1. Genetics – Part 2 Quantitative Traits Individual Performance BW 72 205
205
700
100 ratio
365
1,375 SC 40
18 Mo.
BW=birth weight, 205=205 day weight, sc=scotal circumfirence, epd=expected progeny difference, acc=accurancy, ced=calving ease direct (how easy this calves are born, bw=birth weight, ww=weaning weight, YW=yearling weight, MCE=maternal calving ease, milk=milking ability of daughter expressed as increased weaning weight of calf, MWW=maternal weaning weight=1/2 x WW +Milk, Stay = The expected difference in probability of daughter staying in the herd to at least the age of six years.
TRAIT
CED BW WW YW
MCE
MILK
MWW
STAY
EPD
+5.7
+3.1
+62.1
+114.9
+6.4
+2.2
+33.2
+19.1
ACC
.44
.81
.76
.74
.18
.60
.61
.19

2. Quantitative Traits
Genotype + Environment = Phenotype
The phenotype is what we see or measure.
Genotype: Total effect of the animal’s genetics

3. What are examples of environmental effects?
Environment: the cause of any phenotypic variation that can not be attributed to genetics. Environmental effects randomly affect specific animals.
What are examples of environmental effects?
Examples of environmental effects include: weather, temperature, stress, feed   quality, error in measurements, random exposure to disease, lightning strike,   training, human-animal interactions, etc.

4. Example: Milk production for a lactation.
Each upper case allele is worth +500 lbs. milk such that:
“A,” “B,” or “C” = +500 lbs. and
“a,” “b,” or “c” = +0 lbs.

5. Assume 7 different individuals with genetic values ranging from 21,000 to 24,000 lbs. of milk
Genotype
Lbs. Milk
Environment
Phenotype
AA BB CC:
24,000 + =
Aa BB CC:
23,500
Aa Bb CC:
23,000
Aa Bb Cc:
22,500
aa Bb Cc:
22,000
aa bb Cc:
21,500
aa bb cc:
21,000

6. We want to rank these cows on lbs. milk but we do not know the genotype. Selection is based on phenotype.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 + =
Elle AaBBCC:
23,500
Latte AaBbCC:
23,000
Betz AaBbCc:
22,500
May aaBbCc:
22,000
Queen aabbCc:
21,500
Dee aabbcc:
21,000

7. Ada had a very difficult calving in the middle of the night and got off to a very slow start.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
Latte AaBbCC:
23,000
Betz AaBbCc:
May aaBbCc:
22,000
Queen aabbCc:
21,500
Dee aabbcc:
21,000

8. Elle freshened during cool weather and lush pastures
Elle freshened during cool weather and lush pastures. She also had the cleanest barn stall.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
4500
28,000
Latte AaBbCC:
23,000
Betz AaBbCc:
May aaBbCc:
22,000
Queen aabbCc:
21,500
Dee aabbcc:
21,000

9. There was an error in the milk recording that incorrectly gave Latte an extra 1000 lbs.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
4500
28,000
Latte AaBbCC:
23,000
1000
Betz AaBbCc:
May aaBbCc:
22,000
Queen aabbCc:
21,500
Dee aabbcc:
21,000

10. Betz had really bad pneumonia as a calf which led to permanent lung damage.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
4500
28,000
Latte AaBbCC:
23,000
1000
Betz AaBbCc:
-3500
19,000
May aaBbCc:
22,000
Queen aabbCc:
21,500
Dee aabbcc:
21,000

11. May was a favorite “show” animal so was fed premium feed and talked nice to each day.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
4500
28,000
Latte AaBbCC:
23,000
1000
Betz AaBbCc:
-3500
19,000
May aaBbCc:
22,000
4000
26,000
Queen aabbCc:
21,500
Dee aabbcc:
21,000

12. Queen freshened in really hot weather which led to an extended bout of ketosis.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
4500
28,000
Latte AaBbCC:
23,000
1000
Betz AaBbCc:
-3500
19,000
May aaBbCc:
22,000
4000
26,000
Queen aabbCc:
21,500
18,000
Dee aabbcc:
21,000

13. Dee had many environmental influences; good and bad that averaged out to “-500”.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
22,500
Elle AaBBCC:
23,500
4500
28,000
Latte AaBbCC:
23,000
1000
Betz AaBbCc:
-3500
19,000
May aaBbCc:
22,000
4000
26,000
Queen aabbCc:
21,500
18,000
Dee aabbcc:
21,000
-500
20,500

14. Note that the cows do not rank the same phenotypically as they did genetically.
Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
4th 22,500
Elle AaBBCC:
23,500
4500
1st 28,000
Latte AaBbCC:
23,000
1000
3rd 24,000
Betz AaBbCc:
22,500
-3500
5th 19,000
May aaBbCc:
22,000
4000
2nd 26,000
Queen aabbCc:
21,500
7th 18,000
Dee aabbcc:
21,000
-500
6th 20,500
This is what makes animal breeding frustrating and challenging in that the animals with the best phenotypes may not be the best genetically.

