BIL 250 2007 Midterm III Form 1

BIL 250 2007 Midterm III Form 1 - Spring 2007 Midterm III,...

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Unformatted text preview: Spring 2007 Midterm III, Form 1 Dr Alex Wilson NAME: C#: COMPREHENSION QUESTIONS (1 point each) ./ 1. Individuals with Down syndrome, Turner syndrome, and Klinefelter syndrome are aneuploldr )(D A; Y 6) True P‘ b. False 2. The alkaloid colchicine can be used to produce polyploid organisms. A @ True b. False x. "3:2 9. Reciprocal translocations involve an exchange of chromosome segments between two homologous chromosomes. a True H B ® False . Ionizing radiation such as gamma rays and X-rays can cause chromosome breakage. True . b. False _ I .( A normal human female heterozygous for two X-linked alleles will express only one, or the other, of those alleles in any given cell in her body. ._ I, V "- @ True }‘ ,r ..T b. False An inducible gene is transcribed when a specific substance is present. . x l All the ABO blood type alleles map to a single gene locus. “'4' f K All human ABO blood group alleles are co-dominant. V .,< a. True - . ® False Epistasis involves intra-allelic gene interaction. ‘ ‘I - A a. True \, b. False l: PVT" :_. c 10. In diploids, single recessive alleles can be expressed in diploids via . pseudodominance or hemizygosity. ~\ ‘1 @ True 9% b. False Spring 2007 Midterm 111, Form 1 Dr Alex Wilson NAME: C#: MULTIPLE CHOICE QUESTIONS - SELECT E OF THE FOLLOWING ' ANSWERS (1 point each) 11. can cause genes to move from one linkage group to another. a. Inversions b. Deletions jag X-rays Cl @ Transiocations e. Unequal crossing over fie. of the gametes produced by an individual heterozygous for a ‘6 translocation will be nonvlable. b 1/2 c None d. 1/4 e. US For questions 13-14, choose the type of control illustrated by each example. 13. E. céll lac operon control bUac I negative inducible _ .. I ‘ r . ‘ b. negative repressible .‘flwe. ‘- ' V v Q c. positive inducible K" 'v r ' - ' " d. positive repressible -- 14. E. coll lac dperon control by CAP a. negative inducible b. negative repressible C Q positive inducible ‘/ cl. positive repressible V' 15. What is the function of CAMP in regulation of the lac operon? a. activates a repressor protein (ET. activates an activator protein \3 i inactivates a repressor protein V “5. inactivates an activator protein 16.1n a mammal how many inactivated X chromosomes (Barr bodies) would be present in cells of an individuals who were ALANHO a. b. d. e. 17. The chromosomes constitution of an®etrapioid can be represented by a. n1 + n; ' ® 2n1+2n2 '\ \O C. 2H1 Cl. 211: e. H1 + 2n; 18. In a triplod genotype B/b/b, what proportion of gametes will be 5? a. 1/4 I . b P} l? 7 2f ’L 6 b. \0' ' * 1/6 lo. .1 a. 3-” . > b l. 6b. ‘91 Elo‘ \M {—It M11773 (5"? / Spring 2007 NAME: Midterm III, Form 1 Dr Alex Wilson C#:_ 19. What is the chromosome aberration of the type ABCD. EEIEGH called? (Note: the “dot” denotes the centromere.) a. Deletion b. Translocation @ Duplication d. Pericentric inversion e. Paracentric inversion 20. A new Neurospora mutant was tested for auxotrophy. The mutant grows on minimal medium(M)+arginine(A)+histidine(H), M+proline(P)-I-‘H, M+A+H+P, but not on M, or on M+A+P. The mutant requires a. Arginine @ Histidine \3 c. Prolifie cl. Histidine and Proline Site-directed mutagenesis is used to change a codon form AAA lysine to AGA b % arginine but the phenotype produced is still wild type. The type of mutation is called a. Suppression b. Silent @ L.) c. Nonsensex . Missense e. Fran’i’eshift 22.4w individual cell homozygous for a mutant allele of the gene for human enzyme phenylalanine hydroxylase contains no detectable activity for that enzyme. The mutation is best described as a. Dominant b. Recessive 33C Prototrophic é Gain of function mm H1? (_rlL-IIHIII{' Q + 'r o; »_._,_,_‘. Null __ _, 2g. A null repressor muation 0-) results In X No transcription *- r, 33" Inducible transcription l. .l g; Transcription but no translation . No translation Constitutive expression 24.A - _rnoter muation (P-) results in No transcription b. Inducible transcription c. Transcription but no translation d. No translation e. Constitutive expression 25. In the purple penguin, a series of alleles occurs at the p locus on an autosome. All alleles affect the color of feathers: p” = dark-purple, p'" = medium-purple, p’ = light- purple and ,0M = very pale purple (almost white). The order of dominance is p” > p’” > p" > ,0”. If a light-purple female, heterozygous for very pale purple, is crossed to a dark-purple male, heterozygous for medium-purple, the ratio of phenotypes expected c. 1 dark:1 medium:1 light:1 very pale among the baby penguins would be X X '1 cl 3“ d. 1 medium:1 light \ 1'. a. 2 dark:1 medium:1 light (5) 1 dark:1 medium e. 2 dark:1 light:1 very pale ' n].