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goldfiish [28.3K]
4 years ago
10

The table gives the average temperatures and elevations of the inner canyon and the north rim of the Grand Canyon. Inner Canyon

North Rim average elevation 2,100 feet 8,000 feet average temperature in July 107°F 77°F average temperature in January 56°F 37°F Why is there such a vast climate difference in these two areas of the Grand Canyon?
Biology
2 answers:
Yuki888 [10]4 years ago
8 0

Answer:

C for my plato peoples

Explanation:

sergiy2304 [10]4 years ago
4 0
The north rim is higher so it'll be cooler
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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

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).

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Explanation: Hope this helps luv :)

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