In a random mating population an autosomal recessive disorder is present in 60 individual out of 6000 what is the number of homozygous normal (AA) OFFSPRING in POPULATION

Question

Arun · Answer

The ratio for BB:Bb:bb is 1:2:1.....so number of homozygous (BB) individuals is equal to no. of recessive (bb) individuals . AS BOTH RATIOS ARE EQUAL, NO. OF BB INDIVIDUALS=no. of bb individuals=60 so your answer is 60

Vikas TU · Answer

Dear student Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%). The frequency of the dominant allele. Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%). The frequency of heterozygous individuals.

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In a random mating population an autosomal recessive disorder is present in 60 individual out of 6000 what is the number of homozygous normal (AA) OFFSPRING in POPULATION

In a random mating population an autosomal recessive disorder is present in 60 individual out of 6000 what is the number of homozygous normal (AA) OFFSPRING
in POPULATION

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In a random mating population an autosomal recessive disorder is present in 60 individual out of 6000 what is the number of homozygous normal (AA) OFFSPRING in POPULATION

In a random mating population an autosomal recessive disorder is present in 60 individual out of 6000 what is the number of homozygous normal (AA) OFFSPRINGin POPULATION

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In a random mating population an autosomal recessive disorder is present in 60 individual out of 6000 what is the number of homozygous normal (AA) OFFSPRING
in POPULATION