Respuesta :
Answer:
a) p=0.665
b) q=0.335
c) 2pq=0.445
d) (0.665^2)+(0.335^2)+(2*0.665*0.335)=1
Explanation:
For solve this exercise I will use the Hardy-Weinberg principle. We have the following:
a) and b)
number of purple flowers=142
number of white flowers=18
total number of flowers=160
frequency of white flowers=q^2=18/160=0.112
q=0.335
if
p+q=1
Clearing p:
p=1-0.335=0.665
c) the heterozygous is 2pq:
frequency of heterozygous allele: 2pq=2*0.665*0.335=0.445
d) p^2+2pq+q^2=1
(0.665^2)+(0.335^2)+(2*0.665*0.335)=1
Answer:
A) p = 0.1125
B) q = 0.8875
C) The frequency of heterozygous plants in the field = 0.1997
D) p² + 2pq + q² = 0.0127 + 0.1997 + 0.7876 = 1
Explanation:
In a pea plants, Flowers can be purple (p) or white (q). Purple flowers (p) is the dominant allele and white flowers (q) are the recessive allele. They are 18 plants with white flowers, and 142 plants with purple flowers
A) According Hardy-Weinberg equilibrium equation:
p² + 2pq + q² = 1 and p + q = 1
Where p is the frequency of the dominant allele and q is the frequency of the recessive allele.
Since they are 18 plants with white flowers, and 142 plants with purple flowers, the total number of plants = (18 + 142) = 160
Therefore p = number of plants with purple flower/ total number of plants = 18/160 = 0.1125
p = 0.1125
B) q = number of plants with white flower/ total number of plants = 142/160 = 0.8875
q = 0.8875
C) The frequency of heterozygous plants in the field = 2pq = 2 × 0.8875 × 0.1125 = 0.1997
frequency of heterozygous plants in the field = 0.1997
D) Given that p² + 2pq + q² = 1
∴ (0.1125)² + 0.1997 + (0.8875)² = 0.0127 + 0.1997 + 0.7876 = 1
