenine-thymine C. guanine-thymine D. guanine-cytosine E. The same number of hydrogen bonds in formed by all of these. 4. The wavelength of light absorbed maximally by DNA in solution is _________ nm, and the amount of light absorbed at this wavelength is __________ when the DNA is single-stranded than when it is double-stranded. A. 280 / greater B. 260 / less C. 280 / less D. 260 / greater E. 360 / the same 5. In ordinary DNA replication in cells, a primer is needed. The primer is… A. double-stranded DNA. B. single-stranded DNA. C. double-stranded RNA. D. single-stranded RNA. E. none of these. 6. During DNA replication, new nucleotides are added to which end of already existing polynucleotide chains? A. the 3' end B. the 5' end 7. The major DNA-replication enzyme in E. coli, the one responsible for chain elongation, is… A. DNA polymerase I. B. DNA polymerase II. C. DNA polymerase III. D. DNA polymerase IV. E. DNA polymerase V. 8. Which one of the following DNA sequences is a palindromic sequence that might be recognized by a restriction enzyme? A. GGACCT B. GGATCC C. GGAAGG D. GGATTC E. more than one of these 9. (Refer to above question.) Assuming a random DNA sequence and equal proportions of each of the four bases, what is the expected average size of fragments produced by a restriction enzyme recognizing this (these) sequence(s)? (Choose closest answer.) A. 1 kb B. 2 kb C. 4 kb D. 8 kb E. 16 kb 10. Southern blotting is a method designed to detect all the fragments produced by a given restriction enzyme. A. true B. false 11. The following is the sequence of a primer chosen for use in a PCR reaction designed to amplify a particular gene from E. coli, whose genome consists of 4.7 x 106 np. GGTACCTGAGCT How often would you expect this sequence to occur by chance in the E. coli genome? (Choose best answer.) A. only once B. 5 times C. 10 times D. 20 times E. 100 times 12. (Use information contained in the preceding question.) All the genes of E. coli are contained in a single DNA molecule. How long is this molecule? (Choose closest answer.) A. 0.0001 mm B. 0.001 mm C. 0.01 mm D. 0.1 mm E. 1 mm 13. A set of dideoxy chain-termination sequencing reactions was carried out on a segment of DNA. The newly synthesized fragments were subjected to gel electrophoresis. A small portion of the gel is shown. The direction of migration is from top to bottom. What is the sequence of the template strand (written according to the usual convention)? A. CTTGAG B. GAACTC C. GAGTTC D. CTCAAG G — A T C — — 14. The drug AZT, which is used to inhibit HIV replication in AIDS patients, is a dideoxynucleoside. A. true B. false 15. In what direction does a DNA polymerase move along the template strand? A. 5' ® 3' B. 3' ® 5' — — — 16. Human mitochondrial DNA consists of a circular piece of double-stranded DNA with a contour length of 5.6 µm. About how many genes could it contain? (Assume the genes are of average size, and consist of nothing but coding sequences. Choose best answer.) A. only one or two B. ten or twenty C. about one- or two hundred D. a few thousand E. a few tens of thousands 17. Human beings (along with other mammals) have more DNA than any other species of plant or animal on earth. A. true B. false 18. Very large fragments of DNA, or whole DNA molecules (e.g., greater than 20 kb) can be physically separated by means of… A. molecular hybridization. B. Southern blotting. C. pulsed-field gel electrophoresis. D. autoradiography. E. renaturation kinetic analysis. 19. The chromatin (DNA + protein) in the nuclei of eukaryotic cells is organized into nucleosomes, consisting of DNA and… A. enzymes. B. polymerases. C. polysaccharides. D. histones. E. more than one of the above. 20. The amount of DNA in a nucleosome is roughly equivalent to the amount in an average-sized gene. A. true B. false 21. The DNA of higher organisms includes repetitive sequences. Because it consists of many highly similar or identical sequences, fragments of repetitive DNA renature more readily and more rapidly than fragments of unique-sequence (non-repetitive) DNA from the same genome. A. true B. false This graph (taken from the journal Genetics, vol.63, page 865) shows the renaturation kinetics of DNA from the fruit fly Drosophila melanogaster. Refer to this graph in answering the next 3 questions. 