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antiseptic1488 [7]
4 years ago
12

marsupials, such as kangaroos, koalas, and wombats, are mammals that care for their newborn young in a pouch. all marsupials tha

t are found in australia are thought to have evolved from a single species that migrated to the island and then adapted to the various niches. what does this example illustrate? behavioral adaptation adaptive radiation geographic isolation population explosion
Biology
2 answers:
Allisa [31]4 years ago
7 0

Correct answer on edg. is

Adaptive Radiation :)

docker41 [41]4 years ago
3 0
The answer is <span>geographic isolation.</span>

<span>When a new species form by geographic isolation, it is called geographical speciation or allopatric speciation. Geographical isolation is a physical separation which prevents gene flow between distinct populations (on the island and on continent). The island populations will eventually diverge from a single species that migrated to the island. The members of the populations, which once belonged to the same species, will no longer be able to mate and produce offspring. Eventually, the formation of new species will occur, for example kangaroos, koala, and wombats, form that one species.</span>

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I'm pretty sure the answer is Pollen

Explanation:

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Which of the following correctly describes connective tissue
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There are no multiple choices

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Incomplete Dominance <br> I need help
Dafna11 [192]

Answer:

1.

Red = rr

Blue = bb

Purple = rb

2.

a) Genotype - rb

Phenotype - purple

b) 0%

c) 100%

d) 0%

3.

a) Genotypes - rr, rb, bb

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b) 25%

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3 years ago
Assume a population of purple and white flower pea plants (2000 total plants) is in Hardy-Weinberg equilibrium. The frequency of
lys-0071 [83]

Answer:

a) 0.48

b)

p= 0.4\\q=0.6

c) 720

Explanation:

Given ,

Frequency of white flower plant (p^2) is 0.16

Thus, frequency of dominant white allele (p) is equal to \sqrt{0.16} \\= 0.4

As we that, as per  Hardy-Weinberg equilibrium equation

p+q=1

Substituting the value of p in above equation, we get

0.4+q=1\\q=1-0.4\\q=0.6

a) As per Hardy-Weinberg equilibrium equation

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

Substituting the given and calculated values in above equation, we get -

0.4^2+0.6^2+2pq=1\\0.16+0.36+2pq=1\\2pq=1-0.16-0.36\\2pq=0.48

b) The frequency (decimal form) of the dominant allele

p= 0.4\\q=0.6

c) number of homozygous purple flower plants

= frequency of purple flower plants (q^2) * Number of flowers

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