Biology General Genetics Midterm 2002

Biology General Genetics Midterm 2002 - Biology 250...

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Unformatted text preview: Biology 250 (General Genetics) Spring, 2002 FIRST MID-TERM EXAMINATION February 21, 2002 P 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: • There are 42 questions. In answering the questions below, always choose the best answer from among the choices. • The questions that might take the most time are 1, 2, 3, 17, and 39-40. • • • • • • • • The following information may be useful: Avogadro’s number = 6.022 x 1023 Average “molecular weight” of a nucleotide pair in DNA = 617 daltons (approx. 600) Distance between successive nucleotide pairs along the axis of the double helix = 0.34 nm Average number of amino acids in a polypeptide chain = about 300-400 A human sperm cell contains 3.1 x 10-12 grams of DNA. A “genome” is the basic set of genes found in an organism. The G+C content of human DNA is about 40%. The code table is on the last page. 1. The single-celled diploid eukaryotic organism Saccharomyces cerevisiae (yeast) has about 0.025 pg of DNA per cell. What is its approximate genome size, in number of nucleotide-pairs? A. 6 million B. 12 million C. 24 million D. 48 million E. 96 million 2. (Refer to previous question.) The introns in the genes of this organism, as well as the spaces between genes, are on average about the same size as the exons. About how many genes does this organism have in its genome? A. 500 B. 1000 C. 3000 D. 6000 E. 12,000 3. This single DNA strand segment contains a complete open reading frame (ORF). Which reading frame has the ORF? --atggcgtcaacgttagaaacttgacgt-A. the reading frame beginning with the first adenine B. the reading frame beginning with the first thymine C. the reading frame beginning with the first guanine 4. The majority of prokaryotic genomes are circular DNA molecules. A. true B. false 5. In some organisms, small genes are sometimes embedded in the introns of larger genes. A. true B. false 6. In a human or house-cat or mouse, a skin cell just after mitosis contains how much DNA, relative to the amount of DNA in a sperm cell from the same species? A. half as much B. the same amount C. twice as much D. 4x as much E. 8x as much 7. The plant Paeonia japonica (Japanese peony) has a diploid number of 10 (5 pairs of homologous chromosomes). Assuming that genetic differences exist between the homologs of each pair of chromosomes, how many different chromosome combinations could there be in the pollen nuclei after meiosis? A. 2 B. 4 C. 8 D. 16 E. 32 8. In human oocytes at the stage of meiosis where homologous chromosomes are paired, what is the total number of chromatids present? A. 23 B. 46 C. 92 D. 184 E. 368 9. Albinism in humans is inherited as a simple recessive trait. If two normally pigmented parents have five children, one of which is an albino, what are the genotypes of the parents? A. Cc and Cc B. CC and Cc C. CC and CC D. CC and cc E. Cc and cc 10. Pigeons exhibit a checkered or plain feather pattern. In a series of controlled matings, the following data were obtained: (a) checkered x checkered F1 offspring all checkered; (b) checkered x plain F1 offspring all checkered; (c) plain x plain F1 offspring all plain. In these crosses, the allele responsible for the checkered phenotype is inherited as a . . . A. dominant. B. recessive. 11. (Refer to the previous question.) Predict the results of an F1 x F1 mating from the cross (b). A. all checkered B. all plain C. checkered & plain in a ratio of 3:1 D. checkered & plain in a ratio of 1:3 E. checkered & plain in a ratio of 1:1 12. Parents with the genotypes AABB and aabb are mated with each other, and the F1’s are interbred. If, in a series of test crosses, you wanted to determine the genotypes of individual F2 progeny, what genotype would you use to mate them with? A. aabb B. AABB C. AaBb D. None of these are suitable. E. Any of these could be used. 13. If a cross is carried out between two parents with the genotype AaBbCc, how many kinds of gamete can be produced by each parent? A. 3 B. 6 C. 8 D. 9 E. 27 14. (Refer to previous question.) How many different genotypes would you expect among the progeny? A. 3 B. 8 C. 16 D. 21 E. 27 15. (Refer to question 13 again.) Assuming the upper-case and lower-case letters refer to dominant and recessive alleles, how many different phenotypes would you expect among the progeny? A. 3 B. 6 C. 8 D. 9 E. more than 9. 16. (Refer to question 13 again.) If the upper-case alleles of each of these three genes independently contributes an equal amount to the same phenotypic trait (rather than to three different phenotypic traits), now how many different phenotypes would you expect among the offspring? A. 3 B. 6 C. 7 D. 8 E. 9 17. A cross is carried out between two parents with the genotype AaBb. If the upper-case alleles of each of these two genes independently contributes an equal amount to the same phenotypic trait (rather than to two different phenotypic traits, what proportion of the offspring would be expected to have a phenotype different from the parents? (Choose closest answer.) A. 62% B. 55% C. 50% D. 45% E. 38% 18. In humans, how many alleles determine the common “ABO” blood group phenotypes (type A, type B, type AB, and type O)? A. 2 B. 3 C. 4 D. 5 E. 6 19. A woman (whose father was type B) who has type A blood marries a man who has type AB blood. What are the possible blood types of their children? A. types A and B B. types A, B, and O C. types A and AB D. types A, B, and AB E. types A, B, AB, and O 20. A woman (with