BIOB51 Lecture 9 and 10 Notes Final Exam notes UTSC Evolutionary BIO

BIOB51 – Lecture 9


Mechanisms of Evolution:
What are the mechanisms that produce evolutionary change in populations?

• Individuals don’t evolve, instead populations evolve over time
  
• Is there a way to monitor alleles in populations to predict whether evolution might occur?
  o Populations: are interbreeding groups of organisms and their offspring Population genetics:
  
  • Is a theory that allows us to track the fate of alleles and genotypes across generations, which is a tool for studying evolution
    
  • Quantitative predictions can be made:
    o Can help tell if the hypothesis is reasonable
    Mechanisms of evolution:
    
    • These are processes that can lead to changes in the allele frequencies across generations
      
    • After Hardy-Weinberg equations, the following can occur:
      o Natural selection o Mutation
      o Migration
      o Genetic drift
      Importance of population genetics:
      
      • Helps give an insight in the natural world
        o Helps us understand population genetics
        
      • Gives an insight of human diseases and conditions o For example, HIV
        
      • Gives an insight of how human actions affect populations
        o How do cows produce milk?
        o How does trophy hunting affect animal populations?
        ▪ For example, the big-horned sheep, or even elephant tusks Population Genetics Problem:
        

• We expect there to be variation in biological systems
o For example, the variation in the timing of reproduction leads to fluctuations in
allele frequencies
Hardy-Weinberg Equilibrium:

• Shows the calculation of expected allele and genotype frequencies if evolution hasn’t occurred
  
• Highlights processes that lead to evolution
  

• The H-W equilibrium occurs if the assumptions are met: o No selection
  ▪ No alleles removed o No mutation
  ▪ No alleles removed or added o No migration
  ▪ No alleles added o No genetic drift
  ▪ No random loss of alleles
  
• H-W equilibrium also says that random mating occurs, which is a predictable genotype
  frequency
  Life Cycle Stages:
  
  • In this example, we will start off with adults who are diploid individuals (2n). The adults put copies of all their alleles in a gene pool population.
    
  • They are equally likely to bump into another gamete (n), which is random mating
    
  • This creates zygotes, which are diploid (2n). These zygotes become adults, and the cycle starts again
    
  • An example of this life cycle is:
    

o Everyone in the room is
wearing black or white shoes. Some people are wearing only black shoes or only white shoes. Others are wearing one black shoe and one white. These are the genotype frequencies.
o The gene pool is formed when all the shoes in the room are thrown in a pile
o Now, pairs of shoes are chosen at random for each person. Every person in the
room gets to wear shoes again, but they might not be the same colour as they originally put in the pile.
▪ A person originally might’ve had 2 black shoes and now has 1 white and 1 black.

• ▪  The genotype of the shoes have changed
  
• ▪  However, the number of the shoes haven’t changed (the allele frequencies
  haven’t changed)
  HW Equilibrium – General Case:
  

• Assuming the H-W assumptions are true, there are two alleles: A1 and A2. o There are 3 possible genotypes:
▪ A1A1 ▪ A1A2

▪ A2A2
o The gene pool:
▪ The initial frequency (A1) = p
• The probability of picking A1 gamete
▪ The initial frequency (A2) = q
• The probability of picking A2 gamete
▪ Since we’re only looking at 2 alleles in one locus: • p +q =1
o Random mating:
▪ A punnet square is
used to determine the frequency of the genotypes and what the zygotes will be
• The probability of
getting pq = 2 o Next generation:
▪ To find the gene pool,
you need to follow the equations:
• Frequency (A1):
o p2 + p (1-p) = p
• Frequency (A2):
o (1-q)q + q2 = q

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H-W Equilibrium – Conclusions:
• When the assumptions are met:
o The allele frequencies in a population will not change generation after generation

o If allele frequencies are given by p and q, genotype frequencies are: (shown in diagram below)
▪ There is random mating assumed here
o Genotype frequencies can change while allele frequencies remain the same ▪ Thus, we cannot just measure genotypes in each generation
▪ As a result, we need Hardy-Weinberg
What use is Hardy-Weinberg?

• A study was done regarding the case of polymorphic Myoglobin
  o It is a protein that supplies oxygen to the muscles
  o Human populations show polymorphism in myoglobin alleles o A1 and A2 differ by 2 base paits
  o In White, Black, and Hispanic populations:
  ▪ f(A1) = 0.41 – 0.46
  
• Polymorphism: is the measure of the number of different alleles in a population
  
• In this study, a population of 100 people in Japan were sampled.
  o The frequency of A1 = 0.755, which is quite high
  ▪ It is different than other measured populations, by p < 0.05
  o The frequency of A2 = 0.245
  
• Hypothesis: does the selection for A1 in Japanese populations lead to increased
  frequency?
  
