The answer is 49.92%
Let's use the <span>Hardy-Weinberg principle:
p + q = 1
p</span>² + 2pq + q² = 1
<span>
where:
p - the frequency of dominant allele G
q - </span>the frequency of recessive allele g
p² - the frequency of homozygous dominant individuals GG with colour green
2pq - the frequency of heterozygous individuals Gg with colour green
p² - the frequency of homozygous recessive individuals gg with color brown
23% of the population is brown: p² = 23% = 0.23
p = √(p²) = √0.23 = 0.48
p = 0.48
p + q = 1
0.48 + q = 1
q = 1 - 0.48 = 0.52
<span>The percentage of the population that is expected to be heterozygous is 2pq:
2pq = 2 * p * q = 2 * 0.48 * 0.52 = 0.4992 = 49.92%</span>
Answer:
I would expect that the sample has only 50% of the original 235 U.
Explanation:
That is because 235 U has a half-life of 700 million years and I expect that my sample be 350 million years old. 350 is half of 700.
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