Population Genetics

The Hardy-Weinberg Law

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.

That’s 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:

• Reproduction/mating is random (this is usually not the case in humans).
• Mutations (of the gene under study) do not occur or occur in equal numbers in both directions.
• The population is large enough so that changes in gene frequency don’t occur (there are stable percentages of the alleles — the ratios are not fluctuating wildly.).
• Migration into/out of the population is negligible, and is not contributing to changes in the frequencies of the alleles.
• Natural selection is not putting selective pressure on any of the phenotypes coded for by the gene in question.
• The generations are non-overlapping, so the parental generation is not able to or does not breed with the offspring.
• Each pair of parents produces a number of offspring.

If all of these conditions are fulfilled, then it is possible to calculate/predict the frequency of each allele in a population. The Hardy-Weinberg Law (1908) says that, for some gene with alleles A and a, if the frequency/proportion of A is “p” and the frequency/proportion of a is “q,” then:

• If these are the only two choices/alleles, then p + q = 1. The probability of any given gene can range from 0 to 1 (100%). If there are two or more possibilities, the sum of all probabilities must equal 1.
• Remembering that an organism has two alleles for each gene, the possible genotypes are AA, Aa, or aa. Since the genotype AA is made up of two A alleles, the probability of that genotype is the probability of A times the probability of A or (p × p) = p². Likewise, the probability of aa is q × q = q². Remember that there are two ways to get Aa: either Aa or aA, so the probability of genotype Aa is 2pq.
• Since the sum of the probabilities for all possible outcomes must be equal to 1, and if AA, Aa, and aa are the only choices for possible genotypes this means that
  p² + 2pq + q² = 1
  and you may recall from algebra that
  p² + 2pq + q² = (p + q)².
  Note that it makes sense that if p + q = 1 that (p + q)² = 1, too.

However, if all of these conditions are not fulfilled and there are other factors like mutation, migration, or natural selection influencing the gene pool, then those factors must also be taken into account (as explained below).

Example:

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 = ²⁵⁄₄₉ ≈ ½. If both parents are Tt, then from a regular Punnett square, remember that the probability of them having a tt child is ¼. This means that the probability of two tasters having a non-taster child is ~½ × ¼ = ~⅛ = 0.128 or 12.8%.

Problem:

1. Consider the gene for Rh factor. Let R be the allele for Rh⁺ and r be the allele for Rh^–, such that RR and Rr are Rh⁺ and rr is Rh^–. Assume a large population in equilibrium in which 16% of the individuals are Rh^–.
   1. What is the calculated frequency of the r allele? Of the R allele?
   2. What is the calculated frequency of the genotypes in this population?
   3. An Rh⁺ man marries an Rh^– woman. What is the probability that the man is heterozygous (Rr)? That he is homozygous (RR)? What is the probability that this couple’s first child will be Rh^–?

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Non-Random Breeding:

The Hardy-Weinberg Law can also be used to determine if breeding is, indeed, random.

Example:

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:

320	MM	therefore	640	M alleles
800	MN	therefore	800	M alleles
		and	800	N alleles
880	NN	therefore	1760	N alleles
2000	people	therefore	4000	alleles

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 χ² comparison with the actual numbers indicates that there is not random breeding (or the two sets of numbers would be more nearly the same):

O	E	(O–E)2/E
320	260	13.846
800	920	15.652
880	820	4.390
Σ = 33.888			= χ²calc

The degrees of freedom (df) = 3 – 1 = 2, so χ²_tab at the 0.05 level = 5.991. Thus, χ²_calc is greater than χ²_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.

Problem:

2. The MN blood group in humans includes the following genotypes and phenotypes: MM = type M, MN = type MN, and NN = type N.
   1. What is the gene frequency for M and N in a population of people with the following blood types: 40 with type M, 300 with MN, and 660 with N. Is this population breeding at random?
   2. What is the gene frequency for M and N in another population of people with the following blood types: 280 type M, 500 type MN, and 220 type N. Is this population breeding at random?

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Sex-Linked Genes:

The Hardy-Weinberg Law can also be used in cases of sex-linked genes.

Example:

In humans, red-green colorblindness is a sex-linked recessive trait. A colorblind male has the genotype X^bY and a normal male has the genotype X^BY, thus for males, the genotype and phenotype frequencies are equal. Females could be X^BX^B (normal), X^BX^b (carrier), or X^bX^b (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.

