Chapter 7 The Genetic Model for Quantitative Traits

Transcription

1 Chapter 7 The Genetic Model for Quantitative Traits I. The Basic Model II. Breeding Value III. Gene Combination Value IV. Producing Ability

2 Chapter 7 The Genetic Model for Quantitative Traits Learning Objective: To understand the Genetic Model for Quantitative Traits that involves the concepts of Breeding Value and Gene Combination Value. Applications include the use of Expected Progeny Differences (EPD s) and Producing Ability (PA) for improving traits of economic importance.

3 I. The Basic Model P = μ + G + E where P = the phenotypic value or performance of an individual animal for a trait, μ = (the Greek letter mu) the population mean or average phenotypic value for the trait for all the animals in the population, G = the genotypic value of the individual for the trait (i.e., the effect of the animal s genes, singly and in combination), and E = the environmental effect on the individual s performance for the trait (i.e., external, non-genetic factors).

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5 II. Breeding Value Definition: Breeding Value (BV) - The value of an individual as a genetic parent to its progeny s performance. The aim is to accurately identify those animals with the best set of genes that constitute BV. - Each gene imparts a numeric independent effect towards the animal s aggregate breeding value for the trait. - Each gene imparts a value, which is manifested at the physiological level ( Genetics is the basis of physiological expressions of traits ; a simple example is dwarfism involving a deficiency of growth hormone).

6 II. Breeding Value Example using a Single-Locus Model: B = 10 g; b = -10 g; complete dominance; trait is mature body weight in mice. Genotype BB Bb bb BV 20 g 0 g -20 g

7 II. Breeding Value Example using a Single-Locus Model: B = 10 g; b = -10 g; complete dominance; trait is mature body weight in mice. Genotype BV EPD BB 20 g +10 g Bb 0 g 0 g bb -20 g -10 g EPD = ½BV ( Expected Progeny Difference )

8 II. Breeding Value Example using a Single-Locus Model: B = 10 g; b = -10 g; complete dominance; trait is mature body weight in mice. Genotype BV (G ) BB 20 g 20 g Bb 0 g 20 g bb -20 g -20 g When there is complete dominance, genotypic values (G) are the same for BB and Bb. Further, G is the value of an individual s genes to its own performance.

9 II. Breeding Value Definition for EPD: Half an animal s estimated breeding value the expected difference between the mean performance of the individual s progeny and the mean performance of all progeny (assuming randomly chosen mates). ^ ^ PD = ½BV ^ ^ ^ BV S + BV D BV O = 2 (An interim estimate until an animal s own records become available [e.g., ET calves].)

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12 III. Gene Combination Value Definition: The part of an animal s genotype (G) that is due to the effects of gene combination (dominance and epistasis) and cannot, therefore, be transmitted from parent to offspring. G = BV + GCV BV is the transmissible part of G, whereas GCV is the non-transmissible part of G.

13 III. Gene Combination Value Example using a Single-Locus Model: B = 10 g; b = -10 g; complete dominance; trait is mature body weight in mice. Genotype BV G GCV BB 20 g 20 g 0 g Bb 0 g 20 g 20 g bb -20 g -20 g 0 g {G = BV + GCV or GCV = G - BV}

14 III. Gene Combination Value Genotype BV G GCV BB 20 g 20 g 0 g Bb 0 g 20 g 20 g bb -20 g -20 g 0 g Comments: Homozygotes breed as good as or poorly as they look, whereas Heterozygotes tend to look better than they breed! Further, seedstock breeders rely on and actively promote BV s, whereas commercial producers market performance based on G (BV is important, while GCV plays a major role in heterosis expression, a benefit of crossbreeding).

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17 All genes have independent effects (BV), but genes may also contribute to GCV effects. III. Gene Combination Value Comments: GCV O ½ (GCV S + GCV D ) GCV is also called the non-additive genetic value, whereas BV is the additive genetic value.

18 Review: P = μ + G + E G = BV + GCV G = BV + D + I BV factors into heritability (Chapter 9) GCV factors (D (dominance) and I (epistasis)) into heterosis (Chapter 18)

19 Genetic and environmental influences on livestock traits: P = μ + BV + GCV + E Repro/ Health Production Yield Heritability Heterosis Environment (BV) (GCV) (E) Low Most High (0-15%) Moderate Some Moderate (15-40%) High (>40%) Least Low

20 Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + G + E; {G = BV + GCV} (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G E P AABBCCDD AABbccDD aabbccdd

21 Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + G + E; {G = BV + GCV} (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G E P AABBCCDD +24 AABbccDD +6 aabbccdd -12

