1. The goal of a livestock system
including goats is to produce a quantity of quality
products with maximum efficiency. A component in achieving
this goal is the genetic improvement of goats in the
areas of quantity, quality, and efficiency. Genetic
improvement is achieved by selection. The rate of
improvement is directly related to the accuracy with
which the goats are ranked, the intensity with which
they are selected, the amount of genetic variation
available in the trait(s), and the generation interval.
Once goats have been selected to become parents of
the next generation, one must consider alternative
mating plans. Various mating strategies differ in
their goals, and the consequences of each should be
understood when considering programs for genetic improvement.
2. Traits of goats can be considered
either to be qualitative (simply inherited) or quantitative.
Most economically important traits are quantitative.
3. Genetic Parameters Quantitative
traits of goats are those which are influenced by
genes at many different loci (gene sites on a chromosome),
each contributing a relatively small amount to the
total expression of the trait. A second characteristic
of quantitative traits is that their expression is
influenced to some degree by the environment in which
the goat performs.
4. The phenotype of a goat is
the observable expression of some trait, e.g., pounds
of milk produced in a lactation. The phenotype (P)
for a trait can be defined as the sum of the goat's
genetic merit for that trait (G), the influence of
the environment (E) on the record, and the population
mean for the trait (M). If one looks at phenotype
of several goats for a given trait, one can also determine
their average performance and a certain variation
from animal to animal called variance. The sum of
the variances due to genetic and environmental influences
makes up the total phenotypic variance, from which
the standard deviation can be extracted; i.e., the
standard deviation is the square root of the variance.
5. Average performance and standard
deviation describe a trait in a given population.
If a trait is normally distributed along a bell-curve,
then 500f the trait's records will lie between -0.67
and +0.67 standard deviations and 950f the records
between -1.96 and +1.96 standard deviations in a particular
population (Figure 1). The ratio of the additive genetic
variance over the phenotypic variance is the important
parameter called heritability (h+/-), which can take
on values from 0 to 1. A value of 0 means more of
the variation in the trait is genetic, and a value
of 1 means all the observed variation is genetic.
Few economically important traits in goats have values
exceeding 0.5. General characterizations for selecting
traits are:
low heritability
less than 0.15
moderate heritability
0.15 to 0.30
high heritability
more than 0.30
6. Reproductive traits have low heritabilities. Milk
and fat yields are examples of traits with moderate
heritability. Milk composition and most growth-related
traits have high heritabilities.
7. Heritability has many uses in
goat genetics. For example, heritability can be used
to estimate the breeding value (genetic merit) (BV)
of a goat. Assume the average production of a certain
breed of goats is 1000 lb (P) after adjusting the
records for influences of age of doe, season of kidding,
etc. A certain doe in that breed produced 1100 lb
milk (P). Hence, her breed superiority (P - P) is
100 lb. The portion of this phenotypic superiority
due to her genetic merit is 100 x 0.25, if the heritability
for milk production in dairy goats is 0.25. Thus,
her Estimated Breeding Value (EBV) is 25 lb.
8. The true breeding value (BV)
of goats is never known and they are compared on estimates
of BV, which are subject to the variance or error
in estimation. Accuracy denotes how well the BV of
a goat has been estimated. The more information available
on an individual, either in terms of repeated records
or information on relatives, the more accurate the
estimated BV's (EBV), and the less likely the comparison
of individuals is in error.
9. The EBV's can be used to rank
goats comparatively for selection. In the example,
the goat (A) producing 1100 lb of milk was 25 lb genetically
superior to the breed average doe. Another doe (B)
producing 900 lb milk is 25 lb inferior to the breed
average (-100 x 0.25). Thus, goat A is 50 lb milk
genetically superior to goat B.
10. The EBV of an individual
represents its own genetic merit but greater interest
lies in the merit of that individual's progeny. ''How
much of an in dividual's breeding value or superiority
will be transmitted to its progeny?'' is the question.
