2007+DA+finalKey - Bio 97 2007 Final Exam Section D version...

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Unformatted text preview: Bio 97 2007 Final Exam Section D version A – pg. 3 of 11 Name _______________________________ ID # ______________________ 1 Which section are you in? A B C D (hint: this is the correct answer) Which exam version do you have? A B C D A woman’s maternal grandmother (i.e. her mother’s mother) is affected by an x-linked recessive condition, but no one else in the family is. What are the chances that she is a carrier of this condition? A. 100% B. 67% (2/3) C. 50% (1/2) D. 25% (1/4) E. 0% I. II. Cancer is the result of an uncontrolled proliferation of cells. Although none of your other siblings or other relatives have cancer, you have a greater than 50% chance of developing lung cancer because your twin sister has lung cancer. III. Loss of heterozygosity refers to loss of the functional allele of a tumor suppressor gene in a somatic cell. The above statements are: A. I. True II. True B. I. False II. True C. I. True II. False D. I. False II. False E. I. True II. True III. True III. False III. True III. False III. False 2 3 4 5 Through at least one mutation, a cancer cell develops the ability to invade blood vessels. This allows it to: A. vascularize B. get more oxygen C. grow faster D. become immortal E. metastasize 3 Bio 97 2007 Final Exam 6 Section D version A – pg. 4 of 11 This is a two-part true/false question (i) A missense mutation results in the change of one amino acid for another. (ii) A silent mutation is likely to be more damaging to a protein’s function than a frameshift. A. (i) = TRUE; (ii) = TRUE B. (i) = TRUE; (ii) = FALSE C. (i) = FALSE; (ii) = TRUE D. (i) = FALSE; (ii) = FALSE Fill in the blanks: There are several ways for bacteria to acquire new genes. _____________ refers to the process of taking in DNA from the environment. Packaging of bacterial DNA into phage coat proteins results in ________________. Both processes require ________________ for the new genes to become part of the chromosome. A. B. C. D. E. Transduction, lysis, conjugation Transformation, transduction, recombination Conjugation, lysis, transformation Transformation, transduction, conjugation Conjugation, transduction, transposase 7 8 Wilson disease is a rare autosomal-recessive copper overload disorder caused by mutations of the Wilson disease gene ATP7BA located on chromosome 13. A man and a woman, neither of whom is affected by Wilson’s disease, have a daughter who has Wilson’s disease. The woman then gets pregnant with a second child. If the second child is a girl, what are the chances that she will have Wilson’s disease? If the second child is a boy, what are the chances that he will have Wilson’s disease? A. girl = 50%; boy = 50% B. girl = 25%; boy = 25% C. girl = 25%; boy = 50% D. girl = 67%; boy = 100% E. girl = 50%; boy = 100% 9 1 2 3 7 4 5 8 6 9 10 11 12 The pedigree shows the inheritance of a genetic condition in four generations of a family. Based on the pedigree, the mode of inheritance of this condition could be… A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. A or B 4 Bio 97 2007 Final Exam 10 Section D version A – pg. 5 of 11 Assume the condition in the previous pedigree is very rare in the general population. Then the mode of inheritance of this condition is most likely… A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. D or E I. II. III. Mutation of BRCA1 increases the likelihood of cancer because telomerase is activated. p53 is the universal guardian against cancer. One homolog of chromosome 11 from cells in a tumor has a longer than normal arm. The gene affected by this is probably an oncogene. The above statements are: A. I. True II. True B. I. True II. True C. I. False II. True D. I. True II. False E. I. False II. True III. True III. False III. True III. True III. False 11 12 Which statement is FALSE? Homeotic genes A. regulate transcription of other genes. B. are not in humans. C. can be mutated resulting in homeotic transformation. D. are generally found in clusters. E. specify segment identity. Consider the allelic frequency of the alleles of the bigmouth gene in a population of 10 people. In this