normal color-vision) whose father was colorblind married a man who was also colorblind (“I want a guy just like the guy who married dear old mom!” she said). What are the chances that any one of her children will be colorblind? A. 0% B. 25% C. 50% D. 75% E. 100% 21. (Refer to previous question.) Does it make any difference to the probability, in this case, whether the child in question is a boy or a girl? A. yes B. no 22. Consider the following pedigree, involving a single human trait (solid symbol). Ignoring possible complications such as rare mutations, multiple genes determining the trait, etc., how is the inheritance of the trait best described? A. dominant and X-linked B. dominant and autosomal C. recessive and X-linked D. recessive and autosomal E. more than one of the above is possible. ❍----❏ ________ _ __ ❏❏ 23. Consider the following pedigree, involving a single human trait different from that in the previous question. Ignoring the same possible complications as in the previous question, how is the inheritance of the trait best described? A. dominant and X-linked B. dominant and autosomal C. recessive and X-linked D. recessive and autosomal E. more than one of the above is possible. ❍----❏ ________ _ __ ❍❏ 24. “Complementation analysis” refers to a genetic study designed to determine A. whether two proteins determined by two genes interact at the biochemical level. B. whether two independently-isolated mutations are allelic or not. C. whether the same mutant gene acts differently in males vs. females. D. whether the same mutant gene has more than one phenotypic effect. E. whether genetic or environmental factors are more important in determining the trait. 25. Non-disjunction of the sex chromosomes in human male meiosis (either in the first meiotic division or in the second meiotic division) could give rise to which sex-chromosome complements in individual sperm cells? A. XX B. XY C. no sex chromosomes D. all three of these. E. only two of these. 26. In humans, the “triplo-X” condition (47 chromosomes total, with three X chromosomes) can arise by . . . A. non-disjunction in a spermatocyte in the father. B. non-disjunction in an oocyte in the mother. C. non-disjunction in either parent. 27. In humans, the “47, XYY” condition can arise by . . . A. non-disjunction in a spermatocyte in the father. B. non-disjunction in an oocyte in the mother. C. non-disjunction in either parent. 28. The making of an RNA copy from a DNA template is called . . . A. transcription. B. translation. C. translocation. D. splicing. E. replication. 29. During mRNA synthesis, new ribonucleotides are added to which end of the already existing polynucleotide chain? A. the 3' end B. the 5' end 30. A sample of DNA from a species of sea urchin has 17% cytosine. (on a molar basis). How much adenine does it have? A.17% B. 33% C. 50% D. 66% E. can’t estimate without information on at least one other base. 31. Guanine is a . . . A. purine. B. pyrimidine 32. In experiments on the reassociation kinetics of DNA, which reassociates faster? A. DNA with repeated sequences. B. DNA with only unique sequences. 33. All the members of the human “globin gene family” (which includes the _- and _hemoglobin chains of adult hemoglobin) are found on the same chromosome. A. true B. false 34. The _-chain of adult human hemoglobin is 146 amino acids in length after the cell removes the initial methionine. What is the number of nucleotides in the coding region of the mRNA for this polypeptide chain? (Remember there must also be some mechanism for stopping the assembly of amino acids into a chain.) A. The number is an odd number. B. The number is an even number. 35. The codon GCA corresponds to what amino acid? A. lysine B. glutamic acid C. threonine D. serine E. alanine 36. Could a mutation that changes an asparagine to a serine in a protein result from a change in a single base-pair in the corresponding gene? A. yes B. no 37. The DNA sequence that is chosen (by general agreement) to represent “the gene” is the sequence on . . . A. the template strand of DNA. B. the non-template strand of DNA. 38. In a gene’s DNA sequence as conventionally represented, “downstream” is in what direction? A. toward the 5'-end B. toward the 3'-end 39. The amino acid sequence -ile-gly-his- is known to correspond to this segment of a gene: -GTGGCCGAT-CACCGGCTAWhich strand of this DNA segment is used as the template for mRNA synthesis? A. the top strand B. the bottom strand C. both the top and bottom strands 40. (Refer to previous question.) Which of the following represents the polarities of the two DNA strands shown? A. 5'---------3' 3'---------5' B. 3'---------3' 5'---------5' C. 5'---------5' 3'---------3' D. 3'---------5' 5'---------3' 41. Following the usual convention of writing all nucleic acid molecules (of whatever kind) in the same orientation, which of the following might be an appropriate anticodon in tRNAThr ? A. GCA B. ACG C. CGU D. UGC E. none of these 42. Which kind of RNA molecule typically has about 75 nucleotides? A. mRNA B. tRNA C. rRNA Answers to First Mid-Term Exam (Feb. 21, 2002) 1.B 11.C 21.B 31.A 41.C 2.D 12.A 22.D 32.A 42.B 3.C 13.C 23.E 33.B 4.A 14.E 24.B 34.B 5.A 15.C 25.D 35.E 6.C 16.C 26.C 36.A 7.E 17.A 27.A 37.B 8.C 18.B 28.A 38.B 9.A 19.D 29.A 39.A 10.A 20.C 30.B 40.A An explanation of the answers to some of the most commonly missed questions is given below. It’s clear that some questions were commonly missed simply as a result of not doing the reading. A few of the questions on the exam were factual questions, taken directly from the reading (but not