Respuesta :
Answer:
Thus, the frequency of people who are heterozygous for this disease is 0.21
Explanation:
Given,
The frequency of gene p is [tex]0.3[/tex]
As per Hardy- Weinberg equation,
Sum of allele frequencies for all allele at locus [tex]= 1\\[/tex]
Which means sum of allele frequencies of recessive and dominant allele is 1
[tex]p + q = 1\\[/tex]
Substituting the value of p in above equation, we get -
[tex]0.3 + q = 1\\q = 1 - 0.3\\q = 0.7\\[/tex]
Another equation of Hardy- Weinberg is
[tex]p^2 + 2pq + q^2 = 1[/tex]
Substituting the value of p and q in above equation, we get -
[tex](0.3)^2 + (0.7)^2 + 2pq = 1\\2pq = 1 - 0.09 - 0.49\\pq = \frac{0.42}{2} \\pq = 0.21[/tex]
Thus, the frequency of people who are heterozygous for this disease is 0.21
Answer: Frequency of heterozygous people is 0,42
Explanation: taking into account Hardy-Weinberg equations for allele frequencies it is known that the sum of the fequencies of the alleles is 1
p + q = 1
If we know that p allele has a frequency of 0,3, the frequency for P allele will be 0,7 since
1- q = p or 1 - 0,3 = 0,7
Now, the other equation p2 +2pq+q2 = 1 express that sum of genotype frequencies is 1, where p2 is homozygous dominant frequency, 2pq is heterozygous frequency and q2 is homozygous recesive frequency.
As we know the values for p and q we can calculate fequency for heterozygous:
2pq = 2 (0,7) (0,3) = 0,42
