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:
No, If that was the case, then we should only be eating carbohydrates.
The number of people who are carriers (heterozygous) for PKU if there are 33 of the 300,000 people in Corpus Christi, TX, have PKU is 15. Heterozygous or also called as zygosity refers to having the unlike genes or different genes.
Answer:
The cell would not split in two
Explanation:
Each daughter cell has the same number and kind of chromosomes as the parent nucleus If mitosis were not accompanied by cytoplasmic division, there would only be one cell with two identical nuclei because the division did not occur for there to be two different cells.
