Allopatric v. Sympatric Speciation

 

Allopatric speciation refers to speciation that has arisen due to geographic or habitat isolation. The name derives from the Greek word “allo” meaning “other” and “patric” meaning “country. An excellent example of this comes from the Great Lakes in Africa.  A barrier sprang up about 5 000 years ago. Today in the newly created lake there are six species of cichlids. One species is also found in the larger “parent” lake, but there are also five species found nowhere else. They had to have evolved from the original cichlid species that was trapped by the physical barrier.

 

Sympatric speciation refers to speciation through RIMs other than ecological isolation. The name is derived from the Greek meaning “same country”. It may include behavioural mechanisms or improper cell division leading to extra sets of chromosomes (polyploidy in plants). Bread wheat is a hexaploid with 42 chromosomes believed to have evolved as a hybrid of a 28 chromosome cultivated wheat and a 14 chromosome wild wheat.  Researchers are trying to cultivate new polyploids in hopes to create more disease-resistant, high-yield, and hardy plant species.

 

Hardy-Weinberg Equilibrium and Equation

 

A non-evolving population is said to be at equilibrium, that is, the allele frequency will not change from generation to generation.

 

Given a population of wildflowers that is diploid and comprised of red and white flowers. Red flowers are the result of a dominant allele ‘A’, white flowers are coded by the recessive allele ‘a’. The population size is 500 flowers. 320 are homozygous red (AA), 160 are heterozygous red (Aa), and 20 are white (aa).

 

Since each individual flower has two alleles for colour, there are 1000 alleles in the population.

Q. How many of these alleles are dominant?

A. 640 from the homozygous red flowers and 160 from the heterozygous red flowers, for a total of 800.

 

Q. How many of these alleles are recessive?

A. 160 from the heterozygotes and 40 from the white flowers, for a total of 200.

(Easier: subtract 800 from 1000)

 

 

 

Q. What are the frequencies of these alleles?

 

A. The dominant allele, A, has a frequency of 800/1000 or 80%. The recessive allele has a frequency of 20% (200/1000).

 

The frequencies of the genotypes of the next generation can be calculated using allele frequencies. We’ll use A= 0.8 and a = 0.2.

 

NOTE: The allele frequencies must add to 1. The allele frequencies in the population will be reflected equally in both sexes. AND the alleles segregate during meiosis. What does this mean?

 

Gametes produced by either the males or the females in the population will contain A alleles at a frequency of 0.8 and a alleles at a frequency of 0.2.

 

If mating is random:

 

a) The probability that a zygote will receive two A alleles is 0.8 x 0.8 or 0.64

In other words, we expect that 64% of the next generation will be AA.

 

b) The probability that a zygote will receive one A from the male parent and one a allele from the female parent is 0.8 x 0.2 or 0.16. The probability that a zygote will receive one A from the female parent and one a allele from the male parent is 0.8 x 0.2 or 0.16. There are two ways to produce a heterozygote, so we expect that 32% (2 x 0.16) of the next generation will be Aa.

 

c) The probability that a zygote will receive two a alleles is 0.2 x 0.2 or 0.04. We expect that 4% of the next generation will be aa.

 

The frequency of the genotypes must add to 1.

 

Hardy-Weinberg rules

 

Let the dominant allele be represented by the letter p

Let the recessive allele be represented by the letter q

 

For allelic frequency           p + q = 1

 

When gametes fuse to join a zygote, the probability of producing the AA genotype can be calculated by p X p or p².  The probability of producing the Aa genotype is 2pq. The probability of producing the aa genotype can be calculated by q X q or q².

 

For genotypic frequencies

 

(p + q) (p + q) = p² + 2pq + q² = 1

 

The equation predicts the genotypic frequency in a non-evolving population. If our observed frequencies are significantly different, then we do not have equilibrium, and therefore, we do not have a non-evolving population. This means that one or more of the five conditions are not being met.

 

 

Let’s go back to our second generation of wildflowers to see if we can derive the allelic frequencies from the genotypic frequency.

 

Let’s say the wildflowers are particularly “fruitful” and produce 2000 offspring in the next generation.

64 % are AA, so 1280 are AA (2560 A alleles).

32% are Aa , so 640 are Aa (640 A alleles and 640 a alleles)

4% are aa, so 80 are aa (160 a alleles).

 

Total number of alleles is 4000.

 

Frequency of the A allele                 (2560 + 640)/4000 = 0.8

 

Frequency of the a allele                 (640 + 160)/4000 = 0.2

 

An easier way? WE’RE SCIENTISTS! Of course, there’s an easier way.

 

Frequency of the A allele         0.64 + ½ (0.32) = 0.8

 

Frequency of the a allele         ½ (0.32) + 0.4 = 0.2

 

Yippee!

 

What can we do with these?

 

If you know p, you can find q.

If you know q, you can find p.

 

If you know the frequency of the recessives, you know what q² is. If you know what q² is, you can find q. And if you know what q is, you can find p.

 

If you know the frequency of the dominant phenotype, you can find the frequency of the recessive phenotype. Then you know q², then you find q, then you can find p.

 

If you know the frequency of the heterozygotes, you know squat. Why? Because the heterozygotes are represented by 2pq. There are two unknowns, therefore you cannot solve the problem.

 

Clear as mud yet? We’ll go over every conceivable combination of known and unknown values in class (eight problems).