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2 - Foundation of animal health > Week 8- Evaluating Response to Selection and Genetic Gain > Flashcards

Week 8- Evaluating Response to Selection and Genetic Gain Flashcards

1
Q

What do you evaluate the effectiveness of selection

A
  • Change in gene frequency (Δp)

* Expected genetic change (Δ G) in the mean of a population as a result of one generation of selection

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2
Q

Why do you evaluate the effectiveness of selection within-breed selection

A

Selection within breeds is intended to increase the average level of genetic merit (breeding value) of the population.

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3
Q

What are the step involved in within-breed selection

A

• Decide what to improve (define the breeding goal)
• Decide what to measure and select for (determine the selection criteria)
– In some cases the selection criteria will be the same as the breeding goal
• Indirect measurements: when a trait in the breeding goal can only be measured in one sex or after slaughter.
– For example, measures of backfat thickness and ribeye area to
• Design the breeding program
– Decide the number of breeding males and females to be selected each year
– Decide the age at mating
– Mating system to be followed
• Implement the breeding program
– Data recording
• ID numbers, date of birth, trait measurements, etc.
– Evaluate potential breeding animals using EBVs, EPDs, ratios, indexes, etc.
– Mating of selected males and females
– Monitor genetic progress and redesign the breeding program, if necessary

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4
Q

How do you calculate the effectiveness of selection

A 
• Δ G = h2 x SD
– Δ G = expected genetic change
resulting from one generation of selection
– h(square) = heritability of the trait
– SD = selection differential
SD = ps - p
ps = mean of the selected parents
p = mean of the population from which the parent were selected
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5
Q

How are parents selected for breeding

A

Only if their phenotypic value is greater than a truncation point in the frequency- phenotypic values graph

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6
Q

What is selection differential (S)

A

is the phenotypic superiority of selected parents ( the difference in value more than the truncation point)

– Selection differential tells us how superior the parents are to the rest of the population

– Heritability tells us how much of that superiority is genetic

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7
Q

What is selection differential called as well

A

reach,

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8
Q

What is the magnitude of the selection differential depend on

A

depends on proportion of the population saved to be parents
• The smaller the proportion saved, the higher the selection differential
• Selection differential usually larger for male parents than for female parents
• May need only 5% of the males for breeding, whereas
50% of the females.

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9
Q

What are the Three major factors that determine the

amount of genetic response from one generation of selection

A

– Heritability of the trait
– Proportion of population saved to be parents
– Amount of variation in the population

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10
Q

What is the expected fleece weight in the next generation if we assume that the heritability of fleece
weight = 0.60 in this flock

Flock average is 7kg for ewes, 7 for rams
Average selected sheep is 10 and 12 respectively
The effect of herittability if the heritability of fleece weight is only 0.10

A

Average selection differential = (3+5)/2 = 4kg
Delta G = H(square) x SD = (0.6)x4 = 2.4 kg
The expected fleece weight in the next generation will be =7 + 2.4 = 9.4 kg

Effect of inheritability
Delta G = Average selection differential x heritability
= 0.1 x 4 = 0.4kg
Expected fleece weight in next gen = 7 + 0.4 =7.4kg

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11
Q

What can Delta G be use for

A

To predict the average breeding value in the next generation.

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12
Q

What does H(square) (Ps - P) represent

A

the average genetic superiority of selected parents

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13
Q

How do you predicte the average breeding value for the next generation (G^)

A

G^ = G-) + h(square) (ps- p)

G^ = predicted average breeding value for the next generation 
G- = mean breeding value of the herd or population in this generation

Same goes for individual animal

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14
Q

The formula for estimate breeding value(EBV) for an individual animal

A

EBV = p + h(square)(P - p)

p = herd average
P individual’s own phenotype for a trait
h2 = heritability of the trait

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15
Q

What is breeding value

A

value of an animal in a breeding program
• An animal’s genetic superiority or inferiority, idea about type of progeny expected to be produced
• Measure of an animal’s expected progeny performance relative to the population mean (EPD)

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16
Q

Estimation of breeding value
The average weaning weight in a herd of cattle = 500 kg
• A heifer with a weaning weight of 520 kg is selected
• The h2 of weaning weight in this herd = 0.40

A

EBV = 5-00 + 0.4(520 - 500) = 500 + 8 = 508

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17
Q

What is the Accuracy of Selection (rGP)

A
  • Important factor that influences genetic progress due to selection (ΔG)
  • The higher the correlation between the breeding value and the variables on which selection is based, the more accurate is selection and the greater ΔG
  • Accuracy to be as high as possible
  • Accuracy can range from 0.0 (breeding value is guessed) to 1.0 (breeding value is known exactly)

sqroot of heritability

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18
Q

How to increase accuracy of selection (rGP)?