discussed in lecture). Some people in the class may think that if it’s not covered in lecture, it’s not important and won’t appear on an exam. That’s not correct. Questions 1 and 2: (Calculate the genome size of yeast, and estimate the number of genes.) Many of you probably did the basic calculation correctly, but failed to take into account that 0.025 pg is the amount of DNA in the diploid cell, so that the amount of DNA in the genome would be half that. Of course, if you got that wrong, then you’d get the estimate of the number of genes in the genome wrong too, in the next question. The picture of the genome described in words in Question 2, where introns and intergenic spacers are about the same sizes as exons, can be represented this way (where == stands for exons, --- stands for introns, and .... stands for intergenic spacers): ....==---==---==.... ==---==---==....==---==---==.... ==---==---==....==---==---== This would mean that the amount of DNA in exons, the coding DNA, is about half the total DNA in the genome. The total coding DNA (exons) divided by the average number of codons per gene gives the total number of genes. Question 3: “Open reading frames” were discussed at some length in the book, and also in one of the files on Cellarius. If you are given any fragment of DNA, you don’t know in which groups of 3 bases the mRNA copied from it might be read; also, a random fragment, assuming it’s part of an exon, most of the time won’t contain the ATG triplet corresponding to the initiation codon, so just because you see an ATG somewhere you can’t assume it corresponds to the actual initiation codon. The sequence --atggcgtcaacgttagaaacttgacgt-could theoretically be read in any of the following ways: -- atg gcg tca acg tta gaa act tga cgt ---a tgg cgt caa cgt tag aaa ctt gac gt- - -at ggc gtc aac gtt aga aac ttg acg t-The first two have stop codons (or, more precisely, the DNA triplets corresponding to stop codons), therefore by definition can’t be open reading frames. Only the last one has triplets without stop codons. Question 16: This is the case of polygenic inheritance: aabbcc has no contributing (upper case) alleles; Aabbcc, aaBbcc, aabbCc each have one contributing allele; AAbbcc, AaBbcc, aabbCC, etc., each have two contributing alleles, etc. With 6 “slots” to fill with a contributing allele, the different total numbers of contributing alleles can be 0, 1, 2, 3, 4, 5, or 6. That’s seven phenotypes. Question 17: Another example of polygenic inheritance, but a more difficult question. This time you have to take into account the relative numbers of offspring with various numbers of contributing alleles, from the cross AaBb x AaBb. If you work out the offspring with a Punnett square or by means of the “forked-line” method, you will see that the relative numbers of offspring with 0, 1, 2, 3, 4 contributing alleles are, respectively, 1, 4, 6, 4, 1. These numbers show that there are 6 offspring (out of 16) with 2 contributing alleles, like each of the parents. So there must be 10 offspring that are phenotypically different from the parents: 10/16 = 62.5% (answer A). Question 34: The _-chain of adult human hemoglobin is 146 amino acids in length. That would correspond to 146 codons. But there are also the initiation (met) and termination (stop) codons. Therefore the total number of codons for this protein is 146 + 1 + 1 = 148. Question 36: The codon for asparagine is AAY (= AAU or AAC). You could change AAY to AGY (= AGU or AGC), with a single base-pair change in DNA. The codon would now code for serine. SECOND MID-TERM EXAMINATION April 18, 2002 Code table at bottom 1. In E. coli, the lactose operon is repressible, while the tryptophan operon is inducible. A. true B. false 2. In the presence of lactose, in E. coli cells with the genotype I+OcZ+/F' I+O+Z-, the synthesis of the enzyme _-galactosidase is A. constitutive B. permanently repressed C. inducible D. repressible E. absent 3. In the lac operon of E. coli, the repressor . . . A. activates the _-galactosidase gene. B. binds the operator C. binds the co-repressor. D. is the product of the promoter gene. E. binds RNA polymerase. 4. Here is a segment of the leader sequence in the attenuator region of the tryptophan operon of E. coli. What would be the expected consequence of a mutation that changed the guanine at the arrow to an adenine? ↓ ..tcgacaatgaaagcaattttcgtactgaaaggttggtggcgcacttcctgaaacgggcag.. A. The enzymes of the trp operon would be synthesized at very low rates no matter what the level of tryptophan in the medium. B. The enzymes of the trp operon would be synthesized at maximal rates no matter what the level of tryptophan in the medium. C. The enzymes of the trp operon would be synthesized constitutively only in the absence of tryptophan in the medium. D. The enzymes of the trp operon would be synthesized constitutively, whether or not there was any tryptophan in the medium. E. None of the enzymes of the trp operon would be synthesized under any conditions. 5. (See the beginning of the previous question.) What would be the expected consequence of a mutation that changed this guanine to a cytosine? A. There would probably be no change in the regulation of the trp operon. B. The enzymes of the trp operon would probably be synthesized in amounts higher than what would be needed for a given level of tryptophan in the medium. C. The enzymes of the trp operon would probably be synthesized in amounts lower than what would be needed for a given level of tryptophan in the medium. D. The enzymes of the trp operon would be synthesized constitutively in the presence of tryptophan in the medium. E. None of the enzymes of the trp operon would be synthesized under any conditions. 