15. Genotype
lbs. Milk
Environment
Phenotype
Ada AABBCC:
24,000 +
-1500 =
4th 22,500
Elle AaBBCC:
23,500
4500
1st 28,000
Latte AaBbCC:
23,000
1000
3rd 24,000
Betz AaBbCc:
22,500
-3500
6th 19,000
May aaBbCc:
22,000
4000
2nd 26,000
Queen aabbCc:
21,500
7th 18,000
Dee aabbcc:
21,000
-500
5th 20,500
This is what makes animal breeding frustrating and challenging in that the animals with the best phenotypes may not always be the best genetically.
In the above example there is a correlation between the genotype and phenotype of .5 which corresponds to a heritability of .25.

16. Heritability
B. Heritability = h2 = % of differences between animals in the “same” environment that is due to genetics. (The animals are in the same herd, but each animal has its own individual environment.)
h2 is the symbol for heritability. It is not heritability squared.

17. Heritability Category of Traits Average h2 Conformation 35-50%
Performance
20-35%
Disease Resistance
5-15%
Reproduction
1-5%

19. Example:
There are a group of bulls raised in a pen. The average daily gain (ADG) is 3.5 lbs. The top bull gains 4.5 lbs. per day. h2 for ADG is What % of the top bulls superiority for ADG would be expected to be due to genetics?

20. Example:
There are a group of bulls raised in a pen. The average daily gain (ADG) is 3.5 lbs. The top bull gains 4.5 lbs. per day. h2 for ADG is What % of the top bull’s superiority for ADG would be expected to be due to genetics?
Answer: 25% - - h2 is the differences between animals that is due to genetics.

21. Example:
There are a group of bulls raised in a pen. The average daily gain (ADG) is 3.5 lbs. The top bull gains 4.5 lbs. per day. h2 for ADG is .25. What is this bull’s genetic value for ADG?

22. There are a group of bulls raised in a pen
There are a group of bulls raised in a pen. The average daily gain (ADG) is 3.5 lbs. The top bull gains 4.5 lbs. per day. h2 for ADG is .25. What is this bull’s genetic value for ADG?
The bull is +1 lb. better than penmates.
In animal breeding we always deviate records from pen or herdmates to remove as much of the environmental influences as possible.

23. Bull’s difference from herd mates.
+ 1 lb.{

24. } } + 1 lb.{ 75% due to the environment 25% due to genetics
Bull’s difference from herd mates. }
75% due to the environment
+ 1 lb.{ }
25% due to genetics

25. } } + 1 lb.{ 75% due to the environment 25% due to genetics
Bull’s difference from herd mates. }
75% due to the environment
+.75 lb
+ 1 lb.{ }
25% due to genetics
+.25 lb

26. Bull’s genetic value for ADG = +.25 lb.}
75% due to the environment
+.75 lb
+ 1 lb.{ }
25% due to genetics
+.25 lb

27. +1 lb. x heritability of .25 = +.25 lb genetic superiority for ADG}
75% due to the environment
+.75 lb
+ 1 lb.{ }
25% due to genetics
+.25 lb

28. Genetic Value = h2 x phenotypic deviation.

29. Genetic Value = h2 x phenotypic deviation.
Genetic Value = GV
= Estimated Breeding Value = EBV

30. Genetic Value = h2 x phenotypic deviation.
Genetic Value = GV
= Estimated Breeding Value = EBV
Estimated Transmitting Ability = ETA = ½ EBV

31. The bull would be expected to transmit ½ of this genetic superiority to his offspring.}
75% due to the environment
+.75 lb
+ 1 lb.{ }
25% due to genetics
EPD = lb.

32. The bull would be expected to transmit ½ of this genetic superiority to his offspring.}
75% due to the environment
+.75 lb
+ 1 lb.{ }
25% due to genetics
EPD = lb.
EPD = Expected Progeny Differences; terminology use for cattle, swine, sheep.

