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# Name_

Published on: Mar 3, 2016
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• 1. Name________________________ AS 332, Principles of Animal Breeding Problem Set 10 Please write your name above and on the back. 1. Consider a dairy cattle population with a normal (bell-shaped) distribution for milk production. Currently, the average annual milk production per cow is 20,000 lb and the phenotypic standard deviation is 1500 lb. The top 2 % of males and top 30 % of females will be saved for breeding. The average generation interval is 3.0 years for males and 5.0 years for females. Assume h2=.30. a. Give the selection intensity (i from table) for males, females and averaged over sexes. i♂ = 2.42 i♀ = 1.16 iavg = 1.79 b. Calculate the predicted selection differential averaged over sexes. Savg = 1500*1.79=2,685 lbs c. Calculate the predicted change in the population mean milk yield resulting from one generation of selection for increased milk yield. R/gen= 2685*.3=805.5 lbs d. Calculate the expected change in the population mean resulting from 10 years of selection for increased milk yield. R/year = 805.5/4=201.375 lbs Milk Yield = 2013.75 lbs 2. Consider a cattle population with a normal distribution for yearling weight. The current average yearling weight is 900 lb (adjusted over sexes) and the standard deviation is 60 lb. Assume h2=.3. Consider the three situations below. i. The top 5 % of males and top 70 % of females will be saved for breeding; generation interval = 4 years. ii. The top 10 % of males and top 70 % of females will be saved for breeding; generation interval = 5 years. iii. The top 10 % of males and top 80 % of females will be saved for breeding; generation interval = 6 years. a. Show which of the three situations above is expected to result in the greatest increase in yearling weight per generation? i. 5 % males (i=2.06)and 70 % females(i=0.5), iaver=1.28, S= i x σ = 76.8 lbs, R/gen= 23.04 lbs. (Sit.1) ii. 10 % males (i=1.76) and 70 % females (i=0.5), iaver= 1.13, S= i x σ = 67.8 lbs, R/gen=20.34 lbs. iii. 10 % males (i=1.76)and 80 % females (i=0.35), iaver= 1.055, S= i x σ = 63.3 lbs, R/gen= 18.99 lbs. b. Show which of the three situations is expected to result in the greatest increase in yearling weight per year? i. R/gen= 23.04 lbs, L=4 years, R/year= 5.76 lbs. (Sit. 1) ii. R/gen=20.34 lbs, L= 5 years, R/year= 4.068 lbs. iii. R/gen= 18.99 lbs, L= 6 years, R/year= 3.165 lbs. 3. How can a breeder improve the heritability and repeatability of traits of interest? Lowering the phenotypic variation by increasing the environmental uniformity, by managing the animals in contemporary groups, taking more accurate measurements, adjusting the data for fixed effects. 4. How would it affect the farm profitability? The increase on heritability would lead to a greater genetic improvement and generate offspring more profitable. Especially for animal breeders the increase on selection accuracy would be very interesting if the major income is to sell genetic material (sires, semen or heifers). Commercial producers, which the major income is a commodity, should be worried in maintaining an accurate data record, but many times (in general) the best management for increasing the farm efficiency is to split the herd in groups of animals with different production levels or growth rates in order to balance a diet specific for each group, for example.
• 2. 5. Why is important to correct the weights at birth, weaning and yearling to the age of the dam? The age of the cow affects its performance because of the necessity of using part of the nutrients for growth in young ages (heifers) and due the decrease of physiological efficiency on old cows. It has a big impact on BW, WW and YW of the offspring and consequently some animals will be heavier weight at birth, weaning or at one year of age not because of its genetic potential but due the effect of the cow’s age. In order to select the animals with higher breeding value it’s necessary to take this effect into account. 6. What is Contemporary Group? What is its importance on Genetic Evaluation? It’s a group of animals of the same sex and similar ages that were managed alike. These animals experienced a similar environment so phenotypic differences are due genotypic differences. In other words they had the same chance /possibility /opportunity to express their genetic potential. It’s the basic unit for genetic evaluation. Once animals belonging to the same contemporary group experienced the same environment its easier to identify animals with higher breeding value. 7. What are Genotypic and Breeding Values? What is the difference between them? Genotypic value is the value of the animal’s genes on itself or the effect of the animal’s genes on the expression of his phenotype. It includes additive, dominance and epistatic effects. Breeding value is the value of the animal’s genes on the offspring or the effect of the animal’s genes on the expression of his progeny phenotype. It’s the value of an animal as a parent. It includes only additive gene effect, which has an effect on phenotype independent of the alleles inherited from the other parent. 8. What is the expected progeny difference and predicted transmitting ability. What is their utility? They are predictions of the difference of the individual’s progeny in relation to the mean of all progenies. They are useful as a tool to identify animals that will transmit the desired phenotypes for traits of interest. 9. What is accuracy? What is its importance? It’s the correlation between the estimated and the true breeding value. It reflects the level of confidence that the EPD closely approximates true PD and consequently which phenotype is going to be transmitted to the offspring. 10. How can one improve the EPD and ACC of a given animal? The EPD of an animal cannot be improved. The genes of an animal cannot be changed and so the EPD. In other words, the ability of an animal to generate improved offspring doesn’t change. Its possible to change the genetic potential of the population over the generations, but not the animal’s EPD. It’s possible to increase the accuracy of breeding value estimation by increasing the environmental uniformity, by managing the animals in contemporary groups, taking more accurate measurements, adjusting the data for fixed effects. The incorporation of information of related animals (parents, sibs,…) greatly affects the accuracy of an EBV. Ultimately the generation of offspring is what show the value of an animal as a parent or “prove” an animal allowing the accuracy to reach values close to 0.99. 11. Define Reference Sire. How is it useful? A reference sire is an animal that have generated offspring in different herds, usually by the utilization of Artificial Insemination and so serves as a benchmark for comparisons between animals belonging to different herds/ contemporary groups. The presence of animals with IBD genes (from the reference sire) in different CG allows the estimation of the environmental effects and the estimation of population parameter such as heritability and correlation. .