Problem:

3. If the frequency of red-green colorblind males in a population is 0.20, what is the frequency of red-green colorblind females in this population? What proportion of the females could have red-green colorblind sons?
   

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Multiple Alleles:

A slightly more complicated case is that of multiple alleles.

Example:

In the human ABO blood group, there are three alleles for this gene: I^A, I^B, and i. A person can have any two of the three alleles, so
I^AI^A or I^Ai make type A blood,
I^BI^B or I^Bi make type B blood,
I^AI^B makes type AB blood, and
ii makes type O blood.
Let p represent the frequency of I^A,
q represent the frequency of I^B, 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 I^AI^A and 2pr is the probability of I^Ai
(thus probability of type A = p² + 2pr),
q² is the probability of I^BI^B and 2qr is the probability of I^Bi
(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 I^AI^B
(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” I^B, 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 = √0.815 = 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) = √0.577 = 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.

Problems:

4. For the ABO blood group, what blood type percentages would you find in a population in which the frequency of I^A = 0.1 and I^B = 0.4?
   

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5. Calculate the gene frequencies of I^A, I^B and i for the following population: 490 have type O blood, 150 have type A, 320 type B, and 40 type AB.
   

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Multiple Genes:

The Hardy-Weinberg Law also applies for multiple genes — either linked or non-linked.

Example:

For a gene with alleles A and a, p_A + q_a = 1 and p²_A + 2p_Aq_a + q²_a = 1
and for some other gene with alleles B and b, p_B + q_b = 1 and p²_B + 2p_Bq_b + q²_b = 1.
Note that, assuming these are not linked genes, p_A and q_a are totally unrelated to and independent of p_B and q_b.
If, for example, it is desired to find the frequency of AaBb, multiply the frequencies needed (2p_Aq_a × 2p_Bq_b).
If p_A = 0.2 (thus q_a = 0.8) and if p_B = 0.1 (thus q_b = 0.9),
then the probability of AaBb should be 2 × 0.2 × 0.8 × 2 × 0.1 × 0.9 = 0.058.

Problem:

6. What will the frequency of the genotype AaBB be at equilibrium if the frequency of a is 0.60 and the frequency of b is 0.20? What genotype will be most frequent in this population?

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Some More Complicated Examples

Effects of Mutation:

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 p_A starts out at 0.96, after 693,000 generations, p_A 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.

Example:

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, Δq = gain – loss = up – vq. When equilibrium is established, gain = loss, or Δq = 0 and up = vq. Thus, at equilibrium:

vq	=	up
	=	u(1 – q)
	=	u – uq
vq + uq	=	u
q(u + v)	=	u
q	=	u/(u + v)

To distinguish this as being the equilibrium value of q, it is written as q̂  . Similarly, p̂  = v/(u + v).

Problems:

7. If the mutation rate of K to k is 1.5 × 10^–6, the reverse mutation does not occur, and thereis no selective advantage or disadvantage to either allele, what will the gene frequency be after many generations if the initial frequencies are p_K = 0.80 and q_k = 0.20?

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8. What will the gene frequencies be at equilibrium in a population in which the rate of mutation of M to m is 0.000015 and the rate of the reverse mutation is 0.000030?

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Effects of Migration:

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.

Example:

If we let q_init be the q value of the original population under consideration for a certain allele, q_m be the q value for the migratory population, q_fin be the frequency after immigration has taken place and m be the fraction of the total population which are new migrants , then q_fin is comprised of the probability of that allele in the migrants (q_m) times the proportion of migrants (m) plus the frequency of that allele in the non-migrants (q_init) times the proportion of non-migrants (1 – m), or:

qfin	=	mqm + (1 – m)qinit
	=	mqm – mqinit + qinit
	=	m(qm – qinit) + qinit.

When the original population and the population after immigration are considered,

Δq	=	qfin – qinit
	=	m(qm – qinit).

Thus, the effect of immigration on the frequency of that allele in the population (Δq) depends on the number of individuals that migrate (m) and the difference in q between the migrating population and the original population (q_m – q_init). Note that if the frequencies of the allele in the initial population and in the migrating population are the same (q_init = q_m), then there is no change in gene frequency (Δq = 0), but if they are different, then the number of individuals that migrated is also important.

