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Exam 2 Solutions

by: Annie Burton

Exam 2 Solutions 220

Marketplace > Chemical Engineering > 220 > Exam 2 Solutions
Annie Burton
GPA 3.45
Biological Systems
Shelly Peyton

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Chem Eng 220 with Shelly Peyton Exam 2 Solutions
Biological Systems
Shelly Peyton
Study Guide
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This 8 page Study Guide was uploaded by Annie Burton on Friday October 9, 2015. The Study Guide belongs to 220 at a university taught by Shelly Peyton in Fall. Since its upload, it has received 162 views.

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Date Created: 10/09/15
1 EXAM 2 NAME ID ChE 220 Chemical Engineering Principles of Biological Systems October 3039quot 2014 This exam is open book You may use any of your class materials andfor textbooks to help you solve the exam You will need a calculator Average molecular weight of a DNA nucleotide base A T G C is taoglmol Other Flules 1 Box all your final answers If your final answer is not clear you will not be given the benefit of the doubt 2 Write governing equations and illustrations to obtain partial credit if you are running out of time or stumped 3 You will be penalized for improper use of significant digits and missing units 4 Keep your eyes at your desk If you are looking around the room again you will not be given the benefit of the doubt For your first offense you will be given a warning Second offense results in a zero on your test Put your name on every page 0quot PROBLEM 1 20 POINTS POSSIBLE PROBLEM 2 40 POINTS POSSIBLE D PROBLEM 3 40 POINTS POSSIBLE O y TOTAL 100 POINTS POSSIBLE l EXAM 2 NAME i h ID a ONE 220 Chemical Engineering Principles of Biological Systems October so 2M4 1i Short Answers 20 points 6 a You are using a nucleotide sequence of DNA to predict a protein sequence Imagine that one strand of your DNA reaas 539AATG39TCAGGTACCTGC3 What is the sequence of the complementary strand for this segment of DNA 339 TTACAAQTCC ATQQACa s39 b What are the corresponding rnFlNA strands iotr at There are wo 4 s10 93 UUACAAqUCCAUGQAcq5 2 ElAAUQUUECAQGIUACCUQC B MET c Given what you discovered in a and bin what do you think is the corresponding amino acid sequencer and why Crieri0ta WW WW ii 15 0 my WP Codons NQStds 0de 0L Ming M mum as am imrq is 0m QSSUWUI in HM sequence 33 Mom sum you OWQ Mott mENA in ilrk Y39Iah39l triad19m 51413 BMW Strands COtQ Cd cxtnol 1 R b are Options x SD 8 POSSlblh39HQS mg i slop commas no MU iii nicely Thr mop ix39flihM 0 SWP Dd0M 0 knits MEF MELJPWQ AYTTLQU 39 8h MPSJWQCLM 39TO WWK mist 185W Stop Clowquotm it ID EXAM 2 NAME 7 ChE 220 Chemical Engineering Principles of Biological Systems October 3039quot 2014 d DNA electrophoresis You are designing a way to run DNA molecules through a gel in order to separate DNA fragments by weight If you add DNA to wells at the top of a gel should you piace the positive electrode anode at the top or at the bottom of the gel Explain your choice DNA is Magenta 0M4 PhOSPhOdui Ojrourps on meritth 80 you ShO uuol de 4M mom on at bottom at m to 8b We bNA mews mwuld 1 1 EXAM 2 NAME ID alt ChE 220 Chemical Engineering Principles of Biollogical Systems October sot 2014 2 40 points P39i S QQQh POW139 r39ou have engineered an enzyme in the tab that is supposed to break down protein A into polypeptide B The ratio of moies of protein A consumed to polypeptide B produced is 39i 10 a Write a chemicai reaction equation for this reaction like we did in class eg A B C if you assume that the reaction is irreversibie Don39t liorget to include the stoichiome iry A9lcg enAtE WOB b dust by looking at the stoichiometry you accounted forin awhat is the assumed reaction order ZErO Qeotct39ton l8 Wreye rable V2 credit For otdd ir starch tr gcudina it or l2 c In your separate experimental analysisquot you actually discover thai the reaction is rst order