In a population with two alleles, B and b, and the allele frequency of b is 0.4. Calculate the allele frequency of heterozygotes if the population is in Hardy–Weinberg equilibrium? Show your work.
Thanks for any help in advance!
b=0.4
B+b=1, so
1-b=B
B=1-0.4=0.6
(B+b)^2=B^2+2Bb+b^2=1
(0.6)^2+[2(0.4)(0.6)]+(0.4)^2=1
Bb=heterozygous
Bb=2(0.4)(0.6)=0.48
But if you need to know,
BB=0.36
bb=0.16
To calculate the allele frequency of heterozygotes (Bb), we can use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
where:
p^2 = frequency of homozygous dominant individuals (BB)
2pq = frequency of heterozygous individuals (Bb)
q^2 = frequency of homozygous recessive individuals (bb)
and p = frequency of the B allele, q = frequency of the b allele.
Given that the allele frequency of b (q) is 0.4, we can substitute this value into the equation.
p^2 + 2pq + q^2 = 1
p^2 + 2p(0.4) + (0.4)^2 = 1
p^2 + 0.8p + 0.16 = 1
p^2 + 0.8p = 0.84
Now, let's solve this quadratic equation.
p^2 + 0.8p - 0.84 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. Let's use the quadratic formula:
p = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = 0.8, and c = -0.84.
p = (-0.8 ± √((0.8)^2 - 4(1)(-0.84))) / (2(1))
p = (-0.8 ± √(0.64 + 3.36)) / 2
p = (-0.8 ± √4) / 2
p = (-0.8 + 2) / 2 or p = (-0.8 - 2) / 2
p = 1 / 2 or p = -2.8 / 2
p = 0.5 or p = -1.4
Since allele frequencies cannot be negative, we discard the negative solution. Therefore, the allele frequency of the B allele (p) is 0.5.
Now, substitute this value of p into the equation p^2 + 0.8p = 0.84 to find the value of p^2:
(0.5)^2 + 0.8(0.5) = 0.84
0.25 + 0.4 = 0.84
0.65 = 0.84
The equation is not balanced, indicating that there is an error in the calculation. It is likely that our derived value of p is incorrect.
Please check your initial information, and let me know if there are any additional details that need to be considered.
To calculate the allele frequency of heterozygotes (Bb) in a population in Hardy-Weinberg equilibrium, we will use the formula:
p^2 + 2pq + q^2 = 1
Given that the allele frequency of b is 0.4, we can determine that q^2 (the frequency of bb genotype) is 0.4.
Therefore, q = √(0.4) = 0.632.
Since we know that p + q = 1 (given that there are only two alleles, B and b), we can calculate p as follows:
p = 1 - q = 1 - 0.632 = 0.368.
Now we have both p and q values, and we can substitute them into the formula to find the frequency of heterozygotes (2pq):
2pq = 2 * 0.368 * 0.632 = 0.4656.
So, the allele frequency of heterozygotes is 0.4656 or approximately 0.466.