Genetic Variation in Populations Flashcards

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1
Q

Allele and genotype frequencies

A

Frequency of A in population (p) - Number of A alleles/total number of alleles

Frequency of a in population (q) - Number of a alleles/ Total number of alleles

Think of p and q as GAMETE genotype frequencies across a population

p+q=1

Genotype frequencies:
AA - p2
aa - q2
Aa = 2pq

p2 + q2 +2pq = 1

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2
Q

When to use hardy weinberg

A

Mating random

Population stable

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3
Q

Using hardy-Weinberg to calculate cystic fibrosis carrier frequency

A

Carrier frequency = 2pq

Incidence of CF affected individials - q2- 1/2000 = 0.0005

q = Square root of 0.0005 = 0.022

p+q = 1 therefore p=1-0.022 = 0.978

2pq = 2 x 0.978 x 0.022 = 0.043 or 1 in 23

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4
Q

Calculating probability of having child with CF

A
  1. Unrelated partner - no CF history
    0.043 squared (probability of both being carriers) x 0.25 (Probability of two carriers of having an affected child) = 4.6x10^-4 = approx 1 in 2160
  2. Sibling with CF
    0.043 x 2/3 (parents carriers, individual unaffected) x 0.25 = 0.072 = approx 1 in 140
  • consider testing of partner
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5
Q

Assumptions of hardy weinberg rule

A
  • Mating random
  • No inbreeding
  • Allele frequencies remain constant across generations - Mutation, selection e.g. heterozygote advantages, genetic drift (random changes in population)
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6
Q

Measures of inbreeding

A
  • Coefficient of relationship (R) - Proportion of alleles shared by two people by having common ancestors (identity by descent)

sum of (½)n where n = number of links through a common ancestor between two individuals
for full siblings: (½)2 + (½)2 = ½

Coefficient of inbreeding (F): proportion of loci at which individual is expected to be homozygous
½R of parents
if parents are cousins F = ½(½4 + ½4) = 1/16 = 0.0625

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7
Q

Inbreeding and risk of recessive disease

A

For first cousins (Janet and John) and CF:
chance of John being carrier for cystic fibrosis = 0.043 (q = 0.022)
chance of Janet being carrier if John is a carrier = 1/8 (R for cousins)
chance of both being carriers and having an affected child = 0.043 x 0.125 x 0.25 = 0.0013 or 1 in 744 (compared to 1 in 2160 if they were unrelated)

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8
Q

Inbreeding and fitness - general

A

5 categories:
reproductive success, risky behaviours, cognitive ability,
body size,
health

Especially males

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9
Q

Mutations

A
  • 1 per 30 million base pairs per generation (100 per genome)

Average of one protein-coding gene per generation

Persistence depends on:
- Type of mutation (dominant, recessive, X-linked)
- Selection (positive, neutral, negative)

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10
Q

Persistence of deleterious mutations

A
  • Dominant lethal - single generation
  • Dominant conditions affecting reproductive success: one t0 a few generations
  • Late onset dominant (e.g. Huntington’s): many generations
  • X-linked lethal: 1/3 lost per generation
  • Autosomal recessive: several-many generations but mostly eventually lost, depending on pop size, selective disadvantage

Balanced by de novo mutations

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11
Q

Random drift: Founder effect

A

Source population - Contains mutant alleles
->
Founder of new populations
->
New population - Reduced variability and increases frequency of previously rare alleles

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12
Q

Multifactorial traits

A

e.g. height, skin colour, susceptibility to disease, response to drugs

Number of genes:
Single gene -> Polygenic

Vp = Vg + Ve -> Heritability -> h2 = Vg/Vp

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13
Q

Heritability

A

How much of the observed variation in a trait is caused by genetics

Highly penetrant, single gene disorders should have heritability of 0 (no variation)

In reality – nearly always some variability (effects of modifier genes + environment)

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14
Q

Complex genetic diseases

A
  • Unlike Mendelian disorders, there is no clear pattern of inheritance
    But tend to “run” in families
    Few large pedigrees of multiply affected individuals
    Most people have no known family history
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15
Q
A
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