population, one has the genotype BB, two have the genotype bb, and the rest are heterozygotes. What is the frequency p of the wild-type allele B? A. p = 0.55 B. p = 0.50 C. p = 0.45 D. p = 0.2 E. The allele frequencies can only be determined if the population is in Hardy-Weinberg equilibrium. Marfan syndrome is an autosomal recessive disease with an incidence of 1/10,000. What is the approximate frequency of heterozygote carriers for this disease? Assume Hardy-Weinberg conditions apply. A. 1/1000, or 0.1% B. 1/100, or 1% C. 1/50, or 2% D. 1/10, or 10% E. 1/4, or 25% Which of the following statements is FALSE? A. Bacteria isolated from clinical infections are often resistant to multiple antibiotics. B. Some antibiotic resistant bacteria can pump the antibiotic out. C. Antibiotic resistance genes encode proteins that can function to neutralize the antibiotic. D. Antibiotic resistant bacteria generally contain R plasmids. E. Because viruses are not susceptible to antibiotics, they are often the source of antibiotic-resistance genes. 13 14 15 5 Bio 97 2007 Final Exam 16 Section D version A – pg. 6 of 11 Blood pressure is a multifactorial trait with a heritability of 0.6. A study is conducted on identical (monozygotic, or MZ) twins that were separated at birth and reared apart. They are compared to fraternal (dizygotic, or DZ) twins reared apart (NOTE: This often happens with twins given up for adoption). The most likely outcome of this study is that... A. MZ twins will have almost identical blood pressures, much more so than DZ twins B. MZ twins will have more similar blood pressures than DZ twins, but there will still be significant variation between MZ twins due to environmental factors C. Both MZ and DZ twins will have very similar blood pressures D. DZ twins will have more similar blood pressures than MZ twins E. Environmental factors are more important than genetic factors with regard to blood pressure Which of the following statements most closely reflects the U.S. Army’s policy on genetic discrimination towards those who have served less than 8 years and become ill? A. If your heath problem is “your fault”, either because of your genes* or your behavior (e.g. smoking), then you will likely lose health benefits B. If your genes* clearly caused your illness, then you may lose health benefits C. Even if your genes* are only thought to increase the risk of developing your illness, you may lose your health benefits D. Only those who have sustained battlefield injuries are eligible for heath benefits, so genetic discrimination is really not an issue E. Genetic discrimination with regard to health benefits is prohibited in the U.S. Army * meaning your alleles 17 18 Consider the following statement: “At some times and in some places, the government has forcibly sterilized some people with the goal of keeping their ‘bad’ alleles out of the next generation’s gene pool.” Which of the following is most TRUE regarding this statement. A. This has never happened in the U.S. B. Although this has occurred in the U.S. a long time ago, the practice has been banned for over 150 years. C. This has occurred in the U.S. within the past 50 years. D. If this was done on a major scale, it would rapidly eliminate low-fitness recessive alleles from the gene pool. E. None of the above statements (A-D) are true. The reason that many scientists are interested in using _____ with human embryonic stem cells is to ______ A. B. C. D. E. therapeutic cloning; limit immune rejection transgenic technology; make them herbicide resistant reproductive cloning; grow organs antibodies; make gene libraries gene therapy; make designer babies 19 20 Relative to random mating, inbreeding results in: A. Increased frequency of heterozygotes B. Increased frequency of homozygotes C. Increased frequency of rare recessive diseases D. B and C E. None of the above 6 Bio 97 2007 Final Exam 21 Section D version A – pg. 7 of 11 In recombinant DNA work(i.e. “cloning a gene”) one of the following three reagents is typically used before the other, and one of the three is usually not used at all: Ligase (L) Restriction Enzyme (R) Exonuclease (E) Which of the following best describes how these reagents are typically used? A. L