A

• Compare animals under controlled environmental conditions
– Bulls, rams, or boars at a central test station
Use statistical adjustments to remove some of the non-genetic variation
– Correction factors to adjust the records for age of animal, age of dam, sex of the offspring, etc.

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19
Q

How to Reduce environmental sources of variation

A
  • Reduce environmental sources of variation,
  • Increase h2 by reducing the environmental variance in the denominator

h(square) = σG2/(σGsq + σEsq + σGEsq
• High h2 , predicting breeding values from individual phenotypes is more accurate,
• Low h2 is low, lot of errors in predicting breeding values

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20
Q

What are the Aides to Individual Selection

A 
• hsq of a trait is too low, the accuracy of selection can be improved by using
– Measurements on relatives
– Repeated measurements
– Pedigree selection
– Progeny testing
– Measurements of correlated traits
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21
Q

What genes are transmitted to offspring and which does not

A 
• Only the additive portion of the genotype is
transmitted from parent to offspring
– Dominance and epistasis not passed on
to the offspring,
– Gene combinations broken up during
segregation and recombination
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22
Q

Which parents should be selected

A

• Select parents with best additive genetic values or breeding values
– Breeding values are predicted from the available information, such as the performance of the individual animal and the performance of his/her relatives
– Know how close the relationship is between the individual and his relative

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23
Q

What are the relationship in the family tree

A 
Relationship of an individual with...
Relationship
Himself 1.0
One of his parents ½
A full-sib ½
A half-sib ¼
A grandparent ¼
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24
Q

How much does Full-sibs have in common

A

Should have ½ of their genes in common
 Same parents, but will not have exactly the same genes,
 The probability of a given gene being passed from parent to offspring is 1/2
• Full-sibs are individuals that have both parents in common

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25
Q

What is Half-sibs

A

• Half-sibs are individuals that have one parent in common
– Paternal half-sibs have the same sire, but different dams
– Maternal half-sibs have the same dam,
but different sires

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26
Q

What is Collateral Relatives

A

An individual’s collateral relatives to estimate his/her
breeding value
– Collateral relatives are the individual’s full- and half-sibs, aunts and uncles, cousins
– Should have some genes in common with that individual
– Collateral relatives have outstanding performance, animal may have same good genes and is likely to have a high
breeding value

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27
Q

What are the ancestors for

A

An individual’s ancestors (sire and dam, grandparents,
etc) to predict his/her breeding value
– The individual has genes from his/her ancestors
• Use their performance to predict his breeding value
• Ancestors had outstanding performance, they probably had some good genes that they passed on to him

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28
Q

Why are older ancestor have less valuable information

A

• The further back in the pedigree, the less valuable is the information from ancestors in predicting breeding values
– Less likely to contribute a significant number of genes to the individual
– Normally, do not use ancestors further back in the pedigree than the grandparents in predicting the breeding value

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29
Q

Sources of Information for Predicting Breeding Values

A

• Use a combination of several of these sources of information to estimate breeding values

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30
Q

What is the second aid to selection

A

Use of Repeated Records

• For some traits, such as milk production in dairy cows or litter size of sows, we can take repeated measurements
• Improves the accuracy of selection for traits that are
influenced by temporary environmental effects

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31
Q

What is the use of Repeatability

A

• Useful for traits expressed several times during an animal’s lifetime
– Milk production of dairy cows over several lactations
– Litter sizes of sows over several farrowings
– Weaning weights of calves from cows over several years

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32
Q

What is repeatability

A

Repeatability is defined as the fraction of
phenotypic variance that is due to
permanent effects

33
Q

Why are traits not 100% repeatable?

