Population Genetics, Summaries of Genetics

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Population Genetics

Section 4 (1.5 hours)

Learning Objectives

• Understand the importance of Hardy Weinberg equilibrium and how to calculate deviance.
• Describe population substructure and how it can confound results. Also understand methods for adjusting for it in analysis.

Single mating pair and offspring

¼ (AA) + 2 / 4 (Aa) + ¼ (aa)

Expected genotype combinations

The Hardy-Weinberg principle

The Hardy-Weinberg law under the

assumption of non-evolving allele frequencies

• The Hardy-Weinberg Law provides two equations allowing us to relate the expected allele and genotype frequencies to each other
• Assume a SNP with alleles A (frequency p) alleles a (frequency q)
• p+q=1 (allele frequencies)
• p 2 +2qp+q 2 =1 (genotype frequencies)

HWE example

• Assume 100 cats (200 alleles) with alleles B and b. 16 of the cats are white (genotype bb). If you assume HWE, what are the allele (B,b) and genotype (BB, Bb, bb) frequencies?
• p+q=
• p 2 +2qp+q 2 =
• q 2 =0.

HWE example

• Assume 100 cats (200 alleles) with alleles B and b. 16 of the cats are white (genotype bb). If you assume HWE, what are the allele (B,b) and genotype (BB, Bb, bb) frequencies?
• p+q=
• p 2 +2qp+q 2 =
• q 2 =0.
• q=0.4, p=0.

HWE example

• Assume 100 cats (200 alleles) with alleles B and b. 16 of the cats are white (genotype bb). If you assume HWE, what are the allele (B,b) and genotype (BB, Bb, bb) frequencies?
• p+q=
• p 2 +2qp+q 2 =
• q 2 =0.
• q=0.4, p=0.
• p 2 =0.
• 2pq=2x0.6x0.4=0.

HWE example

• Assume 100 cats (200 alleles) with alleles B and b. 16 of the cats are white (genotype bb). If you assume HWE, what are the allele (B,b) and genotype (BB, Bb, bb) frequencies?
• p+q=
• p 2 +2qp+q 2 =
• q 2 =0.
• q=0.4, p=0.
• p 2 =0.
• 2pq=2x0.6x0.4=0.
• 16 white cats and 84 black cats (28 Bb, 36 BB)

Calculate expected genotype frequencies

• Lay out what we know:
• q 2 = 65/
• p 2 + 2pq = 150/
• p + q = 1
• p 2 + 2pq + q 2 =

Calculate expected genotype frequencies

• Lay out what we know:
• q 2 = 65/
• p 2 + 2pq = 150/
• p + q = 1
• q = sqrt(65/215) = 0.

How many are expected to be TT vs TC?

• Lay out what we know:
• q 2 = 65/215 = 0.
• p 2 + 2pq = 150/
• p + q = 1
• q = sqrt(65/215) = 0.
• p + 0.55 = 1
• P = 1- 0.55 = 0.
  • p 2 = (0.45) 2 = 0.
  • 2pq =2(0.450.55) = 0.
  • TT = 0.20*215 = 43
  • TC = 0.50 *215 = 107

How different are the frequencies?

Compare expected and observed genotype frequencies: 𝑥 2 = ෎

2 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 Compare to 𝑥 2 for degree of freedom, p < 0.05 = 3. If < 3.841 then population is not out of HWE If > 3.841 then population IS out of HWE