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satela [25.4K]
3 years ago
12

Reflect: write just a few words about

Biology
1 answer:
Elina [12.6K]3 years ago
8 0

Answer:

Glucose: is a simple sugar which is an important energy source in living organisms and is a component of many carbohydrates.

Fructose: or fruit sugar, is a simple ketonic monosaccharide found in many plants.

Deoxyribose: sugar derived from ribose by replacing a hydroxyl group with hydrogen.

Ribose: is a kind of sugar that is produced by the body. It is used as a medicine. Ribose is used for heart disease, mental function, athletic performance, chronic fatigue syndrome (CFS), fibromyalgia, and many other conditions, but there is no good scientific evidence to support these uses.

Glyceraldehyde (glyceral) is a triose monosaccharide with chemical formula

Explanation:

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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]

Answer:

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

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