Answer:
0.5
Explanation:
According to Hardy-Weinberg equilibrium p2+2pq+q2=1 (p+q=1)
p2 is frequency of the dominant homozygous genotype
2pq is the frequency of the heterozygous genotype
q2 is the frequency of the recessive homozygous genotype
In the example above 80 chickens have bare legs (ff)-recessive homozygous which means that the frequency of that genotype is 80/240+80=0.25 (q2), frequency of the recessive allele is
=0.5. This means that the frequency of the dominant allele (p) is 1-0.5=0.5
So, the frequency of the heterozygous genotype (2pq) is 2*0.5*0.5=0.5
The frequency of the dominant homozygous genotype is p2=0.25
Answer:
Explanation:
The plate were supposed to streak by D. radiodurans but unfortunately it was streaked by E.coli.
As there is no other petri plate available, this plate was exposed to the UV for the 30 minutes. The plate was not left uncovered and because of that the total E.coli could not die and another streak of D. radiodurans was done on the same plate.
As a result of incubation there will be growth of both of the organism in the petri dish.
The answer is DNA replication
Answer:
It doesnt only have to apply to a child. It can apply to anyone any age.
Reaches earth by Radiation