22. What percentage of total Drosophila DNA consists of repeated sequences? (Choose closest answer.) A. 90% B. 75% C. 50% D. 25% E. 10% 23. What is the approximate degree of repetition of the repetitive DNA fraction in Drosophila? (Choose closest answer.) A. 300 B. 3000 C. 30,000 D. 300,000 E. 3,000,000 24. The genome of Drosophila contains 1.65 x 108 np. If the repetitive DNA consisted of just one sequence repeated many times, how long would this sequence be? (Choose closest answer; the answer to this question depends on the answer to the previous question.) A. 5 np B. 50 np C. 500 np D. 5000 np E. 50,000 np 25. Which of these two DNA fragments has the lower melting temperature? A. AGAACTGCTTAGTA B. AGGACCTGCTCGAC TCTTGACGAATCAT TCCTGGACGAGCTG 26. Which DNA fragments move faster in gel electrophoresis, small fragments or large fragments? A. small B. large 27. A chromosome with its centromere very close to one end, so that one chromosome arm is very small, is known as… A. a telocentric chromosome. B. an acrocentric chromosome. C. a submetacentric chromosome. D. a metacentric chromosome. E. a sex chromosome. 28. One of the reasons polyploidy is rare in animals (unlike plants) is that regular meiotic segregation of sex chromosomes in parents is likely to give offspring with sex-chromosome constitutions unlike either parent. A. true B. false 29. A fertile sexually reproducing allotetraploid plant has its origin from hybridization between two ancestral species whose gametes contain 4 and 5 chromosomes, respectively. The number of bivalents formed in meiosis in the allotetraploid species is… A. 9 B. 18 C. 27 D. 36 E. 45 30. The condition of a diploid's having one extra chromosome of a particular kind in called… A. polysomy. B. aneuploidy. C. trisomy. D. none of the above. E. more than one of the above. 31. The gene in humans which, in its mutant form, causes Bruton agammaglobulinemia is located at Xq21. This means that it is found at band 21 on the _________ arm of the X chromosome. A. short B. long 32. In humans, having three copies of chromosome 21, instead of two, causes… A. Turner syndrome. B. Klinefelter syndrome. C. Down syndrome. D. Edward syndrome. E. none of the above. 33. In humans, having three sex chromosomes, all X's, instead of one or two X's, causes… A. Turner syndrome. B. Klinefelter syndrome. C. Down syndrome. D. Edward syndrome. E. none of the above. 34. Non-disjunction in meiosis in humans, leading to the production of aneuploid gametes and embryos, is more likely to occur in… A. the female. B. the male. 35. The frequency of Down syndrome in live births increases with the age of the mother. A. true B. false 36. There are on record about a dozen cases of young Down syndrome (DS) women impregnated by men (with normal chromosome constitutions), and later giving birth. Among the children of these women, what is the expected ratio of normal to DS individuals? A. 100:0 B. 75:25 C. 50:50 D. 25:75 E. 0:100 37. Female human beings are mosaics for X-linked genes, while female kangaroos are not. A. true B. false Answers 1. 2. 3. 4. 5. 6. 7. A A D D D A C 8. B 9. C 10. B 11. A 12. E 13. D 14. A 15. B 16. B 17. B 18. C 19. D 20. B 21. A 22. E 23. A 24. E 25. A 26. A 27. A or B* 28. A 29. A 30. E 31. B 32. C 33. E 34. A 35. A 36. C 37. A * see explanation below 11. A sequence with 12 bases has a probability of approximately 1/(4^12), which is about 1/(1.7 x 10^7); that is, by chance, once in approximately every 1.7 x 10^7 bases. But this number of bases is more than 3 times longer than the whole E. coli genome. Therefore if the sequence occurs at all (and we assume it does, because it was chosen from a known gene), it is unlikely to occur more than once. 12. This is really a fairly easy question. If there are 4.7 x 10^6 np in the E. coli genome, and 0.34 nm between each successive np, then the total length is 4.7 x 10^6 x 0.34 nm, which is 1.6 x 10^6 nm, or 1.6 x 10^3 um, or 1.6 mm. The best answer is E. 13. Shorter polynucleotide chains, nearer the bottom of the sequencing gel, have their chainterminating bases closer to the 5'-end. Therefore the correct way to read the NEWLY SYNTHESIZED chains, in the 5' -- 3' direction, is from bottom to top, 5'-CTTGAG-3'. The TEMPLATE strand, however, will have the opposite polarity and the complementary bases; i.e., 3'-GAACTC-5' or, written in conventional 5' -- 3' direction, CTCAAG. 