2008_04_25_17_52_46

2008_04_25_17_52_46 - BICD 100, Genetics Page I of 5 Ist...

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Unformatted text preview: BICD 100, Genetics Page I of 5 Ist Midterm Exam Spring 2008 April 23, 2008 TEAR OFF THE LAST PAGE ( ANSWER SHEET ) AND FILL IT OUT. HAND ONLY IT IN. YOU MAY KEEP THE EXAM ITSELF. NO CREDIT FOR ANSWERS LEFT ON THE EXAM AND NOT PUT ON THE ANSWER SHEET. I. (25 pix) Each of the following phrases is the definition of a common genetic term, either a noun or adjective. Supply the term on your answer sheet . a) a condition in which a single mutation affects more than 1 trait b) observable characteristics of an organism c) a measure of distance on a genetic map (1) in female meiosis, a cell that does not develop into a gamete e) in male meiosis. a cell that undergoes 2 meiotic divisions 11. (40 prs.) The following in either higher plants or animals occurs in a stage or stages, give only 1 of the appropriate stages; distinguish between meiosis and mitosis, i.e. you can‘tjust say anaphase. PUT THE ANSWERS ON THE ANSWER SHEET. ()0 pm each) a) cells with homologues but no sister chromatids in‘l‘F'FPh-‘SC b) cells with neither sister chromatids nor homologues {.g fophnjf II c) cells with sister chromatids but without homologuesz Prop“ an _I: d) cellswith both sister chromatids and homologues 1119;“?th I 999999999999999999999999999999999999999999999999999999999999999999999999999999999 \ Brojj jiUOf BICD 100, Genetics Page 2 of 5 lst Midterm Exam Spring 2008 April 23, 2008 III. (56 pts, 8 pts ea.) Multiple Choice Questions — % correct answer only! List answers on your answer sheet only. 1. The following growth responses of Neurospora mutants #1-4 were seen on the related biosynthetic intermediates A, B. C, D and E. Assume all intermediates are able to enter the cell, that each mutant carries only one mutation. and that mutants affect steps after B in the pathway. 3““ Growth on Mutant ® 1 0 (+ means growth; 0 means no growth) 000+ ODOOCU 2 3 4 Which of the scheme below fits best with the preceding data, with regard to the biosynthetic pathway? 93 x a C / E —I-—A—I-—D B—h—A—r—D Ema-AX B—{bA—h-E—zb-D \E \E \C—b-D \C 2. The following is a pedigree of tWo families for deaf-mutism in Northern Ireland. This trait is inherited as: ' A) Dominant _ B) Recessive C) Can’t tell from this pedigree 3. If III-8 marries [119, and 6 children are born that are not deaf, the most likely genetic explanation is: fl not enough progeny observed in that generation mutations in different genes epistatic interactions BICD 100, Genetics Page 3 of 5 lst Midterm Exam Spring 2008 April 23, 2008 4. While one gene usually specifies one enzyme, one of the following is NOT true: One enzyme may be determined by two genes, if the enzyme has two different kinds of polypeptide chains. 7E) One mutation may cause multiple nutritional requirements if it interferes with the synthesis of an intermediate common to several pathways. Two mutations in the same gene can never have different effects, even when studied in great . biochemical detail. Cg) Recombination is possible between two point mutations within a gene. 5. In certain haploid fungi, similar to Neurospora, the resistance to a specific inhibitor, oligomycin, can occur by mutation from oil's to olir either in the nuclear genome or the mitochondrial genome. A cross is made between an olir strain and an oiis strain and no oiir progeny (haploid) are found. The most likely explanatiou for this olir mutation is: A) oil? was in the nucleus, and was the male parent; B) (Julir was in the mitochondria, and was the male parent; C) 00’ was in the nucleus, and was the female parent; D) oli" was in the mitochondria, and was the female parent. Using just the information available in the pedigrees below, answer the following two questions. Both pedigrees depict th_e same autosomal (rare) recessive abnormality. ‘ I 2/; ’2, /_l {I V" I AF- R L ‘2 at ' I / Z If 1 2 ..