Mendelian Genetics in Populations I

Hardy-Weinberg Genotype Frequency Equilibrium Assumptions

1)   All individuals, regardless of genotype, mate randomly.

2)   No natural selection.

3)   Populations are infinitely large (or just really big).

4)   No individuals enter or leave the population.

5)   No mutation.

 

If these assumptions are true, we can show that the frequency of an allele in the parent population will not change from one generation to the next – i.e., there is no evolution.

 

F(A1)  = F(A1A1) + F(A1A2)

F(A2)  =F(A1A2) + F(A2A2)

 

 

F(A1A1) = F(A1) x F(A1) = F(A1)2

 

F(A1A2) = 2[F(A1) x F(A2)]

 

F(A2A2) = F(A2) x F(A2) = F(A2)2

Genotype Frequency with

Random Mating and Autosomal Genes

 

Define

In Adult Generation:  U = F(A1A1)

                                  V = F(A1A2)

                                  W = F(A2A2)

 

 

Offspring

 

Parent 1

X

Parent 2

Frequency

A1A1

A1A2

A2A2

 

A1A1

 

A1A1

U2

U2

—

—

A1A1

 

A1A2

2UV

UV

UV

—

A1A1

 

A2A2

2UW

 

2UW

—

A1A2

 

A1A2

V2

1/4V2

1/2V2

1/4V2

A1A2

 

A2A2

2VW

—

VW

VW

A2A2

 

A2A2

W2

—

—

W2

 

 

 

 

U’

V’

W’

 

In the next (i.e., offspring) generation:

U’ = U2 + UV + 1/4V2

     = (U + 1/2V)2 = F(A1)2

V’ = UV + 2UW + 1/2V2 + VW

     = 2[(U + 1/2V)(W + 1/2V)] = 2[F(A1) x F(A2)]

U’ = W2 + VW + 1/4V2

     = (W + 1/2V)2 = F(A2)2

 

REMARKABLE! If HW assumptions are met we know a lot of things!

 


Example of

Hardy-Weinberg Equilibrium

Genotype Frequencies

 

N = total number of individuals in the population

A1 = allele 1

A2 = allele 2

 

                                     Observed Number

A1A1

A1A2

A2A2

N

 

17,062

 

1,295

 

28

 

18,385

 

 

F(A1) =  =  = 0.9633

 

F(A2) =  =  = 0.0367

 

Expected F(A1A1) = (0.9633)2 = 0.9279

Expected Number A1A1 = F(A1A1) x N = 0.927 x 18,385

                                                          = 17,059

Expected F(A1A2) = 2(0.9633)(0.0367) = 0.0707

Expected Number A1A2 = F(A1A2) x N = 0.0707 x 18,385

                                                          = 1,299

Expected F(A2A2) = (0.0367)2 = 0.0014

Expected Number A2A2 = F(A2A2) x N = 0.0014 x 18,385

                                                          = 26

 

Genotype

Observed

Expected

A1A1

17,062

17,059

A1A2

1,295

1,299

A2A2

28

26

 

 

Hardy-Weinberg Genotype Frequency Equilibrium Assumptions

1)   All individuals, regardless of genotype, mate randomly.

2)   No natural selection.

3)   Populations are infinitely large (or just really big).

4)   No individuals enter or leave the population.

5)   No mutation.

 

 

 

 

 

Offspring

 

Parent 1

X

Parent 2

Frequency

A1A1

A1A2

A2A2

 

A1A1

 

A1A1

U2

U2

—

—

A1A1

 

A1A2

2UV

UV

UV

—

A1A1

 

A2A2

2UW

 

2UW

—

A1A2

 

A1A2

V2

1/4V2

1/2V2

1/4V2

A1A2

 

A2A2

2VW

—

VW

VW

A2A2

 

A2A2

W2

—

—

W2

 

 

 

 

U’

V’

W’

Deviations to

Hardy-Weinberg Equilibrium

Genotype Frequencies

 

 

                                     Observed Number

A1A1

A1A2

A2A2

N

 

625

 

750

 

3,625

 

5,000

 

 

F(A1) =  =  = 0.20

 

F(A2) =  =  = 0.80

 

Expected F(A1A1) = (0.20)2 = 0.04

Expected Number A1A1 = F(A1A1) x N = 0.04 x 5,000

                                                          = 200

Expected F(A1A2) = 2(0.20)(0.80) = 0.32

Expected Number A1A2 = F(A1A2) x N = 0.32 x 5,000

                                                          = 1,600

Expected F(A2A2) = (0.80)2 = 0.64

Expected Number A2A2 = F(A2A2) x N = 0.64 x 5,000

                                                          = 3,200

 

 

Genotype

Observed

Expected

A1A1

625

200

A1A2

750

1,600

A2A2

3,625

3,200

 

Significance of deviation can be calculated with Chi-square test.
Evolution by Natural Selection
I)    Background
A)  Natural Selection is not the same as Evolution

1)   Evolution is a two step process: variation and change

A)  Natural Selection does not always lead to Evolution

B)  Natural Selection is only effective if phenotype is genetically based

II)  Consequences of Natural Selection

A)  Relationship between Genotype and Phenotype

B)  Relationship between Phenotype and Fitness

C)  Therefore there is a relationship between Genotype and Fitness

III)        Fitness

A)  Fitness – Average per capita contribution of a genotype to the population after 1 or more generations

B)  Calculation of Fitness

                                          Genotype               A                         B

                                         Fecundity          60 eggs                40 eggs

                                     Survivorship             0.05                     0.10

 

                        Net Growth Rate (R)               3                         4

 

Frequency of genotype in Population             0.20                     0.80

 

            Population Growth Rate () =        (3 x 0.20)    +      (4 x 0.80)

                                                        = 3.8

 

                           Relative Fitness (Wi)           3/4                       1.0

Average Fitness

                                                     =        0.2 x 0.75    +      0.8 x 1.0

                                                        = 0.95

Coefficient of Selection (s)                 =         1.0 – s                    1.0

                                                      s = 0.25

C) 

1 — Viability
 

5 – Compatibility
 
Components of Fitness/Selection

2 – Sexual
 

4 – Gametic
 

3 – Fecundity
 

 


1 – Survival to Reproduction

 

2 – Mating Success

 

3 – Number of Gametes Produced

 

4 – Segregation and Viability of Gametes

 

5 – Fertilization Success

 

D)  Models of Selection

a.    Recessive, Deleterious Allele

 

1)   General Model

 

                      Genotype       AA         Aa        aa

                          Fitness        1            1       1 – s

 

0       ≤ s ≤ 1.0

 

          e.g., if s = 0.10, 10% fewer individuals survive to reproduce

 

Define F(A) = p and the F(a) = q

 

                      Genotype       AA         Aa        aa

At Birth
 
                    Frequency        p2         2pq       q2

                          Fitness        1            1       1 – s

                          Fitness      W11        W12      W22

             Adult Frequency     p2         2pq  q2(1 – s)

 

 

Remember p2 + 2pq + q2 = 1.0

 

Therefore

p2 + 2pq + q2(1 – s) < 1.0

 

Need to normalize adult population frequencies to new maximum

 

*  = Average Adult Population Fitness

 

* = p2 + 2pq + q2(1 – s)         expand

     = p2 + 2pq + q2 – q2s

     = (p2 + 2pq + q2) – q2s       simplify

     = 1 – q2s

 

2)   Genotype Frequencies in the Next Generation

 

In Adults
 
                      Genotype                AA             Aa              aa

                    Frequency                      

 

3)   Allele Frequencies in the Next Generation

 

Define:

U’ = F(AA) in next generation

V’ = F(Aa) in next generation

 

p’ = F(A) in next generation

 

p’ = U’ +

 

     =  +

 

     =

 

     =

 

     =

 

4)   Change in Allele Frequency from One Generation to the Next

Dp  = p’ – p

                                =  – p

                                =  –

                                =  –

                                =

                                =

Remember:

                      p, q, and s are all ≥ 0.0 (i.e., positive) and always ≤ 1.0

 

Therefore:      Dp ≥ 0.0 (i.e., always positive)

 

p = Frequency of the non-deleterious allele (A) and is always increasing in successive generations and eventually the frequency of the A allele will equal 1.0

 

Note also:

 

* = 1 – q2s and as p increases q decreases and the average population fitness is always increasing (i.e., 1 is being reduced by a smaller and smaller amount as p increases).