In the Hardy-Weinberg equation, the value of the term $\displaystyle p^2$ is the percentage of homozygous dominant (BB) individuals in a population. Thus, to answer this question we will need to solve for $\displaystyle p^2$.

$\displaystyle p^{2}+ 2pq + q^{2} = 1$

With the information we are given, we can calculate $\displaystyle q^2$, which is the percentage of the population that is homozygous recessive (bb). We know the total number of homozygous recessive rabbits (306 white rabbits) and the total number of rabbits in the population (306 + 544 = 850). We can thus divide the number of white rabbits by the population total to find $\displaystyle q^2$:

$\displaystyle q^{2} = \frac{bb(white)}{Total} = \frac{306}{850} = 0.36$

36% of the population is homozygous recessive. We then take the square root of $\displaystyle q^2$ to find $\displaystyle q$:

$\displaystyle q = \sqrt{q^{2}} = \sqrt{0.36} = 0.60$

This value tells us that 60% of all the alleles in the population are recessive (b).  

The Hardy-Weinberg equation is based off the idea that the total number of alleles in a population is the sum of the dominant and recessive alleles, as shown by this equation:

$\displaystyle p + q = 1$

We can use this equation to solve for $\displaystyle p$ using $\displaystyle q$:

$\displaystyle p = (1-q) = (1-0.60) = 0.40$

This value tells us that 40% of all the alleles in the population are dominant (B).

We can now square $\displaystyle p$ to obtain $\displaystyle p^2$. Recall that this will be the frequency of homozygous dominant individuals in the population:

$\displaystyle p^{2} = (0.40)^{2} = 0.16$

The result is 16%, meaning 16% of the rabbits in the population are homozygous dominant. We can then multiply 16% by the total population to find the number of rabbits who are homozygous dominant:

$\displaystyle BB = (p^{2} \cdot Total) = (0.16 \cdot 850) = 136$

Thus, 136 homozygous dominant rabbits would be expected in this population.

In the Hardy-Weinberg equation, the value of the term $\displaystyle 2pq$ is the percentage of heterozygous (Bb) individuals in a population. Thus, to answer this question we will need to solve for $\displaystyle 2pq$.

$\displaystyle p^{2}+ 2pq + q^{2} = 1$

With the information we are given, we can calculate $\displaystyle q^2$, which is the percentage of the population that is homozygous recessive (bb). We know the total number of homozygous recessive rabbits (76 without spots) and the total number of rabbits in the population (76+ 399= 475). We can thus divide the number of frogs without spots by the population total to find $\displaystyle q^2$:

$\displaystyle q^{2} = \frac{bb(spotless)}{Total} = \frac{76}{475} = 0.16$

16% of the population is homozygous recessive. We then take the square root of $\displaystyle q^2$ to find $\displaystyle q$:

$\displaystyle q = \sqrt{q^{2}} = \sqrt{0.16} = 0.40$

This value tells us that 40% of all the alleles in the population are recessive (b).  

The Hardy-Weinberg equation is based off the idea that the total number of alleles in a population is the sum of the dominant and recessive alleles, as shown by this equation:

$\displaystyle p + q = 1$

We can use this equation to solve for $\displaystyle p$ using $\displaystyle q$:

$\displaystyle p = (1-q) = (1-0.40) = 0.60$

This value tells us that 60% of all the alleles in the population are dominant (B).

Now that we have $\displaystyle p$ and $\displaystyle q$, we can calculate $\displaystyle 2pq$. Recall that this will be the frequency of heterozygous individuals in the population:

$\displaystyle 2pq = 2 (0.60) (0.40) = 0.48$

The result is 48%, meaning 48% of the frogs in the population are heterozygous.  We can then multiply 48% by the total population to find the number of frogs who are heterozygous::

$\displaystyle Bb = (2pq \cdot Total) = (0.48)( 475) = 228$

Thus, 228 heterozygous frogs would be expected in this population.

There are two equations for Hardy-Weinberg equilibrium:

$\displaystyle A + a = 1$

$\displaystyle A^2 + 2Aa + a^2 = 1$

Using the first equation, we can substitute in 0.85 for $\displaystyle A$ and solve to get $\displaystyle a = 0.15$.

We can then use the second equation to find the frequencies of each genotype.

$\displaystyle A^2$ = frequency of $\displaystyle AA$ genotype

$\displaystyle 2Aa$ = frequency of $\displaystyle Aa$ genotype

(Why do we multiply by 2? Because we must count both $\displaystyle Aa$ and $\displaystyle aA$ as heterozygotes.)

$\displaystyle a^2$ = frequency of $\displaystyle aa$ genotype

Thus we get $\displaystyle 2Aa=2(0.85)(0.15) = 0.255$.

Recall that under Hardy-Weinberg conditions, the allele frequencies remain constant. The formulas for Hardy-Weinberg equilibrium are:

$\displaystyle p+q=1$

and

$\displaystyle p^2+2pq+q^2=1$

Here, $\displaystyle p^2$ is the frequency of the homozygous dominant genotype, $\displaystyle 2pq$ is the frequency of the heterozygous genotype, and $\displaystyle q^2$ is the frequency of the homozygous recessive genotype. Since we have all the variables we need, and we want to find the genotypic frequencies (not the phenotypic) we plug into the second equation.

$\displaystyle 0.9^2+2(0.9*0.1)+0.1^2=1$

$\displaystyle p^2=0.81$

$\displaystyle 2pq=0.18$

$\displaystyle q^2=0.01$

Be sure to check to make sure these frequencies add up to 1.

Recall that under Hardy-Weinberg conditions, the allele frequencies remain constant. The formulas for Hardy-Weinberg equilibrium are:

$\displaystyle p+q=1$

and

$\displaystyle p^2 + 2pq + q^2 = 1$

Here, $\displaystyle p^2$ is the frequency of the homozygous dominant genotype, $\displaystyle 2pq$ is the frequency of the heterozygous genotype, and $\displaystyle q^2$ is the frequency of the homozygous recessive genotype. Since we have all the variables we need, and we want to find the genotypic frequencies (not the phenotypic) we plug into the second equation.

$\displaystyle 0.3^2+2(0.3*0.7)+0.7^2=1$

$\displaystyle p^2=0.09$

$\displaystyle 2pq=0.42$

$\displaystyle q^2=0.49$

Be sure to check to make sure these frequencies add up to 1.

To begin, we must know that we are working with the Hardy-Weinberg equations:

$\displaystyle p+q=1$

$\displaystyle p^{^{2}}+2pq + q^{2}=1$

We can designate the dominant allele (L) as $\displaystyle p$ and the recessive allele (l) as $\displaystyle q$. 

Since the frequency of L was given (0.4) we know the the frequency of l must be:

$\displaystyle 1 - 0.4 =0.6$

Now we have both allele frequencies and can plug them into the equation:

$\displaystyle (0.4)^{^{2}}+2(0.4)(0.6)+(0.6)^{2} = 1$

So for $\displaystyle p^{^{2}}$, LL (homozygous dominant), is equal to 0.16

for $\displaystyle 2pq$, Ll (heterozygous), is equal to 0.48

for $\displaystyle q^{2}$, ll (homozygous recessive), is equal to 0.36