methods-2020-exam

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Studocu is not sponsored or endorsed by any college or university Methods 2020 exam Scientific Method in Biology (The University of Western Ontario) Studocu is not sponsored or endorsed by any college or university Methods 2020 exam Scientific Method in Biology (The University of Western Ontario) Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

THE UNIVERSITY OF WESTERN ONTARIO DEPARTMENT OF BIOLOGY Biology 2290F Final Exam Wednesday, December 16 th , 2020 Time: 9:00 am – 12:05 pm Timed exam administered through OWL Part 1: Experimentation and Communications Unit – 4 questions, 50 marks (10%) Part 2: Genetics Unit – 6 questions, 30 marks (10%) Part 3: Instrumentation Unit – 2 questions, 31 marks (10%) Answer boxes are limited to the space provided. Use concise language. I understand that group work, discussion of questions with others, and/or sharing of answers in any way during the exam is absolutely prohibited. By entering my name below, I acknowledge that I will receive a zero on this exam if I’m found to have done so. STUDENT NAME: Jennifer Pham STUDENT NUMBER: 251084684 not guaranteed that the answers will be correct but can give you an idea of what questions to expect!! Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

Biology 2290F Experimentation and Communication Unit Final Exam (10%) 50 Points (Marks) Total 1. The following sentences were taken from a scientific poster in which the authors studied the ecophysiology of Carpobrotus . Beside each sentence place an “I”, “M”, “R”, or “D” to indicate whether the sentence belongs to the Introduction, Methods, Results, or Discussion section of the poster. Each letter can be used more than once. 1.1 [1 pt] Root to shoot ratio (RSR) decreased by 8% in plants grown in the LN, but not in plants growing in the HN substrate ( F = 8.1, P = 0.006, for the interaction nutrients ° competition). R 1.2 [1 pt] Human-mediated exchange of species has intensified greatly during the past centuries, and an ever- increasing number of exotic plant species are disrupting natural communities and ecosystem functioning. I 1.3 [1 pt] Our study shows that both taxa diverged markedly in their reproductive strategies because C. edulis reproduced both sexually and asexually, whereas C . aff. acinaciformis only reproduced asexually during the length of the experiment. D 1.4 [1 pt] Reflectance spectra (wavelengths from 306 to 1136 nm) were calculated for each leaf by dividing the spectral radiance of the leaf by the radiance of a reflective white standard (Spectralon Reflectance Standard, Labsphere, North Sutton, NH). M 1.5 [1 pt] The study findings demonstrated that the taxonomic differentiation between taxa correspond to ecophysiological differentiation, warranting detailed examination of all trades-offs existing to predict the long term outcome of the interaction between these taxa. D 1.6 [1 pt] Differences in both PRI 531 and CHL 550 increased in the taxa, with higher mean values in C . aff. acinaciformis than in C. edulis . R 1.7 [1 pt] The initial dry mass was estimated from the fresh mass, by applying the relationship between dry mass and fresh mass calculated in a subset of 80 spare clonal fragments. M 1.8 [1 pt] We aimed to determine whether the taxonomic differentiation between taxa coincides with ecophysiological differentiation by testing differences between both morphotypes in functional traits related to competitive ability and resource-use efficiency. I Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

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2. A study was conducted to examine the reproductive biology of native, introduced, and hybrid species of Phragmites . The goal was to identify the degree to which hybridization between lineages is ongoing. Ten seeds from each of 10 panicles (a panicle is a flowering structure) ( n = 100 seeds per lineage stand) were removed. Seeds were planted in cell trays in a soil mixture of equal parts coarse sand and peat moss. Two seeds were planted in each cell at a depth of less than 1 cm, for a total of five cells per panicle. Trays were maintained in a greenhouse and germination was determined as visible shoot growth above the soil surface. To evaluate the effect of cold stratification on seed germination for each lineage, the same sets of panicles from the above experiment were stored at 4±C for 2 months to mimic winter conditions. The planting, germination, and harvesting process as described above was repeated with the cold-treated seeds. The data were analyzed using a one-way Anova and Tukey tests. The results of this experiment are presented below in Fig. 1. FIGURE 1. Dormancy in Phragmites australis lineages illustrated by mean percent germination – SE responses of seeds to control and cold stratification. Different letters above bars denote significant differences among lineages within a treatment ( P < 0.05) according to Tukey test. 2.1 [1 pt] How many replications were used in this study? 10 2.2. [3 pts] Describe one replicate in the hybrid control treatment 1 panicle of a hybrid Phragmite australis plant with 10 seeds planted 2.3. [5 pts] What was/were the explanatory variable(s) in this study? The 3 Phragmites australis lineages and the germination temperature 0 5 10 15 20 25 30 35 40 45 Native Introduced Hybrid Control a b b Native Introduced Hybrid Cold stratification b b a Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