6. In Zea mays (maize), brown midrib, bm, is a (recessive) leaf trait, and floury, fl, is a (recessive) trait of the corn kernel. Plants that are heterozygous for both traits, with normal midrib and normal kernel, were crossed to brown-midrib, floury plants, with the following results in the next growing season: brown midrib, normal kernel 37% normal midrib, floury kernel 34% brown midrib, floury kernel 13% normal midrib, normal kernel 16% Are the genes for these two traits located on the same chromosome? A. yes B. no 7. (Refer to question 6.) What is the recombination frequency between these two genes? (Choose closest answer.) A. 15% B. 30% C. 35% D. 39% E. 50% 8. (Refer to question 6.) Does the double-heterozygote parent plant have the alleles in the cis, or the trans, arrangement? A. cis B. trans 9. In linkage studies, recombination frequencies usually underestimate the actual distance between two genes, because . . . A. the closer the two genes are, the more multiple crossovers occur. B. undetected crossovers occur, particularly when the two genes being studied are far apart. C. the general rule that the farther apart two genes are, the greater the likelihood of crossing over between them, does not hold for many, perhaps most, organisms. D. frequencies of parental types can never exceed 50%. E. many gene-pairs assort independently. 10. Three linked genes, A, B, and C, are involved in this cross. Upper-case alleles are dominant, lower-case alleles are recessive. Both of the original parents were homozygous at all three loci. One parent had the recessive phenotype for the B gene (AAbbCC), while the other parent had the recessive phenotype for the other two genes (aaBBcc). All the offspring of these parents were, of course, triple heterozygotes. When these offspring were test-crossed, their progeny showed phenotypes of the following kinds and numbers. (There are 1000 offspring total.) ABC abc aBC Abc ABc abC aBc AbC 115 101 35 41 13 11 359 325 What are the linkage relationships between these three genes? A. A and B are linked, but C is not linked to either. B. A and C are linked, but B is not linked to either. C. B and C are linked, but A is not linked to either. D. A, B, and C are all linked to each other. E. None of these genes are linked to any of the other two. 11. (Refer to question 10.) What is the approximate map distance between genes A and C? (Choose closest answer.) A. 10 cM B. 15 cM C. 20 cM D. 25 cM E. 30 cM 12. (Refer to question 10.) What is the order of these genes along the chromosome(s) they are located on? A. A-B-C B. B-A-C C. A-C-B D. The question cannot be answered because two different chromosomes are involved. E. The question cannot be answered because three different chromosomes are involved. 13. The recognition site for the restriction enzyme MarI is ˇAGCT. The recognition site for the enzyme BbrI is AˇAGCTT, where the four internal bases are identical to the MarI site. Does this mean that the single-stranded ends produced by these two enzymes are identical? A. yes B. no C. sometimes, but not always 14. (Refer to question 13.) Could a fragment of human DNA cut with MarI be inserted into a plasmid vector cut with BbrI? A. yes B. no C. sometimes, but not always 15. (Refer to question 14.) Could the human DNA inserted into the vector this way be cut out again with the use of MarI? A. yes B. no C. sometimes, but not always 16. (Refer to question 14.) Could the human DNA inserted into the vector this way be cut out again with the use of BbrI? A. yes B. no C. sometimes, but not always 17. (Refer to question 13.) Which of these two restriction enzymes would be expected to give the larger fragments of human DNA? A. MarI B. BbrI 18. You are given a cDNA library of human genes prepared in a bacterial plasmid vector. You are also given the cloned EF-1a yeast gene, one that is highly conserved (but not identical) among eukaryotes. Which one of the following methods would be the first and most direct one to use to find the human EF-1a gene? A. fluorescent in situ hybridization, using a synthetic probe at least 16 nucleotides in length. B. Restriction enzyme digestion, electrophoresis, and Southern blotting. C. Reverse transcriptase and Northern blotting. D. Construction of artificial primers, polymerase chain reaction. E. colony-DNA hybridization, using yeast EF-1a as a probe. 19. Determining the specific, detailed sequence of a DNA segment depends on . . . A. the breakage of the DNA segment at specific sites using restriction enzymes, and reconstructing the segment with the aid of a restriction map. B. the microscopic analysis of fluorescence patterns produced by specific binding of fluorochromes to specific nucleotides in the DNA segment. C. amino acid analysis of the protein “product” of the gene, with the use of the code table to determine the nucleotide sequence. D. the analysis of the sizes of fragments of DNA produced by premature termination during a DNA synthesis reaction. E. electrophoresis and Southern blotting with specific artificially synthesized probes of known sequence. 20. After sequencing and proofreading the DNA sequence of a genome, the next step in understanding the genetic information it contains is called . . . A. compilation. B. classification. C. underlining. D. annotation. E. shotgun cloning. 