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
Answer:
10
Explanation:
(6−
6
3
)2−
(4)(3)
2
=(6−2)2−
(4)(3)
2
=42−
(4)(3)
2
=16−
(4)(3)
2
=16−
12
2
=16−6
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<span>C. Gametes are the sex cells, and each (sperm and ovum) contain half of the parent's genetic material. These cells will fuse to produce a zygote, which will usually contain the full number of genes required by an organism to live. If there is a chromosomal abnormality, such as a gamete with an extra chromosome, the offspring can either die in development or sometimes be born with physiological and/or developmental difficulties. An example of this phenomenon is trisomy 21, also known as Down Syndrome, in which the 21st chromosome bears 3 copies instead of the regular 2.</span>
The correct answer for the above question is the ejaculatory ducts. The male urethra is divided into three regions, one of them is the prostatic urethra which is the proximal portion that passes through the prostrate gland. It receives the ejaculatory duct, which contains sperm and secretions from the seminal vesicles, and numerous ducts from the prostrate glands.
