The major architects of Neo-Darwinian theory
(=fusion of Mendel and Darwin theory)
R.A. Fisher, Sewell Wright, J.B.S. Haldane
A comprehensive theory of evolution, the modern synthesis, took form in the early 1940s.
� It integrated discoveries and ideas from paleontology, taxonomy, biogeography, and
population genetics.
The first archite
Bottleneck Effect
the form of genetic drift that occurs
when the population size is drastically reduced.
Founder Effect
when a population is initially established
by a small number of breeding individuals, they may appear
genetically and phenotypically different than the original
population it was derived from.
Genetic Drift
random change in allelic frequency due to
chance.
Migration:
exchange of genes with other population and
therefore alter the gene frequency of the existing alleles.
Natural Selection:
struggle for survival and the survival
depend on the hereditary constitution of offspring.
Population Genetics
A genetic branch to study genetic
variation among individuals within groups, and the genetic
basis for evolutionary change, seeks to understand how
patterns vary geographically and through time.
The Hard-Weinberg Law:
The genotype frequencies and
gene frequencies of a large, randomly mating population
remain constant provided immigration; mutation and
selection do not take place.
Allelic frequency
the frequency of one allele in a
population.
Genotypic frequency
Count individuals with one
genotype and divide by total number of individuals
p2 + 2pq + q2 = 1
The Hardy-Weinberg Law
Assumptions:
� Population is infinitely large
� Mating is random
� No natural selection
� No mutation.
� No migration.
HWE---Predictions:
* Allele frequencies do not change over generations.
*Genotype frequencies can be predicted from alleles
� At equilibration: p2 + 2pq + q2 = 1
o p = allelic frequency of A
o q = allelic frequency of a
o p2 frequency of AA
o 2pq frequency of Aa
o q2 freque
For three alleles (A, B, and C) with frequencies p, q, and
r:
(p + q + r)2 = p2(AA) + 2pq(AB) + q2(BB) + 2pr(AC) +
2qr(BC) + r2(CC)
For four alleles (A, B, C, and D) with frequencies p, q, r,
and s:
(p + q + r + s)2 = p2(AA) + 2pq(AB) + q2 (BB) + 2pr(AC) +
2qr(BC) + r2 (CC) + 2ps(AD) + 2qs(BD) + 2rs(CD) + s2
(DD)
For X-linked alleles on males
p + q = 1
Migration:
-Migrants from population I:
-Residents from population II:
-After migration:
1) Px
2) Py
3) f(A) = m px + (1-m)py
?p = m(px -py); m = migrants/(migrants + residents)
Allelic Frequency
Dominant Allele
F(A)= (2xcount of AA)+(1 Count of Aa)/ 2 x Total # of indviduals = (Freq of AA)+(1/2 x Freq. of Aa)
P+Q=1
HWE frequencies def. of how calc
Calculate the freq of particular alleles based on freq of specific phenotype (A =p and a=q)
Natural Selection
The principle force that alters gene frequencies (evolution) and differential success of dissimilar individuals (nonrandom)
Based on Darwin's Four Principles:
1. Heritable differences among individuals in population
2. Heritable differences passed to offspring
3. More offspring are born than survive
4. Some offspring are more "fit" than others in a given
environment
Bottom line: Changes in relative abundance of
� Frequency of recessive genetic disorder
q2 frequency
q = square root of q2 frequency
� p + q = 1 SO
p=1-q
With p and q frequencies you can calculate
frequencies of three genotypes (p2 + 2pq + q2 = 1)
homozygous dominant = p2
heterozygotes = 2pq
homozygous recessive = q2
Can Predict Occurrence of Rare Diseases
q2 (cc) = 1/2500 = 0.0004
We can calculate:
Diseased allele frequency: q = 0.02
Normal allele frequency: p = 1 - q = 0.98
Heterozygous frequency: 2 pq = 2(0.98)(0.02) = 0.039
SO q = c allele = 0.02
Relation Between Allele & Genotype Frequencies
Draw the graph
Why Are Deleterious Alleles Maintained in
Population?� Two theories:
1. Mutation-selection balance hypothesis
- homozygous recessive lost from population
- new mutations arise at same rate
2. Heterozygote superiority hypothesis
- defective allele provides an advantage
- heterozygotes (+/-) has best of both
� Sickle-cell an
Fitness
Natural selection is dependent on the fitness of individuals or
The quantifiable measure of an individual's genetic
contribution to the future.
