ANSC 3500 Exam 3 Biase - First Name OVMY it,Last Name Kim...

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Unformatted text preview: First Name: OVMY it ,Last Name: Kim U‘j“ 2016/10/21 ANSC 3500 -— Animal Breeding Exam “1- 90 points Write your answers on the iines whenever presented to you, and complete the tables with your answers wherever available. Present the mathematical logic that led you to an answer wherever calculations are needed. Fill in the blank {1 point each). 1. The range of values for a correlation coefficient. ‘1 ‘W \ J 2. The range of vaiues for a heritability estimate. 0 ‘1"! \ / LO F5 3. The range of values for accuracy of selection. 0 > i 0 'h \ O +0 1—" halo maria l - - / 4. The range of values for repeatability estimates. l 2 5. The average ratio of a contemporary group. \ 0 O Add the ietters to the right column according to the corresponding item on the left column (1 point each]. / 6. eritabuity bwthe effect of genes in the dam of an individual that influence the 7. (to) accuracy of seiection performance of the individual through the environment provided by the dam. 8. Mrepeatabiiity L/a measure oi strength of the relationShip between ma records for a 9. (d) contemporary group trait in a population 10. {e} breeding value a} ”genetic phenomenon of a singie gene affecting more than one trait. lifi'genetic correlation g/the difference between the mean phenotype of those individuals selected to 12.,lg-lrselection intensity 13Watemal effect 14.,fi1’ple‘rotro py be parents and the average phenotype of ail potential parents, expressed in standard deviation units of the phenotype. is measure of the strength of the reiationship between phenotyoic values and genotypic values for a trait in a population. £measures the strength ofthe relationship between breeding values in one trait and breeding values in another trait. / Q the proportion of an individuai's ganotvpic value that is due to independent and therefore transmittabie gene effects. / a group of animals that have experienced a similar environment with respect to the expression of a trait. bye measure of strength of the reiat‘ronship between true breeding vaiue and their predictor. Page 1 of 6 First Name: ( All): t ,Last Name: RA “(BE 3 2016/10/21 0\ 15. (9 paints) Considering the adjustment factor for weaning weight (lb) and accompanying information presented below: Adiusimeni (Eb) _ 4 _ A E'he Formula for adjustment of weaning weight is: Age of dam (yr) Males mai measured Wt. -biri:h We . 2 50 54 (W) x205 + birth Wt. +c1ge_ciamfactor E.__~_4G 36 Average of adjusted weaning weight = 686 lb 4 20 ‘8 Variance of adjusted weaning weight = 5,792 lb2 . 5m! 5 D D Variance of breeding value of adjusted weaning weight = 1,564 IlaZ i1+ 20 1 And the data beiow from animals maintained in a contemporary group, fill in the agar; cefls: i l f—Adjusted ' Weaning Fredisted Age of clam Birth Weaning 3 Age ofweighing Calf Sex weaning weight breeding value (year) weight a weight (days) i ‘ i ‘ ‘ weight 1 ratio *** ‘ A , r 81 ‘ 221 - i 1i Male m 721 680 (small NIH-‘79 1.6%) i i 8 Male 4 67‘ 543 221 C)77777_ 1” :1 Ml‘li‘ . 1-, ff: Only calculate the values for the emgty ceiEs in this tahia icelis not crossed}. :- W Predict breedin value usin the deviation of the ad‘usted weight in this contemgerag group. EBV =h2x(P——;2) 7 z ***Siiowyourcaiculationson pageZ/3*** 'n 3 J—dz V 5 P U“? i" (0‘50"8‘wa‘é48‘4un: UNLU/ : 10,94 11“ 91911. / 1' 011/ [Nib .u x, _ Mi , met: We t Ewwxw—n r 0.11 x Mae—ma '— Lem. '. Ski - / CM % UAW?" Shag" mi 220 : awn Wrin— {we 1 111/ Effigy PageZofE ii?“ (6%.? men?) : "ii .‘Léio Firsmame; 0“ Vi ,Last Name: R/LAU‘“ 2016/10/21 Page 3 of 6 First Name: U“ h I; , Last Name: IQ (1W 2016f10/21 16. (5 points: What is the effect of reducing the variance of contemporary group on heritability? Explain briefly and f, legiblywhy? “(MUM in; valuing ii a wmmiimm Gimme WM) m Mtg-i M irmm‘mw / mix—limix.fi M (Mam Haiti/Wit UH. \(Liii‘nuiit traits iii a (limit: .m mm. \ “Uiimbk. catfish ~211th Niel. gum cue. ”p 17. (10 points) Calculate the rate of genetic change in feed conversion in a' swine population given the following: Heritability: 0.35; Phenotypic standard deviation: 0.2 ib; accuracy of male selection: (18* accuracy of female selection: 0.5; Intensity of male selection: 72.4‘ intensity of female selection: —1.5; generation intervai for males: 1.8 years; - -—‘ ‘ _..... eneratlon interval forfemales: 1.8 ears. ' 1‘ "- g V ., h - U ‘3‘ iii/i: u L Lm A L+ L0 emit-M3361“ 2 Li0<%--1~\\+to.9 . “mil Mg r 51m 4‘41”! / Lg ’Hfi (Evens / - ’zia'i I (MW —O.’5lSDb , _ _*-~--.