Sure, let’s solve this problem step-by-step.
We start with a population of 200 individuals, in which 86 express the recessive phenotype.
### Step 1: Determine the frequency of the recessive phenotype (q²)
The recessive phenotype indicates homozygosity for the recessive allele (aa). Hence, the frequency of this phenotype (q²) in the population is given by:
[tex]\[ q^2 = \frac{\text{number of individuals with recessive phenotype}}{\text{total population}} \][/tex]
[tex]\[ q^2 = \frac{86}{200} = 0.43 \][/tex]
### Step 2: Calculate the frequency of the recessive allele (q)
To find the frequency of the recessive allele (q), we take the square root of q²:
[tex]\[ q = \sqrt{0.43} \approx 0.6557 \][/tex]
### Step 3: Determine the frequency of the dominant allele (p)
According to the Hardy-Weinberg principle, the sum of the frequencies of the dominant (p) and recessive (q) alleles is 1:
[tex]\[ p + q = 1 \][/tex]
The frequency of the dominant allele (p) can be computed as:
[tex]\[ p = 1 - q \][/tex]
[tex]\[ p = 1 - 0.6557 \approx 0.3443 \][/tex]
### Step 4: Calculate the heterozygous frequency (2pq)
The frequency of heterozygous individuals (Aa) in the population is given by:
[tex]\[ 2pq \][/tex]
Substitute the values of p and q:
[tex]\[ 2pq = 2 \times 0.3443 \times 0.6557 \approx 0.4515 \][/tex]
### Step 5: Determine the percentage of heterozygous individuals
To find the percentage of the population that is heterozygous, we multiply the heterozygous frequency by 100:
[tex]\[ 0.4515 \times 100 \approx 45 \% \][/tex]
Therefore, the percent of the population predicted to be heterozygous is
[tex]\[ \boxed{45 \%} \][/tex]
