BICD 100 Final 2006 key

BICD 100 Final 2006 key - BICD100 Final Exam 3/20/06 KEY to...

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Unformatted text preview: BICD100 Final Exam 3/20/06 KEY to version A (1) Hospitals often experience problems with bacteria that acquire resistance to multiple antibiotics, which cause infections that are very difficult to treat. You are a microbiologist in a hospital experiencing an outbreak of infections caused by a multiply antibiotic resistant strain of E. coli. You isolate the strain and find that it is resistant to streptomycin (strR), tetracycline (tetR), ampicillin (ampR), and kanamycin (kanR), but is sensitive to a less widely used antibiotic called nalidixic acid (nalS ). Your task is to find out more about the genetic basis of antibiotic resistance in this strain to help develop a strategy for combating it. Experiment I: You transform the multiply antibiotic resistant E. coli strain with an F plasmid and are successful in isolating an Hfr version of the strain, which is strR, tetR, ampR, kanR, nalS . This strain is mated with a tamer E. coli strain that is F- and nalR, but strS , tetS , ampS , kanS . The two strains are mated; mating pairs disrupted at 5 minute intervals and plated on media containing nalidixic acid plus one other antibiotic. The following results are obtained: time 5 min. 10 min. 15 min. 20 min. 25 min. 30 min. 35 min. 40 min. 45 min. # of colonies growing on plates containing the following antibiotics (per 100 Hfr cells plated): nal + kan nal + amp nal + tet nal + strep 0 0 0 0 0 0 0 0 0 0 0 0 5 6 0 0 29 35 0 3 60 61 0 25 63 64 0 48 64 65 0 52 65 66 0 54 (data were collected every 5 minutes for another hour, but the results were never significantly different from the result at 45 minutes) a. On a "time of transfer map" of the multiply antibiotic resistant strain's chromosome, how many minutes would there be between the Hfr origin of transfer and the strepr gene? (5 pts) 25 minutes (the first timepoint at which the donor allele strepR appears) b. How could you explain why tetR nalR colonies never appear? (5 points) 3 answers accepted for full credit: tetR too far away from F insertion to be transferred, tetR located after nalS, tetR very close to nalS. Experiment II. You mate the Hfr strR, tetR, ampR, kanR nalS donor strain with the F- nalR, strS , tetS , ampS , kanS recipient strain again. After 45 minutes, the mating pairs are disrupted and plated on media containing nalidixic acid and streptinomycin. 500 colonies that grew on these plates were then tested for their ability to grow on plates containing ampicillin (amp) or kanamycin (kan), with the following results: 388 colonies grew on both amp plates and kan plates 91 colonies grew on neither amp plates nor on kan plates 19 colonies grew on kan plates but not on amp plates 2 colonies grew on amp plates but not on kan plates c. Use the results of experiments I and II (given on the previous page) to draw a map of the E. coli chromosome in the multiply antibiotic resistant strain showing the locations of the F insertion (put this at "zero minutes") and the strR, tetR, ampR, kanR and nalS genes. Show the orientation of the F insertion using an arrow to indicate the direction of transfer. Show the location of each gene on the map as precisely as the data allow, indicating locations in minutes where possible. For genes whose exact location in minutes cannot be determined, show their relative locations. (15 pts) (See map next page) 1 +1 for showing chromosome as a circle +2 for F insertion at 0 min. with arrow pointing right direction relative to resistance genes +2 for ampR/ kanR at ~20 min. +6 for correct relative positions of ampR, kanR and strepR 20 min. +2 for showing nalS after kanR, strR and ampR R S S +2 for showing tet close to nal , after nal , or more than halfway around circle from F insertion 25 min. 0 min. F insertion ampR kanRR strep tetR nals (2) Drosophila melanogaster caught in the wild display one of two different larval foraging behaviors. "Rovers" show longer foraging trails on food and tend to wander from one patch of food to another. "Sitters" show shorter foraging trails on food and tend to remain on a single patch of food (see illustration). This behavioral difference is seen only in the presence of food, and is controlled by a single gene, for, with two alleles fors and forr. In a particular wild Drosophila population, 714 individuals are fors homozygotes, 264 are