High Fitness: Genotypes with high rate of survival
and/or high level of reproduction (genes are passed on)
Low
Selection can Change Natural Populations?
Three types of Selection-induced changes in a population
Stabilizing Selection
Directional Selection
Disruptive Selection
Applications of Hardy-Weinberg
1. It can be used to estimate no. of heterozygous individuals
2. It can be used to predict frequencies of systems of multiple alleles
3. It can be used to predict frequencies of systems with X-linked traits
Selection Pressure (s):
0
Stabilizing Selection
When selective pressures select against the two extremes of a trait, the population experiences stabilizing selection. For example, plant height might be acted on by stabilizing selection. A plant that is too short may not be able to compete with other pl
Directional Selection
In directional selection, one extreme of the trait distribution experiences selection against it. The result is that the population's trait distribution shifts toward the other extreme. In the case of such selection, the mean of the population graph shift
Disruptive Selection
In disruptive selection, selection pressures act against individuals in the middle of the trait distribution. The result is a bimodal, or two-peaked, curve in which the two extremes of the curve create their own smaller curves. For example, imagine a plan
Given a one locus, two allele system displaying codominance where selection acts on diploids, the initial allelic frequencies are equal, is in Hardy-Weinberg equilibrium other than its being under selection, and diplays selection coefficients of 0.5, 1.0,
Note first that the allele frequencies are 0.5 for both alleles (since they are equally frequent). Name the alleles A and a (or whatever turns you on). This means the genotypes are (with selection coefficients denoted parenthetically): AA (0.5), Aa (1.0),
Heterozygote superiority (balanced polymorphism)
Totally unfit alleles are not immediately removed from the population
because in the heterozygotic condition they provide an advantage
*Sickle cell anemia
Heterozygotes ("carriers') survive and reproduce better than normal or sickle cell individuals in re
Changes in population structure (No H-W equilibrium)
1 Non-random mating:
Inbreeding (consanguineous mating)
=occurs whenever mates are genetically related
=allele freq. remain the same but genotypic frequencies change
= results in rare alleles becoming homozygous
2.. Genetic Drift
0
Hardy-Weinberg Equilibrium
0
HWE HISTORY
*G.H. Hardy and G. Weinberg, both in 1908, made their mark on evolutionary biology by pointing out that genotype frequency could vary in the absence of any change in allelic frequency. Up until that point in time scholars were under the mistaken impressio
Give five conditions that must be met in order for Hardy-Weinberg equilibrium to hold? Bonus points for a sixth condition which makes such equilibria meaningful:
(i) infinite population, (ii) random mating, (iii) no selection, (iv) no migration, (v) no mutation, (vi-bonus) genetic polymorphism. Note that no genetic drift and infinite population mean the same thing.
Given (i) two allele, one loci, diploid genetics (A and a), (ii) conditions appropriate for the establishment of Hardy-Weinberg equilibrium, and (iii) a frequency of A of 0.15, what are the genotype frequencies?
f(A) = 0.15, f(a) = 0.85, f(AA) = f(A)
f(A) = 0.023, f(Aa) = 2
f(A)
f(a) = 0.255, f(aa) = f(a)
f(a) = 0.723.
