B. I think thats right :/
Answer:
The correct option is;
D. 45%
Explanation:
From the Hardy- Weinberg law, we have;
p² + 2·p·q + q² = 1
p + q = 1
Where:
p = Dom inant allele frequency in the population
q = Recessive allele frequency in the population
p² = The percentage of individuals in the population that are hom ozygous dominant
q² = The percentage of individuals in the population that are homo zygous recessive
2×p×q = The percentage of hete rozyous individuals in the population
The number of individuals that express the recessive phenotype = 86
The number of individuals in the population = 200
The percentage of individuals that express the recessive phenotype, q² = 86/200 = 0.43
Therefore;
q = √0.43 = 0.656
p + q = 1
p = 1 - q = 1 - 0.656= 0.344
∴ The frequency of individuals that express the do minant phe notype, p = 0.344
The percentage of heterozyous individuals in the population = 2×p×q × 100 = 2 × 0.656 × 0.344 × 100 = 45.15% ≈ 45%
Answer:
Phosphorylation within the nuclear export signal interferes with the function of the signal.
Explanation:
In biochemistry, phosphorylation is the addition of a phosphate group (PO4) to a protein or other molecule. Phosphorylation is a major player in protein regulation mechanisms, preventing protein-catalyzed reaction product from accumulating in the body causing problems.
However, in some cases phosphorylation may cause nuclear accumulation of a protein in the nucleus of the cell. An example of this is the protein shown in the question above. In this case, phosphorylation in the nuclear export signal interferes with the signal function, resulting in protein accumulation in the nucleus.