• Observed genotype frequency: o f (A1A1) = 0.59
  o f (A1A2) = 0.33
  o f (A2A2) = 0.08
  
• The researchers asked whether the observed frequency of A1 is higher than expected
  under H-W
  o It is not a H-W population, there will be no selection on the myoglobin gene
  Effect of Mutation and H-W:
  
  • There is a population of mice:
    o AA mice have dark fur colour
    ▪ = 0.81 genotype frequency o aa have yellow fur colour
    ▪ = 0.01 genotype frequency o Aaisamix
    ▪ = 0.18 genotype frequency
    
  • The gamete frequencies are:
    o A = 0.9 o a=0.1
    

• Mutation is introduced:
o ‘A’ are transformed to ‘a’ alleles at the rate of μ
o Some of the gametes have changed
o μ mutation rate: is the rate at which one allele is converted to another allele, per
generation
▪ Aa converts at a rate of 10-4
• Every 1 of 10 000 gametes gets this conversion from ‘A’ to ‘a’ ▪ freq(A) = 0.9 = p

• • •
▪ freq(a) •
•
•

p* = p - μp
p* = 0.9 – (10-4)(0.9) = 0.89991
= 0.1 = q
q* = q + μp
q* = 0.1 + (10-4)(0.9) = 0.10009

o Random mating then occurs:
▪ Freq (AA) = (p*)2 = 0.80984
▪ Freq (Aa) = 2(p*)(q*) = 0.18014 ▪ Freq (aa) = (q*)2 = 0.01002
o One generation:

• ▪  p* = p - μp
  
• ▪  = 0.89991
  
• ▪  0.90
  
• ▪  So the mutations can change allele frequencies slightly, but it isn’t
  significant p > 0.05
  

o More time:

• ▪  Mutation alone will take a very long time to produce significant allele frequency change
  
• ▪  Without mutation, evolution wouldn’t be possible
  
• ▪  Mutation alone will rarely change allele frequencies
  

• But mutation + selection = potent evolutionary force
• Zygotes:
o Zygotes are then created, and have new frequencies
Mutation and selection:
• Mutation creates new allele: o If beneficial,
▪ There will be selection for mutant allele

▪ The allele increases in frequency

• It is very beneficial and may spread to fixation
  
• It shows evolutionary change
  o If deleterious,
  ▪ Selection for normal allele
  

• Allele lost
▪ It very deleterious, may be lost in one generation
▪ It will not appear in population level frequency at all

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BIOB51 – Lecture 10 Mutation and Selection:

• When using the life cycle diagram, when there is no evolution occurring, there will be the same allele frequencies at the end of the cycle unless one of the known methods of evolution is used.
  
• When selection is added to the life cycle diagram, the population of adults vary in fur colour.
  

o When selection is added during the gamete phase, not all the individuals contribute gametes because they might not all survive and reproduce
o When selection is added during the gene pool phase, not all gametes are equally likely to form a zygote in the next generation
▪ This might be because they might carry mutations that don’t let them fertilize other gametes
o When selection is added during the zygote phase, not all of the zygotes are able to survive till maturity and become a reproductive adult
• When selection is added at any point, quantitative predictions can be made about the offspring.
Lenski’s long-term evolution experiment:

• Took a lot of single cell E.coli (with one genome) and put them in a nutrient-limited medium. This allowed selection to exploit available nutrients faster than competitors
  o The E.coli that had been reproduced are in competition with each other
  
• Mutation is the only source of genetic information; there was no recombination for this
  particular strain of E.coli
  
• In the experiment, 1 cell was placed into a low nutrient medium in order to reproduce.
  Every day, a sample of these cells was produced and put into a new low nutrient medium. o A lot of these are randomly chosen by selection and then the cycle is repeated.
  o Every day there were 7 generations produced
  o This was repeated for 10 000 generations, which took 1 500 days.
  o Some samples of each day and generation were taken and frozen
  
• The data resulted in the relative fitness of one line being under selection, when compared
  to the ancestral line.
  o The ancestral line cells were adapted to living in the lab for generations
  o When the ancestral and new generation lines were competed regarding the
  relative fitness, the fitness gradually increases, although there are periods where it doesn’t change, and other times when it jumps in change.
  
