EXAM 2 answeres

EXAM 2 answeres - Biology 313 Fall 2010 Exam 2 Name Score...

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Unformatted text preview: Biology 313, Fall 2010 Exam 2, Oct 18, 2010 Name Score: This exam has 100 total points Multiple Choice (2 points each): Circle the correct answer. 1. What is the name for the interaction that occurs between the alleles of a single gene whereby the phenotype of a heterozygote falls Within the range of phenotypes of the two homozygotes? a. Complete dominance i- Epistasis Incomplete dominance ’i' Penetrance e. Co—dominance 2. Which of these is true about the gene (“a”) for the CFTR protein? a. the gene has null “c” alleles that have no protein activity b. heterozygotes (c+/c) that express the normal and mutant proteins are co- dominant at the molecular level, but the “0” allele is recessive in heterozygotes at the physiological level 0. the CFTR protein is a transmembrane conductance regulator of chloride ions (1. mutations in this gene are responsible for cystic fibrosis 1 of the above are true ' ‘3. A phenotype produced by an environmental factor that is the same as that produced by a mutant genotype is called: a. A coincidence of extraordinary magnitude b. Conditional allele - Genomic imprinting 4. “The phenotype of the offspring is determined by the nuclear genotype of the maternal parent”. This is the definition of: a. A sex-linked characteristic . A sex—limited characteristic. enetic maternal effect A genocopy e. Epigenetics 5. The first genetic map was derived for the Drosophila X chromosome by “ TH Morgan Sturtevant ' . Beau Bridges (1. “Dutch” Waardenburg e. Gregor Mendel Multiple Choice (4 points each) 6. Which of the following is NOT true? A person with blood type A: a. has an 1A1A or IAi genotype b. produces antibodies to the B antigen only c. has an A terminal sugar attached to Compound H d. can donate red blood cells to a person with blood types A and AB only an receive blood fiom a person with blood type 0 only if 7. Which of the below is NOT true about X—linked recessive traits? . More males than females are affected. oes not skip generations. " Approximately one-half of the sons of a female carrier are affected. (1. They cannot be passed from father to son. 8. Look at the data below from a study in which the percentages of lung cancer and manic depression were determined in a population of monozygotic and dizygotic twins. What do these data say about the genetic and environmental influence on each characteristic? ' (M2 = monozygotic twins; DZ = dizygotic twins.) ! Concordance (%) Trait MZ DZ Lung cancer 20 19 Manic depression 80 20 j he role of genetic factors is great in determining individual differences for anic depression. b. The role of genetic factors is great in determining individual differences for lung cancer. 0. The role of genetic factors is great in determining individual differences for both manic depression and lung cancer. d. Environment does not play a role in determining individual differences for manic depression. e. These data are conflicting and adoption studies need to be done. 9. In a particular cross you calculate that the co—efficient of coincidence = +1.6. You scratch your head, re-check your math and again you calculate a value of +1 .6. What do you conclude? . a. A crossover in one region of the chromosome interferes with additional crossovers in the same region. b. The interference .= - 0.4 rossing over in this region of the chromosome is enhanced by the occurrence of other crossovers in this region. (1. Fewer single crossovers took place than expected. e. The occurrence of one crossing over does not depend on the occurrence of another crossing over; they are independent events. PROBLEMS: 10. (5 points) You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short—ossicone (1) allele. However the short ossicone phenotype is only 60% penetrant. You cross two "heterozygous giraffes. What is the probability that the first offspring will be a male with ' short-ossicones? Show your work. ' A! , a“? 