Covalent Bonds: A Hydrogen Example. A covalent bond is a chemical bond that comes from the sharing of one or more electron pairs between two atoms. Hydrogen is an example of an extremely simple covalent compound. A hydrogen example.
Answer: Gravity is by far the weakest force we know.
Gravity and weight are not the same thing.
Gravity makes waves that move at light speed.
Explanation: I hope this helps!!
Parietal cells<span> also called oxyntic </span>cells<span> </span>
The lichens show mutualism and they are found in every biome of the Earth. The lichens have algae and fungus associated with them and they live in a mutualistic relationship with each other. Lichens can grow on soil where no other species can grow like rocks, barren soils and they enrich the soil on dying where other plants and trees can grow.
Lichens being pioneer species change the environment by building up the soil and liberating nutrients. Also the lichens contribute to the chemical weathering of the soil releasing phosphate along with erosion and weather conditions.
When the simple plants die, the wildflowers and other plants start growing replacing lichens or mosses.
From the above explanation it can be concluded that lichens alter the abiotic environment and favor other species of plants to grow by breaking down rocks from soil and providing nutrients.
The frequency of homozygous dominant individuals is: 42%
Hardy–Weinberg equilibrium (HWE) is a null model of the relationship between allele and genotype frequencies, both within and between generations, assuming no mutation, no migration, no selection, random mating, and infinite population size. To find the frequency of the recessive allele, we must first find the frequency of the dominant allele (p). According to the Hardy-Weinberg principle, the square root of the homozygous genotype frequency is equal to the allele frequency. The Hardy-Weinberg equation used to determine genotype frequencies is: p2 + 2pq + q2 = 1. Where 'p2' represents the frequency of the homozygous dominant genotype (AA), '2pq' the frequency of the heterozygous genotype (Aa) and 'q2' the frequency of the homozygous recessive genotype (aa).
Calculation:
q2 = 49
q= 0.7
According to Hardy–Weinberg equilibrium p + q= 1
p = 1 - q
1-0.7= 0.3
p2+ 2pq+q2=1
One can substitute the values
2pq= 2(0.3) (0.7) = 0.42
42% is the answer.
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