Given a frequency of Aa of 0.70 and of AA of 0.30. If the selection coefficients associated with AA, Aa, and aa are 0.50, 0.60, and 1.0, respectively, what (i) are allelic frequencies in the current generation and (ii) the expected allelic frequencies in
Current generation: f(A) = (2
f(AA) + f(Aa)) / 2 = 0.65; f(a) = 1 - f(A) = 0.35; Next generation: f'(A) = ((f(Aa)
0.60) + (2
f(AA)
0.5)) / 2 = .36; f'(a) = ((f(Aa)
0.60) + (2
f(aa) * 1.0)) / 2 = .21; f(A) = f'(A) / (f'(A) + f'(a)) = 0.36 / (0.36 + 0.21) =
Given frequencies of Aa of 0.30, of AA of 0.60, and aa of 0.10. If the selection coefficients associated with AA, Aa, and aa are 1.00, 1.00, and 0.50, respectively, what (i) are allelic frequencies in the current generation and (ii) the expected allelic f
Current generation: f(A) = (2
f(AA) + f(Aa)) / 2 = 0.75; f(a) = 1 - f(A) = 0.25; Next generation: f'(A) = ((f(Aa)
1.00) + (2
f(AA)
1.00)) / 2 = 0.75; f'(a) = ((f(Aa)
1.00) + (2
f(aa) * 0.50)) / 2 = 0.20; f(A) = f'(A) / (f'(A) + f'(a)) = 0.75 / (0.75 + 0.2
How, if at all, does the mathematical treatment of the survival component of fitness differ from the mathematical treatment of the replication component of fitness?
there is no difference
A large, isolated, randomly mating population of brown and blue eyed individuals is 20% blue eyed. Keeping in mind that blue eyes is the homozygous recessive condition, in this population what is the frequency of the brown eyed allele (i.e., brown yes res
ii, 0.55. That is, the frequency of the brown eyed allele is equal to p which is equal to 1 - q. We are given q2 as 0.2 (i.e., 20% of the population has blue eyes). q thus is equal to the square root of 0.2, which is approximately 0.45 (which is the frequ
A recessive allele is present at a rate of 1 in 5000. What is the frequency of the heterozygote? (choose best answer)
one in 5000
one in 10,000
one in 25,000
0.0002
0.0001
0.0004
vi, 0.0004.
Given a one locus, three allele system (with allele frequencies of 0.1, 0.3, and 0.6), what is the frequency of the most prevalent heterozygote? (choose best answer) [PEEK]
0.12.
0.24.
0.36.
0.48.
0.60.
0.72.
iii, 0.36.
Given a one locus, three allele system (with allele frequencies of 0.1, 0.3, and 0.6), what is the frequency of the most prevalent genotype? (choose best answer)
0.12.
0.24.
0.36.
0.48.
0.60.
0.72.
iii, 0.36
A plant population consists of two flower morphs, white and red, controlled by a single locus and two alleles. The white morph represents the recessive homozygote. The red morph consists of some combination of heterozygotes and dominant homozygotes. Assum
the frequency of a is 0.707 (the square root of 0.5); the frequency of A is 0.293 (= 1 - 0.707); the frequency of aa is 0.5, the frequency of AA is 0.086; the frequency of Aa is 0.414 (note that I had to use three significant figures to avoid rounding err
A population of 40 guinea pigs undergoing natural selection is reduced to 20 individuals in only five generations. At this point genotype frequencies are 0.25, 0.5, and 0.25 for the genotypes AA, Aa, and aa, respectively. Assume that natural selection con
No, it is being subjected to natural selection and does not have a large size
Given a locus with three alleles and allele frequencies of 0.4, 0.4, and 0.2, what is the frequency of the least prevalent heterozygote? [PEEK]
0.06
0.16
0.26
0.36
0.10
0.08
(2) 2
0.4
0.2 = 0.16
Given a population that up to now had been in Hardy-Weinberg equilibrium. Assume two alleles, one locus, p = 0.5, and distinctly different (and unambiguous) phenotypes associated with each genotype. Now assume internal fertilization and that all matings o
AA parents are 25% of population and produce only AA children; aa parents are 25% of the population and produce only aa children; Aa parents are 50% of the population and produce children who are 25% AA, 50% Aa, and 25% aa. Since these latter children mak
What is a balanced polymorphism?
a stably-existing polymorphism
Relative fitnesses that vary as a function of genotype frequency is an example of what?
frequency dependent selection
You have a population that, before selection, consists of 0.5 AA individuals, 0.25 Aa individuals, and 0.25 aa individuals. The relative fitnesses associated with each of these three genotypes are 0.3, 0.5, and 1.0, respectively. Define p, the frequency o
p = 0.625 [(0.5
2 + 0.25
1)/2] prior to selection and [(0.5 * 0.3 * 2) + (0.25 * 0.5 * 1)] * 1 / [(0.5
0.3
2) + (0.25
0.5
1 / [(0.5
.5*2 + 0.25*1)/2] prior to selection and [(0.5 * 0.3 * 2) + (0.25 * 0.5 * 1)] * 1 / [(0.5 * 0.3 * 2) + (0.25 * 0.5
2) + (0.