  • ▪  During these “jumps”, a rare beneficial mutation occurs and spreads through the population
    
  • ▪  During the periods of no change, the mutations don’t survive because selection is operating
    
  • ▪  If relative fitness was equal to the ancestral line, it would equal 1
    

• ▪  If the relative fitness was greater than the ancestral line, it would be greater than 1
  
• ▪  If the relative fitness was less than the ancestral line, it would be less than 1
  

• In this experiment, the fitness increases as time passed
o However, there is variation among lines in relative fitness.
o The mutations that arise aren’t the same mutations for each line. Different fitness levels show that there was more competition within the experiment

• When there is a stable environment, the mutations aren’t deleterious since competition is introduced
  
• When there are generations, the likelihood of beneficial mutations occurring increases
  o They ran the experiment for 10 000 human generations, which is about 200 000 –
  300 000 years.
  Selection:
  

• If we assume that phenotypes fall into discrete categories that are determined by genotypes, we can assign relative lifetime reproductive success to each genotype and use the H-W equilibrium
o For example, there are 2 alleles: R and r ▪ The intitial freq(R) = 0.5
▪ There are about 1000 individuals in the population ▪ The selection of initial population:

• All RR adults survive
  
• 90Z% of Rr adults survive
  
• 60% of rr adults survive
  o Do allele frequencies change in the next generation? By how much? ▪ Follow the life cycle:
  

• Allele frequencies: gametes (n)
  
• Genotype frequencies: zygotes (2n)
  
• # of individuals: population
  
• Action on individuals (fitness): who survives/reproduces
  
• Calculate genotype frequencies: among parents (2n)
  o Sum of genotype frequencies:
  ▪ For example, freq(RR) + freq(rr) = 1.0
  

▪ Every individual is one of these genotypes • Calculate allele frequencies: gametes (n)
o Sum of allele frequencies = 1.0
▪ For example, if there are only 2 alleles, the freq(R)
= 1 – freq(r)
▪ Every gamete carries one or the other allele
o Calculated Example:
▪ The initial population: n = 1000 ▪ Allele frequencies:

• Let freq(R) = p = 0.5
  
• Freq(r) = q = 1 - p
  o =0.5
  ▪ Before selection – genotype frequencies:
  
  • RR=p2=0.25
    
  • Rr = 2pq = 0.50
    
  • rr=q2=0.25
    
  • # of individuals: genotype freq x n = freq x 1000
    

o RR=250
o Rr = 500
o rr=250
o Total = 1000
▪ After selection:

• RR=250(1.0)=250
  
• Rr = 500 (0.9) = 450
  
• rr = 250 (0.6) = 150
  
• Total = 850
  ▪ New genotype frequency:
  
  • RR = 250/850 = 0.2941
    
  • Rr = 500/850 = 0.5294
    
  • rr = 250/850 = 0.1765 ▪ New allele frequency:
    
    • R = 0.2941 + 0.5294 (0.5) = 0.559
      
    • r = 0.5 (0.529) + 0.176 = 0.441
      ▪ The allele frequency has changed due to selection:
      
      • Freq(R) has increased by 0.059
        
      • Freq(r) has decreased by 0.059
        
      • In just one generation, p < 0.05
        Computer Simulation:
        

• The alleles started from a very low frequency
  
• The consistent selection leads to consistent directional change
  

• The stronger the selection, the more rapid change occurs
  
• Significant changes in allele frequency can occur rapidly
  Planned Experiment:
  

• An experiment was conducted regarding Drosophilia melanogaster
o It was exposed to ethanol which is a poisonous, which is broken down by alcohol
dehydrogenase (which is known as Adh)
o The D. melanogaster population has two alleles: AdhF or AdhS
o AdhF homozygote breaks down ethanol 2 times faster than AdhS homozygote o There was significant allele frequency change, just within 6 generations

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Summary:

• Hardy Weinberg:
  o It is a null model, with no evolution o It assumes that:
  ▪ There are no mechanisms of evolution that can operate
  ▪ The gametes associate randomly, which is random mating
  
• Mutation: alone can cause very little evolutionary change
  o It is very slow evolutionary change
  
• Selection on mutant alleles:
  o It can cause significant change, which is more rapid change The case of polymorphic Myoglobin:
  
  • There is a population of 100 people in Japan o f(A1) = 0.755
    o f(A2) = 0.245
    
  • Observed genotype frequency:
    o f(A1A1) = 0.59 o f(A1A2) = 0.33 o f(A2A2) = 0.08
    
  • H-W expectation:
    o f(A1A1) = 0.57 = p2
    o f(A1A2) = 0.37 = 2pqo f(A2A2) = 0.06 = q2 Supertree of human populations
    
    • It shows the human populations using data from 200 studies
    Point mutations and pain:
    • Point mutation is autosomal recessive in the SCN9A gene.
    o It codes for sub-unit in sodium channel in pain sensing neurons
    o This gene causes an inability to feel pain. There are no other side effects