16. Mitochondrial DNA with a length of 5.6 um (= 5600 nm) has 5600/0.34 = about 16,000 np. If a gene consisted of only coding DNA, it would be about 1000-2000 np long. Therefore mitochondrial DNA is long enough to contain 8-16 such genes. 22-24. The graph shows that the fastest-reassociating DNA is only about 10% of the total; most of the DNA reassociates slowly. The fastest-reassociating DNA is only 1/9 as concentrated as the slow DNA (10% : 90%), so to compare the Cot(1/2) values, the fast DNA can be imagined as 9x more concentrated than it really was in the experimental situation (to make it comparable in concentation to the slow DNA). If it were 9x more concentrated, it would reassociate 9x faster, and therefore have an "adjusted" reassociation-time 9x shorter. The observed half-reassociation-time (Cot(1/2)) of the fast DNA is about 3 (from the graph). Its "adjusted" Cot(1/2) value would then be 3/9 = 0.33. This value can now be compared with the Cot(1/2) value for the slow DNA, which from the graph is a little bit more than 100; it looks as it it might be about 105, but we can use 100 as an approximation. The fast DNA (using its "adjusted" value) is reassociating about 100/0.33 times faster than the slow DNA; therefore the degree of repeition must be about 100/0.33 = 300. The repetitious DNA of Drosophila is 10% of the total, or 1.65 x 10^7 np altogether. One sequence that could add up to this amount of DNA, if repeated 300 times, would be a single sequence that has (1.65 x 10^7)/300 = 55,000 np (answer E). 27. The correct answer is "B" (acrocentric). However, because the book makes a misleading error in the italicized remarks on p. 271, "A" will also be accepted (even though it's wrong). The definition of a telocentric chromosome is one in which the centromere is AT THE END of the chromosome; no second arm is detectable in the microscope. Each of the small "telocentrics" mentioned on p. 271 does in fact have a small second arm visible in the microscope, and all human geneticists these days call them "acrocentrics" for that reason. In the early days of human cytogenetics (e.g., in 1959, the year that the paper on Down syndrome by Le Jeune et al. (p.271) was published), the chromosome preparations were often not good enough to see a small second arm, and so in those days those small human chromosomes were sometimes called "telocentrics." 28. The "parents" referred to are of course the polyploid parents. If ancestors are XX females and XY males, the polyploid parents are likely to be XXXX females and XXYY males. Likely eggs from XXXX females are XXX and X as well as XX. Likely sperm from XXYY males are XXY, XYY, and XY, as well as XX and YY. When sperm meets egg, many offspring sex-chromosome combinations besides the parental XXXX and XXYY will result, likely giving a high proportion of sterile offspring. 33. Turner's is XO, Klinefelter's is XXY, Down's is trisomy 21, and Edwards' is trisomy for another autosome. The chromosome constitution 47,XXX is none of these. (It is called simply "triplo-X".) ----------------------------------------------------------------------------------------------EXAM I Biology 250 (General Genetics) FIRST MID-TERM EXAMINATION September 18, 1998 Fall, 1998 (Answers at bottom) Make sure you are sitting in your assigned seat. Before starting to fill in your answer sheet, fill in the boxes and circles corresponding to: Last name, First name. (Upper left) I.D. number. (Lower left) Sign CONSENT statement and fill in "YES" circle if you agree. Sign HONOR PLEDGE. (Upper right) FAILURE TO FOLLOW THE ABOVE INSTRUCTIONS WILL RESULT IN A DEDUCTION OF ONE POINT FROM YOUR EXAM SCORE. Notes: · In answering the questions below, always choose the best answer from among the choices. · There are 50 questions. · The questions which might take the most time are 36, 38, 44, 46, 49 and 50. 