21. In scanning long DNA sequences for the presence of genes, computer programs often are designed to find . . . (choose best answer) A. enhancer sequences. B. poly-A signals. C. ORFs. D. the DNA triplet ATG. E. the DNA triplet TAG. 22. Duchenne muscular dystrophy (DMD) is inherited as an X-linked recessive disease, often fatal in males who have it: very few of them become fathers. A woman (marked with a star below) whose younger brother has the disease (black symbol) wants to know if she is a carrier of the defective allele. A DNA analysis of a closely linked RFLP is carried out on her, her parents, and her siblings. The results are shown below, with the probed fragments shown under each individual. ❍----❏ ___________________ __ ___ ❍ ❍* ❏ ❏ _ _ _ _ _ _ _ _ _ _ _ _ _ Is the woman a carrier of the defective form of the gene? _ _ A. yes B. no 23. Allele-specific oligonucleotides (ASOs) are used to examine the genotypes of different individuals (some of them not yet born) with regard to the _-hemoglobin gene: its normal allele and the allele responsible (in the homozygous condition) for sickle-cell anemia. Which of the following corresponds to what you would expect to see for (left to right) the homozygous normal individual, the heterozygous carrier, and the individual with sickle-cell anemia (or who will develop it later)? Hybridization of the ASO with a DNA sample is indicated by a dark spot, failure to hybridize by an open circle. The ASO for the normal allele was applied to the DNA samples on the top row, and for the mutant allele on the bottom row. A B C D E ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ 24. DNA fingerprints of a mother (M), her child (C), and three(F1, F2, F3) are shown. Who is the father of the child? A. F1 B. F2 C. F3 D. none of these men. E. all of these men. M C F1 F2 F3 _ _ __ _ _ _ __ _ _ __ _ _ _ _ _ _ _ _ _ __ _ ___ __ _ _ _ _ _ _ _ _ 25. What was the first human gene product to be made by using recombinant DNA and licensed for therapeutic use? A. human growth hormone B. estrogen C. adenosine deaminase D. insulin E. Factor VIII 26. In humans, there is an RFLP at the D7S477 locus on chromosome 7, which can be detected by Southern blotting using BamHI 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 cannot be broken at that site by BamHI (allele "1"). In a survey of 200 people in Charleston, Virginia, 84 had only allele 1, 25 had only allele 2, and 91 had both alleles. What is the gene frequency for allele 2? (Choose closest answer.) A. 0.17 B. 0.34 C. 0.51 D. 0.65 E. 0.78 27. (Refer to previous question.) Allele “2” in the population can be identified with a particular short DNA sequence, while allele “1” may actually include several different short DNA sequences. A. true B. false 28. Hemochromatosis is a human genetic disease affect the metabolism of iron. It is inherited as an autosomal (chromosome 6) recessive, and is often fatal if not recognized (but is easily treated when correctly diagnosed). Homozygotes occur at a frequency of about 0.3%. What is the approximate frequency of the recessive allele? (Choose closest answer.) A. 1% B. 3% C. 5% D. 10% E. 20% 29. What is the approximate frequency of carriers of the recessive allele? (Choose closest answer.) A. 1% B. 3% C. 5% D. 10% E. 20% 30. As you know, human ABO blood groups are determined by the “ABO gene,” which for any one individual may have any two of the three alleles, IA, IB, and i. Among Iranians, the frequency of blood types is A, 31.2%; B, 25.3%, AB, 8.1%, and O, 35.4%. What is the frequency of the i allele in Iran? (Choose closest answer.) A. 0.2 B. 0.3 C. 0.4 D. 0.5 E. 0.6 31. (Refer to data given in the previous question.) What is the frequency of the IB allele? (Choose closest answer.) A. 0.2 B. 0.3 C. 0.4 D. 0.5 E. 0.6 32. Ocular albinism (type 1), an X-linked recessive condition, occurs at a frequency of about 1 in 30,000 males. What is the expected frequency of women who are carriers of the allele causing the condition? (Choose closest answer.) A. 1/30,000 B. 1/15,000 C. 1/10,000 D. 1/5000 E. 1/2500 33. Spontaneous mutation rates appear to be higher in mice and humans than in bacteria. A. true B. false 34. Tautomeric shifts in the DNA bases can lead to transition, but not transversion, mutations. A. true B. false 35. Of transition mutations that alter the 3rd codon position in the 61 codons for amino acids, how many will be missense mutations? (Choose the range within which the correct answer lies.) A. one only B. two or three C. four to six D. six to ten E. ten to fifty-eight 36. In general, for coding regions of genes, mutations that involve base-pair substitutions are likely to be more detrimental than mutations that add or delete a single base-pair. A. true B. false 37. The “thymine dimers” that are often formed in DNA as a result of UV irradiation are dimers between . . . A. two thymines (nearly) facing each other on opposite strands of the doublehelical DNA molecule. B. two thymines adjacent to one another on the same DNA strand. 38. If the human genome contains 35,000 genes, and the mutation rate for each gene averages 10-5 per gamete formed, what is the average number of new mutations carried by each individual? (Choose best answer.) A. less than one B. between one and three C. between 5 and 10 D. between 25 and 50. E. significantly more than 50. 39. “Site-directed mutagenesis” is . . . A. the use of mutagenic agents (such as UV, or alkylating agents), directed at one particular tissue or cell type, in the treatment of cancer and other somatic genetic diseases. B. the use of mutagenic agents that affect several genes on only one chromosome, or one chromosome arm. C. a procedure in which a particular base-pair (or small number of base-pairs) in the DNA of a particular gene can be altered at will. D. spontaneous mutation that is not entirely random, but affect DNA stretches with certain higher- or lower-than-average base compositions. E. a procedure in which spontaneous mutations in a certain gene are selected on the basis of a nutritional deficiency or some other biochemical defect. 