Given a population with genotype frequencies of 0.2, 0.2, and 0.6 for genotypes AA, Aa, and aa, what should be the genotype frequencies assuming Hardy-Weinberg equilibrium?
p = 0.3; q = 0.7; p2 = 0.09, 2pq = 0.42, q2 = 0.49
What is assortative mating and what does it do to a population's gene pool?
It is non-random mating in which phenotypically similar individuals preferentially mate. It subdivides the population into more than one gene pool; alternatively, you might want to think of assortative mating's effect on the gene pool as one of subdividin
You have a haploid population (hey, a bunch of T4 phage!) and one loci, two allele genetics. 50% of the A individuals die per round of replication but all of the a individuals survive. Survivors reproduce at the same rate, independent of genotype. You beg
p = 0.5; following selection the p = (0.5
0.5) / [(0.5
0.5) + 0.5] = 0.33; the ratio is either 0.33 or 3. If 70% of A died then p = (0.5
0.7) / [(0.5
0.7) + 0.5].
The B allele is dominant to the b allele. The phenotype associated with the former is brown eyes, while blue eyes is the phenotype associated with the latter. The brown eye allele is present in the population at a frequency of 0.2. Given Hardy-Weinberg eq
p = 0.2; q = 0.8; p2 = 0.04; 2pq = 0.32; q2 = 0.64; the frequency of brown-eyed people (men or women) is 0.04 + 0.32 = 0.36; the frequency of blue-eyed people (men or women) is 0.64; the probability that a brown-eyed man will marry a blue eyed woman there
Given a one locus, two allele system in which the relative fitness of AA is 0.5, Aa is 1.0, and aa is 0.0, this is an example of what?
heterozygote advantage
Draw a population of 10 giraffes following strong (i.e., highly effective) diversifying selection. Try to avoid being ambiguous in your drawings.
the giraffes should be distributed into two, preferably quantitative phenotypes such as with one set of giraffes having short necks and the other group having long necks
What is the maximum number of alleles that a diploid individual can have at each locus? Consider only loci found on autosomal chromosomes
2
A population is in Hardy-Weinberg equilibrium. Consider only a single locus and two alleles found at that locus. If the frequency of the a allele is 0.4, what is the frequency of all of the possible genotypes at this locus? Call the other allele A. Assume
f(aa) = 0.16, f(Aa) = 0.48, f(AA) = 0.36; 0.16 + 0.48 + 0.36 = 1.0
A population consists of 200 aa individuals and is in Hardy-Weinberg equilibrium. Assuming one-locus, two-allele genetics, what is the frequency of the a allele if the population consists of a total of 1,000 individuals? What if the 200 aa individual popu
square-root of 0.2 in a population of 1000 = 0.45; square-root of 0.02 in a population of 10,000 = 0.14
A population is in Hardy-Weinberg equilibrium. The frequency of the dominant phenotype is 0.99. What fraction of individuals that carry at least one recessive allele are homozygous at this locus? Assume one-locus, two-allele genetics.
1-0.99 = frequency of the recessive homozygote = 0.01; The frequency of the recessive allele is 0.1; the frequency of the heterozygote therefore is 2
0.1
(1-0.1) = 0.18; the fraction of individuals that carry the recessive alleles that are homozygous for
Inbreeding and assortative mating are both examples of what?
Inbreeding mating
How does heterozygous advantage contribute to the maintenance of polymorphisms, i.e., balanced polymorphisms?
with heterozygous advantage, two different alleles are required to display the phenotype/genotype with the greatest associated fitness. Thus, selection favors the maintenance of two alleles per that locus and the maintenance of two or more alleles per giv
What characteristic of the two parent populations is typically a necessary prerequisite for the occurrence of hybrid vigor?