The concept of Estimated Transmitting Ability (ETA)
equals one half the EBV of an individual goat, since
one half of her genes are represented in her progeny;
the other half being supplied by the other parent.
Genes obtained by one particular progeny from its
parent are a random sample. A progeny may receive
in the extreme a sample of the parent's best or its
worst genes only. This explains why poor progeny sometimes
may result from good parents and good ones from poor
parents.
11. The genetic parameter closely
related to heritability is repeatability (r). It is
also a ratio of variances, namely the variance for
permanent environmental influence (e.g. injury) plus
the total genetic variance, not just the additive,
over the phenotypic variance. Repeatability is equal
to or greater than heritability by definition. Repeatability
can be used to predict the future performance of a
goat based on her past performance.
12. This Most Probable Producing
Ability (MPPA) can be calculated as: P'+r(P-P') when
the goat has one record. The doe (A) producing 1100
lb (P) in the population averaging 1000 lbs (P') has
an MPPA of 1050, if r=0.50. When the goat had n records
the calculation for MPPA is:
nr
P' +
r (P-P').
1+(n-1)r
13. All domestic livestock have
several traits of economic importance, and their relationships
to each other are critical for selection programs.
A correlation describes the relationship between two
traits. There are phenotypic, genetic, and environmental
correlations. A correlation can have values between
-1.0 and +1.0, with zero meaning no relationship.
The nearer the correlation is to +1.0 or -1.0, the
closer the relationship is between the two traits.
A positive correlation (+) indicates high measures
of one trait tend to occur with high values of the
second and low values for the first with low values
for the second. A negative correlation (-) indicates
a tendency for high values of one trait to occur with
low values of the second. For example, milk and fat
yields of goats are positively correlated. As pounds
of milk per lactation increase so do the pounds of
fat produced. However, milk yield and milk fat percentage
are negatively correlated. As pounds of milk increase,
the percentage of fat in the milk of goats tends to
decrease.
14. Genetic correlations are
important in selection and have two biological causes:
pleiotropy and linkage. Pleiotropy is the result of
one gene contributing to the phenotype of more than
one trait. Linkage means a gene (or set of genes)
is in close proximity on a chromosome to a gene for
a second trait. Being close together on the same chromosome,
they are passed on to the progeny together and cause
the genetic correlation. Thus selection for one trait
will alter also the performance of the population
for all other traits which are genetically correlated
to the trait under direct selection. That change in
a correlated trait is called a correlated response.
Some correlated responses can be beneficial in terms
of improving the total productivity of goats; however,
others may be detrimental. Genetic correlations as
well as phenotypic correlations may be used in indexing
animals for simultaneous selection of more than one
trait.
15. Selection Response The first
step in the selection process is to define the goals
of the program, e.g., which trait or traits are desired
in selection. The appropriate records need to be collected
on the selection candidates and their relatives. From
these records, the BV's of the individuals are estimated
and the goats ranked from best to worst. The breeder
must now decide how many goats are needed for both
sexes, and selection is then simply keeping the top
ranked animals. Fewer bucks are required to maintain
the population than females, therefore the intensity
of selection for males can be much greater. This points
out that more progress can be made by concentrating
efforts on buck selection.
16. A selection differential
is the phenotypic average difference of the selected
parent animals (Ps) from the population average (P').
Selection intensity (i) is the selection differential
expressed in terms of phenotypic standard deviations
(s); i.e. the ratio of (Ps-P') over s. Selection is
used in predicting genetic response due to selection
because it can be related to that percent of the population
saved as parents (see Table 1). If the top 100f goats
available are used for selection, their mean phenotypic
superiority due to selection intensity is 1.75 standard
deviations above the population mean. If the top 70are
selected then the selection intensity will be only
0.5 standard deviations above average.
17. Table 1. Selection intensities
(i) for different percentages of individuals selected
to be parents from a large population.