2.4. [3 pts] State the reason why the authors chose to report SE values. SE values were used because there are 10 values (seeds) per replicate and the mean of those 10 seeds are taken per replicate 2.5. [2 pts] Which lineage exhibited the highest percent germination? Native 2.6. [1 pt] Did the response variable exhibit the same pattern (i.e. qualitative trend) with respect to the independent variable(s)? Indicate YES OR NO Yes 2.7. [1 pt] Which of the two treatments exhibited a better response in percent germination? Cold stratification 2.8. [4 pts] “T he introduced lineage in this study did not benefit as much from a cold treatment .” Do you agree with this statement? Indicate YES or NO and provide a reason to support your answer. NO because the percent germination of the introduced lineage has the largest increase from the control (greenhouse treatment) to the cold treatment. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

3. A study was conducted to compare reproductive traits of two plant species, Baptisia arachnifera (a rare species) and B. lanceolata (a common widely distributed species) and soil texture of their habitats. Table 1 . Comparison of mean reproductive traits (  SE) and soil magnesium (  SE) of Baptisia arachnifera (rare) and B. lanceolata (common) sampled from various populations. Traits/Soil nutrient B. arachnifera B. lanceolata Pod production per plant 18.81  4.51 a 12.22  2.31 a Pod size (cm 3 ) 0.65  0.04 b 2.75  0.20 a Damaged pods (%) 52.38  5.45 a 26.31  4.54 b Seed production (seeds per pod) 2.00  0.26 b 5.40  0.63 a Seed abortion (%) 40.79  6.03 a 30.73  4.46 a Soil magnesium(g/m 2 ) 0.33  0.05 b 2.32  0.81 a Note: Means in each row followed by the same letter are not significantly different ( P <0.05) according to the Student’s t test. Examine Table 1 and state whether the hypotheses of the authors, listed below, were supported (S) or disproved (D) . 3.1. [2 pts] The rare B. arachnifera has significantly lower pod and seed production. Pod production D Seed production S 3.2. [2 pts] There is no significant difference between Baptisia species in terms of pod size and seed abortion. Pod size D Seed abortion S 3.3 [1 pt] Pod damage by insects is not significantly different between the two species. Damaged pods D 3.4 [1 pt] There is no significant difference in soil magnesium of Baptisia arachnifera and B. lanceolata habitats. Soil magnesium D Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

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4. Closely related species (congeners) occurring in sympatry (same geographical area) may experience the negative consequences of interspecific pollen transfer if reproductive isolation (RI) barriers are not in place. Intraspecific (within a species) and interspecific (between species) crosses were conducted in a greenhouse to evaluate post‐pollination barriers to hybrid fruit set and potential costs of hybridization. Plants, of each species, served as a pollen donors and pollen recipients for each of two congeners. For analyses, fruit set (fruits/flowers) was used as a proxy of hybrid viability. χ 2 = 10.5, P = 0.005 χ 2 = 21.8, P < 0.0001 χ 2 = 5.2, P = 0.075 FIGURE 2. Mean fruit set (– SE) obtained from intra- and interspecific crosses between three Achimenes species. Letters indicate species and shapes indicate maternal species for each cross: A /diamond = A. anthirrhina , F /triangle = A. flava , P /circle = A. patens . Different letters over symbols indicate significant differences between treatments (P<0.05) within each maternal species according to Chi-square test Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