33. C. Predicting Genetic Change
∆G = h2 x Average Selection Differential
The “average selection differential” is the average phenotypic difference of selected breeding animals from their herdmates.
∆ = Delta = scientific symbol for change

34. C. Predicting Genetic Change ∆G = h2 x Average Selection Differential
Example: A beef producer wants to select for increased yearling weight. Current adjusted average is 1050 lbs. Selected bulls average Selected heifers average Heritability is .3
Adjusted average means that we have adjusted for gender differences, age of dam, etc.

35. Change in genetic value = .3 x (50 + 10)/2 = 9 lbs.
∆G = h2 x Average Selection Differential
Example: A beef producer wants to select for increased yearling weight. Current adjusted average is 1050 lbs. Selected bulls average Selected heifers average Heritability is .3
Change in genetic value = .3 x ( )/2 = 9 lbs.
Next generation, you should expect yearling weights to be 1059 lbs. ( )

37. Review Problem 1
A bull is heterozygous for the polled gene (Pp). He is mated to a herd of cows where the frequency of the polled gene is .2. What % of the offspring would be expected to be polled?

38. .8 p Pp pp .2 P PP Pp .5 P .5 p Review Problem 1
A bull is heterozygous for the polled gene (Pp). He is mated to a herd of cows where the frequency of the polled gene is .2. What % of the offspring would be expected to be polled?
.5 P p
.2 P PP Pp
.8 p Pp pp

39. .6 or 60% of the offspring would be expected to be polled.
.5 P p
.2 P PP Pp
.8 p Pp pp

40. 40% would be horned.

41. What is the frequency of the gene?
Review problem 2
Suppose that 1 out of 400 animals are born with a recessive genetic defect.
What is the frequency of the gene?
What is the frequency of the carriers?
The photo is from the Pipestone, MN vet clinic. Notice the crooked front legs in this lamb. Spider Lamb Syndrome is a genetic disease. Lambs are normal at birth but quickly develop limb deformities.

42. What is the frequency of the gene? = q
Genotype: NN Nn nn
Frequency: p pq + q2 = 1
Suppose that 1 out of 400 animals are born with a recessive genetic defect. 1/400 = = q2
What is the frequency of the gene?
= q
What is the frequency of the carriers? = 2pq

43. What is the frequency of the gene? = q = .0025 = .05 or 5%
Genotype: NN Nn nn
Frequency: p pq q2 = 1
Suppose that 1 out of 400 animals are born with a recessive genetic defect. 1/400 = = q2
What is the frequency of the gene?
= q = = .05 or 5%
What is the frequency of the carriers? = 2pq = 2 x .95 x .05 = .095
or 9.5%

44. Review problem 3
You are a seed stock producer selecting for leaner hogs. Boars selected are .2” leaner than average. Gilts are .1” leaner than average. Average back fat thickness is 1.1” h2 = .4.
What is the expected
fat thickness of the next
generation?

45. Review problem 3
You are a seed stock producer selecting for leaner hogs. Boars selected are .2” leaner than average. Gilts are .1” leaner than average. Average back fat thickness is 1.1”
h2 = .4.
∆G = h2 x Average Selection Differential
What is the expected fat thickness of the next generation?

46. Boars selected are .2” leaner than average.
Gilts are .1” leaner than average. Average back fat thickness is 1.1”
h2 = .4.
∆G = h2 x Average Selection Differential
= .4 x (-.2” + -.1”)/2 = .4 x -.15” = -.06”
The next generation would be expected to be -.06” leaner than the previous.

47. Which boar would you choose?
IV. Methods of Selection When Selecting for Multiple Traits.
Boar
Backfat
ADG A
0.75
2.35 B
0.80
2.42 C
0.83
2.28 D
0.86
2.58 E
0.90
2.23 F
0.94
2.55 G
0.96
2.45
Average
2.41
Which boar would you choose?

48. IV. Methods of Selection When Selecting for Multiple Traits.
A. Tandem – one behind the other; in single file.

49. IV. Methods of Selection When Selecting for Multiple Traits.
A. Tandem –
1. Decide which trait is most important.
2. Select for that trait until desired performance is achieved. Then the second trait is selected for.

50. IV. Methods of Selection When Selecting for Multiple Traits.
A. Tandem –
1. Decide which trait is most important.
2. Select for that trait until desired performance is achieved. Then the second trait is selected for.
In the boar selection case, let’s say that backfat is the more important trait.

