As you know by now, random mutations occasionally occur within any population of organisms. Each mutation makes the organism in which it occurs more or less fit — more or less suited to its particular environment. Mutations may be beneficial, neutral, or disadvantageous depending on the selective pressure put on the organism by the environment in which it is living. Due to that mutation, the organism may be better or worse able to compete for resources, or in other words, natural selection is acting on that organism. Recall that during the coal-powered Industrial Revolution in England, the melanistic Peppered Moths were better camouflaged against the sooty, black, city tree trunks, and thus, lived to reproduce. Perhaps, however, they were initially less fit with respect to some of their other genes. Over time, more mutations would occur and be acted upon by natural selection such that, over time, the gene pool of the city-dwelling Peppered Moths would gradually change to improve the reproductive potential of the black moths, to make them more fit in/for that particular environment.
Thats the idea behind adaptation. While disadvantageous mutations are weeded out, other, beneficial and neutral mutations are retained, resulting in increased variation within the gene pool of the population. The beneficial and neutral mutations are passed on to the next generation through sexual reproduction, resulting in an overall change in the percentages of the various alleles within that population.
Because different populations are under different selective pressures, different alleles, both good and bad, are present in different populations in different percentages. In humans, consider the uneven distribution within the various ethnic groups of skin-color alleles, eye-color alleles, or alleles for genetic disorders like sickle-cell or Tay-Sachs.
Given a suitable environment and enough elapsed time following the original mutation, the gene pool of a specific population will usually reach some equilibrium point where the proportions of the various alleles become stable. Thus, initially, we need to make several background assumptions when examining the percentages of alleles in a population:
For PTC (phenylthiocarbamide) paper, about 70% of the population can taste the paper
(the dominant trait) and about 30% cannot. This means that genotypes TT and Tt (together since we can’t tell them apart) make up about
0.70 of the population, so we can say that p2 + 2pq = 0.70, and genotype tt makes up about 0.30 of the population, so we can say that q2 = 0.30.
If q2 = 0.30,
then by taking the square root of both sides, q = 0.55.
Since p + q must = 1, then p = 1 – q = 1 – 0.55 = 0.45.
This means that p2 = (0.45)2 = 0.20
and 2pq = 2 × 0.55 × 0.45 = 0.50.
Note that p2 + 2pq + q2 = 0.20 + 0.50 + 0.30 = 1.
Then, the probability of a dominant phenotype (taster) equals the sum of all possible ways to have a dominant genotype (TT and Tt), so the probability of a taster is equal to the probability of TT + the probability of Tt,
or 0.20 + 0.50 = 0.70
(to double check, note that this is equal to the 70% statistic noted above). Suppose you want to determine the probability of two tasters having a child who is a non-taster. For the child to be a non-taster, his/her genotype must be tt, and the only way this could be is if both parents are genotype Tt. Out of all people with dominant phenotypes (TT and Tt), 0.20/0.70 = 2/7 of them have the genotype TT and 0.50/0.70 = 5/7 of them have genotype Tt. Thus, for this population, the probability of both parents being Tt, is 5/7 × 5/7. If both parents are Tt, then from a regular Punnett square, remember that the probability of them having a tt child is 1/4. This means that the probability of two tasters having a non-taster child is
5/7 × 5/7 × 1/4 = 25/196 = 0.128 or 12.8%.
The Hardy-Weinberg Law can also be used to determine if breeding is, indeed, random.
For the human MN blood group, a person can have either type M (genotype MM) or type
MN (genotype MN) or type N (genotype NN). From a population in Australia there were 320 type
M people, 800 type MN, and 880 type N, so:
Note that these 4000 alleles include 1440 M alleles and 1440/4000 = 0.36 = 36% of the alleles.
Similarly, there are 2560 N alleles, so 2560/4000 = 0.64 = 64% of the alleles.
Thus, p = 0.36 and q = 0.64.
If p and q have these values and there is random breeding, then we should expect to see
p2 = 0.13, 2pq = 0.46, and q2 = 0.41.
This means that in a population of 2000 individuals we should then expect to see
0.13 × 2000 = 260 type M people,
0.46 × 2000 = 920 type MN people, and
0.41 × 2000 = 820 type N people.
A comparison with the actual numbers indicates that there is not random breeding (or the two sets of numbers would be more nearly the same):
|= 33.888||= calc|
The degrees of freedom (df) = 3 – 1 = 2, so tab at the 0.05 level = 5.991. Thus, calc is greater than tab, so the null hypothesis can be rejected: there is not random breeding.
