Lecture 25

Question 1

Selective agents on Adh:

Answer 1

• Alcohol
• Temperature
• Alcohol and temperature
• Something else?

Question 2

Clarke Protocol:

Answer 2

5. Difference in molecular function must be reflected in fitness differences in postulated selective regime
6. Natural populations must be re-examined to seek a comprehensive explanation for observed allele frequency distributions in space and time
7. Test your hypothesis by perturbing (changing) allele frequencies in a natural population

Question 3

Fitness in the presence of ethanol:

Answer 3

• Population cages AdhF frequency increases to fixation when select with lower concentrations of ethanol
• Mixed success

Question 4

Fitness in the absence of ethanol:

Answer 4

• Population cages frequencies converge on an intermediate frequency

Question 5

Chateau Tahbilk Winery:

Answer 5

• Populations of dros. in the cellar and outside are separate for most of the year
• Once a year there is massive unidirectional gene flow from outside via the fermentation area in the cellar
• This replenishes cellar population

Question 6

What are the conditions in the Tahilk winery?

Answer 6

• Range from 8 - 22 degrees
• The population is temperature dependent, and stop breeding at 15 degrees
• Most flies die over winter and there are tiny sub-populations on each vat

Question 7

Evidence for natural selection:

Answer 7

• Parallel clines over 5 continents of the S allele decreasing further away from the equator
• Selection for the fast allele in low alcohol environments in cages, selection for a polymorphism even in the absence of alcohol
• Lack of fluctuation in allele frequencies when population size crashes

Question 8

Chateau Tahbilk Overview:

Answer 8

• Returned to the pre-perturbation levels

- Returned at a rate which was proportional to temperature/ADH activity relationship

Question 9

Perturbation Experiment 1:

Answer 9

• Release a large population of flies into the cellar 1 with an increased frequency of Adh-F
• Same thing done in cellar 2 with a decreased frequency of Adh-F

Question 10

Balancing selection maintains F/S polymorphism:

Answer 10

• Lack of noise in allele frequencies in face of potential for drift
• Perturbation experiments
• Parallel clines
• Environmental heterogeneity

Question 11

Problems with the conclusions drawn:

Answer 11

Unable to generate consistent fitness estimates for the three genotypes

• Type of balancing selection unclear
• Cannot define the selective agent

Question 12

For other systems:

Answer 12

• Parallel clines can be found
• Selective agents are generally unidentified
• Fitness differences between genotypes are extremely difficult to demonstrate
• Detailed studies of inividual populations has not been done

Question 13

Clarke approach:

Answer 13

1. Choose gene

2. Attempt to find the selective agent

Question 14

The reverse to the Clark approach may be more useful:

Answer 14

1. Choose a selective agent which has an impact on viability eg) heat stress, insecticides, ethanol tolerance
2. Identify gene(s) responsible for adaptation (mapping/DNA seq, candidate gene approach)

Question 15

Benefits of using bacteria for studies:

Answer 15

• Allozymic variation exists
• Short generation time
• Population size is near infinite (drift effects are minor)
• Defined cultural medium, so the environment is controlled and manipulated
• Different alleles can be introduces into an identical background using transduction
• Haploid, no heterozygotes

Question 16

Chemostat:

Answer 16

• Volume of 30ml, this is a big population size
• 600 generations
• Strains with different alleles are competed pairwise
• T5 bacteriophage resistance is used as a genetic marker
• Nutrient medium enters at top, the same amount of bacteria drip out the bottom, allowing us to sample a population through time (lacz1 and lacz2)

Question 17

Growth rate = fitness:

In(pn/qn) = ln(po/q0)+n x Inw

Answer 17

• Growth rate of two strains in chemostat is a measure of relative fitness: This formula is used to work out the relative fitness of two strains
• po = initial proportion of strain A
• qo = initial proportion of strain B
• n = number of generations
• pn = proportion of Strain A after n generations
  qn = proportion of strain B after n generations
• w = w/1 = relative fitness of strain A (relative to strain B)

Question 18

Five pgi alleles were compared pairwise:

Answer 18

• Activity of PGI allozymes differed when measured under chemostat conditions
• pgi converts glucose-6-p to fructose-6-p and reverse

Question 19

Observations from pgi experiment:

Answer 19

• within the limits of detection but no selection

- one allele promotes decreased growth with limiting factors

Question 20

Comparisons in fitness between different types of sugars:

Answer 20

• Strains with different ‘natural’ lac operons were competed for lactose or one of three different B-galactosidase substrates which were not normally encountered
• Very small fitness differences when the substrate was lactose
• Very big fitness differences when different B-gal substrates were used.