Chapter 26: Population and Evolutionary Genetics
● The study of genetic variation in populations and how it changes over time.
● Patterns of genetic variation within and among groups of interbreeding individuals.
● Microevolution – change within populations of a species.
● Macroevolution – emergence of a new species.
Genetic variation is present in most populations and species
● Population – group of individuals who share a common set of genes, live in the same area
and can interbreed.
● Gene pool – all the alleles shared by these individuals.
The Hardy – Weinberg Law describes allele frequencies in populations
● Populations are dynamic – birth and death rates, migration or contact with other
populations.
● Often some individuals produce more offspring than others, contributing a disproportionate
fraction of their alleles to the next generation.
● Differential reproduction leads to changes in allele and genotypic frequencies.
● Changes in allele frequencies = microevolution.
● If we examine a single genetic locus in a population, we find that the distribution of alleles at
this locus produces individuals with different genotypes.
● Hardy – Weinberg Law – describes what happens to alleles and genotypes in an ideal
population (relative proportions of alleles in the gene pool and the frequencies of different
genotypes) that is infinitely large with random mating and experiences no evolutionary
forces (mutation, migration, selection).
● Makes 2 predictions:
1. Frequencies of alleles in gene pool do not change over time.
2. If 2 alleles at a locus, A and a, after one generation of random mating, the frequencies of
the genotypes AA:Aa:aa can be calculated as p2 + 2pq + q2 = 1.
● It is rare for genotypic frequencies do not go unchanged for generations.
● A single autosomal locus with 2 alleles, A and a, where the frequency of A is 0.7 and a is 0.3.
(0.7 + 0.3 = 1).
o Probability that both will contain A is 0.7 x 0.7 = 0.49. Therefore AA will occur 49% of
the time.
o Aa is 0.7 x 0.3 = 0.21. Therefore Aa will occur 42% of the time (0.21 x 0.21) because
the A or a could come from the male or the female.
o Probability of aa is 0.3 x 0.3 = 0.09. Therefore frequency is 9%.
o 0.49 + 0.42 + 0.09 = 1.
● All genotypes have equal rates of survival and reproduction. All genotypes contribute equally
to new gene pool of the next generation.
● Half of the alleles in Aa individuals carry allele A and the other half carry allele a.
o Frequency of allele A = 0.49 + ½ (0.42) = 0.7
o Frequency of allele a = 0.09 + ½ (0.42) = 0.3
● p+q=1
1
, ● Distribution of genotypes among zygotes: p2 + 2pq + q2 = 1.
● Neutral genes – not being operated on by evolution.
● Dominant traits do not necessarily increase from one generation to the next.
● Genetic variability can be maintains in a population since allele frequencies remain
unchanged.
The Hardy – Weinberg Law can be applied to human populations
● Some people who have unprotected sex with a HIV – positive partner remain unaffected =
homozygous for mutant allele of the gene called CCR5.
Calculating allele frequency:
● The CCR5 gene encodes a protein called C-C chemokine receptor – 5. Receptor for strains of
HIV-1, allowing it to gain entry into cells. Contains a 32 – bp deletion in a coding region, making it
shorter and non-functional.
● In individuals homozygous for this mutation, HIV-1 cannot enter the cell.
● Normal allele = CCR51 (1).
● Mutant allele = CCR5 - ∆32 (∆32).
● Homozygous ∆32/∆32 are resistant to HIV – 1.
● Heterozygous 1/ ∆32 are susceptible to HIV but progree more slowly to AIDS.
● Frequency of allele = number of alleles/total x 100.
Testing for Hardy – Weinberg equilibrium:
● Determine whether population genotypes are in equilibrium.
● Determine the frequencies of genotypes.
● Calculate allele frequencies from genotype frequencies.
o A has a frequency of 0.89 and a has a frequency of 0.11.
o AA = p2 = (0.89)2 = 0.792
o Aa = 2pq = 2(0.89)(0.11) = 0.196
o Aa = q2 = (0.11)2 = 0.012
● Compare expected frequencies to observed frequencies.
● Do a chi square analysis.
● Df = k – 1 – m
● K = number of genotypes.
● M = number of independent allele frequencies.
Calculating frequencies for multiple alleles in Hardy – Weinberg populations:
● We commonly find several alleles of a single gene in a population – ABO blood group.
● Locus I has 3 alleles 1A , IB , IO with 6 possible combinations. A and B are codominant and are
both dominant over O.
● Therefore sometimes phenotypes are identical but genotypes differ (AA and AO).
● p+q+r=1
● p2 + q2 + r2 + 2pq + 2pr + 2 pq = 1
● For the recessive gene, O, the frequency = r (therefore square root).
