Don't forget to review the two handouts distributed with this lab (the list of human discrete traits and the genetic terminology review sheet)!

I.  Human Variation: Discrete Traits.
  These are physical traits that can be scored as present or absent, or which are manifested as one of a small number of possibilities.  All of these traits have some genetic basis, although in recent years the exact nature of some of them has come into question – some are polygenic rather than monogenic, and some may have an environmental component – but most can be usefully modeled as single-gene traits inherited in simple Mendelian fashion (i.e., one dominant form and one recessive).

     The discrete traits we looked at in this lab are listed on the discrete trait handout, with probable mode of inheritance and pattern of dominance where such information is reasonably well known.

II.  ABO and Rh Blood Groups.
    The ABO and Rh blood group systems are a more complicated case of the discrete trait phenomena mentioned above.  In the case of the ABO system, there are three rather than two alleles (Ia, Ib, and Io), and the first two are co-dominant.   Although there are many other blood group systems, these two are the most important in clinical settings, ABO for blood transfusions and Rh for fetal-maternal compatibility.

    The ABO blood group system is a case where there are four phenotypes (A, B, AB, and O) and six genotypes (IaIa, IaIo; IbIb, IbIo; IaIb; IoIo).  From this information we can tell that the Ia and Ib alleles are codominant and the Io allele is recessive.  This means that the Ia and Ib traits will always mask the Io trait, and if an individual inherits both an Ia and an Ib allele, both will be expressed in the phenotype.  How does this work?

    Essentially, most red blood cells have antigens on their surfaces.  By observing a reaction called agglutination when bloods were mixed, early researchers such as Landsteiner deduced that most red blood cells have certain antigens on their surfaces, and most plasma (the liquid component) has antibodies.  Type A blood, for instance, has A antigens on the erythrocytes (red cells) and anti-B antibodies in the plasma; thus, if Type B blood is mixed with it, the B antigens on the B erythrocytes will react with the anti-B antibodies and cause damage (and vice-versa).

    AB blood has both types of antigen and none of the antibodies, while type O blood has no antigens and both of the antibodies in the plasma.  Thus, type O blood can be given to any blood type, as there are no antigens on the surface of its erythrocytes; similarly, individuals with type AB blood can receive a donation of any blood type, as they have no antibodies to react with the A and B antigens.  Note that the alleles in question here actually code for the presence or absence of antigens on the erythrocytes.

    The Rh system is somewhat more complicated; up to 8 alleles are involved, but here we consider only two, Rh negative and Rh positive.  If you have the antigen, you are Rh positive.  Unlike the case with ABO, an Rh negative person by default does not have anti-Rh antibodies present in the serum, but it is possible to produce them through sensitization (as when, for instance, an Rh- person receives an Rh+ transfusion -- this is why the doctors on "ER" always ask for a unit of "O-neg"; although one Rh+ transfusion won't hurt an Rh- person, a second such transfusion could cause complications, so O negative is the preferred blood for emergency use).  The most commonly discussed instance of sensitization, however, occurs when an Rh- mother carries a fetus that is Rh+.  In this case, some fetal Rh antigens may enter the mother’s system, and antibodies produced (as if the mother were being "vaccinated" against Rh).  Normally this is not a problem for the child, as it takes time for enough reaction to occur to impact fetal health.  However, if the mother later becomes pregnant with another Rh+ fetus, the anti-Rh antibodies in her blood may attack the Rh antigens in the new fetus, casing agglutination and hemolysis of the fetus’ red blood cells.  In severe cases the baby's blood is removed and replaced with Rh negative blood to reduce the level of anti-Rh antibodies.

III.  Population Genetics

    If we are interested in evolutionary processes, we want to be able to examine allele frequencies – although of course selection in the form of environmental pressures must act on the phenotype, we expect that the underlying alleles will also be either selected for or against; both dominant and recessive alleles can be favored in this way.  The example we’ll focus on is the sickle allele.  This case is not only a good example of a two-allele system which has been extensively studied, it also represents a case of genetic adaptation, one of the two ways in which people may adapt to an environmental stressor (the other way is a biological adaptation, which is a response by an individual to a particular stressor).