51. Which boar would you choose?
Boar A would be our choice. Unfortunately, he is below average for ADG.
Boar
Backfat
ADG A
0.75
2.35 B
0.80
2.42 C
0.83
2.28 D
0.86
2.58 E
0.90
2.23 F
0.94
2.55 G
0.96
2.45
Average
2.41
There is a genetic correlation or .35 between backfat and ADG in pigs – meaning that faster growing pigs tend to be fatter.

52. Tandem – Key Points
1. Only one trait is select for at a time.
2. Simple and easy to understand.
3. If there is a genetic antagonism between traits, as progress is being made in one, the others may be getting worse.

53. B. Independent Culling Levels
1. Set minimum standards for each trait.
For example – above average in all traits.
2. Adjust standards for each trait to have a desired number of animals making the standards.

54. Let’s apply the minimum standard that the animals must be better than average in all traits.
Boar
Backfat
ADG A
0.75
2.35 B
0.80
2.42 C
0.83
2.28 D
0.86
2.58 E
0.90
2.23 F
0.94
2.55 G
0.96
2.45
Average
2.41
There is a genetic correlation or .35 between backfat and ADG in pigs – meaning that faster growing pigs tend to be fatter.

55. Let’s apply the minimum standard that the animals must be better than average in all traits.
Boar
Backfat
ADG A
0.75
2.35 B
0.80
2.42 C
0.83
2.28 D
0.86
2.58 E
0.90
2.23 F
0.94
2.55 G
0.96
2.45
Average
2.41 X X
There is a genetic correlation or .35 between backfat and ADG in pigs – meaning that faster growing pigs tend to be fatter. X X

56. Let’s apply the minimum standard that the animals must be better than average in all traits.
Boar
Backfat
ADG A
0.75
2.35 B
0.80
2.42 C
0.83
2.28 D
0.86
2.58 E
0.90
2.23 F
0.94
2.55 G
0.96
2.45
Average
2.41 X X X X X
There is a genetic correlation or .35 between backfat and ADG in pigs – meaning that faster growing pigs tend to be fatter. X X

57. B is the selected boar in that he meets the selected criteria for both traits.
Backfat
ADG A
0.75
2.35 B
0.80
2.42 C
0.83
2.28 D
0.86
2.58 E
0.90
2.23 F
0.94
2.55 G
0.96
2.45
Average
2.41 X X X X X
There is a genetic correlation or .35 between backfat and ADG in pigs – meaning that faster growing pigs tend to be fatter. X X

58. Index = (ADG + -2 x Backfat) x 10
C. Selection Index – weighting is given to each trait based on economics, h2, and genetic correlation with other traits.
A swine index once used by the swine industry was similar to:
Index = (ADG + -2 x Backfat) x 10
Time does not allow us to go into details here as far as how these exact weightings in the index are determined.

59. Boar
Backfat
ADG
Index A
0.75
2.35
8.5 B
0.80
2.42
8.2 C
0.83
2.28
6.2 D
0.86
2.58
8.6 E
0.90
2.23
4.3 F
0.94
2.55
6.7 G
0.96
2.45
5.3
Average
2.41

60. The Selection Index allows us to rank all animals from top to bottom based on economic worth.

61. Boar
Backfat
ADG
Index D
0.86
2.58
8.6 A
0.75
2.35
8.5 B
0.80
2.42
8.2 F
0.94
2.55
6.7 C
0.83
2.28
6.2 G
0.96
2.45
5.3 E
0.90
2.23
4.3
Average
2.41
6.8
D is the top ranked boar, followed by A,B,F,etc.

62. D. Combination of methods – in practice many people use a combination of the above mentioned methods. The animals may be ranked on an index combining several traits but than a minimum standard set for another trait.

63. Leg Boar Backfat ADG Index Score D 0.86 2.58 8.6 4 A 0.75 2.35 8.5 2 B
0.80
2.42
8.2 5 F
0.94
2.55
6.7 1 C
0.83
2.28
6.2 G
0.96
2.45
5.3 E
0.90
2.23
4.3
Average
2.41
6.8
Leg score refers to leg soundness score. Pigs scored 1 or 2 may have mobility problems in the future.