1. Which one(s) of the following kinds of observation indicate that the physical nature of the gene is the DNA molecule? A. transformation in the bacterium Streptococcus. B. transfer of radioactive material from parent bacteriophage T2 to progeny bacteriophage T2. C. the strict correlation between base changes in DNA and specific genetic diseases in humans. D. two of the above. E. all three of the above. 2. What radioactive atom was used in the Hershey-Chase "blender" experiment to demonstrate that the gene is made of DNA? A. iron B. phosphorus C. sulfur D. carbon E. nitrogen 3. The process of transcription produces what kind of RNA? A. transfer RNA B. ribosomal RNA C. messenger RNA D. two of the above E. all three of the above 4. A person's DNA contains information for the construction of… A. nucleic acid molecules. B. protein molecules. C. enzyme molecules D. two of the above E. all three of the above 5. In double-helical DNA, the base guanine is always paired with… A. guanine B. adenine C. cytosine D. uracil E. thymine 6. The process by which information in linear sequences of nucleic acid bases becomes information in linear sequences of amino acids is known as… A. transformation B. transcription C. transfection D. translation E. transmigration 7. The term "pleiotropy" refers to the fact that… A. one gene can affect more than one trait. B. one trait can be affected by more than one gene. C. in some cases, some genes are not expressed even though they may be present. D. a gene may be expressed to varying degrees, depending on environmental variables. E. recessive genes may "skip" a generation. 8. The term "epistasis" refers to the fact that… A. one gene can affect more than one trait. B. one trait can be affected by more than one gene. C. in some cases, some genes are not expressed even though they may be present. D. a gene may be expressed to varying degrees, depending on environmental variables. E. recessive genes may "skip" a generation. 9. Sickle-cell anemia is the result of the substitution, in normal hemoglobin, of a single amino acid for another. A. true B. false 10. Is anemia (in general) caused by heredity or environment? A. heredity B. environment C. both D. neither E. There is no simple answer to this question. 11. Bacteria belong to a kingdom of cells referred to as prokaryotes, which lack a membranebounded nucleus. A. true B. false 12. A synthetic RNA molecule consists of the repeating base sequence --UCUCUCUCUCU… When this is used in vitro to generate polypeptide chains, each chain consists of… A. one amino acid repeated over and over. B. two amino acids. C. three amino acids. D. a random array of many amino acids. D. no amino acids. 13. In human DNA, the adenine content is 30%. What is the cytosine content? A. 50% B. 40% C. 30% D. 25% E. 20% 14. The DNA extracted from a virus consists of 24% adenine, 30% thymine, 20% guanine, and 26% cytosine. What can you reasonably conclude from this information? A. It is really RNA, not DNA. B. It is circular DNA, not linear DNA. C. It cannot be DNA from the virus, but must be DNA from the infected host cell. D. It is not double-stranded DNA, but single-stranded DNA. E. None of the above are reasonable conclusions; further information is required. 15. Can a gene with a mutation in its coding region give a polypeptide gene-product with the same amino acid sequence as the nonmutant gene? A. yes B. no 16. Mating of two diploid organisms produces a 1:1 ratio of phenotypes in the progeny. What are the genotypes of the two parents? A. AA and AA B. aa and aa C. Aa and aa D. Aa and Aa E. AA and aa 17. In the haploid unicellular green alga Chlamydomonas, there are several mutations that give the non-motile phenotype paralyzed flagella (pf). In 1988 a new strain with this mutant phenotype, called pf36, was induced by UV irradiation. Which of the following findings would allow the conclusion that strain pf36 has a mutation in a gene not discovered before, rather than simply another allele of a gene in one of the previously obtained pf strains? A. In crosses with all previously known pf strains, some mutant (non-motile) progeny were always obtained in each cross. B. In crosses with all previously known pf strains, some wild-type (nonmutant, motile) progeny were always obtained in each cross. C. In crosses of different pf36 cells with each other, only mutant (non-motile) progeny were obtained in each cross. D. In crosses of different pf36 cells with each other, only wild-type (nonmutant, motile) progeny were obtained in each cross. 