    Naturally, if we want to examine a population for evidence of microevolutionary change, we must have some idea of what would be expected in the default condition – the null hypothesis must be no evolutionary change.  The best-case scenario for such studies would be repeated sampling of a population over a long time period; ideally sampling would occur at least once per generation for several generations.  Repeated samples would enable us to determine if allele frequencies were changing (the biological definition of evolution) or not.  We might, for instance, test a tropical population for their hemoglobin alleles at two points in time.  A represents the allele for normal hemoglobin, and S represents the allele for the sickle trait:

           Time 1               Time 2
AA         50                       35
AS        100                     100
SS           50                       10

    We want to calculate the allele frequencies for the A and S alleles at both Time 1 and Time 2.  Allele frequency is the total number of copies of an allele divided by the number of alleles in the population (which is twice the number of individuals in the population).  So at Time 1, the frequency of the A allele is:

            A allele frequency = p = [(50x2) + 100]/400 = .5
                (Multiply 50, the number of A homozygotes, by two because each has two A alleles.)

We could calculate the frequency of the S allele the same way, or (in this instance) we could subtract the frequency of the A allele from 1, because the frequencies must always sum to 1.  Clearly, the frequency of the S allele = q = .5.  (By convention, we use p for the frequency of a dominant allele and q for the frequency of the recessive, in a two-allele system.  Note that these frequencies also represent the probability of randomly choosing a dominant -- p -- or recessive -- q -- from the allele population.)

    By Time 2, things have changed.  Now, we see that p = .59 and q = .41:

        p = [(35x2) + 100]/400 = .59
        q = [(10x2) + 100]/400 = .41

    It is obvious that microevolution is occurring – there is a shift in allele frequencies.  We know the reason for this shift in this case: Once a population is exposed to malaria, any individual possessing a copy of the mutated S allele (whether a new mutation or inherited) will have higher fitness than someone without such an allele, as it gives that individual some resistance to the malaria parasite.  However, two copies of this allele result in sickle cell anemia, which is often fatal in childhood, especially in traditional societies without Western medicine.  Even if we did not know the mechanism, which was the case at one time, we would be able to tell that heterozygotes were becoming more numerous at the expense of both homozygotes, and that homozygous S was more strongly selected against than homozygous A.  Note that this is a hypothetical example; in fact, the first S allele would have been introduced through mutation, and in a malarial environment, its frequency would have increased until heterozygotes became numerous enough that their matings began to produce S homozygotes, who have sickle cell anemia; selection would have removed these individuals, decreasing the frequency of S to a population-level balance.  The selection against S homozygotes is strong enough that the frequency of the S allele would never reach .5.

    Unfortunately, it is not usually possible to obtain repeated samples over many years; it is usually difficult enough to sample twice, and sometimes only one sample is available.  In these cases, we need a way to determine from a single genetic snapshot of a trait whether or not evolution is occurring.

    Fortunately, we do have such a tool at our disposal: the Hardy-Weinberg equilibrium formulae.  Essentially, the Hardy-Weinberg equations allow us to mathematically state a null hypothesis of no change—they allow us to obtain allele frequencies and genotype frequencies in a hypothetical situation of no change, to be compared to real-world data to see if change is taking place, and if so, in which direction.  The Hardy-Weinberg equilibrium is the point at which allele frequencies would undergo no further change assuming (1) random mating and (2) that no evolutionary forces are impacting the population – no gene flow, no mutation, no genetic drift, and no natural selection.  In the example we have been considering, we could use the Hardy-Weinberg formula to calculate the probabilities of an individual having each genotype based on the allele frequencies we observe in our one data-collection session, and the above two assumptions.