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● The study of genetic variation in populations and how it changes over time.
● Patterns of genetic variation within and among groups of interbreeding individuals.
● Microevolution – change within populations of a species.
● Macroevolution – emergence of a new species.
Genetic variation is present in most populations and species
● Population – group of individuals who share a common set of genes, live in the same area
and can interbreed.
● Gene pool – all the alleles shared by these individuals.
The Hardy – Weinberg Law describes allele frequencies in populations
● Populations are dynamic – birth and death rates, migration or contact with other
populations.
● Often some individuals produce more offspring than others, contributing a disproportionate
fraction of their alleles to the next generation.
● Differential reproduction leads to changes in allele and genotypic frequencies.
● Changes in allele frequencies = microevolution.
● If we examine a single genetic locus in a population, we find that the distribution of alleles at
this locus produces individuals with different genotypes.
● Hardy – Weinberg Law – describes what happens to alleles and genotypes in an ideal
population (relative proportions of alleles in the gene pool and the frequencies of different
genotypes) that is infinitely large with random mating and experiences no evolutionary
forces (mutation, migration, selection).
● Makes 2 predictions:
1. Frequencies of alleles in gene pool do not change over time.
2. If 2 alleles at a locus, A and a, after one generation of random mating, the frequencies of
the genotypes AA:Aa:aa can be calculated as p2 + 2pq + q2 = 1.
● It is rare for genotypic frequencies do not go unchanged for generations.
● A single autosomal locus with 2 alleles, A and a, where the frequency of A is 0.7 and a is 0.3.
(0.7 + 0.3 = 1).
o Probability that both will contain A is 0.7 x 0.7 = 0.49. Therefore AA will occur 49% of
the time.
o Aa is 0.7 x 0.3 = 0.21. Therefore Aa will occur 42% of the time (0.21 x 0.21) because
the A or a could come from the male or the female.
o Probability of aa is 0.3 x 0.3 = 0.09. Therefore frequency is 9%.
o 0.49 + 0.42 + 0.09 = 1.
● All genotypes have equal rates of survival and reproduction. All genotypes contribute equally
to new gene pool of the next generation.
● Half of the alleles in Aa individuals carry allele A and the other half carry allele a.
o Frequency of allele A = 0.49 + ½ (0.42) = 0.7
o Frequency of allele a = 0.09 + ½ (0.42) = 0.3
● p+q=1
1
, ● Distribution of genotypes among zygotes: p2 + 2pq + q2 = 1.
● Neutral genes – not being operated on by evolution.
● Dominant traits do not necessarily increase from one generation to the next.
● Genetic variability can be maintains in a population since allele frequencies remain
unchanged.
The Hardy – Weinberg Law can be applied to human populations
● Some people who have unprotected sex with a HIV – positive partner remain unaffected =
homozygous for mutant allele of the gene called CCR5.
Calculating allele frequency:
● The CCR5 gene encodes a protein called C-C chemokine receptor – 5. Receptor for strains of
HIV-1, allowing it to gain entry into cells. Contains a 32 – bp deletion in a coding region, making it
shorter and non-functional.
● In individuals homozygous for this mutation, HIV-1 cannot enter the cell.
● Normal allele = CCR51 (1).
● Mutant allele = CCR5 - ∆32 (∆32).
● Homozygous ∆32/∆32 are resistant to HIV – 1.
● Heterozygous 1/ ∆32 are susceptible to HIV but progree more slowly to AIDS.
● Frequency of allele = number of alleles/total x 100.
Testing for Hardy – Weinberg equilibrium:
● Determine whether population genotypes are in equilibrium.
● Determine the frequencies of genotypes.
● Calculate allele frequencies from genotype frequencies.
o A has a frequency of 0.89 and a has a frequency of 0.11.
o AA = p2 = (0.89)2 = 0.792
o Aa = 2pq = 2(0.89)(0.11) = 0.196
o Aa = q2 = (0.11)2 = 0.012
● Compare expected frequencies to observed frequencies.
● Do a chi square analysis.
● Df = k – 1 – m
● K = number of genotypes.
● M = number of independent allele frequencies.
Calculating frequencies for multiple alleles in Hardy – Weinberg populations:
● We commonly find several alleles of a single gene in a population – ABO blood group.
● Locus I has 3 alleles 1A , IB , IO with 6 possible combinations. A and B are codominant and are
both dominant over O.
● Therefore sometimes phenotypes are identical but genotypes differ (AA and AO).
● p+q+r=1
● p2 + q2 + r2 + 2pq + 2pr + 2 pq = 1
● For the recessive gene, O, the frequency = r (therefore square root).
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