40. The energy of X-rays is greater than that of UV light, and therefore X-rays are more likely to cause breaks in DNA and chromosomes. A. true B. false 41. The organism in which transposable elements were first discovered was . . . A. mice. B. humans. C. yeast. D. fruit fly. E. maize (corn). 42. The “inverted terminal repeats” that characterize insertion sequences (for example, in transposable elements) are best represented by which of the following diagrams (where the arrows indicate the 5'-3' direction of the same short sequence of DNA)? A B C -------------------------------------------------______________________________________________________________________________ U C A G _________________________________ U F phe S ser Y tyr C cys U F phe S ser Y tyr C cys C L leu S ser * * A L leu S ser * W trp G C L leu P pro H his R arg U L leu P pro H his R arg C L leu P pro Q gln R arg A L leu P pro Q gln R arg G A I ile T thr N asn S ser U I ile T thr N asn S ser C I ile T thr K lys R arg A M met T thr K lys R arg G G V val A ala D asp G gly U V val A ala D asp G gly C V val A ala E glu G gly A V val A ala E glu G gly G Answers to Second Mid-Term Exam (Apr. 18, 2002) (Yes, this was a hard exam.) 1.B 2.A 3.B 4.E 5.C 6.A 7.B 8.B 9.B 10.D 11.A 12.B 21.C 22.B 31.A 32.B 41.E 42.A 13.A 23.A 33.A 14.A 24.B 34.A 15.A 25.D 35.B 16.C 26.B 36.B 17.B 27.A 37.B 18.E 28.C 38.A 19.D 29.D 39.C 20.D 30.E 40.A Explanations of Answers to the Questions most frequently missed: Questions 4 and 5, on the trp attenuator in E. coli: The attenuator system works by the ribosome’s stalling (or not) on the leader sequence in the mRNA, which contains two trp codons one right after the other (in the DNA, -tggtgg-). If there is already plenty of tryptophan in the cell, the ribosome won’t stall; it moves along easily through the trp codons, allowing the mRNA to assume a “transcription-termination” fold right at the beginning; therefore transcription stops, and the trp-synthetase enzymes don’t get synthesized (which is appropriate, because there’s already plenty of tryptophan around). The mutation in Question 4 changes the second DNA tgg triplet (corresponding to the second trp codon) to tga. This will be a stop codon in the mRNA’s leader sequence, so even if the mRNA is made, no translation of the coding regions for the enzymes will occur, and the enzymes won’t be made, whether or not there’s any tryptophan around (answer E.) The mutation in Question 5 changes the second DNA tgg triplet to a tgc. This corresponds to a cysteine codon, so now there’s only one trp codon instead of two, and the cell’s trp operon should be less responsive to tryptophan levels. Ribosomes will move more easily through the leader region, as if there were more tryptophan around than there really is. This will mean that the cell will stop making the enzymes even when there isn’t much tryptophan around (answer C). Question 10, 11, and 12, the “3-point” linkage/mapping question: This was a straightforward question. Note that in the triple heterozygote, the alleles of the A-C gene pair were in the cis arrangement, while those of the A-B and B-C gene pairs were in the trans arrangement. (This is shown by the original parents’ genotypes, and also by the allele arrangements in the most-common “parental” types among the offspring.) This is important in defining what specific allele combinations represent recombinant types. The most direct and sure way to answer this question (and the next two, which depend on it), is to calculate the recombination frequencies for the three possible gene-pairs: A - B: recombinants are AB and ab: 115 + 101 + 13 + 11 = 240 = 24.0% B - C: recombinants are BC and bc: 115 + 101 + 35 + 41 = 292 = 29.2% A - C: recombinants are Ac and aC: 35 + 41 + 13 + 11 = 100 = 10.0% All are less than 50%, so all are linked. B and C must be the farthest apart (largest recombination frequency), with A in between, closer to C: B-------------A----C. The recombination frequency between A and C is approximately equal to the map distance, 10 cM. Question 23, on the use of allele-specific oligonucleotides (taken directly from Figs. 19-7, 198, and Problem 8 on p. 413): Two duplicate samples of denatured (single-stranded) total DNA (or DNA amplified by PCR from a specific region) are spotted onto a filter, one above the other. The spots are then hybridized with an oligonucleotide specific for one allele or the other (ASO). In this kind of experiment, the question being asked is, Does the DNA from one individual contain just one allele, or both alleles, or just the other allele? The results would indicate one kind of homozygote, a heterozygote, and the other kind of homozygote, respectively. If the ASO for the normal allele (say, allele 1) is applied to the top row of spots, and that for the mutant allele (say, allele 2) to the bottom row of spots, then the three double-spot patterns corresponding to the three genotypes will be: 11 12 22 ___ ___ In other words, the “11” homozygote DNA binds the ASO for allele 1, but not the ASO for allele 2. The “22” homozygote DNA binds the ASO for allele 2, but not the ASO for allele 1. The “12” heterozygotes binds both ASOs. Question 27, on what is an “allele” at an RFLP site: An RFLP (“restriction-fragment length polymorphism”) site is a particular