Question

A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is

• A. \(0.99\)
• B. \(0.01\)
• C. \(0.02\)
• D. \(0.10\)

Hint:

• Always pay close attention to whether the question asks for an allele frequency (\(p\) or \(q\)) or a genotype/phenotype frequency (\(p^2\), \(2pq\), or \(q^2\)).

• To find the other allele's frequency: \(q = 1 - p = 1 - 0.1 = 0.9\).

• The heterozygous frequency (\(Aa\)) would be \(2pq = 2 \times 0.1 \times 0.9 = 0.18\).

Answer 1

The Correct Option is B

Step 1: Read what is given.
The population is in Hardy-Weinberg equilibrium and the frequency of allele $A$ is given as $p = 0.1$. We are asked for the frequency of the genotype $AA$.
Step 2: Recall the basic rule.
For a gene with two alleles $A$ and $a$, we write $p$ for the frequency of $A$ and $q$ for the frequency of $a$, with $p + q = 1$.
Step 3: Write the genotype distribution.
Hardy-Weinberg gives $p^2 + 2pq + q^2 = 1$, where $p^2$ is the share of $AA$, $2pq$ is the share of $Aa$, and $q^2$ is the share of $aa$.
Step 4: Pick the term we need.
The homozygous dominant genotype $AA$ corresponds simply to the term $p^2$.
Step 5: Plug in the number.
Since $p = 0.1$, we get $p^2 = (0.1)^2 = 0.01$.
Step 6: Match to the options.
The value $0.01$ matches option (2). A quick sanity check: a low allele frequency of $0.1$ should give an even smaller homozygous frequency, and $0.01$ fits that nicely.
\[ \boxed{0.01} \]