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snow_tiger [21]

3 years ago

15

¿Cada gen determina un caracter de un ser no vivo?

1 answer:

Leya [2.2K]3 years ago

5 0

Answer:

Explanation:

Putos

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What animals have been dominant in the Cenozoic era

vodomira [7]

Explanation:

Cave lions, sabre-toothed cats, cave bears, giant deer, woolly rhinoceroses, and woolly mammoths were prevailing species of the Quaternary period. Without the dinosaurs, plant life had an opportunity to flourish during the Cenozoic era.

4 0

2 years ago

Which is the term for a hormone that acts on the same cell that secreted it? A : limited hormone B : paracrine C : autocrine D :

suter [353]

Answer:

C: autocrine

Explanation:

There are three types of hormones according to where their target cells are.

Endocrine: Their target cells are distributed far and they travel via blood to reach the target cells.
Paracrine: Their target cells are often their neighbors so they often diffuse to reach them
Autocrine: Their secretion acts on themselves. Hence they are their own targets.

3 0

2 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

PLEASE HELP!!!!!

LekaFEV [45]

Answer:

C)

Explanation:

1000+22-8+6

= 1020

=2% increase

8 0

2 years ago

Read 2 more answers

You are giving ventilations to a 5-year-old child through a resuscitation mask, you should give 1 ventilation about every

Yuki888 [10]

You should typically give ventilation’s to a young child every 30 seconds.

Hope I could help! :)

7 0

3 years ago