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NikAS [45]
3 years ago
8

What pigment are. Give two example

Biology
1 answer:
Sphinxa [80]3 years ago
6 0
<span>An organic compound that gives a characteristic color to plant or animal tissues and is involved in vital processes. Chlorophyll, which gives a green color to plants, and hemoglobin, which gives blood its red color, are examples of pigments.
<span>
A substance or material used as coloring.

that would be my answer

</span></span>
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Muscle cells, which need a lot of ATP, would be expected to have a lot of _______, the organelles involved in cellular respirati
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Its <span>mitochondria im sure 

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8 0
3 years ago
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For thousands of years, people have operated farms at the bases of active volcanoes in spite of the risks. what might be the rea
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Hello There!

As well as the obvious negatives of volcanic eruptions, there are benefits too.
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5 0
4 years ago
Which best describes how to classify water?
Anastaziya [24]

Answer:

b)It is a compound because it is made of a single kind of molecule.

Explanation:

8 0
3 years ago
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At what stage of the cycle does the cell copy its DNA?
Murljashka [212]

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S-Phase

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5 0
3 years ago
A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
Amanda [17]
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
6 0
3 years ago
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