heterozygotes, and 22 are forr homozygotes. a. Use the Chi Squared test to determine whether this population is in Hardy Weinberg equilibrium. Include in your answer your hypothesis, the value of chi squared, the corresponding p value, and your conclusion. See chi squared table, top of next page. (12 pts) Hypothesis: population is in H-W equilibrium. If so, then f(aa) = q2 , f(Aa) = 2pq, f(AA) = p2 Call fors a and forr A. Observed values for f(aa) = 0.714, f(Aa) = 0.264, f(AA) = 0.022. q = f(a) = f(aa) + 1/2f(Aa) = 0.714 + 0.132 = 0.846 p = 1 q = 0.154 Expected values are f(aa) = q2 = 0.716, f(Aa) = 2pq = 0.261, f(AA) = p2 = 0.024 2 = [(O-E)2 /E] = (714 - 716)2 /716 + (264 261)2 /261 + (22 24)2 /24 = 0.206 At one degree of freedom, 0.9 > p > 0.5 null hypothesis cannot be rejected Conclusion: population is in Hardy Weinberg equilibrium. 2 b. If the population size remains at >1000, the fitness of rovers and sitters is equal, the mutation rate at the for gene is negligible, there is no migration in or out of the population, and mating is random, what will be the frequency of fors and forr alleles in subsequent generations? (5 pts) This describes Hardy-Weinberg equilibrium frequencies will stay the same. c. Through inbreeding, true-breeding rover and sitter lines are established, and then crossed together. F1 progeny of this cross are all rovers. When these F1 rovers are crossed together, 3/4 of the progeny are rovers and 1/4 are sitters. What can you conclude from these results (one sentence)? (5 pts) Rover is dominant to sitter. Sitter is recessive to rover. Either answer is sufficient. d. Identification of the for gene showed that it encodes a cyclic GMP-dependent protein kinase (PKG) and is expressed in the central nervous system (CNS) of developing larvae. The PKG enzyme isolated from the CNS of rover flies is about 12% more active than that from sitter flies. A scientist wants to use RNAi to determine the phenotype of flies lacking PKG enzyme activity. Circle each of the following things that are crucial to the success of this strategy: (9 pts) >>i. double stranded for RNA ii. expression of the for gene in tissues where it would not normally be expressed iii. disruption of the for gene by insertion of a neomycin resistance gene >>iv. cleavage of double stranded RNA by the enzyme "dicer" >>v. formation of protein-RNA complexes that cause the degradation of endogenous for mRNA vi. flies homozygous for a transgene consisting of two copies of the for gene arranged in an inverted repeat configuration (note flies don't have to be homozygous!) 3 e. The result of the RNAi experiment is that larvae lacking PKG activity are immobile and die before forming adults. Circle each of the following statements that is consistent with this and all other information provided earlier (9 pts): i. Foraging behavior is a quantitative trait. No trait shows two discrete phenotypes and the difference between the two phenotypes is controlled by alleles of a single gene. >>ii. The sitter phenotype is due to a subtle loss of function mutation in the PKG gene. iii. The sitter phenotype is due to a subtle antimorphic (dominant negative) mutation in the PKG gene. No in this case sitter would be semi-dominant or dominant. iv. The rover phenotype is due to a subtle loss of function mutation in the PKG gene. No not consistent with enzyme activity results. v. The rover phenotype is due to a subtle antimorphic (dominant negative) mutation in the PKG gene. No - this describes a loss of function mutation, which is not consistent with the enzyme activity results. >>vi. The rover phenotype is due to a subtle gain of function mutation in the PKG gene. (3) The Mennonites of Lancaster County, Pennsylvania, have been the subject of many genetic studies. They are descended from a small founder population of ~100 people who migrated to North America from Switzerland in the 1700s, and have lived in nearly complete isolation ever since. The Mennonites have extensive genealogical records (i.e. pedigrees), and are also of interest to human geneticists because they have unusually high frequencies of some genetic diseases, including Maple Syrup Urine Disease (MSUD), which is caused by a single recessive mutation. MSUD is a metabolic disorder. Without treatment, children with the disease invariably die, but the disease can be managed by control of the diet such that affected individuals live normal lives. The gene responsible for MSUD is known and Mennonites are