Inbreeding or, at least, the fixing of deleterious alleles not shared with the other parental population such that masking of deleterious alleles in the offspring may occur
In stabilizing selection, what categories of phenotypes is selection "editing out" of the population?
both extremes of the phenotype, i.e., in terms of quantitative , the low-end expression of the trait as well as the high-end expression
In many human societies both genders are responsible for the care of offspring though with the duties typically separated such that the female is more responsible for meeting the physical needs of offspring while the male is more responsible for meeting t
sexual selection
In addition to no mutation, no natural selection, no genetic drift, and no migration, what further criteria must be met for a population to be in Hardy-Weinberg equilibrium?
random mating between individuals
What is assortative mating?
choosing a mate on the basis of resemblance to that individual
Considering eye color, if B is the brown-eye allele and b is the blue-eye allele, given a population consisting of the following, 323 BB, 23 Bb, and 45 bb, what are the frequencies of allele B and of allele b? Is this population in Hardy-Weinberg equilibr
frequency of B is equal to [(323 * 2) + (23 * 1) + (45 * 0)] / [(323
2) + (23
2) + (45 * 2)] = 669 / 782 = 0.855; frequency of b is equal to 1 - 0.855 = 0.145; f(BB) = (0.855)2 = 0.73, f(Bb) = 0.25, f(bb) = 0.021, which corresponds to actual numb
0)] / [(
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
-The frequency of the "aa" genotype:
-The frequency of the "a" allele:
-The frequency of the "A
-36%, as given in the problem itself.
- The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.
-Since q = 0.6, and p
Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individua
9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).
There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and
-Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%).
-Since q = 0.2, and p + q = 1, then p = 0.8 (80%).
-The frequency of heterozygous individuals is equal to 2pq. In this case,
Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:
-The perc
The first thing you'll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square
A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:
-The allele frequencies of each allele.
-The expected genotype frequencies
Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we don't have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396
A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population.
Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the fr
After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population total
Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with
Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
-The frequency of the recessive allele in the population.
-The frequency of the dominant allele
Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
Answer: The frequen
In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blo
Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency o
The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies.
Answer: First, lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Taking the square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1 so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT,
In a study of the Hopi, a Native American tribe of central Arizona, Woolf and
Dukepoo (1959) found 26 albino individuals in a total population of 6000. This form of
albinism is controlled by a single gene with two alleles: albinism is recessive to normal
a. Because you can't tell who might be a carrier just by
looking.
b. If we assume that the population's in H-W equilibrium,
then the frequency of individuals with the albino
genotype is the square of the frequency of the albino
allele. In other words, fre
A wildflower native to California, the dwarf lupin (Lupinus nanus) normally bears blue
flowers. Occasionally, plants with pink flowers are observed in wild populations. Flower
color is controlled at a single locus, with the pink allele completely recessiv
a. Let B be the blue allele and b be the pink allele, so
that p = frequency (B) and q = frequency (b).
The frequency of the bb genotype = 25/3316 = q2, so q =
?(0.00754) = 0.0868.
p = 1 - q, so p = 0.913
freq (BB) = p2 = 0.834
freq (Bb) = 2pq = 0.158
b. F
Cooke and Ryder (1971) studied the nestlings of Ross's goose, a small Arctic nesting
goose. Goslings (baby geese) exist in two color morphs, grey or yellow. Cooke and
Ryder reported that a population of geese at Karrack Lake, Canada included 263 yellow
go
a. For both of these calculations, p = frequency of
dominant allele, and q = frequency of recessive
allele. If grey is dominant:
q2 = 263 / 676 = 0.389
q = ? (0.389) = 0.624 = frequency of yellow allele
p = 1 - q = 0.376 = frequency of grey allele
Predict
4. A 1970 study of 93 house mice (Mus musculus) in a single barn in Texas focused on a
single locus (the gene for a certain enzyme) with two alleles, A and A'.
