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General Genetics

by: Dr. Harrison Metz

General Genetics BIOLOGY 283

Dr. Harrison Metz
GPA 3.55


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This 29 page Class Notes was uploaded by Dr. Harrison Metz on Friday October 30, 2015. The Class Notes belongs to BIOLOGY 283 at University of Massachusetts taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/232247/biology-283-university-of-massachusetts in Biology at University of Massachusetts.


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Date Created: 10/30/15
General Genetics Biol 283 Prof Walker Spring 2014 Homework 1 25 points 1 5 points Consider the experiments carried out by Griffith using Streptococcus pneumoniae When Griffith injected mice with the virulent strain of S pneumoniae the mice died When he injected them with the avirulent strain they lived When he heated the virulent strain to kill it and then injected it the mice lived We can tabulate these data as follows reatment none treatment a 3 points The key experiment performed by Griffith was to mix heat treated virulent cells with untreated avirulent cells What did Griffith probably would result when he injected the mixture into mice Explain briefly use 1 2 sentences NOTE I do NOT want to know what actually happened I want to know what Griffith thought would happen b 3 points What hypothesis did Griffith make to explain what he actually observed as a result from the experiment in part a Be brief use 2 4 sentences to explain the hypothesis Page 1 2 5 points Hershey and Chase used radiolabeled compounds to physically trace the genetic material as it moved from bacteriophage into E coli cells a 1 point BRIEFLY outline how a phage infects E coli A diagram will work best Be sure to point out the placetime wherewhen the genetic material is transferred b 2 points In Hershey and Chase s experiment what was the evidence that genetic material had moved from the phage into the E coli cells Be specific You can correctly answer in one two sentences c 1 point Hershey and Chase used a centrifuge to separate the phage or at least their empty husks from the E coli cells The E coli formed a pellet at the bottom of the tube while the phage remained in the supernatant the liquid part If the phage were labeled with 358 which would be radioactive in HampC s experiment the supernatant or the pellet Page 2 3 6 points Six mutant strains of N crassa have been identified that are unable to synthesize Vitamin B6 Results of all possible complementation tests for the six strains are shown below where indicates complementation and indicates lack of complementation A B C D E a 2 points When the complementation tests were carried out what type of medium were the heterokaryons grown on b 2 points How many complementation groups were observed c 2 points Number your complementation groups The order of the numbering does not matter State which mutants belong to each complementation group Hint EVERY mutant strain belongs to a complementation group Page 3 4 9 points A mutant screen is canied out in N crassa for mutants that are unable to synthesize the amino acid glutamic acid GLU A number of mutants are isolated and classified into four groups according to their ability to grow or not grow in minimal medium supplemented with possible intermediates citrulline C glutamic semialdehyde GS arginine A and ornithine O The data are shown in the table below Mutant Supplement added to growth media Complete the diagram shown below Each arrow indicates one or more biochemical reactions Within each circle write the class of mutants 14 whose products contribute to the reactions symbolized by the arrow and in the squares write the name of the intermediate or amino acid C GS A O or GLU at that position in the pathway HINT This problem is very like 128 in your textbook Check that problem and the answer in the back of the book for practice Page 4 N ame Student ID Gene Geneues Em 233 Pm Walker Sprmg 2m 14 Huntsvime as game mum IEI mm m mu ms m Draw bands In Lb appmpnane planes m gel Isms 25 below w 41m ikh 21m um Embp lane 2 Urdxgesmd agmem pEGFPNl lane a Fragment digesfed M01qud Fragment digesfed ll h Xba lane 5 angmenmgesed ll h 3m Page 1 Student ID 2 4 points Below is