all genotyped now at birth for their alleles at this locus. The incidence of MSUD in the present day Mennonite population is 1/358 and the frequency of carriers is 1/13. In the ancestral European population from which the Mennonites are descended, the incidence of MSUD is thought to have been 1/200,000 (because this is the frequency of the disease in this part of Europe today). In answering each of the following questions, show the key steps in your work (correct final answers showing no work will not receive full credit). a. Assuming that mating was random within the ancestral European population, what was the frequency of the MSUD mutation in this population? (5 pts) f(aa) = 1/200,000 If mating is random, f(aa) = q2 so q = 0.0022 Unfortunately many students misinterpreted this question and thought I was asking for , the frequency at which the wild type allele mutates to the MSUD allele. There was not enough information provided to answer this question so there is no "correct" answer to this question. Because this was such a common problem, this question will be thrown out and the 5 points redistributed elsewhere. b. What is the frequency of the MSUD allele in the present day Mennonite population (do not assume that mating is random)? (5 pts) f(a) = f(aa) + 1/2f(Aa) = 1/358 + (1/2)(1/13) = 0.04125 4 The high incidence of MSUD in the Mennonite population could in principle be caused by two different phenomena. One is inbreeding, which is a common occurrence in relatively small, isolated populations. The other is the so called "founder effect" in which the very small group of individuals who founded the Mennonite population in the 1700s by chance had a high frequency of MSUD carriers and that random fluctuations in allele frequency occurring within the initially small population further increased the frequency of the MSUD allele. c. What is the inbreeding coefficient for the present day Mennonite population? (6 pts) H = observed frequency f(Aa) = 1/13 = 0.077; q = 0.041 (from part b); p = 1 q = 0.959 F = 1 (H/2pq) = 1 (0.077/2(0.959)(0.041) = 0.025 You could also answer this starting with f(Aa) = 2pq(1 F). If you do the algebra and rearrange this equation to solve for F, you get F = 1 f(Aa)/2pq. Another equally correct but less straightforward route to the answer is to use the equation f(aa) = qF + q2 (1 F). The algebra here is a little more involved, but if you rearrange this equation to solve for F you get F = [f(aa) q2 ]/[q q2 ] and this also gives 0.025. d. Can inbreeding alone explain the high incidence of MSUD in the present day Mennonite population compared to the ancestral population? Why or why not (one sentence)? (6 pts) No. Inbreeding increases the frequency of homozygotes and decreases the frequency of heterozygotes, but does not by itself increase or decrease allele frequencies. The frequency of the MSUD allele has increased from 0.0022 in the ancestral population to 0.04125 in the present day Mennonite population. Even if you misinterpreted question 3a so did not calculated there that f(a) in ancestral population was 0.0022, you could have calculated it here to be able to make this comparison. e. In 1927, shortly after the invention of the automobile, the Mennonites of Lancaster County split into two groups the Weaverland Mennonites (aka "horse and buggy Mennonites", who shunned automobiles) and the Groffdale Mennonites (aka "black bumper Mennonites", who accepted automobiles but thought that chrome bumpers were an unnecessary adornment and therefore painted their bumpers black). These two groups have not intermated, and now have markedly different frequencies of MSUD in the Weaverland Mennonites, the frequency is 1/686 (carrier frequency 1/17) and in the Groffdale Mennonites, it is 1/271 (carrier frequency 1/10). If 500 individuals from each group joined together to form a new group called the Reunited Mennonites, who intermate with each other but not with either the Groffdales or Weaverlands, what would be the expected frequency of individuals with MSUD in the next generation of Reunited Mennonites? Assume that the inbreeding coefficient within the reunited group is the same as what you calculated in part c. If you don't have a value for F, make one up but clearly indicate what value you are assuming. Also assume that MSUD symptoms are managed so that affected individuals have the same fitness as unaffected individuals, and ignore the effects of mutation. (12 pts) In this problem, we have migration followed by non-random mating (inbreeding). First calculate the new value of q after migration: Let pi = p for Weaverlands qi = f(aa) + 1/2f(Aa) = 1/686 + (1/2)(1/17) = 0.031 pi = 1 0.031 = 0.969 Let pm = p for Groffdales qm = 1/271 + (1/2)(1/10) = 0.054 pm = 1 0.054 = 0.946 p' = mpm + (1 m)pi = (0.5)(0.946) + (0.5)(0.969) = 0.9575 q' = 1 p' = 1 0.9575 = 0.0425 Note that you could also calculate the q' as = #a alleles/total #alleles in new population = [2f(aa)Groffdale x 500 + f(Aa)Groffdale x 500 + 2f(aa)Weaverland x 500 + f(Aa)Weaverland x 500]/2000 = 0.0425 f(aa) = qF + q2 (1 F) = (0.0425)(0.025) + (0.0018)(0.975) = 0.0034 or approx. 