• 3. 12. Complete the table below by calculating adjusted 205-day weights and adjusted 365-day weights for 3 bull calves. Assume that the month of February has 28 days. Use the BIF standard adjustment factors for age of dam from your handout. Birth weaning yearling dam actual adjusted actual adjusted Calf age date wt, lb date wt, lb wt, lb date wt, lb wt, lb 1 8 3/28 100 10/20 630 627.43 3/30 1150 1144.2 2 3 4/17 90 10/20 560 648.01 3/30 1100 1184.66 3 12 3/10 80 10/20 550 530.13 3/30 1050 1027.02 WW 1- (206d) => ((630-100)/206 x 205) +100 + 0= 627.43 lbs 2- (186d) => ((560-90)/196 x 205) +90 + 40= 639.49 lbs 3- (224d) => ((550-80)/224 x 205) +80 + 20= 450.13 lbs YW 1- (161d) => ((1150-630)x 160)/161 +627.43 = 1144.2 lbs 2- (161d) => ((1100-560)x 160)/161 +648.01 = 1184.66 lbs 3- (161d) => ((1050-550)x 160)/161 +530.13 = 1027.02 lbs 13. The data below are from a national breed association genetic evaluation program for beef cattle. Maternal Birth wt, lb Direct weaning wt, lb Yearling wt, lb Milk, lb (weaning wean wt wt) Bull EPD ACC EPD ACC EPD ACC EPD ACC EPD A -1.5 .77 24.0 .79 24.5 .70 16 .50 28.0 B 8.8 .28 50.0 .29 72.7 .25 12.0 PE 37.0 C 4.5 .56 60.0 .59 67.0 .52 10.0 .26 40.0 a. What is the milk EPD for Bull A (remember, Maternal WW EPD = Milk EPD + ½ Direct WW EPD)? 28=MEPD+24/2 MilkEPD= 16 lbs b. Offspring from Bull C are expected to average ___6_____ lb MORE or less (circle one) than offspring from Bull A for birth weight. c. Offspring from daughters of Bull C are expected to average ___3__ lb MORE or less (circle one) than offspring from daughters of Bull B for weaning weight. d. Offspring from Bull A are expected to average ____48.2____ lb more or LESS (circle one) than offspring from Bull B for yearling weight. e. Which bull's ranking for direct weaning weight EPD is likely to change the most as more data becomes available? Bull B f. Which bull's daughters are expected to give the most milk, on average? Bull A g. Which bull's daughters are expected to wean the heaviest calves, on average? Bull C h. Which bull likely has no daughters with progeny weaning weight records? Bull B i. Which bull’s progeny are expected to be the heaviest at birth, on average? Bull B
• 4. 14. The data below are from the American Angus Association national genetic evaluation program. The data are adjusted to a common slaughter age. Bull Carcass wt, lb Marbling score Rib-eye area, in2 Fat thickness, in G 0 +.15 -.07 .29 H -2 -.03 -.03 -.02 I +21 +.03 +.23 +.03 a. Progeny of Bull G are expected to have carcass weights that are ____2____ lb HEAVIER or lighter (circle one) than progeny of Bull H, on average. b. Progeny of Bull H are expected to produce carcasses with ___.26___ in2 larger or SMALLER (circle one) rib-eye area than progeny of Bull I, on average. c. On average, which bull's progeny are expected to have the most marbling in the ribeye muscle? Bull G d. On average, which bull's progeny are expected to be the most heavily muscled? Bull I e. On average, which bull's progeny are expected to have the fastest carcass growth from birth to slaughter? Bull I 15. BelowarePredictedTransmittingAbility (PTA)datafromadairysiresumm forfourbulls. ary PTA YIELD, LB PTA % REL Bull MILK PRO FAT PRO FAT % A 200 2 -12 -.02 -.08 94 B 633 41 21 -.03 -.15 85 C 2208 72 79 .03 .00 87 D 491 34 41 .08 .10 86 PRO=protein,REL=reliability. a. Of these bulls,daughters