the sequence of a small gene The start and stop codons in this gene are indicated in bold face There is a small intron in this gene shown as underlined sequence As you can see the intron contains a simple sequence repeat region SSR that varies among different alleles of the gene Only one allele is shown of course 5 I GGTCAAAGCAACCAAACACATAAAAGAGAGATTTAATACAAAAGAAAGAGAAAAAAGAAAGATATGGCAGGACT CATCAACAAGATCGGAGACGCACTCCACATTGGAGGAGGCAACAAGGAAGGTGAGCACAAGAAGGAAGAGGAAC ACAAGAAACACGTTGACGAGCACAAGAGAGAGAGAGAGAGAGAGAGAGAGAGATCAAAGACAAGATCCACGGTG GTGAAGGTAAAAGCCACGACGGAGAAGGCAAAAGCCACGACGGTquotquotquotquot MAMur mr mr r nr mr mr r m AAGAAACATCATGATGATGGTCACCACAGCAGCAGCAGTGACAGCGACAGCGATTAAGGTGAGGAAGTGAGGAG GATCGCTTGAATAAAACAGATCTGGTTCTGGCTATTATTAATTAATGTTGCTGTATGTTCTTATCATCTTAGAG AGAGGTTAAAGACAGGAGAACCG 3 39 1 point Circle the SSR that is in the intron of this gene Design PCR primers that will allow you to amplify this gene so that it is possible to visualize different alleles that occur when the SSR shrinks and expands Use the following rules Your primers must not anneal within the SSR region Your primers cannot be more than 25 nt long or less than 18 nt long Underline the positions of the primers in the sequence above and write the sequence of each primer below PLEASE NOTICE THAT YOU MUST HAVE THE 5 ENDS ON THE LEFT AS INDICATED Left Primer 5 3 1 point Right Primer 5 3 1 point 1 point Indicate the size of the fragment that you will amplify from the allele shown above base pairs Page 2 Name tu out 3 6 points The following diagram shows the map of a piece of genomic DNA and the position of two probes A and B that can be used to detect this DNA on a Southern blot Probe A Probe B EamHI ECORI BamHl EmRI ECoRl BamHl BamHl l I I I I l 0 1000 2000 3000 4000 5000 6000 Genomic DNA that includes the DNA shown aboVe is digested with EcoRI loaded in lane 2 BamHI loaded in lane 3 and both EcoRI and BamHI at the same time loaded in lane 4 and then subjected to agarose gel electrophoresis and blotted to a membrane Draw the band you expect to see after hybridization and autoradiography using Probe A using Probe B Page 3 Student ID 4 In humans earlobes are either free or attached The free earlobe trait is dominant over attached earlobes a 1 point A man with free earlobes marries a woman with attached earlobes They have eight children all with free earlobes What are the probable genotypes of the individuals of the family Use proper symbols to explain your answers b 1 point A man with free earlobes marries a woman with attached earlobes Of their four children one has free earlobes and three have attached earlobes Give the genotypes of all members of the family Use proper symbols to explain your answers c 1 point Could a man and a woman both with attached earlobes have a child with free earlobes d 1 point Can a couple one of whom is homozygous for the free earlobe trait and the other homozygous for the attached earlobe trait have a child with attached earlobes Use proper symbols to explain your answers e 1 point Can a man and a woman both with free earlobes have children with attached earlobes Use proper symbols to explain your answers Page 4 Student ID 5 A maize pollen grain of genotype a B D e fertilizes a an ovule of genotype A b d e What are the possibilities for the genotypes of a 2 points the zygote b 2 points a gamete ofthe plant that develops from that zygote c 2 points Ifthe zygote mentioned in part a grows into a plant and the plant is self pollinated crossed to itself what is the probability of obtaining progeny that phenotypically show all the recessive traits a b d e Page 5 General Genetics BIOL 283 Spring 2014 Markstein Homework 1 DUE Friday April 4 YOU MUST BE IN CLASS ON FRIDAY APRIL 4 TO COMPLETE THIS HOMEWORK Late homework not accepted TOTAL POINTS 50 40 Takehome 10 To be completed in class You must be here YOUR LAST NAME ONLY YOUR STUDENT ID PRINT YOUR LAST NAME HERE Question 1 Family Pedigrees 10 points total You have worked with six patterns of gene transmission 1Autosomal Recess39ve 2 Autosomal Dominant 3 X linked Recessive 4 X linked Dominant 5 YlInked 6 Mitochondrial Is it possible for the transmission of an Autosomal Recessive trait in a family pedigree to be distinguished 39om the other 5 forms oftransmission