Problem:

9. A genetics book I have says that among the Utes, a Native American population from Montana, 2.6% of the population had type A blood, 0.0% have type B, 0.0% have type AB, and 97.4% have type O. Thus, for that group, p_A = 0.013, q_B = 0, and r_O = 0.987.
   This book also cites a similar study done among English people from London. In that group, 42.4% had type A blood, 8.3% have type B, 1.4% have type AB, and 47.9% have type O, so p_A = 0.250, q_B = 0.050, and r_O = 0.692.
   Suppose that a group of 200 Utes were living in a given area of Montana, and a group of 30 English colonists arrived and settled there. If these people intermarried, how would the arrival of the English change the frequencies of each of the alleles and phenotypes in this population?

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Natural Selection:

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:

	AA	Aa	aa
w =	1.00	1.00	0.99

then there is selection against aa, so for every 100 AA individuals produced, only 99 aa are produced. Note also that in this example, for aa, s = 1 – w = 0.01. In another example:

	AA	Aa	aa
w =	0.90	0.90	1.00

there is a selection coefficient of 0.10 against both dominant phenotypes.

Examples:

Three different cases are possible:

Case 1 — Selection Against Dominant:

If the initial value of p (p_i) = 0.8, the initial value of q (q_i) = 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:

	AA	Aa	aa
w =	0.90	0.95	1.00
s =	0.10	0.05	0.00

Case 2 — Selection Against Recessive:

In this case, the general formula would be:

	AA	Aa	aa
w =	1.00	1.00	1–s

In the simplest (most extreme) example, s = 1 so w = 0 for aa. If a population would start with 51% A– phenotype and 49% aa, then q_i = √0.49 = 0.70, and p_i = 0.30, so p² = 0.09, 2pq = 0.42, and q² = 0.49. If all of the aa genotypes die or become sterile for whatever reason, the proportions of AA and Aa remaining in the population become:
for AA, 0.09/(0.09 + 0.42) = 0.176 and
for Aa, 0.42/(0.09 + 0.42) = 0.823.
The possible matings between the remaining Aa and Aa individuals and the probabilities of each 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

Note that, for AA × Aa, the combined total is 0.176 × 0.823 × 2 = 0.290.

The frequencies/probabilities of the possible progeny from each of these matings would be:

	AA	Aa	aa
AA × AA	1.000 × 0.031 = 0.031	–none–	–none–
AA × Aa	0.500 × 0.290 = 0.145	0.500 × 0.290 = 0.145	–none–
Aa × Aa	0.250 × 0.677 = 0.169	0.500 × 0.677 = 0.338	0.250 × 0.677 = 0.169

At this point, since the proportion of aa individuals is 0.169, then q_f² = 0.169, so q_f = 0.41, which is only 59% of the original value of 0.70 — q, the frequency of a, has decreased nearly half in only one generation.

If s = 1.00 against aa (all of them die) and we start with q_i = p_i = 0.50, the following chart can be constructed:

Note especially the column for the ratio of Aa:aa. This means that for every aa that is produced (then dies or becomes sterile), there are 2n number of Aa produced that could have aa offspring (for example, for the 100^th generation, 200 Aa are produced for every aa), thus the more generations that occur, the less effective continued selection actually is.
If, however, s = 0.10 against aa, then:

Note that with this smaller value for s, less selection takes place. The general expressions for the gametes produced are:
AA genotypes contribute p² × 1 of the gametes, all of which are A.
Aa genotypes contribute 2pq × 1 of the gametes, which are half A and half a.
aa genotypes contribute q² × (1 – s) of the gametes, all of which are a.

Note that this means there are p² + pq of the A gametes produced
and pq + q² × (1 – s) = pq + q² – sq² of the a gametes produced.
Then, the total number of gametes produced is (p² + 2pq) + q²(1 – s) = p² + 2pq + q² – sq².
Remember that p² + 2pq + q² = 1, so p² + 2pq + q² – sq² can be simplified to 1 – sq².

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 (Δq) as a result of selection:

Δq = 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 =	Δq = qf – qi
1	0.500	–0.0128
2	0.487	–0.0125
3	0.475	–0.0121

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:

	AA	Aa	aa
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.

For example, if p_i = q_i = 0.500, s_A = 0.200, and s_a = 0.600 (w_A = 0.800 and w_a = 0.400), then:

Eventually, the population will reach equilibrium. The number of generations required will be determined by s_A and s_a. At equilibrium,

pn+1	=	pn
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,        p̂	=	sa/(sA + sa)
and similarly,        q̂	=	sA/(sA + sa).

To continue the above example: p̂   = 0.600/(0.200 + 0.600) = 0.750
and q̂   = 0.200/(0.200 + 0.600) = 0.250.