            The Hardy-Weinberg equation is: p² + 2pq + q² = 1

You will notice that this is the equation obtained by squaring both sides of the allele frequency equation (p + q = 1); in this case, the terms on the left side of the equation represent genotype frequencies in the human population rather than allele frequencies in the allele population.  Using logic, we see that:

        p² = expected frequency of AA
        2pq = expected frequency of AS
        q² = expected frequency of SS

Note that these expected frequencies also represent the probability of a given individual selected at random from the population having each genotype.  This makes sense, because to determine the chance of an individual "drawing" genotype AA from the allele pool, we would need to multiply the chance of the first allele being A (p) by the chance of the second being A (also p).  The probability of "drawing" SS (hence the expected frequency in the population) is calculated the same way.  We multiply pq by 2 in the case of the heterozygote because there are two ways to "draw" this combination (the A could come first or second).  So, if we have only the data from Time 2 above, we would calculate the following expected genotype frequencies:

        AA = p² = (.59)² = .3481
        AS = 2pq = 2(.59)(.41) = .4838
        SS = q² = (.41)² = .1681

To determine if evolution is occurring, we calculate the number of individuals expected per genotype given the same size population as we observed at Time 2 (145 individuals):

Genotype           Expected Frequency                           Observed Frequency
    AA                      .3481(145) = 50                                            35
    AS                       .4838(145) = 70                                            100
    SS                        .1681(145) = 24                                            10

    Clearly, we have many more heterozygotes than expected and many fewer homozygotes of both types; homozygous recessives are particularly few.  Even if we didn't know the precise reasons, we would be able to determine that there was selection against both homozygotes, and that this selection was particularly strong against SS.

    Hardy-Weinberg can also be used to calculate expected genotype frequencies from phenotype data alone, provided that the trait in question is a simple Mendelian trait (i.e., only two alleles, one dominant and one recessive).  In this instance, we would first calculate the frequency of the recessive phenotype in the population, which would give us q² (because each person manifesting the recessive phenotype has definitely "drawn" two copies of the recessive allele in the genetic lottery).  The square root of this number would be q, the frequency of recessive alleles in the allele population.  It is now a simple matter to determine the proportion p of the dominant allele and use Hardy-Weinberg to obtain the genotype frequencies.  For instance, imagine we have a sample of 100 individuals, 85 of whom habitually place their left thumb on top when folding their hands.  If we assume that this trait is selectively neutral, that placing the left thumb on top is a dominant trait coded for by allele F, and that placing the right thumb on top is coded for by allele f, we can determine the expected proportion of heterozygotes:

        First, we know that the frequency of having the phenotype of left thumb on top is .85; it follows that the frequency of phenotype right thumb on top is .15.   This number (.15) is q², the proportion of homozygous recessives (because anyone who places the right thumb on top definitely has two copies of the recessive f allele).  We take the square root of .15 to obtain q, the allele frequency of f = .387.

        Now, we obtain p, the allele frequency of F:

                    p + q = 1
                    p + .387 = 1
                    p = .613
 

        Using the Hardy-Weinberg formula given above, we can determine the proportion of heterozygotes:

        Frequency Ff = 2pq = 2(.387)(.613) = .474

        If we wanted the frequency of homozygous F, we could either use p² or calculate it by subtracting the frequencies of the other two genotypes from 1 (we already know the frequency of homozygous f):  .376.  If we wanted estimated numbers of individuals carrying each genotype, this would be obtained by multiplying the frequencies by the size of the population; in this case, since there are 100 individuals, we would expect about 38 homozygous FF, 47 heterozygotes, and 15 homozygous ff.

    Please note that we would only want to use Hardy-Weinberg to determine genotype frequencies from phenotype data like this IF we could be reasonably certain that the trait under consideration was not subject to selection (that it was selectively neutral), otherwise the assumption of no selection would be violated and our genotype frequencies would be likely to be biased.  When we suspect that selection MAY be taking place, we would want to obtain actual allele and genotype frequencies and compare these to the frequencies that would be expected under Hardy-Weinberg equilibrium conditions: if the two differ, we could reasonably conclude that the population was under selection for that trait.