---- rr 6 1| r (- (D yg— R1- 6. If 11-1 and 11-6 produce a child, the probability it is phenotypically normal will be: A) 0 D) 3M B) U6 @none of the above C) 9J3 7. If 11-2 and 11-4 produce a child, the probability that it will be abnormal is: A) US R T D) 1K3 (B? we E) 3x4 C) H4 R F) none ofthese I 2 //A \ 2- ... _____ -- r /“ /3 C /!1 ) ’ l p / DDDDHHI II ll I1 ll IULJl ll'lEll Jill ll Til lDEll Iljljf lflDl‘ll—lDDDDDUDDDDDDDDDDD BICD 100, Genetics Page 4 of 5 lst Midterm Exam Spring 2008 April 23, 2008 IV. Seven new point mutants the mméion ofJ'4\wcre found to map as follows: I; i 5 _._____.l 1—-—--.._’&_________. 20? 37. 12 93 22. 36 (7 [4—0.10—1» lac—020.»le—ozo—tit—mo—tlar-oso-J 1—O.10—) ]\ PLfl E’A—H“ “I 'gmucs ! D 1 gem?“ legion M W; rogion (distances in map units) a. (25 prs) A new magical reagent, called Benzerwhiz is used to isolate deletion mutants (A-E) entirely in this n] region. Each deletion is exactbz 0.25 map units long. Five of these deletions were crossed against the point mutants above and gave the following results: .5, j. 1 9 “I f 3 “B 7 12 22 37 86 93 207 e—Mutant # Hf+ EH r+ r fi' r I I+ A ’] Lu" is: rt E9 1* r+ :0 B 61*" __ r+ rt rt rt 1-" l— C ) deletions alt M g..- J r n r” r+ ' 1— - 1'" 1— 1'" " On your answer sheet indicate where these 5 deletions lie on the map. b. (5 pts) Compicmcntation tests are done between the point mutants and the following pattern resulted. Note that + = compicmentation. and t} = no complementation. Mutant # On your map (answer sheet), draw a line indicating roughly the division line between the 2 cistrons of this region, and designate this line with an arrow. DDDDDDDDDDDDDDDUI IDDDI It I! IDDHI IL IDDl—If IDDDDUDBDDDDDD BICD 100, Genetics Page 5 of 5 lst Midterm Exam Spring 2008 April 23, 2008 V. (48 pts.) An organism shows 3 linked genes with a map of this type: <— 10m.u -t s— 20 m.u —> a I b 11 c The organism of a’jb' c+ / a" b' c+ is crossed to 21+ b+ c' / a+rb+ c' and the resulting tri-hybrid is then test-crossed. Indicate the 8 phenotypes in the progeny and their % from this test cross. Put the answers in the box on your answer sheet, assume no interference. Also :1+ is dominant to a', etc. (Pb-ct an: m- dc"): M HWY? 0" 5+ 5’ mnmm . ype T.A. Disc A01 Monday 10:00a - 10:50a HSS 1128A Graham Disc A02 Monday 11:00a — 11:50a HSS 1128A Beth Disc A03 Monday 2:00p — 2:50p CI‘ R 207 Alison Disc A04 Monday 4:00p — 4:50p CI‘R 201 Jing Disc A05. Monday 5:00p — 5:50p cm 201 Jing Disc A06 Wednesday 10:00a — 10:50a CTR 201 Amanda Disc A07 Wednesday 1 1:00a - 1 1:50a CI‘R 201 Amanda Disc A08 Wednesday 12:00p — 12:50p CI‘ R 201 Nick Disc A09 Thursday 10:00a — 10:50a CTR 207 Jennifer Disc A10 Thursday 1 1:00a - 11:50a CTR 207 Sophia Disc Al 1 Thursday 12:00p — 12:50p CI‘ R 207 Charlene Disc A12 Thursday 1:00p — 1:50p CI‘ R 207 Burt ...
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2008_04_25_17_52_46 - BICD 100, Genetics Page I of 5 Ist...

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