64. Leg Boar Backfat ADG Index Score D 0.86 2.58 8.6 4 A 0.75 2.35 8.5 2 B
A and F are eliminated because of poor leg soundness score.
Leg
Boar
Backfat
ADG
Index
Score D
0.86
2.58
8.6 4 A
0.75
2.35
8.5 2 B
0.80
2.42
8.2 5 F
0.94
2.55
6.7 1 C
0.83
2.28
6.2 G
0.96
2.45
5.3 E
0.90
2.23
4.3
Average
2.41
6.8
A common maximum standard is price of the animal. D might be out of our price range.

65. V. Basis for Selection
A. Pedigree – Selection based on the performance or estimated transmitting abilities of ancestors.
Joey and Ben are two young calves that we have no information on except for their pedigree.

66. Joey’s pedigree index for PTA milk is +750.
V. Basis for Selection
A. Pedigree – Selection based on the performance or estimated transmitting abilities of ancestors.
Joey’s pedigree index for PTA milk is +750.
Ben’s is
PTA milk
Joey and Ben are two young calves that we have no information on except for their pedigree.
PTA milk
Ben
PTA milk
PTA milk

67. Advantages:
1. selection is early in animal’s life
2. very cheap
Disadvantages:
  1. relatively low accuracy because you do not know what sample of genes you may get from each
parent. For example, there can be a lot of genetic variation between full sibs.

68. B. Individual performance (mass selection)
Selection based on phenotypes of individuals

69. Advantages:
1. do not need pedigrees
2. may be able to measure large
numbers of animals
Disadvantages:
1. not accurate if heritability is low
2. traits must be measured on
each individual.

70. C. Progeny Testing -
Selection based on performance of progeny

71. Advantages:
1. can be very accurate – can achieve 99% accuracy if there are enough progeny. This is true even for lowly heritable traits.
2. can be used for sex-limited traits

72. Disadvantages:
1. time consuming and expensive – takes 5 years and may cost $60,000 to progeny test one dairy bull
2. a limited number of individuals can be progeny tested – There are more than 9,000,000 dairy cows in the U.S. but only about 1500 bulls are progeny tested each year.

73. D. Calculation of Estimated Breeding Values (EBV’s)
An estimated breeding value is calculated for each individual using information on all relatives.

74. Generally report as transmitting abilities, (transmitting ability = ½ BV)
Terminology for different species:
ETA’s: poultry, some breeds of horses and dogs
PTA’s: dairy, dairy goats
EPD’s: beef, swine, sheep

75. ETA’s: poultry, some breeds of horses and dogs
ETA = Estimated Transmitting Ability

76. PTA’s: dairy, dairy goats
PTA = Predicted Transmitting Ability

77. EPD’s: beef, swine, sheep
EPD = Expected Progeny Differences

78. Advantages:
Most accurate way to utilize all information available
Disadvantages:
Need sophisticated computer program
Need excellent pedigree records

79. Genomics – refers to using genetic markers or single nucleotide polymorphisms to help estimate an animal’s genetic value.

80. Genetic marker: string of base pairs (A, C, G, T) associated with a trait

81. Single nucleotide polymorphisms (SNPS):
a single DNA test can identify differences in 1000’s of single bases spread across the genome.

82. Genotype Data for Elevation Chromosome 1
0 = AA, 1 = AT, 2 = TT
0 = homozygous for first allele ; 1 = heterozygous ; 2 = homozygous for second allele
This shows 864 alleles. (out of over 2000) If we were to show all of the base pairs for the first chromosome, I would have to show you 1 slide, per second for over 3 days.
Vast portions of the chromosome are homozygous, and not interesting,, ie all animals have the same DNA sequence. Looking at a foundation bull such as elevation shows many alleles are heterozygous.
Pedigree analysis by Holstein Association 956 bulls in the April 2008 sire guide red book had 15.2% genes in common with Elevation. Seven generations back, showed on average Elevation appearing 3.5 times in these bulls pedigrees. 8 generations 2 times. Nine generations showed Elevation an average of once per bull.
So what does this really mean?

83. SNP Effects for Milk

84. Advantages of Genomics:
Can add to the accuracy of other methods.
Can be used to estimate genetic values on traits not usually measured
(feed efficiency, disease resistance)
3. DNA test can be done at young age.

85. Disadvantages:
Because of cost, use limited many to dairy and beef industries
Need a lot of data to accurately estimate “SNP” effects.

86. Thumbelina, a miniature mare at 17
Thumbelina, a miniature mare at 17.5 inches at the withers (2006) and Radar, a Belgian, 79.5 inches. Tallest and shortest horses.

87. The End – Genetics Part 2