(Solved) Hardy weinberg p and q, frequencies of alleles

Jessc

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7 years ago

Hardy weinberg p and q, frequencies of alleles

In a population of 1000 individuals with the following genotype counts: AA = 100, Aa = 400, aa = 500, what are p and q, the frequencies of the A and a alleles?

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bio_man

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7 years ago

Hi there

Here's a similar question with the answer!

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In a population of 1000 individuals with the following genotype counts: AA = 300, Aa = 500, aa = 200, what are p and q, the frequencies of the A and a alleles?

Answer:

The frequency of genotype, AA = 300/1000 = 0.3

The frequency of allele A = Square root of 0.3

= 0.55

The frequency of genotype, aa = 200/1000 = 0.2

The frequency of allele A = Square root of 0.2

= 0.45

"p" is the dominant allele that is A

"q" is the recessive allele that is a

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Jessc Author

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7 years ago

So is this the correct work..........

The frequency of genotype, AA = 100/1000 = 0.1
The frequency of allele A = Square root of 0.1
= 0.32
The frequency of genotype, aa = 500/1000 = 0.5
The frequency of allele A = Square root of 0.5
= 0.71
"p" is the dominant allele that is a
"q" is the recessive allele that is A

Or is this........

Frequency of A = p = (AA/2 +Aa)/total individuals
= {(100/2) + 400}/1000
= 450/1000
= 0.45
According to Hardy Weinberg law, p+q= 1
Therefore, 0.45+q= 1
q = 0.55

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bio_man

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Hardy weinberg p and q, frequencies of alleles

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