Population Genetics

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Population genetics is the study of how Mendel's laws and other genetic principles apply to an entire population.

1. What forces does the environment (selection) exert on the frequency of certain alleles. See Figure 27.4.

2. Population genetics involves measuring the frequency of different alleles for a given population, often comparing what is found with what might be expected.

Genetic Polymorphism

Determining Allelic Frequencies

1. If you can determine the phenotype directly from the population, then e.g. MN blood types (number of M alleles)/(total number of M and N alleles).

  • Example: 4,000 M, 3000 MN, and 756 N.
    2X4000 = 8000
    1X3000 = 3000
    total M = 11,000
  • Total of all alleles = (4,000 + 3,000 + 756) X 2 = 15,512 11,000/15,512 = 0.709 for M

    Work out N the same way. However, remembering your p's&q's:

    What do you guess for the frequency of the N allele? p+q = 1 If p=M then N = .291

    In this case M=p and N=q. If you know either p or q, you can easily calculate the other.

    1-q = p; 1-p = q

    Another method for determining the frequency:

    freq. A = freq. AA + (freq. Aa/2)

    or

    freq. a = freq. aa + (freq. Aa/2)

    Once one is determined, the other will be easily calculable.

    E.g.: Freq AA = 1,000/2,000 = 0.50
    Freq Aa = 800/2,000 = 0.40
    Freq aa = 200/2,000 = 0.10

    All added together it equals 1.00.

    (Sometime, genotypic frequencies are all that are given, making this method much more attractive.)

    Hardy-Weinberg Equilibrium

    The allelic frequencies of a population will remain in a stable equilibrium if there is no selection, genetic drift, mutation and migration, a condition known as the Hardy-Weinberg Equilibrium. It also requires that there is random mating for the gene in question; and, to a lesser extent, within the population in question.

    What if the heterozygous members of a population cannot be distinguished from the homozygous dominant members. How might we estimate the gene frequency?

    1. A population in equilibrium is one where the frequencies of the alleles do not change from generation to generation.

    2. This means no environmental or social selection; or, non-random mating. Selection causes discrimination against or preference for alleles. Non-random mating might increase homozygousity, or theoretically, heterozygousity in a population above what would be predicted by the Hardy-Weinberg equilibrium.

    3. This also means no mutation. Over short periods of time mutation frequency typically has little effect on frequency of alleles in a population.

    4. This means no genetic drift. Small populations are subject to random fluctuations in allelic frequencies. Why do you think this would be the case.

    5. The population is large enough to prevent inbreeding at such a level as to significantly increase homozygousity in the population.

    The equation for Hardy-Weinberg Equilibrium:

    (p+q)2 = 1 (As in (p + q) squared.)

    Blue eyes in the class:

    If 8 students have blue eyes, and 42 have brown eyes, then 8/50 = 0.16 = q2. The frequency of q is 0.4, which is the frequency of the blue eyed allele. Then, p = 1-q = 0.6, which is the frequency of the brown eyed allele.

    How many heterozygotes do we have in the population?

    (p+q) x (p+q) = p2 + 2pq + q2 = 1

    The 2pq is the frequency of the heterozygotes.

    Therefore, 2 x (0.4 x 0.6) = 0.24; 0.24 x 50 students = 12 students are heterozygotes.

    Calculate the other frequencies and number of students. Do they add up to 50? What does this prove?

    Why do (p+q)2 = 1 ? (1x1=1)

    p = frequency of one allele in the population.

    q = frequency of alternate allele in the population.

    Therefore, we expect that gametes will either have A or a, in the same p, q frequency.

    pp + pq + pq + qq

    p2 + 2pq + q2

    (p + q)2

     The standard proof for the maintenance of Hardy-Weinberg equilibrium is as follows:

    female
    AA

    .25
    Aa

    .5
    aa

    .25
    male
    AA

    .25
    1

    .062
    2

    .125
    3

    .062
    Aa

    .5
    4

    .125
    5

    .25
    6

    .125
    aa

    .25
    7

    .062
    8

    .125
    9

    .062

     

    1. AAxAA Cells 1 = 0.062  
    2. AAxAa Cells 2, 4 0.125 + 0.125 = 0.25  
    3. AAxaa Cells 3, 7 0.0625 + 0.0625 = 0.125  
    4. AaxAa Cells 5   0.25  
    5. Aaxaa Cells 6, 8 0.125 + 0.125 = 0.25  
    6. aaxaa Cells 9 = 0.062  
           
     =
    		 1.0

     