As an example of non-random breeding, consider plants that are self-pollinating. In this case, AA can only mate with AA, Aa can only mate with Aa (producing AA, Aa, and aa offspring), and aa can only mate with aa. Thus, over time, the number of AA and aa individuals in the population increase relative to the number of Aa individuals.
The Hardy-Weinberg Law can also be used in cases of sex-linked genes.
In humans, red-green colorblindness is a sex-linked recessive trait. A colorblind male has
the genotype XbY and a normal male has the genotype XBY, thus for males, the genotype and phenotype
frequencies are equal. Females could be XBXB (normal), XBXb (carrier), or XbXb (colorblind), thus
the normal Hardy-Weinberg Law applies. If 10% (0.10) of the males are colorblind and 90% (0.90)
are normal, then p = 0.90 and q = 0.10 because each male has only one allele for this gene. From
this the phenotype frequencies for the females can be calculated:
p2 = (0.90)2 = 0.81 (or 81%) normal,
2pq = 2 × 0.90 × 0.10 = 0.18 (or 18%) carriers, and
q2 = (0.10)2 = 0.01 (or 1%) colorblind.
A slightly more complicated case is that of multiple alleles.
In the human ABO blood group, there are three alleles for this gene: IA, IB, and i. A
person can have any two of the three alleles, so
IAIA or IAi make type A blood,
IBIB or IBi make type B blood,
IAIB makes type AB blood, and
ii makes type O blood.
Let p represent the frequency of IA,
q represent the frequency of IB, and
r represent the frequency of i.
Remember that p + q + r = 1.
The Hardy-Weinberg Law says that at equilibrium, (p + q + r)2 = p2 + q2 + r2 + 2pq + 2rp + 2qr = 1
where p2 is the probability of IAIA and 2pr is the probability of IAi (thus probability of type A = p2 + 2pr),
q2 is the probability of IBIB and 2qr is the probability of IBi (thus probability of type B = q2 + 2qr),
r2 is the probability of ii (thus probability of type O = r2),
and 2pq is the probability of IAIB (thus probability of type AB = 2pq).
If out of 2000 people, 37.8% are type A, 14.0% type B, 4.5% type AB, and 43.7% type O,
then r2 = 0.437 so r = 0.661.
Temporarily ignoring IB, the frequency of type A blood + frequency of type O blood = p2 + 2pr + r2 = (p + r)2.
We know that there are 37.8% with type A blood (p2 + 2pr) and 43.7% with type O blood (r2),
so (p + r)2 = 0.378 + 0.437 = 0.815.
Then p + r = = 0.903,
so p = 0.903 – r = 0.903 – 0.661 = 0.242.
Also, the frequency of type B + frequency of type O = q2 + 2qr + r2 = (q + r)2 = 0.140 + 0.437 = 0.577,
so (q + r) = = 0.760.
Therefore, q = 0.760 – r = 0.760 – 0.661 = 0.099.
To check, p + q + r = 0.242 + 0.099 + 0.661 = 1.0, so that is correct.
Also, p2 + q2 + r2 + 2pq + 2pr + 2qr = 0.059 + 0.010 + 0.437 + 0.048 + 0.320 + 0.131 = 1.005, so that is correct.
The Hardy-Weinberg Law also applies for multiple genes — either linked or non-linked.
For a gene with alleles A and a, pA + qa = 1 and p2A + 2pAqa + q2a = 1
and for some other gene with alleles B and b, pB + qb = 1 and p2B + 2pBqb + q2b = 1.
Note that, assuming these are not linked genes, pA and qa are totally unrelated to and independent of pB and qb.
If, for example, it is desired to find the frequency of AaBb, multiply the frequencies needed (2pAqa × 2pBqb).
If pA = 0.2 (thus qa = 0.8) and pB = 0.1 (thus qb = 0.9),
then the probability of AaBb should be 2 × 0.2 × 0.8 × 2 × 0.1 × 0.9 = 0.058.