1/300 5 (4) Another disease that occurs at a high frequency in the Mennonite population is Hirschsprung's Disease (HSCR), which is characterized by gastrointestinal problems resulting from the failure of the colon to become properly innervated during embryonic development. It is highly heritable, but is influenced by alleles of at least 3 genes. A gene increasing the risk of HSCR in the Mennonite population is located on chromsome 13 in the 13q22 region. The following diagram shows the locations of four equally spaced DNA markers in the 13q22 region (D13S160, 317, 170 and 264) that were analyzed for linkage disequilibrium with Hirschsprung's Disease in the Mennonite population. For each marker, the probability that the marker is not in linkage disequilibrium with HSCR is given. D13S160 D13S317 D13S170 DS13S264 p = 0.09 p < 0.001 p < 0.001 p = 0.196 a. According to this data, which of the following is the most likely location for a mutation increasing the risk of HSCR (circle one)? (5 pts) i. To the left of D13S160 >>iii. Between D13S317 and D13S170 v. To the right of DS13S264 ii. Between D13S160 and D13S317 iv. Between D13S170 and DS13S264 vi. Either to the left of D13S160 or to the right of DS13S264 Further mapping studies led to the identification of a single gene, EDNRB, as a candidate for the gene in the 13q22 region increasing the risk of Hirschsprung's Disease in the Mennonites. This gene encodes a receptor for endothelins, a group of secreted proteins having diverse biological effects. In 90 Mennonite individuals who did not have HSCR or a family history of HSCR, amino acid 276 in the EDNRB protein was usually found to be a tryptophan. In contrast, the EDNRB alleles found in Mennonites with HSCR encode a cysteine at position 276 instead of tryptophan, but are otherwise the same as the alleles of unaffected individuals. Further studies showed that EDNRB protein with cysteine at position 276 doesn't function as well as an endothelin receptor as does EDNRB protein with tryptophan at position 276. b. Circle each of the following statements that is consistent with this information. >>i. A point mutation in the EDNRB gene increases the risk of HSCR. ii. A nonsense mutation in the EDNRB gene increases the risk of HSCR. iii. A silent mutation in the EDNRB increases the risk of HSCR. >>iv. A missense mutation in the EDNRB increases the risk of HSCR. >>v. A loss of function mutation in the EDNRB gene increases the risk of HSCR. vi. A gain of function mutation in the EDNRB increases the risk of HSCR. 6 c. To further investigate EDNRB as a candidate for a gene contributing to Hirschsprung's Disease, a group of scientists set out to determine the phenotype resulting from a loss of EDNRB function in mice. One approach taken was to knockout the wild type EDNRB gene via a gene replacement strategy, and the other was to express a dominant negative form of the EDNRB receptor containing a endothelin binding domain but lacking transmembrane and intracellular domains. Indicate which of the things listed below would be needed to implement each of these approaches (write the numbers for each necessary thing in the blanks provided): (12 pts) Knockout/gene replacement approach_______3, 7, 8, 9_________________________________________ Dominant negative approach______________4, 5, 8___________________________________________ 1. Synthesis of double stranded RNA for a portion of the EDNRB gene 2. Insertion of a thymidine kinase gene into the EDNRB gene (note -"INTO" IS NOT NEXT TO!) 3. Insertion of a neomycin resistance gene into the EDNRB gene 4. Truncation of the EDNRB gene to remove the portions encoding the transmembrane and intracellular domains of the endothelin receptor protein 5. Microinjection of a mutant version of the EDNRB gene into mouse zygotes 6. Infection of embryonic stem