Following the Hardy-Weinberg equilibrium theory, the frequency of the heter0zyg0us genotype is 2pq. In the exposed example, 2pq = 0.48.
<h3>
Hardy-Winberg equilibrium</h3>
The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q.
Assuming a diallelic gene,
→ The allelic frequencies are
- p is the frequency of the dominant allele,
- q is the frequency of the recessive allele.
→ The genotypic frequencies after one generation are
- p² (H0m0zyg0us dominant genotypic frequency),
- 2pq (Heter0zyg0us genotypic frequency),
- q² (H0m0zyg0us recessive genotypic frequency).
If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.
The addition of the allelic frequencies equals 1 ⇒ p + q = 1.
The sum of genotypic frequencies equals 1 ⇒ p² + 2pq + q² = 1
If the allele A has a frequency of 0.6, and the allele B has a frequency of 0.4, then the frequency of the heter0zyg0us genotype is
2pq = 2 x 0.6 x 0.4 =<u> 0.48</u>
You can learn more about the Hardy-Weinberg equilibrium at
brainly.com/question/3406634
Stop the treatment immediately, if the medication given needs antagonist to counteract it's effects, then it should be given. reassess the diagnosis and exclude all the differential diagnosis. conduct investigations relevant to the histories including serological test, chest CT and MRi. inform the patient the situation (if he is awake) and make sure the next treatment is the best after arriving to the correct diagnosis.
Perhaps the bird’s calls could be less attracting and disorienting to predators and the ones with the said distinctive calls could have survived to repopulate and pass down the genes.
