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algol [13]
2 years ago
9

Name two pond populations that could not survive in a forest ecosystem

Biology
1 answer:
Rus_ich [418]2 years ago
7 0

Answer:

Name two pond populations that could not survive in a forest ecosystem,  monkeys, ants, and birds.

Explanation:

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Use an example from nature to explain how natural selection causes populations to change as their environment change include the
ivanzaharov [21]

Answer:

<u><em>Galapagos finches</em></u><u> have various beak sizes that make foraging for food more successful.</u>

Explanation:

Organisms evolve over time due to changes in their genome. These are pontaneous, and occur in DNA at random. These changes are called mutations and form alleles or different forms of a gene.

Over time within a population, the number alleles increase the variation of the population. These variants may confer specific traits within an individual, that may confer a biological advantage.

Thus, the trait may make the organisms more capable of obtaining food, shelter a mate etc. or ensure survival, i.e. they are able to pass on their genes to the next generation.

5 0
2 years ago
Why does the nuclear membrane need to dissolve at the start of mitosis?
Molodets [167]

Answer:

The nuclear envelope does not disappear in metaphase of mitosis, because it already did in prophase. ... The nuclear envelope needs to be broken apart so that the chromosomes can be found, aligned in the middle of the cell, and then pulled apart.

Explanation:

4 0
3 years ago
small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

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

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

7 0
3 years ago
54:58
jarptica [38.1K]

Answer:

CORRECT ANSWER IS FALSE.

Explanation:

I JUST DID THE QUIZ

E2021

4 0
2 years ago
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Is bison national animal ? If so,which coutry ?​
Nana76 [90]

Answer:

The American bison is the national animal of USA..

3 0
2 years ago
Read 2 more answers
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