Question

matching question
in pea plants, the purple flower allele is dominant over the white flower allele. if p represents the dominant allele and q represents the recessive allele, what are the
correct genotypic ratios for flower color in a population that contains 90 purple flowering plants and 10 white flowering plants?
instructions
0.46
0.10
0.44
2pq
q²
p²

Explanation:

Step1: Calculate the frequency of white - flowered plants

The frequency of white - flowered plants (homozygous recessive) is \(q^{2}\). Given that there are 10 white - flowered plants out of 100 (\(n = 100\)), so \(q^{2}=\frac{10}{100}=0.10\)

Step2: Calculate the frequency of the recessive allele \(q\)

If \(q^{2}=0.10\), then \(q=\sqrt{0.10}\approx0.32\). But we can also use the Hardy - Weinberg principle \(p + q=1\). Since \(q^{2} = 0.10\), we know that \(q=\sqrt{0.10}\approx0.32\), and \(p = 1 - q\). But another way:
We know that the frequency of homozygous dominant (\(p^{2}\)):
The number of homozygous dominant and heterozygous plants is 90. Let's assume the frequency of homozygous dominant is \(p^{2}\) and heterozygous is \(2pq\).
We also know that \(p^{2}+2pq + q^{2}=1\). Since \(q^{2}=0.10\), then \(p^{2}+2pq=0.90\)
If we assume \(p^{2}=0.46\) and \(2pq = 0.44\) (by trial - and - error or more accurate calculation based on the fact that \(p^{2}+2pq+q^{2}=(p + q)^{2}=1\))

Answer:

\(p^{2}=0.46\), \(2pq = 0.44\), \(q^{2}=0.10\)