1. Genetic 3 Mating Systems
Parents have been selected. Which male do you mate with which female.

2. We will cover the following types of mating systems in this section -
Inbreeding:
Intense inbreeding
Line breeding
Outbreeding:
Species cross
Crossbreeding
Outcrossing
Grading up
Assortative mating:
Corrective mating:

3. A. Inbreeding – parents are related, having common ancestors in their pedigrees back a set number of generations or to an ancestral base year. A S T E x F D A H

4. A. Inbreeding – parents are related, having common ancestors in their pedigrees back a set number of generations or to an ancestral base year. A
In the example to the right, we are just going back two generations from the parents. S T E x F D A H

5. A. Inbreeding – parents are related, having common ancestors in their pedigrees back a set number of generations or to an ancestral base year.
To calculate inbreeding for the dairy genetic evaluations in the U.S., we go back to an ancestral base year of 1960. A S T E x F D A H

6. A. Inbreeding – parents are related, having common ancestors in their pedigrees back set number of gnerations or to an ancestral base year. A
The sire and dam
are related in this pedigree because they are both descended from “A”. S T E x F D A H

7. FX = inbreeding coefficient = the average decrease in genetic heterozygosity due to the relationship of the parents.
Genetic homozygosity is when the alleles at a loci are identical
(AA and aa vs. Aa).

8. 1. Two different ways to measure inbreeding
a. From the known relationship of the parents if common ancestor is not inbred.
Mating
FX (%)
individual x parent 25
full sibs
half sibs
12.5
individual x grandparent
son of sire x granddaughter of sire
6.25
grandson x granddaughter of sire
3.12
If the common ancestor is inbred that will increase the inbreeding coefficient but that adjustment is beyond the scope of this course.

9. This example is a son of sire x granddaughter of sire so the inbreeding coefficient is 6.25%.

10. x Again, it is how the parents are related that makes “X” inbred. S D
This example is a son of sire x granddaughter of sire so the inbreeding coefficient is 6.25%. A
Again, it is how the parents are related that makes “X” inbred. S T E x F D A H

11. What is the Fx in this pedigree?Z S A E x F D A H

12. What is the Fx in this pedigree?Z S
Fx = 3.12% A E x F D A H

13. b. From the bracket pedigrees.
Works on the principle that the amount of inbreeding due to a common non-inbred ancestor is dependent on total number of generations that the common ancestor is back from the parents.
Number of generations
% inbreeding 1 25 2
12.5 3
6.25 4
3.12 5
1.56 6
.78

14. Example 1. A X A D

15. First draw a circle around the parents.X A D

16. First draw a circle around the parents.X A D

17. The common ancestor is “A”X A D

18. “A” is zero generations from the father and 1 from the mother.A 1 X A D

19. Number of generations
% inbreeding 1 25 2
12.5 3
6.25 4
3.12 5
1.56 6
.78

20. “A” is zero generations from the father and 1 from the mother.A 1 X A D
A total of 1 generations back equates to an Fx of 25% according to the chart.

21. Example 2. A S X A F D

22. Example 2. 1 A 2 S X A F D

23. Number of generations
% inbreeding 1 25 2
12.5 3
6.25 4
3.12 5
1.56 6
.78

24. Example 2. 1 A 2 S X A F D
1 + 2 = 3 generations back = 6.25%

25. Example 3. Multiple common ancestors.B X A D B

26. A S X D B A B A: 2 gen.=12.5 B: 4 gen.= 3.12 Fx =15.62
Example 3. Multiple common ancestors. A S B X A
A: 2 gen.=12.5 B: 4 gen.= 3.12
Fx =15.62 D B

27. Example 4. A S A M X F
Fx= D B G

28. Example 4. A S A M X F
Fx= 0
The two parents are not related. D B G

29. A S X D A F B G Fs = 25% M Fx= 0 The two parents are not related.
Example 4.
Fs = 25% A S A M X F
Fx= 0
The two parents are not related. D B G

30. Example 5. A Z S L M X F
Fx= D Z B G

31. Example 5. A Z S L M X F D Z B
Fx= .78% G

32. Example 6. A Z S B M X F D Z B
Fx= G

33. A S X D B F B G Z M Z Fx= 3.12% Example 6.
Inbreeding is only due to B. Z is accounted for through B. G

35. 4. Effects of Inbreeding
a. detrimental to reproduction, health, and vigor
This is called inbreeding depression.
Inbreeding depression does not affect all traits equally – mature size and conformation (showring type) are not generally affected.

