1. Below are the genotypes for two individuals. Loci A and B are two different microsatellites. The numbers tell us which allele of the microsatellite they carry. Locus A 8/9 Locus A 8/8 5/7 3/5 Person 1 Locus B Person 2 Locus B Draw the chromosome, labelled with alleles for Person 1 and 2. Assume Locus A and B are located close together on the same chromosome.
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- . E. coli chromosomes in which every nitrogen atom is labeled (that is, every nitrogen atom is the heavy isotope15N instead of the normal isotope 14N) are allowed to replicate in an environment in which all the nitrogen is 14N.Using a solid line to represent a heavy polynucleotidechain and a dashed line for a light chain, sketch each ofthe following descriptions:a. The heavy parental chromosome and the productsof the first replication after transfer to a 14N medium,assuming that the chromosome is one DNA doublehelix and that replication is semiconservative.b. Repeat part a, but now assume that replication isconservative.c. If the daughter chromosomes from the first divisionin 14N are spun in a cesium chloride density gradientand a single band is obtained, which of the possibilitiesin parts a and b can be ruled out? Reconsider theMeselson and Stahl experiment: What does it prove?2c) If the whole potoroo genome is 4.2 x 10' bp, and the highlyrepetitive DNA in the potoroo genome is composed entirely ofcopies of the sequence 5'AAGACT' and its complement, howmany copies of this sequence are present in the potoroogenome?(i) For the chromatogram below, what is the sequence of the template DNA from base 115 to 125? CTGTGTGAAATTGT TA T CCGC T CA CA AT T C CACA CA A CATA CGAGC CGGAAG CA TA A 110 120 130 140 150 160 (ii) An allele of a gene has the following change in it's sequence ATG GTG CÁC CTG ACT CCT GTG GAG AAG TCT compared to the wild type ATG GTG CAC CTG ACT CT GAG GAG AAG TCT With reference to the sequence; there is a codon, resulting in a change from is a mutation in the to which mutation.
- Please consider the figure below, parts A and B. A В Gene B Gene C Gene B Normal Gene A Chromosome 12 Normal Chromosome 1 Please consider figure A. a. Do these chromosomes come from a dividing or non-dividing cell? Give a reason for your answer. b. How many molecules of double stranded DNA will be present at anaphase I of a cell from this organism? Please consider figure B. A potentially carcinogenic mutation occurred on one of the chromosomes. The gene affected by the mutation codes for a protein involved in the repair of DNA damage. c. What is the correct term used to describe this chromosomal mutation? d. In terms of the development of cancer, is this a dominant or recessive mutation? e. What is the consequence of this mutation if it occurs during meiosis and is inherited by the offspring? Explain your answer in terms of the function of the protein.The results of a paternity test are shown in the table below. Numbers indicate the number of short tandem repeats for loci tested. Whos the daddy? How sure are you?The completely synthetic yeast chromosome Syn IIIcontains a loxP site in the 3′ UTR of every gene thatis potentially nonessential to yeast survival. As youwill recall from Chapter 6, loxP sites are targets ofsite-specific recombination. The researchers who constructed Syn III included these loxP sites as a way to“scramble” the chromosome, meaning that parts ofthe chromosome could easily be deleted or rearranged.The goal of these investigations is to drive the evolution of Syn III so as to define a minimal genome thatcan support the life of this organism. Outline the experiment the researchers would do to scramble Syn IIIin order to define a minimal genome.
- Summary table of data for one yeast DNA microarray. Shows 14 of the 6,200 genes on the full microarray. Also contained in the data is the location of each spot on the microarray. Block Column Row Red 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Gene Name tubl tub2 sec1 sec2 sec3 act1 act2 fus 1 idp2 idp1 idh1 idh2 erd1 erd2 2,345 3,589 4,109 1,500 1,246 1,937 2,561 2,962 3,585 2,796 2,170 1,896 1,023 1,698 Green 2,467 2,158 1,469 3,589 1,258 2,104 1,562 3,012 1,209 1,005 4,245 2,996 3,354 2,896 Red: Green Ratio 0.95 1.66 2.80 0.42 0.99 0.92 1.64 0.98 2.97 2.78 0.51 0.63 0.31 0.59 Induced or Repressed? In the table above, write down if the gene is induced, repressed, or equally expressed.When the interrupted mating technique was used withfive different strains of Hfr bacteria, the following orders ofgene entry and recombination were observed. On the basisof these data, draw a map of the bacterial chromosome.Do the data support the concept of circularity?HfrStrain Order1 T C H R O2 H R O M B3 M O R H C4 M B A K T5 C T K A BA molecular biologist is investigating homologous recombination. One aim of this study is to reconstitute stages of the process in vitro. Draw diagrams to show how the four synthetic oligonucleotides below could base-pair to form a stable model Holliday junction. W 5’ GATCGCATTGTAGCCGTAGGTCCACTGTAA 3’ X 5’ GTCCCATACGTAGCCGTAGGACATGTACCG 3’ Y 5’ CGGTACATGTCCTACGGCTACAATGCGATC 3’ Z 5’ TTACAGTGGACCTACGGCTACGTATGGGAC 3’
- A DNA fragment of interest is inserted in the BamHI site of plasmid PBR322.The recombinants will be: HindIII ECORI EcoRV BamHI 439 0 29 is 4000 375 Sall PstI, 651 tet amp 1000 PBR322 4361 bp 3000 ori 2000 2295 Ndel A. Resistant to amp an tet B. None of the other options, Xgal is required for selection O C. xensitive to amp, resistant to tet None of the other options, Xgal is required for selection O D. Sensitive to tet, resistant amp O E. Sensitive to amp and tetFig 3.16 EcoRI SacI KpnI |ampR Aval Xmal Smal lacZ BamHI Xbal SalI Accl HincII PstI Sphl HindIII Bam H1 Bam H1 Bam H1 1 2 3 4 The ends of the double-stranded DNA fragment above are blunt ends. The directionality of genes contained within the fragment is from left to right (arrow). After digestion by BamH1, which fragment can be inserted into the vector cut with BamH1 and Sma 1 maintaining the same directionality as the lacz promoter (green segment with arrow on vector)? 3 O 4 O1 227. Given the following plasmid map, list how many fragments would exist and their bp length when treated with Hindll and Nde1 Dail 91 Ben BI 51 EstAPI 179 BemBI 2683 Ndel 183 Kasl - Narl - Sfol 235 Bgll 245 Fspl 256 Pul 276 EcoO1091 2674 Aatll - Zral 2617 BiM 2542 Sspl 2501 Prull 306 Benrl 364 Acll 2297 BæAl 387 Apol - EeoRI 396 Xmnl 2294 lacza haall - Sacl - Ecos3KI 402 Accbs1 - Kpnl 400 Awal- BsoN - Smal- TapM I- Ymal 412 Beal 2215 MCS Scal 2177 haml 417 Xhal 0 Pvul 2066 Accl - Hinl - Sall 29 pl. Alau a sb 434 Avall 2059 PUC19 2,686 bp BerDI 1985 Sphi 441 Hindi 447 Acll 1924. Fspl 1919 Prull 628 Avall 1837 Til 641 EsaXI 659 me AllI 1822 Rgll 1813 Bpal 1784 BSPOI - Sapl 683 BsrFl 1779 Bsal 1766 Thl 781 AII - Peil 806 BerDI 1753 Dal 908 Bmrl 1744 ori Aldl 1694 BkiM 1015 BseYl 1110 AlwNI 1217 BeeN 1292 dy