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Biology 13 Online
OpenStudy (anonymous):

the population of a region is 100,000. there are 250 albinos in the region . the estimated number of normal person carryiing albino gene is : A. 5900 B. 9500 C. 95000 D.59000 E. 159000

OpenStudy (anonymous):

@Abhisar

OpenStudy (abhisar):

Hello @hemarubenY !

OpenStudy (abhisar):

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OpenStudy (abhisar):

Which do you think it is? Why or why not? It would be easier to tutor you if you explained what about the question is giving you trouble.

OpenStudy (abhisar):

In order to make u understand this i'll work out an other example related to this question !

OpenStudy (abhisar):

\(\bf Formulas~Used\) \(\bullet\) P+Q =1 \(\bullet\) and \[P ^{2}+2PQ+Q ^{2}=1\]

OpenStudy (abhisar):

Here, P=frequency of A Q=frequency of a 2PQ = frequency of Aa

OpenStudy (abhisar):

Now let's see an example \(\color{red}{\text{In a random population frequency of recessive phenotype is 0.09.}}\)\(\color{red}{\text{What is the frequency of heterozygous genotype ?}}\)

OpenStudy (abhisar):

\(\huge\color{blue}{\text{Solution}}\) Let the recessive allele be a and dominant trait be A. Now frequency of a recessive phenotype = 0.09 i.e frequency of aa = 0.09.

OpenStudy (abhisar):

This implies that frequency of a = \(\sqrt{0.09}\) = 0.03 Now using the formula P+Q = 1, we get that frequency of A = 1-0.03 = 0.7

OpenStudy (abhisar):

Frequency of heterozygote i.e Aa = 2PQ = 2*0.7*0.03 = 0.42

OpenStudy (abhisar):

This means in a population of 100 individuals, 42 will have a genotype of Aa

OpenStudy (abhisar):

You can workout ur question in same manner. Just remember that Albinism is a recessive disease. They have given uh the frequency of aa and wants u to calculate Aa

OpenStudy (abhisar):

You can workout ur question in same manner. Just remember that Albinism is a recessive disease. They have given u the number of aa so and wants u to calculate number of Aa. Frequency of aa will be 250/100000.

OpenStudy (abhisar):

Hope it helps ! \(\color{green}{\huge\ddot\smile}\)

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