The effects of mutation can also be calculated. If mutations of allele A to allele A' (not
necessarily A to a — could be a to A) recur at a given rate, and if this is the only mutation of A that
takes place, then the frequency of A can go from 100% down to 0% while the frequency of A' goes
from 0% to 100%. We can symbolize the mutation rate as u, and for this example, let it be
1 × 10–6. There is a concept that is sort of like the idea of half-life in radioactive isotopes — the
number of generations (t) needed to reduce the frequency of A to 0.5 of its original value. The
equation for this is
t = –ln(0.5)/u, where ln(0.5) = –0.693,
so if u = 1 × 10–6, t = –(–0.693)/(1 × 10–6) = 6.93 × 10–1+6 = 6.93 × 105 = 693,000 generations. Thus, if pA starts out at 0.96, after 693,000 generations, pA will be 0.48 and after another 693,000 generations (= 1,386,000 total) it will be 0.24.
If the reverse mutation also occurs, eventually, the population will reach an equilibrium point where the frequencies of A and A' alleles are stable.
For a gene with alleles A and a, p is the frequency of A and q is the frequency of a. Let
u be the rate of mutation from A a
and v be the rate of mutation from a A. For allele a, the net
rate of change in frequency, q = gain – loss = up – vq. When equilibrium is established, gain
= loss, or q = 0 and up = vq. Thus, at equilibrium:
|=||u(1 – q)|
|=||u – uq|
|vq + uq||=||u|
|q(u + v)||=||u|
|q||=||u/(u + v)|
The effects of migration can also be calculated. If immigration is sporadic, we say there is gene exchange between populations. If it happens more often, there is gene flow.
If we let qinit be the q value of the original population under consideration for a certain allele,
qm be the q value for the migratory population, qfin be the frequency after immigration has taken place
and m be the fraction of the total population which are new migrants, then
qfin is comprised of the probability of that allele in the migrants (qm) times the proportion of migrants (m) plus the frequency of that allele in the non-migrants (qinit) times the proportion of non-migrants (1 – m), or:
|qfin||=||mqm + (1 – m)qinit|
|=||mqm – mqinit + qinit|
|=||m(qm – qinit) + qinit.|
|q||=||qfin – qinit|
|=||m(qm – qinit).|
Natural selection affects different genotypes (AA, Aa, aa) differently — selection will be for
or against each one of these genotypes. Let w = adaptive value and s = selection coefficient, such
that w + s = 1. Note that as w increases, s decreases and visa versa. These values are assigned
to each genotype (irrespective of the other genotypes). W = 1 if the genotype is the most adaptive
(able to adapt to new conditions). If w is less than 1, there is selection against that genotype. Also,
the larger s is, the more selection will take place. For example, if we have:
Three different cases are possible:
Case 1 — Selection Against Dominant:
If the initial value of p (pi) = 0.8, the initial value of q (qi) = 0.2, and s = 0.1, over the
course of a number of generations, p will decrease until it becomes 0 and q will increase until it
reaches 1. How long this takes depends on what s is — if s = 1, then AA and Aa will not produce
any offspring. For example, a population could have:
Case 2 — Selection Against Recessive:
In this case, the general formula would be:
|AA (0.176)||Aa (0.823)|
|AA (0.176)|| AA × AA |
0.176 × 0.176 = 0.031
| AA × Aa |
0.176 × 0.823 = 0.145
|Aa (0.823)|| Aa × AA |
0.176 × 0.823 = 0.145
| Aa × Aa |
0.823 × 0.823 = 0.677
The frequencies/probabilities of the possible progeny from each of these matings would be:
|AA × AA|| 1.000 × 0.031|
| –none– || –none– |
|AA × Aa|| 0.500 × 0.290|
| 0.500 × 0.290|
| –none– |
|Aa × Aa|| 0.250 × 0.677|
| 0.500 × 0.677|
| 0.250 × 0.677|
If s = 1.00 against aa (all of them die) and we start with qi = pi = 0.50, the following chart can be constructed:
|gen.||qa init||qa2||pA2||2pAqa|| ratio of |
| proportion |
of Aa left
(2pAqa)/(pA2 + 2pAqa)
| prob Aa×Aa |
(Aa left)×(Aa left)
| probaa offsp |
= (qa fin)2
|1||0.50|| 0.50 × 0.50 |
| 0.50 × 0.50 |
| 2 × 0.50 × 0.50 |
| 0.50/(0.50 + 0.25) |
| 0.67 × .067 |
| 0.25 × 0.44 |
|100||9.9 × 10–3||9.8 × 10–5||9.8 × 10–1||1.96 × 10–2||200:1||1.96 × 10–2||3.84 × 10–4||9.61 × 10–5||9.8 × 10–3|
If, however, s = 0.10 against aa, then:
= # of
= # of
= # of
= # of
= # of
A & a
= # of
=1 – sq2
|# of A|
|# of a|
Thus, q for the next generation (n + 1) can be calculated from the number of a gametes produced divided by the total
number of gametes produced or:
|qn+1||=||[qn2(1 – s) + 0.5×(2pnqn)]/(1 – sqn2)|
|=||[qn2 – sqn2 + pnqn]/(1 – sqn2)|
|=||[qn2 – sqn2 + (1 – qn) × qn]/(1 – sqn2)|
|=||[qn2 – sqn2 + qn – qn2]/(1 – sqn2)|
|=||[qn – sqn2]/(1 – sqn2)|
We can, then, calculate the change in q (q) as a result of selection:
|q = qn+1 – qn||=||[qn – sqn2]/(1 – sqn2) – qn|
|=||[qn – sqn2]/(1 – sqn2) – [qn(1 – sqn2)]/(1 – sqn2)|
|=||[qn – sqn2]/(1 – sqn2) – [qn – sqn3]/(1 – sqn2)|
|=||[qn – sqn2 – qn + sqn3]/(1 – sqn2)|
|=||[sqn3 – sqn2]/(1 – sqn2)|
|=||–sqn2(1 – qn)/(1 – sqn2) where, if it helps, 1 – qn = pn|
So, for the above example,
|gen||q =||q = qf – qi|
Case 3 — Selection Against Both Homozygotes:
In this example, Aa is superior to either AA or aa, and this is called overdominance or
balanced polymorphism. An example of this would be the gene for sickle-cell anemia and
resistance to malaria in humans. Thus, for example:
|w =||1 – sA||1.00||1 – sa|
After a generation of selection, there will be
|p2(1 – sA)||from AA,|
|+||2pq||from Aa, and|
|+||q2(1 – sa)||from aa, so|
|=||1 – sAp2 – saq2||total gametes produced.|
|gen||pi||qi||p2||q2||2pq||p2(1–sA)||q2(1–sa)||total|| # A |
p2(1–sA) + pq
| # a |
q2(1–sa) + pq
|% of A||% of a|
Eventually, the population will reach equilibrium. The number of generations required will be determined by sA and sa. At equilibrium,
|so p||=||[p2(1 – sA) + ½(2pq)] / [p2(1 – sA) + (2pq) + q2(1 – sa)]|
|and q||=||1 – p|
|so p||=||[p2(1 – sA) + p(1 – p)] / [p2(1 – sA) + 2p(1 – p) + (1 – p)2(1 – sa)]|
|1||=||[p(1 – sA) + (1 – p)] / [p2(1 – sA) + 2p(1 – p) + (1 – p)2(1 – sa)]|
|[p(1 – sA) + (1 – p)]||=||[p2(1 – sA) + 2p(1 – p) + (1 – p)2(1 – sa)]|
|0||=||(p2 – p)(1 – sA) + (1 – p)(2p – 1) + (1 – p)2(1 – sa)|
|0||=||–p(1 – sA) + (2p – 1) + (1 – p)(1 – sa)|
|0||=||–p + psA + 2p – 1 + 1 – p – sa + psa|
|0||=||psA – sa + psa|
|sa||=||p(sA + sa)|
|thus,||=||sa/(sA + sa)|
|and similarly,||=||sA/(sA + sa).|
Several interactions between these two factors are possible.
|1.||w =||1 – s||1 – s||1||and||selection & mutation work in same direction|
|2.||w =||1 – s||1 – s||1||and||selection & mutation working against each other — eventually, equilibrium will be established|
|3.||w =||1||1||1 – s||and||selection & mutation working against each other here too|
Recall that, due to mutation, qmut = up – vq and because of selection, qsel = –sq2(1 – q)/(1 – sq2). Equilibrium will be
reached when the effects of mutation and selection cancel out each other — when they are working in exactly the opposite direction, or qmut = –qsel so:
|up – vq||=||–[–sq2(1 – q)/(1 – sq2)]|
|u(1 – q) – vq||=||sq2(1 – q)/(1 – sq2)|
|u(1 – q)||sq2(1 – q)|
There are also several cases of mutation from recessive to dominant with selection against
the dominant. A semidominant lethal refers first to the expression of the gene and secondly to the
vitality of the organism. A gene is semilethal if the organism is born alive, lives for a while, then
dies. A gene is subvital if the organism is not as viable. In general, these can all be expressed as:
|4.||w =||1 – s||1||1|
|4b.||w =||0||1 – s||1|
Copyright © 1998 by J. Stein Carter. All rights reserved.
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