4.1. [13 pts] Results: Write the scientific story (text) as shown by the data presented in Fig. 2. Response should NOT EXCEED A MAXIMUM OF 4 SENTENCES. The mean fruit set was significantly lower for hybrid species of A. anthirrhina when compared to the purebred species (Fig. 2, P < 0.005). For A. flava, the mean fruit set was significantly lower for hybrid species crossed with A. anthirrhina when compared to the purebred species and the species crossed with A. patens (Fig. 2, P<0.0001). There is no significant difference between the species for A. patens (Fig. 2, P<0.075). Overall, there is no meaningful trend observed between the fruit set and the type of pollen transfer (intraspecific or interspecific) in each species . 4.2 [3 pts] Which interspecific cross showed the highest fruit set? Which plant was the pollen donor and which plant was the pollen recipient? The interspecific cross of A. flava and A. patens showed the highest fruit set. The pollen donor is A patens and the pollen recipient is A. flava. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

. Genetics Unit Final Exam (10%) December 16 th , 2020 30 marks total Scenario: We want to clone a protein called Anti-COVID2290 which destroys the COVID-19 virus. We want to ligate the and insert that codes for Anti-COVID 2290 into a plasmid named pUWO so that we can express the insert in E. Coli JM101 thus producing Anti-COVID 2990. As part of this research, it has been found by another research group that incubating 100  L of pUWO with 5  L of a chemical called NCB301 for 1 hour prior to beginning the restriction digest improves both transformation and ligation efficiencies. The tube of 100  L pUWO and 5  L NCB301 is used as thee tube of plasmid for the restriction digest. We are performing an experiment to see if we can re-produce these results. All questions are based on this scenario. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

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Below, you will find the sequence of Anti-COVID2290 and a plasmid map of pUWO Anti-COVID2290 Sequence: Start Codon: AGT Stop Codon: TAG Restriction Enzymes (/ indicates cut): HindIII – A/AGCTT BamHI – G/GATCC BglII – A/GATCT XbaI – T/CTAGA PstI – CTGCA/G AAGTACCATGT GGATCC ACTTGATGGAGGATGGAGTTCAT AAGCTT AGTG AGT TGGAGTGCTTGCTCTTCGAGTTGTCATCGGTATA GGACGAGAGCATGTACCGTTCCACCTCCGATGAATGGAGGACAGCCATGTTTTGGAGATGATCTGATGACACAAGAGTGTCCGGCA CAGCTTTGCACTGCGGACTCCTCTCGGATAGTAATCAGTGATACTGCTGTTTATGGATCAGTTGCATCTATTTTCATTGTTGCCTCATTT ATTCTTGCAATTCTTGCAATGTTCTGCTGTAAACGTGGCAATTCAAAA AAGCTT CGAAACCATTGAAACCGCAGAAAATTAATTCGGA AAAAGCAGGAGGAATCTACTATTCTGAGCCTCCAGGTGTTCGAAGATTACTTTTGGAGCATCAACATGGAACATTGCTCGGAGAGAA AATATCATCGTGTTCTCAATATTTTGAGCCTCCTCCTCTACCACATTCAACAACTCTTCGATCTGGAAAAAGTGCATTCAGTGGATATTC TTCAACAAGAAACGCTGGCTCACGAGCTGCTTTAATTCAAGAATGTAGTTCGAGTAGTTCAGGATCTGGTGGAAAACGGACAATGCT TAGAACTTCAAGCTCAAATTG TCTAGA CGGATGACGATAATTATGCTACGTTATATGATTATATGGAAGACAAGAGTGTTCTTGGATT AGACACTTCACAAAATATTGTTGCTGCGCAGATTGATAGCAATGGAGCACGATTGTCTTTATCAAAAAGTGGAGCAAGATTAATTGTA CCTGAATTGGCAGTGGAAGGAGAGAAAATGCTT TAG ACCTCG CTGCAG GCTGTTTCGGA AGATCT TAACA Plasmid Map: pUWO 3000 bp Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