first, R second, E not used B. R first, E second, L not used C. E first, L second, R not used D. R first, L second, E not used E. None of the above Which of the following statements is FALSE? A. The frequency of recombination over a long interval of a chromosome is always smaller than the map distance over the same interval. B. Map distance between two genes can be expressed in cM or map units. C. A length of heterochromatin will appear shorter in the genetic map than an equal length of euchromatin. D. Some tissues like skin are mosaic due to meiotic recombination. E. Recombination never occurs in Drosophila males. Which of the following statements is TRUE? A. Recombination occurs randomly with more than 50 events per chromosome. B. A coefficient of coincidence of one means that, in a given region, all of the double crossovers were observed that were expected. C. If a crossover between two genes occurs in every meiosis, this would result in a recombination frequency of 100%. D. Interference is the result of crossing over between sister chromatids. E. Alleles in trans often show a higher frequency of recombination. Fill in the blanks: An R plasmid is a(n) ________________ plasmid that contains ________________ elements. These encode proteins that provide _________________to antibiotics. A. F, insertion, sensitivity B. non-conjugative, transposable, resistance C. conjugative, transposable, resistance D. F, transposable, sensitivity E. conjugative, Hfr, resistance Which of the following statements is TRUE? A. Checkpoints keep non-dividing cells such as muscle cells from dividing. B. Once cancer cells begin to grow as a tumor, they are genetically identical. C. Cancer cells are not cycling cells and so don’t need to inactivate checkpoints. D. Checkpoint failure contributes to cancer by promoting genetic instability. E. Checkpoints may not be inactivated in a cancer cell if DNA repair enzymes have already been inactivated. Which of the following transgenic or genetically-modified organisms has not yet been created: A. A rabbit that glows green B. A soybean that is resistant to a potent herbicide (herbicide= plant killing chemical) C. A transgenic fish over ten times normal size D. A “supermouse” that can run twice as far as a normal mouse E. A human cured of a genetic disease by germline gene therapy 22 23 24 25 26 7 Bio 97 2007 Final Exam 27 Section D version A – pg. 8 of 11 This is a two-part true/false question (i) On average, each new kid born carries one new mutant allele that was not present in the somatic cells of either parent. (ii) Most mutations result from replication errors. A. (i) = TRUE; (ii) = TRUE B. (i) = TRUE; (ii) = FALSE C. (i) = FALSE; (ii) = TRUE D. (i) = FALSE; (ii) = FALSE The following statements refer to genetically-modified (GM) crops. Which of the following statements (A-D) are TRUE? (Pick E if you think more than one of A-D are true) A. GM organisms are banned in the U.S. B. GM animals are banned in the U.S., but not GM crops C. GM crops pose essentially no risk to farmland ecology D. GM crops are widespread in the U.S. food supply E. More than one of the above statements (A-D) is true. Molecular geneticists have performed experiments in which they altered the number of copies of the bicoid gene in flies, affecting the amount of Bicoid protein produced. Answer the two questions in the table below, choosing the best answer from the 4 possibilities below. Write the number of the answer into the table. 0.5 points each 1. 2. 3. 4. Enlargement of anterior structures No effect Reduction of anterior structures Duplication of anterior structures 1 28 29 What would be the effect on development of an increased number of copies of the bicoid gene? What would be the effect on development of a decrease in bicoid protein? 3 30 In tomatoes, tall vine (d+) is dominant over dwarf (d), spherical fruit shape (p+) is dominant over pear shape (p), and red fruit (y+) is dominant over yellow fruit (y). Genes d and p are 12 map units apart in the same autosomal linkage group. Gene y is in a different autosomal linkage group. Toby performs a testcross with a tall plant with spherical shaped yellow fruit. The linked alleles are in trans. What percentage of the progeny are tall plants with pear shaped yellow fruit? 