Answer this question by devising a family pedigree ofyour choosing Yourfamily can have up to 10 people and it can span up to three generations Construct the tree in stages 5 2 points Make sure to construct a tree that is consistent with transmission of an 85 a perms Create patte 39 Rama 9quoter llnk r1 REPS 1 a and n Xilmka D m mani x7 u de m the space allotted R aw E p re n E8 E Milochondrral inherilance 2 points if 5 So lifr E PRINT YOUR LAST NAME HERE Question 2 Sexlinkage 10 points A 25 points Which of a woman s grandparents could not be the source of any of the genes on either of her Xchromosomes Assume normal transmission of chromosomes and genes 1 Her mother s father 2 Her father s mother 3 Her mother s mother hc i39 3 hffl l 5 Each her grandparents could be the source of her X chromosome genes This is because X chromosomes are not passed down the lineage from father to son Instead the Y chromosome is passed down the lineage from father to son B 25 points Which of a man s grandparents could be the source of genes on his Xchromosome 1 His paternal grandmother only 2 Both his paternal grandfather and grandmother 3 His maternal grandmother only 4 His maternal randfather onl L l iliw This is because females receive their X chromosomes from both parents The man most likely received an X chromosome from his mom that is a recombinant of her mother and father s X chromosome Using the terminology of Lecture 9 April 4 he most likely inherited X chromosome haplotypes from both his maternal grandmother and his maternal grandfather PRINT YOUR LAST NAME HERE Question 2 Continued C 5 points This is a true story A young woman in a family like the one shown was worried that the men in her family were cursed with a gene that made them extremely violent Two of her brothers one of her male cousins and her grandfather each found themselves in jail for extremely violent crimes She approached her family doctor for help 15 years later scientists mapped the gene Defense lawyers have used this finding to argue that people with mutations in this gene should not be held accountable for their violent acts 101 2 3D 4 5 6 7 8 9 10 11 12 13 14 15 2 points When the scientists first mapped the gene they reported to the press that the likelihood that they were right about finding the gene was 100001 What they meant was that the odds of linkage between the socalled violence gene and a marker Marker M at thetaO was 100001 What is the LOD score for linkage between Marker M and the violence gene 39v a quot Tim 7 l T l n lirr 3977 10quot4 10000 3 points What chromosome is Marker M most liker on A The Y chromosome 13 T33 kg v 135 may C n autosome D Mitochondria E There is not enough information to speculate L llll wll ll The pedigree is a classic example of a recessive Xlinked trait it appears to skip a generation because the female carriers don t show it but half their sons do It does not make sense to be most likely autosomal recessive because it would necessitate that people 3 and 7 who are outside the family are also carriers As we discussed in class it is rare to have people be carriers for diseases PRINT YOUR LAST NAME HERE Question 3 LOD score analysis 10 points of us never r birlh main an mat quot last minule But you still have to do it Below is the LCD Score data for 5 nurkers on chromosome 22 that you have tested for linkage to the gene for being punctual and organized PUNC Beiow are LOD SCORES piutted lor 5 markers 1 MAMEL H i N T H E quotJ g i 7 i3 Tug Lab Scan Fall 39 39K l 5 39 Lod score FROM TH TA Titamgreu iLIRKKEK l l 5 39 r Theta Recombination Frequency 10 C M 5 d 3L L Incovicluslla 39 A 5 points To the right of the LOD score graph draw a map or chromosome 22 showing the relative pastuon oithe markers 1 2 3 4 5 relative to PUNC When you do lhis p it 7 exnlaln B 5 points Theta ranges from 0 to 50 Explain why it is never over 50 Your answer should not be longerthan 3 sentences and should discuss independent assorlme meta i anumei m Itquot 39 39 39 39 distance The farthest two genes or markers can be ph icalu is on two separate chromosomes which by the law of independent assortment we know are inherited 50 ui 39 L quot quot 39 R or arge than 50 r 39 39 exhibit RF greaterthan 50 because as the distance between