    Progeny Genotypic Frequency After Next Generation

      Mating Proportion AA Aa aa    
    1. AA X AA 0.0625 0.0625        
    2. AA X Aa 0.25 0.125 0.125      
    3. AA X aa 0.125   0.125      
    4. Aa X Aa 0.25 0.0625 0.125 0.0625    
    5. Aa X aa 0.25   0.125 0.125  
    6. aa X aa 0.0625 ______  _____ 0.0625  
           0.25  0.5 0.25
    =
    1

    • So p+q = 1; p2 + 2pq + q2=1; (p+q)2=1.
    • What if q2=0.4. This means 40% of the population has the recessive phenotype. What proportion of recessive alleles are in the heterozygous population?
      • 'V-.4=.63; p=.37; q=.63. 2(.37)(.63)=2pq=.46; .23+.4=.63; .23/.63=.365=proportion of recessive alleles in the heterozygotes; .4/.63=.635=proportion of recessive alleles in the homozygotes.

    • What if q2=0.004. What proportion of a genes are in the heterozygous population?
      • 'V-.004=.063; p=.937; q=.063; 2(.063)(.937)=.118; .059+.004=.063;
      • .059/.063=.94=proportion of recessive alleles in the heterozygotes;
      • .004/.063=.06=proportion of recessive alleles in the homozygotes.
    • From the above, the difficulty of using selection by culling to eliminate undesired recessive alleles from the population becomes obvious. The lower the frequency of the recessive allele the greater the proportion is found in the heterozygotes, which cannot be distinguished from the homozygous dominants.

      Multiple Alleles in Hardy-Weinberg Equilibrium

      p + q + r = 1

      pp + pq + pr + pq + qq + qr + pr + qr + rr

      = pp + 2pq + 2pr + 2qr + qq + rr = 1

      = (p+q+r)2 =1

      For ABO blood type, AB and O can be estimated from the population.

      r = 'V-O; p = 1 - 'V-(B+O); q = 1 - 'V-(A+O)

      This will usually not equal 1 so a correction value must be estimated.

      Correction value (CV) = 1 - p + q + r from above.

      Corrected r = (r + CV/2) X (1 + CV/2)

      Corrected p = p X (1 + CV/2)

      Corrected q = q X (1 + CV/2)

    Sex-Linked Genes

    Sex Linked Genes: Males have much higher frequency of the recessive phenotype in the population than females.

    Males: p = A; q = a. We can estimate gene frequency directly from the male population, if survival is equal for both phenotypes.

    This would make multiple allele estimates much easier.

    (Calculate several different phenotypic frequencies for males versus females.)

    Changes in Allelic Frequencies

    • Migration (Gene Flow)
    • 1. Population supplying migrants is called donor population.
    • 2. Population receiving is called recipient population.
    • m= proportion of the total population that is from migrant population, (m + recipient) population.
    • q1 = q0 (1-m) + mQ
    • Q = Frequency of allele "a".
    • q1 = population's genotypic frequency after one generation of random mating. (Is one generation enough to establish Hardy-Weinberg?)
    • q0 = Q in generation previous to migration.
    • Change in allelic frequency is q1-q0.
    • See Figure 27.15.

    Genetic Drift

    • Allelic frequencies randomly shift when there is a small number of parents or progeny. See Figure 27.12.
    • Founder Effect: small group from large population begins a new population. There is a change in the genetic composition from the original population. See Figure 27.13.
    • Bottleneck: population is drastically reduced, then reproduces to original size.
    •  

     

    Mutations in Populations

    Constant mutations of an allele would slowly change the frequency towards the mutated allele in the absence of back mutations and selection.

    Can classify: 1) deleterious; 2) neutral; 3) favorable

    Rate of change is from 10-4 to 10-6 / generation.

    This is low.

    Selection

    Selection is one of the major forces for the change of allelic frequencies, acting through differential reproduction.

    Fitness - measured by a genotype's success in transmitting gametes to the next generation relative to other genotypes.

    Example in the readings:

    Genotype
    Survival Rate
    		  
    Proportion of Survivors
    AA
    		 4,563 (after)/6,084 (before)   0.75
    Aa
    		 2,574 (after)/3,432 (before)   0.75
    aa
    		 261 (after)/484 (before)   0.54

    Relative fitness (W) - measures the relative success of a genotype compared to others. Equals the ratio of survival rates with largest survival rate as denominator.