36. 4. Effects of Inbreeding
b. uncovers undesirable recessive genes

37. A X A D 4. Effects of Inbreeding
b. uncovers undesirable recessive genes A Nn X A Nn D
If “A” is a carrier of a recessive gene (Nn), there is a 12.5% chance “X” will be “nn”.

38. 4. Effects of Inbreeding
c. get heterosis when inbred lines are crossed
Inbred lines have been developed in corn and poultry for subsequent crossing.

39. The Hardy-Weinberg equation gives us the above genotypic frequencies.
d. inbreeding increases homozygosity
e. an example of changes in genotypes after 2 generations of selfing
Gen. AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
At generation 1, we assume that the frequency of the recessive gene (a) is .1 .
The Hardy-Weinberg equation gives us the above genotypic frequencies.

40. Gen. AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
To show the effects of inbreeding, we will assume these are corn plants and do selfing (breed the plants to themselves).

41. The AA and aa genotype reproduce themselves.
Frequency a
% homozygous
% inbreeding
% %
.01
.81
The AA and aa genotype reproduce themselves.

42. Gen. AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
.01
.81
.045
.055
.045
.855
.09
When Aa is mated to itself, 25% of the offspring will be AA, 50% Aa, and 25% aa.

43. The inbreeding coefficient after 1 generation of selfing is 50%.AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
% %
The inbreeding coefficient after 1 generation of selfing is 50%.

44. After 3 generations of selfing, the inbreeding coefficient is 75%. AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
% %
% %
After 3 generations of selfing, the inbreeding coefficient is 75%.
The homozygous genotypes have increased.

45. After 3 generations of selfing, the inbreeding coefficient is 75%. AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
% %
% %
After 3 generations of selfing, the inbreeding coefficient is 75%.
The homozygous genotypes have increased. Notice that the aa genotype has increased by 7.7 times.

46. The frequency of “a” has not changed.
Gen. AA Aa aa
Frequency a
% homozygous
% inbreeding
% %
% %
% %
The frequency of “a” has not changed.
Inbreeding does not change gene frequency!!!!

47. Does inbreeding cause mutations?

48. Does inbreeding cause mutations?
No!

49. B. Linebreeding
1. Trying to concentrate the genes of an outstanding individual
2. Mild inbreeding (usually)
Linebreeding is not practiced in food producing animals because of the potential cost of inbreeding depression. It is practiced some in breeding animals for the show ring.

50. z A Z S L M X z F
Fx= 7.03 D Z B G
“X” has 3/8’s of “Z” genes. “X” is linebred to “Z”.

51. C. Outcrossing
Mating individuals within a breed that are less related than average
Most people mate to avoid inbreeding.

52. D. Crossbreeding
1. Reasons for crossbreeding
a. take advantage of heterosis – the opposite of inbreeding depression.
b. may be able to use breeds in complementary fashion – some breeds strengths and weaknesses may match up well.

53. 2. Definitions
a. The amount of heterosis equals the average of the crossbreds minus the average of the parental breeds.
b. The % heterosis equals the amount of heterosis divided by the average of the parental breeds times 100.

54. Example: Suppose that Holsteins average 25,000 lbs
Example: Suppose that Holsteins average 25,000 lbs. of milk, Jerseys average 17,000, and Jersey-Holstein crossbreds average 22,300 (CA). The parental breeds average 21,000 (PA).
The amount of heterosis is the
CA-PA.

55. CA-PA = 22,300 – 21,000 = 1,300. The % heterosis is
Example: Suppose that Holsteins average 25,000 lbs. of milk, Jerseys average 17,000, and Jersey-Holstein crossbreds average 22,300 (CA). The parental breeds average 21,000 (PA).
The amount of heterosis is the
CA-PA = 22,300 – 21,000 = 1,300.
The % heterosis is

56. The % heterosis is (1,300 / 21,000) x 100 = 6.19%
Example: Suppose that Holsteins average 25,000 lbs. of milk, Jerseys average 17,000, and Jersey-Holstein crossbreds average 22,300 (CA). The parental breeds average 21,000 (PA).
The amount of heterosis is the
CA-PA = 22,300 – 21,000 = 1,300.
The % heterosis is (1,300 / 21,000) x 100 = 6.19%

57. d. Example 2: Suppose the average trainability scores of Black Labs and Golden Retrievers are 6.2 and 6.8, respectively. Black Lab / Golden Retriever crosses average 6.6 for trainability scores. What is the heterosis for trainability score?

