R1 12 kb R1 4 kb R1. R1 10 kb R1 2 kb R1 4 kb R1

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1 Bcor101 Sample questions Midterm 3 1. The maps of the sites for restriction enzyme EcoR1 (R1) in the wild type and mutated cystic fibrosis genes are shown below: Wild Type R1 12 kb R1 4 kb R1 _ _ CF probe Mutant R1 10 kb R1 2 kb R1 4 kb R1 _ _ Samples of DNA obtained from a fetus (F) and her parents (M and P) were analyzed by gel electrophoresis followed by the Southern blot technique and hybridization with the radioactively labeled probed designated "CF probe" in the above figure. The autoradiographic results are shown in the following figure: _P F M _ Given that cystic fibrosis is a recessive mutation, will the fetus be affected? Explain. Both parents are heterozygous contain 4 EcoR1 fragments (one of which (4 kb) is common between the wild type and mutant gene). The fetus has this common fragment and one other fragment (12 kb) that corresponds to the size found in the wt gene. Therefore the fetus is homozygous wt and will not be affected.

2 2. Several different haploid and diploid genotypes for the lac operon are given below. For each genotype, determine whether b-galactosidase and permease will be produced under conditions of noninduction (no lactose present) and induction (lactose present). Fill in the table with a (+) if the enzyme is produced and a (-) if the enzyme is not produced. b-galactosidase Permease Genotype no lactose lactose no lactose lactose (a) i + p - o + z + y (b) i + p + o c z + y (c) i + p + o c z + y - / i - p + o+ z - y + (d) i s p + o + z- y + / i + p + o + z + y - (e) i + p + o c z + y + / i s p + o + z - y - 3) A linear DNA molecule 1000 base-pairs long is digested with the following restriction enzymes with the following results: EcoRI BglII EcoRI + BglII 400 bp, 600 bp 250 bp, 750 bp 250, 350, 400 bp EcoR1 BglII 400bp 350 bp 250 bp

3 Determine the restriction map of this DNA G A T C (top of gel) 4. Identify the open-reading frame represented in this DNA sequence (bottom of gel)

4 DNA sequence: 5 CTGACTGCTTACCTAGGTACCGCATGGCATGCGATTTGTACATA3 met-ala-cys-asp-leu-tyr-ileu 5) A short DNA fragment with the sequence 3 -CTGATAAGGCTTTG-5 is used as a template for sequencing by the dideoxynucleotide method. Each of the four reaction vessels contains DNA polymerase, a primer with the sequence 5 -GACTA-3, labelled deoxynucleotide triphosphates and a small amount of either ddatp, ddttp, ddctp, ddgtp. Draw the dideoxy sequencing gel pattern expected for this template. G A T C bottom of gel 6. You are studying Neurospora mutants that are proline auxotrophs (unable to grow in the absence of proline). One of your students isolates a revertant which is able to grow on minimal media. You cross the revertant with a wild type strain of Neurospora in order to determine the nature of the mutation in the revertant. What proportion of the progeny from this cross would be proline independent if a) the reversion precisely reversed the original change that produced the pro - mutant allele? 100%

5 b) the revertant phenotype was produced by a suppressor mutation in a second gene located on a different chromosome? 75% - Will get four classes of genotypes in equal frequency: pro-sup- -same genotype as the revertant parent proline independent pro-sup+ - recombinant doesn t contain the suppressor mutation proline dependent pro+ supindependent recombinant sup- mutation has no phenotype on its own proline pro+sup+ - same as wild type parent proline independent c) the suppressor mutation occurred in a second gene located 15 map units from the pro - locus on the same chromosome? 92.5% 42.5%: pro-sup- -same genotype as the revertant parent proline independent 7.5%: pro-sup+ - recombinant doesn t contain the suppressor mutation proline dependent 7.5%: pro+ sup- recombinant sup- mutation has no phenotype on its own proline independent 42,5%: pro+sup+ - same as wild type parent proline independent

6 7. A plant was transformed using Ti plasmid which contains a kanamycin-resistance gene between the left and right borders of the T-DNA region. Two kanamycin-resistant plants were identified for further study. The plants were allowed to self, and the results were as follows: Progeny from selfing plant A: 3/4 progeny were resistant to kanamycin and 1/4 progeny were sensitive to kanamycin Progeny from selfing plant B: 15/16 of the progeny were resistant to kanamycin and 1/16 of the progeny were sensitive to kanamycin. a) Explain the molecular and genetic basis for these two different ratios. Plant A: A single insertion event of the T-DNA into a plant chromosome the parent plant is hemizygous for the insertion. When this plant is selfed, you will get a 1:2:1 segregation for kan-resistance, since this is a dominant marker. Therefore, 3/4 will be kanamycin-resistant and 1/4 will be sensitive to kanamycin Plant B: Two independent insertions of the T-DNA occurred on different chromosomes. Therefore the two T-DNA insertions are unlinked. When plant B is selfed, the two loci will segregate 9:3:3:1, with 15 individuals that are kanamycin resistant and 1 individual out of 16 that is kanamycin sensistive. b) what ratios would you expect from these two plants if they were used in a test cross? Plant A: 50% kanamycin resistant;50% kanamycin sensitive Plant B: 75% kanamycin resistant; 25% kanamycin sensitive