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BIOL-202: Basic Genetics Study Notes

Instructor Information

• TA: Andy Garcia Gonzalez
• Email: andy.garciagonzalez@mail.mcgill.ca

Question 14: Relative Fitness and Mutation

Concepts:

• Relative Fitness of three genotypes:
  • wA/A: 1.0
  • wA/a: 1.0
  • wa/a: 0.7

Part A: Next Generation Allele Frequency

• Given: The initial allele frequency $ ext{p} = 0.5$
• Calculation of Mean Fitness ($W$):
  • Formula:
    $W = (p<em>A^2)w</em>{A/A} + 2(p<em>A)(p</em>a)w<em>{A/a} + (p</em>a^2)w_{a/a}$
  • Where: $ ext{p}A = 0.5$, $ ext{p}a = 0.5$ (since $ ext{p}a = 1 - ext{p}A$)
  • Calculation Steps:
  • $W = (0.5^2)(1) + 2(0.5)(0.5)(1) + (0.5^2)(0.7)$
  • $W = 0.25 + 0.5 + 0.175$
  • $W = 0.925$
• Next Generation Allele Frequency ($p'$):
  • Formula:
    $p' = \frac{(p<em>A^2)w</em>{A/A} + (p<em>A)(p</em>a)w_{A/a}}{W}$
  • Calculation Steps:
  • $p' = \frac{(0.5^2)(1) + (0.5)(0.5)(1)}{0.925}$
  • $p' = \frac{0.25 + 0.25}{0.925}$
  • $p' = \frac{0.5}{0.925}$
  • $p' ext{ (next generation allele frequency)} o 0.54$

Part B: Equilibrium Allele Frequency

• Mutation Rate from A to a: $2 imes 10^{-5}$
• Calculation of Equilibrium Allele Frequency ($ ilde{p}$):
  • Formula:
    $ilde{p} = ext{sqrt}\bigg(\frac{ ext{Mutation rate}}{1 - ext{frequency of a}}\bigg)$
  • Here, the frequency of a is $1 - 0.7 = 0.3$
  • Calculation Steps:
  • $ilde{p} = ext{sqrt}\bigg(\frac{2 imes 10^{-5}}{0.3}\bigg)$
  • $ilde{p} = 0.008$

Question 16: Semidominance and Fitness Changes

Concepts:

• In semidominance, the relative fitness of a heterozygote is intermediate between the two homozygotes.
• Example relative fitness values:
  • wA/A: 1.0
  • wA/a: 0.9
  • wa/a: 0.8

Part A: Deleterious Recessive Allele

• Change Needed: Alter a fitness value so that a becomes a deleterious recessive allele.
• Degree of Dominance: Represented by $h$.
• Recall: If $h = 0$, the deleterious allele behaves as fully recessive.
• Fitness Formula for heterozygote:$w_{A/a} = 1 - hs$
  • If $h = 0$, then $w_{A/a} = 1$

Part B: Favored Dominant Allele

• Change Needed: Adjust a fitness value so that A becomes a favored dominant allele.
• Condition: For A to be favored, the fitnesses of A/A and A/a must be equal and higher than that of a/a.
• Adjusted fitness value:
  • Change from $wA/a = 0.9$ to $wA/a = 1.0$

Question 26: Inbreeding Coefficient from Pedigree

Pedigree Analysis

• Scenario: Half-sib mating
• Figure 18-12 shows the pedigree for analysis.

Part A: Known Inbreeding Coefficient for Common Ancestor A

• Given: $F_A = 1/2$
• Goal: Find inbreeding coefficient of Individual I ($F_I$)
• Formula:$F<em>I = \bigg(\frac{1}{2}\bigg)^n (1 + F</em>A)$
  • Where $n$ is the number of individuals in the inbreeding loop (not counting I).
• Result Calculation:
  • $F_I = \bigg(\frac{1}{2}\bigg)^3(1 + \frac{1}{2}) = \frac{1}{8}(1.5) = \frac{3}{16}$

Part B: Known Inbreeding Coefficient for Individual I

• Given: $F_I = 1/8$
• Goal: Find inbreeding coefficient of Common Ancestor A ($F_A$)
• Rearranged Formula:
  $F<em>I = \bigg(\frac{1}{2}\bigg)^n (1 + F</em>A)$
• Result Calculation:
  • $\frac{1}{8} = \bigg(\frac{1}{2}\bigg)^3(1 + F_A)$
  • By solving: $F_A = 0$

Question 32: Inbreeding Coefficient for Various Matings

Task Overview

• Use pedigree diagrams to calculate the inbreeding coefficient (F) for offspring from various mating types:
  • Parent-offspring matings
  • First-cousin matings
  • Aunt-nephew or uncle-niece matings
  • Self-fertilization of a hermaphrodite

Part A: Parent-Offspring Matings

• Calculation of $F$:
  • Formula derived:
    $F = (1/2)^2 = 1/4$

Part B: First-Cousin Matings

• Calculation of $F$:
  • Formula:
    $F_I = (1/2)^5 + (1/2)^5 = 1/32 + 1/32 = 1/16$

Part C: Aunt-Nephew or Uncle-Niece Matings

• Calculation of $F$:
  • Formula:
    $F_I = (1/2)^4 + (1/2)^4 = 1/8$

Part D: Self-Fertilization of a Hermaphrodite

• Calculation of $F$:
  • For self-fertilization, the formula is simple:
    $F_I = \frac{1}{2}$