20160128124614-1 - Name k[NM Nth"mam Section Q— ANSC...

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Unformatted text preview: Name: k [NM Nth] "mam Section: Q— ANSC 3500 — Spring 2015 Exam | worth 100 points Please show all work for partial credit. Write legibly Multiple Choice. Please circle the letter of the most correct answer. Each worth 2 points each. 1. Who is the father of modern genetics? a. Robert Bakewell b. Charles Darwin E Gregor Mendel d. Charles Henderson 2. What is any observable or measurable characteristic of an individual called? 3. Biological Type b. Environmental Effect c. Phenotype Trait 3. An individual possessing functionally different genes at a locus is known as: a. Homolog Heterozygote c. Homozygote cl. Allele 4. What is the name of the process that produces gametes in mammals? a. Mitosis b. Gene Assortment E Meiosis d. Gene Segregation 5. What is an observed category or measured level of performance in an individual called? a. Biological Type b. Environmental Effect E Phenotype d. Trait 6. What tool(s) do(es) animal breeders use to find parents for the next generation? a. Selection b. Mutation c. Mating d. Migration A&C f. A&D 7. Which of the following trait would not be affected by environment? a. Racing Times of Quarter Horses b. Fleece weight in Ramboulliet ewes @Coat color in Labrador dogs d. Horn length in Longhorn cattle 8. In which phase of gamete formation does crossing over generally occur? a. Interphase l [a Prophase | c. Metaphase l d. Prophase |l . What causes an exception to Mendel’s Law of Independent Assortment? a. Crossing over Sinile nucleotide polymorphisms d. Sex Linked Inheritance 10. What are the assumptions of Hardy-Weinburg Equilibrium a. Populations undergoing selection, mutation, migration and genetic grift @ Large populations, random mating and no forces affecting gene frequency c. Selection and mating of parents d. Independent Assortment and Segregation H 0 d9“ IlfianmrThe type of inheritance by which the expression of the heterozygote is exactly midway of the homozygous genotypes. E _ Fill in the blank (2 points each) i “(odommanoe/ i exsw/ influenced % W. JCOW‘Q «J WINK? M€QLSfl§W “I J, a... sex lament: . £9. «Jamar 1 Ammam. (N (’Y’ W0 ~6€X1llhltc6l ‘lfmmfle 'Bieeaiflfl 9‘ The type of inheritance by which the phenotypic expression is limited to one sex. The type of inheritance by which the expression of the heterozygote is identical to the expression of the homozygous dominant genotype. An interaction among genes at different loci such that the expression of genes at one locus is dependent on genes present at one or more other loci. The type of inheritance by which the mode of gene expression differs based on the sex of the individual. The type of inheritance by which the expression of the heterozygote more closely resembles the homozygous dominant genotype. The type of inheritance by which the expression of the heterozygote is outside the range defined by the expression of the homozygous dominant genotypes. The type of inheritance by which the mode of gene expression is based on genes residing on the sex chromosomes. The genetic makeup of the individual. The value of an individual as a parent. Short Answer (4 points each) What is the model describing phenotype for simply inherited traits? Please define all model 21. (8:9 cfénpisthm \J darmmm Mi (1an ,W domimm X MR N'\\\ w. exqvemot. Nun w acne to mimic ma, DH H m“ WNW a qclmflwg 0-( M W fm \“r \‘U (MW/1M M and do if it! Vtche. ($3; a 22. What is the model describing phenotype for quantitative traits? Please define all model components not described in the revious question. MW (N (1th vv- await: can mm 0m alitmvx § 3) WW 01mm Mt (mm—(a and environ/um com “EN (1 VOW N WCM. @h*@ 23. Why does the model differ for qualitative and quantitative traits? imam m 4VMH‘MNC (\finllW mmmmm) ,OV‘N (W qem u (H mm 0km m qualms-tram (priming)J ”(U \leolt 0\(V\N (G‘MC ivm View. For questions 24 to 27, show all work. If no work is shown, no partial credit is possible 24. (16 points) In pigs, black (B — Hampshire black) coat color is completely dominant to BP (Berkshire black) and b (Duroc reddish-brown). BP is codominant with red (gives black spots on a reddish- brown skin). At another locus, the l allele is completely dominant to the i allele. The i allele inhibits the expression of the alleles at the black-red locus and results in a white pig 3) What are the colors of pigs with the following genotypes: “dammit Wm. CY 6 «mm -wm m. an m W. ' mew—WW 3X1,“ mm mm: All whit “”33 = 9/10 5 6070 x A“ \0\&C\¢ Q1038 : 41m) : 26%) All it] blaccwm on realm anin unm : WHO -. 