two genes increases the likelihood of multiple crossovers increases PRINTYOUR LAST NAME HERE Question 4 Recombination and GWAS 10 points A 5 points In humans the genetic map is 4400 cM in females and 2700 cM in males Although females and males have different areas of recombination hot spots in general meiosis in which sex is more likely to produce lucky breaks like the ones that we discussed in class that enable nemapping of genes on autosomes Circle your answer 2 Males 3 Lucky breaks are equally likely female and male germ lines The take home message is that genetic distance and physical distance are not the same thing The extreme example is that a stretch of chromosome that is the same physical distance in females and males will often have different degrees of recombination in the two sexes in females there tendsto be more recombination than in males Thus the genetic map in females is greaterthan in males there are more cM in the female genetic map than in the male genetic map because there tends to be more crossovers The more crossovers you have the better your chances of getting a lucky break between the gene you are mapping and the markers you are mapping it against B 5 points lfthe human genome evolved to have 30000 non recombining chromosomes with one gene per chromosome could you ever map genes PRINT YOUR LAST NAME HERE Question 5 TO BE COMPLETED IN CLASS 10 points Question 5 part 1 5 points Explain your answer to Question ZB See explanation that I already wrote by the answer Also many of you drew out a pedigree and that was also a ne approach Question 5 art 2 5 oints For Question 3A Which map is most accurate for Markers 1 amp 3 relative to PUNC 1 3 A C PUNC They are 3 1 equally B l PUNC accurate The markers can be either to the left or right of PUNC There is nothing indicating orientation lfyou found the explanation of LCD scores on Moo le TLDR then y u have mistakenly thought that left and right made a difference It doesn t in the current example Name Student D 4 You are a criminal lawyer defending Willy Winn tr suspect in a murder case Early on in the case there were two suspects but unfortunately the other suspect Harry quotLuckyquot Bottoms was knifed while awaiting questioning and died A bureaucratic foulup allowed his body to be cremated before tissue and blood samples were taken Hence Lucky39s blood type is unknown Both Willy and Lucky have been placed by witnesses at the scene of the crime Tissue samples taken from under the victim s ngernails indicated that his assailant was blood type 0 Willy s blood type is type 0 Somehow you have to prove that Lucky could be the murderer that will cast reasonable doubt as to whether Willy was the real culprit You must nd out whether Lucky could have been type 0 if so he could have been the murderer Fortunately Lucky s parents and a few of siblings are aliveand will consent to giving a sample forty Ing k on The results are musl b M a r 9 l K Lucky s mother A 39 Lox lt p 3 39 r 9 7 I Lucky s father B in l39 b I 39 a 39 b Lucky39s stemAand A39B mw 39 be 2 1 6 4 I 39 7 b Lucky s brother B g 1 5c 39 c 935 I 39 X a 2 points Could Lucky be the murder3r Ia I rule L outquot s 39 x 39 t5 I t l R quot M can fagb 39 It b 2 points The Jury aren39t buying it they all unked genetics They in istthatthe ehmcersio j lwgkg being type 0 considering his parents39 blood types is practicglly G 1 Explain how Lucky could have had type 0 blood r quot We LagJ kJK rum 3 I 1 35 ight 8 M u I 039 Who it C 7V v 39 quot S A a 30 A pen5 c 2 points Late in the trial a woman comes forward who to child The mother39s bl type is A and so is the chll39d39 0 ran 39 v9 7 LP aquot 0 have nr a 7 g M IS on d 2 points Youswin the case Willy goes free Who I quot Lucky39s other lllegitimeteehitd comes forward 39 Should Willy make amt forth border 39 m Name Student ID 6 Albino cquotV black CB cream 3 and sepiacs are all coat colors of guinea pigs Individual animals not necessarily from pure lines showing these colors were mated the results are tabulated a 8 points Write the genotypes of the parents for each of the crosses underneath their phenotypes If a genotype cannot be determined indicate any unknown alleles using cquot b 4 points Write the dominance ranking of the albino CA black CB cream