Problems:

10. Consider the following situation:

    	AA	Aa	aa
    w =	1.0	1.0	0.8

    Is selection against the dominant or recessive? What is the value of s, the selection coefficient?

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11. Consider the following situation:

    	AA	Aa	aa
    w =	0.6	1.0	0.8

    What are the values of p and q at equilibrium?

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Effects of Mutation and Selection Together:

Several interactions between these two factors are possible.

Examples:

Consider the following examples:

		AA	Aa	aa
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

In this third example, selection is against the recessive, but mutation is toward it.

Recall that, due to mutation, Δq_mut = up – vq and because of selection, Δq_sel = –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 Δq_mut = –Δq_sel so:

up – vq	=	–[–sq2(1 – q)/(1 – sq2)]
u(1 – q) – vq	=	sq2(1 – q)/(1 – sq2)

Note that as q approaches 0, qv approaches 0 (because, in this case, v is so small, too) and 1 – sq² approaches 1, thus if q is nearly 0,

u(1 – q)	≃	sq2(1 – q)
u	≃	sq2

So q̂  ² ≃ u/s and q̂   ≃ √u/s. If the aa genotype is lethal, s = 1.00 so q̂   ≃ √u.

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:

		AA	Aa	aa
4.	w =	1 – s	1	1

Within this are several special cases:

		AA	Aa	aa
4a.	w =	0	1	1

An example of this is brachydactyly in humans. The homozygote is not viable, but Aa is not affected. Note that the frequency of A = p̂   = √v.

		AA	Aa	aa
4b.	w =	0	1 – s	1

An example in humans is achondroplasia. The homozygote is not viable, while the heterozygote has abnormal cartilage and is a dwarf. Note that the frequency of A = p̂   = v/s.

		AA	Aa	aa
4c.	w =	0	0	1

Neither the homozygote nor the heterozygote live. The homozygote is lethal and the heterozygote is born alive but soon dies. In humans, this is represented by retinoblastoma, a condition in which the retinas develop fatal tumors just after birth (not fatal if the eyes are removed shortly after birth to stop the formation/spread of tumors). Note that no Aa genotypes live to reproduce (unless their eyes are removed), thus no AA offspring can be produced (unless two Aa people, who survived because they both had their eyes removed as babies, managed to find each other, fall in love, get married, and have children), and it is extremely unlikely that a mutation of a → A would occur in both parents to form an AA offspring. In this case, then, p̂   = v because all born are either Aa or aa. For example, in Michigan from 1936 to 1945, out of 1,054,985 people, only 49 cases of retinoblastoma were reported — this is 49 dominant alleles out of 2 × 1,054,985 total alleles, so v = 49/(2 × 1,054,985) = 2.32 × 10^–5, which, in this case also equals p̂  . Since w = 0 (s = 1) for both AA and Aa, the only ones to live and reproduce are aa. Thus 100% of the gametes come from aa parents and any A alleles in the population are due totally to mutation. The rate of mutation of a → A is v (and u = 0), thus the number of alleles that have mutated (v) is equal to the number of A alleles at equilibrium (p̂  ). Also, if p̂   = v/s and s = 1, then p̂   = v.

Problems:

12. In the American white population, the gene frequency for Rh^– (r) is 0.4 as determined by the fact that about 16% of the population is Rh^–. Additionally, about one birth in 250 is a baby who is erythroblastotic. This is a disorder characterized by the abnormal presence of erythroblasts in the circulating blood (they are normally found in the bone marrow) usually as a result of trying to compensate for an Rh incompatibility with an Rh^– mother, and thus, affects infants who are heterozygous for Rh factor. Although, if these babies survive to birth they can be given a total blood transfusion to alleviate the symptoms, assume for this problem that these infants do not survive to adulthood.
    1. How many R alleles are found in a population of 1000 newborn babies per generation? How many r alleles?
    2. How many heterozygotes will be eliminated? How many R alleles? How many r alleles?
    3. Under these circumstances, what will tend to happen over a number of generations to the frequencies of R and r?
    4. What would tend to happen to the frequencies of R and r in a population in which p_R = 0.4 and q_r = 0.6?

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13. Chondrodystrophic dwarfism (a dominant semilethal) was found to afflict 8 children of 94,075 born to normal parents in Denmark. What is the mutation rate for d to D?

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14. If the rate of mutation from A to a is 0.0000049 and the selection coefficient against a is 0.001, what will be the frequency of the genotype aa at equilibrium? What will be the frequency of gene a?

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