Lecture 5 - Quantitative variation 3 and Mutation Flashcards

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1
Q

What is artificial selection?

A

The process by which humans breed plants and animals to select for certain traits

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2
Q

What are the features of natural selection?

A
  • reponse to selection can lead to phenotypes beyond natural variation
  • there are limits to the response (no dog has six legs)
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3
Q

What was Johannsen’s unsuccessful selection experiment?

A

1926

  • started with an inbred line of seeds
  • bred heavy seeds and light seeds
  • there was no response to selection
  • later repeated the experiment with a mixture of seeds which led to a response
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4
Q

What does a response to selection require?

A
  • genetic variability of the trait

i. e. the trait needs to be heritable

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5
Q

What factors will influence the strength of selection ?

A
  • the greater the heritability, the stronger the response to selection
  • the strength of selection will also affect (the difference in fitness between individuals in the parental population)
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6
Q

How is the response to selection limited and by what?

A
Often plataeus
-no further genetic variation 
-traits may be linked:
on the chromosome (physical linkage)
through negative pleiotropic effects
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7
Q

What is pleiotropy?

A

when one gene influences multiple, seemingly unrelated phenotypic traits

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8
Q

What do response to selection graphs help to visualise?

A

the different factors that determine the magnitude of the response to selection

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9
Q

What is the formula for the s (selection differential)?

A

s = Ms - M

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10
Q

What is the formula for the response to selection?

A

R = M’ - M

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11
Q

How does the mean in the offspring generation often differ from the mean of the selected parents?

A

The mean in the offspring generation is usually smaller than the mean of the selected parents
-this means that the response is less strong than selection

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12
Q

Why is the reponse to selective breeding in the offspring generation often less strong than selection?

A
  • not all phenotypic variation within the parental generation is due to genetic variation i.e. h2 <1
  • heritability is therefore useful for predicting the response to selection
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13
Q

How can heritability help to predict the response to selection?

A
  • not all phenotypic variation within the parental generation is due to genetic variation i.e. h2 <1
  • through the breeders equation
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14
Q

What is the breeders equation?

A

R = h^2*s

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15
Q

How does a response to selection graph look if h2 = 0?

A

Use breeders equation
R=0
M’ and M are the same

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16
Q

How does a response to selection graph look if h2 = 1?

A

Use breeders equation
R=s
M’ = Ms

17
Q

How was heritability and the response to selection used to predict selection on beak size in the Galapagous finches?

A

Finches hatched in 1976, before the drought
-mean bill depth - (M) = 9.0mm
Birds that survived the drought (parents)
-(Ms) = 10.2mm
Next generation
-(M’) = 9.7mm
Selection differential
Mean selected parents - population mean before drought
s = Ms - M
=10.2mm - 9mm = 1.2m
Predicted response to selection
R = h^2s
R = 0.8
1.2 = 0.96mm
Predicted mean bill depth next generation, M + R
= 9mm +0.96 = 9.96
Actual measurement 9.7mm
Actual response
-New mean - previous mean = 9.7 - 9 = 0.7

18
Q

What are the features of mutations?

A
  • help create variation
  • understanding the mutation process helps understanding of nucleotide composition of genomes and the difference in it among species (e.g. elevated TA or GC content)
  • occur at very low frequencies (although highly variable)
19
Q

What are the different types of mutation?

A
Point mutations
Insertions and deletions
-gene duplication 
-copy number variation 
-repeat sequences i.e. microsatellites
Inversions
Lateral gene transfer
20
Q

Why use point mutations?

A
  • they are very common
  • their mutation process is very well understood
  • their effect is reasonably well understood
21
Q

How much variation is there in SNPs in the human population?

A

-Human mutation rate = ~ 1 X 10^-9 muations/position/generation
-Diploid genome is ~2X3.2X10^9 nucleotides
-~6.4 mutations/diploid genome/generation
-7 billion people
-7 billion X 6.5 mutations/generation/position
= 44.8 X 10^9 mutations per generation

Large genomes and population sizes mean that there are a lot of mutations in a population

22
Q

What is the human mutation rate?

A

-Human mutation rate = ~ 1 X 10^-9 muations/position/generation

23
Q

How many nucleotides are there in a diplod human genome?

A

-Diploid genome is ~2X3.2X10^9 nucleotides

24
Q

What are the two types of effects of mutations on allele frequencies?

A

Mutations can be

  • irreversible
  • reversible
25
Q

What is the result of an irreversible mutation in the absence of other forces?

A

eventually leads to the loss of the initial allele

26
Q

What is the formula for the frequency of A (original allele in an irreversible mutation) after time t?

A

Pt = Po (1 - μ (mutation rate of A to a/generation))^t

27
Q

What are the results of allele frequencies with a reversible mutation?

A

A: initial allele with frequency p
Reversible mutation to a (has an initial frequency of q)
-results in an equilibrium of allele frequencies

28
Q

Can the equilibrium allele frequences of a reversible mutation be calculated?

A
Mutation rate A to a: μ
Mutation rate a to A: v
Frequency of A after t generations:
pt = pt-1(1- μ) + (1-pt-1)*v
At the equilibrium, p=pt=pt-1

P=(v/( μ+v))