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WITCHER [35]
4 years ago
6

Despite the high-level of biodiversity, this biome has infertile soils, principally oxisols, to which the vegetation has roots s

ystems that capture nutrients from litter decay.
Biology
1 answer:
irina [24]4 years ago
3 0
The biome that has infertile soil, principally oxisols, to which the vegetation <span>has roots systems that capture nutrients from litter decay, is the tropical rainforest.

Soil types are highly variable in the tropics and are the result of a combination of several elements or variables which includes climate, vegetation, topographic position, parent material and also the soil age. Most of the tropical soils are described by leaching and poor nutrients. It also has two classifications: namely the ultisols and oxisols. Ultisols are known as well-weathered, acidic red clays soils. They are deficient primarily in major nutrients such as potassium and calcium. Oxisols are also an acidic type of tropical soil. However, they are well drained compared to the ultisols.
</span>
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A cell from a turkey wing contains 40 chromosomes, how many chromosomes are there in a unfertilized chicken egg
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4 years ago
A population of 100 tigers has 30 with horizontal stripes. Using pedigrees, you ascertain that horizontal striping is caused by
bixtya [17]

Assuming that this population is in Hardy-Weinberg equilibrium, we can sayt that  <em>the</em><em> probability</em><em> that, given enough time, all tigers in the population will have horizontal stripes is</em><em> 30% = 0.3.</em>

--------------------------

<u>Available data:</u>

  • Population size, N = 100 tigers
  • Number of Tigers with horizontal stripes = 30
  • Horizontal stripes is caused by an autosomal recessive mutation.

We need to know the probability that all tigers in the population will have horizontal stripes, which is the recessive phenotype.

We assume that the population is in Hardy-Weinberg equilibrium, so there should be no evolution.

Since the population is in H-W equilibrium, the allelic, genotypic and phenotypic frequencies will remain the same generation after generation.

Now, we will calculate the allelic and genotypic frequencies of horizontal striped individuals in the population.

There are 100 individuals, and only 30 have horizontal stripes.

So, the phenotypic frequency, F(HS) is 30/100 = 0.3 = 30%

Since this is a recessive phenotype, this value equals the genotypic frequency, F(hh) = 30%

Finally, we can get the allelic frequency by taking the square root of this value.

F(hh) = q² = 0.3

f(h) = q = √0.3 = 0.55 = 55%

According to these calcs, the probability for the fixation of the recessive allele is 55%, and the probability that all individuals express the recessive horizontal stripes phenotype is 30%.

--------------------------

You can learn more about Hardy-Weinberg equilibrium at

brainly.com/question/12724120?referrer=searchResults

brainly.com/question/9870841?referrer=searchResults

brainly.com/question/419732?referrer=searchResults

brainly.com/question/13603001?referrer=searchResults

brainly.com/question/5028378?referrer=searchResults

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