Processing math: 100%
+ - 0:00:00
Notes for current slide
Notes for next slide

Population Values and Means

Jinliang Yang

Mar. 18, 2024

1 / 31

Basic Genetic Model

P=G+E

  • P: Phenotypic value
  • G: Genotypic value
  • E: Environmental deviation
2 / 31

Basic Genetic Model

P=G+E

  • P: Phenotypic value
  • G: Genotypic value
  • E: Environmental deviation

If we replicate a genotype in a large number of environments

  • Average environmental deviation would be zero
  • And phenotypic average would be equal to the genotypic value of that genotype
3 / 31

Basic Genetic Model

P=G+E

  • P: Phenotypic value
  • G: Genotypic value
  • E: Environmental deviation

If we replicate a genotype in a large number of environments

  • Average environmental deviation would be zero
  • And phenotypic average would be equal to the genotypic value of that genotype

Genotypic Value (G)

When a single locus is under consideration,

G=A+D

  • A: Additive genetic value
  • D: Dominance deviation
4 / 31

Genotypic value

  • We need to assign arbitrary values to genotypes to dissect the genotypic value of an individual
  • Midpoint between A1A1 and A2A2 is commonly standardized to 0. Or, represented as m.
5 / 31

Genotypic value

  • We need to assign arbitrary values to genotypes to dissect the genotypic value of an individual
  • Midpoint between A1A1 and A2A2 is commonly standardized to 0. Or, represented as m.

Gene action (or mode of inheritance):

  • Value of A1A2 indicates gene action at the locus
    • d=0: additive gene action
    • d=a or a: complete dominance or complete recessive
    • d>a or d<a: overdominance or underdominance
    • Any other values of d besides those above values are partial dominance
6 / 31

Population Mean

In a population in HWE:

Genotype Freq. Value
A1A1 p2 +a
A1A2 2pq d
A2A2 q2 a

What is the Population mean ( M ) or the expected genotypic value?

7 / 31

Population Mean

In a population in HWE:

Genotype Freq. Value
A1A1 p2 +a
A1A2 2pq d
A2A2 q2 a

What is the Population mean ( M ) or the expected genotypic value?

According to Formula (1) in the STAT Note:

E(f(X))=ki=1f(xi)Pr(X=xi)M=p2a+2pqdq2a=a(pq)+2pqd

8 / 31

Population Mean

In a population in HWE:

Genotype Freq. Value
A1A1 p2 +a
A1A2 2pq d
A2A2 q2 a

What is the Population mean ( M ) or the expected genotypic value?

According to Formula (1) in the STAT Note:

E(f(X))=ki=1f(xi)Pr(X=xi)M=p2a+2pqdq2a=a(pq)+2pqd where M is the deviation from the midpoint of the homozygotes.

  • The first term comes from the homozygote values
  • The second from the value of the heterozygote
9 / 31

B Locus Example

Booroola (B) locus influences fecundity in Merino sheep

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25
10 / 31

B Locus Example

Booroola (B) locus influences fecundity in Merino sheep

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25

What is the M?

  • First, determine midpoint:

  • then, determine a and d:

  • And finally calculate the population mean

11 / 31

B Locus Example

Booroola (B) locus influences fecundity in Merino sheep

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25

What is the M?

  • First, determine midpoint:
    • (2.66+1.48)/2=2.07
  • then, determine a and d:
    • a=2.662.07=0.59
    • d=2.462.07=0.39
  • And finally calculate the population mean M=a(pq)+2pqd=0.59×(0.250.75)+2×0.25×0.75×0.39=0.149

Note that this is the deviation from the midpoint, 2.07.

12 / 31

Multiple Loci

When multiple loci are acting independently and contributing to overall value, the mean is:

M=ki=1ai(piqi)+2ki=1piqidi

  • Summation over all the loci

  • k is the number of loci affecting the genotypic value of a trait

13 / 31

Average effect of an allele

  • Parents pass on alleles, not genotypes

  • The average effect of an allele depends on

    • the genotypic values ( a, d )
    • and allele frequency ( p, q )
  • Thus, an average effect can be unique for each population because of differences in allele freq between populations.

14 / 31

Average effect of an allele

  • Parents pass on alleles, not genotypes

  • The average effect of an allele depends on

    • the genotypic values ( a, d )
    • and allele frequency ( p, q )
  • Thus, an average effect can be unique for each population because of differences in allele freq between populations.

    Average effect of A1

  • Mean deviation from the population mean of individuals who received that allele from one parent

  • The other allele is received at random

    • i.e., according to the allele freq, q, from the population
15 / 31

The avarage effect of A1

  • Consider the A1 allele. Under HWE, the probability an A1 allele combines with another A1 allele to form an A1A1 genotype is p. The value of the A1A1 genotype is a.
  • The probability an A1 allele combines with an A2 allele to form a heterozygote, with value d, is q.
16 / 31

The avarage effect of A1

  • Consider the A1 allele. Under HWE, the probability an A1 allele combines with another A1 allele to form an A1A1 genotype is p. The value of the A1A1 genotype is a.
  • The probability an A1 allele combines with an A2 allele to form a heterozygote, with value d, is q.

  • The upshot of this is that

    • allele A1 forms genotypes A1A1 with value of a at a frequency of p,
    • and forms genotypes A1A2 with a value of d at a frequency of q.
  • Thus, the mean value of individuals that received A1 is pa+qd

    • and, the average effect of allele A1 is (recall definition above):
17 / 31

The avarage effect of A1

  • Consider the A1 allele. Under HWE, the probability an A1 allele combines with another A1 allele to form an A1A1 genotype is p. The value of the A1A1 genotype is a.
  • The probability an A1 allele combines with an A2 allele to form a heterozygote, with value d, is q.

  • The upshot of this is that

    • allele A1 forms genotypes A1A1 with value of a at a frequency of p,
    • and forms genotypes A1A2 with a value of d at a frequency of q.
  • Thus, the mean value of individuals that received A1 is pa+qd

    • and, the average effect of allele A1 is (recall definition above):

α1=pa+dqM=pa+dq(a(pq)+2pqd)=q(a+d(qp))

18 / 31

Allele Substitution

19 / 31

Allele Substitution

What is effect of substituting an A1 allele for A2?
  • This can be simply expressed as the difference in average effect of the two alleles:

α=α1α2=q(a+d(qp))(p(a+d(qp)))=a+d(qp)

20 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the average effect of allele B1?

21 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the average effect of allele B1?

  • Mean of individuals receiving a B1 allele (others coming from random) is: pa+qd

    • 0.25×0.59+0.75×0.39=0.44
  • Averge effect of B1:

    • α1=0.44M=0.44(0.149)=0.589
22 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the average effect of allele substitution α?

23 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the average effect of allele substitution α?

α=a+d(qp)=0.59+0.39×(0.750.25)=0.785

24 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the average effect of allele substitution α?

α=a+d(qp)=0.59+0.39×(0.750.25)=0.785

The average effect of allele B2 or α2?

α=α1α2α2=α1α=0.5890.785=0.196

25 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the M and α values if B1=0.85?

26 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the M and α values if B1=0.85?

M=a(pq)+2pqd=0.59×(0.850.15)+2×0.85×0.15×0.39=0.512α=a+d(qp)=0.59+0.39×(0.150.85)=0.317

27 / 31

Revisit the Booroola locus example

Genotype Mean litter size
B1B1 2.66
B1B2 2.46
B2B2 1.48
  • Freq of B1=0.25, p=1q=0.25
  • Recall that a=0.59, d=0.39, and M=0.149

What is the M and α values if B1=0.85?

M=a(pq)+2pqd=0.59×(0.850.15)+2×0.85×0.15×0.39=0.512α=a+d(qp)=0.59+0.39×(0.150.85)=0.317 As population mean increases, average effect of the allele substitution decreases.

28 / 31

Avg effect of A1 vs. allele freq

α1=q(a+d(qp))

When d=0, α1=q(a+d(qp))=(1p)a=apa

Average effect of an allele plotted against its frequency in the population.

29 / 31

Avg effect vs. allele freq

α1=q(a+d(qp))α=a+d(qp)α1=qα

  • If dominant is absent (i.e. d=0), A1A2 is right in the middle in any case, so α is just equal to a.
30 / 31

Avg effect vs. allele freq

α1=q(a+d(qp))α=a+d(qp)α1=qα

  • If dominant is absent (i.e. d=0), A1A2 is right in the middle in any case, so α is just equal to a.
  • If dominant is present, randomly change A2 alleles to A1 alleles will have a greater effect when A1 alleles are less frequent.

α=a+d(qp)=a+d(1pp)=(a+d)2d×p

31 / 31

Basic Genetic Model

P=G+E

  • P: Phenotypic value
  • G: Genotypic value
  • E: Environmental deviation
2 / 31
Paused

Help

Keyboard shortcuts

, , Pg Up, k Go to previous slide
, , Pg Dn, Space, j Go to next slide
Home Go to first slide
End Go to last slide
Number + Return Go to specific slide
b / m / f Toggle blackout / mirrored / fullscreen mode
c Clone slideshow
p Toggle presenter mode
t Restart the presentation timer
?, h Toggle this help
Esc Back to slideshow