and only depends on the concentration of the protein A your enzyme is llillieiy in excess to cease this Write down the reaction rate equation given this new information dA r ch Cit A d Calculate the time required to consume 99 of the precursor protein A if the rate of formation is 001s Assume you start with 1 mole of protein A in a liter flask K 6 Jo quotquot it quotit ud n ct if it g JqCmeaa koqad GOVMWOD isi tauoi gt t5 4L9 i H is i EXAM 2 NAME ID ChE 220 Chemical Engineering Principles of Biological Systems October 30 2014 EXTRA SPACE FOR PROBLEM 2 EXAM 2 NAME ID ChE 220 Chemical Engineering Principles of Biological Systems October 30 quot 2014 3 PCFI 40 points You re a bioengineer at a new startup company in Cambridge MA that is hunting down new genes responsible for early onset dementia and Alzheimer39s Your very first assignment as an entrylevei Chemical Engineer with training in Bioengineering from UMass is to run PCFi for a few candidate sequences that might be promising gene targets Candidate protein 1 gene 1 40mer amino acid sequence Candidate protein 2 gene 2 100rrier protein sequence About atth 03093 cm um are ch Fluid You are given mFtNA samples 5ug from ten different patients each This is arout 1x105 cells Remember human cells are dipioiijii Five that have been diagnosed with Alzheimer39s Group 1 and five other patients serving as negative controls no diagnosis of Alzherimer39s Group 2 You also already have primers for the candidate genes iiisted above tie How many ug of cDNA wiii you have for each of the candidate genes after you run 22 PCFI cyclies P md 0 GENEXAQ mer 0mm Otcidkr with x tilt 05199 r ZQhrorrosumsiv 39 245 08 24xlO39i39bdsas x 220ij x222 i Diquot Olim iKOSOZX lo b r1 W1 x so We 2 m l0 3 01r22pji GENE2 r umer Aimgvtxmhztoxioims x2 9 quot O L002el023x lSO frneQ 5amp43elo j I f Eb When you get the PCR results you find that Group 3911 patients are overexpressing producing much much more of 0 gene 1 than Group 2 patients and both Group 1 and Group 2 patients express about the same amount of gene 2 You decide to follow up these results by measuring these genes at the protein ievel You find again much much more of protein 1 in the Group 1 patients compared to Group 2 However interestingly you find no detectable protein 2 in Group 1 and normal amounts in Group 2 How might you explain your results for both of these genes 1 Miqu 1 tremors 0 be importwrt for Aithimer s onset or progrQs aon loic ou Consistenita moi r uprtzguioigd 0o both lt3 n9 and proion lizth in W groom is group 2 Wrights Q imam 2 is more CD From opinion i005 in at Cnpig but Hxi MLN39tmr s points okOhLi MN hhchorw WW Mi M Wt u Mtaidlcit ii KSVWVOWJ Sim 39 Wgs i L FARM Nuts WPVQ SSW EXAM 2 NAME ID alt ChE 220 Chemical Engineering Principles of Biological Systems October 30 quot 2014 l 0 PTS 0 A few months later a new colleague joins the company and her first task is to repeat the results you found in part a She is having issues repeating your results for gene 1 she39s not getting any detectible amplification out of the PCR When you go through her protocol with her you realize that she is having issues with her machine and it39s only heating up to about 85 degrees during the very first cycle She is still able to reproduce your results for gene 2 but not for gene 1 When you look closely at the sequences for these genes you realize that gene 1 is much more GC rich than gene 2 How might you explain to her what39s happening 0th h s It le Pct madamquot dozer act it imitaka ml prDPQX llj NAbV QDN shat glob kiwi durtrxnj tm WWW mm W UMP65 0m stud C391ka Mules cud w PCB LSUR W cowl crypt l 3mm W 1 l8 tQ CeC two it lS Stack 4mm gyrth GtC mas mute bh5Qmquot b8n Mm An lm TNS l3 860 ngg 2 but mt W 13 W 9 W gm m with WOCWh l EXAM 2 NAME a ID ChE 220 Chemical Engineering Principles of Biological Systems October 3039quot 2014 EXTRA SPACE FOR PROBLEM 3


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