short DNA sequence recognized by a restriction enzyme, but one in which one or more nucleotide-pairs may differ from chromosome to chromosome in the population. This means that in some chromosomes the site is not recognized by the enzyme, and so is not cut, thus giving rise to variation in length of fragments (recognized by an appropriate probe). The only thing that can be recognized at an RFLP site is the difference in length of DNA fragments, reflecting a DNA segment’s being cut or not cut by the enzyme. To take a particular example, the D7S477 BamHI site mentioned in the previous question: “Allele 2” (arbitrarily so named) is the form of the DNA that is cut by the enzyme; it must have the -GGATCC- palindromic sequence recognized by the enzyme. “Allele 1” is the form of the DNA at that site that is not cut by the enzyme. What sequence is that? . . . any variant of the -GGATCC- sequence that fails to be recognized by BamHI! It could be -AGATCC-, -GTATCC-, -GGTTCC-, -GGAACC-, etc., etc. So “allele 1” could be any one or more of those sequences. (You could in fact calculate that there are 46 = 4096 possible different 6-base-pair sequences at a given site. Only one of them could be cut by BamHI; the other 4065 could not. So “allele 2” is one particular sequence, while “allele 1” would be all the other DNA sequences that exist in the population at this chromosome site.) Sample Exam Questions from recent semesters, corresponding to material covered up to the First Mid-Term Exam, Fall 2002 (answers at bottom). Note: These are only examples. These questions were compiled from more than one exam; the number of questions does not necessarily correspond to what you will see on the First Mid-Term exam this semester. Emphasis varies from semester to semester, so the content of these questions may not correspond exactly to what you will see on the First Mid-Term exam this semester. 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. 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. 8. In human DNA, the adenine content is 30%. What is the cytosine content? A. 50% B. 40% C. 30% D. 25% E. 20% 9. 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. 10. 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 11. 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 12. 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. 13. 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 14. 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 15. 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 16. 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 17. 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. 18. Any strain that "breeds true" for a particular trait has individuals whole alleles are ______ for that trait. A. homozygous B. heterozygous 19. How many different kinds of egg could be formed by a female with the genotype AaBbCCDdee? A. 2 B. 4 C. 8 D. 16 E. 32 20. 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 21. 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 22. 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. 23. The diploid number for humans is 46. How many bivalents are present in spermatocytes? A. 23 B. 46 C. 92 D. 184 24. 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 (in humans) with 4 alleles of an X-linked gene? A. 8 B. 10 C. 12 D. 14 E. more than 14 25. 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. 26. 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. 27. 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 28. In what direction does a DNA polymerase move along the template strand? A. 5' → 3' B. 3' 5' → 29. 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 30. 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. 31. The amount of DNA in a nucleosome is roughly equivalent to the amount in an averagesized gene. A. true B. false 32. 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 33. Which of these two DNA fragments has the lower melting temperature? A. AGAACTGCTTAGTA B. AGGACCTGCTCGAC TCTTGACGAATCAT TCCTGGACGAGCTG 34. Which DNA fragments move faster in gel electrophoresis, small fragments or large fragments? A. small B. large 35. 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. 36. 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. 37. Which enzyme functions to polymerize DNA from an RNA template? A. DNA polymerase B. RNA polymerase C. Reverse transcriptase D. Ligase E. Endonuclease 38. What amino acid is specified by the codon AAU? (See code table.) A. isoleucine (ile) B. asparagine (asn) C. phenylalanine (phe) D. arginine (arg) E. lysine (lys) 39. 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 40. (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 41. 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 42. 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---43. 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. 44. 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. 45. 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. 