Answer:
The answer is "0.11"
Explanation:
The population is [tex]181[/tex] persons with genotype [tex]A_2 A_2[/tex] . Persons with [tex]A_2 A_2[/tex] are transferred to the mattress pool of 11, Throughout this new population [tex]181+11=192[/tex] are also the number of people.
The genotypic frequency [tex](A_1A_1 + (\frac{1}{2}) \ to \ A_1A_2)[/tex] alone is the [tex]A_1[/tex] frequencies:
[tex]= (\frac{181}{192})+(\frac{1}{2})\times 0\\\\= (\frac{181}{192})+(0.5)\times 0\\\\= 0.9427+0[/tex]
Allelic frequency of [tex]A_2\ is\ 1- A_1 = 1- 0.9427=0.0573[/tex]
[tex]A_1 \ Person \ A_2[/tex] frequency is [tex]2 \times A_1 \times A_2= 2 \times 0.9427 \times 0.0573 =0.10803342[/tex]
The frequency of [tex]A_1A_2[/tex] individual up to two decimal places is [tex]0.11[/tex]
The frequency of A1A2 individuals in the offspring is 0.11 Hz.
The number of times a repeated event occurs per unit of time is known as frequency.
It is calculated in Hz.
Given the population is 181 A2A2 individuals
The number of migrated individuals is 11 (A1A2)
So, the total population is 181 + 11 = 192
Now, the genotype frequency is [tex]\bold{(A_1A_1 + (\frac{1}{2} )\times A_1A_2)}[/tex]
[tex]\dfrac{182}{192} + \dfrac{1}{2} \times 0 = 0.9427[/tex]
Now, we will calculate the allele frequency of [tex]\bold{A_2 = 1 - A_1}[/tex]
[tex]\bold{A_2 = 1 -0.9427 = 0.0573}[/tex]
The frequency is:
[tex]\bold= {2\times A_1 \times A_2}[/tex]
[tex]\bold= {2\times 0.9427\times 0.0573 = 0.1080}[/tex]
Thus, the frequency is 0.11 Hz.
Learn more about frequency, here:
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