GN 301 Module 7: Population Genetics and Hardy-Weinberg Equilibrium

Define:

• Population Genetics: study of inherited variation in populations over time and space
• Population: local group of a species among which mating occurs
• Gene Frequency: the ratio of a particular allele to the total of all other alleles of the same gene in a given population.

Calculating frequencies: Example

Calculate allele (gene) frequencies and genotype frequencies from population data

• Genotypic Frequencies:

f(GG) = 300/1000 = 0.3

f(GS) = 600/1000 = 0.6

f(SS) =  100/1000 = 0.1

• Allele Frequencies: (note: allele frequency and gene frequency are synonymous terms)

f(G) = (2(300) + (600)) / 2(1000) = 1200 / 2000 = 0.6

f(S) = 1 - f(G) = 0.4

• Gametic Array: 0.6G + 0.4S Genotypic Array:  0.3GG + 0.6GS + 0.1SS

Calculate allele frequencies from genotypic array: Given a genotypic array of 0.3 GG + 0.6 GS + 0.1 SS

q^2 + 2pq = f(SS) + f(GS)0.5 = 0.1 + (0.60.5) = 0.4

(0.5 is used to account for the fact that the S allele can come from either parent in the GS genotype)

Therefore, q = sqrt(0.4) = 0.632

To calculate the frequency of the G allele (p), we can use the formula:

p + q = 1

Therefore, p = 1 - q = 1 - 0.632 = 0.368

So, the allele frequencies are:

f(G) = 0.368

f(S) = 0.632

Given the progeny generation below, calculate the genotypic and allele frequencies

f(GG) = 0.3          f(GS) = 0.6       f(SS) = 0.1

f(G) = 0.6, f(S) = 0.4

Hardy-Weinberg Law

• Hardy-Weinberg Law: After one generation of mating, allele frequencies will remain at equilibrium.

• Conditions:

  • large population
  • no selection
  • no immigration
  • no mutation
  • panmictic

• Frequencies at equilibrium: for population that has gametic array of pA + qa, the genotypic array at equilibrium will be p2 AA + 2pq Aa + q2 aa

• NOTE: p + q = 1 and p2 + 2pq + q2 = 1 ALWAYS!

Hardy-Weinberg Law Predictions eg: Gametic Array of 0.6 A + 0.4 a. What will be the genotypic array if this population is allowed to mate at random with all conditions for HWE being met?

0.6 A and 0.4 a

Eg. Assuming HWE, calculate allele and genotype frequencies in a population

The frequency of albinism (inherited as an autosomal recessive disorder) in a certain population is 1 in 10,000. What proportion of the population is expected to be heterozygous for albinism?

p2 AA + 2pq Aa + q2 aa

q2 = 1/10,000 = 0.0001, q= sqrt(1/10,000) = 0.01

q2 = 0.0001

q = 0.01

1-q = p, 1 - 0.01 = 0.99

2(0.99)(0.01) = 0.0198

So 198/10,000 are heterozygous for albinism. \n

Practice Problem: A population was scored for the MN blood locus. 31% are type M, 42% are type MN

And 27% are type N. What is the frequency of the M allele?

MM = 31%  = 0.31

MN = 42% = 0.42

NN = 27% = 0.27

f(M) = f(MM) + ½(MN)

0.31 + 0.21 = 0.52 \n

Sample Problem: Hitchhiker’s thumb is due to the presence of a dominant allele (H). Individuals without hitchhiker’s thumb are genotype hh. In a population of 1250 students, there were 282 with Hitchhiker’s thumb and 968 without hitchhiker’s thumb. If this population is in Hardy-Weinberg equilibrium frequencies with regards to this locus, how many people are expected to be heterozygous (genotype Hh)? How many people are expected to be genotype HH?

Hitchhikers: Hh, HH, Not hitchhiker= hh

hh = 968 = 0.7744 out of 1250

HH and Hh = 282 = 0.2256 out of 1250

Total individuals: 1250

Frequency of H allele (p) = (282(2)) / 2(1250) = 0.226

q = 1 - p = 1 - 0.226 = 0.774

p^2 + 2pq + q^2 = 1

HH = q^2

sqrt(0.7744) = 0.88 = q

1 - 0.88 = 0.12 = p

p2 = (0.12)^2 = 0.0144

(0.0144)(1250) = 18 \n \n - It is estimated that each of us carries about__2-3__ alleles for debilitating recessive disorders.