Question

Please show how to works with type. Thank you.

Gene Frequencies We will let p designate the frequency (=proportion) of gene B in a population, and we will let q designate the frequency of gene b. Since these are the only two alleles for that characteristic, then p + q = 1. (When all possible outcome frequencies are added, the sum of their frequencies must equal one (certainty)]. Now let us assume that completely random mating and offspring survival occurs with respect to these two alleles. What this means is that these alleles do not affect survival or choice of mate. Therefore, reproduction will be random with respect to these two genes. When reproduction and survival are random, we can reason as follows: 1. The probability that an offspring will get one gene B is equal to p. Using the product rule of probability, the probability that an offspring will get two genes B will be equal to p x p. So the frequency of BB offspring in the next generation should be p?. 2. Using the same argument for the bb offspring, their frequency will be q?. 3. The frequency of heterozygous offspring in the new generation is 2pq. This can be derived easily, as follows: The two gene frequencies are: P + 4 = 1 If they combine randomly in all possible combinations, then (p + 2)2 = 12 and by expansion p+ 2pq + q² = 1 These relationships allow us to make predictions about the next generation: Example: The frequency of allele b in the population is known to be 0.3. What phenotypic proportions are expected in the next generation? What proportion of the next generation should be heterozygous? Answer: The frequency of recessives (bb) will be 0.3 x 0.3, or 0.09. This means that 9% of the population will be recessives. The remaining 91% will have the dominant phenotype. The frequency of heterozygotes will be 2pq, which means that 2 x 0.7 x 0.3 = 0.42. (As a check, also note the frequency of BB individuals will be 0.7 x 0.7 = 0.49. So: 0.49 +0.42 + 0.09 = 1) It is important to emphasize one last time that the assumption of random reproduction and survival underlies all of these computations.

Answer 1

Answer: Hardy-Weinberg principle is explained in the give example.

Before applying the principle to a population, following conditions need to be satisfied.

· Population should be large

· There should be random mating between the individuals of the species.

· No genetic drift taking place because of migration, mutation etc.

· No selection of a particular allele

If all these conditions are satisfied then the population is said to be in H-W equilibrium, according to this principle the allele frequencies in the population will remain the same over the generations

If we consider any character and if there are two alleles for the character

Let’s consider the alleles as B and b

If B allele represents the dominant character then it will expressed in homozygous BB as well as heterozygous condition Bb

And the recessive allele will express itself only in homozygous condition i.e. bb

The allele frequencies are given by the equation

p²+2pq+q² =1   i.e. p+q =1

p² = BB= frequency of the homozygous dominant genotype.

2pq = Bb= frequency of heterozygotes in the population.

q² = bb= frequency of the homozygous recessive genotype

p = frequency of the dominant allele, note that dominant allele B will appear in homozygous as well as in heterozygous condition.

q = frequency of the recessive allele, note that recessive allele b will appear in homozygous as well as in heterozygous condition.

Now to start with the calculation, recessive character is considered first as it will appear only in homozygous condition.

If the question is asked, make sure what is given in the question, is it the frequency of the individuals with the particular character or the frequency of the allele is mentioned. Accordingly fill the values of p and q

In the given example they have given the frequency of the recessive allele that means q is given

q= 0.3 therefore the frequency of the homozygous recessive individuals will be

q² that will be equal to 0.3×0.3 =0.09, if you multiply this with 100, you will get the percentage of individuals in the population who are showing the recessive character (recessive phenotype)

Now same way value of p can be calculated by using the formula p+q=1

Therefore p=1-0.3 that is p=0.7 this is the frequency of B allele in the population.

To calculate the homozygous dominant individuals BB, you need to calculate p² that will be equal to 0.7×0.7 that is 0.49. Hence the percentage of homozygous individuals in the population is 49%

To calculate the heterozygous population, you need to consider 2pq that will be equal to 2×0.7×0.3=0.42, hence the percentage of heterozygous individuals in the population is 42%

If you add the both the above percentages (as dominant character is expressed in homozygous as well as heterozygous condition)

it will be equal to 91%, the same value which we get if consider the percentage of recessive individuals as 9% which we got from the first calculation.

To summarise

In the given population, if the phenotypes are considered then the number of individuals expressing the dominant phenotype = 91%

And the number of individuals expressing the recessive phenotype =9%

If the genotypes are considered then the frequency of B allele in the population is 0.7

And frequency of b allele in the population is 0.3.

You can cross check by putting the values in the equation p²+2pq+q² =1  

Note: suppose in the problem, the number of individuals are given then you have to consider p² or q² depending on whether dominant or recessive trait is considered as in the following example, only the first line of the problem I have mentioned

The number of individuals in the population expressing the recessive character is 36%

By reading this it can be concluded that q² is given which is 0.36 therefore p² in this case will be 64% that is 0.64

Now the further calculations can be done with this information.