1 point. 44% What is the genotype of the tall plants with pear shaped yellow fruit? Use proper nomenclature. 1 point d+ p/d p y/y You needed to indicate that y was on a separate chromosome. 8 Bio 97 2007 Final Exam 31 Section D version A – pg. 9 of 11 (3 point short answer) An 8 kb plasmid is digested with EcoRI (E) and/or BamHI (B), and the digests are run on an agarose gel and stained. The results are shown below; molecular size standards are shown. Based on the results of the gel, show on the plasmid on the right (using the letter B) where the BamHI sites must be. The position of the EcoRI site has already been put in; the bars denote 1 Kb intervals. If you think there is more than one correct answer, just show a single correct answer. B E B+E E 8 7 6 kb 5 kb 4 kb 3 kb 2 kb 1 kb kb 8 kb 7 6 kb 5 kb 4 kb 3 kb 2 kb 1 kb kb One Bam site directly across from E and the other either one Kb to the right or left of E. B E E B OR B B 9 Bio 97 2007 Final Exam 32 Section D version A – pg. 10 of 11 (3 points short answer) Murder most foul! A crime has been committed, and a DNA samples are taken from the crime scene and from the victim. The police believe the crime scene sample is a mixture of the DNA of the victim and the DNA of the murderer(s). The police also round up three suspects and collect DNA from them. All DNA samples are then subjected to the DNA fingerprinting technique, but unfortunately the crime lab can only afford to use one pair of PCR primers for their analysis (so the results are not as conclusive as they could have been). Nevertheless, if we assume that at least one of the three suspects was at the crime scene and that any suspect(s) at the crime scene contributed DNA to the crime scene sample, then, based upon the DNA evidence… Which suspect(s) cannot have been at the crime scene? DJ Which suspect(s) might have been at the crime scene? Which suspect(s) must have been at the crime scene? OJ AJ 10 Bio 97 2007 Final Exam 33 Section D version A – pg. 11 of 11 In a cross of px • ry • cn / + • + • + females X px • ry • cn / px • ry • cn males, the following progeny were counted (+ is used to denote the wild-type alleles, cn+, ry+ and px+): px• ry• cn px• ry• + px• +• + px• +• cn +• ry• cn +• ry• + +• +• + +• +• cn / / / / / / / / px• ry• cn px• ry• cn px• ry• cn px• ry• cn px• ry• cn px• ry• cn px• ry• cn px• ry• cn 300 75 101 11 89 10 329 85 A. What is one of the maternal gamete genotypes that occur from a single crossover only between px and ry? Use correct nomenclature (hint: your answer should not have any “dots” in it). 2 points. px+ ry cn or px ry+ cn+ B. How many double recombinants are expected (round to nearest whole number)? 2 points. px to ry, 101+11+89+10 = 211/1000 = .211 ry to cn, 75+11+10+85 =181/1000 = .181 multiply .181 X .211 X 1000 = expected DCO = 38 C. What is the interference in the region? 1 point. cc = 21/38 = 0.55 or 0.6 i = 1-cc = 0.45 or .5 or .4 34 Consider the diagram below of chromosome 2 from a heterozygous individual. A b R A b R a B r Must draw in the sister chromatids for each chromosome and then show a crossover between b and r. A R a a b B B r r Draw in the meiotic recombination event(s) that would give result in a gamete of the genotype A B r. 1 point. 11 Bio 97 2007 Final Exam 35 Section D version A – pg. 12 of 11 Two streptomycin-sensitive prototrophic Hfr strains were compared in matings with a streptomycin-resistant F- strain auxotrophic for leu, phe, pro, trp, and ala, no order implied. With HfrT, the markers leu+, pro+ and ala+ were transmitted at 16, 19 and 29 minutes, respectively. With HfrC, markers leu+ and ala+ were transmitted at 16 and 3 minutes, respectively. A. What time of entry would you expect for marker pro+ in the HfrC strain? 1 point. 13 minutes B. What selective media was used to identify pro+ recombinants in this experiment? 2 points. minimal media + leucine, phenylalanine, tryptophan, alanine AND, streptomycin 12 ...
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This note was uploaded on 02/09/2011 for the course BIO 97 taught by Professor Edinger during the Spring '10 term at UC Irvine.

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