Answer:
The genotypic frequency = 1:1
The phenotypic frequency = 1:1
Explanation:
Given that:
The allele → R = Red beetles
The allele → B = Blue beetles
Since the gene color shows a codominant allele
The Red Beetle = RR
The blue beetles will be = BB
The heterozygous beetle will be = RB
∴
The punnet square showing the crossing of RB × RR is:
R B
R RR RB
R RR RB
The result shows that we have two red beetles and two heterozygous beetles.
Hence;
The genotypic frequency = 1:1
The phenotypic frequency = 1:1
I guess it is aluminium in cationic form
The dialysis bag will remain the same size in a isotonic solution.
One way is to see if it responds to stimulants or if it reproduces
