Answer:
In the F1 generation
RR = 0%
Rr = 50% (or 0.5)
rr = 50% (or 0.5)
Explanation:
A pink flowering plant has the genotype Rr. It is heterozygous for the allele. The alleles for this gene appears to show incomplete dominance, as the heterozygous phenotype is a blend of the two homozygous genotypes.
A white flowering plant has the genotype rr. It is homozyogous for the white allele
A punnet square of the cross is shown.
The resulting punnet square shows that only Rr and rr genotypes are possible, at a ratio of 50:50 (or 1:1). Therefore, the genotype frequency of Rr is 50%, and rr is 50% in the F1 generation. This can also be written as 0.5. It is not possible to get a red plant, as the genotype RR can not come from this cross
1.317 × 10^25 lbs i believe.
Answer:
The correct option is 3. "At pH 6.5 the enyzme is 50% active"
Explanation:
For the titratable group to be protonated and cause the enzyme to be in the active state, it needs to have gained a hydrogen cation (H+). In order for that to happen, there must be enough hydrogen cations in the environment of the enzyme, and hence, an acidic pH is required in this case.
Answer:
B. It provides the code for the protein
Explanation:
C. amnion
if you need explanation lmk