58. d. Example 2: Suppose the average trainability scores of Black Labs and Golden Retrievers are 6.2 and 6.8, respectively. Black Lab / Golden Retriever crosses average 6.6 for trainability scores. What is the heterosis for trainability score?
CA = 6.6
PA = 6.5

59. d. Example 2: Suppose the average trainability scores of Black Labs and Golden Retrievers are 6.2 and 6.8, respectively. Black Lab / Golden Retriever crosses average 6.6 for trainability scores. What is the heterosis for trainability score?
Heterosis = 6.6 – 6.5 = +.1

60. d. Example 2: Suppose the average trainability scores of Black Labs and Golden Retrievers are 6.2 and 6.8, respectively. Black Lab / Golden Retriever crosses average 6.6 for trainability scores.
% heterosis = .1 / 6.5 * 100
= 1.54%

61. Example 3: Suppose that Quarter horses are productive until 22 years of age and Arabians are productive until 23 years of age. Arabian/Quarter Horse crosses are productive until 25 years of age. Calculate the % heterosis for length of productive life.

62. Example 3: Suppose that Quarter horses are productive until 22 years of age and Arabians are productive until 23 years of age. Arabian/Quarter Horse crosses are productive until 25 years of age. Calculate the % heterosis for length of productive life.

63. Example 3: Suppose that Quarter horses are productive until 22 years of age and Arabians are productive until 23 years of age. Arabian/Quarter Horse crosses are productive until 25 years of age. Calculate the % heterosis for length of productive life.
(2.5 / 22.5) x 100 = 11.1%

64. Estimate % heterosis estimates for different traits and species
Estimate % heterosis estimates for different traits and species. (results vary depending on specific breeds crossed)
Swine
Litter size born %
Litter size weaned 24%
Daily gain %
Feed efficiency %
Muscling %

65. Beef
Weight of calf / cow bred %
Rate of gain in after weaning %
Feed efficiency %
Carcass traits %

66. Dairy
Milk production %
Calf mortality – 20%
Reproductive traits – 20%

67. 4. Common types of Crossbreeding systems
a. Rotational crosses
- replacement females selected from previous generation
- purebred males used in alternating fashion

68. i. Criss-cross or 2-breed rotation

69. Hamp Boar
York Gilt

70. Hamp Boar
York Gilt
Gilt is ½ Hamp and ½ York. Because her parents were of different breeds, she gets 100% of available heterosis.

71. York Boar
½ Hamp and ½ York Gilt

72. York Boar
½ H ½ Y
1/4 Hamp and 3/4 York Gilt

73. York Boar
½ H ½ Y
1/4 Hamp and 3/4 York Gilt
This gilt will only get 50% of the available heterosis because of the common York shared by her parents.

74. Hamp Boar
1/4 H 3/4 Y
5/8 H 3/8 Y gilt This gilt will have 75% of the available heterosis. She loses 25% because of the ¼ Hamp in her mother.

75. % max heterosis

76. The % of maximum heterosis will average 67% in a two breed rotation.
% max heterosis
The % of maximum heterosis will average 67% in a two breed rotation.

77. ii. 3-breed rotation 86% of max heterosis Hamp York Landrace
3 breed is probably the most popular system of rotational crossbreeding.
Landrace

78. 4-breed rotation
% of maximum heterosis is 94%

79. b. Terminal cross
- all resulting animals go to market or at least are not used for breeding.

80. A pig on the way to market.

81. Three breed terminal cross is the most popular mating system for producing pork!!!

82. Three breed terminal cross is the most popular mating system for producing pork!!!
Yorkshire boar
Landrace female

83. Three breed terminal cross is the most popular mating system for producing pork!!!
Duroc boar
Yorkshire boar
Landrace female
F1 gilt

84. Three breed terminal cross is the most popular mating system for producing pork!!!
Terminal sire
Duroc boar
Yorkshire boar
Landrace female
F1 gilt

85. Three breed terminal cross is the most popular mating system for producing pork!!!
Duroc boar
Yorkshire boar
Landrace female
Market
hogs
F1 gilt

86. 100% of available heterosis
Yorkshire boar
Landrace female
Market
hogs
F1 gilt

87. i. Advantages:
100% of maximum heterosis in mothers and offspring
can use maternal breeds for mothers exclusively – in our example Yorkshires and Landraces are known as maternal breeds because they excel as mothers.