A

Because of temporary environmental effects

σTEsq

34
Q

What is the use of Repeatability

A

Repeatability of a trait helps us determine how many
records we need to accurately select animals

– If repeatability is high, the first record will be a good
indicator of future records

– If repeatability is low, we do not want to select based
on a single record, because the first record is not a
good indicator of future records

• Repeatability of a trait can be useful in making
culling decisions.

Examples:
• While culling cows based on milk production, a cow
with poor milk production in the first lactation should
be culled given that repeatability of yield is high (r =
aprox 0.5).
• A cow with descend milk yield but long service
period should be retained for another lactation since
the traits, calving interval and services per
conception, are not very repeatable (r = aprox 0.15).

35
Q

What is pedigree selection (3rd way to aid selection)

A

• Use the ancestors of an individual as an aid to selection to identify the genetically superior animals
– The pedigree is useful only if the objective performance records on the ancestors are available
– Pedigree selection is helpful for traits expressed late in life
Example, post weaning feedlot gain
and post weaning feed efficiency

36
Q

What is the advantages of pedigree selection

A

• Pedigree selection is helpful for carcass traits

– No need to harvest the animal to get the data
– Selection for carcass traits based on the carcass traits of his ancestors
• Pedigree selection is helpful for sex-limited traits such as milk production that are only expressed in one sex
– Select males for milk production based on their dams and granddams milk production

37
Q

What is progeny testing (4th way to aid selection)

A

• Use progeny to improve the accuracy of the breeding value estimates
• In order to justify the time and cost progeny testing requires, need to be able to make widespread use of the genetically superior animals once identified
– AI allows extensive use of outstanding
sires once we identify them through progeny testing

38
Q

What is the advantages of progeny testing

A

Advantages
• Progeny testing provides a very accurate estimate of the breeding value of an animal even if the heritability of
the trait is low
• Progeny testing allows us to select males for sex-limited traits such as milk production (we can measure the milk production of the daughters)

39
Q

What is the disadvantages of progeny testing

A

– Increased generation interval (it takes time to produce the progeny and evaluate them)
– Decreases the selection differential, because it increases the proportion saved and decreases the proportion culled
– Have to test more animals than we intend to actually use for breeding
– Costly, need to produce a lot of progeny, feed and evaluate

40
Q

Why is Progeny testing is not used nearly as much for females as for males

A

– Not as many progeny from females as from males.
– By the time get enough progeny from a cow or ewe or mare to have an effective progeny test, their productive lifetime may be about over
– Selection of females relies heavily on individual performance

41
Q

What are the 4 aids to selection

A
  • Ancestor
  • Repeatability
  • Progeny testing
  • Pedigree selection
42
Q

What are the Factors affecting response to selection (Δ G)

A
  1. Genetic Variability (σG)
  2. Heritability
  3. Generation Interval
  4. Genetic Correlations
43
Q

Why genetic variability affect response to selection

A

• Is a characteristic of the trait itself and is hard to change.
Example
• h2 of % fat and % protein = 0.55
• σG = 0.30 for % fat and 0.20 for % protein
• Expect more genetic progress for % fat than for % protein:
• The larger σG, the more progress from selection
– Genetic variation in a population is good

44
Q

Why Heritability affect response to selection

A

• Heritability is a measure of the strength of the
relationship between performance (phenotypic values)
and breeding values for a trait in a population.
• Heritability is the fraction of total phenotypic variance that is due to hereditary or genetic effects (breeding value).
• An extremely important population parameter used
for:
• Predicting breeding values
• Predicting expected response to selection.

45
Q

What are the two kinds of heritability

A

– Heritability in the broad sense

– Heritability in the narrow sense

46
Q

What is the symbol for heritability in broad sense

A

The symbol for heritability in broad sense is hB2 = Total
genotypic variance divided by total phenotypic variance:
hB2 = σG2 / σP2

47
Q

What is the symbol for heritability in narrow sense

A

• The symbol is hN2
Is the fraction of the total phenotypic variance that is due to the additive effects of genes:
hN2 = σA2/ σP2

48
Q

Which heritability is more useful

A

• hN2
is more useful than hB2, since only the additive effects
of genes are passed from parent to offspring
– hN2 is the fraction of total phenotypic variation in
parents that is passed on to offspring
– Variation due to dominance and epistasis is not passed from parent to offspring
– hB2 must be > hN2