The genotype frequencies found were:
AA
0.226
AA'
0.400
A'A' 0.374
a) Calculate the allele freq
Quick and easy way:
Freq (A) = p = 0.226 + (0.400 / 2) = 0.426
Freq (a) = q = 0.374 + (0.400 / 2) = 0.574
Predicted freq (AA) = p2 = 0.181
Predicted freq (AA') = 2pq = 0.489
Predicted freq (A'A') = q2 = 0.329
Could be several things, but notice in particu
5. The geneticist P. M. Sheppard (1959) carried out a selection experiment on a
laboratory population of the fruit fly Drosophila melanogaster. The stubble allele, which
affects bristle shape of the fly, is dominant to the wild-type allele. Flies that are
Let S be the stubble allele and s be the normal
allele.
Freq (SS) = 0
Freq (Ss) = 0.14
Freq (ss) = 0.86
Freq (S) = p = 0 + (0.14 / 2) = 0.07
Freq (s) = q = 0.86 + (0.14 / 2) = 0.93
wSS = 0 (because all flies with this genotype die)
wSs = 1
wss = 1
Just be
P. D. N. Hebert studied the frequencies of alleles for the gene that codes for the
enzyme malate dehydrogenase (Mdh) in the "water flea," Daphnia magna, living in
ponds near Cambridge, England. There are three alleles of the Mdh gene, abbreviated S,
M and
Easy. 114 individuals = 228 alleles. Frequency of S =
(3*2 + 8 + 19)/228, i.e. 3 individuals with two S
alleles and 8+19 individuals with one each, all
divided by 228. Freq(S) = 0.145.
You calculate freq(M) in the same way: (15*2 + 8 +
37)/228 = 0.329, an
8. Avena fatua is a species of wild oat (a type of grass). Jain and Marshall studied wild
oat population genetics in California. One of the traits they examined was the pubescence
(hairiness) of the leaf sheath, which is controlled by a single locus with
This oughta be easy by now. The quick way:
p = freq(L) = 0.571 + (0.071/2) = 0.606
q = freq(l) = 0.358 + (0.071/2) = 0.394
Ho hum. .
genotype
LL
Ll
ll
observed
57.1%
7.1%
35.8%
Looks a lot like inbreeding, doesn't it? Again, you've
got that decrease in he
9. The biologist B. Battaglia raised the marine copepod Tisbe reticulata (a small free-
swimming marine crustacean) under crowded conditions. T. reticulata has one gene with
two alleles, Vv and Vm, showing incomplete dominance. In one of his tanks, Battag
Let p = freq(Vv). Then p=(353
2 + 1069) / (1751
2) =
0.507. And q = freq(Vm) = 1-p = 0.493. If the
population were in H-W, then the frequency of VvVv
individuals would be p2, or 0.257; in reality it is
353/1751 = 0.202. The frequency of VmVm individuals
w
11. Fundulus heteroclitus (common name: mummichog) is a small fish that lives in bays
and estuaries along the east coast of North America, from Newfoundland to Florida. It's
been extensively used in evolutionary studies.1
Northern populations of F. hetero
For simplicity, I'll call the two alleles A and B.
Northern populations are virtually all AA, and
Southern populations are virtually all BB.
500 AA fish and 500 BB fish means that the frequency
of both alleles is 0.5. So let p = frequency(A) and q
= frequ
12. In 1973, Francisco Ayala and colleagues sampled a population of deep-sea starfish of
the species Nearchaster aciculosus, living at a depth of 1244 meters off the coast of San
Diego, California. They used electrophoresis to determine the frequencies of
Oooh! Four alleles! The formula for the four-allele
Hardy-Weinberg equilibrium is
(p + q + r + s)2 = 1
which comes to
p2+2pq+2pr+2ps+q2+2qr+2qs+r2+2rs+s2 = 1
Each term of the equation gives the predicted
frequency for a particular genotype; p2 is the
pred
Mendel vs Darwin
Darwin proposed a mechanism for change in species over time.
When Mendel's research was rediscovered in the early 20th century, many geneticists
believed that his laws of inheritance conflicted with Darwin's theory of natural selection.
� Darwin emphasize
gene pool.
The total aggregate of genes in a population at any one time