c39 and sepiacs alleles S C H CBgt gtc gtc below Cross Parental phenotypes Phenotypes of progeny albino black cream sepia l Bblack X black Mi 3 d L C X c c 2 black X albino 392 2 Lacs 3 r C 3 eregm X cream A 3 C C C c X c C 4 sepia X cream A A 391 392 L SL q Xe C 5 black X albino l2 1 d C8 C X Lab 6 black X cream u H b C I2 1 CBC K CCCCo ocof A I 7 Bb39agkxseplg 7 gm la a c5lc s or c 12 C L r X C C 39 S H acoui be c c on 8 blagk X sepisa l 14 c 39A r c ms 0 C Av C as on I n L CM 39 fcll Afghan 9 sepi Xsepia 39A 34 LS a XCSC 0M 0 quotquot1 V Cc 5quot I It 4 he my Mall Ac 10 cream X albino 39o 12 a A A C CH X c 6 Name Student ID b 1 points Write the dominance ranking ofthe albino CA black CB cream cc and sepiacs alleles 3 6 points In sweet peas the intensity of purple pigment in the flowers varies widely depending on the genetic background of the plant Generally sweet peas could be categorized as one of these dark purple purple pale purple white In sweet peas the gene p encodes an enzyme called P that converts white precursor into a purple pigment This pigment is responsible for the typical purple color of sweet pea flowers In other words plants homozygous for the WT allele pp have purple flowers Notice that the flowers are not dark purple they are merely purple The activity of the enzyme can be measured in tissue extracts and in mixtures of tissue extracts In the table tissue extracts were prepared from flowers of plants that have particular genotypes see first column The activity of the enzyme P was measured in each extract and is given as percent activity In the third column 1 ml of each tissue extract was mixed with 1 ml of extract from a pp flower and the activity of P was measured Genotype Percent activity Percent activity when mixed 5050 with pp extract pp 100 100 pZpz 0 50 p3p3 300 200 a 2 points What flower color phenotype do you expect for a plant that has the genotype pZp Since the p2 allele does not produce a functional enzyme 0 activity in the table and since the mixture of P and p2 extracts has only 50 activity I expect about 50 of the normal pigment amount which would give me a pale purple color b 2 points What flower color phenotype do you expect for a plant that has the genotype p3p Since the p3 allele produces a functional enzyme with extra high activity 300 percent and since the mixture of p and p2 extracts has excess activity I expect about 200 of the normal pigment amount which would give me a dark purple color c 2 points What flower color phenotype do you expect for a plant that has the genotype pZpz Since the p2 allele does not produce a functional enzyme 0 activity in the table I expect that p2 homozygotes would have no pigment Thus they would be white Page 3 I39IUIIIHWUI K 0 1 Maw3954 mw hub 7 ln com colored aleurone R is dominant to colorless r and green plant color G is dominant to yellow 3 Two plants each heterozygous for both characteristics are testcmssed to homozy gous recessives and their progeny are combined to produce the following totals colored green 0 39 A no rm l colored yellow 97 Iquot Acf o l LA I L 6 id colorless green 103 I colorlessyellow 100 1 for Q L C 5395 a Used squamanalysis to hestthesedataiu r J th 39 8 points Note The table for Chisquare P valst is on the last page 1 l 0550 Pedc 0 64 O SJ w 0 0 o 0 4 q q I a Q 3 q 100 I O 3 I 0 0 3 a 4 m 1 o 0 409 0 0 o L go 1 oo 0 15 Z BDF NOT SIGNQlFlLQNTLY DIFFEAEAT r n u A n j b When the progeny of each of the two heterozygous plants are scoredos e parately the followzpb a ing results are obtained PHENOTYPES PLANT 1 PLANT 2 colored green 88 12 colored yellow 12 85 colorless green 8 9S colorless yellow 92 8 Use chisquare analysis to rest each data set for independent assortment 8 points PIAllquot O E 394 caquot 7 A I 283 1 7 Q MA r lt15 2 2311 3W fml1710quotmw 4 50 3579 quotU MWH7 AL 3 MN m h Cl 50 3932 ZZ 5gorl393 thquot Z 1 0939 E H WW 1 3950 2861 WVW Zul qg So 0 5 N0 WWW 64 E o 35 3 ISOFM 7373322 c BxplaintlLeresultsofmemmed umanalyaes l v u 4 Lwc P M H wad M ii Safe 5022 a nkaj 6 PIMH 39 M lm 2 R5 quot3 r 7 General Genetics BIOL 283 Spring 2014 Markstein Homework 2 DUE MONDAY April 28 YOU MUST BE IN CLASS MONDAY APRIL 28 TO COMPLETE THIS HOMEWORK Late homework not accepted TOTAL POINTS 50 40 Takehome 10 To be