Lecture 7: Selection

Question 1

History of measuring selection (3)

Answer 1

Until 1930’s, biologists assumed that selection was too weak to measure.

• Discovery of (a) industrial melanism in peppered moths & (b) heavy metal tolerance in plants changed this view.
• Followed by many studies of polymorphisms

Question 2

polymorphisms

Answer 2

discrete phenotypes controlled by few genes

Question 3

Selection equation

Answer 3

-s=1-ω of least fit phenotype

ω = lifetime fittness

Question 4

Viability fittness

Answer 4

ωViability of X = proportion of x that survive / highest proportion of phenotype that survived

ex: blue pidgeons excape 10% of attacks, white pidgeons escape 90%

ωViability (white) = 0.9 / 0.9 = 1

ωViability (blue) = 0.1 / 0.9 = 0.11

Question 5

Viability

Answer 5

measure of the number of individuals surviving in one phenotypic class relative to another

Question 6

Reporductive fittness

Answer 6

ωReproduction of X = reproductive output / highest reporductive output

ex: blue pidgeon clutch size is 4 chicks per year, white pidgeon clutch size is 8 chicks per year

ωReproduction ( blue) = 8/8 = 1

ωReproduction (white) = 4/8 = 0.5

Question 7

Lifetime fittness

Answer 7

ωLifetime of x = ωViability x ωReproduction / The Largest (ωViability x ωReproduction) within the population

ex from our pidgeons:

ωLifetime (white) = 1 x 0.5 / 0.5 = 1

ωLifetime (blue) = 0.11 x 1/ 0.5 = 0.22

Question 8

Population mean fittness

Answer 8

ωBAR = “population mean fitness”

ωBAR = p^2ωA1A1 + 2pq ωA1A2 + q^2ωA2A

Where:

(in this case) A1A1 is the white pidgeon (recessive allele) = ωLifetime (white)

(in this case) A2A2 is the blue pidgeon (dominate allele) = ωLifetime (blue)

And A1A2 is the heterozygote. Note this will always be the same ωLifetime of the dominant allele. (because it will be blue in reality so its fittness is the same as the blue)

Question 9

Frequency after selection

Answer 9

Frequency after selection =

p^2ωA1A1 / wBAr for homo A1

2pq ωA1A2 / wBAR for hetero

q^2ωA2A /wBAR for homo A2

Question 10

Starting Allele frequency (2nd generation)

Answer 10

p’ = (freq after selection of A1) + (1/2 freq after selection of A1A2)

q’ = (freq after selection of A2) + (1/2 freq after selection of A1A2)

Question 11

positive selection

Answer 11

when a trait is strongly favored by selection and will increase over generations

Question 12

Directional selection

Answer 12

shifting the average (bell curve) toward an extreme

Question 13

Stabelizing selection

Answer 13

Moves average away from extremes towards middle

Question 14

Disrupting selection

Answer 14

Moves average towards both extremes

Question 15

Overdominance (2)

Answer 15

-Overdominance occurs when a heterozygote has a higher fitness than both homozygotes.

-This results in more heterozygotes than expeced

Question 16

Negative frequency advantage (2)

Answer 16

-where Fitness of each phenotype negatively related to its frequency

-color polymorphism in the grove snail, where predators like thrushes tend to focus on the most common shell pattern, leading to higher predation pressure on that pattern and thus favoring the rarer color variations; essentially, the rarer the snail’s color, the better its chances of survival against the predator’s “search image” for the common pattern.

Question 17

Multiple optimal phenotypes (4)

Answer 17

-when the best phenotype for one task is not necessarily the best for another.

-Lowers variation within populations, may maintain variation within species

-underdominant = hetero disadvantage

ex Traits that make males more attractive to females may also attract attention from predators and parasites