Answer:
5.) The possible genotypes should be: RR RB BB ( if the could for the same proteins)
6.) Frequency of R: 0.5
7.)Frequency of R: 0.5
8.)Frequency of R: 0.375
Explanation:
5.) If R and B code for the different forms of the same protein then they are only a few possible out comes
6.) First calculate the total alleles in population
allele R= 40
Allele B=40
total allele = R+B= 40+40
=80
Now to find allele R frequency is:
(Total R alleles) / (Total allele in pop)
40/80=
0.5
7.)Calculate the total alleles
Alleles from for R
RR= 10
R=10x 2= 20
Multiply the value by 2 because there are 2 R alleles present in
RB=20
R=20
Number doesn't change there is only 1 R allele
Total R=20+20
=40
Alleles For B
BB=10
B=10×2
=20
Same thing here, two B alleles together so multiple by 2
RB=20
B=20
Total B= 20+20
=40
Total alleles in pop add
40+40
=80
Frequency of R
Total of R/Total Alleles
=40/80
=0.5
8.)Repeat the same thing in 7 but use different numbers
RR=10
R=10x2
=20
RB=10
R=10
Total R= 20+10
=30
Answer: "mutations" .
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Answer:
Chemical bonds contain potential energy.
Explanation:
Chemical bonds always contain potential energy. The atoms of the bond want to move to a lower energy to become more stable.. The energy for breaking bonds only comes when stronger bonds are formed. This energy is used to tear apart the bonds holding the Hydrogen atoms together. The strength of the covalent bonds depend on the overlap between the valence orbitals of the bonded Atom.
Answer:
you won't find the nucleus in prokaryotic cells