—..___.. —: ..—————- a .— I M m. -l 0 oerwl, 18. (10 points) Calculate the estimated progeny difference value fur birth weight of a steer given the following pieces lo ofintormation: Maggot phenotype torjjgioffspring ésfla‘heritabilitv: 0.32 EH) : 9W \Qtwm , Li, a a 20:5 / L‘iflrhh“ Lida—“Lo m) d 511% , S . . . - . 199 ‘0.”5072’ 19.6%: 19. (5 points) Caiculate the accuracy of an estimated breeding value of a steer given the following pieces of information: Average of phenotype of five half sits is 55m. heritability: 0.25 W —. HAW)» ,3; , ktiflmvhw I lit». «BANG-Km — W : 0.19 / Page 4 of 6 FirstName: (AG-WE ,LastName: EQMAQ _ 2016/10/21 Use the tabie below to answer the next two questionsi "TabEe 3‘ Estimaees oi genetic (below diagonal), phenotypic (above diagonal) correiations, heritabilities (diagonai) and their standard error of estimates among performance traits." um ri‘zlgsx'afiu um 10.": . ' .: “42M- -. 9. 16:33.52 magma; ' ' fi33:&fl‘ (1:13.64 - (Marina 7 autumn '7 mini M. mm W ma Birth weigh! Mgr WKWSNHQ migh (keii Mum-Hubs” «elem Elm: ENDWnEnd Want mgr we: iiwmga daily gain {yd} UFD-Uinaswnd lucid-E dlmh (mmi- um irmurd loin dapm (mm “WSW. iama'id EMF; Fciand immersion ruin. ‘rm signiflam mmaans m boim (Rom) dii-i D,i371/fiimii.pune.fli mmmaa 0313 MM “Nine nmfim‘fln miNed "um Minr er al. 2011 Dim are NEW/lieumaimales.nrgfpiosaaeta'tlc‘eiid=1m137111mimai pone 0123105 g 28 (5 points) )Which trait would you expect to have the] owest rate of genetic change? i0W€$‘\' weaning weight (WNW) @361 mwzgnmNDW) c. feed conversion (FER) d. uitresound iFN'E (UIMF) ’ i.“'\ 13-??? ~0 7A) “'0.'1{a g 29 (5 points] )Which piece of information did you use to choose your answerfor question 28? inikmg {H NKWBMWWS W . W M William aim iowui uaiUc I 0.01. “a Use the table below to answer the next two questions. Table 3 Esfz'mm efgmerie mrrefafizms chew: and phenetygzic correlations below diugmmi fiemm'n body candidate 5cm (8C8), days refiys: semis; {DPS} mmaretnm mfg (NR8) lactation somaéic cell scare {LSCS}, )smd 305 «jay yidde ef méflc (MILK) fat, {FA}? and pmz‘ein {PROT} foggflier with their sfmjma’ ermrsi (5 e.) m P1 L pmwthcsis. Regain; m. $9175.51} en man-memm wwdei mfianw canipunenfi whine i207: method scs NER m _\ + +\ 433- ens wee, Am .639 %'=éa65 $24 4122 .612 STY‘mflflflS-i POS%\\0L2, r013. $27 3-12,, 9.13. {:13 ea; 4+; 0 \ouim 190551 W Body canditicn score 3:35 Non return rate NRR (3:02 Days to first service DES ~G-M I Lactation SDmafiC ceii score LSCS 4392 $32 005 305mg yieid Bf milk MILK 43.15 "0.12 9.15 fat EAT ~01}. 4%}? 0:09 protein SWOT 42-11 —04i}9 0113 i Page 3 of 6 pWW'i‘jipiL -S First Name: U {h R , Last Name: KIA-W 2016j10/21 E gs '3 "-5 w 3C! {5 points} Which two traits would you expect to have the lowest ff: Non Return Mt: [Nam WM ”Md 42:"; wt. .and W fig; 31 (5 points) Which piece of information did you use to choose your answer for question 30? is 6 MW, 094 at ”Mt “Cf” \Itihq Oilago me Valve 0“), Wig ‘ I I a a ‘ o ’v . v" I I agfié i ' v ' ' Looking at this portion of a sire summary with EPDs and accuracies (Act) we will designate each of three sites: A, B, and C. -- m I. II. II. lfldfiiflm _ +1§ {-35 +3 439 Q13 $1.56 -IIIIIIIII BW: birth weight, WW: weaning weight, YW: yearling weight, 5C: scrotal circumference. 5 32 (5 point} Considering b'rth weighg which sire has the EPD estimate that best approximate to the true breedingvalue?w A Pt‘ 0 ‘1‘ Pa 0 rare (“mom 33 {5 point) What was the rationale for choosing your answer for question 32? 5 give A has m may“ GLcum ”Umlflcfl'l Utmm: 0m 7.: ewes V‘L m Scum; wma-iui (1ch em 34 (5 pointslw t Is the expected difference between the avenge pefigcmance of progeny sired by A and the average performance of progeny sired by Sire C for weaning weight? 2% J 5% ~23 ”3 1% Page 6 of 6 ...
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