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

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%

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 %

180-142=38 for the acute angle next to the entry

90+38=128

180-128=52

a=52

step-by-step explanation: