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
When peanut <span>seedlings were grown under identical conditions, the factor that could count for differences in height among the seedlings is GENETICS.</span>
Answer:
Nothing is there can you repost it?
Explanation:
Answer:
Cell A is laterally being inhibited by other neighbouring cells.
Explanation:
Cell A showing lower activity compare to cell B despite both having high simulation is because Cell A is surrounded by neighbouring cells and these neighbouring cells hinder the action potential of cell A through their numerous activity they perform around cell A.
Therefore they restrain the activities of cell A.
Abiotic
Abiotic
Abiotic
Biotic
Biotic
Biotic