- e a A Spring 2007 Midterm III, Form 1 Dr Alex Wilson NAME: C#: 26. In sweet peas the allele C is needed for color expression (c results in no color - white). The precise color expressed is determined by the alleles R (red) and :- (blue). A cross between red and blue plants resulted in the following & ro en : p g g/Sred Cc—R-‘Z K (LEVY, C;W}\.H 3/8 blue R . v“ (- 1/4 white I“ 8!. ‘1: V. @c What were the genotypes of the plants crossed? r R V 1 "i i y. r If t I. Ic a. C/C;R/R x C/c;R/r ‘x'm H r‘ - "3‘ "“ «~- __ if. b. C/c;r/er/C;R/r *4- CC ‘ c. C/c;R/r x C/c;R/r ' Q) C/c;R/rx C/C;r/r i/a . 4 a. c/c;R/rx C/c;R/r 27. In chickens the dominant allele Cr produces the creeper phenotype (having extremely short legs). However, the creeper allele is lethal in the homozygous condition. If two creepers are mated, what proportion of the living progeny will be creepers? .— i. U m .L f 'p ... -. , I I C) a. 1/4 C E] I C“ ‘ " ’ k“ b. 1/2 C L . r y L l' "at - - L C y (4 C C c. 3/4 C C L C E9) d. 1B 7.; _ _ Q _ ,{o If _._. 2/3‘1 Fa... ._. 1‘.» _ 1.3:” _ i; C ( Q3“ 8 In I osophila the recessive alleles for brown an scarlet eyes (of two a independent genes) interact so that bw/bw;st/st is white. If a pure-breeding brown is crossed to a pure-breeding scarlet, what proportion of the F2 will be i? " white? bxo S 5 ,( g5 l3 ‘1. "A ' a. 1/16 L1 6 \0 __ , _“ ,I If: E‘ bio'iis b-Ua SS \‘N‘. 3- 2:16 55.. a . . .3 . ._ , E) e. 13/16 b D b0 5 5 ~ 29.A ura- yeast cell is transformed to ura+ by a cloned fragment that replaces \ {$335 - S 5 the endogenous gene. When this transformant is crossed to wild-type 9 (standard ura+), what is the proportion of ura- progeny? L __ r“ _l , . r. 'r ‘3) g Clo/o i I f I] I _ 'l. 6 W3 b. 25% w W W D r" r” " " o\ )g 89 3 c. 50% Wk - d. 75°/o . e. 100% 30.Which of the following is not involved with the initiation of transcription of g r I 1 human genes? is 2‘ a. TATA-binding protein b. RNA polymerase DNA polymerase . Activators e. CoactivatOrs L/N Bio * 3 3 K5 ®Q§D§§s RF 3 (55 b3 bj Spring 2007 Midterm 111, Form 1 Dr Alex Wilson NAME: C#: SHORT ANSWER QUESTIONS 31.The normal sequence of nine genes on a certain Drosophila chromosome is 123 0 456789, where the dot represents the centromere. Some fruit flies were found to have aberrant chromosomes with the structures listed in the table. Name each type of chromosomal rearrangement by completing the table below (4 points). .. .2, ' II J 1230476539 'Or(r({w.'ir' C rr'x'ffl'Stc-n/ / I 123 046789 016 i? i , I 123 0 4566789 0N / I 32. One of the jobs of the Hiroshima-Nagasaki Atomic Bomb Casualty Commission was to assess the genetic consequences of the blast. One of the first things that they studied was the sex ratio in the offspring of the survivors. Why do you suppose they did this? (4 points) ON 0/ mo; (OM/Tidfl 7n; 0/ a.__ kaJ mu {cg/eon #1 a lac/Ala {cm :5 b] Moi/(inc: 0.4“ Jr)” {misfit /fl #3") r { Mr [or an ("0.5a Hog Ma" _ A)“ W) -/ I» .ajf tn. '5 X /q/fpn'f X‘PC’QJ-JJl/e 7(fl} 7/ ( t I L I I Ii 1/ ’CTS{ a /6#"5L/ flu ; I Obfrf [(ELIX if X (NW “W m x .‘ r J, 01: (ré‘fli’ N“ m I _ C ‘_ J; ,1 5.“ z I _. I ’ INK 61/ i /’ th. frfirfingg (Amy 0.1% ’l/ (#1; II f a [fill‘ _/-!/’.2 f \ JL/l K L / II-fw' !.; s/rvAJ-J I, ‘1’“! [/ng Am IC?/ 7! € /_ In! , 2/7 3 U " {J' _ :J / 4;”,(1 /r’j /c arc/f 5 Spring 2007 Midterm III, Form 1 Dr Alex Wilson NAME: C#: 33. Mutants that are led” retain the capacity to synthesize [il-galactosidase. However, even though the lac! gene is still intact, B-galactosidase can no longer be induced by adding lactose to the medium. How can you explain this? In you answer be sure to describe the product produced by the iacY gene, the function of that product and to explain why fi-galactosidase will not be induced. (4 points) 7/1! /0( -/ jtfl.) €h~C ,f /;7/ {399; yr};- ; r / . _ 4 a} mic 9. r Am :3 M%crz;x'//U r464“ mat/WM” / / t’ l l ' ' ' H 1w 5 L :74" f' r 7' ’" ’ {J / (Gr-hf? rah/1C“ 26 “ft ‘ . (a .‘r +0 Xvi-50': Cfi 0'“ f' “r L/ /6h(;€r" m 1'? " " Q! ‘ .-" .. . . , . r ,r "' "l '- J ‘ rimr' :53. 1- _.€ _- _.u_..*’ 34. In the maternity ward, four babies have been accidentally mixed up. The . ABO types of the four babies are known to be 0, A, B and AB. The ABC) types of the four sets of parents have also been determined. Indicate which baby belongs to each set of parents by completing the table. (4 points) Ch14 Q5 A A A0 5- .mlnm .. wrl'lf'L‘ ...
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This document was uploaded on 10/27/2011 for the course BIL 250 at University of Miami.

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BIL 250 2007 Midterm III Form 1 - Spring 2007 Midterm III,...

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