completed in class You must be here Plus 5 points extra credit are possible See last page YOUR LAST NAME ONLY YOUR STUDENT ID PRINT YOUR LAST NAME HERE 1 SNP genotyping and personal genomics 10 points A clinical lab just produced a SNP chip to identify SNPs in the BRCA1 locus region ofthe genome where BRCA1 is located that has been associated with breast cancer Three of these SNPs are causative meaning that they are inside the BRCA1 gene itselfand cause a change to the protein A 3 points There are many kinds of mutations that can occur in people and every other life form such as insertions deletions single nucleotide polymorphisms inversions and translocations See pages 38788 in your Sander s textbook for review What kinds of mutations can be caused by SNPs Circle all that are correct i 395 tntsaiwta mttt an itwww n ti i5tif tst mntia m it 3 frameshift mutations B 2 points If you are a woman and nd that you do not have the 3 SNPS on the chip that are causative for breast cancer nor any of the other SNPs on their chip that correlate with breast cancer does this mean that you don t have to worry about getting breast cancer Support your answer with two arguments quot A m are called somatic mutations because they occur In your regula than having been passed on to you through either ofyour parent s germlines C 25 points What are advantages of SNP genotyping Pick one answer 1 It is relatively inexpensive 2 You can use it to anal ze many genes at once 3 It can identify all mutations known to date m 5 All of the above D 25 points What would be the most informative way to have your BRCA1 locus analyzed Pick one answer attiitii 2 SNP genotyping the BRCA1 locus 2 Karyotyping WI h 25t PRINT YOUR LAST NAME HERE 2 SNP genotyping designs 5 points A 25 points A new SNP rs1000 wasjust discovered in BRCA1 that is associated with breast cancer These are the alleles RS1000 A Allele 5 ATTTTGCATAATAACCCAgt3 RS1000 G Allele 5 ATTTTGCATGATAACCCA gt339 1 Write the sequence ofthe primer you would design to detect the A allele and indicate which end is 5 and which is 3 2 Write the sequence ofthe primer you would design to detect the G allele and indicate which end is 5 and which is 3 C 25 points Your friend wants to start a new personal genomics company so that people can know whether or not they have inherited an allele for Huntington s disease As you know from Biol 283 Huntington s Disease is trinucleotide repeat disorder caused by having too many CAG repeats in the coding part ofthe gene Here is a fragment ofthe Huntington s Gene the anking sequences are made up for this example Most people have 2030 CAG repeats and are healthy People with 40 or more CAG repeats will have Huntington s Disease AATCTGAACTTACAG CAGn CAGTAATACCGATA Your friend wants you to design a SNP Chip to detect alleles that cause Huntington s disease Can you do this lfso draw your primer design below If not describe a method to detect HD alleles that would not depend on sequencing 9 Design your primers outside of the repeat and them amplify the DNA between the primers We discussed this in the first section the amplification of SMALL SEQUENCE REPEATS as highly polymorphic molecular markers for mapping genes MOST SSRs occur between genes not in genes and are benign Huntington s disease is a very special case ofan SSR that has an effect PRINT YOUR LAST NAME HERE 3 Cloning DNA 5 points You have been assigned the job of cloning your advisor s Favorite Gene Your advisor gave you a head start by giving you a tube of AFG already digested with the restriction enzyme EcoRI You can therefore easily paste yourAFG gene into the EcoRI site of a plasmid For review see pages 537539 ofyourtextbook have the Ampicillin resistance gene mp asmid B contains the Lac gen which encodes an enz me that converts the substrate Xgal into a blue substance Plasmid 0 contains a Letha Gene LG which kills bacten39a Insertion ofAFG into the EcoRI site in LacZ will disrupt the LacZ gene and insertion of DNA into the EcoRI site of LG wi disrupt LG You have a choice ofthree different plasmids in which to clone AFG A plasmids ey EcoRl EcoRl EcoRI LacZ LG Plasmid A Plasmid E W Amp W A 25 points Which plasmid would enable you to select s eci call for bacteria that have taken up a plasmid with AFG