BIO325 Exam 3 - GENETICS - BI0325 Spring 2011 Prof: Janice...

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Unformatted text preview: GENETICS - BI0325 Spring 2011 Prof: Janice Fischer EXAM #3 March 31, 2011 Your Name 0 No questions will be answered during the exam. 0 Please tear off the last page of the exam and use it for reference. It contains the Genetic Code Table and the Wobble Rules. 0 Please read the questions CAREFULLY and answer the questions that I ask - not other questions. 0 You MUST indicate the 5'-ends of all codons and anticodons - otherwise you will get no credit for your answers. 0 CIRCLE the answers when you are asked to do so. 0 The numbers in parentheses next to each question indicate the points, out of 100, that the question is worth. l. A. Taking into account the Wobble Rules and the Genetic Code Table, what is the minimal number of distinct tRNA genes required to recognize all Leucine codons? (Your answer needs to be a single number — circle it.) (10 pts.) W anfiifi 0'15! ‘— 5uuA 514M s’uué } s’ I r u 7 Mfg“ 5 W E‘s/CHA 4wg%u@i I scflé B. What are the anticodon sequences of the tRNAs that the genes in part A encode? (You must indicate the 5' ends to get credit for your answer.) (10 pts.) ANY owe 0? "mm ArJSUA’FS flfiovc WAS AKCPMGLE, 2. A tRNA with the anticodon sequence S'UAU cannot exist. Why not? (One sentence answer - 10 words maximum. This means that I’m reading only the first 10 words you write.) CIRCLE YOUR ANSWER (15 pts.) ,Mé-Md The dnh‘c‘odon WOU - fflOleZd Cod-0n: V Mgfi and Ila. 3. A mutant bacterium has a nonsense suppressor tRNA that inserts glutamine (Gin) to match a S'UAG codon, but not other nonsense codons. The nonsense suppressor mutation is a single nucleotide change from wild-type. A. What is the anticodon of the mutant suppressor tRNA? (Indicate the 5’ end and circle your answer.) (10 pts.) The mange/«EC <L(/.,pf€§§or IS '1. ,P B. What is the anticodon of the wild-type tRNAG'" that was mutated to generate the nonsense suppressor? (Indicate the 5' end and circle your answer.) (5 pts.) 0‘ 4. The sequence of a cloned fragment of DNA was determined using the dideoxy method. A part of the autoradiogram of the sequencing gel is shown here. The larger molecules are at the top (BIG) and smaller ones at the bottom (SMALL). 5MNJ. Deduce the nucleotide sequence of the single-stranded DNA nucleotide chain that was the template For DNA synthesis. Label the 5’ and 3’ ends. (10 pts.) DNA §qnlllésiggdg 5’ TTC 6AM GGTCVACCCCT CquAcm 7 TAth ’I’e MP (ale DNA 3 , fl b/ Mac/177% ACTGGGGMUGWMNCTE \‘ ‘7 r ‘ ‘ 7 << 7 H 7 -‘ '(A’, 5. Cystic fibrous is caused by autosomal recessive loss-oF-Function mutations in the CF gene. One common disease allele is a point mutation that results in an RFLP, as shown below (R=Eco RI): R R wow—i—_+_ <f~---”iarq£c,£¢fil "rags: {i‘ : CFpmbe — <__. qwi (a, i“: fiagmi I j_ M ifff c;1 1:: \NT, aim/g, i- - - —-l__ c€’allel¢ Samples of DNA obtained from a fetus (F) and her parents (M and P) were analyzed by gel electrophoresis Followed by Southern blotting and hybridization with a radioactively labeled probe designated “CF probe” in the diagram above. The autoradiographic results are shown in the figure above. A. Do either of the parents have the disease? YES (circle one) (5 pts.) 8. Which parent is heterozygous For the disease allele? P M circle one) (5 pts.) C. Will the Fetus have the disease? @NO (circle one) (5 pts.) D. If these people have another child, what is the chance that it will have the disease? (your answer should be a number - circle it) (5 pts.) ' Exi‘L c€' of” , , w z vi cw 1/4 m : , 6. A thief broke a window and left some of his blood in the shattered glass. Four Suspects with bloody hands are rounded up. You are called in to identify the criminal. You prepare genomic DNA from the blood Found at the crime scene and also from blood samples from each the Four suspects. One of the experiments you do is to examine the alleles at the particular VNTR locus shown below for each DNA sample. You do this by analyzing genomic blots of each DNA sample digested to completion with Eco RI, using the unique genomic sequence probe (probe 1) indicated in the diagram. The results are indicated below to the right of the VNTR locus diagram. VNTR locus on chromosome 13 Eco RI ml: allele; window 80 10 nt repeats suspect 1 AA suspect 2 BD _ — suspect 3 AD probe 1 probe 2 suspect 4 CC A. Show the positions of bands in each of the 5 samples on the autoradiogram below labeled A. (10 pts.) 8. Show below on the autoradiogram labled 8 what the results are likely to look-like if the probe correSponds to the 10 nt VNTR repeats instead of to unique genomic DNA sequence. (5 pts.) so“ so") § § suspects 1 2 3 4 suspects C. What additional information about the alleles would you need in order to determine, based solely on this data, how likely it is that suspect 2 committed the crime? CIRCLE YOUR ANSWER (5 pts.) il'i/Usll.‘{ir’;7:/F,fi"i' (in? ,7) ‘37. 'r‘ Wink} ~' " ‘ 3 | BONUS QUESTION You have generated 3 diFFerent recessive mutant alleles oF the Fruit Fly wing/e55 (wg) gene. The molecular nature oF each oF the 3 mutations is shown in the diagram below, which depicts region oF genomic DNA containing the wg gene. Fl <— 5000 —> Ruooo+n <— 6000 ——> R probe The molecular lesions in the three wg alleles are: wgl : A single base pair change at the position shown above in exon 1. ng : Deletion of 100 base pairs in exon 2 as shown above. wg3 : Deletion oF 500 base pairs in exon 3 as shown above. On the diagrams below, show the positions of the bands that you would see on (A) blots oF genomic DNA cut to completion with restriction enzyme R and (8) blots oF total mRNA. For each blot, the DNA or RNA was isolated From wild-type Flies and From Flies homozygous For each oF the three wg mutant alleles. Also, the same probe was used For each blot, the extent of which is indicated in the diagram above. wt wg1 wg2 wgS big small genomic DNA blot mRNA blot A B ...
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BIO325 Exam 3 - GENETICS - BI0325 Spring 2011 Prof: Janice...

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