Hardy-Weinberg Law for Population Genetics

As stated in the introduction to population genetics, the Hardy-Weinberg Law states that under the following conditions both phenotypic and allelic frequencies remain constant from generation to generation in sexually reproducing populations, a condition known as Hardy-Weinberg equilibrium.
1.large population size
2.no mutation
3.no immigration or emigration
4.random mating
5.random reproductive success

If we mate two individuals that are heterozygous (e.g., Bb) for a trait, we find that
25% of their offspring are homozygous for the dominant allele (BB)
50% are heterozygous like their parents (Bb) and
25% are homozygous for the recessive allele (bb) and thus, unlike their parents, express the recessive phenotype.

This is what Mendel found when he crossed monohybrids. It occurs because Meiosis separates the two alleles of each heterozygous parent so that 50% of the gametes will carry one allele and 50% the other.

When the gametes are brought together at random, each B (or b)-carrying egg will have a 1 in 2 probability of being fertilized by a sperm carrying B (or b).

Results of random union of the two gametes produced by two individuals, each heterozygous for a given trait. As a result of meiosis, half the gametes produced by each parent with carry allele B; the other half allele b.

Results of random union of the gametes produced by an entire population with a gene pool containing 80% B and 20% b.

-	0.5 B	0.5 b	-------------------------------------------------------------	-	0.8 B	0.2 b
0.5 B	0.25 BB	0.25 Bb	-------------------------------------------------------------	0.8 B	0.64 BB	0.16 Bb
0.5 b	0.25 Bb	0.25 bb	-------------------------------------------------------------	0.2 b	0.16 Bb	0.04 bb

But the frequency of two alleles in an entire population of organisms is unlikely to be exactly the same. Let us take as a hypothetical case, a population of hamsters in which
80% of all the gametes in the population carry a dominant allele for black coat (B) and
20% carry the recessive allele for gray coat (b).
Random union of these gametes (right table) will produce a generation:
64% homozygous for BB (0.8 x 0.8 = 0.64)
32% Bb heterozygotes (0.8 x 0.2 x 2 = 0.32)
4% homozygous (bb) for gray coat (0.2 x 0.2 = 0.04)
So 96% of this generation will have black coats; only 4% gray coats.
Will gray coated hamsters eventually disappear?

No. Let's see why not.
All the gametes formed by BB hamsters will contain allele B as will one-half the gametes formed by heterozygous (Bb) hamsters.
So, 80% (0.64 + .5*0.32) of the pool of gametes formed by this generation with contain B.
All the gametes of the gray (bb) hamsters (4%) will contain b but
one-half of the gametes of the heterozygous hamsters will as well.
So 20% (0.04 + .5*0.32) of the gametes will contain b.
So we have duplicated the initial situation exactly. The proportion of allele b in the population has remained the same. The heterozygous hamsters ensure that each generation will contain 4% gray hamsters.

Now let us look at an algebraic analysis of the same problem using the expansion of the binomial (p+q)2.
(p+q)2 = p2 + 2pq + q2
The total number of genes in a population is its gene pool.
Let p represent the frequency of one gene in the pool and q the frequency of its single allele.
So, p + q = 1
p2 = the fraction of the population homozygous for p
q2 = the fraction homozygous for q
2pq = the fraction of heterozygotes
In our example, p = 0.8, q = 0.2, and thus
(0.8 + 0.2)2 = (0.8)2 + 2(0.8)(0.2) + (0.2)2 = 0.64 + 0.32 + 0.04

The algebraic method enables us to work backward as well as forward. In fact, because we chose to make B fully dominant, the only way that the frequency of B and b in the gene pool could be known is by determining the frequency of the recessive phenotype (gray) and computing from it the value of q.

q2 = 0.04, so q = 0.2, the frequency of the b allele in the gene pool. Since p + q = 1, p = 0.8 and allele B makes up 80% of the gene pool. Because B is completely dominant over b, we cannot distinguish the Bb hamsters from the BB ones by their phenotype. But substituting in the middle term (2pq) of the expansion gives the percentage of heterozygous hamsters. 2pq = (2)(0.8)(0.2) = 0.32

So, recessive genes do not tend to be lost from a population no matter how small their representation

Ooooooooops....Found Somewhat Difficult...!! But they ask this sometimes in CM ( PSM ) as Spotting or in Example...Have to Understand once..