    WAA fitness of AA is 0.75/0.75

    WAa fitness of Aa is 0.75/0.75

    Waa fitness of aa is 0.54/0.75

    1 - W=s; s=Obverse of fitness = selection coefficient.

    sAA = 1-1 = 0

    sAa = 1-1 = 0

    saa = 1-.72=.28

    Complete Selection Against a Recessive Allele

    • Selection against a dominant trait is easy. Eliminate all those that express the trait. The allele will be eliminated in just one generation.
    • Selection against a recessive allele is another story. The more we eliminate those that express the trait, the more the allele is masked by having higher proportions in the heterozygous members versus the homozygous recessives.

    • Eugenics: The application of principles of artificial selection to human populations.
    • Positive Eugenics: The encouragement of individuals with desired genetic traits to have more children than average.
    • Negative Eugenics: The discouragement of individuals with undesired traits from having children. Often, it is selection against a recessive trait.

    • If there is complete selection against recessives, then s = 1 (fitness value = 0) and p2 + 2pq = total remaining post-selection population, and all remaining recessives are in the heterozygotes.
    • Only het X het matings will generate the next generation's homozygous recessives.
    • The proportion of hets in the new population will equal: 2pq / (p2 + 2pq).
      • Therefore the probability of a Het X Het mating is = (2pq / (p2 + 2pq))2.
      • (2pq / (p2 + 2pq))2 X (1/4) = proportion of these mating generating homozygous recessives.
    • We can substitute p for 1-q, because p+q=1. The (2pq / (p2 + 2pq))2 becomes (q / (1+ q))2, which is the frequency of homozygous recessives generated in the next generation after total selection against the homozygous recessives.
    • The allelic frequency will be the square root of the above equation: q / (1+ q); i.e., the frequency of alleles of the recessive type remaining after complete selection against homozygous recessives in the population.
    • As the frequency of the recessive allele decreases in the population, greater is the proportion found in the carriers of the population (as the square function would indicate).
    • Expected frequency after 1 or more generations (n generations) of complete selection:
      • The frequency expected for the homozygous recessives in the population after n generations of continual complete selection against homozygous recessives: (q / (1+ nq))2.
      • The allelic frequency after n generations of continual complete selection against homozygous recessives: q / (1+ nq) = qn.
      • To determine the change in allelic frequency, you simply calculate the expected new frequency using the equation and subtract it from the initial frequency.
    • The above discussion was for total selection against a recessive allele by eliminating all homozygous recessives.

     

    Partial Selection Against Recessive Allele

    Then s < 1. The frequency q after one generation of partial selection is: q1 = ((1-s) q2 + pq) / (1 - sq2).

    See Figure 27.9.

    See Figure 27.10.

    Genetic Diseases in Humans

    • There is approximately a 10% chance of giving birth to a child with serious genetic defects. Although each detrimental allele for a gene might be relatively low in frequency in the population, there are many genes that might be affected. Each human has 30,000 to 40,000 genes. (The actual number of genes a human has is being determined presently.)
    • Recently, many cancers, disease susceptibilities, heart disease, and other ailments have been traced to alleles. It is therefor likely that each of us carry many detrimental alleles. This strengthens the arguments against interbreeding.
    • The environment has a great effect on the rate of damage to genes. Some human populations have been genetically ruined.
    • Humanity could technically launch a brave new world solution to genetic diseases.
    • The technologies of in vitro fertilization, preimplantation genetic diagnosis, embryo freezing, and transplantation could be used to salvage the human gene pool, if mankind inadvertently poisoned it to the point where survival became questionable. See Figure 27.7 and 27.18.

    Polygenes and Quantitative Inheritance

    Quantitative Inheritance is the genetics of plant and animal breeders. It functions from the perspective of those interested in selecting for or against traits important to breeders.

    Traits discussed in the population genetics section are often discontinuous, either yes or no; one or the other.

    1. Quantitative traits are more continuous, a scale with individuals ranging from one extreme to the other.

    2. Quantitative traits are influenced by more than one locus. Often, many genes affect the phenotype.

    See Figure 25.1.

    Polygenes, the genetic basis of quantitative genetics:
    AABB is dark.

    • AaBB, AABb are not as dark as above.
    • AaBb, AAbb, aaBB are lighter than the above.
    • Aabb, aaBb are even lighter yet.
    • Those that are aabb lightest.
    • Each locus can either add or not to the phenotype. Each dominant allele can add more color. Each acts like an incomplete dominant gene, if each is additive.

    The genes influencing survival are often polygenes. See Figure 25.7.

    • The number of genes involved can sometimes be calculated from the extreme case frequency.
    • If the lightest equals 1/64 in the F2 population, then (1/4)z = 1/64; z = 3. See Figure 25.5.
    • When too many different genes are involved, it becomes difficult to determine distinct classes.
    • We then resort to other means.

