Answer:
980,100
Explanation:
let the recessive condition (aa) be represented as q² which is the genotypic frequency = 100/1,000,000 = 0.0001
allelic frequency of a = q = √q² = √0.0001 = 0.01. thus q = 0.01
From the formula p + q = 1 where p is the allelic frequency of A.
since q = 0.01
p = 1 - 0.01 = 0.99. The allelic frequency of A (p) = 0.99
p² = 0.99² = 0.9801. Genotypic frequency of AA= 0.9801
= 0.9801 x 1,000,000 = 980,100 individuals with AA (homozygous normal)
For 2pq genotypic frequency for hetrozygous (Aa).
Using the formula p² + 2pq + q² = 1. Since p² = 0.9801 and q² = 0.0001
2pq = 1 - (p² + q²)
= 1 - (0.9801 + 0.0001)
= 1 - (0.9802)
= 0.0198 = 0.0198 x 1,000,000 = 19,800 individuals with Aa
The measurements that are normal for a client is a blood pressure of 108/80, a pulse of 112 beats per minute and respiratory rate of 15 breaths per minute. Option C
<h3>What is the normal blood pressure?</h3>
The blood pressure refers to the rate at which the heart pumps blood to the other parts of the body. The blood pressure would rise if the blood vessels are blocked and the heart would have to do more work to pump blood.
The measurements that are normal for a client is a blood pressure of 108/80, a pulse of 112 beats per minute and respiratory rate of 15 breaths per minute.
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These are similar because they both included desired characteristics. They are different because artificial selection has to do with humans being involved. where natural selection is with out humans being involved and is life or death.
C i think...................
Answer:
D. Limit further logging and clearcutting of pine trees.
Explanation:
