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
Phylum. The classification system goes: Kingdom, Phylum, Class, Order, Family, Genus, Species. Anything that shares a class also share anything above it.
<span>The correct answer is b: Coenzyme A.</span>
During the process of pyruvate oxidation, acetyl CoA molecule is produced. Pyruvate oxidation is the link between glycolysis and Krebs cycle and it converts pyruvate (three-carbon molecule) into acetyl-CoA ( two-carbon molecule attached to Coenzyme A). NADH is produced and one CO2 released. AcetylCoA is the substrate for the next stage of cellular respiration, citric acid cycle.