46. The 3' end of mRNA corresponds to the carboxyl terminus of the protein. A. true B. false 47. For a protein to function properly, each amino acid at each position has to be a certain one. A. true B. false 48. Transcript RNA processing involves removal of exons. A. true B. false 49. Newly synthesized RNA molecules, like newly synthesized DNA molecules, are made from the 5'-end to the 3'-end. A. true B. false 50. Termination of transcription depends on a consensus sequence in DNA being recognized by the RNA polymerase. A. true B. false 51. Polycistronic messenger RNA is found in prokaryotes, but not in most eukaryotes. A. true B. false 52. The ribosome binding site lies at the 3' end of mRNA. A. true B. false 53. The anticodon of one tRNA can usually bind with several codons. A. true B. false 54. All messenger RNA's are polyadenylated. Answers: 1. E 2. B 3. E 4. E 5. C 6. D 7. B 8. E 9. D 10. A 11. C 12. B 13. C 14. E 15. D 16. C 17. B 18. A 19. C 20. D 21. A 22. D 23. A 24. D 25. A 26. D 27. D 28. B 29. B 30. D 31. B 32. A 33. A 34. A 35. C 36. E 37. C 38. B 39. B 40. B A. true B. false 41. C 42. A 43. E 44. E 45. C 46. A 47. B 48. B 49. A 50. B 51. A 52. B 53. A 54. A Sample Questions for Exam II. Notes: • These are only examples, more or less related to some of the material covered in class since the last exam. Some of them may be easier and some of them may be harder than what you will see on the Second Mid-Term exam this semester; but you have been given the information necessary to do them all. • The topics represented by these questions may not include everything that you will see on the Second MidTerm exam this semester. • These questions were compiled from more than one exam. • The relative number of questions in each area does not necessarily correspond to what you will see on the Second Mid-Term exam this semester. • Since emphasis varied from semester to semester, some of the questions may be easier and some of them may be harder than what you will see on the Second Mid-Term exam this semester; but you have been given the information necessary to do them all. 1. Sickle-cell anemia is the result of the substitution, in normal hemoglobin, of a single amino acid for another. A. true B. false 2. 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 3. Which of these two DNA fragments has the lower melting temperature? A. AGAACTGCTTAGTA B. AGGACCTGCTCGAC TCTTGACGAATCAT TCCTGGACGAGCTG 4. Recombination frequencies decrease with increasing distance of the genes on a chromosome. A. true B. false 5. 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 6. 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 7. 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. 8. 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 9. (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 10. 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 11. 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 12. (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 13. Southern blotting is a method designed to detect all the fragments produced by a given restriction enzyme. A. true B. false 14. 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 15. 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 GATC — — — — — — 16. Which DNA fragments move faster in gel electrophoresis, small fragments or large fragments? A. small B. large 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. The restriction enzymes for which of the following target sites would be expected to produce the largest DNA fragments from human DNA? A. AGCT B. GCGC C. GCCGGC D. ATATAT 19. 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. 20. 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? (Note: human DNA has about 40% G+C.) (Choose closest answer.) A. 400 kb B. 65 kb C. 4 kb D. 1.0 kb E. 0.25 kb 21. 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. 22. 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 23. Which enzyme functions to polymerize DNA from an RNA template? A. DNA polymerase B. RNA polymerase C. Reverse transcriptase D. Ligase E. Endonuclease 24. Constitutive gene expression refers to… A. constant 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. 25. An operator mutation may give rise to constitutive gene expression. A. true B. false 26. 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. 27. The tryptophan operon is… A. inducible B. repressible. 28. 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. 29. 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. 30. 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. 31. In sickle-cell anemia, the glutamic acid at position 6 in normal hemoglobin is replaced by valine. This is the result of . . . A. a transition mutation B. a transversion mutation. 32. 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. 33. Frameshift mutations are caused by insertions or deletions. A. true B. false 34. 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. 35. 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 36. Most spontaneous mutations, after they have initially occurred in a cell, are corrected by DNA repair mechanisms in the cell. A. true B. false 37. Mutations caused by base substitutions undergo reversion readily, compared to large deletions, which almost never revert. A. true B. false 38. 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 39. 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 40. 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 41. 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% 42. (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% 43. 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 44. (Refer to previous question.) What is the frequency of female carriers in the population? A. less than 1/25,000 B. 1/25,000 C. 1/2500 D. 1/250 E. more than 1/250 45. 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% Answers: 1. A 2. A 3. A 4. B 5. A 6. C 7. A 8. C 9. B 10. B 11. B 12. C 13. B 14. A 15. D 16. A 17. E 18. C 19. A 20. A 21. D 22. A 23. C 24. A 25. A 26. D 27. B 28. E 29. A 30. B P 31. B 32. C 33. A 34. D 35. C 36. A 37. A 38. B 39. E 40. D 41. B 42. D 43. B 44. B 45. C ...
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