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timofeeve [1]
2 years ago
11

which of the following provides the best analogy for an electron in an atomic orbital? a. a bee moving from flower to flower in

a garden b. a bird resting on a tree branch c. an ant crawling on the surface of a leaf d. a bee trying to escape from a closed jar
Biology
1 answer:
DochEvi [55]2 years ago
5 0
The answer is d. a bee trying to escape from a closed jar

An electron in an atomic orbital moves very very fast and in a limited area.
So, the only possible analogy is a bee trying to escape from a closed jar. The bee is in the limited area and probably moves very fast because it is agitated and tries to escape.

Choice a cannot be analogy because a bee will not rush from flower to flower.
Choice b. cannot be analogy because a bird is not moving.
Choice c. cannot be analogy because an ant is moving really slow.
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Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfw
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Answer:

Question 1.

The mean relative fitness is 0.85

Question 2.

The expected allele frequency change of A, p = 0.63: AA, p2 =0.63 x 0.63 = 0.39.

Explanation:

Mean relative fitness is given as

p2W(AA) + 2pqW(Aa) + q2W(aa).

Where W(AA)= 1.00, W(Aa) = 0.80, W(aa) = 0.60

Firstly, find p2, 2pq, q2 which are the frequencies of AA, Aa and aa respectively.

AA= 500,Aa=250 ,aa=250. Total is 1000

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250/2)+500= 625.

625/1000= 0.625.

Therefore A, p is 0.625

p2 = 0.625x0.625= 0.39

For q, since p+q =1

q= 1 - 0.625 = 0.375

q2= 0.375x0.375=0.14.

2pq= 2x0.625x0.375= 0.47.

Mean relative fitness is

p2W(AA) + 2pqW(Aa) + q2W(aa).

0.39x1+ (0.47x0.80) + (0.14x0.60)=0.85.

Question 2.

From Hardy-Weinberg equilibrium

F = p2 + 2pq + q2.

Frequency of A, p= 0.625 and AA, p2 is 0.625 x 0. 0625= 0.39.

3 0
3 years ago
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