1. To begin the cloning experiment with pUWO and Anti-COVID2290, we need to perform a restriction digest. Keeping in mind the goal is to express the Anti-COVID 2290 protein, would you choose to do a single digest (1 restriction enzyme) or a double digest (2 restriction enzymes). Explain why. (3 marks) You would choose to do a double digest because you want the insert to go into the plasmid in the correct orientation and a double digest would ensure that. You would use HindIII or BamHI and PstI. They are all found in the MCS and have cut sites before the start codon and after the stop codon, allowing the whole sequence to be cut. 2. Looking at the plasmid map and the sequence of Anti-COVID2290, which restriction enzyme(s) would you choose. Explain why. (3 marks) 3. Below, you will find the outline of an experimental diagram in the form of a chart. Fill in the diagram for each sample and control that you will need. Each column is a sample and each row is a step in the cloning experiment. Indicate, if NCB301 is applied to the plasmid before restriction digest or not, plasmid or TE in the restriction digest, insert or H2O in the ligation and what type of plate each is plated on and what type of sample it is (it is okay to use the short forms indicated at the bottom of the diagram. You may or may not need to fill in each sample (column) listed depending on how many technical controls you have. If a cell is not relevant leave it blank. (10 marks) NCB301 Yes or No? Yes No Yes No Yes No Yes No Restriction Digest with? Plasmid Plasmid Plasmid Plasmid TE TE Plasmid Plasmid Ligation with? Insert Insert Water Water Water Water No No Transformation with ligation or plasmid only Ligation Ligation Ligation Ligation Ligation Ligation Plasmid Plasmid Plate type(s) LB+amp LB+amp LB+amp LB+amp LB+amp LB LB+ amp LB LB+amp LB+amp Type of Sample SOI, EC, TCL, TCT* SOI EC TCL TCL TCT TCT TCT TCT *SOI = Sample of interest EC = Experimental control TCL = Technical control ligation TCT = Technical control transformation Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

4. Below you will find a table of some of the results of the experiment. Based on the numbers in the table, do our results support the findings that NCB301 improves transformation and ligation efficiencies? Justify your answer with any appropriate calculations or reasoning (6 marks). # Blue colonies # white colonies + NCB301 38 21 -NCB301 116 32 LB only (diluted to 10 -6 ) 57 NCB301 improves ligation effieciency as the ligation effiency for (+) NCB301 is 0.356, which is greater than the ligation efficiency of (-) NCB301 of 0.216. However, NCB301 does not improve transformation efficiency as the transformation efficiency of (-) NCB301, which is 2.60 x 10^-6, is greater than the transformation efficency of the (+) NCB301, which is 1.04 x 10^-6. 5. The results also showed that when performing a ligation with H2O instead of insert there were only 1 or 2 colonies on the plate. What colour do you expect these colonies to show? Propose an explanation for why there are so few colonies on the plates? (5 marks) The colonies should be blue as there is no insert so B galactosidase is expressed, making all colonies blue. Without the insert, there is no ampilicin resistance gene in the bacteria. Thus, when plated on an LB + amp plate, most of the bacteria should not be able to survive due to the presence of ampicilin, resulting in only a few colonies. 6. What colour are the colonies found on the LB only plate? Explain why. (3 marks) This is a positive transformation control. There is no plasmid in this control and it is used purely to see if the E.coli survive the transformation protocol. So, the colonies should be white, cream or light yellow. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

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Instrumentation Unit Final Exam (10%) 31 Marks Total 1. The following is an excerpt of a methods and results section from a paper: Methods: To determine the decolorization potential of AgNP a slightly modified protocol of a previous paper was adopted. The sample was prepared in 50 mmol/ sodium citrate buffer whi ch contained dye solution [M ethyl red (50 mg/l)], AgNP (1 mg/ml), Biomass extract (2 mg/ml) and AgNO 3 (1 mM Samples were taken at 24 h interval and centrifuged at 8000 rpm for 20 min). The decrease of color intensity was analyzed spectrophotometrically at 430 nm (absorbance maximum f or the orange color of methyl red). Methyl red percent dye decolorization was calculated according to the formula: D=100(Aini – Aobs)/Aini where, D is decolorization (in%), Aini, initial absorbance and Aobs, observed absorbance. Results: F ig. 3. A.B. Decolorization of organic dye methyl red by AgNP and Microchaete after 2   h of incubation. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

a. In the above methods, what would the researchers have used as a blank? Explain your reasoning (4 marks) The blank would be the sodium citrate buffer because a blank should contain everything except the substance that is being tested. Since the substances were prepared in a sodium citrate buffer, the buffer should be used as a blank. b. Looking at Fig.3 B, does AgNP or biomass have a higher observed absorbance reading at 48 hrs? Explain your reasoning. (3 marks) Biomass has a higher observed absorbance reading at 48 hrs because it has a lower % decolorization which means the Aobs is higher resulting in a smaller difference between Aini and Aobs. c. Looking at Fig 3A, what kind of control do you think “control Methyl Red” is? What w ould be in this control and why? (3 marks) It is a biological, experimental control because the you are testing the condition without the treatments of AgNP and Biomass to obtain a point of comparison when there is no decolorization. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

2. A BCA assay is a commonly used colorimetric assay for accurately determining protein concentration (direct measurement at A280 is not accurate). This assay consists of Bovine Serum Albumin (BSA) at 2 mg/mL (this is the stock solution) which is a protein standard and Colour Reagent which produces more purple colour when in the presence of more protein. The Colour Reagent is added directly to each standard and unknown sample before being read at 562 nm on a plate reader. The assay has a detection range of 20-2000 ²g/mL a. Fill in the following table with the volumes of H2O and stock BSA required for each standard (dilutions are made with H2O) such that the final volume adds up to the volume indicated in the 4 th column (9 marks). Tube Volume of H2O Volume of BSA Final Volume in Tube (²L) (BSA+H2O) Final BSA Concentration (²g/mL) A 0²L 300²L of stock 300 2000 B 125²L 375²L of stock 500 1500 C 325²L 325²L of stock 650 1000 D 175²L 175²L of tube B dilution 350 750 E 325²L 325²L of tube C dilution 650 500 F 325²L 325²L of tube E dilution 650 250 G 325²L 325²L of tube F dilution 650 125 H 400²L 100²L of tube G dilution 500 25 I 400²L 0²L of stock 400 0 = Blank On the next page you will find a schematic of how a microplate was loaded with the standards indicated above and three protein samples of unknown concentration. There is also a figure of the standard curve produced and a chart of the absorbance values of each standard and protein sample which matches the order of the microplate schematic Microplate Schematic: Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

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Tube A Tube A Tube A Sample 1 Sample 1 Sample 1 Tube B Tube B Tube B Sample 2 Sample 2 Sample 2 Tube C Tube C Tube C Sample 3 Sample 3 Sample 3 Tube D Tube D Tube D Tube E Tube E Tube E Tube F Tube F Tube F Tube G Tube G Tube G Tube H Tube H Tube H Tube I Tube I Tube I Absorbance Values : 1.5 1.6 1.3 0.6 1.1 0.2 1.1 1.15 1.1 0.48 0.5 0.51 0.8 0.77 0.79 1.9 2.0 1.9 0.6 0.55 0.53 0.38 0.38 0.35 0.2 0.22 0.19 0.08 0.1 0.1 0.05 0.05 0.04 0.01 0 0.02 Standard Curve: Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

b. Looking at the standard curve only, describe what you see (describe what is good and/or bad) in a qualitative manner for the standard curve using the absorbance data and the information on the graph (you are making a qualitative assessment). Based on your qualitative assessment of the standard curve data, would you use this curve to determine the concentrations of unknown samples? (3 marks) The standard curve looks good because the values used to plot the curve are very close to the curve as shown in the R^2 value. The curve also shows a linear progression and passes through 0,0 which follows Beer’s Law. Thus, I would use this curve to determine the concentration of the unknown samples. c. Now make a qualitative assessment of the absorbance values for each of the samples (describe what is good and/or bad) and then calculate the concentration of each sample. For each sample comment on how confident you are that the calculated concentration of the sample is accurate. (9 marks) Sample 1: Mean = (0.6 + 1.1 + 0.2)/3 = 0.63, 0.63 = 0.0007x + 0.0167, x = 881 ug/mL Since the absorbance values of sample 1 are not very close, they do not show precision, thus, I am not confident that the calculated concentration is accurate. Sample 2: Mean = (0.48 + 0.5 + 0.51)/3 = 0.50, 0.50 = 0.0007x + 0.0167, x = 686 ug/mL Since the absorbance values of sample 2 are close to each other, they show precision, thus, I am confident that the calculated concentration is accurate. Sample 3: Mean = (1.9 + 2.0 + 1.9)/3 = 1.93, 1.93 = 0.0007x + 0.0167, x = 2738 ug/mL Since the absorbance values of sample 2 are close to each other, they show precision, thus, I am confident that the calculated concentration is accurate. Downloaded by Lucy Yang (lucyyaang@gmail.com) lOMoARcPSD|17129263

methods-2020-exam

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