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2 years ago

What is a pioneer species? (Will give brainiest)

Biology

1 answer:

inna [77]2 years ago

8 0

Answer:

The first species to colonize an ecosystem that is previously damaged or disrupted. Supplement. Pioneer species is the first to colonize an ecosystem previously disturbed or damaged. For example, sand that has become barren is colonized by a pioneer species of lyme grass (Leymus arenarius)

Explanation:

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List the following objects in order from smallest to largest: star, galaxy, universe, planet, nebula, asteroid

avanturin [10]

Answer:asteroid,planet,star,nebula,galaxy, and universe

Explanation:I’m not actually sure im just smart

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2 years ago

Qué sucede con el agua en la respiración celular??

Katarina [22]

Se produce en la mitocondria. La respiración celular, como componente del metabolismo, es un proceso catabólico, en el cual la energía contenida en los sustratos usados como combustible es liberada de manera controlada.

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2 years ago

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kykrilka [37]

Answer:

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Explanation:

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2 years ago

How do sea anemone catch prey?

V125BC [204]

Answer:

Tentacles have specialised stinging cells called nematocysts. They use these to immobilise their prey so that the tentacles are then able to move the food into the mouth.

Explanation:

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2 years ago

When pink sweet peas were self-pollinated and the seeds were collected and sown, the following flower colors were obtained: Red

Pavlova-9 [17]

Answer:

c. 1:2:1

The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

Explanation:

If flower color were determined by a gene showing incomplete dominance, the possible genotypes and phenotypes are as follows:

RR- red
ww - white
Rw - pink

If pink sweet peas are self-pollinated, then a cross between two heterozygous individuals is done (Rw x Rw).

<u>From this cross the expected ratios are:</u>

1/4 RR (red)
2/4 Rw (pink)
1/4 ww (white)

So the null hypothesis is that the observed results exhibit a 1:2:1 ratio.

<h3><u>Chi square test</u></h3>

$X^{2} = \sum \frac{(Observed - Expected)^2}{Expected}$

<u>The observed frequencies were:</u>

34 Red
76 Pink
40 White

Total 150

<u>The expected frequencies for our null hypothesis are:</u>

1/4 x 150 = 37.5 Red
2/4 x 150 = 75 Pink
1/4 x 150 = 37.5 white

$X^{2} = \frac{(34- 37.5)^2}{37.5} + \frac{(76- 75)^2}{75} + \frac{(40- 37.5)^2}{37.5}$

$X^2=0.5067$

The degrees of freedom (DF) are calculated as number of phenotypes - 1; in this case DF = 3-1 = 2.

If we look at the Chi square table, for 2 DF and a probability of p0.05, the critical value is 5.991

Our X^2 value of 0.5067 is less than the critical value, so we do not reject the null hypothesis. The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

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2 years ago