    Many traits of agricultural importance are of a quantitative nature. We discussed artificial selection on tandem repeated genes before. Many traits in humans are also quantitative. Height, hair characteristics, facial features, intelligence, and many more.

    One way to look at the outcomes for polygenes that add equally to the traits is using the binomial distrubution.

    Means / Variances

    Means and variances are used to analyze quantitative traits.

    • Mean is a measure of central tendency. The mean is the sum of the individual measurements (x) divided by the number of measurements (n).
    • E = sum, x is the measured feature, and n is number of measurements
    • Variation is a measure of the amount of difference between the individuals. Variation will be due to the environment or genetic factors. See Figure 25.4.
    • We can use the variance values to compare different populations.
      • s2 = variance = the sum of the squared differences between the individual measurements and the mean, then divided by the value: number of observations - 1.
      • s2 = (x1 - x)2 + (x2 - x)2 + ...(xn-x)2/n-1.
      • A short cut is to use s2 = (Ex2 - ((Ex)2/n)) / n - 1.
      • Ex2 is the sum of each squared observation; (Ex)2 is the square of the sum of all the observations.
    • To find out what caused the observed variation, compare the variation found in the F1 generation with that found in the F2 generation. The F1 variation is assigned to the environment. Any additional variation measured in the F2 population is said to be due to genetic factors.
    • The standard deviation is useful for getting a better feel for the significance of the variation found, because the units are more relevant to those of the observed values.
    • Standard deviation = square root of s2 = s.
    • Looking at the area a curve covers, the mean, + or - 1.98 times the s, will contain 95% of the individual observations.
    • The most frequent classes will fall + or - 1s (or 1 times the s).
    • See Figure 25.8.
    • Sometimes we must compare populations that are quite different. Variance and standard deviation has little meaning without making adjustments relative to the means of each population.
    • The adjustment is the term: Coefficient of Variance (C.V.). The Coefficient of Variance (C.V.) = standard deviation / mean times 100.

     

    There are both additive genes and dominant and recessive genes to consider in quantitative inheritance. Dominant genes tend to move the average value toward one of the extremes so that the F1 hets have a value that exceeds the midpoint of the continuum.

     

    bb__________________________I_____________Bb___________BB

    Midpoint

    • The midpoint value is (bb + BB) / 2.
    • a= difference between midpoint and BB value.
    • d= a measure of dominant effect is Bb - midpoint value.
    • d/a = magnitude of the dominance effect.

    Estimating Genetic Variation

    How do we separate out the environmental variation from the genetic variation?

    1. Both parental lines and the F1 are considered to have no genetic variation within each group.

    2. Any variation that exists in these populations is said to be due to environmental variance, VE. The average of all three populations is often used to estimate the environmental variation. (VP1 + VP2 + VF1)/3 = VE.

    3. The larger F2 variation is due to the segregation and recombination of alleles, which occurs in the F2 generation.

    • F2 variance is due to both environmental and genetic factors. VF2 = VP = VG + VE
    • VP = Phenotypic variation.
    • VG = Genetic variation.

    4. Genotypic variation is estimated by:
    VF2 - VF1 = (VG + VE) - VE = VG

    5. Each single allele in a gene pair that might add or subtract from the total effect is a portion of the VA of the total genetic variation.

    6. Deviation from the midpoint of the values of the P1 and P2, that value, d, represents the VD of the genetic variation. If d = 0, then VD = 0.

    7. VG = VD + VA

    8. Therefore, VF2 = VP = VA + VD + VE

    9. An estimation of VD, so it can be subtracted from the VG to get the VA (the useful value) is found from using F1 X P1 and F1 X P2 backcrosses: VA = 2VF2 - (VB1 + VB2).

    10. Most of the predictable progress that can be made in breeding programs is in the VA portion of the variation.

    Heritability

    • When genetic variation is high and the environmental variation is small, then offspring tend to resemble their parents to a high degree.
    • Dominant genes add to the uncertainty of what the progeny will be like. This is because masked recessive genes appear which can generate a very different phenotype.
    • The additive portion of the genotype is the most important to the predictability.
    • The additive variability, VA, is used to calculate the H, heritability, H = VA/VP = VA / (VA + VD + VE).
    • Higher the estimate, the more the progress that can be made by the breeder.
    • One problem with the focus on the additive genes is that the potential benefit of quantitative dominant genes is some what ignored. Modern molecular biology and breeding can address this limitation.

    Threshold Traits

    1. Traits, which are all or none, and are polygenes, (quantitative genes) are called threshold traits.

    2. It is one phenotype with a certain proportion of alleles, then switches to an alternate phenotype when the proportion changes over a certain point. See Figure 25.8.