In a large, randomly mating population with no forces acting to change gene frequencies, the frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000, or 0.09. What percentage of the population carries this trait but displays the dominant phenotype, short eyelashes

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 %

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%