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ivanzaharov [21]
3 years ago
8

Helen was reading about the variety of forms of aquatic organisms. She noted that plants that grow near the surface of the sea a

re very different from those below the surface. She concluded that the difference was because of the pressure of water on submerged plants. Which is a reasonable alternative to Helen's conclusion? A) The temperature increases with depth below the surface. B) Only very small plants can grow below the surface of water. C) Very few animals feed on plants deep below the surface of water. D) Plants below the surface receive less sunlight than those on the surface.
Biology
2 answers:
liq [111]3 years ago
4 0
D) Plants below the surface receive less sunlight than those on the surface
egoroff_w [7]3 years ago
4 0

Answer:

D) Plants below the surface receive less sunlight than those on the surface.

Explanation:

Since the light cannot penetrate deep in sea, the plants present at the surface differ from those presence below the surface. The plants present below the surface carry different type of pigment that is able to trap the light available below the surface.  

Example: Red algae are found in relatively deep water since it has pigment phycoerythrin that can trap the blue light available in deep sea.  

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In RNA molecules, adenine is complementary to
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It would be uracil (C)
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In a population of cats, the phenotypic frequency of black cats is 91%, and the phenotypic frequency of white cats is 9%. Assumi
Talja [164]

Answer:

1. Allele frequency of b = 0.09 (or 9%)

2. Allele frequency of B = 0.91 (0.91%)

3. Genotype frequency of BB = 0.8281 (or 82.81%)

4. Genotype frequency of Bb = 0.1638 (or 16.38%)

Explanation:

Given that:

p = the frequency of the dominant allele (represented here by B)  = 0.91

q = the frequency of the recessive allele (represented here by b)  = 0.09

For a population in genetic equilibrium:

p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)

(p + q)^2 = 1

Therefore:

p^2 + 2pq + q^2 = 1

in which:  

p^2 = frequency of BB (homozygous dominant)

2pq = frequency of Bb (heterozygous)

q^2 = frequency of bb (homozygous recessive)

p^2 = 0.91^2 = 0.8281

2pq = 2(0.91)(0.9) = 0.1638

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Answer:

leukocyte

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