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Llana [10]
3 years ago
5

The half life of C14 is 5730 years. I have a sample that I dug up from a sea cucumber fossil. I expected .00015ppm of C14 in my

living sample, but instead I found that there was exactly1/8th of the expected ratio. How old is my specimen
Biology
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

The answer is 17190 years

Explanation:

Original amount was 5730. To become 1/8 of the original amount, three (3)  half life periods will demand because after first half life the amount will become 1/2 and after second half life, amount become 1/2 again (1/2 x 1/2 = 1/4).  Third half life, the amount become half 1/2 again (1/2 x 1/2 x 1/2 = 1/8).

For the 3 half life periods we have original amount x numbers of half life periods = 5730 years x 3   =  17190 years

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Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advant
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Answer:

0.09

Explanation:

<h3><u>Before selection</u></h3>

Total number in population = 1000

Genotype frequencies

Genotype frequency of BB = 500/1000 = 0.5

Genotype frequency of Bb = 250/1000 = 0.25

Genotype frequency of bb = 250/1000 = 0.25

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375

<h3><u>After selection</u></h3>

We are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).

Now the total number of individuals is 500 + 125 + 250 = 875.

Genotype frequencies

Genotype frequency of BB = 500/875 = 0.57

Genotype frequency of Bb = 250/875 = 0.29

Genotype frequency of bb = 125/875 = 0.14

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285

<h3><u>Change in frequency of B after 1 generation</u></h3>

0.715 - 0.625 = 0.09

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So the genotype will be tt.

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