Percent Saved
i
1
2.67
5
2.06
10
1.75
30
1.16
50
0.80
70
0.50
90
0.19
100
0
18. The parameter of heritability
is used to calculate selection response. The square
root of heritability is called accuracy (h). The formula
for selection response after using a certain superior
male in a goat herd is one-half of the product of
accuracy, times selection intensity, times the additive
genetic variance for the trait. The reason for ''one-half''
is, of course, that only half of the sire's genetic
superiority is passed on to his progeny. Selection
response is the genetic change due to selection in
one generation. Many times, interest lies in the genetic
change per year. To obtain this estimate, one must
divide by the generation interval (t), which is the
average age of the parents when their progeny are
born. For example, if the heritability of yearling
weight is 0.49, then the accuracy is the square root
or 0.7. The generation interval of goats is two years.
According to the above formula, the answer in the
case of 10 election intensity (equal to 1.75 standard
deviations) would be 1/2(0.7) (1.75)/2=0.31 standard
deviations selection response per year for strictly
male selection only. If however, 700f all females
selected is the selection intensity program, then
the selection progress becomes, according to the same
formula, 1/2(0.7) (0.5)/2=0.09 standard deviations
selection response per year above herd average. If
both programs are combined, the answers are combined
and the selection response for yearling weight becomes
0.4 standard deviations progress per year in average.
19. Selection for more than one
trait based on independent culling levels is accomplished
by ranking candidates for both traits and requiring
the selected animals to meet a minimum standard in
both traits. This process is repeated each year with
only minimum standards adjusted for progress made
by selection.
20. Figure 2 shows independent
culling plotted for a large number of animals for
two traits. Goats in the upper righthand quadrant
meet the standards for both traits and are chosen
to become parents of the next generation. Two points
in Figure 2 are goats A and B for the two traits.
Goat A is superior to the population for trait 1 but
falls just below the culling minimum level for trait
2; so this animal would be culled. Goat B has a performance
level just above the standards for both traits and
is kept as a parent. It can be seen that Goat A could
probably be a more appropriate parent than B. Their
difference in performance for trait 2 is minimal while
goat A is far superior to B for trait 1.
21. Compromises have to be made
for goats which are superior for the one trait but
just below borderline for the other trait. This has
led to the procedure of Selection Index which allows
ranking animals simultaneously for two or more traits
in one single index.
22. A selection index is calculated
from the sum of trait means each multiplied by appropriate
factors which weigh their relative importance genetically
and economically. An index provides the opportunity
for multiple trait selection. For example, a goat
breeder considers trait 1 two times as important as
trait 2. Hence, trait 1 would have a weighting factor
double that of trait 2. In figure 2, goat A would
now be selected rather than goat B because of the
index method. Selection index, under certain assumptions,
can maximize selection response for both traits.
23. Breeding Systems Random mating
is a system where an individual goat has an equal
opportunity to be mated with any other individual
of the opposite sex. The mating of two particular
goats occurs essentially by chance in that no breeder
decision was made to join them as mates.
24. Assortative mating is based
on phenotypic performance or characteristics. Mating
individuals of like performance is positive assortative
mating, e.g., large goats to large mates and small
to small. Mating individuals of unlike performance
is negative assortative mating, e.g., large goats
to small mates. Positive assortative mating tends
to cause more variation in the total population of
progeny than would occur from random mating, and negative
assortative mating tends to reduce the variation.
Assortative mating is practiced for type (conformation)
traits. For example, ''corrective'' mating for progeny
of a doe with certain physical weaknesses to a buck
with strengths in those attributes would be negative
assortative mating. Perpetuating strengths of a given
line by selecting a buck also strong in those characteristics
would be positive assortative mating. Assortative
mating deviates from random mating in that decisions
by the breeder based on phenotypes exclude or reduce
the possibility of some matings.
25. The concept of relationship
is important in deciding on certain breeding systems.