18. The test in the previous question is known as… A. a complementation test. B. a mutant screen test. C. an irradiation test. D. the anthocyanin test. E. the homozygosity test. 19. How many common alleles are known for the human blood-group gene that determines the A, B, AB, and O blood-group phenotypes? A. one B. two C. three D. four E. five 20. In human populations around the world, approximately how many different defective alleles have been discovered that cause the inherited recessive disease cystic fibrosis? A. one B. two C. three D. four E. more than a hundred 21. (Refer to information contained in the previous question.) If a normal man and a normal woman each carrying a defective cystic fibrosis allele marry, what is the probability that their first child will have cystic fibrosis? A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 22. Two monohybrid pea-plants are crossed. What is the expected genotypic ratio among the progeny? A. 1:1 B. 2:1 C. 3:1 D. 4:1 E. 1:2:1 23. Two dihybrid pea-plants (carrying one dominant and one recessive allele at each locus) are crossed. What is the expected phenotypic ratio among the progeny? A. 1:1 B. 1:1:1:1 C. 12:4:1 D. 9:3:3:1 E. none of these 24. Two trihybrid pea-plants (carrying one dominant and one recessive allele at each locus) are crossed. How many different phenotypic combinations are expected among their progeny? A. 3 B. 6 C. 8 D. 9 E. 10 or more 25. In human pedigrees, circles stand for _________, while squares stand for _________. A. males, females B. females, males C. children, adults D. adults, children E. different kinds of dances. 26. Huntington's disease is caused by a _________ allele. A. dominant B. recessive 27. Any strain that "breeds true" for a particular trait has individuals whole alleles are ______ for that trait. A. homozygous B. heterozygous 28. How many different kinds of eggs could be formed by a female with the genotype AaBbCCDdee? A. 2 B. 4 C. 8 D. 16 E. 32 29. One parent has the dominant phenotype and the other has the recessive phenotype for a particular trait. Two offspring are produced, and both have the dominant phenotype. What genotype(s) is (are) possible for the parent with the dominant phenotype? A. AA only B. Aa only C. aa only D. AA or Aa E. AA or aa 30. A normally pigmented Hopi Indian man marries a normally pigmented Hopi Indian woman. Each of them has an albino parent. If they have two children, what is the probability that both of the children will have normal skin pigmentation? A. 9/16 B. 3/4 C. 1/4 D. 1/2 E. 13/16 31. A bivalent in the pachytene stage of meiosis consists of… A. two sister chromatids. B. two non-sister chromatids. C. a pair of homologous but unduplicated chromosomes. D. four chromatids. E. none of the above. 32. In meiosis, the two sister centromeres of one homolog (either maternal or paternal) separate from each other in… A. pachytene. B. diplotene. C. anaphase I. D. anaphase II. 33. In meiosis, crossing-over occurs in… A. leptotene. B. pachytene. C. diakinesis. D. metaphase I. E. metaphase II. 34. Independent assortment was first demonstrated cytologically (i.e., under the microscope, as opposed to genetically) by Eleanor Carothers in 1913, in her study of an (equal; unequal) bivalent in a (fruit fly; grasshopper). A. equal; fruit fly B. equal; grasshopper C. unequal; fruit fly D. unequal; grasshopper 35. The diploid number for humans is 46. How many bivalents are present in spermatocytes? A. 23 B. 46 C. 92 D. 184 36. In humans, for an autosomal gene with 4 alleles, there are theoretically 10 possible genotypes, 6 heterozygotes and 4 homozygotes. How many genotypes are possible with 4 alleles of an Xlinked gene? A. 8 B. 10 C. 12 D. 14 E. more than 14 37. In corn (Zea mays), a plant with the recessive phenotype zebra leaves (z) was crossed with another plant having the recessive phenotype red aleurone (r), to produce an F1 double heterozygote. In this heterozygote, the two genes are in… A. the cis arrangement. B. the trans arrangement. 