# Lecture 3 Resemblance Between Relatives by dffhrtcv3

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```									     Lecture 2:
Basic Population and
Quantitative Genetics
Allele and Genotype Frequencies
Given genotype frequencies, we can always compute allele
frequencies, e.g.,

1X
p i = freq(A i ) = freq(Ai Ai ) +    freq(Ai A j )
2
i6 j
=

The converse is not true: given allele frequencies we
cannot uniquely determine the genotype frequencies

For n alleles, there are n(n+1)/2 genotypes

If we are willing to assume random mating,
ž
p2
i       for i = j          Hardy-Weinberg
freq(Ai Aj ) =                                   proportions
2pi pj         =
for i 6 j
Hardy-Weinberg
• Prediction of genotype frequencies from allele freqs

• Allele frequencies remain unchanged over generations,
provided:
• Infinite population size (no genetic drift)
• No mutation
• No selection
• No migration
• Under HW conditions, a single generation of random
mating gives genotype frequencies in Hardy-Weinberg
proportions, and they remain forever in these proportions
Gametes and Gamete Frequencies
When we consider two (or more) loci, we follow gametes

Under random mating, gametes combine at random, e.g.

freq(AAB B ) = freq(AB jfat her) freq(AB jmot her)

freq(AaB B ) = freq(AB jfat her) freq(aB jmot her)
+ freq(aB jfat her) freq(AB jmot her)

Major complication: Even under HW conditions, gamete
frequencies can change over time
AB                  ab
AB                  ab

AB        ab

AB        In the F1, 50% AB gametes
ab        50 % ab gametes

If A and B are unlinked, the F2 gamete frequencies are

AB 25%       ab 25%       Ab 25%      aB 25%

Thus, even under HW conditions, gamete frequencies change
Random mating and recombination eventually changes
gamete frequencies so that they are in linkage equilibrium (LE).
once in LE, gamete frequencies do not change (unless acted
on by other forces)

At LE, alleles in gametes are independent of each other:
When linkage disequilibrium (LD) present, alleles are no
freq(AB --- freq(A) freq(B )
longer independent ) =knowing that one allele is in the
freq(ABC) = freq(A) freq(B) freq(C)
gamete provides information on alleles at other loci
The disequilibrium between alleles A and B is given by
=
freq(AB ) 6 freq(A) freq(B )
D A B = freq(AB ) ° freq(A) freq(B )
The frequency of the AB gamete is given by

freq(AB ) = freq(A) freq(B ) + D A B
Departure from
A
If recombination frequency between theLE and B loci
LE value
is c, the disequilibrium in generation t is

D (t ) = D (0)(1 ° c) t
Initial although
Note that D(t) -> zero, LD value the approach can be
slow when c is very small
Contribution of a locus to a trait
Basic model: P = G + E
Genotypic -- we value
Phenotypic valuevalue will occasionally
Environmental
also use z value for that
G = average phenotypic for this value genotype
if we are able to replicate it over the universe
of environmental values

G - E covariance -- higher performing animals
may be disproportionately rewarded

G x E interaction --- G values are different
across environments. Basic model now
becomes P = G + E + GE
Alternative parameterizations of
Genotypic values

Q1Q1                 Q2Q1               Q2Q2

C                   C + a(1+k)         C + 2a
C                   C+a+d              C + 2a
C -a                C+d                C+a

d = ak =G(Q1Q2 G(Q2 with d ) = 1Q1 1Q1 ]/2
d measures dominance, Q )2QG(Q0 if) the)heterozygote
2a = ) - [G(Q - 2 + G(Q
2
is exactly intermediate to the two homozygotes

k = d/a is a scaled measure of the dominance
Example: Booroola (B) gene

Genotype                   bb     Bb         BB

Average Litter size       1.48 2.17         2.66

2a = G(BB) - G(bb) = 2.66 -1.46 --> a = 0.59

ak =d = G(Bb) - [ G(BB)+G(bb)]/2 = 0.10
k = d/a = 0.17
Fisher’s Decomposition of G
One of Fisher’s key insights was that the genotypic value
consists of a fraction that can be passed from parent to
offspring and a fraction that cannot.

G i j = š G + Æ + Æ + ±i j
i   j
X
Average contribution --- genotypic j valueto their
The genotypic value along = difference (forallele
Since parents with to the
Mean value, pass š G singleG i the fori genotype
Dominance deviationspredicted from¢freq(Q Q ) i
alleles individual
j
allelic effectsthei genotypic value predicted from
Aj) between is (the b
Aioffspring, the athus average effect of allele i) the
two single alleles and the actualš G + Æ value,j
G i j = genotypic + Æ
represent these contributions             i

b
G i j ° Gi j = ±i j
Fisher’s decomposition is a Regression

G i j = š G + Æ + Æ + ±i j
i   j

A notational change clearly showsResidual error
Predicted value this is a regression,