12,-S0/0 , All ‘ , = 7 : 12's? . \9 WNW WON“ J'lwwmflm we 400 WM‘W‘ 4” ”"64 ”MUM“ b) If a boar (Bbii) is mated to sows which have the genotype (BPbli). What are the expected 0(IT genotypic and phenotypic frequencies among their offspring? VM “0- . If d) Another black boar in question is mated to 5 white sows and 3 Berkshire Black sows. The l l Berkshire black sows are known carriers of the i gene. Still assuming 10 piglets per litter, how \ H \7 confident are you that this boar does not carry the i gene? ‘ \_ \l\ l 1 P621 : V (m V (mm) + WWW“ 3 ¥ .0 . 3 \(fl/ QED] :\- [0* oH'lfl‘ol Djibflfit (:mm(£7*(lfl (6m \, ((0.03me )LO‘OKWD : \» 0.00MQ7X _ pip”) WM = w F\ Ankh )fl hbh 25. (8 points) In sheep, expression of the allele for horns (h+) is a sex influenced trait. Males possessing one h+ gene are horned, while it takes two h+ genes for a female to be horned. Dorset sheep are horned and the breed only contains the h+ gene. Suffolk sheep are polled and have a genotype of hh. Two F1 Dorset-Suffolk progeny are mated. What proportion of the resulting ram lambs would you expect to be polled? What proportion of the ewe lambs would you expect to be polled? w. w w o R ‘\\l\0_DOlJt’t it Qcmale and if Mammau X ff “f l HQ 6 All mam ave palm W“ “0‘1 :.- All imam ave nomad. XWY-M: fl) ,9 WWW“ 4w villa is mm am if WWW“ W” m": are = % ng‘ii ___ All imam we. WW & x“ Win" N All Maw ale MW- h X fix” W 26. (8 points) In Simmental cattle, black (B) is completely dominant to red (b). There is now a DNA test that allows exact genotype differentiation between BB and Bb animals. Eight A.|. bulls (all 8 are Bb) are each mated to the same number of cows. In the cows, p(B) = 0.90 and q(b) = .10. a) What are the expected genotypic frequencies (of P, arid Q) and phenotypic frequencies among the calves produced from these matin ' b) What are the gene frequencies in the calves? ”Eb >4 lob , (5g — -- fix 5/ l7. ’ E 3% mm 3 (3b @011! I; 3 l 12 : b\o 27. (16 points) In the following table are 2 individuals with 5 known loci affecting a quantitative trait. Assume the following: Complete dominance at each loci Capital letters represents the dominant allele No epistatsis involved u = 550 lbs. For homozygous combinations, genotypic values are equal to breeding values The independent effect of each dominant gene is +8 lbs. . 1' Am; \ The independent effect of each recessive gene is +1 lbs. \VV‘ VWUJ W (10 l . a) Fill in the table with the appropriate values \M [N ' Genotype X Which individual is the heaviest? so \ win We Nomi ~ .u- b) Aa - 9 \\\/L G QY) .‘ 6 * (C ; a 40le DH. :@ (2c 5 6 GET—9 c) Which individual would produce the heaviest offspring on average? Why? \m (1&1qu 4N tow WM «w woman YDV ZUlloJ $2, fie Extra Credit 1. (2 points) Proof that you gave blood (or tried to give blood) or bought a hamburger from CCM/CCW on Friday, February 13th. 2. (10 points) Show how P[Dn] = 1 (PBB + 3/4P3b + Vszb)" results in n = log (1- P[Dn])/|og(PBB + %P + Vszb) Equation Sheet 1. No. of unique gametes = 2n 2. No. of unique zygotes = 3" x 2m One offspring per mating ‘ Multiple offspring per mating Pan1= 1 — (Pan + 3/4PBb + yzphh)n Panm1= 1 ' (P33 + (Viimpab + (y3)mpbb)n PIDn] = 1 " ”(P33 + 34PM: + 1513m: )n | P[Dn1= 1 ' ”(P33 + (%)'" P31: + (yzim Pbb )n = '08 {1' P[Dn])/I0g{PBB + 3/4PBb + Vzpbb) | n = [03 (1' PiDnm])/|03{PBB + (%}mPBb + {mmpbb} 3. Fitness and Dominance 4. PDorTA=VzBV 5 Poffspring=l1 + yZ BVsire+ +y2 BVDam 6. PA=BV+GCV+Ep ...
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