cells with Agrobacterium carrying the transgene on a modified Ti plasmid 7. Identification of embryonic stem cells in which the endogenous, wild type EDNRB gene is replaced with a null loss of function version of the EDNRB gene on one chromosome 8. Identification of mice having one or more copies of the transgene integrated into their chromosome(s) 9. A series of crosses to produce mice that are homozygous for the transgene 10. Cleavage of EDNRB mRNA by the enzyme "dicer" d. Mice homozygous for a null mutation in the EDNRB gene were found to have colon innervation and gastrointestinal problems resembling those seen in Hirschsprung's Disease patients and die at a young age. Interestingly, these mice also have a piebald spotting phenotype with white spots on their coats. This phenotype is similar to that of mice homozygous for a mutation described long ago called "piebald lethal". EDNRB null heterozygotes were crossed to piebald lethal heterozygotes, and 1/4 of the progeny were found to have the piebald lethal/HSCR-like phenotype. Circle each of the following statements that is clearly supported by these findings: (6 pts) i. The piebald lethal mutation complements the EDNRB null mutation >>ii. The piebald lethal mutation fails to complement the EDNRB null mutation >>iii. The piebald lethal phenotype is caused by a mutation in the EDNRB gene iv. The piebald lethal phenotype is caused by a mutation in another gene (not EDNRB) 7 Below are pedigrees for three Mennonite families affected by Hirschsprung's Disease. For each individual, the genotype for the tryptophan -> cysteine (W -> C) mutation in the EDNRB gene described on page 6 is given (M indicates the mutant allele and + indicates the wild type allele). People with HSCR are shown as filled symbols. Similar to the EDNRB knockout mice, people with HSCR (or who have close relatives with HSCR) often have pigmentation defects affecting skin, hair and eyes, as indicated ("irides" is the plural of iris the colored part of the eye). In addition, they are sometimes deaf. e. Circle each of the following statements that is clearly supported by the information in this pedigree. (6 pts) >>i. The W -> C mutation in the EDNRB gene shows incomplete penetrance ii. The W -> C mutation in the EDNRB gene shows incomplete dominance >>iii. The W -> C mutation in the EDNRB gene is pleiotropic iv. The W -> C mutation in the EDNRB gene shows epistasis If the individuals labeled as A and B marry and have a child, assuming that they are unrelated... f. What is the probability that the child will be a carrier for HSCR? (5 pts) g. What is the probability that the child will have HSCR? (8 pts) Probability that child is homozygous mutant is 1/4. % penetrance is 5/17. So probability that the child will be affected is 5/17 X 1/4 = 5/68 or 0.0735 1/2 8 (5) The size of tomato fruits has been increased dramatically through conventional plant breeding. a. From a genetically heterogeneous population of tomatoes with a mean fruit size of 3 cm, a subset of plants with a mean fruit size of 5 cm was selected and crossed together. In the progeny of this cross, mean fruit size was 4 cm. If a subset of plants in this progeny group with a mean fruit size of 6 cm are chosen and crossed together, what do you expect to be the mean fruit size in the next generation? (8 pts) hN = (Yo Y)/ (Yp Y) = (4 3)/(5 3) = 0.5 For next round of selective mating, Yo = Y+ hN(Yp-Y) = 4 + (0.5)(6 4) = 5 cm The wild ancestor of the "domesticated" tomato we eat (Lycopersicon esculentum) is thought to have had fruits of about 1 cm in diameter, resembling those of various Lycopersicon species found in the wild today, such as the South American variety L. chmielewski. L. esculentum and L. chmielewski are sufficiently closely related that they can be crossed together to generate fertile F1 hybrids, facilitating the genetic analysis of tomato fruit size. b. An inbred line of L. esculentum with a mean fruit diameter of 5 cm was crossed to an inbred line of L. chmielewski with a mean fruit diameter of 1 cm to produce an F1 hybrid line with a mean fruit diameter of 2 cm. The standard deviation in fruit size for both parental lines and the F1 hybrid line when grown under the same conditions is 0.5 cm. The F1 hybrids were allowed to self-fertilize, generating a heterogeneous F2 population with a mean fruit size of 2 cm and a standard deviation of 1.3 cm. What is the broad sense heritability of tomato fruit diameter in this population? (8 