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3 years ago

For water to enter the ground there must be?

Biology

2 answers:

nalin [4]3 years ago

4 0

Answer: Percolation.

elena-s [515]3 years ago

3 0

Answer: Percolation

Explanation: It is the ability of water to sip in into the ground. Percolation is the process wherein the ground absorbs water up to the amount it can hold. When the soil or ground is saturated with water it loses its ability to absorb water and causes it to run off. Unsaturated ground can absorb more water through percolation process.

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The graph shows that corals in the Caribbean reefs are declining

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3 years ago

The means for transmission of disease-causing microorganism is provided by the ____.

Sergio039 [100]

Answer: The means for transmission of disease-causing microorganism is provided by the direct or indirect contact.

Microorganisms can cause disease only once they are transferred to the body. The disease causing microorganisms are termed as pathogens which are transmitted by several ways such as from skin to skin, by nuclei droplets, through blood and body fluids or via air. In vector transmissions the disease is carried by the parasitic insects via animals, air borne transmission occurs when microorganisms move through air or the dust particles, droplet transmission occurs by coughing, sneezing or talking by the person who is infected while indirect transmission occurs by physical contact or by touching contaminated objects.

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3 years ago

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MaRussiya [10]

Answer:

Explanation:

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3 years ago

In the 3rd line of defense blood vessels expand,

julia-pushkina [17]

The answer is true!!

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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. <u>If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive)</u>, 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