G i j = š G + 2Æ + (Æ ° Æ )N + ±i j
1    2   1
Intercept 8 Regression = #
Independent (predictor) variable Nslopeof Q2 alleles
Regression residual
> 2Æ
<   1              forN = 0; e.g, Q 1 Q 1
2Æ + (Æ ° Æ )N = Æ + Æ
1    2   1                       forN = 1; e.g, Q 1 Q 2
> 1
:
1
2Æ1              forN = 2; e.g, Q 2 Q 2
Both Q2 and Q2 a2 > a1
Allele Q11 common,frequent, a1 = a2 = 0
common, a1   2

G21         Slope = a2 - a1

G                                G22

G11

0           1          2
N
Consider a diallelic locus, where p1 = freq(Q1)

Genotype         Q1Q1          Q2Q1          Q2Q2

Genotypic           0          a(1+k)             2a
value

Mean     š G = 2p2 a(1 + p1 k)
Allelic effects
Æ = p1 a [ 1 + k ( p1 ° p2 ) ]
2
Æ = ° p2 a [1 + k ( p1 ° p2 ) ]
1
Dominance deviations   ±i j = G i j ° š G ° Æ ° Æ
i   j
Average effects and Breeding Values
The a values are the average effects of an allele
Breeders focus on breeding value (BV)
X • (k)
n            ¥
B V(Gi j ) =ÆÆ++ (kÆ
BV =      i i
k= 1
(
k)
Æ j        )
Why all the fuss over the BV?

Consider the offspring of a QxQy sire mated
to a random dam. What is the expected value
of the offspring?
Genot ype            Frequency                  Value
Q x Qw                 1/ 4               šG   + Æ +
x     Æ + ±x w
w
Q x Qz                 1/ 4               šG   + Æ +
x     Æ + ±x z
z
Qy Qw                  1/ 4               šG   + Æ +
y     Æ + ±y w
w
Qy Qz                  1/ 4               šG   + Æ +
y     Æ + ±y z
z

The expected value of an offspring is the expected value of
µ            Ž       µ            Ž
Æ +Æ
x   y               Æ + Æ
w   z         ±x w + ±x z + ±yw + ±yz
šO = šG +                    +                    +
2                    2                       4
For random these have expected value
For a random dam, w and z alleles, this has an 0
expected value of zero
Hence,             µ                  Ž
Æ + Æ
x   y             B V(Sire)
šO ° šG =                               =
2                   2
We can thus estimate the BV for a sire by twice
the deviation of his offspring from the pop mean,

B V(Sire) = 2(š 0 ° š G )
More generally, the expected value of an offspring
is the average breeding value of its parents,

B V(Sire) B V(Dam)
š0 ° šG    =          +
2          2
Genetic Variances
G i j = š g + (Æ + Æ ) + ±i j
i   j

æ2 (G) = æ2 (š g X (Æ + Æ ) + ±i j ) = æ2n i + Æ ) + æ2 (±i j )
n
+ i      j             X(Æ     j
( k)   (k)              ( k)
æ2 (G) =          æ2 (Æ    + Æ )+          æ2 (±i j )
i      As Cov(a,d) = 0
j
k= 1                     k= 1

2
æG         =    2
æA    +    2
æD
Dominance Genetic Variance
(or simply dominance
m
X                  Q1Q1     Q1Q2       Q2Q2
æ2 = 2E [Æ ] = 2
2                2
Æ pi    Since E[a] = 0,
Var(a)0 E[(aa(1+k)] = E[a2a
=      -ma)2      2]
A                         i
i= 1

One locus, 2 alleles:       2
æA = 2p 1 p2 a2 [ 1 + k ( p 1 ° p2 ) ]2
When dominance present,
Dominance effects
m
X Xm         asymmetric function of allele
2 = 2E [±2 ] =
æD                   ±i j i jFrequencies
2 p p

i= 1 j= 1

One locus, 2 alleles:         2
æD = (2p 1 p2 ak)2
Equals zero if k = 0
This is a symmetric function of
allele frequencies
Additive variance, VA, with no dominance (k = 0)

VA

Allele frequency, p
Complete dominance (k = 1)

VA

VD

Allele frequency, p
Overdominance (k = 2)

VA
VD

variance

Allele frequency, p

Allele frequency, p
Epistasis
G i j kl = š G + (Æ + Æ + Æ + Æ ) + (±i j + ±k j )
i   j   k   l

Æ          Æ       Æ        Æ
+ (Æ i k + Æ i l + Æ j k + Æ j l )
±          ±        ±        ±
+ (Æ i k l + Æ j k l + Æ ki j + Æ l i j )
+ (±±i j k l )
= š G + A + D + AA + AD + D

Dominance xvalue interactions
Dominancedominance interaction
Breeding value
(or orthogonal), interaction between the dominance
the so that between a single allele
interactions between an allele locus
between
interactionsthe two alleles at a at one
at onewith the genotype at the dominance
locus locus with a single allele at another
deviation at one locus with another, e.g.
2 allele2 i andanother. B2
+ æD at genotype kj
A
æ2 = æA deviation+ æA A + æA D + æ2 D
G
2
D

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