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babunello [35]
3 years ago
12

In the DESERT BIOME, which of the following could be an example of mutualism?

Biology
2 answers:
DENIUS [597]3 years ago
7 0

The answer is B:

A chipmunk takes shelter in and oak tree and is protected from predators, and in return disperses seed around the area

Llana [10]3 years ago
5 0
<h2>Answer:</h2>

The mutualism example which is apt for desert biome is c<u>hipmunk and oak trees</u>.

<h2>Explanation:</h2>

The mutualism refers to the ecological interaction in which the species involved get benefited either ways. This ecological interaction is mostly found in most of the ecological community. This interaction is of two types which includes cooperation (working together) and symbiosis (parasitic, mutualistic, commensal or symbiotic).

The other options given does not come under mutualism. Though the desert coyote comes in desert biome it is not a correct answer because desert coyote does not eat fruit.

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

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

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