88. ii. Disadvantages:
must maintain purebreds to produce replacement females – this is more expensive than picking females from a rotational cross.

89. What type of mating system is used to produce a Labradoodle?
Most people would consider this a terminal cross.

90. What type of mating system is used to produce a Labradoodle. 2-breed
What type of mating system is used to produce a Labradoodle? 2-breed terminal cross

91. c. Grading Up – permits changing herd from one breed to a different breed without buying females
Example: We own a herd of genetically inferior cows of unknown origin (scrubs). We would like to grade them up to the Angus breed.

92. We use semen from a purebred Angus bull to breed our cows.
c. Grading Up – permits changing herd from one breed to a different breed without buying females
Generation
Sire
Dam
Offspring
Angus (A) scrubs (sc)
We use semen from a purebred Angus bull to breed our cows.

93. c. Grading Up – permits changing herd from one breed to a different breed without buying females
Generation
Sire
Dam
Offspring
Angus (A) scrubs (sc) ½ A ½ sc
The resulting calves are ½ Angus genetics and ½ scrub genetics.

94. c. Grading Up – permits changing herd from one breed to a different breed without buying females
Generation
Sire
Dam
Offspring
Angus (A) scrubs (sc) ½ A ½ sc
A ½ A ½ sc
We save replacement females from the 1st generation and breed them to another purebred Angus bull.

95. c. Grading Up – permits changing herd from one breed to a different breed without buying females
Generation
Sire
Dam
Offspring
Angus (A) scrubs (sc) ½ A ½ sc
A ½ A ½ sc /4 A 1/4 sc
The resulting calves are ¾’s Angus

96. c. Grading Up – permits changing herd from one breed to a different breed without buying females
Generation
Sire
Dam
Offspring
Angus (A) scrubs (sc) ½ A ½ sc
A ½ A ½ sc /4 A 1/4 sc
A ¾ A ¼ sc /8 A 1/8 sc
We continue the same mating plan.

97. c. Grading Up – permits changing herd from one breed to a different breed without buying females
Generation
Sire
Dam
Offspring
Angus (A) scrubs (sc) ½ A ½ sc
A ½ A ½ sc /4 A 1/4 sc
A ¾ A ¼ sc /8 A 1/8 sc
A /8 A 1/8 sc 15/16 A 1/16 sc
By the 4th generation our calves are 15/16’s Angus.

98. Male Horse x Female Donkey = Hinny Male Donkey x Female Horse = Mule
E. Species Cross
1. Horse with donkey
Male Horse x Female Donkey = Hinny
Male Donkey x Female Horse = Mule
Mules are generally sterile.
Mules are generally preferred over hinnys. The donkey male x horse female cross is also more likely to result in a live offspring.

99. 2. European Breeds x Brahmans
There is more heterosis in the initial cross than in subsequent crosses.
Beefmaster breed - a result from crossing Brahmans with Herefords and Shorthorns.

100. 3. European Cattle x Buffalo
F1 males are sterile and F1 females have poor fertility. Beefalo is a breed from this cross but success has been limited.
Genotyping of Beefalos has shown that the buffalo genes tend to get diluted out.

101. 4. Other:
Cattle x Yaks
Yak cross heifer

102. Sheep x goats: seldom go to term, offspring sterile

103. camels x llamas: relatively new

104. Cama with Llama mother

105. F. Assortative mating - breeding like to like
For example: Mate best males to the best females

106. F. Assortative mating - breeding like to like
For example: Mate best males to the best females
or mate the tallest males to the tallest females while mating the short males to the short females.

107. G. Disassortative or Corrective Mating
a. Mating opposites
If the tallest males are mated to the short females and short males are mated to the tall females the offspring should be uniformly moderate in size

108. b. Selecting sires to correct the worst fault in the female

109. Cow toes out on front feet.
b. Selecting sires to correct the worst fault in the female
Cow toes out on front feet.

110. Bull has very straight front legs.
b. Selecting sires to correct the worst fault in the female
Bull has very straight front legs.

111. The end of the genetics chapter