Calcium is a chemical element with symbol Ca and atomic number 20. Calcium is a soft gray Group 2 alkaline earth metal, fifth-most-abundant element by mass in the Earth's crust. The ion Ca2+ is also the fifth-most-abundant dissolved ion in seawater by both molarity and mass, after sodium, chloride, magnesium, and sulfate.[5] Free calcium metal is too reactive to occur in nature. Calcium is produced in supernova nucleosynthesis.
Answer:
Firstly, the partial pressure of nitrogen (78%) is crucial to breathing purposes. Without this pressure, the lungs will burst and animals cannot survive.
Secondly, nitrogen is required for the formation of amino acids (building blocks of proteins) and other organic compounds that are necessary for the survival of living organisms. Principally, in the atmosphere, nitrogen is present in the form of molecular nitrogen (N2). N2 is fixed by nitrogen-fixing bacteria that form nitrates and nitrites. These molecules are then used in biochemical processes to produce proteins (amino acids) and other organic compounds. In the absence of nitrogen, these processes could become seize of limited significantly thus affecting life overall.
Thirdly, nitrogen and its derivatives act as greenhouse gases that maintain the Earth's temperature within a range that supports life. Yes, the increased abundance of nitrous oxides is not good because of acid rain and other issues, however, still, the presence of nitrogen is important for life on this planet.
u can use con doms to stop it.
The frequency <em>p</em> of the yellow (A) allele is <em>p</em>= 0.3
The frequency <em>q</em> of the blue (a) allele is <em>q= </em><em>0.7</em>
Hardy–Weinberg equilibrium, states that allele and genotype frequencies in a population will remain constant from generation to generation. Equilibrium is reached in the absence of selection, mutation, genetic drift and other forces and allele frequencies p and q are constant between generations. In the simplest case of a single locus with two alleles denoted A and a with frequencies f(A) = p and f(a) = q, the expected genotype frequencies under random mating are f(AA) = p² for the AA homozygotes, f(aa) = q² for the aa homozygotes, and f(Aa) = 2pq for the heterozygotes.
p²+2*p*q+q²= 1 p+q= 1 q= 1-p
yellow (p²)= 9%= 0.09 p= √0.09= 0.3
green (2*p*q)= 42%= 0.42
blue (q²)=49%= 0.49 q=1-0.3= 0.7 <em>or</em> q= √0.49= 0.7
Amino Group, Carboxyl Group, and the R group
