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2 years ago

What conditions make for a violent volcanic eruption?

Biology

2 answers:

krok68 [10]2 years ago

8 0

Answer:

the two main factors that influence how a volcano will erupt are viscosity and gas content. Both are related to the composition of the magma . Hawaiian volcanoes tend to erupt basalt , which is low in viscosity and low in gas content (about 0.5 weight percent). so OB

Explanation:

Natalija [7]2 years ago

8 0

Answer: B

Explanation:

Magma with low viscosity

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Explain how my model can be used to predict how changing the Earth’s tilt would impact seasons

Sholpan [36]

Earth is always rotating around the sun that means if it were winter on one side of the world, (that is not on the side of the sun) the other side of the world would be summer, ( the side that the sun is shining on)

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2 years ago

Gene control mechanisms, or "gene expression," control _____.

Luda [366]

Answer:

I believe the correct answer is C. mRnA translation.

Explanation:

Because the gene expression is controlled on two levels. First, transcription is controlled by limiting the amount of mRNA that is produced from a particular gene. The second level of control is through post-transcriptional events that regulate the translation of mRNA into proteins.

3 0

2 years ago

2. Just as in humans, systems interact in plants to maintain homeostasis and accomplish the functions of transport, reproduction

Xelga [282]

Answer:

osmosis

Explanation:

plants use osmosis whic is moving water though a semipermial membrane to draw up water. It moves from places of high conetartions of water (dirt) has to is less water (roots). When it does this the plant mains homeostasis by getting water up useing their xylem (tubes in stems) to the leaves.

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2 years ago

HELP HELP!! Why might you find fewer photosynthetic organisms at deeper levels of an ocean? Group of answer choices

riadik2000 [5.3K]

1.
B
Photosynthetic organisms require light, in which the sun’s rays can only go so far
2.
C
The energy flow pyramid goes from decomposers to producers, meaning that if producers are dying out due to lack of nutrients, there must be something wrong with the decomposer population

4 0

3 years ago

Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe

Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive), the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

$p+q=1\\(p+q)^{2} = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1$

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, $P(aa)=q^{2}=0.000588$ . And with that we can calculate the value of q,

$P(a)=q=\sqrt{0.000588}=0.024$

And since we know that p+q=1, we can find out the value of p.

$p+0.024=1\\1-0.024=p\\p=0.976$

Next, we find out the genotypic frequency of the genotype AA:

$P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95$

Now, we can find out the genotypic frequency of the genotype Aa:

$P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046$

Notice than:

$p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1$

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

5 0

3 years ago