15mm “lid LL” # . Fennmwg, fl” Mr» (90% i can" «at? 3 .v .1, ‘. J;le 2% Qua: ofla " 223% 11 (8 points). The exquisite Hotel Patti in Perry, Iowa, has served as the site of the “National Tongue Rolling Contest” for a number of years; last year’s title was won by Francine Parks, a mether of five from Waterloo. The ability to roll the tongue is caused by a dominant allele T. Francine is a “roller,” but one of her parents is not. Francine’s significant other, Josh from Boone, is a “nonroller.” a. (3 points) Determine Francine’s genotype. Show your work. Tate. um Ft’mndm, Must have. tam: emi- Je T" {mm or“. unread? (NWA my}. “% Ti FVQ’W‘WQ. {3mm “chem Amarailer’ WWW b. (3 points) What is the probability that the next child of Francine and Josh will be a male roller? Show your work. Francine. , Tusk V (333,; ii T‘k K “gal”: (it; Tt:w: air. c. (2 pains) Why do you think some people are better rollers than others? ' ongue rolling is an example of how the expressivity of a trait can differ among eople who have the same genotype. b. This is classic example of penetrance. c. “T” is an example of a conditional allele whereby something in the environment in Waterloo affects tongue rolling ability; this factor is not found in Boone. d. Francine’s tongue rolling ability is likely caused by maternal genetic imprinting. e. From the data presented it is safe to assume that tongue rolling is a sex-limited trait and expressed only in females. 12. (16 points) Tomatoes have white, red or orange fruit depending on the type of carotenoid they accumulate. A true-breeding variety of tomato that produces white fruit is crossed with a true-breeding variety that has orange fruit. All the F 1 progeny have orange fruit. When the F1 plants are self-fertilized (plants have male and female parts), the following F2 progeny are found: 1008 plants with orange fruit 448 plants with white fruit 336 plants with red fi'uit l “l ‘i it. a. (5 points) Using Table 5-2, what is the phenotypic ratio of the plants in the F2? What type of gene interaction do you propose underlies this ratio? Show your work. OVQMQ, 3 “WE? ‘Xlétilfi‘i'iwi‘ 6% wists, '= “tilt? mistrust: ‘i Mi 3% r; atlt’ifits3 ‘lf‘ti‘%: finmttiv‘a «esp t r imam W b. (5 points) Assign genotypes to the phenotypes in the F2. Designate the genotypes as in Table 5-2 (A__ B_; A__ bb; aa B_ and aa bb) Ci Rm E m 1 Orange; 3 iii W i9 i3 1 Feud mm. E) m. mam. in is) Pervertttva ace»... W latte. c. (6 points) Assume that A and B are genes for enzymes. Draw a biochemical pathway that is consistent with the data, showing what steps A and B catalyze, then clearly describe how your scheme would produce the different types of fruit. ('3 in) 2N 3r e, “Emmi? “imam cl, :3 0m“ 3%; ~ “ Clad. ” “Custodial at? 9 Mid-lee" ti o tee.le “PM? “‘1 may l” {scrawny-k mlth “in FNMA. i loaning” mil, @5th lat-L, it’ekufifitr‘i‘aém A“ ( mm. a m omit «sensible (gewxflwwml / (Qua? $wa f , l‘ was" mecumwla’vefi veil, since new“ i {erratum i” Whig iif'l""“l‘§i _ FDVMK-‘nbx a'l‘" Vsfixai rag mots. pl“ ding» 53ft at: no, law Table 5.2 Modified dihvbrid phenotypio ratios due to gene interaction amettxnen} QT‘DJWJVL, $3 ,3 Genotype Ratio* 11. B_ [L Db v 33 IL aa bb Type of Interaction Example None 9:3:321 9 Seed shape and endosperm color in peas fees 24' “ “ “ ' ' 'rL'rfneeégswaiépisasist:, : :Jeéal-eamifin Lgb'raaéri? ~ 12 :3 : 1 Dominant epistasis Color in squash g1 , '2 " , , ,nggnsgarecessi'va' iAtbinieminenails . L . s . , ’ epistasis, _' Duplicate interaction -— -“ 5:,Duptibate‘dominant W ‘ f , ' epistasi3‘*"' * ‘ Dominant and recessive ~— epistasis ‘Each ratio is produced by a dihybrid cross (Aa Bb x As Bb). Shaded bars represent combinations of genotypes that give the same phenotype. Table 5~2 Genetics-A Conceptual Approach, Third Edition 6 2009 W.H. Freeman and Company 13. (10 points) A series of two—point crosses were carried out among five loci (a, b, c, d, e), producing the following recombination fiequencies. Map the five loci, showing their linkage groups, the order of the loci in each linkage group, and distances between the loci of each linkage group. Loci ' % Recombination a & b 50 a & c 50 n a & d N\<l§u\g&w {3mg}? 