22 Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + G + E; {G = BV + GCV} (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G E P AABBCCDD AABbccDD aabbccdd -12 0

23 Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + G + E; {G = BV + GCV} (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G E P AABBCCDD AABbccDD aabbccdd

24 Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + G + E; {G = BV + GCV} (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G E P AABBCCDD AABbccDD aabbccdd

25 Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + G + E; {G = BV + GCV} (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G E P AABBCCDD AABbccDD aabbccdd

26 Genotype BV GCV G E P AABBCCDD AABbccDD aabbccdd ) Which individual lays the most eggs? Explain why. 2) Which individual as a parent would be expected to produce offspring that lay the most eggs? 3) Give possible reasons why hen #2 has an E value of -10, while hen #3 has an E value of +12.

27 Example 2: In mature, male Proghorn antelope, figure that horn length is only affected by three loci. The horn measurement is the total length of one horn. Assume symmetry with regards to the genetic model is as follows: P = μ + BV + GCV + E (μ = 6 inches) Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for dominant homozygotes and heterozygotes. The independent effect of each gene symbol which is capitalized is +2 inches. The independent effect of each gene symbol which is not capitalized is +1 inches. Genotype BV GCV G E P AABBCC +12 AaBbCc +9 AaBBCc +10 Aabbcc +7

28 Example 2: In mature, male Proghorn antelope, figure that horn length is only affected by three loci. The horn measurement is the total length of one horn. Assume symmetry with regards to the genetic model is as follows: P = μ + BV + GCV + E (μ = 6 inches) Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for dominant homozygotes and heterozygotes. The independent effect of each gene symbol which is capitalized is +2 inches. The independent effect of each gene symbol which is not capitalized is +1 inches. Genotype BV GCV G E P AABBCC AaBbCc AaBBCc Aabbcc +7 +8

29 Example 2: In mature, male Proghorn antelope, figure that horn length is only affected by three loci. The horn measurement is the total length of one horn. Assume symmetry with regards to the genetic model is as follows: P = μ + BV + GCV + E (μ = 6 inches) Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for dominant homozygotes and heterozygotes. The independent effect of each gene symbol which is capitalized is +2 inches. The independent effect of each gene symbol which is not capitalized is +1 inches. Genotype BV GCV G E P AABBCC AaBbCc AaBBCc Aabbcc

30 Example 2: In mature, male Proghorn antelope, figure that horn length is only affected by three loci. The horn measurement is the total length of one horn. Assume symmetry with regards to the genetic model is as follows: P = μ + BV + GCV + E (μ = 6 inches) Complete dominance at all loci. No epistasis. Genotypic values (G) are equal for dominant homozygotes and heterozygotes. The independent effect of each gene symbol which is capitalized is +2 inches. The independent effect of each gene symbol which is not capitalized is +1 inches. Genotype BV GCV G E P AABBCC AaBbCc AaBBCc Aabbcc

31 Example 2: In mature, male Proghorn antelope, figure that horn length is only affected by three loci. The horn measurement is the total length of one horn. Assume symmetry with regards to the genetic model is as follows: P = μ + BV + GCV + E (μ = 6 inches) Genotype BV GCV G E P AABBCC AaBbCc AaBBCc Aabbcc ) Which individual has the longest horns? Explain why. 2) Which individual as a parent would be expected to produce offspring that has the longest horns? 3) Give possible reasons why animal #2 has an E value of +2, while animal #3 has an E value of -8 inches.

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33 Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E; μ = 600 lbs). Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs for loci A through D, but also for each e gene at the E locus; The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs; Complete dominance exists only at loci B and D; and Epistasis exists only between genotypes aa and EE, which is -20 lbs. Genotype BV D I G E P AABBCCDDEE +68 AaBbCcDdEe +28 AABbCCDDEE +54 aabbccddee -30

34 Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E; μ = 600 lbs). Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs for loci A through D, but also for each e gene at the E locus; The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs; Complete dominance exists only at loci B and D; and Epistasis exists only between genotypes aa and EE, which is -20 lbs. Genotype BV D I G E P AABBCCDDEE AaBbCcDdEe AABbCCDDEE aabbccddee -30 0

35 Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E; μ = 600 lbs). Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs for loci A through D, but also for each e gene at the E locus; The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs; Complete dominance exists only at loci B and D; and Epistasis exists only between genotypes aa and EE, which is -20 lbs. Genotype BV D I G E P AABBCCDDEE AaBbCcDdEe AABbCCDDEE aabbccddee

36 Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E; μ = 600 lbs). Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs for loci A through D, but also for each e gene at the E locus; The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs; Complete dominance exists only at loci B and D; and Epistasis exists only between genotypes aa and EE, which is -20 lbs. Genotype BV D I G E P AABBCCDDEE AaBbCcDdEe AABbCCDDEE aabbccddee

