Answer:
0.483
Explanation:
The given population is in Hardy-Weinberg equilibrium. If the gene has two alleles, the sum total of the frequencies of these two alleles will be one.
Therefore, the total of the frequency of allele B and frequency of allele b will be 1. f(B) + f(b)=1
If the frequency of allele "B" is 0.59, then the frequency of allele "b" will be=1-0.59= 0.41
The frequency of heterozygous genotype in the population= 2pq
p= frequency of the dominant allele
q= frequency of the recessive allele
So, 2pq= 2 x 0.59 x 0.41 = 0.483
In order: 50% 0% 100%? 50%
you have a 50% chance of having healthy offspring b/c they do not possess the gene
0% b/c they all carry the disease
100% b/c all of them have the recessive gene
50% b/c there’s only two that are completely recessive (aa)
A. Ring of fire I hope this hoped u today
Theses are what the Nurse should include:
•sodium or salt
• meat(some kind of protein)
•grains like bread or corn flakes cereal
•exactly one serving
•dairy low in phosphorus
• fruit and juice from concentrate