Bull ___C__ wouldbeexpecte toproduc themostpounds milk 4 of d e of proteinperlactation,onaverage. b. Of these bulls,daughters Bull ___D__ wouldbeexpecte toproduc themostproteinpergallon 4 of d e of milk, onaverage. c. Of these bulls,daughters Bull ___C__ wouldbeexpecte toproduc themostpounds milk fat 4 of d e of perlactation,onaverage. d. Of these bulls,daughters Bull ___D___ wouldbeexpecte toproduc themostfatpergallonof 4 of d e milk, onaverage. e. Of these bulls,Bull ___A___ haslikely hadthemostdaughters production. 4 in f. Daughters Bull A wouldbeexpecte toaverag __2008___ lb perlactationmore LESS(circleone)than of d e or daughters Bull C formilk production. of g. Daughters Bull B wouldbeexpecte toaverag ____20_____ lb perlactationmore LESS(circleone) of d e or thandaughters Bull D formilk fatproduction. of
• 5. h. Daughters Bull A wouldbeexpecte toaverag ____32____ lb perlactationmore LESS(circleone)than of d e or daughters Bull D formilk proteinproduction. of i. If daughters Bull A averag d of e 30,000 of milk in aparticularherd,thendaughters Bull C wouldbe lbs of expecte toaverag ___32,008_milk productionin thatsame d e herd,assuming twobullswere the matedto cows similargeneticmerit. of j. If daughters Bull A averag d of e 2.5%milk fatin aparticularherd,thendaughters Bull C wouldbe of expecte toaverag ___2.58____ % milk fatin thatsam herd,assuming twobullswere d e e the mated cows to of similargeneticmerit. Proportion Selection Selected Intensity 0.01 2.67 0.02 2.42 0.03 2.27 0.05 2.06 0.1 1.76 0.2 1.4 0.3 1.16 0.5 0.8 0.7 0.5 0.8 0.35 1 0
• 6. Table: Accuracy and Associated Possible Change on EPDs Production EPD Carcass EPD Body Composition EPD Acc BW WW Milk YW MW MH SC CW Marb RE Fat %RP %IMF RE Fat %RP 0.05 2.73 11.0 9.21 16.17 39.32 0.74 0.70 15.42 0.25 0.270 0.034 0.53 0.18 0.30 0.020 0.35 0.15 2.44 9.85 8.24 14.47 35.18 0.66 0.62 13.80 0.22 0.25 0.030 0.48 0.16 0.27 0.018 0.32 0.30 2.01 8.12 6.79 11.92 28.97 0.54 0.51 11.36 0.18 0.20 0.025 0.39 0.14 0.22 0.015 0.26 0.35 1.87 7.54 6.30 11.06 26.90 0.51 0.48 10.55 0.17 0.19 0.023 0.36 0.13 0.21 0.014 0.24 0.40 1.72 6.96 5.82 11.21 24.83 0.47 0.44 9.74 0.16 0.17 0.021 0.34 0.12 0.19 0.013 0.22 0.50 1.44 5.80 4.85 8.51 20.69 0.39 0.37 8.12 0.13 0.14 0.018 0.28 0.10 0.16 0.011 0.19 0.55 1.29 5.22 4.36 7.66 18.62 0.35 0.33 7.30 0.12 0.13 0.016 0.25 0.09 0.14 0.010 0.17 0.70 0.86 3.48 2.91 5.11 12.42 0.23 0.22 4.87 0.08 0.09 0.011 0.17 0.06 0.10 0.006 0.11 0.75 0.72 2.90 2.42 4.26 10.35 0.19 0.18 4.06 0.06 0.07 0.009 0.14 0.05 0.08 0.005 0.09 0.80 0.57 2.32 1.94 3.40 8.28 0.16 0.15 3.25 0.05 0.06 0.007 0.11 0.04 0.06 0.004 0.07 0.85 0.43 1.74 1.45 2.55 6.21 0.12 0.11 2.43 0.04 0.04 0.005 0.08 0.03 0.05 0.003 0.06 0.90 0.29 1.16 0.97 1.70 4.14 0.08 0.07 1.62 0.03 0.03 0.004 0.06 0.02 0.03 0.002 0.04 0.95 0.14 0.58 0.48 0.85 2.07 0.04 0.04 0.81 0.01 0.01 0.002 0.03 0.01 0.02 0.001 0.02 Table: Birth weight adjustment factor for age of the dam. AOD (yr) BIF Bulls Heifers ∆G ixσpxh 2 2 8 8 = 3 4 5 2 5 2 year L .5-10 0 0 11+ 3 3 Table: Weaning weight adjustment factor for age of the dam. AOD (yr) BIF Bulls Heifers 2 60 54 3 40 36 4 20 18 .5-10 0 0 11+ 20 18 WW205 = (Actual wt. – birth wt) / age at weighing x 205 + birth wt. + age of dam factor YW365 = (A x 160) / No. of days between Wts. + B A = Actual Final Wt. - Actual Weaning Wt. B = 205-day Weaning Wt. Adj. For Age-of-Da