49
Q

What are the estimation of heritability

A 
• Low heritability = 0 – 20%
– Reproductive traits
• Moderate heritability = 20 – 50%
– Weights, gains, feed efficiency, wool production, milk production
• High heritability = > 50%
– Carcass traits
50
Q

What are the heritability for beef

A 
Birth Weight .45
Weaning wt .25
Yearling wt .60
Frame Size (inches @hip) .60
Ribeye area .70
Fertility .10
51
Q

What are the heritability for dairy

A

Milk Production .25
Butterfat % .55
Fertility .10

52
Q

What are the heritability for sheep

A

Birth wt .35
Weaning wt .25
Fleece wt (fine wool, Medium wool) .40, .40
Staple length (fine wool, Medium wool) .40, .40
Frequency of multiple births .12

53
Q

What are the heritability for swine

A 
ADG .30
Feed efficiency .30
Loineye area .48
Backfat .50
Carcass length .60
% Ham and loin .50
# pigs farrowed
assuming partuition .10
# pigs weaned .10
54
Q

What are the heritability for horses

A

Working performance
Pulling ability Moderate (.25)
Cow sense Moderate to high (.50)

Riding performance
Jumping, 3-day event and dressage :  Moderate (0 to .70)
Racing performance
Running ability Moderate to high (.50)
w/ sulky  Moderate to high (.43)
Flat racer  Moderate to high (.50)
(Thoroughbred)
55
Q

How does Generation Interval affect response to selection

A

• Generation interval is defined as the average age of the parents when their offspring are born
• Can be used to calculate genetic response per year
– Δ G / yr = (h2 x SD) / GI
– Δ G/yr = (i x rGP x σG)/GI
– Δ G/yr = (i x h2 x σP)/GI
i = intensity of selection

56
Q

Generation Interval for Various Species

A 
Male         Female       
Beef cattle  3.0-4.0 yr 4.5-6.0 yr
Dairy cattle 3.0-4.0     4.5-6.0
Sheep         2.0-3.0     4.0-4.5
Swine          1.5-2.0      1.5-2.0
Chickens     1.0-1.5       1.0-1.5
Horses        6.0            6.0
57
Q

Generation Interval for cattle

A

the generation interval is usually shorter for bulls than for cows
• Usually keep a cow until she is 11 or 12 yr old, or even older, as long as she keeps having calves

• Turn over the bulls more quickly than the cows.
Generally use a bull for a couple of years and then replace him with another bull

• Average age of the bulls in the herd is usually less
than the average age of the cows when the calves are born

58
Q

Generation Interval for Swine and Chickens - generation interval is short

A

– They can begin reproducing at a younger age than
cattle and they also have a shorter gestation period than cattle
– They are younger when their offspring are born

59
Q

Generation Interval for horses

A

generally have a long generation interval

– One reason is that horses are often not used for breeding until after they have retired from racing or from the show circuit

60
Q

What is Genetic correlations

A

• If two traits are correlated, selection for one trait will result in change in the other trait
– For example, there is a positive genetic correlation between birth weight and yearling weight
• Selection for increased yearling weight will also increase birth weight, and could increase calving difficulty (dystocia)

61
Q

What is the use of genetic correlations

A

• Use advantage of genetic correlations by selecting for an easy to measure trait such as ADG instead of a hard to measure trait such as feed efficiency
– Most producers do not have the facilities to measure individual feed intake, and cannot select for feed efficiency directly

– Comparatively easy to measure and select for ADG

– If we select for ADG, feed efficiency will also improve, the traits are genetically correlated

62
Q

What are the causes of genetic correlations

A

• Linkage is one cause of genetic correlations between traits
– Say, for example, the A and B loci are linked (i.e., they are on the same chromosome)
• If A affects gain and B affects feed efficiency, gain and feed efficiency will be genetically correlated, and, therefore, selection for one will also affect the other

• Pleiotropy is the primary cause of genetic correlations
– Pleiotropy refers to the situation in which one gene influences more than one trait
– For example, pleiotropy would occur if the same gene influenced both rate of gain and feed efficiency

• An enzyme or hormone produced by a single gene is likely to affect several body processes, and, therefore, is likely to affect several traits (e.g., weights, gains, and carcass traits) and create genetic correlations among these traits