4; a&e 50 M;m.mr°~ b&c 10 3 tiww :3 m, c: b & d 50 l 3 =1 b & e 18 0»- at c & d 50 Lin «3% c & e 26 54‘) CM , '~ ‘ 1 Me SOCTMnK’ I r I new! N are. C b a (:1 it: MW “or; chimed L‘“W”f“““’”““£ 5 Grammar” 22% met, (M310 Olm- 9:3“ mp!» ottoman (~2an is an) 14. (9 points) In the following pedigree, assume fiill penetrance of the trait; no new mutations; and that the trait is rare (non-family members do not have an allele for the trait). ’ 1 0 l 2 I g E D 1 2 III a. (3 points) If the phenotype followed in Pedigree 3 is for an X—linked recessive trait, then the genotype of 11—2 is: a. homozygous dominant eterozygous c. homozygous recessive d. hemizygous dominant e. hemizygous recessive b. (3 points) Ifthe phenotype followed in Pedigree 3 is for an X-linked recessive trait, then the genotype of III-l is: a. homozygous dominant b. heterozygdus c. homozygous recessive d. hemizygous dominant Inizygous recessive c. (3 points) If the phenotype followed in Pedigree 3 is for an autosomal recessive trait, then the genotype of LI is a. homozygous dominant erozygous c. homozygous recessive d. hemizygous dominant e. hemizygous recessive 15. (26 points) Waxy endosperm (wx), shrunken endosperm (sh), and yellow seedling (v) are encoded by three recessive genes in corn that are linked on chromosome 5. A com plant homozygous for all three recessive alleles is crossed with a plant homozygous for all the dominant alleles. The resulting F1 are then crossed with a plant homozygous for the recessive genes in a three—point testcross. Following are the progeny of the testcross 5 L0 wx Sh V 87 Parmi‘acl S, W Sh 94 t m , , I 9 W: Sh if 3479 WKWKgL‘SkVE’i’C b9wa SlMlWU Ofelia} wa sh v 3478 . gC :- Wx sh V 1515 F v ° wx Sh v 1531 , | ‘ . I, , SQ Tr ‘ wx Sh V 292 t “J24 a“; Wx sh v w k d“ total 10,756 x um, \wx Sn .5 C 'l‘tist‘mtf} a. (8 points) Determine the order of these genes on the chromosome. Explain your rationale. \[I S ir\ kucause.’ ‘\ if it 's +i'\L a..\ie..\e. ' ac (.2; \ m W «’2,» versed “i‘g‘flv’u; %W“£:n“i*mi§ +x‘» i" :‘c d“ a A l k $1 as 3 UM V. gt.“ W b. (12 points) Calculate the map distances between the genes. Show your work. ' Retentiin ‘. ibl‘fmmm w. a: i: ’ gime wit \/ SR W75; (m: x) 4% W75 4!" BCO : i} CLM~Wfi .................... mm“ ' T a,” “x too W I: M V (AK Mk)“ '3“; (251‘ (017%: Seeing” disk ’53) r L ‘3 bioxwo-u 3a a!“ WK f2. Butane Lax... \/ *5. $0931 [ax/V 8L it?!) Soon: +- Bw _; '£QE+;ta k CH 4.37 , ,x/ v 3k ' "'E‘efl'mi W” {Gym-c, I $0) WK \/ SK {$7 ‘ r 4 “fl 0.07 Km: m» W»: ‘3 3h 9 c. (6 points) Determine the coefficient of coincidence and the interference among these genes. Show your work. What is the significance of the values you obtained? w x e k in: mm gamer we “Macrame. 3ow\+ 7cm Em» i3“ .~ l ' ewes. “been a 0.3 x13”? X 10,??4. i o m . a (j. 7 CW '3 QQQ X V , w: u » §k - v Coincidence” g (T a: X KOO/o 0% Mac-mm (Ar-(5.” 6&336Nfi. nit Vlwvmwmxrmmlw‘ \ r {:QJ‘ME,£_ 22'. \t W ém‘mfirfl" MEL) ’5‘- i W {2L “é.” $34.3; m V9100“ at than DQD are, not flawed 7 £¢Cfi.E!{§-Q¢¢ Hm." €>¢€xarremnm 05%“ a «9,, Qrwssauaf” {Khugtunt With ‘Hw, T‘Md‘i‘iox ash-ham i“ flask main!» ...
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EXAM 2 answeres - Biology 313 Fall 2010 Exam 2 Name Score...

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