38. (Refer to above question.) The z and r genes are known to be moderately closely linked on chromosome 10. When the F1 double heterozygotes produced as described above are crossed with each other, would you expect more, or fewer, double-recessive phenotypes (showing both z and r) among the F2 offspring than if the two genes were not linked? A. more B. fewer 39. Two genes showing a 50% recombination frequency are… A. located on different chromosomes. B. closely linked. C. far apart on a single chromosome. D. allelic E. (more than one of the above) 40. Recombination frequencies decrease with increasing distance of the genes on a chromosome. A. true B. false 41. Recombination frequencies are the same in test crosses of double heterozygotes (of a given sex), whether the heterozygotes are cis or trans. A. true B. false 42. In a three-point test cross (i.e., one involving a triple heterozygote), the following recombination frequencies were observed: between gene P and gene Q: 6% between gene P and gene R: 9% between gene Q and gene R: 14% What is the order of the genes on the chromosome? A. P-Q-R B. P-R-Q C. Q-P-R 43. Genes that show less than 50% recombination between them are said to belong to the same… A. linkage group. B. complementation group. C. sex. D. mating type. E. translocation group. 44. In the fruit fly Drosophila melanogaster, scarlet eyes (sc) and delta (wing) veins (d) are both inherited as autosomal recessive traits. A female double heterozygote (with wild-type phenotype) was mated with a scarlet-eyed delta-wing-veined male; she produced 100 offspring with the following phenotypes: wild type 11 scarlet-eyed 43 delta 38 scarlet-eyed, delta 8 What is the approximate map distance between the two genes? (Choose closest answer.) A. 10 map units B. 15 map units C. 20 map units D. 30 map units E. 40 map units 45. (Refer to above question.) The parents of the female in the previous question both came from true-breeding lines. One of those parents must have had a wild-type phenotype. A. true B. false 46. In the tomato, mottled leaf (m), oblate fruit shape (o), and white flower color (w) are all inherited as recessive traits. A triple heterozygote plant is test-crossed, and the following progeny are obtained in the percentages given. wild type……………………..17 mottled only………………..…6 oblate only…………………....6 white only………………..….23 mottled and oblate…………..19 mottled and white…………….5 oblate and white……………...9 mottled, oblate, and white..…15 Which genes are linked? A. none of them B. m and o C. m and w D. o and w E. all three of them 47. In the analysis of ordered tetrads, the map distance between a given gene and the centromere of the chromosome it's on is reflected most directly in the frequency of… A. first-division segregation. B. second-division segregation 48. In the bread mold Neurospora crassa, in a cross between a double mutant (f g) and a wild-type strain, the following ordered tetrads were obtained in the numbers shown (100 total). fg fg ++ ++ 67 fg ++ fg ++ 6 ++ fg fg ++ 4 f+ fg ++ +g 10 fg f+ +g ++ 3 +g f+ f+ ++ fg f+ fg +g +g f+ ++ +g 7 2 1 How many tetrads in this experiment are NPD tetrads? A. more than 33 B. between 17 and 33 (inclusive) C. between 2 and 16 (inclusive) D. one E. none 49. (Refer to question 48.) The data show that the two genes are linked. What is the recombination frequency? (Choose closest answer.) A. 4% B. 8% C. 12% D. 16% E. 20% 50. (Refer to question 48.) Which gene is closest to the centromere? A. the f gene B. the g gene -------------------------------------------------------------------------------------------------------------Answers: 1. E 11. * 21. A 31. D 41. A 2. B 12. B 22. E 32. D 42. C 3. E 13. E 23. D 33. B 43. A 4. E 14. D 24. C 34. D 44. C 5. C 15. A 25. B 35. A 45. B 6. D 16. C 26. A 36. D 46. B 7. A 17. B 27. A 