A Plasmid A B Plasmid B m D Each because they all use selection E None because screening is necessary B 25 points A student in class suggested an interesting idea Instead ofusing a vector with Ampicillin you could instead rst fuse AFG tot e gene for Ampicillin it is only 1 kb and clone it into a plasmid that does not have any seectabe markers fyou took this approach would you have to use BlueWhite screening or some other form ofscreening to nd bacterial colonies that have taken up your plasmids with the AFGAmpiciIIin insert One sentence maximum PRINT YOUR LAST NAME HERE 4 Genes and cDNA 5 points A 25 points On average human proteins are 400 amino acids Based on this alone you would expect human genes to be how long Give a number B125 points Human genes on average span about 30000 bp in the genome Typically when people think of genes they think ofthe part that codes for protein Since the average protein is only 400 aa on average what percent of the average human gene codes for protein Give a number 5 Transgenic Mice 5 points When making a knockin construct targeted to your favorite gene it is necessary to use both positive and negative selection If you use the Neomycin resistance selectable marker but forget to use a negative selection in your scheme what do you expect will be true ofthe ES cells you select for neomycin resistance See pages 577579 in your textbook for review i I 61 l i39maw have your construct integrated at your favorite gene C None or few of the cells will have your construct integrated anywhere PRINT YOUR LAST NAME HERE 6 Oncogenes and tumor suppressors 10 points total April 25th lecture You are a geneticist with your own startup company Your mission is to cure cancer One arm of your company is exploring ways to reengineer the human genome so that humans are less likely to get cancer Your investors suggest that you remove all the protooncogenes from the genome You have the technology to do this with the snap of your ngers A 5 points Ethics aside is this a good idea Your answer must discuss protooncogenes 3 sentences maximum B 25 points Mutations in tumor suppressors that result in cancer are typically 2 Dominant gainoffunction mutations 3 Recessive gainoffunction mutations The other arm of your company is involved in drug discovery to nd new drugs for cancer patients You have been given a chemical library called liquid gold composed of 100 different rare earth compounds Some of these might have anticancer properties It turns out that you have just enough of each compound to do a single screen with only one ofyour company s prized cancer models Model 1 is an Oncogene Cancer Model produced by the overexpression ofan oncogene Model 2 is a Tumor Suppressor Model produced by a deletion ofa tumor suppressor gene C 25 points You will be more likely to nd drugs that inhibit tumor growth by screening which ofyour libraries Circle your answer iquot i 2 y 7 r A1 2 Model 2 the tumor suppressor model 3 Both models are equally likely to be sensitive to drugs that can inhibit cancer s l l as n 1 PRINT YOUR LAST NAME HERE 7 In Class Portion 10 points PRINT YOUR LAST NAME HERE EXTRA CREDIT 5 points to be added to your nal grade in Markstein section ofthe course In class we discussed making animal models for Huntington s Disease Yourjob is to nd one paper in which a transgenic mouse was made to model Huntlngton s Disease There is no partial credit for this assignment To get credit do the following act 2 Wnte a maximum oftwo sentences that states what kind oftransgenic mouse was made and what the major result or conclusion was You should be able to get this information simply from reading the abstract Here is an example sentence from a paper that I made up this example 39n mouse was made using the Huntington Gene HTT Jsed to GFP and the result was that it was found to be expressed in all the tissues throughout the adu t mouse THE STU DENT WROTE that nrndlll ru 39 mum Huntington39s Disease gene They found that removing the gene completely was L L L L L I A I u m lessm 251l5 81123 behavioral and morphological changes in heterozygous Naslu Fll Hn KllklRm FnMthmnlM llrle kl M rm in am n u 4 Ma gnuthnrlnfnrmatlnn Ansu acl initial 0 lrlllatl n ganglia


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