37 Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E; μ = 600 lbs). Assume the following: The independent effect of each gene symbol which is capitalized is +10 lbs for loci A through D, but also for each e gene at the E locus; The independent effect of each gene symbol which is not capitalized is -4 lbs for loci A through D, except for each E gene at the E locus which is -6 lbs; Complete dominance exists only at loci B and D; and Epistasis exists only between genotypes aa and EE, which is -20 lbs. Genotype BV D I G E P AABBCCDDEE AaBbCcDdEe AABbCCDDEE aabbccddee

38 Example 3: Consider a hypothetical quantitative trait (e.g., calf weaning weight) affected by five loci. (Genetic model: P = μ + BV + D + I + E; μ = 600 lbs). Genotype BV D I G E P AABBCCDDEE AaBbCcDdEe AABbCCDDEE aabbccddee ) Which individual has the heaviest weight? Explain why. 2) Which individual as a parent would be expected to produce offspring that has the heaviest weight? 3) Give possible reasons why animal #3 has an E value of +22 lbs, while animal #2 has an E value of -15 lbs.

39 IV. Producing Ability Definition: Producing Ability (PA) - The performance potential of an individual for a repeated trait. Repeated trait A trait for which individuals commonly have more than one performance record (e.g., antler score, ease of birthing, egg production, fertility, litter size, milk production, race performance, and wool yield).

40 IV. Producing Ability P = μ + BV + GCV + E p + E t {E = E p + E t } ^ PA = G + E p where E p = An environmental effect that permanently influences an individual s performance for a repeated trait (e.g., an animal born in a severe drought year or in an ample rainfall year), and E t = An environmental effect that only influences a single performance record (e.g., a muddy race track or a track in exceptionally ideal condition).

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42 IV. Producing Ability - Problem 7.7 Consider the Thoroughbred stallions: Raise-A- Ruckus and Presidium. Raise-A-Ruckus s BV for racing time is -8 seconds. He was particularly well trained, having a permanent environmental effect of -6 seconds. Presidium s BV for racing time is -12 seconds, but his permanent environmental effect is +2 seconds. Assume both horses have GCV s of 0. Answer the following: a. Calculate progeny difference (EPD) for each horse {EPD = ½BV}. R R: ½(-8) = -4 sec Presidium: ½(-12) = -6 sec b. Calculate producing ability (PA) for each horse {PA = BV + GCV + Ep}. R R: PA = (-6) = -14 sec Presidium PA = = -10 sec c. Which horse would you bet on in a race? Why? R R because he has the better PA (-14 vs. -10 seconds) (i.e., 4 sec faster). d. Which horse would you breed mares to? Why? Presidium because he has the better BV (-6 vs. -4 seconds) (i.e., his progeny are predicted to be faster by 2 sec).

43 Genotype BV GCV G Ep Et P PA AABBCCDD +7-3 AABbccDD -6-4 aabbccdd Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + BV + GCV + E p + E t ; PA = BV + GCV + E p (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values.

44 Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + BV + GCV + E p + E t ; PA = BV + GCV + E p (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G Ep Et P PA AABBCCDD AABbccDD aabbccdd

45 Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + BV + GCV + E p + E t ; PA = BV + GCV + E p (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G Ep Et P PA AABBCCDD AABbccDD aabbccdd

46 Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + BV + GCV + E p + E t ; PA = BV + GCV + E p (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G Ep Et P PA AABBCCDD AABbccDD aabbccdd

47 Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + BV + GCV + E p + E t ; PA = BV + GCV + E p (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G Ep Et P PA AABBCCDD AABbccDD aabbccdd

48 Example 1: Consider that egg production in Bantams is affected by only four loci. Genetic model: P = μ + BV + GCV + E p + E t ; PA = BV + GCV + E p (μ = 60 eggs) Complete the table below. Assume the following: Complete dominance at all loci. No epistasis. The independent effect of each gene symbol which is capitalized is +3 eggs. The independent effect of each gene symbol which is not capitalized is -3 eggs. For homozygous combinations, genotypic values are equal to breeding values. Genotype BV GCV G Ep Et E P PA AABBCCDD AABbccDD aabbccdd

49 Genotype BV GCV G Ep Et P PA AABBCCDD AABbccDD aabbccdd ) Which individual lays the most eggs? P 2) Which individual as a parent would be expected to produce offspring that lay the most eggs? BV 3) Which hen is predicted to lay more eggs in the future with regards to her producing ability? PA

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