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Hunter-Best [27]
3 years ago
15

The Serengeti region of Africa contains a population of cheetahs and a population of lions studies have shown that the cheetah p

opulation has very low genetic diversity, while the genetic diversity of the lion population is quite high. The cheetah population is also much smaller than the lion population both populations are vulnerable to infection with a particular virus that can be fatal in both species what is most likely the outcome if both populations are infected with this virus?
Biology
1 answer:
BARSIC [14]3 years ago
6 0

High genetic diversity and a large population can protect the lion population from being wiped out completely.  Those lions whose genetics offer them protection from the virus may be immune.  The small genetic diversity and low population of the cheetahs could leave the cheetah population with only a handful of immune individuals.  

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In a species of bird, the allele for the striped trait is recessive to the allele for the speckled trait. Suppose that a coastal
mylen [45]

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The answer is "0.42"

Explanation:

Please find the complete question in the attached file.

Coastal community striped allele intensity = q = 0.62

Landlocked community, strip intensity allele=q'=0.40

Its eliminated allele frequency for movement = q"=0.42 \ in \ indoorscommunity And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts=q_2=0.39

Heritable (striped) allele frecency:

\to q=\sqrt{q_2}

      =\sqrt{0.39} \\\\ =0.624 \\\\ =0.62

(round up to the closest cent) is therefore the only frequency of coastal people.

The genome of pulling function in host population= q'_2=0.16 Heterozygous feature regularly.

Therefore the inland community rate of recessed (striped), allele rate:

\to q'= \sqrt{q'_{2}}= \sqrt{0.16}=0.40

Following migration;

Its percentage of coastal migrants:

m=10\% \\\\

   = \frac{10}{100} \\\\= 0.1

Coastal population Non-immigrant percentage of coastal residents:

=1- m\\\\=1-0.1\\\\ =0.9

Stripping coastal community of allele rate after immigration:

(q")=\text{(stripping of migrant only freq)}\times (m)+\text{(stripping allele baud rate for non-immigrants)} \times ( 1-m)

       = [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42

8 0
3 years ago
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