37. B 47. B 8. B 18. A 28. C 38. B 48. D 9. A 19. C 29. D 39. E 49. C 10. C or E 20. E 30. A 40. B 50. A * ambiguously worded true-false question; credit given for either answer. ------------------------------------------------------------------------------------------------The following questions gave difficulty ----17, 36, 38, 46, 48, 49. Outlines of answers to these questions follow ---17. If pf36 is a mutation in the same gene as one mutated in another strain; e.g., pf1, then a cross between pf36 and pf1, (pf36) x (pf1), could give no wild-type recombinant progeny. On the other hand, if the pf36 mutation is in a gene different from that mutated in pf1, then it would be diagrammed as (pf36 +) x (+ pf1), showing that some recombinant progeny could be produced (e.g., by independent assortment if on different chromosomes, or by crossing-over if on the same chromosome), including wild-type (+ +). Therefore a finding of wild-type progeny in each and every cross between pf36 and other pf strains would mean that the pf36 mutation is in a new gene, different from all the other pf genes discovered up to that point (answer B). 36. 4 alleles: a, b, c, d in all possible combinations in the two X's of females are: ab, ac, ad; bc, bd; cd -- the 6 heterozygotes; aa, bb, cc, dd -- the 4 homozygotes; in males, with only one X (and a Y), the possible genotypes are: a(Y), b(Y), c(Y), d(Y) -- these are the 4 "hemizygotes" (no homozygotes or heterozygotes are possible in normal males). Therefore, for both females and males, there are 10 + 4 = 14 different genotypes in all. 38. From the previous question (#37), the F1's came from the cross zzRR x ZZrr, giving double heterozygotes ZzRr with the genes in the trans arrangement (meiotic diagram Zr/zR). So when F1's are crossed with each other, the gametes produced are Zr and zR (parental types) and ZR and zr (recombinant types). If the two genes are LINKED, then the recombinant zr gametes will be produced at a frequency LESS than the 25% expected from independent assortment. Therefore in the next generation the zzrr genotype (= double-recessive phenotype) will also be less than if the genes were not linked (answer B). 46. Since the progeny were produced by a test-cross, the phenotypes of the progeny (given on the left) correspond to the gamete-genotypes (on the right) that the progeny must have developed from. wild type……………………..17………+++ mottled only………………..…6……….m++ oblate only…………………....6……….+o+ white only………………..….23……….++w mottled and oblate…………..19…….…mo+ mottled and white…………….5…….…m+w oblate and white……………...9……….+ow mottled, oblate, and white..…15………..mow With the genotypes as written on the right, you can now look at the frequencies of the genotypes taken in PAIRS. For the m and o gene-pair, the numbers are, for mo and ++, 17+23+19+15; and for m+ and +o, 6+6+5+9. Even without doing the addition, you can see that reciprocal types are UNEQUAL; therefore m and o are linked. For the o and w gene-pair, the numbers are, for ow and ++, 17+6+9+15; and for o+ and +w, 6+ 23+19+5. Even without doing the addition, you can see that reciprocal types are approximately EQUAL; therefore o and w are not linked. This is enough to answer the question: m and o are linked, and w is linked to neither o nor m (as you can double-check by looking at the frequencies for the m-w gene-pair). (Answer = B) 48. Since all zygotes in this cross must be fg/++ (cis arrangement), nonparental ditype (NPD) tetrads are those with only two types of spore, both non-parental; i.e., f+ and +g. Only the single tetrad on the extreme right is like this. Therefore the answer is "1" (answer D). 49. The recombination frequency is the number of recombinant spores divided by the total number of spores tested; that is, 2 recombinant spores from each tetratype, plus the 4 recombinant spores from the single NPD tetrad, divided by 400. This is 2(10+3+7+2) plus 4, divided by 400, = 48/400 = 0.12, or 12% (answer C). ...
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This document was uploaded on 10/27/2011 for the course BIL 250 at University of Miami.

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