63
Q

What is genotypic selection

A

• Let the individuals with best sets of genes reproduce so that next generation has, on average, more desirable genes than current generation
– In selection we try to choose animals with best breeding values (the animals that will contribute best genes to next generation
– Breeding value – the value of an individual to be a genetic parent

64
Q

What is genomics

A
  • Genomics is the study of the entire set of genes found in living creatures.
  • A genome is the complete set of an organism’s DNA.
  • Genetics may look at a single gene, but genomics look at all genes and how they interact to influence growth and development of an organism.
  • Genomics uses DNA information to predict the genetic merit of bulls and cows, that can be used to predict genomic breeding value.
65
Q

What is Single nucleotide polymorphism (SNP)

A

A variation in a single nucleotide that occurs at a specific position in the genome, where each variation
is present to some appreciable degree within a population (mutation occurred in a single base)

66
Q

What is haplotype

A

refer to the inheritance of a cluster of single SNPs, which are variations at single positions in the DNA sequence among individuals.
• A haplotype is a group of genes within an organism inherited together from a single parent. The word “haplotype” is derived from the word “haploid,” which
describes cells with only one set of chromosomes, and from the word “genotype,” which refers to the genetic makeup of an organism.

67
Q

How does genomics help livestock producers?

A
  • Genetic gain for a particular trait can be enhanced by shortening of the generation interval, increased selection accuracy and increased selection intensity
  • Genomics allows producers to optimize the profitability and yields of their herds by making different animal selection and strategic breeding decisions.

• Provide more accurate genetic prediction to make better decisions while increasing economic returns on their breeding programs

68
Q

Genomic Selection

A

In dairy cattle an optimal breeding design with genomic selection will be more or less as follows:
– Genotype a large number of bull calves from the population
– Calculate the estimated breeding values or EPDs for these calves (accuracy = 0.80)
– Select a team of bull calves based on these EPDs and sell semen from them as soon as they can produce it
• This breeding design will reduce the generation interval from ~4 yr to ~2 yr and the rate of genetic gain will be doubled

69
Q

What is the goal of genomic selection

A
  • The goal of genomic selection is to rank animals based on their genetic merit as accurately and as early in the animal’s life as possible
  • Using genomic selection we can potentially predict the breeding values or EPDs of candidates for selection at birth
  • Consequently we can select animals at a young age
  • In some cases genomic selection is expected to double the rate of genetic improvement per year
70
Q

How much SNPs are used

A

45,000 SNPs are used to genotype bulls and low density 3,000 SNP chips are used to genotype
cows, heifers, and calves on commercial dairy farms for less than $40 per animal

71
Q

How are genetic value predicted

A

Prediction of total genetic value using genomewide dense marker maps. Genetics 157:219-229
– Proposed what is now called genomic selection
– SNP = single nucleotide polymorphism

72
Q

How many SNP chips are available for animals

A 
– Cattle – 750,000 SNPs
– Dogs – 250,000 SNPs
– Sheep – 56,000 SNPs
– Pigs – 60,000 SNPs,
– Horses – 55,000 SNPs
– Chickens – 600,000 SNPS
73
Q

Reproductive technologies to speed up rate of genetic gain

A
  • Artificial Insemination
  • MOET (Multiple Ovulation and Embryo Transfer)
  • JIVET (Juvenile in vitro embryo transfer)
  • Sexing Semen
  • Cloning
  • Genetic engineering – transgenic animals
74
Q

How many steps for genomic selection have

A

two

75
Q

What are the steps for genomic selection

A

– First, the effects of the markers (> 50,000 SNPs) must be estimated in a reference (training) population for which phenotypes and genotypes are available on the animals – Second, this information is then used to predict the breeding values or EPDs of candidates for selection in a test (evaluation) population

76
Q

What are the names of the phases for genomic selection

A

Validation Phase

Application Phase

77
Q

What is Validation phase

A

Test to see if the “DNA fingerprint” predicts a genomic EBV highly correlated with the progeny test EBV in a different group of animals not part of the original research.

78
Q

What is application phase

A

Once the “DNA fingerprint” for a trait is known we can predict the genomic breeding value in a new offspring for, eg. Protein yield, with just a DNA sample. NO production records required. So works equally well for males and females at birth

2 - Foundation of animal health (25 decks)

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