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
               Linkage disequilibrium
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 Decay of Linkage Disequilibrium
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
                 Additive Genetic Variance
                            (or simply dominance
                 (or simply Additive Variance) variance)
                   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
                                   additive variance
 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




     Zero additive
     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

These components are Additivetointeractions -----
           Dominance xvalue interactions
            Additive x Dominant be uncorrelated,
              Dominancedominance interaction
               Breeding value
           Additive x defined -- interaction--
(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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