pts) Variance in F1 population is all environmental, so Ve = (0.5)2 = 0.25 Variance in F2 population is environmental + genotypic, so Vt = (1.3)2 = 1.69 Vt Ve = Vg = 1.69 0.25 = 1.44 HB = Vg /Vt = 1.44/1.69 = 0.85 F2 progeny were scored individually for fruit diameter and also tested by PCR to determine their genotypes for molecular markers distributed throughout the genome. For six of these markers, a significant correlation between genotype and fruit size was found; the corresponding r values are shown to the left of each chromosome on the map below. (see questions about this map, next page...) 9 c. Circle each of the following conclusions that can be drawn from the data shown on the map (bottom of previous page) and your answers to parts a and b: ( 9 pts) i. There are six genes influencing fruit diameter in L. esculentum. Six is the minimum number but most likely there are more genes whose effects are too small and/or genes that are too far away from the closest marker to be picked up by this analysis. ii. CAB1 is closer to a gene controlling fruit diameter than any other marker used in this analysis. Not necessarily value of r is determined both by distance from nearest marker and by magnitude of the effect of the gene on the trait. iii. Chromosomes 3, 4, 5, 8, 10 and 12 have no genes affecting fruit diameter. Can't conclude this see comments for answer i, above. iv. 69% of the variance in fruit diameter in the F2 population is due to genetic differences and the rest is due to environmental effects. >>v. 85% of the variance in fruit diameter in the F2 population is due to genetic differences and the rest is due to environmental effects. 85% is the broad sense heritability for fruit weight = % of variance due to genetic differences. Note that if your answer for broad sense heritability was not 85%, you didn't get credit for selecting answer v. vi. 50% of the variance in fruit diameter in the F2 population is due to genetic differences and the rest is due to environmental effects. d. Fine scale mapping of the QTL linked to CAB1 led to the identification of four candidates for the gene influencing tomato size in this region. When introduced into L. esculentum via Agrobacterium-mediated transformation, one of the four L. chmielewski genes (FSG = fruit size gene) caused about a 25% decrease in fruit size. Analysis of esculentum and chmielewski alleles of FSG showed no differences that would change the amino acid sequence of the gene product. Analysis of gene expression showed that FSG is highly expressed in developing fruits of L. chmielewski, but is expressed at low levels in developing fruits of L. esculentum. Which of the following conclusions is best supported by these results? (Circle one 5 pts) i. Breeding of L. esculentum from its wild ancestor involved selection for a gain-of-function mutation that increased the expression of a gene (FSG) that promotes fruit growth. ii. Breeding of L. esculentum from its wild ancestor involved selection for a gain-of-function mutation that increased the expression of a gene (FSG) that limits fruit growth. iii. Breeding of L. esculentum from its wild ancestor involved selection for a loss-of-function mutation that decreased the expression of a gene (FSG) that promotes fruit growth. >>iv. Breeding of L. esculentum from its wild ancestor involved selection for a loss-of-function mutation that decreased the expression of a gene (FSG) that limits fruit growth. e. Which of the following transgenes, if introduced into L. chmielewski via Agrobacterium-mediated transformation, is most likely to increase the size of L. chmielewski tomatoes? (Circle one 5 pts)) i. A transgene consisting of the L. esculentum FSG promoter and coding region ii. A transgene consisting of the L. chmielewski FSG promoter fused to the L. esculentum FSG coding region >>iii. A transgene consisting of the L. chmielewski FSG promoter fused to two identical fragments of the FSG coding region arranged in an inverted repeat configuration. This describes an RNAi construct. Fruit size in L. chmielewski should increase if FSG expression is suppressed via RNAi. iv. A transgene consisting of the L. esculentum FSG promoter and a truncated version of the FSG coding region. 10 ...
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This test prep was uploaded on 04/07/2008 for the course BICD 100 taught by Professor Nehring during the Winter '08 term at UCSD.

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