The basic principle involved in determining relationship
is that a parent passes one half of its gene complement
to its progeny. Hence, in a noninbred population,
a parent is related to its progeny by 0.5, meaning
500f their genes are in common or they have a relationship
of 50 Other types of relationships in noninbred populations
are calculated as products of this 0.5 or (1/2). For
example, the relationship between paternal half-sibs
(progeny of the same sire mated to different dams)
is 25as can be seen in the following diagram:
dam 1 > progeny 1 < sire
> progeny 2 < dam 2
Each progeny receives a sample
half of the sire's genes. The portion of genes that
progeny 1 received from its sire, which are replicates
of the same genes received by progeny 2, is (1/2)
times (1/2) or 0.25
26. The relationship of a goat
(C) to its grandparent (A) is also 25 because:
great-grandparent (E)
\/
grandparent (A)
\/
parent (B)
\/
progeny (C)
Parent goat B received one half
of its genes from grandparent A and in turn passes
one half of its genes to progeny C. The relationship
is (1/2) times (1/2) or 0.25 x 025
27. The relationship between
C and its great-grandparents (E) would be one half
of 0.25 or 12.5
28. A quick method of calculating
relationships between two individuals in a noninbred
population is to set each ''arrow'' between related
individuals equal to (1/2) and multiply. This is equivalent
to raising (1/2) to the nth power, (1/2)n, where n
represents the number of arrows or generation steps.
The relationship of goat D to goat E having a common
grandparent (A) would be:
grandparent
A
\/
\/
parent B
parent C
\/
\/
progeny D
progeny E
(1/2)4th or 1/16 or 6.25 ince there are 4 arrows or
generation steps connecting D with E
29. The degree of inbreeding
of a particular individual equals one half the relationship
of its parents. This value is called the inbreeding
coefficient (F). The progeny resulting from mating
a sire to his daughters has an F of 0.25, while mating
two paternal half-sibs results in a progeny inbred
12.5
30. The consequence of inbreeding
is increased homozygosity or likeness of genes. If
an individual goat has two unlike genes at a particular
trait locus it is called heterozygous. However, an
inbred animal has more loci in homozygous states than
the average noninbred animal. For a particular trait,
there is a higher number of homozygous individuals
in an inbred population than in a non-inbred one.
Over time with inbreeding, certain genes which were
present in the initial population may get lost in
subsequent generations. Within an inbred line, the
tendency towards fixation of few genes reduces the
amount of genetic variation. In the absence of selection,
genes are lost or fixed at random. The variation between
inbred lines increases, however. Inbreeding can be
used to create diverse lines.
31. An advantage of an inbred
goat as parent is the increased uniformity in its
progeny. Uniformity does not imply superiority, which
is a function of genetic merit. An inbred animal may
or may not be superior. Since inbreeding increases
homozygosity in a population, it follows that this
includes undesirable recessive genes. Hence, with
inbreeding, the risk exists that one would generate
a higher incidence of homozygotes for lethal, sublethal,
and undesirable genotypes than ocurs in random mating
populations.
32. A second negative aspect
of inbreeding is inbreeding depression, which is reduced
performance related to increased homozygosity. The
traits most influenced by inbreeding depression are
in general those which have low heritabilities. Unfortunately,
this includes such traits as viability and reproductive
performance. A realistic threat associated with intensive
inbreeding is producing a population of goats unable
to survive or reproduce well enough to maintain the
population.
33. Inbreeding occurs from breeding
programs designed to mate relatives and in small herds
where introduction of outside breeding stock is rare.
The rate of change per generation in the average level
of inbreeding is not large enough in most herds to
be of concern. For example, if one uses 5 bucks selected
from within a herd of 50 does, the level of inbreeding
increases by 0.0275 per generation. However, in small
herds where, for example, two bucks may be used on
20 does, the increase is 0.069 per generation. A progeny
from two unrelated, inbred parents has an inbreeding
coefficient F of zero. If a breeder feels his herd
is getting too inbred, he can relieve the problem
in the subsequent generation by using unrelated breeding
stock. A breeder can also bring an unrelated, even
though highly inbred sire into his herd without suffering
the consequences of inbreeding depression in the progeny
of that buck, because the progeny of two unrelated,
although inbred, parents has an inbreeding coefficient
of zero.
34. Line breeding maintains a
high degree of relationship of individuals to a superior
ancestor but has less severe consequences than inbreeding.
great grandparent
A
\/
\/
grandparent B
grandparent C
\/
\/
parent D
parent E
\/
\/
progeny Z
Assume goat A was a buck used
on several does in a herd generating daughters B and
C. These daughters are mated to unrelated bucks and
produce a son D and daughter E. The goats D and E
are related to A by 25, but to each other by 6.25
35. Assuming it was recognized
that A was a truly superior buck. If his descendents
were mated each generation to unrelated animals, the
relationship of A to that progeny would continue to
decrease by one half each generation. Progeny Z is
related to great grandparent A not by 12.5but by 25,
because there are both parents D and E related to
Z. Since D and E are related by 6.25, progeny Z has
an inbreeding coefficient of 3.15 The recapture of
some of A's gene combinations is not certain but possible.
The probability can be increased through line breeding
or increased inbreeding with its other consequences.
36. The opposite mating system
is designed to increase heterozygosity. Crossing of
lines within a breed or the crossing of breeds are
examples of such strategies. The fundamental assumption
is that the genes at the various loci differ in the
two parent lines or breeds. For many traits, the crossing
of lines or crossbreeding results in progeny whose
performance exceeds that which was expected from their
parents' performance. This deviation is called heterosis
or hybrid vigor and can be calculated as the difference
from the expected.
37. For example, assume we cross
two goat breeds, one averaging 800 lb of milk per
lactation and the other 1000 lb. The expected progeny
performance would be 900 lb milk while the actual
performance of the crossbred progeny was 950 lb, indicating
5.6 2.256835e+199terosis ((950-90 Progeny exceeding
their expected performance for a trait are not necessarily
superior to both of the parental lines for that trait.
Furthermore, crossbreeding influences all traits so
that if complementary breeds are used, the total merit
of the crossbreds may exceed each parental breed.
38. Heterosis is associated with
dominant gene action. Assume that trait (A) has three
possible genotypes with the following relative phenotypic
values.
genotype:
AA
Aa
aa
value:
100
100
0
Gene (A) is dominant to its recessive
allele (a) and the genotypes (AA) and (Aa) have the
same phenotypic values. At a second trait locus (B)
assume the values for the three genotypes are:
genotype:
BB
Bb
bb
value:
100
50
0
There is no dominance at this
locus since (Bb) is the average of the values of the
two homozygous states. At a third trait locus (C)
it is assumed that the following values exist:
genotype:
CC
Cc
cc
value:
100
60
0
There is incomplete dominance
since Cc has a value exceeding 50 but less than 100.
39. Given these three loci and
their relative values, one can now demonstrate heterosis
from crossing two breeds of goats that have the following
of the above genotypes:
Breed 1
AA
bb
cc
Breed 2
aa
BB
CC
The relative phenotypic value
of each breed is the sum of the values of their genotypes
at each of the loci. Breed 1 would have a value of
100 and breed 2 a value of 200; with an average for
both parents of 150. Any crossbred progeny will be
heterozygous at all three loci. Hence, their value
will be the sum of their heterozygote values: 210.
Heterosis for this particular crossing is 40(210 -
150/150 =0.4). If all three loci had no dominance,
then the heterozygote values at each locus would all
be 50, and the value sum for the crossbred progeny
would be: 150. Hence, with no dominance, the percentage
heterosis is zero
40. Heterosis is essentially
the opposite of inbreeding depression and is also
related to heritability. Those traits with low heritability
usually show the greatest percentage of heterosis.
These include viability and reproductive performance,
both important characteristics in a total production
system.