Answer:
16.38%
Step-by-step explanation:
Given
Frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000
Let p = Probability of homozygous recessive individuals having character extra-long eyelashes = 0.09
p + q = 1 where q = Probability of homozygous recessive individuals not having character extra-long eyelashes
0.09 + p = 1
p = 1 - 0.09
p = 0.91
The frequency of the carrier individual is calculated by npq
Where n = mating partners = 2
Frequency = 2 * 0.09 * 0.91
Frequency = 0.1638 or 16.38%
Answer:
The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals which is
0.42 or 42 %
Step-by-step explanation:
To solve the question, we note that
p²+2pq+q²=1 and p + q =1
where
p = frequency of the dominant allele
q= frequency of the recessive allele
p² = frequency of homozygous dominant allele
q² = frequency of homozygous recessive allele
2·p·q = frequency of hetrozygous individuals
The frequency of the homozygous recessive individuals q² is 0.09
Therefore the frequency of q = √(0.09) = 0.3, therefore p = 1 - 0.3 =0.7
The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals or 2×p×q = Ae = 2*0.3*0.7 = 0.42
→ 0.42× 100 = 42 %
